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188200
https://engoo.com/app/words/word/repel/Mr4YfMRQEeamRiu-nQfTHQ
repel (【Verb】to make something move away ) Meaning, Usage, and Readings | Engoo Words Please refresh the page to get the latest updates.Refresh WORDSTop "repel" Meaning repel /rɪˈpel/ Verb to make something move away "repel" Example Sentences This cream repels insects; you should try it! A magnetic force can both attract and repel. "repel" Related Lesson Material Iraqi helicopter gunships fired on the area to repel the militants as Iraqi tanks rumbled from the south to attack. See Lesson When similarly charged ends of two magnets get close together, they repel or push away from each other. See Lesson She then passes the insect repellent to Nathan. See Lesson Do you always smell like insect repellent? See Lesson Grace, could you pass me the insect repellent, please? See Lesson Sonic repellents, installed by authorities to ward them off, are no longer effective as the birds have acclimated to them. See Lesson Mosquitoes can also spread disease, so use a mosquito net and insect repellent if mosquitoes are common where you're going. See Lesson Police have arrested at least seven individuals, one of whom admitted that his recipe for oplosan, a term for home-brewed alcohol cut with non-food ingredients that have not gone through the distillation process, included mosquito repellent, cough syrup, and pure alcohol. See Lesson Browse words ABCDEFGHIJKLMNOPQRSTUVWXYZ 0 0 0 Presented by Engoo is a service that offers lessons for those learning English. Although the lesson materials can be used for self study, they are intended for use with a teacher. To book a lesson with one of our professional teachers, please visit Engoo. If you would like to use these materials for commercial purposes, please contact us here. © 1998 - 2025. No duplication without prior permission. All Rights Reserved.
188201
https://www.geogebra.org/m/epvE8n8W
Google Classroom GeoGebra Classroom Home Resources Profile Classroom App Downloads Graphing 3D Point Author:Jessica Tentinger Graphing a Point 3D Adjust the sliders and rotate the graph to see how a point is graphed in a 3D plane. Identify each axis as the x, y, or z axis. How many different quadrants are there? New Resources Ringed Polyhedra Colors for GeoGebra 拼砌四邊形 - 工作紙 Transformaciones usando coordenadas 從邊長辨認四邊形 Discover Resources Constructions of Isosceles Triangles Social Science Teaching Aid H15..Inside.Similarity CarolineHan_MeasuringObjects Practice with Division (1-digit divisor) Discover Topics Special Points Isosceles Triangles Statistical Characteristics Differential Calculus Coordinates
188202
https://pmc.ncbi.nlm.nih.gov/articles/PMC4186911/
Mechanisms of abnormal lamellar body secretion and the dysfunctional skin barrier in atopic dermatitis - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. Learn more Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice J Allergy Clin Immunol . Author manuscript; available in PMC: 2015 Oct 1. Published in final edited form as: J Allergy Clin Immunol. 2014 Aug 15;134(4):781–791.e1. doi: 10.1016/j.jaci.2014.05.048 Search in PMC Search in PubMed View in NLM Catalog Add to search Mechanisms of abnormal lamellar body secretion and the dysfunctional skin barrier in atopic dermatitis Peter M Elias Peter M Elias, MD 1 Dermatology Service, Veterans Affairs Medical Center, and Department of Dermatology, University of California, San Francisco, CA Find articles by Peter M Elias 1, Joan Wakefield Joan Wakefield 1 Dermatology Service, Veterans Affairs Medical Center, and Department of Dermatology, University of California, San Francisco, CA Find articles by Joan Wakefield 1 Author information Article notes Copyright and License information 1 Dermatology Service, Veterans Affairs Medical Center, and Department of Dermatology, University of California, San Francisco, CA ✉ Address correspondence to: Peter M. Elias, MD, Dermatology Service (190), VA Medical Center, 4150 Clement Street, San Francisco, CA 94121, TEL: (415) 750-2091, FAX: (415) 750-2106, eliasp@derm.ucsf.edu Issue date 2014 Oct. PMC Copyright notice PMCID: PMC4186911 NIHMSID: NIHMS607762 PMID: 25131691 The publisher's version of this article is available at J Allergy Clin Immunol Abstract We review here how diverse inherited and acquired abnormalities in epidermal structural and enzymatic proteins converge to produce defective permeability barrier function and antimicrobial defense in AD. Although best known are mutations in filaggrin (FLG), mutations in other member of the fused S-100 family of proteins (i.e., hornerin [hrn] and filaggrin 2 [flg-2]); the cornified envelope precursor (e.g., SPRR3); mattrin, encoded by Tmem79, which regulates the assembly of lamellar bodies; SPINK5, which encodes the serine protease inhibitor, LEKTI1; and the fatty acid transporter, FATP4, have all been linked to AD. Yet, these abnormalities often only predispose to AD; additional acquired stressors that further compromise barrier function; e.g., psychological stress, a low ambient humidity, or high pH surfactants, often are required to trigger disease. Th2 cytokines can also compromise barrier function by downregulating expression of multiple epidermal structural proteins, lipid synthetic enzymes and antimicrobial peptides. All of these inherited and acquired abnormalities converge on the lamellar body secretory system, producing abnormalities in lipid composition, secretion and/or extracellular lamellar membrane organization, as well as in antimicrobial defense. Finally, we briefly review therapeutic options that address this new pathogenic paradigm. Keywords: antimicrobial peptides, atopic dermatitis, barrier function, ceramides, cytokines, filaggrin, kallikreins, lamellar bodies, lipid composition, pH, serine protease inhibitors, Th2 cells INTRODUCTION Basis for the Permeability Barrier in Normal Skin 7The epidermis generates a set of protective/defensive functions, mediated by its unique differentiation end-product, the stratum corneum (SC) 1, 2. As they migrate apically, keratinocytes acquire a series of differentiation-specific proteins 3–5 and lipids 6, 7 until cornification occurs. The SC comprises vertical stacks of anucleate corneocytes, embedded in an expanded extracellular matrix, replete with multiple stacks of broad, planar lamellar bilayers, enriched in ceramides, cholesterol, and free fatty acids (FFA) 8. These lamellar arrays of hydrophobic lipid species impede both the outward movement of water, and the inward bombardment of noxious environmental allergens and pathogens. A unique organelle, the epidermal lamellar body, delivers these lipids to the SC interstices as their precursors (e.g., glucosylceramides and phospholipids), along with a set of hydrolytic ‘lipid processing’ enzymes, including β-glucocerebrosidase, acidic sphingomylinase, secretory phospholipase A 2 and steroid sulfatase 9 (Fig. 1). These enzymes generate a family of ceramides (Cer), essential and non-essential free fatty acids (FFA), as well as much of the cholesterol that is required for the supramolecular organization of these non-polar lipid species into mature lamellar membrane structures 10. In parallel, lamellar body-derived, desquamatory proteases and their inhibitors initiate the orderly digestion of corneodesmosomes (= transient intercellular rivets that connect adjacent corneocytes), a process that eventually leads to the desquamation of corneocytes from the skin surface 11, 12 (Fig. 1). Finally, at least two antimicrobial peptides, human β-defensin 2 and the carboxyterminal cathelicidin peptide, LL-37, are delivered to the SC intercellular domains through secretion of lamellar body contents 13–15. In fact, the permeability barrier and the antimicrobial barrier share many features that impede the colonization and invasion of pathogenic organisms, while simultaneously encouraging colonization by non-pathogenic ‘normal’ flora 16. Thus, the physicochemical characteristics of the SC; niche occupancy by resident normal flora; and secreted factors from the normal flora contribute to cutaneous antimicrobial defense. Because these two barriers share so many structural and biochemical features, perturbations in one function inevitably modify the other in parallel 15, 17. Thus, the epidermal lamellar body is a multifunctional organelle, whose contents influence not only permeability barrier status, but also SC cohesion/desquamation and antimicrobial defense. Fig. 1. Multifunctional Impact of Secreted Lamellar Body Contents. Open in a new tab (Modified from106) The Tight Junction (TJ) Controversy How should we interpret an ever-expanding literature that proclaims a potential role for TJ in normal permeability barrier function 18, 19, as well as a potential role for abnormal TJ function in AD 20? We will attempt to navigate this heavily invested subject as follows: First, complex TJ structures, such as those found in the kidney and gastrointestinal tract, do not occur in adult keratinizing epithelia 21. Second, with the exception of these structures in renal collecting tubules, where they comprise multitiered sites of membrane fusion (‘zonulae occludentes’), TJ provide a relatively poor barrier against paracellular water movement. In the author’s opinion, confusion in the skin-related literature has occurred because ‘TJ proteins’ are widely equated with ‘TJ’ 18, 19. Doubtless, the apical-lateral plasma membranes of cells in the outer stratum granulosum of normal epidermis are heavily decorated with multiple TJ proteins, which form ‘kissing points’; i.e., ‘maculae occludentes’ 21. However, as noted above, these focal attachments do not comprise true zonulae occludentes, as occur in tubular epithelia. The most compelling evidence that these putative TJ play no direct role in the paracellular water barrier is that removal of SC lipids by gentle, external lipid solvent treatment completely abrogates the permeability barrier 22. While this observation likely also excludes a possible ‘back-up’ role for TJ-like structures in the water barrier, it still remains possible that these incomplete structures interdict the passage of larger xenobiotes, particularly when the overlying lipid-based barrier becomes defective, as occurs in AD. Yet, these structures, though insufficient to contribute to the water barrier, are nonetheless critical for the development of barrier competence. Transgenic knock-out of the key TJ protein, claudin 1, results in a fatal, post-natal permeability barrier abnormality 23. Indeed, replete TJ are present early in epidermal development, but they become functionally incompetent later in fetal life in parallel with establishment of the lipid-based barrier 24. An acquired reduction in the expression of the TJ protein, claudin 1, has been reported in AD 20, and occludin protein levels decline in FLG-deficient human epidermis 25. It is possible that abnormalities in TJ proteins could result from the Th2 dominant milieu in AD, which simultaneously downregulates many other epidermal differentiation-linked proteins (see below). Since adult epidermis does not generate the types of complex zonulae occludentes necessary to impede paracellular water movement, attention should be focused instead on the possible function(s) of the incomplete junctions (maculae occludentes) in normal epidermis; and how acquired defects in these focal connections could contribute to AD pathogenesis. These structures likely perform important ‘fence functions’ in adult epidermis, including polarizing the direction of lamellar body secretion towards the apex of the outermost granular layer 26. Barrier Dysfunction in AD and Other Atopic Disorders During the pre-genotype era, we and others proposed that the permeability barrier abnormality in AD is not merely an epiphenomenon, but rather the potential ‘driver’ of inflammation in AD (i.e., an ‘outside-to-inside’ view of disease pathogenesis) 27, 28, because i) the extent of the permeability barrier abnormality parallels severity of disease phenotype 29, 30; ii) clinically-uninvolved skin sites display significant barrier abnormalities 30; and iii) sustained barrier abnormalities, regardless of cause, inevitably stimulate a pro-inflammatory cytokine cascade that ‘recruits’ characteristic, disease-specific immunophenotypes 31, 32. Recent studies have shown that the cutaneous barrier abnormality is not only critical for the development of AD, but also other allergic disorders, including asthma, allergic rhinitis, and food allergies 33. The barrier abnormality in AD leads to an increase in pH that activates serine proteases (kallikreins, KLK) in the outer epidermis, with widespread downstream and upstream consequences, as shown in Fig. 2. Yet, these mechanisms also stimulate a series of metabolic responses aimed at restoring permeability barrier homeostasis. Briefly, increased TNFα stimulates differentiation 34, while increased IL-1α enhances epidermal lipid synthesis 35. Nerve growth factor (NGF) and amphiregulin (AR) stimulate epidermal DNA synthesis; and several cytokines enhance antimicrobial peptide activation/production 36. Fig. 2. Cytokine Cascade Leads To Multiple ‘Vicious Cycles’ in AD. Open in a new tab Abbreviations: GM-CSF, granulocyte-macrophage colony-stimulating factor; IL-1, interleukin-1; KLK, kallikrein; NF-κB, nuclear factor kappa B; TNFα, tumor necrosis factor alpha; TSLP, thymic stromal lymphopoietin Related and Unrelated Mutations Impact Barrier Function in AD Recent molecular genetic studies have fortified this ‘outside-to-inside-back- to-outside’ view of disease pathogenesis, because most of the recently-identified mutations associated with AD modify structural and enzymatic proteins that are required for normal barrier function. Filaggrin The initial molecular genetic evidence that a primary structural abnormality underlies the pathogenesis of AD derives from the strong association of loss-of-function mutations in the gene encoding filament aggregating protein (filaggrin, FLG) in AD 37–39. Up to 50% of certain northern European kindreds, and a substantial proportion of Asians, with AD reveal either single or double allele, loss-of-function mutations in the gene encoding FLG. Although more than 40 different FLG mutations have now been associated with AD 37, four predominate in northern and central Europeans 40, 41. The initial product of FLG translation is pro-FLG, a large, histidine-rich, highly cationic phosphoprotein, consisting of ten to twelve FLG repeats, connected by peptide segments enriched in hydrophobic amino acids 42, 43. Pro-FLG contains an amino-terminal sequence, including a calcium-binding A domain as well as a B domain of uncertain function. During cornification in normal, non-atopic humans, pro-FLG is dephosphorylated and proteolytically processed to FLG monomers 42. Then, as the water content of the SC declines in the mid-to-outer stratum corneum, FLG detaches from the cornified envelope, followed by its C-terminal proteolysis by caspase 14, bleomycin hydrolase, and other hydrolases into its constituent amino acids 44. These amino acids subsequently are further deiminated into polycarboxylic acids that account in part for SC hydration and acidification 45 (Fig. 3). Fig. 3. Multiple Downstream Consequences of Filaggrin Deficiency in Atopic Dermatitis. Open in a new tab Trans-urocanic acid (t-UCA) is the most potent endogenous UV-B filter in lightly-pigmented skin. Loss of t-UCA could account for the higher incidence of non-melanoma skin cancers in AD 45. (Modified from Elias & Williams, JID, 2013) Decreased FLG expression results in a paucity of keratohyalin granules, a hallmark of ichthyosis vulgaris (IV) 46, the forme fruste of AD 47. Flg mutations exhibit an allele-dose effect, wherein heterozygous patients with IV show diminished FLG expression and a mild phenotype, as well as abnormalities in surface pH, hydration, and barrier function 25 (Fig. 3). But IV patients with homozygous and compound heterozygous FLG mutations exhibit more severe scaling 40, 48, and more pronounced abnormalities in stratum corneum structure and function 25 (Fig. 4), as well as a further propensity to develop AD 37, 40. Yet importantly, a substantial proportion of these double-allele IV patients still do not exhibit inflammation (i.e., AD), emphasizing the role of exogenous (acquired) stressors in AD pathogenesis (see below). Fig. 4. Abnormalities That Lead to Paracellular Barrier Abnormality in FLG-Deficient Epidermis. Open in a new tab Lanthanum perfusion in double-allele ichthyosis vulgaris (IV) (B) vs. wild-type (A) human epidermis. C: Retraction of cytoskeleton (asterisks) in double-allele IV. D (+ inserts): Impaired loading and secretion of lamellar body cargo in double-allele IV. F–G: Post-secretory abnormalities in lamellar bilayer organization and maturation in double-allele IV (arrows). E: Normal (wild- type) human epidermis. RuO4 post-fixation (Modified from 25). An acquired reduction in epidermal FLG expression also occurs in AD, independent of mutation status 49, due in part to Th2-induced down-regulation of a broad range of proteins associated with epidermal differentiation, including FLG 50,51. Moreover, IgE from AD patients auto-reacts against a variety of keratinocyte antigens, suggesting yet another ‘vicious cycle’ in AD 52. Thus, primary inherited barrier abnormalities in AD ultimately stimulate downstream paracrine mechanisms that likely further compromise permeability barrier function, completing an additional potential ‘outside-inside-outside’ pathogenic loop in AD 53 (Fig. 2). Other Fused S-100 Proteins (Hornerin and Filaggrin-2) The prevalence of FLG mutations in AD patients, though quite high in populations of northern European descent 37, does not account for many cases of AD in such northerners. It can be further assumed that almost all cases of AD in other populations will prove to be associated with other inherited abnormalities that compromise epidermal barrier function. Recent studies suggest an association of AD with mutations in two other members of the fused S-100 proteins; i.e., hornerin (hrn) and filaggrin 2 (Flg-2) 54, 55. Interestingly, FLG-2 mutations are linked to AD in African-Americans 56, 57. While filaggrin, hornerin, and FLG-2 are all differentiation-specific components of the corneocyte envelope 58–62, the specific functions of both hornerin and FLG-2 in normal epidermis remain unknown (Table 1). Hence, how defects in these proteins contribute to a putative barrier abnormality in AD also remains uncertain. Table 1. Consequences of Inherited Barrier Abnormalities in AD | Structural Protein | Immediate Structural Consequences | Downstream Changes → Barrier Defect | :--- | S-100Type | | | | ↓Filaggrin | Attenuated CE → Poor Scaffold | Bilayer disorganization | | | ↑SC pH → ↑KLK | Lamellar body entombment → Lamellar bilayers | | | ↓SC hydration | ↑TEWL, ↑pro-inflammatory cytokines | | ↓Hornerin | ? Attenuated CE | ? | | ↓Filaggrin-2 | ? Attenuated CE | | | Other CE Precursors | | | | SPRR3 | Attenuated CE → Poor Scaffold | | | Enzyme Inhibitor | | | | LEKTI 1 (Netherton syndrome) | ↑KLK activity | Destruction of corneodesmosomes, lipid processing enzymes, LL-37 | | Lipid Metabolism | | | | FATP4 | ↓FFA, FACoA, Esterified FA | ↓Glycerolipids; Detergent effects of excess FFA | | Mattrin | ↓Lamellar body formation/secretion | ?↓Extracellular lipids | Open in a new tab Abbreviations: CE, cornified envelope; FATP4, fatty acid transport protein 4; FFA, free fatty acids; KLK, kallikreins, LEKTI, lympho-epithelial Kazal-type trypsin inhibitor; TEWL, transepidermal water loss SPRR3 is a cornified envelope (CE) precursor protein that is virtually undetectable in normal skin 63. Several types of mutations in SPRR3, including an extra 24 base pair defect in the central domain, as well as additional in-frame deletions and insertions 64 have been associated not only with AD 65, but also with asthma 66. These mutations result in expression of SPRR3 at higher than normal levels in AD 64, 65, likely impacting barrier function through production of a CE scaffold that impairs the supramolecular organization of lamellar body-derived lipids into normal bilayer structures (Table 1). Ultrastructural studies show defective, thinner-than-normal CEs in AD, with decreased extracellular lipids and a poorly cohesive SC, associated with deficiencies not only in SPRR, but also in several other CE precursors in AD, perhaps due to a broad Th2-stimulated down-regulation of these proteins 67,50, 51. TMEM79 Very recent studies have identified non-sense and mis-sense mutations in the gene, TMEM79, which encodes the protein, mattrin, in some Irish AD patients who lack FLG mutations 68. Mutations in the murine orthologue of this gene account for the flaky tail (ma/ma) strain of mice, which develop a spontaneous AD-like dermatitis 69. Mattrin localizes to the cytosol, and more specifically within the trans-Golgi network in the outermost cells of the granular layer 68, 69. Reductions in mattrin levels block the secretion of lamellar body contents, including desquamatory proteases, antiproteases 69, and likely lamellar-derived lipids. Indeed, defective lipid secretion has been demonstrated in flaky tail mice, bearing the mattrin mutation 70. Nonetheless, the association of this mutation with AD eloquently demonstrates that not only inherited deficiencies in structural proteins, but also that mutations which impair the delivery of lamellar body contents can predispose to disease. Protease/Antiprotease Expression Inherited abnormalities that result in excessive serine protease (SP) (kallikreins, KLK) activity predisposes to severe AD, and more importantly, they provide unique insights into the pathogenesis of AD 71, 72 (Fig. 5). The most compelling demonstration for the role of excess SP activity in the pathogenesis of AD comes from Netherton syndrome (NS), an autosomal recessive disorder due to loss-of-function mutations in SPINK5, the gene encoding the SP inhibitor, lymphoepithelial Kazal-type trypsin inhibitor type 1 (LEKTI 1) 73. NS is characterized by a severe AD-like dermatosis, mucosal atopy, and anaphylactic reactions to food antigens. The extent of residual LEKTI expression in humans with NS correlates inversely with excess KLK activity within the outer epidermis 74, and unrestricted KLK activity provokes a severe permeability barrier defect, as well as dramatic thinning of the SC. Both defects can be attributed to KLK-dependent degradation of lipid-processing enzymes and corneodesmosome-constituent proteins, respectively 74. KLK-mediated degradation of the enzymes contributes to the depletion of Cer, a characteristic lipid abnormality in AD 75, 76 (see also below). Likewise, transgenic mice that over-express human KLK7 display a severe AD-like dermatosis 77. In NS, and likely also in AD, one of these KLKs; i.e., KLK5, or the SC tryptic enzyme, binds to the protease activated receptor, type 2 (PAR2), stimulating NFκB-dependent production of the pro-Th2 cytokine, thymic stromal lymphopoietin (TSLP) 78. Fig. 5. Open in a new tab LESSONS FROM NETHERTON SYNDROME: Proposed Roles for Increased pH and KLK Activation in Producing Lipid Abnormalities in AD As a result of these divergent inherited associations, a broad view is emerging that virtually any inherited abnormality that leads to a sustained barrier abnormality can predispose to AD. For example, note the association of AD with loss-of-function mutations in the fatty acid transporter, FATP4, in patients with ichthyosis prematurity syndrome 79. Moreover, many patients with other inherited ichthyoses frequently report severe pruritus 80, although the inflammatory infiltrate in these patients (with the exception of NS) has not yet been characterized (Table 1). Yet, it would not be unreasonable then to query why diseases like psoriasis, which exhibit a well-known barrier abnormality 81, do not develop a Th-2-like immunophenotype. A perhaps too-simple explanation might be that allergen access is impeded by the tenacious scale in psoriasis. How Unrelated Mutations in Epidermal Proteins Converge on the Lamellar Body Secretory System to Provoke a Barrier Abnormality in AD We have shown that both reductions and loss of the cytosolic protein, filaggrin, lead to an extracellular permeability barrier defect, both in filaggrin-deficient IV 25, and in murine models of AD 70, 82. In all these settings, water loss accelerates via an extra- (para-) cellular pathway; i.e., through the extracellular matrix. The link between a defect in the cytosolic protein, Flg, and the extracellular permeability defect was clarified in patients with IV (Figs. 4A&B) 25. Both single- and double-allele patients demonstrate retraction of cytosolic keratin filaments into a perinuclear shell around nuclei of the stratum granulosum (Fig. 4C). This cytoskeletal abnormality appears to impact two cellular processes. First, it results in incomplete loading of cargo into nascent lamellar bodies, evidenced by ‘empty’ micro-vesicles within these organelles (Figs. 4D&E). The resulting deposition of non-lamellar contents in the intercellular spaces then leads to focal defects in the extracellular lamellar bilayer system (Figs. 4F&G), contributing to defective barrier function in IV. Second, the cytoskeletal abnormality impairs secretion of lamellar bodies, resulting in their partial entombment within corneocytes 25, 70 (Fig. 5). This latter pathogenic sequence is similar to that seen in patients with mutations in keratins 1 or 10, where cytoskeletal retraction results in entombment of lamellar bodies, and a paucity of lamellar bilayers 83. Once inflammation is established; i.e., as IV transitions into AD, the pH of the SC increases still further, sufficient to activate a family of KLK in the outer epidermis 84. KLK activation has multiple negative consequences 72, including: i) the degradation of both corneodesmosomes, leading to a poorly cohesive SC, and ii) the destruction of extracellular lipid processing enzymes, β-glucocerebrosidase and acidic sphingomylinase 85. Loss of these two ceramide-generating enzymes (as well as their reduced activity at a high pH) results in failure of lamellar bilayer formation and maturation 25, 82, exactly as occurs in NS 74. A third downstream consequence of increased SP activity is generation of the primary cytokines, IL-1α and IL-1β from their 33kDa pro-forms in human SC 86, which are stored in large quantities in the cytosol of corneocytes 87, 88. This putative pH-induced increase in KLK activity generates the active, 17kDa forms of these cytokines, the first step in the cytokine cascade in AD, which includes downstream production of several additional cytokines, growth factors, chemokines, and adhesion molecules 31, 89, 90, including the TSLP-Th2 cytokine network, described above 78 (Fig. 2). Excess KLK also activates the PAR2 receptor, which localizes to the plasma membrane of granular cells 74. Binding is followed by internalization of the PAR2 receptor, which then down-regulates lamellar body secretion, effectively entombing these organelles within the corneocyte cytosol 91, 92. Thus, even without allergen exposure, an Th2 immunophenotype likely can occur in AD, as described for NS 78 (Fig. 5)! Conversely, applications of either KLK or PAR2 inhibitors, or just acidification of the SC alone 93,94, prevents both the destructive effects of excess KLK activity and PAR2 internalization, normalizing lamellar body secretion and permeability homeostasis 91, 95. These studies demonstrate how multiple, pH-initiated steps in the secretion and post-secretory processing of lamellar body contents leads to AD (Fig. 6). Fig. 6. How Inherited Abnormalities Converge to Produce Defective Permeability and Antimicrobial Barriers in AD. Open in a new tab Abbreviations: FA, fatty acid; Fatp4, fatty acid transport protein 4; Flg, filaggrin; hBD2, human beta-defensin 2; Hrn, hornerin; LB, lamellar body Stressors That Further Aggravate Barrier Dysfunction Can Trigger AD (Fig. 5) Alkaline soaps In ichthyosis vulgaris (IV), even double-allele FLG mutations do not always suffice to provoke inflammation 25, 39, 96, but certain stressors can further aggravate the barrier abnormality; i.e., by provoking an incremental increase in the pH of the SC, leading to a further amplification of SP activity 72. Such a barrier-dependent increase in pH (and SP activity) likely accounts for the precipitation of AD following the use of neutral-to-alkaline soaps, a well-known exogenous stressor of clinical AD 97, 98. Reduced Humidity as a Stressor Prolonged exposure to a reduced environmental humidity, as occurs in radiant-heated homes in temperate climates during the winter, is also a well-known risk factor for AD 99. Under these conditions, transcutaneous water loss accelerates across a defective SC, aggravating the underlying permeability barrier abnormality, while also amplifying cytokine signaling of inflammation 72. Because FLG proteolysis is regulated by changes in external humidity 45, sustained reductions in environmental relative humidities likely further deplete residual FLG in single-allele FLG-deficient patients with AD. Finally, sustained psychological stress (PS) aggravates permeability barrier function in otherwise normal humans 100, 101 and mice 102, 103. PS is also a widely-acknowledged precipitant of AD, but in the case of PS, however, the likely mechanism differs from either high pH surfactant use or decreased environmental humidities. Increased stress in experimental animals induces an increase in endogenous glucocorticoids (GC), which in turn alters permeability barrier homeostasis, SC cohesion and epidermal antimicrobial defense 15, 103, 104. The central role of GC has been demonstrated in two ways: first, exogenous systemic or topical GC recapitulate all of the above, stress-induced functional abnormalities 103, 105. Second, either blockade of GC production with the CRF inhibitor, antalarmin, or peripheral action, with the GC receptor inhibitor, mifeprostone (Ru486), prevent emergence of PS and GC-initiated functional abnormalities 104, 105. A GC-mediated inhibition of synthesis of the three key epidermal lipids that mediate barrier function; i.e., Cer, cholesterol, and FFA, accounts for the negative effects of PS 102, 105. Accordingly, a topical mixture of these three lipids largely normalized all of these functions in mice and humans, even in the face of ongoing PS or GC therapy 102, 106. Basis for Lipid Abnormalities in AD Global Decline in Lipids Filaggrin-associated AD is characterized by profound abnormalities in lipid content, distribution, and lamellar membrane organization in lesional skin 107–109, which result in a paracellular barrier abnormality 25, 70, 82. A moderate impairment of lamellar body secretion, due to a cytoskeletal abnormality, results in entombment of substantial quantities of lamellar bodies within corneocytes, a feature that becomes more prominent once inflammation is established 82. In addition, KLK signaling of the PAR2 downregulates lamellar body secretion 92, likely providing an additional biochemical signal that accounts for entombment of these organelles in nascent corneocytes. Together, these abnormalities result in incomplete delivery of lamellar body-derived cargo, as well as a paucity of extracellular lamellar bilayers, leading to a global reduction in extracellular lipids 72, 108 (Fig. 6). Finally, not only extracellular lamellar bilayers, but also the covalently-bound lipids that form the corneocyte lipid envelope, decline in AD 110, further contributing to the barrier abnormality. Sphingolipid Abnormalities in Atopic Dermatitis The most distinctive hallmark of human AD is the repeatedly-noted, selective reduction in the Cer content of affected SC 75, 76. Several mechanisms likely contribute to the decrease in Cer. As noted above, the barrier-related increase in pH, and a pH-induced increase in KLK activity results in deactivation, and ultimately in accelerated degradation of the Cer-generating enzymes, acidic sphingomylinase and β-glucocerebrosidase (Fig. 5) 74. Moreover, the cytokine cascade, associated with AD and other inflammatory dermatoses with a barrier abnormality, upregulates production of interferon-γ, which downregulates epidermal synthesis of Cer 111. Furthermore, noting that neither sphingomyelin nor glucosylceramides accumulate in the SC of AD, Imokawa, et al. (2009) provided evidence that AD epidermis exhibits novel N-deacylation activities that degrade both sphingomyelin and glucosylceramides. However, the genes for these enzymes have not yet been identified in skin; hence, it remains possible that this deacylase activity could be of bacterial origin. Accordingly, several other microbial pathogens that are known to colonize AD elaborate acidic ceramidase activity 112, 113, which could further decrease Cer content. Yet, the sphingoid base content of the SC in AD is lower (not higher) than in normal SC 114, arguing against an important role for microbial ceramidases in producing Cer deficiency in AD. Finally, it should be noted that abnormalities in the ratio of sphingoid bases, specifically in sphingosine and sphinganine, could modify lamellar membrane permeability, thereby contributing to the barrier defect in AD 115. Increased production of the Th2-derived cytokines, IL-4 and IL-13, also further contributes to the decrease in Cer in AD. In experimental animals, IL-4 not only down-regulates serine palmitoyl transferase, the rate-limiting enzyme of ceramide synthesis, but it also blunts the potential benefits of Th1-derived TNF-α on ceramide-generating enzymes 116, 117. Thus, while Th1 cytokines upregulate Cer production 118, the dominance of Th2 cytokines in AD likely overwhelms this Th1 response, with profound consequences for epidermal structure and function (Fig. 5). Shorter N-Acyl Fatty Acids in AD Researchers in Japan and The Netherlands recently reported that the sum of sphingoid bases plus the N-acyl fatty acids (FA) in ceramides declines, in parallel with a decline in the chain length of free and esterified fatty acids (FA) in lesional AD 119–121. These shorter-chain FA in turn produce abnormalities in lipid organization that likely compromise permeability barrier function in AD 119, 121, 122. The basis for these chain length abnormalities could prove to be reduced expression of two fatty acid elongases, ELOVL1 and ELOVL4, enzymes required to generate the very long chain N-acyl FA in Cer and FFA in AD 111 (Fig. 5). It is intriguing to speculate that the reduced levels of ELOVs could again prove to be an acquired abnormality due to elevated Th2 cytokines. Alternatively, elevated levels of interferon gamma (IFNγ) downregulate ELOV1 and 4 69, which could account for reduced N-acyl chain length 111. Finally, degradation of ELOVs by excessive KLK activity could at least in theory contribute to these abnormalities. CONCLUSION: Clinical and Therapeutic Implications Sustained antigen ingress through a defective barrier leads to a Th2-dominant infiltrate, which then becomes an additional cause of inflammation in AD (Fig. 2).While certain antigens, such as cat dander, preferentially trigger childhood AD in FLG-deficient patients 123, the worst offenders are mites and cockroach antigens, which themselves release and activate KLK, resulting in further damage to the barrier 124. Furthermore, the lipid-depleted barrier in AD may facilitate the penetration of water-soluble haptens, such as nickel. Indeed, nickel-induced, acute allergic dermatitis is more common in humans with FLG-deficient AD than in normals 125,126. Accordingly, correction of the barrier abnormality alone with measures that restore and correct lipid imbalance could prevent and/or ameliorate the barrier abnormality in AD, thereby reducing the inflammatory component in AD 98, 107, 108, 127. There is now emerging evidence that physiologic lipids, if delivered in sufficient quantities, and at appropriate molar ratios, are effective in the treatment of even moderate-to-severe AD, without any of the safety concerns surrounding glucocorticoids and immunomodulators 108. Alternatively, those patients with AD due to either single-allele mutations in FLG, and/or acquired reductions in FLG, become potential candidates for strategies that either upregulate FLG expression 128,129; or enhance the transdermal delivery of FLG monomers to deficient skin 129. Yet, it must be noted that the latter approach, though very elegant in theory, may not be practical for a generalized disease such as AD. Table 2. Key Points 1 A variety of unrelated mutations that compromise epidermal barrier function predispose to the development of atopic dermatitis (AD). 2 An acquired deficiency in filaggrin also occurs, independent of mutation status. 3 These mutations converge on the lamellar body secretory system, producing abnormalities in either lamellar body formation, secretion, or post-secretory processing that compromises extracellular lamellar bilayer structure 4 Theses secretory abnormalities account in large part for the distinctive lipid abnormalities in AD, which included a global decrease in barrier lipids; a further decline in ceramide content; and truncation of the chain lengths of free and esterified fatty acids. 5 The same pathogenic sequence compromising antimicrobial defense accounts at least in part for colonization by S. aureus and other pathogens in AD. 6 Based upon the above, rational therapy should address and correct filaggrin status, and/or the lipid abnormalities in AD. Open in a new tab Acknowledgments This work was supported by NIH grant AR019098, and by the Medical Research Service, Department of Veterans Affairs. These contents are solely the responsibility of the authors and do not necessarily represent the official views of the NIAMS or NIH. Abbreviations AD atopic dermatitis AMP antimicrobial peptides AR amphiregulin Cer ceramides FLG filaggrin FFA free fatty acids GC glucocorticoids hBD human β-defensins IFNγ interferon gamma IV ichthyosis vulgaris KLK kallikreins LEKTI lymphoepithelial Kazal-type trypsin inhibitor NGF nerve growth factor NS Netherton syndrome PAR2 protease activator type 2 receptor PS psychological stress SP serine protease SC stratum corneum TJ tight junction TSLP thymic stromal lymphopoietin Glossary Diverse mechanisms converge on lamellar body secretion, producing the barrier abnormality in atopic dermatitis Lamellar body Small, ovoid, membrane-bound, secretory organelle synthesized by keratinocytes as they reach the stratum spinosum. An intracellular pathway involving the Golgi apparatus produces them. Their lipid molecular contents display a plate like appearance when viewed through high magnification electron microscopy Ceramides A group of amido sphingolipids (ex sphingomyelin and cerebrosides) formed by linking a fatty acid to a sphingoid base (C 18 H 37 NO 2). Antimicrobial peptides Components of the innate immune system capable of inserting into bacterial phospholipids that kill or slow microbial growth. Xenobiotes Chemical compounds foreign to living organisms. Keratohyalin granules Substance synthesized by free ribosomes in a keratinocyte as it passes through the epidermis. These granules expand and then interact with tonofilaments to aggregate keratin in corneocytes. S-100 Calcium binding proteins that modulate both intra- and extracellular processes in the epidermis, including keratinocyte differentiation. TSLP An IL-7-like cytokine that activates dendritic cells to produce pro-Th2 chemokines and primes naïve T cells to differentiate into Th2 cells. pH A scale measurement of acidity/alkalinity on which a value of 7 represents neutrality. Each unit of change represents a 100-fold change in acidity or alkalinity. Skin pH in a healthy adult ranges from 4.5 to 5.5. Washing skin with soap increases skin pH by approximately 3 units. Skin protease activity is enhanced at pH values above 7.5. Humidity Typically expressed as relative humidity: The amount of water vapor in the air at a specified temperature and pressure relative to the total amount it could hold at those values. Footnotes Conflict of Interest Statement: Dr. Elias is a co-developer of this form of treatment for AD, currently licensed from the University of California to PuraCap Pharmaceutical, LLC. (Dr. Elias is a consultant to PuraCap). The authors state no conflicts of interest. 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[DOI] [PubMed] [Google Scholar] ACTIONS View on publisher site PDF (1.1 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract INTRODUCTION CONCLUSION: Clinical and Therapeutic Implications Acknowledgments Abbreviations Glossary Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
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https://en.wikibooks.org/wiki/Nuclear_Physics/Nuclear_Binding_Energies
Nuclear Physics/Nuclear Binding Energies - Wikibooks, open books for an open world Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main Page Help Browse Cookbook Wikijunior Featured books Recent changes Special pages Random book Using Wikibooks Community Reading room forum Community portal Bulletin Board Help out! Policies and guidelines Contact us Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Donations Create account Log in [x] Personal tools Donations Create account Log in [dismiss] The Wikibooks community is developing a policy on the use of generative AI. Please review the draft policy and provide feedback on its talk page. [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 The Binding Energy of a Deuterion 2 H 2 Binding Energies of Isotopes Nuclear Physics/Nuclear Binding Energies [x] Add languages Add links Book Discussion [x] English Read Edit Edit source View history [x] Tools Tools move to sidebar hide Actions Read Edit Edit source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Add interlanguage links Sister projects Wikipedia Wikiversity Wiktionary Wikiquote Wikisource Wikinews Wikivoyage Commons Wikidata MediaWiki Meta-Wiki Print/export Create a collection Download as PDF Printable version In other projects Wikimedia Commons Wikipedia Wikidata item From Wikibooks, open books for an open world <Nuclear Physics This page may need to be reviewed for quality. Nuclear Physics Binding energy is the amount of energy required to separate something down into its base components. In nuclear terms, the energy is derived from the Strong Nuclear Force. The binding energy must come from somewhere, and in the case, because mass and energy are so strongly linked by Einstein's Equation E = mc 2 the Binding Energy is proportional to the Difference in Mass of the Atom, compared to the sum of its parts. This is also called the mass defect. The larger the mass defect, the more stable the isotope is. Example of Binding Energy: The Binding Energy of a Deuterion 2 H [edit | edit source] A deuterion is the nucleus of a deuterium atom, and consists of one proton and one neutron. The masses of the constituents are: m proton = 1.007276 u (where u represents the Atomic mass unit ≈ 1.66053886 x 10-27 kg)m neutron= 1.008665 u m proton + m neutron = 1.007276 + 1.008665 = 2.015941 u The mass of the deuterion is: Atomic Mass 2 H = 2.013553 u The mass difference = 2.015941 - 2.013553 = .002388 u. = 3.9654 x 10-30 kilograms. E{\displaystyle E\,}=m c 2{\displaystyle =mc^{2}\,} =3.9654×10−30 k g×(2.99792458×10 8 m s−1)2{\displaystyle =3.9654\times 10^{-30}kg\times (2.99792458\times 10^{8}ms^{-1})^{2}} =3.5639×10−13 J{\displaystyle =3.5639\times 10^{-13}J} =2.224 M e V{\displaystyle =2.224MeV\,} Binding Energies of Isotopes [edit | edit source] Iron is one of the most stable nuclei around. A picture should really be added here. but if h= 1.007277, then 2h= 2.014554 and the difference between p and n is .001388. when you add this difference you get 2.0028434=2h. try this h=1,007825,n=1.008665 then 2h =2.01565 and 1.008665-1.007825=.00084 then add that difference to change p to n, so 2.01565+.00084=2.01649 now sub. the so call binding energy, 2.01649-.002388= 2.014102 and this is mass of 2d that I find in text books!!! also .002388u mult. by 931.5 = 2.224422 in MeV units! if you follow the above 2.014102 times 2 is 4.028204. and 3 times 4.028204 is 12.084612, this is carbon. 12carbon(12.084612) + 1.007825 = 13nitrogen(13.092437) + .00084 = 13.093277(before the loss of a +beta and neutrino!). 13.093277 - .002388 = 13carbon(13.090889), add 1.007825 to get 14nitrogon(14.098714). again add 1.007825 to get 15oxygen(15.106539). again add .00084 to get 15.107379(again this is before loss of a +beta and neutrino!). again subtrack .002388 to get 15nitrogen(15.104991). now add 1.007825 to get 16oxygen(16.112816). 16oxygen(16.112816) = 12carbon(12.084612) + 4helium(4.028204) and this is the carbon cycle in stars. in David Wayne Ferrin's note: the proton is 1.007277, and Hydrogen is 1.007277+(+beta=.000548)= 1.007825. Also .002388= (2(+beta)+2(-beta)+2(.000098)) or .002388=(.001194+.001194)=((.000548)+(.000548)+(.000098))x2, that is in a world with perfect field breaking symmetry, in fusion this happens 1.007825 + (1.007825 + .000840) = 2.01649. Ok, look this .002388 - .000840 = .001548, .000840 = .000548(first beta) + .000292, and .000292 = (.000098 + .000098 + .000096), and .001548 = (.001096(.000548x2)(second and third) + .000452), now add .000096 + .000452 = .000548(fourth). When betas are formed they do so in pairs (one real, one virtual), so do neutrinos = .000098. Now take 2.016490-1.006985(too light)=1.009505(too heavy). To 1.006985(tl) add .000292= 1.007277 add .000548= 1.007825 add (.000292+.000548)=.000840= 1.008665 add .000840= 1.009505(th). And if 1.006985/3= .3356616667 add .000840= .33650166673= 1.009505. .3356616667931.5= 312.6688425, .3365016667931.5= 313.451325, and 312.6688425 + (2313.4513025)=939.5714476/931.5= 1.008665. Retrieved from " Category: Book:Nuclear Physics This page was last edited on 21 January 2019, at 19:13. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Privacy policy About Wikibooks Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Search Search [x] Toggle the table of contents Nuclear Physics/Nuclear Binding Energies Add languagesAdd topic
188204
https://math.stackexchange.com/questions/3358733/circular-area-maximising-with-polygons-included-help
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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Circular Area maximising with polygons included. Help? Ask Question Asked 6 years ago Modified6 years ago Viewed 106 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. given n concentric circles, each of which contain a point that can move on that circle, with radius’s {r1, r2, …., rn}, find the geometric(angular) arrangement of these points to get the polygon with maximum area. the setup looks like the following: the approach I used was to assign to each point a position vector to get a convenient means to calculate areas using cross-product. It ran into problems because of the following: 1) which vectors should be positioned together 2)the problem seems to be dependent on both, the number of point and their respective radiuses. If there was some general method maybe using some inequality that assigns positions to the vectors or some other method that somehow automatically takes care of the nitty-gritty that would be great. I'm not exactly sure what to field to mark the problem in. All ideas and inputs and solutions are appreciated. [I have a high-school math background] geometry optimization vectors euclidean-geometry circles Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 16, 2019 at 16:53 John_NashJohn_Nash asked Sep 16, 2019 at 16:36 John_NashJohn_Nash 121 6 6 bronze badges 19 Interesting question. Where does it come from? Have you tried solving it for a triangle (three circles)? Four circles?saulspatz –saulspatz 2019-09-16 16:48:59 +00:00 Commented Sep 16, 2019 at 16:48 @saulspatz yeah I did try it but it starts to get complex pretty quick. Initial reaction was to go with a spiral like arrangement based on visual areas. But even with 3 points, changing their radius can possibly change arrangements .problem I devised last year while in math class.John_Nash –John_Nash 2019-09-16 16:50:49 +00:00 Commented Sep 16, 2019 at 16:50 1 Put in a polar coordinate and let the center be the origin. Suppose P i P i is on the circle of radius r i r i, then the directed area of the polygon P 1⋯P r P 1⋯P r is 1 2∑r i=1 r i r i+1 sin θ i 1 2∑i=1 r r i r i+1 sin⁡θ i, where θ i θ i is the difference of the angular coordinate of P i P i and P i+1 P i+1. Then you can maximize this function by Lagrange multiplier under the constraint ∑r i=1 θ i=2 π∑i=1 r θ i=2 π.Hw Chu –Hw Chu 2019-09-16 17:10:27 +00:00 Commented Sep 16, 2019 at 17:10 2 For n=4 n=4, the optimal solution is placing the 4 points on +ve x x-axis, +ve y y-axis, -ve x x-axis and -ve y y-axis (in that order).achille hui –achille hui 2019-09-16 17:18:44 +00:00 Commented Sep 16, 2019 at 17:18 2 @John_Nash For n=4 n=4, area = 1 2|∑i r i r i+1 sin θ i|≤1 2∑i r i r i+1 1 2|∑i r i r i+1 sin⁡θ i|≤1 2∑i r i r i+1 and the arrangement I mentioned before achieve this upper bound.achille hui –achille hui 2019-09-16 17:26:11 +00:00 Commented Sep 16, 2019 at 17:26 |Show 14 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Here is an analysis using HwChu's suggestion of Lagrange multipliers. As usual, Paul's Online Math Notes have a very accessible discussion of this subject. Since the constraint is ∑θ i=2 π∑θ i=2 π, we have, in the notation of the notes, f(θ 1,θ 2,…,θ n)g(θ 1,θ 2,…,θ n)=1 2∑r i r i+1 sin θ i,=∑θ i f(θ 1,θ 2,…,θ n)=1 2∑r i r i+1 sin⁡θ i,g(θ 1,θ 2,…,θ n)=∑θ i (In the first equation, when i=n i=n we interpret i+1 i+1 as meaning 1 1.) We have ∂f∂θ i λ∂g∂θ i=1 2 r i r i+1 cos θ i=λ∂f∂θ i=1 2 r i r i+1 cos⁡θ i λ∂g∂θ i=λ So we see that θ i=arccos(C r i r i+1)θ i=arccos⁡(C r i r i+1) for some constant C C such that ∑θ i=2 π∑θ i=2 π Except in the n=4 n=4 case, where we can take C=0 C=0, I think this will have to be solved numerically. For example, for n=5 n=5 and radii 1,3,5,7,9 1,3,5,7,9WolframAlpha gives C=2.53046 C=2.53046 EDIT There's appears to be an oversight here. Suppose we have the six radii 1,3,5,7,9,15 1,3,5,7,9,15 so that the products of successive radii are 3,15,35,63,135,15 3,15,35,63,135,15 The largest possible value of C C in the above is 3 3, for otherwise arccos(C 3)arccos⁡(C 3) is undefined. Now cos−1(3 3)+cos−1(3 15)+cos−1(3 35)+cos−1(3 63)+cos−1(3 135)+cos−1(3 15)=7.295585>2 π cos−1⁡(3 3)+cos−1⁡(3 15)+cos−1⁡(3 35)+cos−1⁡(3 63)+cos−1⁡(3 135)+cos−1⁡(3 15)=7.295585>2 π Since cos−1 cos−1 is a decreasing function, we see that the condition cannot be satisfied, which means that the maximum doesn't occur when the origin is inside the polygon. This raises two questions. First, when we do find a critical point by the above method, is it possible that it doesn't correspond to a global maximum, and second, how do we deal with the case where we know the maximum-area polygon doesn't contain the center? Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Sep 16, 2019 at 18:04 saulspatzsaulspatz 53.9k 7 7 gold badges 37 37 silver badges 79 79 bronze badges 2 When we put c back into the formula, in what order have we numbered ri and ri+1 (basically to get the theta i) . Would be quite helpful if you could as an example give the theta values corresponding to the respective radiuses using the calculated C John_Nash –John_Nash 2019-09-17 02:25:49 +00:00 Commented Sep 17, 2019 at 2:25 1 With C=2.530463592649571 C=2.530463592649571, I get the following angles (in degrees):32.48971682500606 80.28789817427759 85.85395738640754 87.69803338454467 73.67039416151239 32.48971682500606 80.28789817427759 85.85395738640754 87.69803338454467 73.67039416151239 saulspatz –saulspatz 2019-09-17 12:52:24 +00:00 Commented Sep 17, 2019 at 12:52 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry optimization vectors euclidean-geometry circles See similar questions with these tags. 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188205
https://www.mayoclinic.org/diseases-conditions/respiratory-syncytial-virus/diagnosis-treatment/drc-20353104
Respiratory syncytial virus (RSV) - Diagnosis & treatment - Mayo Clinic Skip to content Care at Mayo Clinic Patient-Centered Care About Mayo Clinic Request Appointment Find a Doctor Locations Clinical Trials Connect to Support Groups Patient & Visitor Guide Billing & Insurance Departments & Centers International Services Contact Us Patient & Visitor Guide Health Library Diseases & Conditions Symptoms Tests & Procedures Drugs & Supplements Healthy Lifestyle Mayo Clinic Health Letter & Books Mayo Clinic Health Letter &Books For Medical Professionals Medical Professional Resources Refer a Patient Continuing Medical Education AskMayoExpert Mayo Clinic Laboratories Video Center Journals & Publications Mayo Clinic Alumni Association Continuing Medical Education Research & Education at Mayo Clinic Research -------- Research at Mayo Clinic Research Faculty Laboratories Core Facilities Centers & Programs Departments & Divisions Clinical Trials Institutional Review Board Postdoctoral Fellowships Training Grant Programs Education --------- Mayo Clinic College of Medicine and Science Mayo Clinic Graduate School of Biomedical Sciences Mayo Clinic Alix School of Medicine Mayo Clinic School of Graduate Medical Education Mayo Clinic School of Health Sciences Mayo Clinic School of Continuous Professional Development Mayo Clinic College of Medicine and Science Giving to Mayo Clinic Give Now Giving to Mayo Clinic Frequently Asked Questions Contact Us to Give Make a Donation Request appointment Log in Search Menu Request appointment Donate Diseases & conditions Find a doctor Care at Mayo Clinic Patient-Centered Care About Mayo Clinic Request Appointment Find a Doctor Locations Clinical Trials Connect to Support Groups Patient & Visitor Guide Billing & Insurance Departments & Centers International Services Contact Us Patient & Visitor Guide Image 8 Health Library Diseases & Conditions Symptoms Tests & Procedures Drugs & Supplements Healthy Lifestyle Mayo Clinic Health Letter & Books Mayo Clinic Health Letter &Books Image 9 For Medical Professionals Medical Professional Resources Refer a Patient Continuing Medical Education AskMayoExpert Mayo Clinic Laboratories Video Center Journals & Publications Mayo Clinic Alumni Association Continuing Medical Education Image 10 Research & Education at Mayo Clinic Research -------- Research at Mayo Clinic Research Faculty Laboratories Core Facilities Centers & Programs Departments & Divisions Clinical Trials Institutional Review Board Postdoctoral Fellowships Training Grant Programs Education --------- Mayo Clinic College of Medicine and Science Mayo Clinic Graduate School of Biomedical Sciences Mayo Clinic Alix School of Medicine Mayo Clinic School of Graduate Medical Education Mayo Clinic School of Health Sciences Mayo Clinic School of Continuous Professional Development Mayo Clinic College of Medicine and Science Giving to Mayo Clinic Give Now Giving to Mayo Clinic Frequently Asked Questions Contact Us to Give Make a Donation Diseases & Conditions Diseases & Conditions Respiratory syncytial virus (RSV) Request an appointment Symptoms & causes Diagnosis & treatment Diagnosis Your doctor may suspect respiratory syncytial virus based on the findings of a physical exam and the time of year the symptoms occur. During the exam, the doctor will listen to the lungs with a stethoscope to check for wheezing or other abnormal sounds. Laboratory and imaging tests aren't usually needed. However, they can help diagnose RSV complications or rule out other conditions that may cause similar symptoms. Tests may include: Blood tests to check white cell counts or to look for viruses, bacteria and other germs Chest X-rays to check for lung inflammation Swab of secretions from inside the mouth or nose to check for signs of the virus Pulse oximetry, a painless skin monitor, to detect lower than normal levels of oxygen in the blood More Information Chest X-rays Treatment Treatment for respiratory syncytial virus generally involves self-care measures to make your child more comfortable (supportive care). But hospital care may be needed if severe symptoms occur. Supportive care Your doctor may recommend an over-the-counter medication such as acetaminophen (Tylenol, others) to reduce fever. (Never give aspirin to a child.) Use of nasal saline drops and suctioning may help clear a stuffy nose. Your doctor may prescribe antibiotics if there's a bacterial complication, such as bacterial pneumonia. Keep your child as comfortable as possible. Offer plenty of fluids and watch for signs of loss of body fluids (dehydration), such as dry mouth, little to no urine output, sunken eyes, and extreme fussiness or sleepiness. Hospital care If the respiratory syncytial virus (RSV)infection is severe, a hospital stay may be necessary. Treatments at the hospital may include: Intravenous (IV) fluids Humidified oxygen A breathing machine (mechanical ventilation), in rare cases An inhaler (bronchodilator) or steroids are not proved to be helpful in treating RSV infection. Request an appointment From Mayo Clinic to your inbox Sign up for free and stay up to date on research advancements, health tips, current health topics, and expertise on managing health. Click here for an email preview. address1 Subscribe! Learn more about Mayo Clinic’s use of data. We use the data you provide to deliver you the content you requested. To provide you with the most relevant and helpful information, we may combine your email and website data with other information we have about you. If you are a Mayo Clinic patient, we will only use your protected health information as outlined in our Notice of Privacy Practices. You may opt out of email communications at any time by clicking on the unsubscribe link in the email. Self care You may not be able to shorten the length of a respiratory syncytial virus (RSV) infection, but you can try to relieve some signs and symptoms. If your child has RSV, do your best to comfort or distract him or her — cuddle, read a book or play a quiet game. Other tips for relieving symptoms are: Create moist air to breathe. Keep the room warm but not overheated. If the air is dry, a cool-mist humidifier or vaporizer can moisten the air and help ease congestion and coughing. Be sure to keep the humidifier clean to prevent the growth of bacteria and molds. Drink fluids. Continue breastfeeding or bottle-feeding your infant as you would normally. For older children and adults, keep a steady supply of cool water at the bedside. Offer warm fluids, such as soup, which may help loosen thickened secretions. Ice pops may be soothing as well. Try saline nasal drops. Over-the-counter (OTC) drops are a safe, effective way to ease congestion, even for young children. Follow your doctor's recommendations and the instructions on the product. Use over-the-counter pain relievers. OTC pain relievers such as acetaminophen (Tylenol, others) may help reduce fever and relieve a sore throat. Ask a doctor for the correct dose for your child's age. Stay away from cigarette smoke. Secondhand smoke can aggravate symptoms. Preparing for your appointment Unless severe symptoms result in an emergency room (ER) visit, you're likely to start by seeing your family doctor or your child's doctor. Here's some information to help you get ready for your appointment, and know what to expect from your doctor. What you can do Before your appointment, you may want to make a list of: Any symptoms you noticed and when they started, even if they seem unrelated to an upper respiratory infection. Key medical information, such as if your child was born prematurely or if he or she has a heart or lung problem. Details about child care, considering other locations where your family may have been exposed to respiratory infections. Questions to ask your doctor. List your questions from most important to least important in case time runs out. Questions to ask your doctor may include: What is likely causing these symptoms? Are there other possible causes? What tests might be needed? How long do symptoms usually last? What is the best treatment? Is medication needed? If you're prescribing a brand-name medication, is there a generic alternative? What can I do to make my child feel better? Are there any brochures or other printed material that I can take home? What websites do you recommend? To what extent should I isolate my child while infected? Don't hesitate to ask any additional questions you may think of during your appointment. What to expect from your doctor Your doctor is likely to ask you a number of questions, such as: When did you first notice symptoms? Do the symptoms come and go or are they continuous? How severe are the symptoms? What, if anything, seems to improve symptoms? What, if anything, appears to worsen symptoms? Is anyone else in the family ill? What symptoms does he or she have? Your doctor will ask additional questions based on your responses, symptoms and needs. Preparing and anticipating questions will help you make the most of your time with the doctor. Request an appointment By Mayo Clinic Staff Oct 04, 2023 Print Show References Kliegman RM, et al. Respiratory syncytial virus. In: Nelson Textbook of Pediatrics. Elsevier; 2020. Accessed Oct. 22, 2020. Ferri FF. Respiratory syncytial virus. In: Ferri's Clinical Advisor 2021. Elsevier; 2021. Accessed Oct. 22, 2020. Hurley LP, et al. Primary care physicians' perspectives on respiratory syncytial virus (RSV) disease in adults and a potential RSV vaccine for adults. Vaccine. 2019; doi:10.1016/j.vaccine.2018.12.031. Chen X, et al. The microbial coinfection in COVID-19. Applied Microbiology and Biotechnology. 2020; doi:10.1007/s00253-020-10814-6. Respiratory syncytial virus infection (RSV): Symptoms and care. Centers for Disease Control and Prevention. Accessed Oct. 22, 2020. Respiratory syncytial virus infection (RSV): RSV prevention. Centers for Disease Control and Prevention. Accessed Aug. 31, 2023. Respiratory syncytial virus infection (RSV): RSV transmission. Centers for Disease Control and Prevention. Accessed Dec. 21, 2020. AskMayoExpert. Respiratory syncytial virus (RSV) infection (child). Mayo Clinic; 2019. AskMayoExpert. Respiratory syncytial virus (RSV) immunoglobulin series. Mayo Clinic; 2020. Nam HH, et al. Respiratory syncytial virus infection in adults. BMJ. 2019; doi:10.1136/bmj.l5021. American Academy of Pediatrics. Policy Statement ― Updated guidance for palivizumab prophylaxis among infants and young children at increased risk of hospitalization for respiratory syncytial virus infection. Pediatrics. 2014; doi:10.1542/peds.2014-1665. Reaffirmed 2019. Respiratory syncytial virus (RSV) and human metapneumovirus infections. Merck Manual Professional Version. Accessed Oct. 22, 2020. Synagis (prescribing information). Medimmune, LLC; 2017. Accessed Oct. 22, 2020. Respiratory syncytial virus infection (RSV): RSV in older adults and adults with chronic medical conditions. Centers for Disease Control and Prevention. Accessed Dec. 29, 2020. Medical review (expert opinion). Mayo Clinic. Dec. 14 and Dec. 30, 2020. Respiratory syncytial virus infection (RSV): RSV in infants and young children. Centers for Disease Control and Prevention. Accessed Dec. 29, 2020. Jefferson JM, et al. Use of nirsevimab for the prevention of respiratory syncytial virus disease among infants and young children: Recommendations of the Advisory Committee on Immunization Practices — United States, 2023. MMWR Morbidity and Mortality Weekly Report, 2023. 72:793-801. FDA approves first vaccine for pregnant individuals to prevent RSV in infants. U.S. Food and Drug Administration. Accessed Oct. 3, 2023. RSV in infants and young children. Centers for Disease Control and Prevention. Accessed Oct. 3, 2023. Respiratory syncytial virus (RSV) preventive antibody: Immunization information statement (IIS). Centers for Disease Control and Prevention. Accessed Oct. 3, 2023. RSV immunization for children 19 months and younger. Centers for Disease Control and Prevention. Accessed Oct. 3, 2023. Should I get the RSV vaccine during pregnancy? American College of Obstetricians and Gynecologists. Accessed Oct. 3, 2023. Frequently asked questions about RSV vaccine for adults. Centers for Disease Control and Prevention. Accessed Oct. 3, 2023. Medical review (expert opinion). Mayo Clinic. Oct. 2, 2023. Medical review (expert opinion). Mayo Clinic. Oct. 3, 2023. Symptoms & causes Diseases & Conditions Respiratory syncytial virus (RSV) - Diagnosis & treatment - Mayo Clinic News from Mayo Clinic (VIDEO) The reality of RSV, who should get vaccinated Associated Procedures Chest X-rays Products & Services A Book: Mayo Clinic Family Health Book Newsletter: Mayo Clinic Health Letter — Digital Edition CON-20309861 Mayo Clinic does not endorse companies or products. Advertising revenue supports our not-for-profit mission. 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188206
https://calculatorlib.com/zh/converter/min-to-hr
分享到 Facebook 分享到 Twitter 分享到 WhatsApp English Español Tiếng Việt हिन्दी 日本語 한국어 Türkiye العربية Pусский Français 中文(简体) 中文(繁體) 广告 目录 从分钟到小时的转换 数学公式: divide the time in 分钟 by 60 示例 将 1 分钟 转换为 小时： 换算表 从分钟到小时的转换 1 分钟 (分钟) 等于 0.0166667 小时 (小时)。 1 分钟 = 0.0166667 小时 或 1 小时 = 60 分钟 数学公式 要将 时间 从 分钟 (分钟) 转换为 小时 (小时)，请将 时间 在 分钟 中除以 60 t(小时) = t(分钟) / 60 示例: 将 1 分钟 转换为 小时： t(小时) = 1分钟 / 60 = 0.0166666667 小时 换算表 | 分钟 (分钟) | 小时 (小时) | --- | | 0.1 分钟 | 0.001666667 小时 | | 0.5 分钟 | 0.008333333 小时 | | 1 分钟 | 0.016666667 小时 | | 3 分钟 | 0.05 小时 | | 5 分钟 | 0.083333333 小时 | | 7 分钟 | 0.116666667 小时 | | 9 分钟 | 0.15 小时 | | 10 分钟 | 0.166666667 小时 | | 30 分钟 | 0.5 小时 | | 50 分钟 | 0.833333333 小时 | | 70 分钟 | 1.166666667 小时 | | 90 分钟 | 1.5 小时 | | 1000 分钟 | 16.666666667 小时 | | 10000 分钟 | 166.666666667 小时 | 单位转换器 面积 数字存储 能量 频率 长度 质量 压力 时间 体积测量
188207
https://english.meta.stackexchange.com/questions/7052/is-google-dictionary-a-valid-definition-reference-in-particular-in-answers
Is Google Dictionary a valid definition reference (in particular in answers)? - English Language & Usage Meta Stack Exchange Join English Language & Usage Meta Joining the parent site gives you automatic access to this meta. Please follow the link below to sign up. Continue to sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Back to English Language & Usage Stack Exchange Return to the main site Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have What's Meta? How Meta is different from the main site About Us Learn more about Stack Overflow the company, and our products current community English Language & Usage helpchat English Language & Usage Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered Tags Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Is Google Dictionary a valid definition reference (in particular in answers)? Ask Question Asked 10 years, 1 month ago Modified3 years, 8 months ago Viewed 25k times This question shows research effort; it is useful and clear 19 Save this question. Show activity on this post. In my answer to this question, another user was adamant that Google is not a citable source and that that is a policy of ELU. The link given as evidence for that was Is TFD a RTFM online source? - a footnote to an answer to a different question so I am not satisfied with that as an authoritative answer to this question. Note that this and many other similar questions/answers were not just LMGTFY. The answer was the word (I didn't Google it to get it). The Google links are just convenient definitions of the word. This question is not about LMGTFY questions and answers. It is only about whether Google Dictionary is a valid reference. I have searched on Help and Meta and cannot find anything that explicitly says it is or isn't. There is the list of recommended dictionaries but it is only a list of recommendations not an approved list where anything else is forbidden. Google isn't on that list but neither is Collins for example. What's up with all these dictionary reference edits? comes closest but there is a disagreement of opinions there. There is speculation about where Google gets their definitions: some suggest they get them from various other online dictionaries (from leeching the definition from the first dictionary link). This doesn't seem to be supported by the wikipedia article on Google Dictionary Google Dictionary was an online dictionary service of Google, originating in its Google Translate service. The Google Dictionary website was terminated on August 5, 2011 after part of its functionality was integrated into Google Search using the define: operator. ...the content now came from another Oxford dictionary, the Oxford American College Dictionary. As of 5 August 2011, Google Dictionary was discontinued as a separate service, although similar results have been incorporated into the Google search sidebar. So it seems that Google have their own dictionary, copied from one or more other dictionaries perhaps but, if so, they undoubtedly have the intellectual rights to do so. They provide this as part of a search (much like searching for "the time now" shows the current time) rather than as a separate URL as they did before. There is of course no plagiarism on their part. They cite no references and don't need to (just like other dictionaries). Personally, I see no reason that Google Dictionary (now incorporated into Google Search) shouldn't be cited as a reference. So long as the definition it gives is copied into the answer (as all such references should) then any volatility of Google's dictionary is hardly an issue (the definition at the time of writing was shown). The source of the definition was Google and was cited. I see no problem with that. There is a potential problem with Google search links in that many users copy a lot of unnecessary junk in the URL but this question only relates to properly sanitised Google search links: e.g. So the question is in two parts: 1) Is there a policy on ELU that Google Dictionary is not acceptable for providing the definitions of words? (If so, please provide the link to where that is explicitly stated. Someone's opinion buried within a question on another subject on meta will not do.) 2) If not, should there be? (And why?) Conclusion I expect everyone who has an opinion has voted and the consensus is quite clear: Google dictionary is not to be referenced. The down votes to the question are curious. Does that mean people feel it shouldn't have been asked or just that it was asked badly? I am glad it has been and I now know where ELU community stands on the question. I don't think it was clear before. People can say Google is not a citable source with a lot more confidence and authority now. discussion reference policy Share Share a link to this question Copy linkCC BY-SA 3.0 Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:38 CommunityBot 1 asked Aug 11, 2015 at 15:51 AvonAvon 5,268 1 1 gold badge 11 11 silver badges 8 8 bronze badges 8 3 There's no approved/forbidden list. There is more vs less respected. Urban Dictionary has a lot of crap in it but sometimes has some good suggestions for slang that don't exist anywhere else. TFD is quick and easy and sources a lot of dictionaries including wiktionary which is still in its infancy. OED, best in the world, still doesn't give every nuance. Google dictionary I'm sure is just fine; everything is up for discussion.Mitch –Mitch 2015-08-11 16:26:27 +00:00 Commented Aug 11, 2015 at 16:26 2 Realted (possible dupe): meta.english.stackexchange.com/q/2267/8019.Tim Lymington –Tim Lymington 2015-08-11 17:27:17 +00:00 Commented Aug 11, 2015 at 17:27 @TimLymington I don't think it is a dupe because this is specifically about Google: whether Google is a special case for whatever reason. If not then that question does provide a general answer. Do I take it from this that you think Google Dictionary isn't a special case?Avon –Avon 2015-08-11 17:44:58 +00:00 Commented Aug 11, 2015 at 17:44 2 You are still eliding 'Google' and 'Google dictionary', which appears to have caused much of the confusion in the original question. Far too often, alas, we get posts saying "'if not then that' [as in your comment] gets 20000 hits on Google, so it must be a recognised phrase; what does it mean?". For that reason Google itself is not citable here as any sort of authority. Google Dictionary can be quoted for what it may be worth: not as much as a properly researched, independent dictionary but more than a random individual's view. And why should it be a special case?Tim Lymington –Tim Lymington 2015-08-11 19:22:28 +00:00 Commented Aug 11, 2015 at 19:22 @TimLymington Google obviously do have a dictionary (cached - like all searches only a lot more static). Google and its dictionary are not the same thing. That they bought it off someone else without exclusive rights seems to be true but it is not withstanding of the fact that they do have a dictionary. If they are quoting a "a properly researched, independent dictionary" as their own (with the permission of that dictionary) then... what is the difference? I don't think its dictionary should be a special case.Avon –Avon 2015-08-11 20:16:44 +00:00 Commented Aug 11, 2015 at 20:16 8 Just yesterday I edited an answer in which the answerer seemed to be quoting a dictionary (or two) for two listed definitions, but didn't provide any credit or link at all. A Google search for the quoted wording yielded one match to Dictionary.com and one match to Free Dictionary, but closer inspection of those two matches revealed that both definitions were actually from Collins Dictionary. In my view, the appropriate links were to Collins, so that's the source I linked to. It doesn't make sense to me to cite a content aggregator when you can cite an original source instead.Sven Yargs –Sven Yargs 2015-08-12 10:27:57 +00:00 Commented Aug 12, 2015 at 10:27 2 Citable is one thing; valid is another, and this parallels the whole Wikipedia thing in academic writing. Only primary sources are citable, and neither a Google definition nor Wikipedia is a primary source and hence are not citable. They are, however, as valid (in the sense of being trustworthy) as the sources they use.Roaring Fish –Roaring Fish 2015-08-22 05:19:15 +00:00 Commented Aug 22, 2015 at 5:19 The phonemic notation used in English Wikipedia is now quite reliable, though it tends to be more American than British. It's much better than the pronunciation in the OED.John Lawler –John Lawler 2022-01-24 01:44:59 +00:00 Commented Jan 24, 2022 at 1:44 Add a comment| 4 Answers 4 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. Meta posts and votes on them define policy, in that (insofar as Stack Exchange allows) the community decides how it should operate. Some aspects of ELU are mandated by the company; one such is the attribution policy. In the absence of anything better, then the footnote on this answer with its upvotes is part of site policy. However, I’ll state it explicitly so it can be voted upon. [The attribution policy] does make for problems when answers simply reproduce the content Google supplies from a define query. Presumably Google have an agreement with their sources where they can omit any reference. I don't think we should endorse that, particularly when it's easy to provide cited references from original dictionary websites. Google is a directory. They have recently taken to providing answers directly on their site, but do so without any attribution. We would not accept a link provided by Google as an answer: an answer should be more than a link and should reference source material. Similarly, we would not accept a Google search link and text from the search results, which is arguably close to providing a dictionary search link and data from the dictionary results. If this site were one where the questions lent themselves to Google’s calculator, we would not accept a bald figure as an answer: the method of calculation would have to be shown. Consequently, no, we should not accept Google’s dictionary definitions at face value either. Google does not acknowledge its sources, despite simply reproducing content†; Stack Exchange policy does not allow unacknowledged material. The work is not Google’s: they have reproduced the work of others, and Stack Exchange policy is that that work should be acknowledged. A side issue, but related, is that answers which reference Google definitions tend — often, but not always — to include only Google’s content. If there is anything which is original at all, it is very little. Stack Exchange answers are supposed to be complete, and are expected to show some originality of thought. Regurgitating Google’s summary of another site is short-changing the asker and providing a bad example to follow. The quality of answers generally goes down, and if saying “No Google” is a way to stem that progress [regress?] then to do so is useful. As I said in my original footnote, it’s easy to provide cited references from original dictionary websites. † This is easily demonstrated by using Google’s own search engine to find the real source of their text. Share Share a link to this answer Copy linkCC BY-SA 3.0 Follow Follow this answer to receive notifications edited Mar 16, 2017 at 16:02 CommunityBot 1 answered Aug 11, 2015 at 19:49 Andrew LeachMod Andrew Leach 104k 3 3 gold badges 50 50 silver badges 114 114 bronze badges 14 You have said quite a lot here so bear with me: In the absence of anything better, then the footnote on this answer with its up-votes is part of site policy. Is that really how it works? A few up-votes and everything it says, even on a point tangential to the answer it is giving, is policy?Avon –Avon 2015-08-11 20:28:03 +00:00 Commented Aug 11, 2015 at 20:28 If we knew the source of Google's dictionary (oxforddictionaries.com as far as I can tell - 100% on the few words I've tried) then would that change matters in your opinion?Avon –Avon 2015-08-11 20:40:33 +00:00 Commented Aug 11, 2015 at 20:40 @Avon: It's obvious you don't agree. Write your own answer with your point of view (rather than simply attempting to rebut mine). Thank you. Specifically, no: we do know that their source is Oxford, but that doesn't change what I wrote here.Andrew Leach –Andrew LeachMod 2015-08-11 20:42:32 +00:00 Commented Aug 11, 2015 at 20:42 1 I am asking questions about your answer. I don't think it would be appropriate to put those in a separate answer.Avon –Avon 2015-08-11 20:44:10 +00:00 Commented Aug 11, 2015 at 20:44 3 To answer your other supplementary: Yes, in the absence of anything better (like this question on specifics) that really is how the site works. But Stack Exchange corporation's views override everything, whether people here upvote them or not.Andrew Leach –Andrew LeachMod 2015-08-11 21:18:59 +00:00 Commented Aug 11, 2015 at 21:18 Unfortunately, this issue highlights the bigger problem in that many answers don't contain any citation or supporting text whatsoever, yet they often have large numbers of up-votes and are accepted by the OPs. I have seen answers given that are incorrect, that is, if you take any notice whatsoever of a dictionary definition, but when they receive in excess of 50 up-votes, one has to assume that dictionaries are not a significant factor! Of course, language is constantly evolving, but I'm not sure what Stack Exchange feels about actively being part of that?Julie Carter –Julie Carter 2015-08-12 23:08:00 +00:00 Commented Aug 12, 2015 at 23:08 @Julie Carter: I'd be interested to see one or two examples of ELU answers with 50+ upvotes which you feel are completely at odds with dictionary definitions. But I would just say there are plenty of neologisms familiar to tens of millions of native speakers which haven't yet made it into the OED (some never will, but we don't know which). No other dictionary comes close to the comprehensiveness of OED in any language, but it certainly ain't (and doesn't list) the last word in English! :)FumbleFingers –FumbleFingers 2015-08-13 00:10:26 +00:00 Commented Aug 13, 2015 at 0:10 @FumbleFingers exampleJulie Carter –Julie Carter 2015-08-13 00:15:27 +00:00 Commented Aug 13, 2015 at 0:15 2 @Julie Carter: OED novicea person who is new to the circumstances in which he or she is placed. I don't see how the answerer's brand new beginning-beginner conflicts with that. But unless other users have the paid-for subscription (or it happens to be the randomly-chosen day of the month when they allow free access) they might not be able to access that definition. (Though they still might, because the free online oxforddictionaries site copies some definitions verbatim.)FumbleFingers –FumbleFingers 2015-08-13 00:31:45 +00:00 Commented Aug 13, 2015 at 0:31 2 ...actually, I just tried googling that definition for you. It's not been replicated by oxforddictionariesonline, but you might wish to note that of several citations I did find, this one clearly has novice as the most basic category, belowa person inexperienced or unskilled in a particular activity (amateur). That usage seems fine to me, but your comments on the linked ELU question suggest you have a problem with it.FumbleFingers –FumbleFingers 2015-08-13 00:37:47 +00:00 Commented Aug 13, 2015 at 0:37 @FumbleFingers this site has problems in maintaining its aims and I'm not wasting any more of my energy over the matter. Not seeing the wood for the trees comes to mind.Julie Carter –Julie Carter 2015-08-13 06:17:25 +00:00 Commented Aug 13, 2015 at 6:17 Google is a lexicon. Google Dictionary is one of the Oxford Dictionaries. oxforddictionaries.com/licensing/googleJesse Ivy –Jesse Ivy 2018-01-13 01:16:55 +00:00 Commented Jan 13, 2018 at 1:16 @AndrewLeach Google owns a massive amount of data (about you and me and all the trees) and they specialize in indexing. And navigation. Correct.Jesse Ivy –Jesse Ivy 2018-01-13 03:48:28 +00:00 Commented Jan 13, 2018 at 3:48 @AndrewLeach What makes you think they would own an inept dictionary?Jesse Ivy –Jesse Ivy 2018-01-13 03:50:53 +00:00 Commented Jan 13, 2018 at 3:50 Add a comment| This answer is useful 12 Save this answer. Show activity on this post. Don't cite Google as a dictionary. Google has licensed two of the dictionaries from Oxford Languages for their search product: The Oxford Dictionary of English The New Oxford American Dictionary These aren't bad dictionaries, mind you. The ODE is a large single-volume dictionary which gives very good coverage of Present-Day English, including many examples taken from the corpus Oxford used to assemble the dictionary. The NOAD is a version of this dictionary, not quite as good, which focuses on American English. The ODE and NOAD are not called by these names online; instead, they've been rebranded as Lexico (formerly Oxford Dictionaries, or Oxford Living Dictionaries). Whatever you call them, it's confusing. The OED is a very different dictionary – a large multi-volume historical dictionary, not specifically focused on Present-Day English – but the acronym is very similar to ODE, so they're easily confused. And the "Oxford Dictionaries" name is likewise confusing, as the OED is the most well-known Oxford dictionary, but it isn't available at the Oxford Dictionaries website. So it's a bit of a mess. But referring to these as the "Google dictionary" just adds to the confusion. Although Google has the license to present these search results without attribution, they haven't compiled their own dictionaries, and if you don't name the actual source, it's difficult for people to learn about the dictionaries themselves – their advantages and disadvantages, how they were compiled, what their purpose and coverage is, and so forth. What's worse, Google doesn't return these results to all users. Many users on this site use Google in languages other than English, but the English dictionary results are not shown to all of these users. If someone cites the "Google dictionary" with a link to a Google search, the link simply won't work for many of the users on this site. That means the link is not a proper citation, and it should not be used on this site. Even worse, imagine Google licensing another dictionary in the future. Apple did this with the Japanese language dictionaries included with its operating system a few years back, dumping one set of dictionaries and replacing them with another when it made financial sense to do so. Since Google doesn't name their sources, you can imagine this could turn out to be terribly confusing! Linking to a Google search is just a bad idea. If you want to link to the ODE or NOAD, please link directly to the Oxford Dictionaries website, Lexico.com. Share Share a link to this answer Copy linkCC BY-SA 4.0 Follow Follow this answer to receive notifications edited Jan 23, 2022 at 13:58 LaurelMod 68k 1 1 gold badge 28 28 silver badges 56 56 bronze badges answered Jun 15, 2018 at 10:02 user28567 user28567 2 Can you comment on the other dictionaries that come up -after- the 'official' google definition (TFD, MW, Collins, etc) and how they relate to ODE and OED? Can you comment on how one -should- reference all these?Mitch –Mitch 2018-06-15 13:29:41 +00:00 Commented Jun 15, 2018 at 13:29 I strongly disagree that the ODE and the NOAD give very good coverage of Present-Day English! Here's my argument against that claim: english.stackexchange.com/questions/499613/… I'm anxious to know if it sounds persuasive to you.dreeves –dreeves 2019-05-25 06:36:43 +00:00 Commented May 25, 2019 at 6:36 Add a comment| This answer is useful -4 Save this answer. Show activity on this post. I think it should be at least as valid as the dictionary that it reproduces: oxforddictionaries.com (an ELU recommended dictionary) it appears. If Google have been given the right to call it their own then it is not unattributed on any meaningful level (morally or legally) - it is attributed to Google as is their right. OED, for example, have a right to publish 'their' definitions because of intellectual property contracts between the company and its staff that produce them. Google has rights by a different route to the works of oxforddictionaries.com's staff. I see no difference. Any other feelings about Google as evidence depends on the subject. This is specifically and solely about definitions of words in Google. The validity of Google for other references or evidence is irrelevant. Share Share a link to this answer Copy linkCC BY-SA 3.0 Follow Follow this answer to receive notifications answered Aug 11, 2015 at 20:59 AvonAvon 5,268 1 1 gold badge 11 11 silver badges 8 8 bronze badges 1 5 "It appears" and "If" are the problem here. Google has very many sources, some good and others bad. Merely citing 'Google' says nothing about the source, so is not a valid citation. If it is from a good source, it is worth finding the origin.Tim Lymington –Tim Lymington 2015-08-11 21:32:50 +00:00 Commented Aug 11, 2015 at 21:32 Add a comment| This answer is useful -6 Save this answer. Show activity on this post. Yes. And no. In my opinion Google Dictionary is a citable source. However, this site requires that all sources be cited with an internet link. As ironic as that in and of itself may be, with regards to the credibility of google I have a link that OED UK emailed me personally upon inquiry. Turns out my suspicion was correct and Google Dictionary is one of the Oxford Dictionaries. It is a great one, which is under the control of nerds, and edited by Oxford. Google and Oxford are partners. Unfortunately we can't use it here because google doesn't appear to provide links to their dictionary directly. I suppose the search results link technically does however satisfy the requirement. I'm very disappointed by the rejection. Share Share a link to this answer Copy linkCC BY-SA 3.0 Follow Follow this answer to receive notifications answered Jan 13, 2018 at 1:23 Jesse IvyJesse Ivy 520 2 2 silver badges 5 5 bronze badges 5 4 It is inaccurate to state that this site strictly requires a link. This is not true. What it — and the entire network — strictly requires is attribution duly provided that can be used to find the actual source when you copy in others’ text. Think of it as an entry in a bibliography. "I found this by Googling it up" is not a valid citation, and because Googling a definition does not say whom it derived its text from, it is impossible to verify. Moreover, the same lookup produces different answers for different people at different times. Define pastatchrist –tchristMod 2018-01-13 02:22:37 +00:00 Commented Jan 13, 2018 at 2:22 3 Please stop using “OED” to mean Oxford Living Dictionaries or Oxford Dictionaries Online. That is not what OED means. OED means this; please do not continue to conflate the two.tchrist –tchristMod 2018-01-13 02:28:11 +00:00 Commented Jan 13, 2018 at 2:28 @tchrist I am neither referring to OED nor ODO, I am referring to Google Dictionary which is a Lexicon. This is the "real" world. They are maintaining a valuable source. Google Dictionary is one of the Oxford Dictionaries. So too are OED, and ODO, as confusing as that is. I guess in some of my writing one might replace the term "OED" Which refers to the difinitive source for English Language Reference with Oxford University Press, it's publisher. Citing "Source: Google Dictionary" is descriptive, refers to an Oxford Dictionary, and I've never had anyone complain about it. It's a good source.Jesse Ivy –Jesse Ivy 2018-01-13 03:32:38 +00:00 Commented Jan 13, 2018 at 3:32 @tchrist And yes. That is who I emailed personally. They replied. They own oxforddictionaries.com and they are scholars.Jesse Ivy –Jesse Ivy 2018-01-13 03:35:07 +00:00 Commented Jan 13, 2018 at 3:35 @tchrist I have provided a link, and I am pleased that it has become meta.Jesse Ivy –Jesse Ivy 2018-01-13 03:44:31 +00:00 Commented Jan 13, 2018 at 3:44 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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https://www.purplemath.com/modules/numtypes.htm
Published Time: Thu, 04 Jan 2024 19:57:48 GMT What are the different types, or kinds, of numbers? | Purplemath Skip to main content Home Lessons HW Guidelines Study Skills Quiz Find Local Tutors Demo MathHelp.com Join MathHelp.com Login Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Search Number Types Purplemath What are the different types of numbers? The different types of numbers are the counting numbers, the natural or whole numbers, the integers, the rationals and irrationals, the real numbers, the imaginary numbers, and the complex numbers. Content Continues Below MathHelp.com Number Types: Rational or Irrational What are the counting numbers? The counting numbers are the numbers that you first learned as a little child, being the numbers that you used to count things; namely, the counting numbers are: 1, 2, 3, 4, 5, 6, … The "…", or "ellipsis", means "and so on and so forth, forever". What are the natural or whole numbers? Affiliate The natural or whole numbers are the counting numbers, together with zero; in other words: 0, 1, 2, 3, 4, 5, 6, … Advertisement Sometimes the counting numbers are called natural numbers. If your textbook or instructor uses slightly different terminology or definitions, make sure that you're clear on which answers will get you full credit; and be aware that a different textbook or instructor might use different terminology or definitions. Mathematicians have a symbol that they use for the set of all natural numbers; namely, ℕ. Historical note: The number zero came to Europe from India via north-African scholars. Zero, also called "the cipher", was originally viewed by European authorities with suspicion; they said that it, along with the other Arabic numerals, were conducive of fraud, and the use of zero was sometimes banned outright. What are the integers? The integers are zero, the counting numbers, and the negatives of the counting numbers; namely, the integers are: …, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, … Mathematicians have a symbol that they use for the set of all integers; namely, ℤ. The double-struck Z stands for "Zahlen", being the German word for "numbers". What are the rational numbers? The rational numbers are the fractions, where the numerators (top numbers) and denominators (bottom numbers) are integers. While counting numbers, natural/whole numbers, and integers all have the property of being one unit apart (so there is always a "next" number of these types), fractions can be any rational distance apart. For any given fraction, there is no "next" fraction that immediately follows it. Mathematicians have a symbol that they use for the set of all rationals; namely, ℚ. The double-struck Q stands for "quotient", which is a helpful reminder that all fractions are divisions. What are the irrational numbers? The irrational numbers are the numbers which cannot be expressed as fractions. For instance, √2 cannot be expressed as a fraction. For any given irrational, there is no "next" number that immediately follows it. There is no symbol for the set of irrationals. What is the difference between rationals and irrationals when they're written as decimals? Fractions (also called rational numbers) can be written as terminating decimals (that is, decimal numbers that ends after some finite number of decimal places) or as repeating decimals (that is, as decimal numbers that have some block of decimal places that repeates forever). Irrational numbers, on the other hand, have decimal forms that never end and never have a continually-repeating block of decimal places. Rationals are nice and neat; irrationals are very much not nice or neat, at least in decimal form. For example, the fraction 1 2\frac{1}{2}2 1​ can be written in decimal form as 0.5, and 19 25\frac{19}{25}25 19​ can be written as 0.76. These two fractions are terminating decimals. The fraction 1 3\frac{1}{3}3 1​ can be written in decimal form as 0.333333…, and 7 13\frac{7}{13}13 7​ can be written as 0.538461538461…. These two fractions are repeating decimals. In the first case, the repeated block is just 3; in the second case, the repeated block is 538461. On the other hand, we have loads of other numbers whose decimal forms are non-repeating, non-terminating decimals; these number are non-rational (that is, they cannot be written as ratios of two integers); this is why they are called the "irrationals". Examples of irrationals in decimal form would be 2\sqrt{2\,}2​=1.41421356…, or the number π=3.14159.... (Yes, you may have used a fraction as an approximation of π, but that fraction did not equal π.) The rationals and the irrationals are two totally separate number types; there is no overlap. Content Continues Below What are the real numbers? The real numbers are the integers, the rationals, and the irrationals. As is the case with rational numbers, there is, for any given real number, no "next" real number that immediately follows it. Mathematicians have a symbol that they use for the set of real numbers; namely, ℝ. (Once you know the symbol for the real numbers, they can use symbols to indicate the set of irrational numbers. The irrationals are the reals, less the rationals; symbolically, the irrationals are ℝ∖ℚ.) How are these number types related to each other? Note that each new type of number is either the opposite of the previous type, or else it contains the previous type within it. The natural numbers are just the counting numbers with zero thrown in. The integers are just the natural numbers with the negatives of the counting numbers thrown in. And the fractions are just the integers with all their divisions thrown in. (Remember that you can turn any integer into a fraction by putting it over the number 1. For example, the integer 4 is also the fraction 4 1\frac{4}{1}1 4​.) The rationals and the irrationals are sort of opposite types of numbers: the rationals can be written as fractions and the irrationals can not be written as fractions. Putting these two sets of numbers together, we get the set of all real numbers. Are there any numbers that aren't real numbers? Affiliate Affordable tutors for hire Find tutors There are numbers that are not real numbers: they are called the imaginary numbers. (No, I'm not making that up.) They are the numbers that include the number i, which is defined as i=−1\small{ i = \sqrt{-1\,} }i=−1​. Imaginaries allow you to take the square roots of negative numbers, and they are crucial to electrical engineering, physics, and other areas of science. You can then add real numbers and imaginary numbers, and get the complex numbers, indicated by ℂ. Complex numbers can be written in different ways; one of them is as a+b i, where a and b are real numbers. If b=0, then you've got a regular real number. If a=0, then you've got a strictly imaginary number. The set of complex numbers is the reals, the imaginaries, and their sums. If you've never seen numbers with i in them, then every number you've ever seen is a real number. The most common question I hear regarding number types is something along the lines of "Is a real number irrational, or is an irrational number real, or neither... or both?" Unless you know about complexes, everything you've ever done has used real numbers. Unless the number has an "i" in it, it's a real number. And counting numbers, natural/whole numbers, rationals, and irrationals are all inside the set of real numbers. Here are some typical number-type questions (assuming that you haven't yet learned about imaginaries and complexes): True or False: An integer is also a rational number. Since any integer can be formatted as a fraction by putting it over 1, then this statement is true. True or False: A rational number is also an integer. Not necessarily; the integer 4 is also the rational number 4 1\frac{4}{1}1 4​ but, for instance, the rational number 2 3\frac{2}{3}3 2​ is not also a integer. So this statement is false. True or False: A number is either a rational number or an irrational number, but not both. True! In decimal form, a number is either non-terminating and non-repeating (so it's an irrational) or else it's not (so it's a rational); there is no overlap between these two number types! Classify according to number type; some numbers may be of more than one type. 0.45 This is a terminating decimal, so it can be written as a fraction: 45 100=9 20\frac{45}{100} = \frac{9}{20}100 45​=20 9​. Since this fraction does not reduce to a whole number, then it's not an integer or a natural. And everything is a real, so the answer is: rational, real 3.14159265358979323846264338327950288419716939937510... You probably recognize this as being π, though this may be more decimal places than you customarily use. The point, however, is that the decimal does not repeat and does not end, so π is an irrational. And everything (that you know about so far) is a real, so the answer is: irrational, real 3.14159 Don't let this fool you! Yes, you often use something like this as an approximation of π, but it isn't π! This is a rounded decimal approximation, and, since this approximation terminates, this is actually a rational, unlike π itself, which is irrational! The answer is: rational, real Affiliate 10 Obviously, this is a counting number. That means it is also a whole number and an integer. Depending on the text and teacher (there is some inconsistency), this may also be counted as a rational, which technically-speaking it is. And of course it's also a real. The answer is: natural, whole, integer, rational (possibly), real 5 3\boldsymbol{\color{green}{\frac{5}{3}}}3 5​ This is a fraction, so it's a rational. It's also a real, so the answer is: rational, real 1 2 3\boldsymbol{\color{green}{1\,\frac{2}{3}}}1 3 2​ This can also be written as 5 3\frac{5}{3}3 5​, which is the same as the previous problem. The answer is: rational, real −81\boldsymbol{\color{green}{-\sqrt{81\,}}}−81​ Your first impulse may be to say that this is irrational, because it's a square root, but notice that this square root simplifies: −81=−9-\sqrt{81\,} = -9−81​=−9, which is just an integer. The answer is: integer, rational, real −9 3\boldsymbol{-\frac{9}{3}}−3 9​ This number is stated a fraction, but notice that it reduces to −3, so this may also count as an integer. The answer is: integer (possibly), rational, real Except for the section in your book where you have to classify numbers according to type, you really won't need to be terribly familiar with this hierarchy. It's more important to know what the terms mean when you hear them. For instance, if your teacher talks about "integers", you should know that the term refers to the counting numbers, their negatives, and zero. (If you're wondering, yes, there are other classifications of numbers. Many other classifications.) URL: Select a Course Below Standardized Test Prep ACCUPLACER Math ACT Math ALEKS Math ASVAB Math CBEST Math CLEP Math FTCE Math GED Math GMAT Math GRE Math HESI Math Math Placement Test NES Math PERT Math PRAXIS Math SAT Math TEAS Math TSI Math VPT Math + more tests K12 Math 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Math College Pre-Algebra Introductory Algebra Intermediate Algebra College Algebra Homeschool Math Pre-Algebra Algebra 1 Geometry Algebra 2 Share This Page Terms of Use Privacy Contact About Purplemath About the Author Tutoring from PM Advertising Linking to PM Site licencing Visit Our Profiles © 2024 Purplemath, Inc.All right reserved.Web Design by
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https://www.youtube.com/watch?v=8vMsMxley8g
Steps to classify conic sections Brian McLogan 1600000 subscribers Description 8426 views Posted: 9 May 2017 Learn how to classify conic sections. A conic section is a figure formed by the intersection of a plane and a cone. A conic section may be a circle, an ellipse, a parabola, or a hyperbola. The general equation of a conic section is given by Ax^2 + By^2 + Cx + Dy + E = 0. When given the general equation of a conic section, we determine the type of conic section using the following characteristics. For a circle, the values of A and B in the equation must be equal. For an ellipse, the values of both A and B must be positive. For a parabola, either A or B is 0, i.e. there can only be one squares term. For a hyperbola, the value of either A or B is negative. conicsections #hyperbolaconicsections conicsections #hyperbolaconicsections 5 comments Transcript: hello okay so in this example what I'd like to do is show you how to classify a conic section now uh when I'm classifying conic sections basically the way that I kind of work my way down from this is I start with um first identifying if it's a parabola then I look to a circle um and then I look into uh if it's an ellipse or a parabola and the next thing I want to do is we're going to justify not only just classify it but we're going to justify it so the main important thing when I'm looking at first thing I want to do to determine if it's a parabola is to look look at my uh look at my variables and determine do I have an X squ and A y^ squ because if you remember back in Algebra 1 Algebra 2 you know we looked at quadratics that gave us shapes of uh parabolas but they're always vertical right and and our two forms of this you guys can notice that the only variable that's squared is X so when we have a parabola X is going to be squared or Y is going to be squared because we learn in con of sections we can have horizontal parabis so the main important thing is I'm going to look up here and I'm going to see that all of my variables I only have one that's going to be x s so therefore you can state that this is a parabola now in our general form uh a x^2 + B y^2 + CX + Dy + e equal Z which is like our general form um there still is a y squar but that y^ 2 means B has to be zero so therefore if I was going to justify it I would say a B is equal to Z and that's going to be true for any Parabola because even though there's not a y^2 there really is a y^ s but just has zero as its coefficient we could even say it look like this you know 0 y^2 + x^2 okay so it is there just B is equal to zero so anytime your coefficients um or really anytime your coefficients um for x s y^ s is going to be zero meaning it's a non-existent term term that's been shown therefore you can justify that it's being a parabola so there's at least a class which is easy and then a little bit of a justification to show thanks
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https://khanacademy.fandom.com/wiki/Optimization
Khan Academy Wiki Sign In Don't have an account? Register Sign In Skip to content Khan Academy Wiki 2,165 pages in: Math exercises, Differential calculus exercises, Differential calculus: Derivative applications Optimization Sign in to edit History Purge Talk (0) | Optimization | | | | Description | | Exercise Name: | Optimization | | Math Missions: | Integral calculus Math Mission | | Types of Problems: | 3 | The exercise Optimization appears under the Differential calculus Math Mission. This exercise teaches users optimization with Calculus. Types of Problems[] There are three types of problems in this exercise: What is the x-coordinate of the point on the given line that is closest to the origin?: This problem asks the user to find out the point on the given line that is closest to the origin and then give the x-coordinate of that point. At what argument does the area of the rectangle achieve its absolute maximum?: This problem gives a graph with a rectangle in it. The user is asked to find in which argument the area of the rectangle achieves its absolute maximum and then write it in the space provided. What is the maximum vertical distance between the given parabola and the given line for the given rule?: This problem asks the user to find out the maximum distance between the given parabola and the given line for the given rule and write it in the box below. Strategies[] Basic knowledge of optimization and calculus is essential to ensure success while doing this exercise. It is easier to sketch the problem and then solve it. Real-life Applications[] Suspension Bridges: The parabolic shape of the cables along suspension bridges help carry the forces acting on the bridge to the top of the towers (which can usually stand flexing, buckling, and oscillation). Parabolic Flight: Also known as weightlessness, this helps future astronauts, and current ones, experience space without ever leaving the Earth. Trajectories: Everything on the planet Earth is subject to gravity. When on object is thrown or forced in a direction not on the ground, eventually gravity grabs it and drags it down. The path the object follows it that of a parabola. Older Spacecraft: Many of the older space missions followed a parabolic path. When launched, the spacecraft never directly went into the atmosphere... that would surely cause it to burn up. Rather, it entered at an angle. While in space, it would continue to fall to the Earth and eventually, could be recovered. Calculus has massive applications to physics, chemistry, biology, economics and many other fields. Categories Community content is available under CC-BY-SA unless otherwise noted.
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https://www.linkedin.com/advice/0/how-can-you-avoid-off-by-one-errors-when-working-arrays-1irhf
How to Avoid Off-by-One Errors with Arrays Agree & Join LinkedIn By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy. Skip to main contentLinkedIn Top Content People Learning Jobs Games Join nowSign in All Engineering Algorithms How can you avoid off-by-one errors when working with arrays? Powered by AI and the LinkedIn community 1 Understand array indexing 2 Use built-in functions and methods 3 Use loops and iterators wisely Be the first to add your personal experience 4 Test your code with edge cases Be the first to add your personal experience 5 Review your code and use comments Be the first to add your personal experience 6 Here’s what else to consider Off-by-one errors are a common source of bugs and frustration when working with arrays. They occur when you use an incorrect index to access or modify an element, usually by one more or one less than the intended position. In this article, you will learn how to avoid off-by-one errors by following some simple tips and best practices. Top experts in this article Selected by the community from 3 contributions. Learn more Richard Piacentini 🚀 CHEQs Just Landed on Base | Co-Founder & CTO at CHEQs | Architecting Blockchain-Based Autonomous Payment Systems for… View contribution 2 See what others are saying 1 Understand array indexing The first step to avoid off-by-one errors is to understand how array indexing works in your programming language. Different languages may use different conventions for the starting index, the ending index, and the length of an array. For example, in Python, the first element of an array is at index 0, the last element is at index -1, and the length is the number of elements. In Java, the first element is also at index 0, but the last element is at index length - 1, and the length is a property of the array object. Knowing these differences can help you avoid confusion and errors when working with arrays. Add your perspective Help others by sharing more (125 characters min.) Cancel Add Save Richard Piacentini 🚀 CHEQs Just Landed on Base | Co-Founder & CTO at CHEQs | Architecting Blockchain-Based Autonomous Payment Systems for AI Agents Copy link to contribution Report contribution Thanks for letting us know! You'll no longer see this contribution 🧪 Always test your code with arrays at their boundary conditions. 🫙 This includes empty arrays, arrays with only one element, and arrays with many elements. 💡 Pay special attention to how your code behaves at the start and end of the array. 🎯 Write unit tests that cover a variety of cases, including edge cases that test the bounds of the array. Automated tests can quickly catch off-by-one errors. …see more Like Like Celebrate Support Love Insightful Funny 2 2 Use built-in functions and methods Many programming languages provide built-in functions and methods that can simplify the manipulation of arrays and avoid off-by-one errors. For example, in Python, you can use the range function to generate a sequence of indices, or the slice operator to extract a subarray. In Java, you can use the Arrays class to perform common operations like sorting, searching, or copying arrays. Using these built-in features can save you time and reduce the chances of making mistakes. Add your perspective Help others by sharing more (125 characters min.) Cancel Add Save Richard Piacentini 🚀 CHEQs Just Landed on Base | Co-Founder & CTO at CHEQs | Architecting Blockchain-Based Autonomous Payment Systems for AI Agents Copy link to contribution Report contribution Thanks for letting us know! You'll no longer see this contribution 🎯 Leveraging built-in functions and methods in programming languages is indeed a powerful strategy to avoid off-by-one errors when working with arrays. Let's expand on this concept: ➡️ Ruby's Enumerable mixin offers a rich collection of iteration methods like each, map, and select that abstract away the complexities of index handling. ➡️ JavaScript's methods like forEach(), map(), filter(), and reduce() handle iteration internally, which avoids the need for manual index management. …see more Like Like Celebrate Support Love Insightful Funny 2 3 Use loops and iterators wisely Another common source of off-by-one errors is using loops and iterators incorrectly. When you use a loop or an iterator to traverse an array, you need to be careful about the boundary conditions and the increment or decrement steps. For example, in Python, you can use a for loop with the in operator to iterate over the elements of an array, without worrying about the indices. In Java, you can use a for-each loop or an iterator object to achieve the same effect. However, if you use a traditional for loop with indices, you need to make sure that you start from the correct index, end at the correct index, and increment or decrement by the correct amount. Add your perspective Help others by sharing more (125 characters min.) Cancel Add Save 4 Test your code with edge cases One of the best ways to avoid off-by-one errors is to test your code with edge cases. Edge cases are scenarios that involve extreme or unusual values, such as empty arrays, single-element arrays, or arrays with the maximum or minimum possible values. These cases can reveal hidden bugs and errors that may not be apparent with normal inputs. You can use debugging tools, unit testing frameworks, or manual testing techniques to check your code with edge cases and fix any issues that arise. Add your perspective Help others by sharing more (125 characters min.) Cancel Add Save 5 Review your code and use comments Finally, you can avoid off-by-one errors by reviewing your code and using comments. Reviewing your code can help you spot any logical or syntactical errors that may cause off-by-one errors. You can use code review tools, peer review processes, or self-review methods to improve your code quality and readability. Using comments can help you document your code and explain your logic and assumptions. Comments can also help you and others understand your code and avoid confusion and errors. Off-by-one errors can be tricky and annoying, but they can be avoided with some care and attention. By following these tips and best practices, you can work with arrays more confidently and efficiently. Add your perspective Help others by sharing more (125 characters min.) Cancel Add Save 6 Here’s what else to consider This is a space to share examples, stories, or insights that don’t fit into any of the previous sections. What else would you like to add? Add your perspective Help others by sharing more (125 characters min.) Cancel Add Save Richard Piacentini 🚀 CHEQs Just Landed on Base | Co-Founder & CTO at CHEQs | Architecting Blockchain-Based Autonomous Payment Systems for AI Agents Copy link to contribution Report contribution Thanks for letting us know! You'll no longer see this contribution In addition to utilizing built-in functions and methods to avoid off-by-one errors, there are a few more insights and considerations that can be beneficial: Debugging Techniques: Developing good debugging skills can be invaluable in identifying and fixing off-by-one errors. Understanding how to use breakpoints and inspect variables in the array during runtime can help you catch where the error occurs. Mentality for Error Checking: Adopt a mindset that anticipates potential errors. This involves regularly questioning your assumptions about how your code works, especially when it comes to array boundaries. …see more Like Like Celebrate Support Love Insightful Funny 2 Algorithms Algorithms + Follow Rate this article We created this article with the help of AI. What do you think of it? It’s greatIt’s not so great Thanks for your feedback Your feedback is private. Like or react to bring the conversation to your network. Tell us more Tell us why you didn’t like this article. It’s not on a professional topic It contains inaccuracies It has offensive language It has harmful advice It contains stereotypes or bias It’s redundant and unclear Translation quality is poor It’s not relevant in my country, region or culture If you think something in this article goes against our Professional Community Policies, please let us know. Report this article We appreciate you letting us know. Though we’re unable to respond directly, your feedback helps us improve this experience for everyone. If you think this goes against our Professional Community Policies, please let us know. Report this article Cancel Submit Done Report this article More articles on Algorithms No more previous content Struggling to align cross-functional teams' visions for algorithm optimization? You're behind on the latest AI trends. How will you adjust your algorithm design to keep up? You're behind on the latest AI trends. How will you adjust your algorithm design to keep up? No more next content See all Explore Other Skills Programming Web Development Agile Methodologies Machine Learning Software Development Data Engineering Data Analytics Data Science Artificial Intelligence (AI) Cloud Computing Show more Show less More relevant reading Programming What are the best practices for interfacing with high level languages and libraries for programming? 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188212
https://www.statisticshowto.com/errors-in-the-trapezoidal-rule-and-simpsons-rule/
Skip to content Errors in the Trapezoidal Rule and Simpson’s Rule Riemann Sums > The trapezoidal rule and Simpson’s rule are an approximate way to calculate the area under a curve (i.e. a definite integral). It’s possible to calculate how well these rules approximate the area with the Error Bounds formula. The “error” is the difference between the actual “true” value and the approximation. Errors in the trapezoidal rule and Simpson’s rule can be calculated with a couple of straightforward formulas; These are useful when we want to increase the accuracy of an approximation. Increasing the number of partitions leads to better and better approximations: the following formulas give you a way to quantify those errors. Errors in the Trapezoidal Rule and Simpson’s Rule: Formula 1. Error Bounds Formula for Trapezoidal Rule The error formula for the trapezoidal rule is: Where: a, b, = the endpoints of the closed interval [a, b]. max|f′′(x)| = least upper bound of the second derivative. n = number of partitions (rectangles) used. Example Question: What is the error using the trapezoidal rule for the function f(x) = x4 with 4 intervals on [0, 4]? Solution: Step 1: Calculate the second derivative: f′′ = 12x2. If the second derivative is not a continuous function, you cannot use the formula. Step 2: Find the least upper bound (the “max”) of the second derivative on the interval (for this example, the interval is [0, 4]. You can do this in two ways: Look at a graph and locate the max on the interval, or Find the critical numbers and evaluate the function for those numbers (including at the endpoints). Looking at a graph of f′′ = 12x2, we can see that the max value is f(x) = 192. Step 3: Set up the formula and solve: Plugging in our numbers, we get: Where: a, b are given in the question as 0, 4, n = 4 (from the question) max|f′′(x)| = LUB from Step 3. The error between the Trapezoid rule and definite integral is 64. Increasing the number of partitions “n” will result in better approximations. Trapezoid Rule Error - Numerical Integration Approximation 2. Error Bounds Formula for Simpson’s Rule The error formula for Simpson’s rule is similar. The main difference is that it uses the max of the fourth derivative f4: Simpson's Rule Error - Numerical Integration Approximation Comments? Need to post a correction? Please Contact Us. Leave a Comment You must be logged in to post a comment.
188213
https://www.osmosis.org/learn/Fatty_acid_oxidation
Skip to the video Fatty acid oxidation Foundational SciencesBiochemistry and nutritionBiochemistryBiochemistry and metabolismFat and cholesterol metabolism 44,127views 00:00 / 00:00 Fatty acid oxidation Biochemistry and metabolism Amino acid metabolism Nitrogen and urea cycle Citric acid cycle Electron transport chain and oxidative phosphorylation Gluconeogenesis Glycogen metabolism Glycolysis Pentose phosphate pathway Physiological changes during exercise Cholesterol metabolism Fatty acid oxidation Fatty acid synthesis Ketone body metabolism Metabolic disorders Alkaptonuria Cystinuria (NORD) Hartnup disease Homocystinuria Maple syrup urine disease Ornithine transcarbamylase deficiency Phenylketonuria (NORD) Essential fructosuria Galactosemia Glucose-6-phosphate dehydrogenase (G6PD) deficiency Hereditary fructose intolerance Lactose intolerance Pyruvate dehydrogenase deficiency Abetalipoproteinemia Familial hypercholesterolemia Hyperlipidemia Hypertriglyceridemia Glycogen storage disease type I Glycogen storage disease type II (NORD) Glycogen storage disease type III Glycogen storage disease type IV Glycogen storage disease type V Mucopolysaccharide storage disease type 1 (Hurler syndrome) (NORD) Mucopolysaccharide storage disease type 2 (Hunter syndrome) (NORD) Fabry disease (NORD) Gaucher disease (NORD) Krabbe disease Leukodystrophy Metachromatic leukodystrophy (NORD) Niemann-Pick disease type C Niemann-Pick disease types A and B (NORD) Tay-Sachs disease (NORD) Cystinosis Disorders of amino acid metabolism: Pathology review Disorders of carbohydrate metabolism: Pathology review Disorders of fatty acid metabolism: Pathology review Dyslipidemias: Pathology review Glycogen storage disorders: Pathology review Lysosomal storage disorders: Pathology review High Yield Notes 10 pages Flashcards Fatty acid oxidation 0 of 9 complete Questions USMLE® Step 1 style questions USMLE 0 of 3 complete Biochemistry researchers are studying the pathway of fatty acid oxidation with students in the laboratory. Which of the following is the rate-limiting step of fatty acid oxidation? Transcript Watch video only Content Reviewers Rishi Desai, MD, MPH, Viviana Popa, MD Contributors Jake Ryan, Salma Ladhani, MD Our bodies are capable of surviving without food for long periods of time, at least 3-4 weeks with hydration! The reason we can do that is that we can store our dietary fuels, and then break them down when needed to make energy in the form of adenosine triphosphate, or ATP. Fat is one of the most important ways we store energy and the term “burning fat”, actually refers to fatty acid oxidation. In fact, if two individuals were stranded in the Andes mountains with no food, the person with more fat content would survive longer - yet another reason to avoid working out. What makes fat such a great source of energy are fatty acids, which are the simplest form of fats, composed of long chains of carbon and hydrogens. The transfer of electrons in the form of hydrogen from these fatty acids to certain molecules, can then be used to generate ATP. Fatty acid oxidation primarily takes place in the mitochondria of heart, skeletal muscle, and liver cells. Before we can oxidize fat, it needs to be moved from storage sites to the cells that can use it. Fat is stored in adipocytes or fat cells as triglycerides, which are 3 fatty acids attached to a glycerol molecule. Triglycerides can be broken down by the enzyme hormone sensitive lipase, into free fatty acids and glycerol. So if you’re starving in the Andes, first your blood glucose level falls. In response, the pancreas secretes a hormone called glucagon which increases the activity of hormone sensitive lipase, and increases the breakdown of triglycerides. Now, the free fatty acids can leave the fat cell, and enter the bloodstream, where they bind to a protein called albumin. Albumin carries the fatty acids to target cells, like liver cells, that are capable of fatty acid oxidation. First, the free fatty acid dissociates from albumin and diffuses into the cell. Once inside the cell, a cytosolic enzyme called fatty acyl-CoA synthetase adds a coenzyme A molecule to the end of the fatty acid, turning it into a metabolically active fatty acyl-CoA. This process requires 2 ATP molecules - so it takes a little energy to make energy, but it’s totally worth it in the end. Now the mitochondria is composed of two membranes - an outer membrane and an inner membrane, with a small space in between - and the mitochondrial matrix at the core. The enzymes required for beta oxidation are located in the mitochondrial matrix - however, the fatty acid can’t cross the inner mitochondrial membrane when CoA is attached to it. To get around this problem, we need to free the CoA from the fatty acid. An enzyme within the outer mitochondrial membrane called carnitine palmitoyltransferase 1, or CPT1 replaces the CoA with a carnitine, making fatty acyl-carnitine and a free CoA, both of which can easily cross the inner mitochondrial membrane. Then, along the inner mitochondrial membrane, another enzyme called carnitine palmitoyltransferase 2, or CPT2, substitutes carnitine and CoA back, therefore regenerating fatty acyl-CoA and free carnitine - which is now within the mitochondrial matrix. This whole slick process is called the carnitine shuttle. Carnitine can cross the inner mitochondrial membrane by itself, so it can go back to the outer membrane to meet the next incoming fatty acid. We can also get more carnitine from our diet, mainly from meat products. Finally, the carnitine shuttle can be regulated by a product of fatty acid synthesis called malonyl-CoA, which specifically inhibits CPT1 slowing down fatty acid oxidation. After all, you don’t want to make and break fatty acids at the same time. Okay, so fatty acid chains can vary in length, from short, medium, long and very long fatty acids, but the process of fatty acid oxidation is the same for any length. We’ll explain fatty acid oxidation with an example of a long chain fatty acid: the 16-carbon long palmitoyl-CoA. The enzymatic modifications that take place in fatty acid oxidation happen on the 2nd and 3rd carbon of the chain, also called the alpha and beta carbons, respectively. Each of those carbons enters fatty acid oxidation with 2 hydrogens bound to it. First, an enzyme called acyl-CoA dehydrogenase removes 1 hydrogen from the 2nd carbon, and one hydrogen from the 3rd. The same enzyme then gives those 2 hydrogens to a nearby flavin adenine dinucleotide molecule, or FAD, making FADH2, and converting the palmitoyl-CoA to enoyl-CoA in the process. Summary Fatty acid oxidation is the process your body uses to break down and uses fatty acids for energy. This process occurs in the mitochondria of your cells. During fatty acid oxidation, a fatty acid is broken down into two molecules of acetyl coenzyme A (CoA). These molecules are then used by the mitochondria to produce energy. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see ourCookie Policy Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. 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https://www.bbc.co.uk/bitesize/articles/z3xx8hv
Adding decimals with a different number of decimal places - Maths - Learning with BBC Bitesize BBC Homepage Skip to content Accessibility Help Your account Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds More menu More menu Search Bitesize Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Close menu Bitesize Menu Home Learn Study support Careers Teachers Parents Trending My Bitesize More England Early years KS1 KS2 KS3 GCSE Functional Skills Northern Ireland Foundation Stage KS1 KS2 KS3 GCSE Scotland Early Level 1st Level 2nd Level 3rd Level 4th Level National 4 National 5 Higher Core Skills An Tràth Ìre A' Chiad Ìre An Dàrna Ìre 3mh ìre 4mh ìre Nàiseanta 4 Nàiseanta 5 Àrd Ìre Wales Foundation Phase KS2 KS3 GCSE WBQ Essential Skills Cyfnod Sylfaen CA2 CA3 CBC TGAU International KS3 IGCSE More from Bitesize About us All subjects All levels Primary games Secondary games Adding decimals with a different number of decimal places Part ofLearn & revise Save to My Bitesize Save to My Bitesize Saving Saved Removing Remove from My Bitesize Save to My Bitesize close panel Jump to Adding decimals Example 1 Example 2 Example 3 Example 4 Activity Quiz Adding decimals The aim of this guide is to show how to add decimals with a different number of decimal places. We will focus on the formal written method. Let's look at some examples. Back to top Example 1 What is 4.5 + 3.26? Set out the calculation, making sure each digit is lined up in the correct column based on its place value and that the decimal points are directly underneath one another. We need to be careful here as the numbers have a different number of places. You can write 4.5 as 4.50. This does not change the value of the number, but it may help when setting out and completing the calculation. Starting with the column on the right with the least place value, i.e. with hundredths in this example, we say: 6 hundredths plus 0 hundredths is 6 hundredths, 5 tenths plus 2 tenths is 7 tenths and 4 ones plus 3 ones is 7 ones. Back to top Example 2 What is 47.48 + 6.759? The first decimal has two decimal places and the second decimal has three decimal places. Set out the calculation, making sure each digit is in the correct column based on its place value and that the decimal points are directly underneath one another. You can write 7.48 as 7.480. Starting with the thousandths column, 0 thousandths plus 9 thousandths is 9 thousandths. Moving to the hundredths column, 8 hundredths plus 5 hundredths is 13 hundredths, which is the same as 1 tenth and 3 hundredths. Moving to the tenths column, 4 tenths plus 7 tenths plus the 1 ‘extra’ tenth is 12 tenths, which is the same as 1 one and 2 tenths. Moving to the ones column, 7 ones plus 6 ones plus the 1 ‘extra’ one is 14 ones, which is the same as 1 ten and 4 ones. Back to top Example 3 What is 6.1 + 2.685 + 7.49? Set out the calculation, making sure each digit is in the correct column based on its place value and that the decimal points are directly underneath one another. You can write 6.1 as 6.100 and 7.49 as 7.490. Starting with the thousandths column, 0 thousandths plus 5 thousandths plus 0 thousandths is 5 thousandths. Moving to the hundredths column, 0 hundredths plus 8 hundredths plus 9 hundredths is 17 hundredths, which is the same as 1 tenth and 7 hundredths. Moving to the tenths column, 1 tenth plus 6 tenths plus 4 tenths plus the 1 ‘extra’ tenth is 12 tenths, which is the same as 1 one and 2 tenths. Moving to the ones column, 6 ones plus 2 ones plus 7 ones plus the 1 ‘extra’ one is 16 ones, which is the same as 1 ten and 6 ones. Back to top Example 4 What is 7.85 + 6 + 9.4? Set out the calculation carefully. Think about the place value of each digit and ensure that the decimal points are directly underneath one another. Remember that 6 is – of course – 6 ones and can be written as 6.0. You can write 6.0 as 6.00 and 9.4 as 9.40. Starting with the hundredths column, 5 hundredths plus 0 hundredths plus 0 hundredths is 5 hundredths. Moving to the tenths column, 8 tenths plus 0 tenths plus 4 tenths is 12 tenths, which is the same as 1 one and 2 tenths. Moving to the ones column, 7 ones plus 6 ones plus 9 ones plus the ‘extra’ one is 23 ones, which is the same as 2 tens and 3 ones. Back to top Activity Quiz Back to top Play our fun maths game Guardians: Defenders of Mathematica. game Play our fun maths game Guardians: Defenders of Mathematica Use your times tables and more maths skills to defeat monsters and reclaim the Kingdom of Mathematica Back to top Language: Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Terms of Use About the BBC Privacy Policy Cookies Accessibility Help Parental Guidance Contact the BBC BBC emails for you Advertise with us Do not share or sell my info Copyright © 2025 BBC. The BBC is not responsible for the content of external sites. Read about our approach to external linking.
188215
https://www.geogebra.org/m/Rfb7Jp4v
Reflections over Parallel Lines: Direction Parte Troisxqrthyy Instructions for the first set of reflections. Instructions for the second set of reflections. Do the following to BOTH sketches and answer the question Create line AA'' in both sketches. Your line AA'' needs to intersect lines f and g. Create the points of intersection between AA'' and lines f and g. Measure the angles formed by AA'' and lines f and g. What is the measurement of the angles you created? Conjecturing Compare the preimage (ABC) and the last image (A''B''C''). Does it look like any of the transformations (translation, rotation, reflection) that we have studied? Describe the movement from the preimage to the last image. What direction does it travel? Hint: AA'' will help with this. New Resources Discover Resources Discover Topics Download our apps here: © 2025 GeoGebra®
188216
https://www.quora.com/How-does-temperature-affect-frequency-of-sound-waves
Something went wrong. Wait a moment and try again. Sound Frequency Air Temperature Science About Wave Sound Waves Wave Phenomena Frequency (physics) Properties of Sound 5 How does temperature affect frequency of sound waves? Thomas Wilson There seems to be an open question to why the pitch of a woodwind instrument (flute, sax, trumpet, pipe organ, etc.) goes down as the temperature decreases, and vice versa. Sound's frequency is independent of temperature, while its speed is directly proportional to temperature. So the pitch cannot go down with the temperature in a simple manner (e.g., does the frequency of someone clapping go down when the air temperature decreases?). It is actually because the sound waves travel more slowly inside a musical instrument, which leads to lower rate at which the standing wave is excited, hence low There seems to be an open question to why the pitch of a woodwind instrument (flute, sax, trumpet, pipe organ, etc.) goes down as the temperature decreases, and vice versa. Sound's frequency is independent of temperature, while its speed is directly proportional to temperature. So the pitch cannot go down with the temperature in a simple manner (e.g., does the frequency of someone clapping go down when the air temperature decreases?). It is actually because the sound waves travel more slowly inside a musical instrument, which leads to lower rate at which the standing wave is excited, hence lower pitch. In other words, shortened wavelength corresponds to more standing waves inside the instrument per unit time (i.e., imagine the slower traveling waves getting squashed together more closely), but how often the standing wave oscillates back and forth like a vibrating string, which ultimately produces the sound and the pitch we hear, is affected by the speed of sound and thus the temperature (i.e., how often those wave fronts add up constructively over time). Jess H. Brewer Physics professor since 1977. · Author has 22.9K answers and 59.5M answer views · 8y Nothing affects the frequency of any wave except a change of reference frame. However, any material will support sound waves of some frequencies more easily (with less attenuation) than others, and that spectrum is affected by the “stiffness” of the medium, which in turn is affected by temperature. So if you are asking why a “struck” body “rings” with different frequencies, it’s because a “strike” is approximately a delta function, which contains all frequencies; some are absorbed and others “ring”. Omar Luis Curetti Former FE-SQC at Schlumberger (company) (1975–2008) · Author has 5.5K answers and 7.1M answer views · 8y No frequency could be changed, the only effects you can expect is wavelenght and attenuation. Sound waves are dispersive meaning that high frequency attenuates closer due to several interference modes within the matter, temperature will affect the attenuation in some matters and affect wavelenght as far it affect density and elasticity of them. Related questions Does increasing the temperature affect the frequency, amplitude, and intensity of a sound wave? How does temperature affect sound? How does temperature affect the speed of sound? What doesn't affect the frequency of a sound wave? Can changes in temperature affect sound waves and their emitted frequencies? Assistant Bot · 1y The frequency of sound waves is influenced by the medium through which they travel, and temperature is a key factor in determining the properties of that medium, especially gases like air. How Temperature Affects Frequency Speed of Sound: The speed of sound in a gas increases with temperature. In air, for example, the speed of sound increases by about 0.6 meters per second for each degree Celsius increase in temperature. The relationship can be approximated by the formula: v=331.3+0.6×T where v is the speed of sound in meters per second and T is the temperature in degrees Celsiu The frequency of sound waves is influenced by the medium through which they travel, and temperature is a key factor in determining the properties of that medium, especially gases like air. How Temperature Affects Frequency Speed of Sound: The speed of sound in a gas increases with temperature. In air, for example, the speed of sound increases by about 0.6 meters per second for each degree Celsius increase in temperature. The relationship can be approximated by the formula: v=331.3+0.6×T where v is the speed of sound in meters per second and T is the temperature in degrees Celsius. Frequency and Wavelength: The frequency (f) of a sound wave is related to its speed (v) and wavelength (λ) by the equation: v=f×λ If the speed of sound increases due to a rise in temperature while the wavelength remains constant, the frequency must also increase. Effect on Pitch: As the frequency increases, the perceived pitch of the sound also increases. This means that on warmer days, sounds may be perceived as slightly higher in pitch than on colder days. Summary Higher Temperature: Increases the speed of sound, which can lead to an increase in frequency (and thus pitch) if the wavelength remains constant. Lower Temperature: Decreases the speed of sound, leading to a decrease in frequency (and pitch). In summary, as temperature rises, the frequency of sound waves typically increases due to the increased speed of sound in the medium. Gary Turner MA (Cantab) in Natural Sciences (Physical), University of Cambridge (Graduated 1979) · Author has 1K answers and 687.7K answer views · 4y Originally Answered: How does temperature affect frequency? · Wien's displacement law states that the black-body radiation curve for different temperatures will peak at different wavelengths that are inversely proportional to the temperature. For spectral flux considered per unit frequency (in hertz), Wien's displacement law describes a peak emission at the optical frequency given by: αkT/h ≈ 5.879 1010 Hz/K where α ≈ 2.8214391... is a constant resulting from the numerical solution of the maximization equation, k is the Boltzmann constant, h is the Planck constant, and T is the temperature (in kelvins). So this frequency is directly proportional to the Wien's displacement law states that the black-body radiation curve for different temperatures will peak at different wavelengths that are inversely proportional to the temperature. For spectral flux considered per unit frequency (in hertz), Wien's displacement law describes a peak emission at the optical frequency given by: αkT/h ≈ 5.879 1010 Hz/K where α ≈ 2.8214391... is a constant resulting from the numerical solution of the maximization equation, k is the Boltzmann constant, h is the Planck constant, and T is the temperature (in kelvins). So this frequency is directly proportional to the temperature. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Jul 31 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix. Don’t waste your time Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix. Don’t waste your time browsing insurance sites for a better deal. A company calledInsurify shows you all your options at once — people who do this save up to $996 per year. If you tell them a bit about yourself and your vehicle, they’ll send you personalized quotes so you can compare them and find the best one for you. Tired of overpaying for car insurance? It takes just five minutes to compare your options with Insurify andsee how much you could save on car insurance. Ask This Company to Get a Big Chunk of Your Debt Forgiven A company calledNational Debt Relief could convince your lenders to simply get rid of a big chunk of what you owe. No bankruptcy, no loans — you don’t even need to have good credit. 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No gotchas.Make the move to SoFi and get paid to upgrade your finances. You Can Become a Real Estate Investor for as Little as $10 Take a look at some of the world’s wealthiest people. What do they have in common? Many invest in large private real estate deals. And here’s the thing: There’s no reason you can’t, too — for as little as $10. An investment called the Fundrise Flagship Fund lets you get started in the world of real estate by giving you access to a low-cost, diversified portfolio of private real estate. The best part? You don’t have to be the landlord. The Flagship Fund does all the heavy lifting. With an initial investment as low as $10, your money will be invested in the Fund, which already owns more than $1 billion worth of real estate around the country, from apartment complexes to the thriving housing rental market to larger last-mile e-commerce logistics centers. Want to invest more? Many investors choose to invest $1,000 or more. 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There is no change infrequency. The speed of sound increases by about 0.5 m/s for each degree Celsius when the air temperature rises. Related questions Can the change in intensity of sound affect temperature? What happens as the frequency of a sound wave increases? How is the frequency of a sound wave related to its time period physics? How do amplitude and frequency affect sound? How can you create low frequency sound waves? Willy Roentgen Registered Professional Engineer, BSAeroE, MSME, MBA, Pilot, · Author has 16.3K answers and 13.5M answer views · 8y It doesn’t affect frequency but the speed of sound is impacted by temperature: v = 331 m/s + 0.6 m/s/C T (for air) So the wavelength of the sound does vary, since w.l. = v/f Sponsored by Amazon Business Meet responsible sourcing goals. Buy from local, eco-friendly, and diverse suppliers. Bruce Hartley PhD in Physics, The University of Western Australia (Graduated 1969) · Author has 321 answers and 207.5K answer views · 6y Originally Answered: How does frequency depend upon the temperature? · Your question is not specific enough to give direct answer. Fundamentally the two are unrelated, but if the property of some object changes with temperature, such as gases expending when heated, this may affect the frequency of vibration of a wind instrument. The relationship is not a direct one, but through some other parameter which determines the frequency of vibration. Kenneth Lundgren M.S. in Electrical and Electronics Engineering, Northwestern University (Graduated 1966) · Author has 13.7K answers and 19.1M answer views · 7y Organ pipes change their pitch (frequency) with temperature because of the change in the speed of sound in air. Stringed instruments change their pitch with temperature because of expansion of the strings. Promoted by Betterbuck Anthony Madden Writer for Betterbuck · Updated Mar 24 I just bought my first house. Homeowners: what are some things you wished you knew when you bought your house? I've been a homeowner for 4 years. These are the biggest things I wish somebody told me on day one. Your home equity is a gold-mine. If you need cash, stop taking out high-interest loans. So many people take out high-interest payday loans - please don’t do this. If you get into trouble you can typically get a relatively low-interest HELOC (a home equity line of credit). Essentially with a HELOC, you’re borrowing against the equity you have in your house and use it for whatever you need (much like a credit card). Typically, you’ll get lower interest rates and more flexible repayment terms compar I've been a homeowner for 4 years. These are the biggest things I wish somebody told me on day one. Your home equity is a gold-mine. If you need cash, stop taking out high-interest loans. So many people take out high-interest payday loans - please don’t do this. If you get into trouble you can typically get a relatively low-interest HELOC (a home equity line of credit). Essentially with a HELOC, you’re borrowing against the equity you have in your house and use it for whatever you need (much like a credit card). Typically, you’ll get lower interest rates and more flexible repayment terms compared to traditional loans. Here’s a calculator you can use to see how much/little you could borrow (link here). Switch auto insurance companies every 6-12 months. If you haven’t compared auto insurance rates in the last 6 months, you’re probably overspending (on average by ~$460/year¹). Example: I cut my car insurance bill by ~$1,300 this year by switching carriers (same exact coverage too) and it took me a whopping 5 minutes. Take two minutes and check a comparison site (I used Coverage.com, Auto-Savings.com is solid too) and compare multiple offers from different companies in one go. Worst case scenario: you stay with what you’ve got. Best case scenario: you save a few hundred dollars a year. Here’s a link to a decent comparison site: link. Stop getting price gouged when you shop online. Big retailers like Amazon know that no one has time to price shop through dozens of sites, so there’s often no incentive for them to offer bargain prices. I typically hate browser extensions with a fiery passion, but Capital One Shopping has always worked well for me and I'd recommend trying it (link here). When you shop online (on Amazon or elsewhere) it will automatically compare prices for you, and auto-apply coupon codes when possible. Grab it here. Get yourself a dang advisor. Most people are under the false impression that financial advisors are just for wealthy people. They absolutely aren’t: if you have a net worth of $100k+, you can typically qualify for an advisor. Having an advisor typically increases your yearly net returns by 3%¹. If you don’t know an advisor personally, use a site like WiserAdvisor to find somebody local with decent reviews. Here’s a link to their site. You don't have to pay off your debt by yourself. Very few people know about it, but if you have $10k+ in debt, you can technically ask a debt relief company to come in and take over the process for you. Typically, it’d save you 23% off your total debt, after fees (according to NDR, a big debt relief company). People who are struggling with debt save 23% on avg. when they ask for help from debt relief companies. They’ll negotiate with your creditors and try to get your debt reduced (then they take their fees from your savings). Here’s a calculator you can use to see how much you’d potentially save: link. Save on home insurance. Some homeowners save $1k+/year just by switching home insurance providers (sometimes saving more than changing auto insurance policies). If it’s been more than a year since you’ve checked your rates, it might be worth taking a few minutes to compare offers. Here’s a home insurance comparison site I’ve used: link here. Robert Scott James Studied Physics at Florida Institute of Technology · Author has 353 answers and 219.6K answer views · 8y It's frequency remains fixed. Temperature controls the density of the airmass, or how far the soundwave’s amplitude (potential energy) will carry. The cooler the better. Peter Hauge Studied Engineering & Mathematics · Author has 2.7K answers and 1.6M answer views · 4y Does temperature affect sound wavelength? This question has been asked and answered on Quora. The question itself is kind of backwards. As if a sound wave is a kind of object that might have a thermal expansion coefficient. No to both of those notions. Temperature does not affect sound directly. What does affect all waves, including electromagnetic waves, is their supporting media. And yes, that includes EM propagation through free space. What the question ought to seek to address therefore is whether temperature affects the supporting medium for sound waves, whether it be solid or fluid. For it is that medium that determines its velocit This question has been asked and answered on Quora. The question itself is kind of backwards. As if a sound wave is a kind of object that might have a thermal expansion coefficient. No to both of those notions. Temperature does not affect sound directly. What does affect all waves, including electromagnetic waves, is their supporting media. And yes, that includes EM propagation through free space. What the question ought to seek to address therefore is whether temperature affects the supporting medium for sound waves, whether it be solid or fluid. For it is that medium that determines its velocity. And finally the answer is … found in the answers to all the previous times the question has been asked. Evert De Ruiter MSc in Applied Physics, Delft University of Technology (Graduated 1971) · Author has 5.2K answers and 2.1M answer views · 3y Originally Answered: How does frequency depend upon the temperature? · That depends on the phenomenon. Sound? Thermal radiation? Car accidents? Please consider what exactly you want to know. Or is it QPP only? Pete Singleton Lives in Rochdale (2017–present) · Author has 1.1K answers and 2.1M answer views · 8y Related If the frequency of a sound wave increased what happens to velocity of sound at the same temperature? In an ‘ordinary’ medium sound velocity is constant. If the frequency goes up the wavelength gets shorter. If the velocity changes with frequency the medium is ‘dispersive’. This is similar to what happens to light at a glass-air interface. prism dispersion - Google Search In the glass the light at lower frequency - Red - isn’t slowed down as much as the light at higher frequency - Violet. Sound can reflect and refract just as light does. At an interface between media a sound can be refracted. If the medium the sound is entering is dispersive (to sound) then different frequencies will be refracted In an ‘ordinary’ medium sound velocity is constant. If the frequency goes up the wavelength gets shorter. If the velocity changes with frequency the medium is ‘dispersive’. This is similar to what happens to light at a glass-air interface. prism dispersion - Google Search In the glass the light at lower frequency - Red - isn’t slowed down as much as the light at higher frequency - Violet. Sound can reflect and refract just as light does. At an interface between media a sound can be refracted. If the medium the sound is entering is dispersive (to sound) then different frequencies will be refracted by differing amounts. Related questions Does increasing the temperature affect the frequency, amplitude, and intensity of a sound wave? How does temperature affect sound? How does temperature affect the speed of sound? What doesn't affect the frequency of a sound wave? Can changes in temperature affect sound waves and their emitted frequencies? Can the change in intensity of sound affect temperature? What happens as the frequency of a sound wave increases? How is the frequency of a sound wave related to its time period physics? How do amplitude and frequency affect sound? How can you create low frequency sound waves? How does the temperature of a medium affect sound waves? What is sound frequency? How are sound waves affected by their frequency? What level frequency/sounds/waves are emitted from repellents? What is a high-frequency sound wave? What are some examples of high-frequency sound waves? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
188217
https://polymathprojects.org/2015/09/22/the-erdos-discrepancy-problem-has-been-solved-by-tao/
The polymath blog September 22, 2015 The Erdős discrepancy problem has been solved by Terence Tao Polymath5 was devoted to the Erdős discrepancy problem. It ran in 2010 and there were a few additional posts in 2012, without reaching a solution. The problem has now been solved by Terry Tao using some observations from the polymath project combined with important recent developments in analytic number theory. See this blog post from Tao’s blog and this concluding blog post from Gowers’s blog. Like Loading... Related Possible new polymath projectIn "polymath proposals" Proposal (Tim Gowers): Erdos’ Discrepancy ProblemIn "polymath proposals" Polymath5: Erdős’s discrepancy problemIn "polymath proposals" Comments (3) 3 Comments » 0 0 i Rate This Comment by Mr NuttELAVer7Ya — May 18, 2024 @ 1:12 pm | Reply 2. 0 0 i Rate This Comment by Anonymous — May 20, 2024 @ 3:39 pm | Reply 3. 0 0 i Rate This Comment by Anonymous — May 20, 2024 @ 3:42 pm | Reply RSS feed for comments on this post. TrackBack URI Leave a comment Cancel reply Recent Comments | | | --- | | | Anonymous on Rota’s Basis Conjecture:… | | | Anonymous on A sort of Polymath on a famous… | | | Anonymous on A new polymath proposal (relat… | | | Anonymous on Polymath projects 2021 | | | Anonymous on Polymath proposal: bounded gap… | | | Anonymous on (Research thread V) Determinst… | | | Anonymous on A new polymath proposal (relat… | | | Anonymous on Polymath8 – A Success… | | | Anonymous on Polymath8 – A Success… | | | Anonymous on Polymath8 – A Success… | | | Anonymous on Polymath8 – A Success… | | | Anonymous on Polymath8 – A Success… | | | Ignora (Not revealin… on Polymath8 – A Success… | | | Anonymous on Polymath proposal: finding sim… | | | Anonymous on Polymath10 is now open | ## Polymath Wiki – most recent changes An error has occurred; the feed is probably down. Try again later. Blogroll Gil Kalai Michael Nielsen Terence Tao The Polymath Wiki Tim Gowers Projects Mini-polymath1: IMO Q6 2009 Polymath1: Density Hales-Jewett Polymath2: Embedded c_0 or l^p in explicit Banach spaces Polymath4: Finding primes deterministically Polymath5: Erdos discrepancy problem Proposals Boshernitzan's problem Cap sets Other proposals Polynomial Hirsch conjecture Proposals on Wordpress Pages About General discussion General polymath rules How to use LaTeX in comments Categories discussion (8) finding primes (8) hot spots (7) Improving Roth bounds (1) mock-up (2) news (9) planning (7) polymath proposals (32) polymath5 (1) research (13) Top Posts Minipolymath4 project: IMO 2012 Q3 Polymath8 - A Success ! Polymath6: improving the bounds for Roth's theorem Polymath proposal: bounded gaps between primes Polymath wiki logo Polymath7 research thread 5: the hot spots conjecture Can Bourgain's argument be usefully modified? Mock-up research thread Draft version of polymath4 paper Polymath proposal: explaining identities for irreducible polynomials Blog at WordPress.com. Comment Reblog Subscribe Subscribed The polymath blog Already have a WordPress.com account? Log in now. The polymath blog Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar
188218
https://www.youtube.com/watch?v=UV_lcPDiEDQ
Addition and Subtraction Properties of Equality Ryne Roper 3520 subscribers 12 likes Description 1246 views Posted: 12 Oct 2023 Use this information to help you in your Pre-Algebra class! 🔥 DON'T FORGET to check out my full Pre-Algebra playlist for help with all things Pre-Algebra! ❤️ Enjoyed this video? Subscribe to my channel and hit the notification bell to never miss a new video — and let me know what you think in the comments! addition #subtraction #property #equality #RyneRoper 6 comments Transcript: hello everyone in this video we're going to talk about the addition and subtraction properties of equality so first of all what are equivalent equations that's our first definition that we see at the top of the screen and it just says when two equations have the same Solution that's all we're talking about with equivalent equations so with the additional addition and subtraction properties of equality we're trying to create equivalent equations so for addition it says we can add the same number to each side of the equation and it will produce an equivalent equation so we see kind of the algebra there it says if a is equal to B then A+ C would be equal to B+ C so basically we are just adding C to both sides of the equation a lot of times we can think about this as keeping the equation balanced so we have to do that so we can do whatever we want to one side of the equation as long as we do it to the other side as well same idea with subtraction but now we're just subtracting the same number from each side of the equation so if a is equal to B then a minus C would be equal to B minus C once again we are subtracting C from both sides so we're doing the same operation we're keeping that equation balanced so let's look at how this is going to look with some some examples so for number one we see x - 5 is equal to -2 so for these equations we would consider them simple equations because we only have to do one step in order to solve for the variable so that's our goal we're trying to figure out what the variable is equal to so we can do this by performing inverse operations to get X isolated or by itself so if I look here I see I have x - 5 so I could add five to both sides of my equation the reason I add five is because on the left side -5 + 5 would be zero or in other words those numbers would cancel each other out so all I'm left with is X which is what I want I want to know what x is equal to and now I can evaluate -2 + 5 and that would be three So my answer for number one would be xal 3 and we could always plug in three for our variable to make sure our equation makes sense 3 - 5 is -2 so we did it correctly X is 3 all right look at number two now we have a + 10.1 is equal to 13.2 so now we want to do the opposite of add 10.1 which would be subtract 10.1 and remember whatever we do to one side we need to do to both sides the reason why we're subtracting is is because it's opposite of add 10.1 and the reason it's 10.1 is because positive 10.1 minus 10.1 would cancel out right it would equal Z so we get a is equal 2 and now we can subtract over here 2 - 1 would be 1 and 13 - 10 would be 3 So my answer for number two would be a is equal to 3.1 and once again we could plug that in for a to check and make sure and it does work so we did it right and the last one number three B minus 45ths is equal to 1/5 so now we can do the opposite of minus 4 fths which would be add four fths and we're going to do that to both sides my four fths on the left side will cancel because they're opposite of each other so we get B is equal 2 and 1/5 + 4 4 fths would be 5 fths which would just equal 1 so for number three B would equal 1 and that is how we can use the addition and subtraction properties of equality
188219
https://uclouvain-cbio.github.io/WSBIM1207/sec-dplyr.html
Chapter 5 Manipulating and analyzing data with dplyr | Introduction to bioinformatics Introduction to bioinformatics Laurent Gatto ▾ Chapters Preamble1 Data organisation with Spreadsheets2 R and RStudio3 Introduction to R4 Starting with data5 Manipulating and analyzing data with dplyr Data Manipulation using dplyr and tidyr What are dplyr and tidyr? Selecting columns and filtering rows Pipes Mutate Pulling a variable Split-apply-combine data analysis Reshaping data Exporting data Additional exercises 6 Data visualization7 Joining tables8 Reproducible research9 Bioinformatics10 Additional programming concepts11 Conclusions12 Annex13 Session information Chapter 5 Manipulating and analyzing data with dplyr Learning Objectives Describe the purpose of the dplyr and tidyr packages. Select certain columns in a data frame with the dplyr function select. Select certain rows in a data frame according to filtering conditions with the dplyr function filter . Link the output of one dplyr function to the input of another function with the ‘pipe’ operator %>% or |>. Add new columns to a data frame that are functions of existing columns with mutate. Use the split-apply-combine concept for data analysis. Use summarize, group_by, and count to split a data frame into groups of observations, apply summary statistics for each group, and then combine the results. Describe the concept of a wide and a long table format and for which purpose those formats are useful. Reshape a data frame from long to wide format and back with the pivot_wider() and pivot_longer() commands from the tidyr package. 5.1 Data Manipulation using dplyr and tidyr Copy link Bracket subsetting is handy, but it can be cumbersome and difficult to read, especially for complicated operations. Enter dplyr. dplyr is a package for making tabular data manipulation easier. It pairs nicely with tidyr which enables you to swiftly convert between different data formats for plotting and analysis. Packages in R are basically sets of additional functions that let you do more stuff. The functions we’ve been using so far, like str() or data.frame(), come built into R; packages give you access to more of them. Before you use a package for the first time you need to install it on your machine, and then you should import it in every subsequent R session when you need it. You should already have installed the tidyverse package. This is an “umbrella-package” that installs several packages useful for data analysis which work together well such as tidyr, dplyr, ggplot2, tibble, etc. The tidyverse packages address 3 common issues that arise when doing data analysis with some of functions that come with R: The results from a base R function sometimes depend on the type of data. Using R expressions in a non standard way, which can be confusing for new learners. Hidden arguments, having default operations that new learners are not aware of. Let’s start by loading several of the tidyverse packages with: library("tidyverse") The Data Transformation Cheat Sheet provides an overview of the dplyr grammar, offering more details and functions that we will see in this chapter. The Tidy Data Tutor is a wonderful tool to visually describe what the tidy data operations do. The following tutorial offers nice animations of mutate(), summarize(), group_by(), and ungroup(). 5.2 What are dplyr and tidyr? Copy link The package dplyr provides easy tools for the most common data manipulation tasks. It is built to work directly with data frames, with many common tasks optimized by being written in a compiled language (C++). An additional feature is the ability to work directly with data stored in an external database. The benefits of doing this are that the data can be managed natively in a relational database, queries can be conducted on that database, and only the results of the query are returned. This addresses a common problem with R in that all operations are conducted in-memory and thus the amount of data you can work with is limited by available memory. The database connections essentially remove that limitation in that you can connect to a database of many hundreds of GB, conduct queries on it directly, and pull back into R only what you need for analysis. The package tidyr addresses the common problem of wanting to reshape your data for plotting and use by different R functions. Sometimes we want data sets where we have one row per measurement. Sometimes we want a data frame where each measurement type has its own column, and rows are instead more aggregated groups - like plots or aquaria. Moving back and forth between these formats is nontrivial, and tidyr gives you tools for this and more sophisticated data manipulation. To learn more about dplyr and tidyr after the workshop, you may want to check out this handy data transformation with dplyr cheatsheet and this one about tidyr. We’ll read in our data using the read_csv() function, from the tidyverse package readr, instead of read.csv(). rna <- read_csv("data/rnaseq.csv") ``` Rows: 32428 Columns: 19 ── Column specification ────────────────────────────────────────────────────────────────────────────────────────────────────── Delimiter: "," chr (14): gene, sample, organism, sex, infection, strain, tissue, product, e... dbl (5): expression, age, time, mouse, ENTREZID ℹ Use spec() to retrieve the full column specification for this data. ℹ Specify the column types or set show_col_types = FALSE to quiet this message. ``` ``` inspect the data str(rna) preview the data View(rna) ``` Notice that the class of the data is now tbl_df This is referred to as a “tibble”. Tibbles tweak some of the behaviors of the data frame objects we introduced in the previous episode. The data structure is very similar to a data frame. For our purposes the only differences are that: In addition to displaying the data type of each column under its name, it only prints the first few rows of data and only as many columns as fit on one screen. Columns of class character are never converted into factors. We’re going to learn some of the most common dplyr functions: select(): subset columns filter(): subset rows on conditions mutate(): create new columns by using information from other columns group_by() and summarize(): create summary statisitcs on grouped data arrange(): sort results count(): count discrete values 5.3 Selecting columns and filtering rows Copy link To select columns of a data frame, use select(). The first argument to this function is the data frame (rna), and the subsequent arguments are the columns to keep. select(rna, gene, sample, tissue, expression) ``` # A tibble: 32,428 × 4 gene sample tissue expression 1 Asl GSM2545336 Cerebellum 1170 2 Apod GSM2545336 Cerebellum 36194 3 Cyp2d22 GSM2545336 Cerebellum 4060 4 Klk6 GSM2545336 Cerebellum 287 5 Fcrls GSM2545336 Cerebellum 85 6 Slc2a4 GSM2545336 Cerebellum 782 7 Exd2 GSM2545336 Cerebellum 1619 8 Gjc2 GSM2545336 Cerebellum 288 9 Plp1 GSM2545336 Cerebellum 43217 10 Gnb4 GSM2545336 Cerebellum 1071 # ℹ 32,418 more rows ``` To select all columns except certain ones, put a “-” in front of the variable to exclude it. select(rna, -organism, -strain) ``` # A tibble: 32,428 × 17 gene sample expression age sex infection time tissue mouse ENTREZID 1 Asl GSM2545… 1170 8 Fema… Influenz… 8 Cereb… 14 109900 2 Apod GSM2545… 36194 8 Fema… Influenz… 8 Cereb… 14 11815 3 Cyp2d22 GSM2545… 4060 8 Fema… Influenz… 8 Cereb… 14 56448 4 Klk6 GSM2545… 287 8 Fema… Influenz… 8 Cereb… 14 19144 5 Fcrls GSM2545… 85 8 Fema… Influenz… 8 Cereb… 14 80891 6 Slc2a4 GSM2545… 782 8 Fema… Influenz… 8 Cereb… 14 20528 7 Exd2 GSM2545… 1619 8 Fema… Influenz… 8 Cereb… 14 97827 8 Gjc2 GSM2545… 288 8 Fema… Influenz… 8 Cereb… 14 118454 9 Plp1 GSM2545… 43217 8 Fema… Influenz… 8 Cereb… 14 18823 10 Gnb4 GSM2545… 1071 8 Fema… Influenz… 8 Cereb… 14 14696 # ℹ 32,418 more rows # ℹ 7 more variables: product , ensembl_gene_id , # external_synonym , chromosome_name , gene_biotype , # phenotype_description , hsapiens_homolog_associated_gene_name ``` This will select all the variables in rna except organism and strain. To choose rows based on a specific criteria, use filter(): filter(rna, sex == "Male") ``` # A tibble: 14,740 × 19 gene sample expression organism age sex infection strain time tissue 1 Asl GSM254… 626 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 2 Apod GSM254… 13021 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 3 Cyp2d22 GSM254… 2171 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 4 Klk6 GSM254… 448 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 5 Fcrls GSM254… 180 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 6 Slc2a4 GSM254… 313 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 7 Exd2 GSM254… 2366 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 8 Gjc2 GSM254… 310 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 9 Plp1 GSM254… 53126 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… 10 Gnb4 GSM254… 1355 Mus mus… 8 Male Influenz… C57BL… 4 Cereb… # ℹ 14,730 more rows # ℹ 9 more variables: mouse , ENTREZID , product , # ensembl_gene_id , external_synonym , chromosome_name , # gene_biotype , phenotype_description , # hsapiens_homolog_associated_gene_name ``` filter(rna, sex == "Male" & infection == "NonInfected") ``` # A tibble: 4,422 × 19 gene sample expression organism age sex infection strain time tissue 1 Asl GSM254… 535 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 2 Apod GSM254… 13668 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 3 Cyp2d22 GSM254… 2008 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 4 Klk6 GSM254… 1101 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 5 Fcrls GSM254… 375 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 6 Slc2a4 GSM254… 249 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 7 Exd2 GSM254… 3126 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 8 Gjc2 GSM254… 791 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 9 Plp1 GSM254… 98658 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… 10 Gnb4 GSM254… 2437 Mus mus… 8 Male NonInfec… C57BL… 0 Cereb… # ℹ 4,412 more rows # ℹ 9 more variables: mouse , ENTREZID , product , # ensembl_gene_id , external_synonym , chromosome_name , # gene_biotype , phenotype_description , # hsapiens_homolog_associated_gene_name ``` 5.4 Pipes Copy link What if you want to select and filter at the same time? There are three ways to do this: use intermediate steps, nested functions, or pipes. With intermediate steps, you create a temporary data frame and use that as input to the next function, like this: rna2 <- filter(rna, sex == "Male") rna3 <- select(rna2, gene, sample, tissue, expression) rna3 ``` # A tibble: 14,740 × 4 gene sample tissue expression 1 Asl GSM2545340 Cerebellum 626 2 Apod GSM2545340 Cerebellum 13021 3 Cyp2d22 GSM2545340 Cerebellum 2171 4 Klk6 GSM2545340 Cerebellum 448 5 Fcrls GSM2545340 Cerebellum 180 6 Slc2a4 GSM2545340 Cerebellum 313 7 Exd2 GSM2545340 Cerebellum 2366 8 Gjc2 GSM2545340 Cerebellum 310 9 Plp1 GSM2545340 Cerebellum 53126 10 Gnb4 GSM2545340 Cerebellum 1355 # ℹ 14,730 more rows ``` This is readable, but can clutter up your workspace with lots of objects that you have to name individually. With multiple steps, that can be hard to keep track of. You can also nest functions (i.e.one function inside of another), like this: rna3 <- select(filter(rna, sex == "Male"), gene, sample, tissue, expression) rna3 ``` # A tibble: 14,740 × 4 gene sample tissue expression 1 Asl GSM2545340 Cerebellum 626 2 Apod GSM2545340 Cerebellum 13021 3 Cyp2d22 GSM2545340 Cerebellum 2171 4 Klk6 GSM2545340 Cerebellum 448 5 Fcrls GSM2545340 Cerebellum 180 6 Slc2a4 GSM2545340 Cerebellum 313 7 Exd2 GSM2545340 Cerebellum 2366 8 Gjc2 GSM2545340 Cerebellum 310 9 Plp1 GSM2545340 Cerebellum 53126 10 Gnb4 GSM2545340 Cerebellum 1355 # ℹ 14,730 more rows ``` This is handy, but can be difficult to read if too many functions are nested, as R evaluates the expression from the inside out (in this case, filtering, then selecting). The last option, pipes, are a recent addition to R. Pipes let you take the output of one function and send it directly to the next, which is useful when you need to do many things to the same dataset. Pipes in R look like %>% (as made available via the magrittr package, installed automatically with dplyr) or |> (as available in base R). If you use RStudio, you can type the pipe with Ctrl + Shift + M if you have a PC or Cmd + Shift + M if you have a Mac. rna |> filter(sex == "Male") |> select(gene, sample, tissue, expression) ``` # A tibble: 14,740 × 4 gene sample tissue expression 1 Asl GSM2545340 Cerebellum 626 2 Apod GSM2545340 Cerebellum 13021 3 Cyp2d22 GSM2545340 Cerebellum 2171 4 Klk6 GSM2545340 Cerebellum 448 5 Fcrls GSM2545340 Cerebellum 180 6 Slc2a4 GSM2545340 Cerebellum 313 7 Exd2 GSM2545340 Cerebellum 2366 8 Gjc2 GSM2545340 Cerebellum 310 9 Plp1 GSM2545340 Cerebellum 53126 10 Gnb4 GSM2545340 Cerebellum 1355 # ℹ 14,730 more rows ``` Or rna |> filter(sex == "Male") |> select(gene, sample, tissue, expression) ``` # A tibble: 14,740 × 4 gene sample tissue expression 1 Asl GSM2545340 Cerebellum 626 2 Apod GSM2545340 Cerebellum 13021 3 Cyp2d22 GSM2545340 Cerebellum 2171 4 Klk6 GSM2545340 Cerebellum 448 5 Fcrls GSM2545340 Cerebellum 180 6 Slc2a4 GSM2545340 Cerebellum 313 7 Exd2 GSM2545340 Cerebellum 2366 8 Gjc2 GSM2545340 Cerebellum 310 9 Plp1 GSM2545340 Cerebellum 53126 10 Gnb4 GSM2545340 Cerebellum 1355 # ℹ 14,730 more rows ``` In the above code, we use the pipe to send the rna dataset first through filter() to keep rows where sex is Male, then through select() to keep only the gene, sample, tissue, and expressioncolumns. Since %>% (and |>) takes the object on its left and passes it as the first argument to the function on its right, we don’t need to explicitly include the data frame as an argument to the filter() and select() functions any more. Some may find it helpful to read the pipe like the word “then”. For instance, in the above example, we took the data frame rna, then we filtered for rows with sex == "Male", then we selected columns gene, sample, tissue, and expression. The dplyr functions by themselves are somewhat simple, but by combining them into linear workflows with the pipe, we can accomplish more complex manipulations of data frames. If we want to create a new object with this smaller version of the data, we can assign it a new name: ``` rna3 <- rna |> filter(sex == "Male") |> select(gene, sample, tissue, expression) rna3 ``` ``` # A tibble: 14,740 × 4 gene sample tissue expression 1 Asl GSM2545340 Cerebellum 626 2 Apod GSM2545340 Cerebellum 13021 3 Cyp2d22 GSM2545340 Cerebellum 2171 4 Klk6 GSM2545340 Cerebellum 448 5 Fcrls GSM2545340 Cerebellum 180 6 Slc2a4 GSM2545340 Cerebellum 313 7 Exd2 GSM2545340 Cerebellum 2366 8 Gjc2 GSM2545340 Cerebellum 310 9 Plp1 GSM2545340 Cerebellum 53126 10 Gnb4 GSM2545340 Cerebellum 1355 # ℹ 14,730 more rows ``` ► Question Using pipes, subset the rna data to genes with an expression higher than 50000 in male mice at time 0, and retain only the columns gene, sample, time, expression and age ► Solution rna |> filter(expression > 50000, sex == "Male", time == 0 ) |> select(gene, sample, time, expression, age) ``` # A tibble: 8 × 5 gene sample time expression age 1 Plp1 GSM2545343 0 98658 8 2 Glul GSM2545343 0 54808 8 3 Atp1b1 GSM2545343 0 60083 8 4 Plp1 GSM2545349 0 82722 8 5 Atp1b1 GSM2545349 0 59094 8 6 Plp1 GSM2545354 0 84540 8 7 Glul GSM2545354 0 50584 8 8 Atp1b1 GSM2545354 0 57409 8 ``` 5.5 Mutate Copy link Frequently you’ll want to create new columns based on the values in existing columns, for example to do unit conversions, or to find the ratio of values in two columns. For this we’ll use mutate(). To create a new column of time in hours: rna |> mutate(time_hours = time 24) |> select(time, time_hours) ``` # A tibble: 32,428 × 2 time time_hours 1 8 192 2 8 192 3 8 192 4 8 192 5 8 192 6 8 192 7 8 192 8 8 192 9 8 192 10 8 192 # ℹ 32,418 more rows ``` You can also create a second new column based on the first new column within the same call of mutate(): rna |> mutate(time_hours = time 24, time_mn = time_hours 60) |> select(time, time_hours, time_mn) ``` # A tibble: 32,428 × 3 time time_hours time_mn 1 8 192 11520 2 8 192 11520 3 8 192 11520 4 8 192 11520 5 8 192 11520 6 8 192 11520 7 8 192 11520 8 8 192 11520 9 8 192 11520 10 8 192 11520 # ℹ 32,418 more rows ``` If this runs off your screen and you just want to see the first few rows, you can use a pipe to view the head() of the data. (Pipes work with non-dplyr functions, too, as long as the dplyr or magrittr package is loaded). rna |> mutate(time_hours = time 24, time_mn = time_hours 60) |> select(time, time_hours, time_mn) |> head() ``` # A tibble: 6 × 3 time time_hours time_mn 1 8 192 11520 2 8 192 11520 3 8 192 11520 4 8 192 11520 5 8 192 11520 6 8 192 11520 ``` Let’s imagine we are interested in the human homologs of the mouse genes analysed in this dataset. This information can be found in the last column of the rna tibble, named hsapiens_homolog_associated_gene_name. rna |> select(gene, hsapiens_homolog_associated_gene_name) ``` # A tibble: 32,428 × 2 gene hsapiens_homolog_associated_gene_name 1 Asl ASL 2 Apod APOD 3 Cyp2d22 CYP2D6 4 Klk6 KLK6 5 Fcrls FCRL2 6 Slc2a4 SLC2A4 7 Exd2 EXD2 8 Gjc2 GJC2 9 Plp1 PLP1 10 Gnb4 GNB4 # ℹ 32,418 more rows ``` Some mouse gene have no human homologs. These can be retrieved using a filter() in the chain, and the is.na() function that determines whether something is an NA. rna |> select(gene, hsapiens_homolog_associated_gene_name) |> filter(is.na(hsapiens_homolog_associated_gene_name)) ``` # A tibble: 4,290 × 2 gene hsapiens_homolog_associated_gene_name 1 Prodh 2 Tssk5 3 Vmn2r1 4 Gm10654 5 Hexa 6 Sult1a1 7 Gm6277 8 Tmem198b 9 Adam1a 10 Ebp # ℹ 4,280 more rows ``` If we want to keep only mouse gene that have a human homolog, we can insert a ! symbol that negates the result, so we’re asking for every row where hsapiens_homolog_associated_gene_nameis not an NA. The first few rows of the output are full of NAs, so if we wanted to remove those we could insert a filter() in the chain: rna |> select(gene, hsapiens_homolog_associated_gene_name) |> filter(!is.na(hsapiens_homolog_associated_gene_name)) ``` # A tibble: 28,138 × 2 gene hsapiens_homolog_associated_gene_name 1 Asl ASL 2 Apod APOD 3 Cyp2d22 CYP2D6 4 Klk6 KLK6 5 Fcrls FCRL2 6 Slc2a4 SLC2A4 7 Exd2 EXD2 8 Gjc2 GJC2 9 Plp1 PLP1 10 Gnb4 GNB4 # ℹ 28,128 more rows ``` ► Question Create a new data frame from the rna data that meets the following criteria: contains only the gene, chromosome_name, phenotype_description, sample, and expression columns and a new column giving the log expression the gene. This data frame must only contain gene located on autosomes and associated with a phenotype_description. Hint: think about how the commands should be ordered to produce this data frame! ► Solution rna |> filter(chromosome_name != "X", chromosome_name != "Y") |> mutate(log_expression = log(expression)) |> select(gene, chromosome_name, phenotype_description, sample, log_expression) |> filter(!is.na(phenotype_description)) ``` # A tibble: 21,054 × 5 gene chromosome_name phenotype_description sample log_expression 1 Asl 5 abnormal circulating amino aci… GSM25… 7.06 2 Apod 16 abnormal lipid homeostasis GSM25… 10.5 3 Cyp2d22 15 abnormal skin morphology GSM25… 8.31 4 Klk6 7 abnormal cytokine level GSM25… 5.66 5 Fcrls 3 decreased CD8-positive alpha-b… GSM25… 4.44 6 Slc2a4 11 abnormal circulating glucose l… GSM25… 6.66 7 Gjc2 11 Purkinje cell degeneration GSM25… 5.66 8 Gnb4 3 decreased anxiety-related resp… GSM25… 6.98 9 Tnc 4 abnormal CNS synaptic transmis… GSM25… 5.39 10 Trf 9 abnormal circulating phosphate… GSM25… 9.18 # ℹ 21,044 more rows ``` 5.6 Pulling a variable Copy link The tidy functions that we have seen (and will see below) always take a tidy table (typically a tibble) as input, and return a new, transformed one, possibly composed of a single column/variables. This is an important feature that enables the piping of commands one into another. Sometimes however, one needs to extract the variable that composes the column of a table. This can be done with the pull() function. Below, note the difference between select(), that here returns a one-column tibble … ``` mini_rna <- head(rna) mini_rna |> select(gene) ``` ``` # A tibble: 6 × 1 gene 1 Asl 2 Apod 3 Cyp2d22 4 Klk6 5 Fcrls 6 Slc2a4 ``` … and pull(), that returns a vector: mini_rna |> pull(gene) ## "Asl" "Apod" "Cyp2d22" "Klk6" "Fcrls" "Slc2a4" After using pull(), on can’t pipe the output into any of the functions that expect to get a tibble. 5.7 Split-apply-combine data analysis Copy link Many data analysis tasks can be approached using the split-apply-combine paradigm: split the data into groups, apply some analysis to each group, and then combine the results. dplyr makes this very easy through the use of the group_by() function. rna |> group_by(gene) ``` # A tibble: 32,428 × 19 # Groups: gene [1,474] gene sample expression organism age sex infection strain time tissue 1 Asl GSM254… 1170 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 2 Apod GSM254… 36194 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 3 Cyp2d22 GSM254… 4060 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 4 Klk6 GSM254… 287 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 5 Fcrls GSM254… 85 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 6 Slc2a4 GSM254… 782 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 7 Exd2 GSM254… 1619 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 8 Gjc2 GSM254… 288 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 9 Plp1 GSM254… 43217 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 10 Gnb4 GSM254… 1071 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… # ℹ 32,418 more rows # ℹ 9 more variables: mouse , ENTREZID , product , # ensembl_gene_id , external_synonym , chromosome_name , # gene_biotype , phenotype_description , # hsapiens_homolog_associated_gene_name ``` The group_by() function doesn’t perform any data processing, it groups the data into subsets: in the example above, our initial tibble of 32428 observations is split into 1474 groups based on the gene variable. Once the data have been combined, subsequent operations will be applied on each group independently. To remove this grouping, simply use the ungroup() function. grouped_rna <- rna |> group_by(gene) grouped_rna ``` # A tibble: 32,428 × 19 # Groups: gene [1,474] gene sample expression organism age sex infection strain time tissue 1 Asl GSM254… 1170 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 2 Apod GSM254… 36194 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 3 Cyp2d22 GSM254… 4060 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 4 Klk6 GSM254… 287 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 5 Fcrls GSM254… 85 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 6 Slc2a4 GSM254… 782 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 7 Exd2 GSM254… 1619 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 8 Gjc2 GSM254… 288 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 9 Plp1 GSM254… 43217 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 10 Gnb4 GSM254… 1071 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… # ℹ 32,418 more rows # ℹ 9 more variables: mouse , ENTREZID , product , # ensembl_gene_id , external_synonym , chromosome_name , # gene_biotype , phenotype_description , # hsapiens_homolog_associated_gene_name ``` ungroup(rna) ``` # A tibble: 32,428 × 19 gene sample expression organism age sex infection strain time tissue 1 Asl GSM254… 1170 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 2 Apod GSM254… 36194 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 3 Cyp2d22 GSM254… 4060 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 4 Klk6 GSM254… 287 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 5 Fcrls GSM254… 85 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 6 Slc2a4 GSM254… 782 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 7 Exd2 GSM254… 1619 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 8 Gjc2 GSM254… 288 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 9 Plp1 GSM254… 43217 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… 10 Gnb4 GSM254… 1071 Mus mus… 8 Fema… Influenz… C57BL… 8 Cereb… # ℹ 32,418 more rows # ℹ 9 more variables: mouse , ENTREZID , product , # ensembl_gene_id , external_synonym , chromosome_name , # gene_biotype , phenotype_description , # hsapiens_homolog_associated_gene_name ``` 5.7.1 The summarize() function Copy link group_by() is often used together with summarize(), which collapses each group into a single-row summary of that group. group_by() takes as arguments the column names that contain the categorical variables for which you want to calculate the summary statistics. So to compute the mean expression by gene: rna |> group_by(gene) |> summarize(mean_expression = mean(expression)) ``` # A tibble: 1,474 × 2 gene mean_expression 1 AI504432 1053. 2 AW046200 131. 3 AW551984 295. 4 Aamp 4751. 5 Abca12 4.55 6 Abcc8 2498. 7 Abhd14a 525. 8 Abi2 4909. 9 Abi3bp 1002. 10 Abl2 2124. # ℹ 1,464 more rows ``` You may also have noticed that the output from these calls doesn’t run off the screen anymore. It’s one of the advantages of tbl_df over data frame. You can also group by multiple columns: rna |> group_by(gene, infection, time) |> summarize(mean_expression = mean(expression)) ``` summarise() has grouped output by 'gene', 'infection'. You can override using the .groups argument. ``` ``` # A tibble: 4,422 × 4 # Groups: gene, infection [2,948] gene infection time mean_expression 1 AI504432 InfluenzaA 4 1104. 2 AI504432 InfluenzaA 8 1014 3 AI504432 NonInfected 0 1034. 4 AW046200 InfluenzaA 4 152. 5 AW046200 InfluenzaA 8 81 6 AW046200 NonInfected 0 155. 7 AW551984 InfluenzaA 4 302. 8 AW551984 InfluenzaA 8 342. 9 AW551984 NonInfected 0 238 10 Aamp InfluenzaA 4 4870 # ℹ 4,412 more rows ``` Here, again, the output from these calls doesn’t run off the screen anymore. If you want to display more data, you can use the print() function at the end of your chain with the argument n specifying the number of rows to display: rna |> group_by(gene, infection, time) |> summarize(mean_expression = mean(expression)) |> print(n = 15) ``` summarise() has grouped output by 'gene', 'infection'. You can override using the .groups argument. ``` ``` # A tibble: 4,422 × 4 # Groups: gene, infection [2,948] gene infection time mean_expression 1 AI504432 InfluenzaA 4 1104. 2 AI504432 InfluenzaA 8 1014 3 AI504432 NonInfected 0 1034. 4 AW046200 InfluenzaA 4 152. 5 AW046200 InfluenzaA 8 81 6 AW046200 NonInfected 0 155. 7 AW551984 InfluenzaA 4 302. 8 AW551984 InfluenzaA 8 342. 9 AW551984 NonInfected 0 238 10 Aamp InfluenzaA 4 4870 11 Aamp InfluenzaA 8 4763. 12 Aamp NonInfected 0 4603. 13 Abca12 InfluenzaA 4 4.25 14 Abca12 InfluenzaA 8 4.14 15 Abca12 NonInfected 0 5.29 # ℹ 4,407 more rows ``` Once the data is grouped, you can also summarize multiple variables at the same time (and not necessarily on the same variable). For instance, we could add columns indicating the median expression by gene and by condition: rna |> group_by(gene, infection, time) |> summarize(mean_expression = mean(expression), median_expression = median(expression)) ``` summarise() has grouped output by 'gene', 'infection'. You can override using the .groups argument. ``` ``` # A tibble: 4,422 × 5 # Groups: gene, infection [2,948] gene infection time mean_expression median_expression 1 AI504432 InfluenzaA 4 1104. 1094. 2 AI504432 InfluenzaA 8 1014 985 3 AI504432 NonInfected 0 1034. 1016 4 AW046200 InfluenzaA 4 152. 144. 5 AW046200 InfluenzaA 8 81 82 6 AW046200 NonInfected 0 155. 163 7 AW551984 InfluenzaA 4 302. 245 8 AW551984 InfluenzaA 8 342. 287 9 AW551984 NonInfected 0 238 265 10 Aamp InfluenzaA 4 4870 4708 # ℹ 4,412 more rows ``` It is sometimes useful to rearrange the result of a query to inspect the values. For instance, we can sort on mean_expression to put the genes lowly expressed first: rna |> group_by(gene, infection, time) |> summarize(mean_expression = mean(expression), median_expression = median(expression)) |> arrange(mean_expression) ``` summarise() has grouped output by 'gene', 'infection'. You can override using the .groups argument. ``` ``` # A tibble: 4,422 × 5 # Groups: gene, infection [2,948] gene infection time mean_expression median_expression 1 Gm5415 InfluenzaA 4 0.375 0 2 Selp NonInfected 0 0.429 0 3 Ascl5 NonInfected 0 0.571 1 4 Gm28178 NonInfected 0 0.571 1 5 Il18r1 NonInfected 0 0.571 0 6 Pdcd1 InfluenzaA 8 0.571 0 7 Rln3 NonInfected 0 0.571 0 8 Gm19637 InfluenzaA 4 0.625 0.5 9 Gm6177 InfluenzaA 4 0.625 1 10 Gm7241 InfluenzaA 4 0.625 0.5 # ℹ 4,412 more rows ``` To sort in descending order, we need to add the desc() function: rna |> group_by(gene, infection, time) |> summarize(mean_expression = mean(expression), median_expression = median(expression)) |> arrange(desc(mean_expression)) ``` summarise() has grouped output by 'gene', 'infection'. You can override using the .groups argument. ``` ``` # A tibble: 4,422 × 5 # Groups: gene, infection [2,948] gene infection time mean_expression median_expression 1 Plp1 NonInfected 0 91103. 96534 2 Glul InfluenzaA 8 73948. 71706 3 Plp1 InfluenzaA 4 67198. 63840 4 Atp1b1 InfluenzaA 4 60364. 56546. 5 Atp1b1 InfluenzaA 8 59229 61672 6 Atp1b1 NonInfected 0 57351. 59094 7 Sparc InfluenzaA 8 56106. 57409 8 Glul InfluenzaA 4 55358. 52836. 9 Glul NonInfected 0 48123. 49099 10 Nrep NonInfected 0 40060. 37493 # ℹ 4,412 more rows ``` 5.7.2 Counting Copy link When working with data, we often want to know the number of observations found for each factor or combination of factors. For this task, dplyr provides count(). For example, if we wanted to count the number of rows of data for each infected and non infected, we would do: rna |> count(infection) ``` # A tibble: 2 × 2 infection n 1 InfluenzaA 22110 2 NonInfected 10318 ``` The count() function is shorthand for something we’ve already seen: grouping by a variable, and summarizing it by counting the number of observations in that group. In other words, rna |> count() is equivalent to: rna |> group_by(infection) |> summarise(count = n()) ``` # A tibble: 2 × 2 infection count 1 InfluenzaA 22110 2 NonInfected 10318 ``` For convenience, count() provides the sort argument: rna |> count(infection, sort = TRUE) ``` # A tibble: 2 × 2 infection n 1 InfluenzaA 22110 2 NonInfected 10318 ``` Previous example shows the use of count() to count the number of rows/observations for one factor (i.e., infection). If we wanted to count combination of factors, such as infection and time, we would specify the first and the second factor as the arguments of count(): rna |> count(infection, time) ``` # A tibble: 3 × 3 infection time n 1 InfluenzaA 4 11792 2 InfluenzaA 8 10318 3 NonInfected 0 10318 ``` With the above code, we can proceed with arrange() to sort the table according to a number of criteria so that we have a better comparison. For instance, we might want to arrange the table above by time: rna |> count(infection, time) |> arrange(time) ``` # A tibble: 3 × 3 infection time n 1 NonInfected 0 10318 2 InfluenzaA 4 11792 3 InfluenzaA 8 10318 ``` or by counts: rna |> count(infection, time) |> arrange(n) ``` # A tibble: 3 × 3 infection time n 1 InfluenzaA 8 10318 2 NonInfected 0 10318 3 InfluenzaA 4 11792 ``` ► Question How many genes/observations were analysed in each sample? Use group_by() and summarize() to evaluate the sequencing depth (the sum of all expressions) in each sample. Which sample has the highest sequencing depth? Calculate the mean expression level of gene “Dok3” by timepoints. Pick one sample and evaluate the number of genes by biotype Identify genes associated with “abnormal DNA methylation” phenotype description, and calculate their mean expression (in log) at time 0, time 4 and time 8. ► Solution How many genes/observations were analysed in each sample? rna |> group_by(sample) |> summarise(n = n()) ``` # A tibble: 22 × 2 sample n 1 GSM2545336 1474 2 GSM2545337 1474 3 GSM2545338 1474 4 GSM2545339 1474 5 GSM2545340 1474 6 GSM2545341 1474 7 GSM2545342 1474 8 GSM2545343 1474 9 GSM2545344 1474 10 GSM2545345 1474 # ℹ 12 more rows ``` rna |> count(sample) ``` # A tibble: 22 × 2 sample n 1 GSM2545336 1474 2 GSM2545337 1474 3 GSM2545338 1474 4 GSM2545339 1474 5 GSM2545340 1474 6 GSM2545341 1474 7 GSM2545342 1474 8 GSM2545343 1474 9 GSM2545344 1474 10 GSM2545345 1474 # ℹ 12 more rows ``` Use group_by() and summarize() to evaluate the sequencing depth (the sum of all counts) in each sample. Which sample has the highest sequencing depth? rna |> group_by(sample) |> summarize(seq_depth = sum(expression)) |> arrange(desc(seq_depth)) ``` # A tibble: 22 × 2 sample seq_depth 1 GSM2545350 3255566 2 GSM2545352 3216163 3 GSM2545343 3105652 4 GSM2545336 3039671 5 GSM2545380 3036098 6 GSM2545353 2953249 7 GSM2545348 2913678 8 GSM2545362 2913517 9 GSM2545351 2782464 10 GSM2545349 2758006 # ℹ 12 more rows ``` Calculate the mean expression level of gene “Dok3” by timepoints. rna |> filter(gene == "Dok3") |> group_by(time) |> summarize(mean = mean(expression)) |> arrange(time) ``` # A tibble: 3 × 2 time mean 1 0 169 2 4 156. 3 8 61 ``` Pick one sample and evaluate the number of genes by biotype rna |> filter(sample == "GSM2545336") |> count(gene_biotype) |> arrange(desc(n)) ``` # A tibble: 13 × 2 gene_biotype n 1 protein_coding 1321 2 lncRNA 69 3 processed_pseudogene 59 4 miRNA 7 5 snoRNA 5 6 TEC 4 7 polymorphic_pseudogene 2 8 unprocessed_pseudogene 2 9 IG_C_gene 1 10 scaRNA 1 11 transcribed_processed_pseudogene 1 12 transcribed_unitary_pseudogene 1 13 transcribed_unprocessed_pseudogene 1 ``` Identify genes associated with “abnormal DNA methylation” phenotype description, and calculate their mean expression (in log) at time 0, time 4 and time 8. rna |> filter(phenotype_description == "abnormal DNA methylation") |> group_by(gene, time) |> summarize(mean_expression = mean(log(expression))) |> arrange() ``` summarise() has grouped output by 'gene'. You can override using the .groups argument. ``` ``` # A tibble: 6 × 3 # Groups: gene gene time mean_expression 1 Xist 0 6.95 2 Xist 4 6.34 3 Xist 8 7.13 4 Zdbf2 0 6.27 5 Zdbf2 4 6.27 6 Zdbf2 8 6.19 ``` It is important to be able to conceptually visualise how these different functions operate on data. Being able to do that allows to mentally run a set of commands and get a feeling whether this will work, rather that randomly try things out. The Tidy Data Tutor is a wonderful tool to do exactly that. 5.8 Reshaping data Copy link In rna, the rows contain expression values that are associated with a combination of 2 other variables: gene and sample. All the other columns correspond to variables describing either the sample (age, sex, organism…) or the gene (gene_biotype, ENTREZ_ID, product…). The variables for the same gene and sample pair have the same value in all the rows. This structure is called a long format, as one column contains all the values, and other columns describe the context of the value. These data also tend to be quite long. In certain cases, the long format is not really human-readable or the most appropriate for the task, and another format, called wide format is preferred. It is also generally as a more compact way of representing the data (although we don’t need to worry about the size of the data here). This is typically the case with gene expression values that scientists are used to look as matrices of quantitative data, were rows represent genes (or more generally features or variables) and columns represent samples. To convert the gene expression values from rna into a wide-format, we need to create a new table where each row is composed of expression values associated with each gene. In practical terms this means the values of the sample column in rna would become the names of column variables, and the cells would contain the expression values measured on each gene. The key point here is that we are still following a tidy data structure (a single value per cell), but we have reshaped the data according to the observations of interest: expression levels per gene instead of recording them per gene and per sample. With this new table, it would become therefore straightforward to explore the relationship between the gene expression levels within, and between, the samples. The opposite transformation would be to transform column names into values of a new variable. We can do both these of transformations with two tidyr functions, pivot_longer() and pivot_wider() (see here for details). 5.8.1 Pivoting the data into a wider format Copy link For simplicity, let’s first select the 3 first columns of rna and use pivot_wider() to transform data in a wide-format. rna_exp <- rna |> select(gene, sample, expression) rna_exp ``` # A tibble: 32,428 × 3 gene sample expression 1 Asl GSM2545336 1170 2 Apod GSM2545336 36194 3 Cyp2d22 GSM2545336 4060 4 Klk6 GSM2545336 287 5 Fcrls GSM2545336 85 6 Slc2a4 GSM2545336 782 7 Exd2 GSM2545336 1619 8 Gjc2 GSM2545336 288 9 Plp1 GSM2545336 43217 10 Gnb4 GSM2545336 1071 # ℹ 32,418 more rows ``` pivot_wider takes the following three main arguments: the data to be transformed; the names_from column name whose values will become new column names; the values_from column name whose values will fill the new columns. pivot_wider() generates a new table with 1474 gene for 22 samples - one row for each gene, one column for each sample. We can also directly pipe the data into the pivot_wider(), as illustrated below: rna_wide <- rna_exp |> pivot_wider(names_from = sample, values_from = expression) rna_wide ``` # A tibble: 1,474 × 23 gene GSM2545336 GSM2545337 GSM2545338 GSM2545339 GSM2545340 GSM2545341 1 Asl 1170 361 400 586 626 988 2 Apod 36194 10347 9173 10620 13021 29594 3 Cyp2d22 4060 1616 1603 1901 2171 3349 4 Klk6 287 629 641 578 448 195 5 Fcrls 85 233 244 237 180 38 6 Slc2a4 782 231 248 265 313 786 7 Exd2 1619 2288 2235 2513 2366 1359 8 Gjc2 288 595 568 551 310 146 9 Plp1 43217 101241 96534 58354 53126 27173 10 Gnb4 1071 1791 1867 1430 1355 798 # ℹ 1,464 more rows # ℹ 16 more variables: GSM2545342 , GSM2545343 , GSM2545344 , # GSM2545345 , GSM2545346 , GSM2545347 , GSM2545348 , # GSM2545349 , GSM2545350 , GSM2545351 , GSM2545352 , # GSM2545353 , GSM2545354 , GSM2545362 , GSM2545363 , # GSM2545380 ``` We can now easily compare the gene expression levels in different samples. Note that the pivot_wider() function comes with an optional values_fill argument that can be useful when dealing with missing values. Let’s imagine that for some reason, we had some missing expression values for some genes in certain samples. In the following example, the gene Cyp2d22 has only one expression value, in GSM2545338 sample. rna_with_missing_values <- rna |> select(gene, sample, expression) |> filter(gene %in% c("Asl", "Apod", "Cyp2d22")) |> filter(sample %in% c("GSM2545336", "GSM2545337", "GSM2545338")) |> arrange(sample) |> filter(!(gene == "Cyp2d22" & sample != "GSM2545338")) rna_with_missing_values ``` # A tibble: 7 × 3 gene sample expression 1 Asl GSM2545336 1170 2 Apod GSM2545336 36194 3 Asl GSM2545337 361 4 Apod GSM2545337 10347 5 Asl GSM2545338 400 6 Apod GSM2545338 9173 7 Cyp2d22 GSM2545338 1603 ``` By default, the pivot_wider() function will add NA for missing values. rna_with_missing_values |> pivot_wider(names_from = sample, values_from = expression) ``` # A tibble: 3 × 4 gene GSM2545336 GSM2545337 GSM2545338 1 Asl 1170 361 400 2 Apod 36194 10347 9173 3 Cyp2d22 NA NA 1603 ``` But in some cases, we may wish to fill in the missing values by setting values_fill to a specific value. rna_with_missing_values |> pivot_wider(names_from = sample, values_from = expression, values_fill = 0) ``` # A tibble: 3 × 4 gene GSM2545336 GSM2545337 GSM2545338 1 Asl 1170 361 400 2 Apod 36194 10347 9173 3 Cyp2d22 0 0 1603 ``` 5.8.2 Pivoting data into a longer format Copy link The opposing situation could occur if we had been provided with data in the form of rna_wide, where the sample IDs are column names, but we wished to treat them as values of a sample variable instead. In this situation we are using the column names and turn them into a pair of new variables and need to arrange the expression values accordingly in a new variable. This can be done with the pivot_longer() function. It takes the following four main arguments: the data to be transformed; the new names_to column we wish to create and populate with the current column names; the new values_to column we wish to create and populate with current values; the names of the columns to be used to populate the names_to and values_to variables (or altarnatively, those to drop using a -). To recreate rna_long from rna_long we would create a key called sample and value called expression and use all columns except gene for the key variable. Here we drop gene column with a minus sign. Notice how the new variable names are to be quoted here. rna_long <- rna_wide |> pivot_longer(names_to = "sample", values_to = "expression", -gene) rna_long ``` # A tibble: 32,428 × 3 gene sample expression 1 Asl GSM2545336 1170 2 Asl GSM2545337 361 3 Asl GSM2545338 400 4 Asl GSM2545339 586 5 Asl GSM2545340 626 6 Asl GSM2545341 988 7 Asl GSM2545342 836 8 Asl GSM2545343 535 9 Asl GSM2545344 586 10 Asl GSM2545345 597 # ℹ 32,418 more rows ``` Note that if we had missing values in the wide-format, the NA would be included in the new wide format. Pivoting to wider and longer formats can be a useful way to balance out a dataset so every replicate has the same composition. wide_with_NA <- rna_with_missing_values |> pivot_wider(names_from = sample, values_from = expression) wide_with_NA ``` # A tibble: 3 × 4 gene GSM2545336 GSM2545337 GSM2545338 1 Asl 1170 361 400 2 Apod 36194 10347 9173 3 Cyp2d22 NA NA 1603 ``` wide_with_NA |> pivot_longer(names_to = "sample", values_to = "expression", -gene) ``` # A tibble: 9 × 3 gene sample expression 1 Asl GSM2545336 1170 2 Asl GSM2545337 361 3 Asl GSM2545338 400 4 Apod GSM2545336 36194 5 Apod GSM2545337 10347 6 Apod GSM2545338 9173 7 Cyp2d22 GSM2545336 NA 8 Cyp2d22 GSM2545337 NA 9 Cyp2d22 GSM2545338 1603 ``` We could also have used a specification for what columns to include. This can be useful if you have a large number of identifying columns, and it’s easier to specify what to gather than what to leave alone. Here the starts_with() function can help to retrieve sample names without having to list them all! Another possibility would be to use the : operator! rna_wide |> pivot_longer(names_to = "sample", values_to = "expression", cols = starts_with("GSM")) ``` # A tibble: 32,428 × 3 gene sample expression 1 Asl GSM2545336 1170 2 Asl GSM2545337 361 3 Asl GSM2545338 400 4 Asl GSM2545339 586 5 Asl GSM2545340 626 6 Asl GSM2545341 988 7 Asl GSM2545342 836 8 Asl GSM2545343 535 9 Asl GSM2545344 586 10 Asl GSM2545345 597 # ℹ 32,418 more rows ``` rna_wide |> pivot_longer(names_to = "sample", values_to = "expression", GSM2545336:GSM2545380) ``` # A tibble: 32,428 × 3 gene sample expression 1 Asl GSM2545336 1170 2 Asl GSM2545337 361 3 Asl GSM2545338 400 4 Asl GSM2545339 586 5 Asl GSM2545340 626 6 Asl GSM2545341 988 7 Asl GSM2545342 836 8 Asl GSM2545343 535 9 Asl GSM2545344 586 10 Asl GSM2545345 597 # ℹ 32,418 more rows ``` ► Question Subset genes located on X and Y chromosomes from the rna data.frame and create a new (wide) data.frame with sex as columns, chromosome_name as rows, containing the mean expression of genes located in each chromosome as the values, as shown below. You will need to summarize before reshaping! ► Solution Let’s have a look to variables of interest rna |> filter(chromosome_name == "Y" | chromosome_name == "X") |> select(gene, sample, sex, expression, chromosome_name) |> arrange(gene) ``` # A tibble: 1,298 × 5 gene sample sex expression chromosome_name 1 Armcx5 GSM2545336 Female 1189 X 2 Armcx5 GSM2545337 Female 513 X 3 Armcx5 GSM2545338 Female 414 X 4 Armcx5 GSM2545339 Female 543 X 5 Armcx5 GSM2545340 Male 666 X 6 Armcx5 GSM2545341 Male 1172 X 7 Armcx5 GSM2545342 Female 857 X 8 Armcx5 GSM2545343 Male 559 X 9 Armcx5 GSM2545344 Female 570 X 10 Armcx5 GSM2545345 Male 545 X # ℹ 1,288 more rows ``` rna_1 <- rna |> filter(chromosome_name == "Y" | chromosome_name == "X") |> group_by(sex, chromosome_name) |> summarize(mean = mean(expression)) |> pivot_wider(names_from = sex, values_from = mean) ``` summarise() has grouped output by 'sex'. You can override using the .groups argument. ``` rna_1 ``` # A tibble: 2 × 3 chromosome_name Female Male 1 X 3504. 2497. 2 Y 3 2117. ``` ► Question Now take that data frame and transform it with pivot_longer() so each row is a unique chromosome_name by gender combination. ► Solution rna_1 |> pivot_longer(names_to = "gender", values_to = "mean", -chromosome_name) ``` # A tibble: 4 × 3 chromosome_name gender mean 1 X Female 3504. 2 X Male 2497. 3 Y Female 3 4 Y Male 2117. ``` ► Question Use the rna dataset to create an new table were each row represents the mean expression levels of genes and columns represent the different timepoints. ► Solution rna |> group_by(gene, time) |> summarize(mean_exp = mean(expression)) |> pivot_wider(names_from = time, values_from = mean_exp) ``` summarise() has grouped output by 'gene'. You can override using the .groups argument. ``` ``` # A tibble: 1,474 × 4 # Groups: gene [1,474] gene 0 4 8 1 AI504432 1034. 1104. 1014 2 AW046200 155. 152. 81 3 AW551984 238 302. 342. 4 Aamp 4603. 4870 4763. 5 Abca12 5.29 4.25 4.14 6 Abcc8 2576. 2609. 2292. 7 Abhd14a 591. 547. 432. 8 Abi2 4881. 4903. 4945. 9 Abi3bp 1175. 1061. 762. 10 Abl2 2170. 2078. 2131. # ℹ 1,464 more rows ``` Notice that this generates a tibble with some column names starting by a number. If we wanted to select the column corresponding to the timepoints, we could not use the column names directly… What happens when we select the colum 4? rna |> group_by(gene, time) |> summarize(mean_exp = mean(expression)) |> pivot_wider(names_from = time, values_from = mean_exp) |> select(gene, 4) ``` summarise() has grouped output by 'gene'. You can override using the .groups argument. ``` ``` # A tibble: 1,474 × 2 # Groups: gene [1,474] gene 8 1 AI504432 1014 2 AW046200 81 3 AW551984 342. 4 Aamp 4763. 5 Abca12 4.14 6 Abcc8 2292. 7 Abhd14a 432. 8 Abi2 4945. 9 Abi3bp 762. 10 Abl2 2131. # ℹ 1,464 more rows ``` To select the timepoint 4, we would have to quote the column name, with backticks “`” rna |> group_by(gene, time) |> summarize(mean_exp = mean(expression)) |> pivot_wider(names_from = time, values_from = mean_exp) |> select(gene, `4`) ``` summarise() has grouped output by 'gene'. You can override using the .groups argument. ``` ``` # A tibble: 1,474 × 2 # Groups: gene [1,474] gene 4 1 AI504432 1104. 2 AW046200 152. 3 AW551984 302. 4 Aamp 4870 5 Abca12 4.25 6 Abcc8 2609. 7 Abhd14a 547. 8 Abi2 4903. 9 Abi3bp 1061. 10 Abl2 2078. # ℹ 1,464 more rows ``` Another possibility would be to rename the column, choosing a name that doesn’t start by a number : rna |> group_by(gene, time) |> summarize(mean_exp = mean(expression)) |> pivot_wider(names_from = time, values_from = mean_exp) |> rename("time0" = `0`, "time4" = `4`, "time8" = `8`) |> select(gene, time4) ``` summarise() has grouped output by 'gene'. You can override using the .groups argument. ``` ``` # A tibble: 1,474 × 2 # Groups: gene [1,474] gene time4 1 AI504432 1104. 2 AW046200 152. 3 AW551984 302. 4 Aamp 4870 5 Abca12 4.25 6 Abcc8 2609. 7 Abhd14a 547. 8 Abi2 4903. 9 Abi3bp 1061. 10 Abl2 2078. # ℹ 1,464 more rows ``` ► Question Use the previous data frame containing mean expression levels per timepoint and create a new column containing fold-changes between timepoint 8 and timepoint 0, and fold-changes between timepoint 8 and timepoint 4. Convert this table in a long-format table gathering the foldchanges calculated. ► Solution rna_fc <- rna |> group_by(gene, time) |> summarize(mean_exp = mean(expression)) |> pivot_wider(names_from = time, values_from = mean_exp) |> mutate(time_8_vs_0 = `8` / `0`, time_8_vs_4 = `8` / `4`) ``` summarise() has grouped output by 'gene'. You can override using the .groups argument. ``` rna_fc ``` # A tibble: 1,474 × 6 # Groups: gene [1,474] gene 0 4 8 time_8_vs_0 time_8_vs_4 1 AI504432 1034. 1104. 1014 0.981 0.918 2 AW046200 155. 152. 81 0.522 0.532 3 AW551984 238 302. 342. 1.44 1.13 4 Aamp 4603. 4870 4763. 1.03 0.978 5 Abca12 5.29 4.25 4.14 0.784 0.975 6 Abcc8 2576. 2609. 2292. 0.889 0.878 7 Abhd14a 591. 547. 432. 0.731 0.791 8 Abi2 4881. 4903. 4945. 1.01 1.01 9 Abi3bp 1175. 1061. 762. 0.649 0.719 10 Abl2 2170. 2078. 2131. 0.982 1.03 # ℹ 1,464 more rows ``` rna_fc |> pivot_longer(names_to = "comparisons", values_to = "Fold_changes", time_8_vs_0:time_8_vs_4) ``` # A tibble: 2,948 × 6 # Groups: gene [1,474] gene 0 4 8 comparisons Fold_changes 1 AI504432 1034. 1104. 1014 time_8_vs_0 0.981 2 AI504432 1034. 1104. 1014 time_8_vs_4 0.918 3 AW046200 155. 152. 81 time_8_vs_0 0.522 4 AW046200 155. 152. 81 time_8_vs_4 0.532 5 AW551984 238 302. 342. time_8_vs_0 1.44 6 AW551984 238 302. 342. time_8_vs_4 1.13 7 Aamp 4603. 4870 4763. time_8_vs_0 1.03 8 Aamp 4603. 4870 4763. time_8_vs_4 0.978 9 Abca12 5.29 4.25 4.14 time_8_vs_0 0.784 10 Abca12 5.29 4.25 4.14 time_8_vs_4 0.975 # ℹ 2,938 more rows ``` 5.9 Exporting data Copy link Now that you have learned how to use dplyr to extract information from or summarize your raw data, you may want to export these new data sets to share them with your collaborators or for archival. Similar to the read_csv() function used for reading CSV files into R, there is a write_csv() function that generates CSV files from data frames. Before using write_csv(), we are going to create a new folder, data_output, in our working directory that will store this generated dataset. We don’t want to write generated datasets in the same directory as our raw data. It’s good practice to keep them separate. The data folder should only contain the raw, unaltered data, and should be left alone to make sure we don’t delete or modify it. In contrast, our script will generate the contents of the data_output directory, so even if the files it contains are deleted, we can always re-generate them. In preparation for our next lesson on plotting, we are going to prepare a table representing for each gene, the fold-changes (in log values) between timepoint 8 and timepoint 0, and the fold-changes between timepoint 8 and timepoint 0. rna_fc <- rna |> mutate(expression_log = log(expression)) |> group_by(gene, time) |> summarize(mean_exp = mean(expression_log)) |> pivot_wider(names_from = time, values_from = mean_exp) |> mutate(time_8_vs_0 = `8` - `0`, time_4_vs_0 = `4` - `0`) |> select(gene, time_8_vs_0, time_4_vs_0) ``` summarise() has grouped output by 'gene'. You can override using the .groups argument. ``` rna_fc ``` # A tibble: 1,474 × 3 # Groups: gene [1,474] gene time_8_vs_0 time_4_vs_0 1 AI504432 -0.0248 0.0523 2 AW046200 -0.650 -0.0348 3 AW551984 0.232 -0.428 4 Aamp 0.0271 0.0522 5 Abca12 -0.116 -0.114 6 Abcc8 -0.114 0.0163 7 Abhd14a -0.309 -0.0800 8 Abi2 0.0110 -0.000894 9 Abi3bp -0.432 -0.107 10 Abl2 -0.0188 -0.0550 # ℹ 1,464 more rows ``` We can save the table as a CSV file in our data_output folder. write_csv(rna_fc, file = "data_output/rna_fc.csv") 5.10 Additional exercises Copy link ► Question We are going to re-analyse beer consumption in 48 individuals using dplyr. The data are available in the rWSBIM1207 package. The data illustrated the fictive beer consumption in litres per year at different age according to gender and employment. Load the rWSBIM1207 package. If the package isn’t installed of its version is older than 0.1.1, install it from the UCLouvain-CBIO/rWSBIM1207 GitHub repository using the BiocManager::install() function. Directly load the data by typing data(beers) Remove observations with missing values. Using the Year, Month and Day columns, create a new column Date using dplyr::mutate and lubridate::ymd. What is the class of Date ? Create a new table, containing observations for women older than 35 years old, employed, and select all columns except Day, Month and Year, and order in descending value of consumption of beers. Export the new table to a csv file. Beer consumption analysis: Does employment status have an impact on beer consumption? Do men drink more than women? Does employment status have an influence on beer consumption according to gender? Do men drink more than women according to age and employment status? As we can see from the last exercise, it become difficult to read and interpret multiple results. In the next chapter, we will see how to complement such analysis questions with visualisations such as the following one, that clearly highlight important patterns in our data. Figure 5.1: Visualisation of beer consumption, highlighting different patterns of beer consumption in employed and unemployed males and females. ► Question The Cancer Genome Atlas (TCGA) is a large scale effort that has collected high throughput biology data from hundreds of patients samples. In this exercise, we are going to analyse the clinical variables recorded for a subset of the patients. Load the rWSBIM1207 package. If the package isn’t installed of its version is older than 0.1.1, install it from the UCLouvain-CBIO/rWSBIM1207' GitHub repository using thedevtools::install_github` function. Using the clinical1.csv() function from rWSBIM1207, find the path the clinical1.csv file and read it to produce a data.frame named clinical. Familiarise yourself with the data. Create a smaller data frame called clinical_mini containing only the columns corresponding to patientID, gender, age_at_diagnosis, smoking_history, number_pack_years_smoked, year_of_tobacco_smoking_onset, and stopped_smoking_year. Calculate the number of males and females in the cohort. Create a new variable years_at_diagnosis corresponding to the age at diagnosis converted from days into years. Calculate the mean and median age at diagnosis (in years). Pay attention to missing values! Calculate the mean and median age at diagnosis for males and females. How many patient were diagnosed before 50 years? Compare the mean age at diagnosis between current smoker and lifelong non-smoker. Select patients who stopped smoking more than 15 years ago and calculate the number of smoking years for these cases. Display only cases for which you were able to calculate the data. How many of them smoked less than 5 years? Try to recreate the following table, reporting the number of smokers and lifelong-non smoker between males and females. Note: the layout can be different. | gender | current smoker | lifelong non-smoker | ---: | female | 51 | 55 | | male | 69 | 20 | ► Question Using the interroA.csv() function from the rWSBIM1207 package to get the path to the spreadsheet file, read the data into R using the read_csv function. This data is in the wide format, with the results of each test stored as a separate column. Using the appropriate pivot function, convert the data into a long table with a column interro informing which test that line refers to and a column result with the student’s mark. ► Question Make sure you have rWSBIM1207 version >= 0.1.16 and load the 2022 Belgian road accidents statistics and the associated metadata, describing the variables. The path to the former as an rds file is available with road_accidents_be_2022.rds(). The road_accidents_be_meta.csv() returns the path to the metadata csv file. The data provides the Number of killed, seriously injured, slightly injured and uninjured victims of road accidents, by age group, type of user, sex and various characteristics of the accident in Belgium in 20222. Using the appropriate functions, load both files into R and familiarise yourself with the data. Compare the numbers for man and women over the hours of the day for all age classes. Ignore any unknown information. Compare the number of victims in the different provinces. Do this comparison for the different type of victims. Ignore any unknown information. Come up with additional question/comparisons that you could ask these data. PreviousNext Page built: 2025-06-16 using R version 4.5.0 (2025-04-11)
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https://connecteam.com/e-24-hour-shift-schedule/
Easily manage your daily operations From scheduling, all the way down to payroll Explore Operations Features Time Clock Smart time tracking with automated timesheets Scheduling Efficient staff scheduling, from planning to distribution Task Management Real-time task delegation and progress tracking Forms & Checklists Instant field reporting with live results and insights One app to keep your team connected Keep your entire workforce informed and engaged from anywhere Explore Communications Features Updates Company newsfeed with engagement tracking Chat Secure and compliant business messaging tool Directory Accessible index for all work contact information Surveys Polls with real-time feedback to make informed decisions Knowledge Base One centralized place for all your company know-how Events Announce, promote, and manage company events Help Desk Internal ticketing system for employee queries Build a skilled and engaged workforce Make HR compliant with mobile training and time off management Explore HR & Skills Features Recognition & Rewards Acknowledge employees, increase motivation and reduce turnover Training & Onboarding Interactive courses with progress tracking, available anywhere Quizzes Evaluate employee knowledge and keep staff aligned Documents Safely store and sort all employee files & certificates Timeline Monitor employee lifecycles and important milestones Time Off Create and enforce absence policies with a clear overview Employee ID Card Manage employee identification cards with a click Org Chart Have a clear visual of your company hierarchy and roles Use Case Time Management Operations Management Employee Onboarding Health & Safety Employee Engagement Company Intranet Compliance Company Culture Industry Construction Cleaning Services Field Services Food & Beverage Homecare Manufacturing Retail All industries Role Operations Human Resources IT Internal Communication Training Integrations Gusto Quickbooks Xero Paychex ADP Zapier Wix Bookings Lightspeed General About Connecteam Why Connecteam Newsroom Trust Center Contact Contact Sales Contact Support Customers Our Customers Case Studies Customer Stories Franchise With Us All Resources Blog HR Glossary Template Directory Guides & eBooks Expert Interviews Marketplace Help Center Topics 23min read 24-hour Shift Schedule: Types, Examples & Tips Updated on June 27, 2024 • 23min read Share Written by Michael Graw Michael is a prolific author in business and B2B tech, whose articles can be found on Business Insider, Entrepreneur, TechRadar Pro, IT Pro Portal, Tom's Guide, and more, covering everything from international tech regulations to corporate finance and emerging tech brands and markets. A successful copywriter and entrepreneur, Michael has worked with dozens of SaaS and tech companies, and has his finger firmly on the pulse of B2B tech, finance and business. Edited by David Ruhm David is Connecteam's Senior Editor. With over a decade of experience as a journalist and content professional, David leads the production of industry-leading content to help professionals take their businesses to the next level. He is passionate about creating engaging, informative, and thought-provoking content that resonates with readers. We explain how to create a 24-hour shift schedule, so your team can work around the clock. If your business needs to operate around the clock, you must build a 24-hour shift schedule your employees are happy with and that’s effective for your business’s operations. Otherwise, you risk absenteeism, employee burnout, or lost productivity. In this guide, we explore common types of 24-hour work schedules. We also explain how to choose the best schedule for your business and implement it. Key Takeaways What Is a 24-hour Shift Schedule? A 24-hour schedule is a type of work schedule that enables your business to operate 24/7 with no breaks in staffing. Twenty-four-hour schedules are frequently used in industries like manufacturing, healthcare, transportation, and emergency services such as firefighter schedules and schedules for flight attendents. A growing number of retail and customer service businesses also need 24/7 staffing. Twenty-four-hour schedules typically require more complexity than just assigning employees to 24-hour shifts. Most employees can’t work 24 hours at a time without burning out. So, round-the-clock work schedules usually involve several teams of employees working in shifts. Some schedules are split into 12-hour day and night shifts. Schedules that call for 8 or 10-hour shifts add a third morning shift or a “swing” shift that typically runs from early afternoon to late evening. 24/7 shift schedules must balance several competing factors, including: Fixed vs. Rotating Shifts Most 24/7 work schedules are broken into shifts, and how you assign employees to these shifts is important. You can assign employees to the same shift consistently—a practice known as fixed shifts. Or you can rotate employees among shifts—a practice known as rotating shifts. With fixed shifts, you’ll have some teams of employees that always work day shifts and other teams of employees that always work night shifts. Fixed shifts are simple to manage and give employees more consistent schedules. However, many businesses opt for rotating shift schedules because they’re more equitable for employees. With these, you don’t have 1 group of employees that always works coveted day shifts and another group of employees locked into less desirable night shifts. In the shift patterns below, we highlight how you can rotate shifts to ensure each schedule is as fair as possible for employees. This Might Interest You Discover our guide on how to create a great rotating schedule, how to create a rotating weekend schedule or read more about the best night shift schedules for the sleep and health of your employees. Common Types of 24-hour Shift Patterns Here are some of the most common types of 24-hour work schedules, plus their pros and cons. 4 on, 3 off shift pattern How it works: Employees work 4 consecutive 10-hour days, then take 3 consecutive days off. Your schedule is broken into 3 shifts—typically, day, swing, and night shifts—and each team transitions from day to swing to night shifts on a weekly basis. The schedule resets after 3 weeks. Typical shift length: 10 hours Number of teams required: 6 Example: Pros: Cons: Best for: This is one of the most popular 24/7 shift schedules. It’s ideal for businesses that have times of peak demand, such as retail businesses that receive more traffic during morning and evening hours. 4 on, 4 off shift pattern How it works: Employees work 4 consecutive 12-hour days, then take 4 consecutive days off. The schedule is broken into 2 shifts, typically day and night. Teams can switch between day and night shifts on any pattern you prefer, such as weekly, biweekly, or monthly. Typical shift length: 12 hours Number of teams required: 4 Example: Pros: Cons: Best for: This is a good alternative to the 4-on, 3-off schedule for businesses with constant demand, such as many manufacturing companies. Since shifts are 12 hours long, it’s best suited for jobs that require intermittent work or less focus. 2-2-3 (Panama) shift pattern How it works: In a 2-2-3 shift pattern, employees work in a cycle consisting of 2 days on, 2 days off, 3 days on, and 3 days off. Shifts are typically 12 hours, so employees work 60 hours one week and 24 hours the next. Employees rotate from day to night shifts every 2 weeks. Typical shift length: 12 hours Number of teams required: 4 Example: Pros: Cons: Best for: The Panama schedule can work well for businesses across various industries, including manufacturing, transportation, and food service. However, the schedule is complex and requires employees to be comfortable with long workweeks, so it works best for companies with strong communication between managers and employees. Did You Know? Another 24/7 schedule that closely resembles the 2-2-3 schedule is the 2-3-2 or Pitman schedule. DuPont Schedule How it works: The DuPont schedule, named after the DuPont manufacturing company, is a 4-week schedule that requires 4 teams working 12-hour shifts. Teams alternate between day and night shifts each time they have a stretch of workdays. In a 4-week period, one team’s schedule will look like this: 4 nights on, 3 off, 3 days on, 1 off, 3 nights on, 3 off, 4 days on, then 7 off. Typical shift length: 12 hours Number of teams required: 4 Example: Pros: Cons: Best for: The DuPont schedule was originally created for manufacturing plants, and it continues to be a good option for manufacturing businesses. It can also work well for any business that has consistent demand and is okay with 12-hour shifts. Continental shift pattern How it works: In a continental shift pattern employees work 7 consecutive 8-hour shifts, transitioning every 2-3 days from day to swing to night shifts. After 7 days on, they get 2-3 days off. The schedule resets every 28 days. Typical shift length: 8 hours Number of teams required: 4 Example: Pros: Cons: Best for: This schedule is a good option for businesses that need to operate 24/7 but have very variable demand between day, swing, and night shifts. Employees work a busy day shift for only 2 days before rotating onto a less busy swing shift and then a quiet night shift. a closely related shift pattern to the continental shift pattern is the southern swing schedule. 24 on, 48 off shift pattern How it works: This schedule doesn’t use shifts. Instead, employees work a full 24-hour day followed by 2 days off. Typical shift length: 24 hours Number of teams required: 3 Example: Pros: Cons: Best for: The 24 on, 48 off shift pattern is widely used in emergency services since employees are working in spurts when emergencies strike rather than working continuously for 24 hours. This schedule can also be used to assign on-call duties for healthcare and customer service businesses. 6 on, 4 off shift pattern How it works: Employees work 6 consecutive 10-hour shifts followed by 4 days off. There are morning, swing, and night shifts, and employees rotate onto a new shift for each 6-day work period. The cycle resets every 30 days. Typical shift length: 10 hours Number of teams required: 5 Example: Pros: Cons: Best for: This schedule works well for many healthcare businesses since it enables teams to overlap between shifts and share information. It also gives employees several days off in a row, which can reduce burnout in high-stress jobs. Did You Know? If you want to learn how employees in different industries feel about specific shift patterns, Reddit is a goldmine. For instance, check out this subreddit to see how police cope with 6 on, 4 off shifts. 5 on, 3 off shift pattern How it works: This is a 25-day cycle in which employees work 10-hour shifts. Employees work 5 consecutive days, then get 3 days off. Then they work 5 consecutive days and get 4 days off. Finally they work 5 consecutive days and get 3 days off. This pattern typically involves 3 shifts that overlap for 6 hours per day. Employees rotate between day, swing, and night shifts during each 5-day work period. Typical shift length: 10 hours Number of teams required: 5 Example: Pros: Cons: Best for: This schedule is complex to manage, so it works best for businesses with highly responsible employees and excellent communication channels in place. It’s most suitable for healthcare, retail, and customer service businesses that can benefit from shift overlap. 6 on, 3 off shift pattern How it works: Most employees work 6 consecutive 8-hour shifts followed by 3 days off. However, you also need a part-time team to fill in shifts 2-3 consecutive days each week. Your schedule is split into 3 shifts, and employees rotate shifts at the start of each new 6-day work period. Typical shift length: 8 hours Number of teams required: 4.5 Example: Pros: Cons: Best for: This schedule isn’t frequently used because it requires a part-time team to fill in scheduling gaps. However, it offers more flexibility since you can ask part-time employees to fill in for full-time employees who are out sick or on vacation. It’s best suited for manufacturing or retail businesses because there’s no overlap between shifts. 3 on, 4 off shift pattern How it works: This shift pattern requires employees to work 3 consecutive days of 12-hour shifts followed by 4 days off. Every 2 weeks, employees switch from day shifts to night shifts. Most shifts have 2 teams working simultaneously, but there are occasionally shifts with only 1 team or 3 teams. Typical shift length: 12 hours Number of teams required: 10 Example: Pros: Cons: Best for: This is an uncommonly used schedule because it requires 10 teams of employees and results in some understaffed and overstaffed shifts. However, it offers employees a predictable schedule and a very large amount of rest between shifts. So, it’s sometimes used in industries like oil and gas that involve highly demanding and dangerous work. 4 on, 6 off shift pattern How it works: Employees work 4 consecutive days of 12-hour shifts followed by 6 days off. Most shifts have 1 team working, but there are several shifts each week staffed by 2 teams. Employees rotate between day and night shifts every 2 weeks. Typical shift length: 12 hours Number of teams required: 6 Example: Pros: Cons: Best for: This is an infrequently used schedule that’s best for high-intensity jobs where employees need lots of rest between work periods. For example, it can work well in the oil and gas industry. 5 Over 7 Shift Pattern for 24/5 Schedules Some businesses in customer service operate on 24/5 schedules, offering round-the-clock service 5 days a week instead of 7. The most popular shift pattern for 24/5 staffing is called a 5 over 7 pattern. How it works: A 5 over 7 schedule mimics the typical workweek in that employees work 5 consecutive 8-hour days, then get 2 days off. Twenty-four-hour days are split into 3 shifts, and employees rotate shifts weekly. Typical shift length: 8 hours Number of teams required: 3 Example: Pros: Cons: Best for: This schedule is widely used by customer service businesses. It’s equitable to employees and gives them consistent weekly schedules without requiring overtime. They might even get weekends off. What To Consider When Choosing a 24-hour Shift Schedule for Your Business There’s a lot to think about when choosing the right 24-hour work schedule for your business. You must balance scheduling demands and employee morale, all while maintaining flexibility and keeping productivity as high as possible. Here are several key things to consider when picking a shift pattern. Shift overlap One of the major differences between the shift patterns we highlighted is that some require teams to overlap for several hours a day, while others don’t. This is important because you need enough manpower to handle peak demand on your business, but you also don’t want to have employees standing around without enough to do. In general, businesses in industries like healthcare, retail, customer service, and food service tend to have more ebbs and flows in how busy they are throughout the day and can benefit from shift overlap. Businesses with more constant demand, like manufacturing and transportation businesses, may prefer schedules without shift overlap. Legal considerations Most industries and most states don’t have laws regulating how long employees can work. However, industries like commercial trucking have regulations limiting how long drivers can spend behind the wheel. In addition, states like California have rules requiring that employees approve alternative schedules that require shifts longer than 10 hours. Make sure the schedule you choose follows all applicable regulations around allowable work hours. It’s a good idea to consult a labor attorney to ensure you don’t violate any regulations. This Might Interest You Check out our state labor law guides for more about regulations that may impact 24-hour scheduling. Overtime pay Paying overtime can be a necessary cost of 24/7 operations. However, it’s still important to think about how much you’ll spend on overtime pay with different shift patterns. Schedules that require non-exempt employees to work a maximum of 48 hours a week will generally cost less than schedules that require up to 60 hours per week. This Might Interest You Healthcare facilities may be able to reduce overtime pay with certain schedules—including the 6 on, 4 off schedule—by using an “8 and 80” system. Under this overtime scheme, employees are paid overtime when they work more than 8 hours in a day or exceed 80 hours in a 14-day period. Employee burnout The type of work your employees do can also influence which shift pattern is best-suited for your business. Employees in high-stress roles, such as in healthcare settings, may need to work shorter shifts or fewer consecutive days in a row to avoid burnout. Employees who work intermittently, such as on-call employees in emergency services, may be able to work 12 or 24-hour shifts without a problem. Employees in jobs that require more focus may not be able to work extended shifts safely or productively. Schedule management Schedules that reset on shorter cycles (such as every 3 weeks) are typically easier for you and your employees to manage than schedules that reset on longer cycles. This can improve employees’ work-life balance, since they’re able to coordinate with family and friends more easily. Shorter, more predictable shift patterns also make it easier to manage employee absences due to sickness or vacation. Employees can fill in for colleagues or trade shifts without having to plan months into the future. Number of teams Shift patterns that require more teams (5 or more) can be significantly more complex to manage than shift patterns requiring fewer teams. If an employee takes time off, it can feel like a puzzle to figure out how to fill in for them without disrupting your entire schedule. You also have to hire and manage a greater number of individual employees, which can increase your business’s overall payroll and benefits costs. Tips for Building an Effective 24-hour Shift Schedule Here are 7 tips to ensure your 24-hour schedule is as effective as possible for your team. Communicate with your employees Communicate with your employees about the type of 24/7 schedule they prefer. Some employees might prefer more intense work periods followed by longer rest periods, while others might prefer working shorter shifts more frequently. Employees might also have commitments outside of work that impact their ability to work different shifts. For example, employees with children might prefer working night shifts because they want to be able to take their kids to and from school. Did You Know? Connecteam makes it easy to survey your employees so you can ask them what shift schedule they prefer. You can also collect feedback from employees after implementing a new work schedule. Balance skills and experience across teams When building teams for your shifts, don’t just consider employee preferences—think about the skills and experience each employee will bring to a team. For example, each team should have a mix of newer employees and more veteran employees who can serve as mentors. Each team should also be staffed to fill all required roles for your operations, such as doctors, nurses, and administrators for a healthcare shift. Transferring employees from one team to another can be disruptive to their schedules, so it’s important to minimize how frequently you need to do this. Consult your supervisors to find out which employees work well together and which should be assigned to different teams. Use scheduling software Shift scheduling software like Connecteam can make managing schedules much easier. You can create a schedule for one full cycle of your shift pattern, then use it as a template to create future schedules more quickly. In addition, scheduling software can alert you if a shift is understaffed or overstaffed or if a non-exempt employee is scheduled to work excess overtime. Scheduling software is especially important for managing employee absences. For instance, Connecteam enables employees to trade shifts with your approval or pick up open shifts with or without your approval—so they can easily fill in for one another when needed. Set up shift reminders It’s a good idea to set up automated shift reminders for your employees, especially if your schedule requires employees to work different days or shifts each week. Reminders can reduce missed work and ensure you’re always fully staffed. Did You Know? Connecteam can automatically send push notifications to employees’ phones to let them know they have a shift coming up. It also enables employees to quickly check their work schedule from any device. Modify standard shift patterns The shift schedules we highlighted are commonly used, but that doesn’t mean they’re set in stone. There are several ways you can consider modifying standard shift patterns to better suit your business: Did You Know? Walmart is an example of a company that’s modified standard 24/7 schedules. It introduced what it calls “core hours.” Employees who have a core hours schedule work the same shift for at least 13 weeks, giving them greater ability to plan their personal lives. They can also easily swap shifts and self-schedule extra shifts for added flexibility. Offer incentives for unpopular shifts If you offer fixed shifts, consider offering higher pay for more difficult or less popular shifts. For example, say you run a manufacturing business with steady work across 3 shifts. In this case, the swing and night shifts are likely to be more of a burden on employees than day shifts, so you could offer a higher pay rate for employees working these shifts. This can make it easier to hire and retain employees for less popular shifts. Otherwise, you risk losing employees from your night shift anytime a job opens for a competitor’s day shift. Higher pay for undesirable shifts also encourages your employees to fill in open shifts on nights and weekends. Without incentives, finding someone who’s willing to take an extra shift for an employee who calls in sick for a night shift can be difficult. Use data to evaluate your schedule After you implement a new work schedule, be sure to track data so you can evaluate whether it’s effective. You can monitor productivity metrics as well as send out monthly employee satisfaction surveys. In your surveys, be sure to ask employees about their work-life balance, focus at work, and whether they feel sufficiently rested when they return to work after a break period. You should also make a note of any changes in employee retention or your company’s ability to hire new employees. If you suspect that your schedule is encouraging employees to leave, ask about this in your exit interviews. Did You Know? Connecteam offers in-depth reporting tools to help you monitor absences, overtime, breaks, and more. You can use this information to determine if your schedule is working for you and your team. Conclusion Building an effective 24/7 shift schedule can enable your business to operate around-the-clock while reducing the risk of employee burnout and limiting your overtime costs. There are many different 24-hour work patterns to choose from with different shift lengths and configurations to meet the needs of different businesses. When deciding which one is right for your business, consult your employees and personalize your schedule to suit your business. Once you choose a schedule, you can use Connecteam to put it into practice, send shift reminders to employees, and enable employees to pick up open shifts. Get started with Connecteam for free today! FAQs Are 24-hour shift schedules legal? 24-hour shift schedules are legal in most states. However, there are some industries, like trucking, that limit how many hours employees can work in a set period of time. In addition, California requires that employees approve alternative schedules requiring shifts longer than 10 hours. It’s a good idea to consult a labor attorney to ensure your schedule follows all applicable regulations. Can employees work 24-hour shifts? Employees are legally allowed to work 24-hour shifts in most industries. However, 24-hour shifts can be exhausting, so they’re best suited for jobs that require intermittent work or for employees who are on call. A 24 on, 48 off shift schedule is commonly used for emergency services. What is the best 24-hour shift schedule for manufacturing? There are many different 24-hour work schedules that manufacturing businesses can use. Some of the most common include 4 on 4 off, DuPont, and 2-2-3 shift patterns. These patterns don’t overlap shifts, which is nice since operational demands are very consistent at most manufacturing companies. Subscribe to our newsletter All Rights Reserved © Connecteam.com Product Company Industries Resources Talk with sales
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http://sarahlambart.com/teaching/mineralogy-07.pdf
Time on task: about 1 hour (Material posted on Sept 28th, Office hours: Oct 12th and 14th) Goals: Calculate a mineral formula from weight percent of elements or oxides • Calculate composition from a mineral • Calculate the proportion of the end‐members in a solid solution • Being able to say if the stoichiometry of the mineral is respected • Being able to: This lecture is an introduction to your Problem set #3 (due on Friday Oct 16th). For a demonstration in Excel, check the associated video. If you are not familiar with Excel, you can start by doing the Excel tutorial associated with this lecture. Chemical analyses are usually reported in weight percent of elements or elemental oxides. Calculate mineral formula requires transforming weight percent into atomic percent or molecular percent. SiO2 TiO2 (+4) • Al2O3 Cr2O3 Fe2O3 (+3) • MgO MnO FeO CaO(+2) • Na2O K2O H2O (+1) • P2O5 (+5) • Ions complexes of important cations (cation valency in parentheses) Example 1: weight percent to formula (1) given (2) given (3) = (4) Calculated (5) Calculated Oxide Wt.% MolWt oxide # #moles of cations #moles of oxygen SiO2 59.83 60.086 0.9957 10.9957=0.9957 1 20.9957=1.9914 MgO 40.49 40.312 1.0044 11.0044=1.0044 1 11.0044=1.0044 Total 100.32 2.9958 3 (1) analysis obtained at the electron probe; (2) molecular weight of the oxide (calculated from the periodic table); (4) number of moles of cations. For instance, in one mole of SiO2, I have one mole of Si => in 0.9957 moles of SiO2, I have 0.9957 moles of Si; (5) number of moles of oxygen. For instance, in one mole of SiO2, I have two mole of O => in 0.9957 moles of SiO2, I have 20.9957 moles of O. We then need to add the number of moles of oxygen to get the total. Mole ratio Mg:Si:O = 1:1:3 Formula: MgSiO3 = enstatite Why the analytical total is different from 100 wt.%? Because each analysis as an analytical uncertainty (due to the apparatus or to the homogeneity of the sample). Those errors will propagate to the total. • Which level of precision should I use for the oxide molecular weight? If you don't use enough significant figures, you might introduce additional uncertainty early in the calculations that will • Questions you should ask yourself. L8: Mineral formula Wednesday, July 22, 2020 18:02 Lecture 8 Page 1 propagate (and become larger) in your final calculations. If you use too much digits, they might not be relevant for the final results, however you would not introduce additional uncertainty. It is always good practice to keep the maximal level of precision in your input data, use "formula algebra in Excel" and decide only in the final calculations what is the needed significant digits. The number of significant figures of the final calculations. It depends on the type of calculations. It is usually defined by the input data with the less good precision (e.g., wt.%). In addition, we are calculating the formula of a pure end‐member (no solid solution), so the number of cations (and oxygen) is an integer. • Lecture 8 Page 2 Kyanite: Al2SiO5 (1) given (2) given (3) = (1)(2) (4) = (3)100/(3)total Oxide # of moles MolWt oxide # of grams of oxides wt. % SiO2 1 60.086 60.086 37.079 Al2O3 1 101.963 101.963 62.921 Total 162.049 100 Always assume that Oxygen is in excess. At the surface of the Earth and in the mantle, it's true. So you just have to look at the cations. The number of significant figures is there define by the molecular weight of the oxide you used. Problem 1: Solid solution weight percent to formula. Alkali Feldspars may exist with any composition between NaAlSi3O8 (Albite) and KAlSi3O8 (Sanidine, Orthoclase and Microcline) The formula has 8 oxygens: (Na,K)AlSi3O8 (the parentheses indicate the solid solution) The alkalis may substitute in any ratio, but total alkalis (Na + K) to Al is 1 to 1. (1) given (2) given (3) = (1)/(2) (4) Calculated (5) (6) = (4)8/(5)total Oxide Wt.% MolWt # moles oxide #moles cations #moles of #moles cations for 8 oxygens SiO2 59.83 60.086 1.1350 11.1350=1.1350 21.1350= 2.9997 3 Al2O3 40.49 101.963 0.1892 20.1892=0.3784 30.1892= 1.0001 1 Na2O 10.20 61.9796 0.1646 20.1646=0.3291 0.1645 0.8699 0.87 K2O 2.32 94.204 0.0246 20.0246=0.0493 0.0246 0.1311 0.13 Total 100 3.0269 the true total was probably not equal to one but analyses are often renormalized to 100 in papers to make the comparison between the analyses easier. Mole ratio Si:Al:(Na+K) = 3:1:1 ‐ the stoichiometry of the alkali feldspars is respected We usually don’t use more than 2 digits in the mineral formula: (K.013Na.87)AlSi3O8 L8: Example 2: Formula to weight percent Tuesday, August 25, 2020 3:01 PM Lecture 8 Page 1 anions are always at the end of the formula • cations are in croissant order of charges 1+ (e.g., Na, K), 2+ (Fe, Mg, Mn, Cu), 3+(Cr, Al), 4+ (Si, Ti) • If two cations have the same charge, the one in the highest molar abundance goes first • Rule to write the formula: Alkali Feldspars: NaAlSi3O8 ‐ KAlSi3O8 • Orthopyroxenes: MgSiO3‐ FeSiO3 Enstatite ‐ Ferrosilite (opx) • MgCaSi2O6‐FeCaSi2O6 Diopside‐Hedenbergite (cpx) • Olivines: Mg2SiO4‐ Fe2SiO4 Forsterite ‐ Fayalite • Garnets: Mg3Al2Si3O12‐ Fe3Al2Si3O12 Pyrope ‐ Almandine • Various simple solid solutions: Given the formula En70Fs30 for an orthopyroxene, calculate the weight percent oxides En = Enstatite = MgSiO3 Fs = Ferrosilite = FeSiO3 The formula is (Mg.7Fe.3)SiO3 (1) given (2) given (3) = (1)(2) (4) = (3)100/(3)total Oxide # of moles MolWt oxide # of grams of oxides wt. % MgO 0.7 40.312 28.2184 25.686 FeO 0.3 71.846 21.5538 19.620 SiO2 1 60.086 60.086 54.694 Total 109.8582 100 Problem 2: Solid‐solution ‐ formula to weight percent Consider a Pyroxene solid solution of 40% Jadeite (NaAlSi2O6) and 60% Aegirine (NaFeSi2O6). Calculate the weight percent oxides. The formula of this pyroxene is: Na(Fe.6Al.4)Si2O6. Which oxides do you need to describe this pyroxene? Check the video for calculations in excel ‐ keys are on the next page Problem 3: Solid‐solution with coupled substitutions‐ formula to wt.% Consider a plagioclase composition with 40% Anorthite and 60% Albite. Calculate Weight percent Oxides. The first step is to write the formula of this plagioclase Check the video for calculations in excel ‐ keys are on the next page L8: Example 3 ‐ Solid solution: formula to wt. % Tuesday, August 25, 2020 3:02 PM Lecture 8 Page 1 Problem #2 • The formula is Na(Fe.6Al.4)Si2O6. Fe and Al are present in the same sites and have the same valency as it’s a simple substitution. Hence, the cation is Fe3+ (1) given (2) given (3) = (1)(2) (4) = (3)100/(3)total Oxide # of moles MolWt oxide # of grams of oxides wt. % Na2O 0.5 61.9796 30.9898 14.121 Fe2O3 0.3 159.692 47.9076 21.830 Al2O3 0.2 101.963 20.3926 9.247 SiO2 2 60.086 120.172 54.802 Total 219.462 100 Problem #3 • 1st step, calculate the formula 40% anorthite, 60% albite = 40%CaAl2Si2O3 + 60% NaAlSi3O8 = (Na.6Ca.4)(Al1.4Si2.6)o8 (1) given (2) given (3) = (1)(2) (4) = (3)100/(3)total Oxide # of moles MolWt oxide # of grams of oxides wt. % Na2O 0.6 61.9796 18.5936 7.29 CaO 0.4 55.96 8.9536 3.51 Al2O3 0.7 101.963 71.3741 27.97 SiO2 2.6 60.086 156.2236 61.23 Total 255.1452 100 L8: Keys ‐ Problems #2&3 Tuesday, August 25, 2020 3:27 PM Lecture 8 Page 1 Example 4: Solid solution: weight percent to formula We are given the following chemical analysis of a pyroxene. Compute its formula: (Hint: Jadeite NaAlSi2O6 – Diopside: CaMgSi2O6) (1) given (2) given (3) = (1)/(2) (4) Calculated (5) Calculated (6) = (4)6/(5)total Oxide Wt.% MolWt # moles oxide #moles cations #moles of #moles cations for 6 oxygens SiO2 56.64 60.086 0.9426 0.9426 1.8852 2 Al2O3 7.21 101.963 0.0707 0.1414 0.2121 0.30 Na2O 4.38 61.9796 0.0707 0.1414 0.0707 0.30 CaO 18.46 55.96 0.3299 0.3299 0.3299 0.70 MgO 13.30 40.312 0.3299 0.3299 0.3299 0.70 Total 100 2.8278 We have substitution between Na, Al, Ca and Mg. The formula is: (Na.3Ca.7)(Mg.7Al.3)Si2O6 L8: Example 4: Solid solution: wt.% to formula Lecture 8 Page 1
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https://www.scribd.com/document/848660924/Elasticity-Practice-Sheet-Answers
Elasticity Practice Sheet Answers | PDF | Elasticity (Economics) | Demand Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 451 views 1 page Elasticity Practice Sheet Answers The document contains practice problems related to elasticity in economics, including calculations for price elasticity of demand and supply, cross-price elasticity, and income elasticity. I… Full description Uploaded by junyuzhanp1 AI-enhanced description Go to previous items Go to next items Download Save Save Elasticity Practice Sheet Answers For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Elasticity Practice Sheet Answers For Later You are on page 1/ 1 Search Fullscreen Micro Topics 2.3, 2.4 and 2.5 Elasticity Practice Sheet Answer th e questio ns. Be su re to exp lain and/or show yo ur work. 1.The price of bu rgers inc rease d from $4.00 to $4.40 and the quanti ty deman ded decre ased from 50 to 30 units per hour. Calculate the coefficient for price elasticity of demand. Show your work. -4 = -40%/+10% = -.4/.1. Change in quantity is -40%. Change of price is +10%. 2.Given you r answer to que stion #1, is t he demand for bu rgers per fectl y elasti c, relat ively ela stic, unit elastic, relatively inelastic, or perfectly inelastic? Explain how you determined your answer. The demand for burgers is relatively elastic because the absolute value of the elasticity of demand coefficient is greater than 1. 3.Assum e, instea d, that the qua ntity of bu rgers dem anded rem ained at 50 (aft er the price ch ange in question #1). Would the demand for burgers be perfectly elastic, relatively elastic, unit elastic, relatively inelastic, or perfectly inelastic? Explain how you determined your answer. The demand would be perfectly inelastic because the quantity demanded stayed the same after the increase in price. The elasticity of demand coefficient would be 0.4.The pric e of burge rs incre ased fro m $4 to $4.4 0 and the qua ntity s upplie d increa sed from 5 0 to 60 units. Calculate the price elasticity of the supply coefficient. Show your work. +2 = +20%/+10%. Change in quantity is +20%. Change of price is +10%. 5.Given the info rmati on in question #4, is the suppl y of burgers perf ectly elas tic, rela tivel y elastic, unit elastic, relatively inelastic, or perfectly inelastic? Explain how you determined your answer. Relatively elastic. The coefficient of supply is greater than 1 and the percent change in quantity changed more than the percent change in price. 6.Suppos e that th e price of bu rgers de creas ed from $4.00 to $3.0 0 and the quan tity de manded o f pizza decreased by 50%. Calculate the cross-price elasticity of demand coefficient between burgers and pizza. Show your work. +2 = -50%/-25% = -.5/-.25 7.Given you r answer to que stion #6, ar e burgers and pi zza comp lemen ts, subs titut es, norm al goods, or inferior goods? Explain how you determined your answer. The two goods are substitutes since the cross-price elasticity of demand coefficient is positive. 8.If the dema nd for pizza is re lative ly inelas tic, wil l an increas e in the price of pizz a cause the tot al revenue for pizza restaurants to increase, decrease, or stay the same. Explain. Increase. Demand is inelastic, so quantity falls only a little and prices increases more. This is the total revenue test.9.Assum e that incom es incre ased by 40% and th e quantit y demande d of hot dogs decr eased by 40%. Calculate the income elasticity of hot dogs. Show your work. -1 = -40%/+40% 10.Given your answer to question #9, are hot dogs a complement, substitute, normal good, or inf erior good? Explain how you determined your answer. Inferior good since the INCOME coefficient is negative and the quantity purchased decreased when income increased.11.Given the information in question #9, is the demand for hot dogs be perfectly elastic, relativ ely elastic, unit elastic, relatively inelastic, or perfectly inelastic? Explain how you determined your answer. This is a trick question. The answer is impossible to determine with the given info. The change in income does not indicate the elasticity of demand (or supply). It can only be used to identify if the good is a normal good or inferior good. © Copyright Jacob Clifford 2020 Teachers- Do NOT use this in your classroom. Contact me if you want to use this resource with your students adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Micro Economics - Unit 1,2 - NEP - Notes No ratings yet Micro Economics - Unit 1,2 - NEP - Notes 31 pages Complete Bundle Principles of Microeconomics 8th Edition Mankiw 100% (1) Complete Bundle Principles of Microeconomics 8th Edition Mankiw 406 pages This Study Resource Was: BC3035: Microeconomic Theory Problem Set #4: Solutions 0% (1) This Study Resource Was: BC3035: Microeconomic Theory Problem Set #4: Solutions 6 pages AP Micro Unit 3 DAY 1 PROBLEM SET Elasticity of Demand No ratings yet AP Micro Unit 3 DAY 1 PROBLEM SET Elasticity of Demand 3 pages Demand, Supply and Elasticity Numericals No ratings yet Demand, Supply and Elasticity Numericals 2 pages UAI Study: Market Analysis & Strategy 100% (1) UAI Study: Market Analysis & Strategy 26 pages Home Work: Submitted To 100% (1) Home Work: Submitted To 10 pages Econ 101 - Lecture 4 - Trial Questions and Answers - Muhadeen No ratings yet Econ 101 - Lecture 4 - Trial Questions and Answers - Muhadeen 13 pages Iit Indore Cse Syllabus No ratings yet Iit Indore Cse Syllabus 199 pages ECON 2100 CH 4 Test Bank No ratings yet ECON 2100 CH 4 Test Bank 14 pages Elasticity Concepts in Microeconomics 100% (1) Elasticity Concepts in Microeconomics 4 pages Tutorial Elasticity No ratings yet Tutorial Elasticity 7 pages Nguyen Danh Thai No ratings yet Nguyen Danh Thai 69 pages Eco7 - LAUNCHING A NEW MOTOR OIL PDF 100% (1) Eco7 - LAUNCHING A NEW MOTOR OIL PDF 21 pages Chapters 5-8 Microeconomics Test Review Questions 63% (8) Chapters 5-8 Microeconomics Test Review Questions 157 pages Assignment Group 1 Microeco No ratings yet Assignment Group 1 Microeco 2 pages Chapter 4 Quiz No ratings yet Chapter 4 Quiz 5 pages Price Elasticity and Demand Analysis No ratings yet Price Elasticity and Demand Analysis 5 pages Economics Practice for Students No ratings yet Economics Practice for Students 25 pages BB 107 (Spring) Tutorial 5(s) No ratings yet BB 107 (Spring) Tutorial 5(s) 4 pages Practice Exam 1 No ratings yet Practice Exam 1 4 pages CH2 CH5 Ä Æ È Ç No ratings yet CH2 CH5 Ä Æ È Ç 23 pages Exercises Elasticity No ratings yet Exercises Elasticity 4 pages Set 3 No ratings yet Set 3 4 pages Managerial Use of Price Discrimination No ratings yet Managerial Use of Price Discrimination 32 pages Microeconomics Questions No ratings yet Microeconomics Questions 11 pages PS2 ECGE 1113 Economics I No ratings yet PS2 ECGE 1113 Economics I 2 pages Customer Engagement Best Practices Guide 75% (4) Customer Engagement Best Practices Guide 16 pages Q A 2 No ratings yet Q A 2 8 pages Microeconomics: Supply & Demand No ratings yet Microeconomics: Supply & Demand 13 pages Economics Problem Set Solutions No ratings yet Economics Problem Set Solutions 2 pages Problem Set On Microeconomics No ratings yet Problem Set On Microeconomics 3 pages Strategic Marketing Essentials No ratings yet Strategic Marketing Essentials 34 pages Handout Ch5 No ratings yet Handout Ch5 12 pages 480276924422947282 - CH5 練習題 eng 學生 No ratings yet 480276924422947282 - CH5 練習題 eng 學生 6 pages Economics for Business Students No ratings yet Economics for Business Students 12 pages Solution Chapter 5 No ratings yet Solution Chapter 5 5 pages Study Exercises - Elasticities 2015 No ratings yet Study Exercises - Elasticities 2015 4 pages Ryerson Finance 601 Test Bank No ratings yet Ryerson Finance 601 Test Bank 4 pages Problem Set 5 No ratings yet Problem Set 5 6 pages Microeconomics Elasticity Practice No ratings yet Microeconomics Elasticity Practice 7 pages Microeconomics - Seminar 5 No ratings yet Microeconomics - Seminar 5 16 pages Estonian Startup Database June 7 2020 No ratings yet Estonian Startup Database June 7 2020 71 pages Micro Tutorial 4 No ratings yet Micro Tutorial 4 3 pages Microeconomics Assignment Analysis No ratings yet Microeconomics Assignment Analysis 15 pages Elasticity Practice for Students No ratings yet Elasticity Practice for Students 3 pages Supply and Demand Forex Trading Strategy With Free PDF No ratings yet Supply and Demand Forex Trading Strategy With Free PDF 14 pages Practice 3 - Elasticity - HSB1010 No ratings yet Practice 3 - Elasticity - HSB1010 7 pages Price Elasticity Quiz No ratings yet Price Elasticity Quiz 5 pages Lerner. 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188223
https://media.coltene.com/DE/DE/media/DOC_IFU_30004756-02-23-IFU-President-PTO-ppspssfps_IND.pdf?sprache=EN
Published Time: Thu, 23 Mar 2023 14:23:47 GMT REV-02-2023 30004756 02.23 PRESIDENT putty / putty soft / putty super soft / fast putty soft The Original Coltène/Whaledent AG Feldwiesenstrasse 20 9450 Altstätten/Switzerland T +41 71 757 53 00 F +41 71 757 53 01 info.ch@coltene.com For SDS see www.coltene.com Glossary Consult instructions for use Keep away from sun light Temperature limitation Notified body registration number Identification for Russia Identification for Ukraine Legal Manufacturer Expiry Date Batch Code RX only Medical Device Manufacturing Date Unique Device Identifier European Authorized Representative Reference Number fast putty soft 2:00 min Reporting obligation All serious incidents occurring in conjunction with this product must be reported immediately to the manufacturer as well as to the competent authority. Instructions for use EN Carefully read the instructions prior to use. Product description PRESIDENT The Original is a two-component impression materi-al based on addition-curing polyvinylsiloxanes. This includes the correction materials (PRESIDENT The Original regular body, light body and Xtra light body) and the tray materials (PRESIDENT The Original putty, putty soft, fast putty soft, putty super soft, heavy body and MonoBody). After manual mixing of base and activator or mixing with the mixing tip, the PRESIDENT The Original materials form pastes which are used individually or in combination as dental impression material, usually together with a standard commercial or individual impression tray and using conventional impression techniques. Intended use PRESIDENT The Original is intended for the recording of the cur-rent physical situation in a patient‘s mouth („dental impression material“) for the purpose of repairing, reshaping or replacing the patient‘s teeth. Composition PRESIDENT The Original putty Addition-curing silicone elastomer, putty. - Polyvinylsiloxanes - Colour: yellow Composition PRESIDENT The Original putty soft Addition-curing silicone elastomer, putty. - Polyvinylsiloxanes - Colour: brown Composition PRESIDENT The Original fast putty soft Addition-curing silicone elastomer, putty. - Polyvinylsiloxanes - Colour: brown Composition PRESIDENT The Original putty super soft Addition-curing silicone elastomer, putty. - Polyvinylsiloxanes - Colour: orange Indications for Use • Impression material for pick-up impression in the double mix technique • Tray material for corrective impression technique Contraindications • Do not use in the case of known allergies to ingredients of the PRESIDENT The Original impression materials. • Loose teeth can be further loosened or extracted by taking an impression. Safety instructions • Impressions that come into contact with the oral mucosa may be contaminated. Observe disinfection instructions. • Only supplied to dentists and dental laboratories or upon their instructions. • Always wear non-latex gloves when working. Skin secretions, latex gloves and surfaces contaminated by latex gloves have a negative effect on the setting process of polyvinylsiloxanes. On-ly touch the material and the surfaces of which the impression is taken (teeth, preparations, retraction cords etc.) with vinyl / nitrile gloves. Compounds containing eugenol or haemostatic substances can also impede perfect setting. If hydrogen per-oxide is used for disinfection, it is recommended to thoroughly rinse with lukewarm water afterwards in order to avoid bubble formation. In order to guarantee good cohesion with the cor-rection material, the primary impression should be carefully cleaned and dried before further use. • High temperatures accelerate setting, low temperatures slow down the setting process. • Material combinations with impression materials of other man-ufacturers are not permitted. • Contains nano materials (bonded particles) Side effects / Interactions No harmful reactions or secondary effects on patients and / or den-tal personnel are known. User / patient group The use of PRESIDENT The Original impression materials in the pa-tient‘s mouth may only be undertaken by specialist personnel such as dentists and dental assistants with special qualifications. Suitable for all patient groups. PREPARATION Trays All impression trays that have been designed for dental use, either prefabricated or individual, can be used for impression taking. For perfect adhesion, we recommend applying or spraying a thin layer of COLTENE Adhesive or Adhesive AC to the trays before use. This prevents detachment of the impression material during removal from the oral cavity. Detailed information on the application can be found in the COLTENE Adhesive or Adhesive AC Instructions for Use. PROPER USE Dosage Measure out level scoopfulls of Base and Catalyst from the con-tainer. Processing times putty / putty soft / putty super soft ≤0:30 min ≤0:40 min ≥3:00 min ≤ 1 : 1 0 m i n ≥ 4 : 1 0 m i n 23 °C / 73 °F 35 °C / 95 °F fast putty soft ≤0:30 min ≤0:40 min ≥2:00 min ≤ 1 : 1 0 m i n ≥ 3 : 1 0 m i n 23 °C / 73 °F 35 °C / 95 °F Mixing Take the Base and Catalyst with the fingertips and start mixing. Knead the compound with the fingertips until the colour is uni-form, but for at least 30 seconds. Using a stopwatch is recommend-ed. If a mechanical mixing device is used, observe the manufac-turer‘s instructions. After successful mixing, form a strand with the material and immediately place it in the impression tray. Impression Insert the impression tray with the impression material into the oral cavity and press for 2-3 seconds and keep in situ until set completely. Test methods for correct application Material residue needs considerably more time to set at room tem-perature than in the patient‘s mouth. You should therefore check intraorally if the material is set before removing from the mouth. If the manual pressure test does not leave a pressure mark in the impression material after the time specified in the chapter on „Pro-cessing times“, the impression can be removed from the patient‘s mouth. However, if the test leaves a visible pressure mark in the material, the material is not yet fully set. CLEANING & DISINFECTION Cleaning of the impression After the initial impression is removed from the patient‘s mouth, it must be rinsed thoroughly before the correction materials are ap-plied. After cleaning, the final impressions can be disinfected with commercially available disinfectant solutions (e.g. Dürr Dental SE, MD 520 impression disinfection) specifically designed for dental impression materials and dried using compressed air. Cleaning of trays The impression can be removed with a blunt instrument. Adhesive residues can be dissolved by placing in commercially available uni-versal solvents or benzine. Manufacture of models The ideal time lies between 30 min and 72 h after taking the im-pression. The surface tension will be reduced and pouring will be facilitated if the impression is briefly washed out with a detergent and rinsed thoroughly in lukewarm clear water afterwards. All in-dustry-standard dental stone model materials (e.g. Fujirock Dental Stone, Hard Rock Dental Stone) can be used. Shelf life and storage • Expiry date: see primary packaging • Storage temperature: 15-23°C / 59-73°F • Relative humidity: 50% ± 10% • Shelf life after first opening: 3 months Protect against exposure to heat and sun. Avoid extreme tempera-ture fluctuations. Close containers immediately after use. Disposal PRESIDENT The Original impression materials can be disposed of in small quantities in household waste. PRESIDENT The Original impression material which comes into contact with the patient‘s mouth must be disposed of in accordance with country-specific regulations. Technical data according to ISO 4823 Measurements were taken at 23°C / 73°F room temperature and 50% relative humidity. Mixing time (15ml): 00:30 min Total working time: 1:10 min Oral setting time (35 °C / 95 °F): putty / putty soft / putty super soft 3:00 min
188224
https://www.onlinemathlearning.com/compare-ratio-tables.html
Comparing Ratios Using Ratio Tables (solutions, examples, worksheets, videos, lesson plans) OnlineMathLearning.com - Do Not Process My Personal Information If you wish to opt-out of the sale, sharing to third parties, or processing of your personal or sensitive information for targeted advertising by us, please use the below opt-out section to confirm your selection. Please note that after your opt-out request is processed you may continue seeing interest-based ads based on personal information utilized by us or personal information disclosed to third parties prior to your opt-out. You may separately opt-out of the further disclosure of your personal information by third parties on the IAB’s list of downstream participants. This information may also be disclosed by us to third parties on the IAB’s List of Downstream Participants that may further disclose it to other third parties. Personal Data Processing Opt Outs CONFIRM × Comparing Ratios Using Ratio Tables Related Topics: Lesson Plans and Worksheets for Grade 6 Lesson Plans and Worksheets for all Grades More Lessons for Grade 6 Common Core For Grade 6 Discover more Mathematics Math Calculus learning materials Geometry tool sets General Science Whiteboards markers erasers STEM toys kits Math workbooks Education Math homework help Share this page to Google Classroom Video Solutions to help Grade 6 students learn how to solve problems by comparing different ratios using two or more ratio tables. New York State Common Core Math Grade 6, Module 1, Lesson 11 Worksheets for Grade 6 Lesson 11 Student Outcomes • Students solve problems by comparing different ratios using two or more ratio tables. Lesson 11 Summary Ratio tables can be used to compare two ratios. Look for equal amounts in a row or column to compare the second amount associated with it. You can also extend the values of the tables in order to get comparable amounts. Another method would be to compare the values of the ratios. Write the values of the ratios as fractions and then use your knowledge of fractions to compare the ratios. When ratios are given in words, students can create a table of equivalent ratios in order to compare the ratios. Discover more Math Mathematics Purchase math workbooks General Science Buy SAT practice tests Worksheet Generators Calculus Courses Purchase biology study guide Maths Buy science project kits Discover more Mathematics Math Maths School supplies Educational posters charts General Science Video lesson library Math workbooks Reference books dictionaries Algebra help resource Lesson 11 Classwork Example 1 Create four equivalent ratios (2 by scaling up and 2 by scaling down) using the ratio 30 to 80. Write a ratio to describe the relationship shown in the table. Exercise 1 The following tables show how many words each person can text in a given amount of time. Compare the rates of texting for each person using the ratio table. Complete the table so that it shows Max has a texting rate of 55 words per minute. Exercise 2: Making Juice (Comparing Juice to Water) a. The tables below show the comparison of the amount of water to the amount of juice concentrate (JC) in grape juice made by three different people. Whose juice has the greatest water-to-juice concentrate ratio, and whose juice would taste strongest? Be sure to justify your answer. Put the juices in order from the juice containing the most water to the juice containing the least water. Explain how you used the values in the table to determine the order. What ratio was used to create each table? Explain how the ratio could help you compare the juices. b. The next day, each of the three people made juice again, but this time they were making apple juice. Whose juice has the greatest water-to-juice concentrate ratio, and whose juice would taste the strongest? Be sure to justify your answer. Put the juices in order from the strongest apple taste to the weakest apple taste. Explain how you used the values in the table to determine the order. What ratio was used to create each table? Explain how the ratio could help you compare the juices. How was this problem different than the grape juice questions? c. Max and Sheila are making orange juice. Max has mixed 15 cups of water with 4 cups of juice concentrate. Sheila has made her juice by mixing 8 cups water with 3 cups of juice concentrate. Compare the ratios of juice concentrate to water using ratio tables. State which beverage has a higher juice concentrate-to-water ratio. d. Victor is making recipes for smoothies. His first recipe calls for 2 cups of strawberries and 7 cups of other ingredients. His second recipe says that 3 cups of strawberries are combined with 9 cups of other ingredients. Which smoothie recipe has more strawberries compared to other ingredients? Use ratio tables to justify your answer. Show Step-by-step Solutions Discover more Math Mathematics Purchase college algebra textbook Purchase physics textbook Maths Buy English grammar book Geometry Software Rent math education software Online Courses Purchase Python programming course Discover more Mathematics Math Purchase calculus online course Math Practice Books Rent math education software Hire math tutor online Buy chemistry lab equipment Purchase physics textbook Learning Software Algebra Textbooks Problem Set Sarah and Eva were swimming. a. Use the ratio tables below to determine who the faster swimmer is. b. Explain the method that you used to determine your answer. 2. A 120 lb. person would weigh about 20 lb. on the moon. A 150 lb. person would weigh 28 lb. on Io, a moon of Jupiter. Use ratio tables to determine which moon would make a person weigh the most. Show Step-by-step Solutions Ratio Word Problem Jane and Bob are running. Use the ratio table below to find the faster runner. Show Step-by-step Solutions Try out our new and fun Fraction Concoction Game. Add and subtract fractions to make exciting fraction concoctions following a recipe. There are four levels of difficulty: Easy, medium, hard and insane. Practice the basics of fraction addition and subtraction or challenge yourself with the insane level. × We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. Discover more Math Mathematics Calculus learning materials Math practice quizzes Textbook rentals purchases Trigonometry study guide Educational software games STEM toys kits Geometry learning tools Whiteboards markers erasers × Back to Top | Interactive Zone | Home Copyright © 2005, 2025 - OnlineMathLearning.com. Embedded content, if any, are copyrights of their respective owners. × Discover more Mathematics Math College prep math Trigonometry study guide Textbook rentals purchases Math fun games Calculus learning materials Home Math By Grades Back Pre-K Kindergarten Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grades 7 & 8 Grades 9 & 10 Grades 11 & 12 Basic Algebra Intermediate Algebra High School Geometry Math By Topics Back Arithmetic Algebra Geometry Trigonometry Statistics Probability Word Problems Pre-Calculus Calculus Set Theory Matrices Vectors Math Curriculum Back NY Common Core Illustrative Math Common Core Standards Singapore Math Free Math Worksheets Back Worksheets by Topics Worksheets by Grades Printable & Online Common Core Worksheets Interactive Zone Math Tests Back SAT Math ACT Math GMAT GRE Math GED Math High School-Regents California Standards Math Fun & Games Back Math Trivia Math Games Fun Games Mousehunt Guide Math Video Lessons Back College Algebra College Calculus Linear Algebra Engineering Math Test/Exam Prep Back SAT Preparation ACT Preparation GMAT Preparation GRE Preparation GED Preparation ESL, IELTS, TOEFL GCSE/IGCSE/A Level Back KS3/CheckPoint 1 Science KS3/CheckPoint 2 Science KS3/CheckPoint 3 Science GCSE/IGCSE Maths GCSE/IGCSE Biology CGSE/IGCSE Chemistry CGSE/IGCSE Physics A-Level Maths Others Back Biology Chemistry Science Projects High School Biology High School Chemistry High School Physics Animal Facts English Help English For Kids Programming Calculators Privacy Policy Tutoring Services What's New Contact Us
188225
https://en.wikipedia.org/wiki/Enumerative_combinatorics
Jump to content Search Contents (Top) 1 Generating functions 1.1 Union 1.2 Pairs 1.3 Sequences 2 Combinatorial structures 2.1 Binary and plane trees 3 See also 4 References Enumerative combinatorics العربية Deutsch Español فارسی Magyar 日本語 Português Русский Українська Tiếng Việt 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Area of combinatorics that deals with the number of ways certain patterns can be formed Enumerative combinatorics is an area of combinatorics that deals with the number of ways that certain patterns can be formed. Two examples of this type of problem are counting combinations and counting permutations. More generally, given an infinite collection of finite sets Si indexed by the natural numbers, enumerative combinatorics seeks to describe a counting function which counts the number of objects in Sn for each n. Although counting the number of elements in a set is a rather broad mathematical problem, many of the problems that arise in applications have a relatively simple combinatorial description. The twelvefold way provides a unified framework for counting permutations, combinations and partitions. The simplest such functions are closed formulas, which can be expressed as a composition of elementary functions such as factorials, powers, and so on. For instance, as shown below, the number of different possible orderings of a deck of n cards is f(n) = n!. The problem of finding a closed formula is known as algebraic enumeration, and frequently involves deriving a recurrence relation or generating function and using this to arrive at the desired closed form. Often, a complicated closed formula yields little insight into the behavior of the counting function as the number of counted objects grows. In these cases, a simple asymptotic approximation may be preferable. A function is an asymptotic approximation to if as . In this case, we write Generating functions [edit] Generating functions are used to describe families of combinatorial objects. Let denote the family of objects and let F(x) be its generating function. Then where denotes the number of combinatorial objects of size n. The number of combinatorial objects of size n is therefore given by the coefficient of . Some common operation on families of combinatorial objects and its effect on the generating function will now be developed. The exponential generating function is also sometimes used. In this case it would have the form Once determined, the generating function yields the information given by the previous approaches. In addition, the various natural operations on generating functions such as addition, multiplication, differentiation, etc., have a combinatorial significance; this allows one to extend results from one combinatorial problem in order to solve others. Union [edit] Given two combinatorial families, and with generating functions F(x) and G(x) respectively, the disjoint union of the two families () has generating function F(x) + G(x). Pairs [edit] For two combinatorial families as above the Cartesian product (pair) of the two families () has generating function F(x)G(x). Sequences [edit] A (finite) sequence generalizes the idea of the pair as defined above. Sequences are arbitrary Cartesian products of a combinatorial object with itself. Formally: To put the above in words: An empty sequence or a sequence of one element or a sequence of two elements or a sequence of three elements, etc. The generating function would be: Combinatorial structures [edit] The above operations can now be used to enumerate common combinatorial objects including trees (binary and plane), Dyck paths and cycles. A combinatorial structure is composed of atoms. For example, with trees the atoms would be the nodes. The atoms which compose the object can either be labeled or unlabeled. Unlabeled atoms are indistinguishable from each other, while labelled atoms are distinct. Therefore, for a combinatorial object consisting of labeled atoms a new object can be formed by simply swapping two or more atoms. Binary and plane trees [edit] Binary and plane trees are examples of an unlabeled combinatorial structure. Trees consist of nodes linked by edges in such a way that there are no cycles. There is generally a node called the root, which has no parent node. In plane trees each node can have an arbitrary number of children. In binary trees, a special case of plane trees, each node can have either two or no children. Let denote the family of all plane trees. Then this family can be recursively defined as follows: In this case represents the family of objects consisting of one node. This has generating function x. Let P(x) denote the generating function . Putting the above description in words: A plane tree consists of a node to which is attached an arbitrary number of subtrees, each of which is also a plane tree. Using the operation on families of combinatorial structures developed earlier, this translates to a recursive generating function: After solving for P(x): An explicit formula for the number of plane trees of size n can now be determined by extracting the coefficient of xn: Note: The notation [xn] f(x) refers to the coefficient of xn in f(x). The series expansion of the square root is based on Newton's generalization of the binomial theorem. To get from the fourth to fifth line manipulations using the generalized binomial coefficient is needed. The expression on the last line is equal to the (n − 1)st Catalan number. Therefore, pn = cn−1. See also [edit] Algebraic combinatorics Asymptotic combinatorics Burnside's lemma Combinatorial explosion Combinatorial game theory Combinatorial principles Combinatorial species Inclusion–exclusion principle Method of distinguished element Pólya enumeration theorem Sieve theory References [edit] ^ (sequence A003763 in the OEIS) ^ (sequence A209077 in the OEIS) Zeilberger, Doron, Enumerative and Algebraic Combinatorics Björner, Anders and Stanley, Richard P., A Combinatorial Miscellany Graham, Ronald L., Grötschel M., and Lovász, László, eds. (1996). Handbook of Combinatorics, Volumes 1 and 2. Elsevier (North-Holland), Amsterdam, and MIT Press, Cambridge, Mass. ISBN 0-262-07169-X. Joseph, George Gheverghese (1994) . The Crest of the Peacock: Non-European Roots of Mathematics (2nd ed.). London: Penguin Books. ISBN 0-14-012529-9. Loehr, Nicholas A. (2011). Bijective Combinatorics. CRC Press. ISBN 143984884X, ISBN 978-1439848845. Stanley, Richard P. (1997, 1999). Enumerative Combinatorics, Volumes 1 and 2. Cambridge University Press. ISBN 0-521-55309-1, ISBN 0-521-56069-1. Goulden, Ian P. and Jackson, David M. (2004). Combinatorial Enumeration. Dover Publications. ISBN 0486435970. Riordan, John (1958). An Introduction to Combinatorial Analysis, Wiley & Sons, New York (republished). Riordan, John (1968). Combinatorial Identities, Wiley & Sons, New York (republished). Wilf, Herbert S. (1994). Generatingfunctionology (2nd ed.). Boston, MA: Academic Press. ISBN 0-12-751956-4. Zbl 0831.05001. Retrieved from " Category: Enumerative combinatorics Hidden categories: Articles with short description Short description is different from Wikidata Enumerative combinatorics Add topic
188226
https://extension.illinois.edu/container-gardens/container-drainage-options
Skip to main content Illinois Extension Container Gardens Home Containers Container Size Container Material Choices Container Drainage Options Vegetable Containers Designs Water Gardens Fish and Other Animals Planting the Water Garden Container Problems - Algae and Mosquitoes Pumps and Fountains Seasonal Maintenance Water Garden Containers Water Garden Plants Hanging Baskets Landscaping with Containers Two Plant Simplicity Plans A Full Sun Patio Pot Bird Bath of Foliage Colorful Foliage Exciting Succulents Exciting Texture for Semi-Shade Focus on Sumac Fragrance and Color in a Trough High Impact Using a Single Plant Type Japanese Maple with Vines Jazzed Up Fence with Baskets Large Patio Container Leaf Interest and Color Lush Pot of White Petunias Petunia Containers for the Landscape Purple Peach Combo Shocking Color Spring Bulbs and Violas in Troughs Subtle Foliage Color Planter Tall Elegance for a Semi-Shady Spot Tall and Short Textural Elegance Traditional Urn Upright Combo Using Grass for Interesting Baskets Variegated Vines for Interest Window Box of Color Yellow Combination of Common Annuals Plants Color Considerations Growing Herbs in Containers Growing Topiaries in Containers Growing Vegetables in Containers Growing Vines in Containers Soil Maintenance Disease and Insect Problems Fertilizing Overwintering Plants in Containers Watering Container Drainage Options A hole at the bottom of the container is critical. It allows water in the soil to drain freely so adequate air is available for the roots. While various kinds of plants have differing drainage needs, few can tolerate sitting in stagnate water. Healthy roots mean healthier plants. So be sure there are holes for drainage. Shoreline plants love wet soil, so if the pot does not drain, consider using them. See the water gardening section for more information. Wet soils favor root rot, because they leave little space for air to get to the roots. Plants rarely recover from root rot. If the pot does not come with a hole in it, figure out a way to make a hole. One way is to drill one. Some decorative resin or plastic pots have pre-punched holes at the bottom for easy removal. Generally, very little soil falls through the hole. To keep soil from falling through large drainage holes some folks use a coffee filter paper over the hole, though this is not necessary. Small holes in the bottom of the pot allow the water to drain out and very little soil media is lost. Sometimes a plant is already planted in a pot with no drainage at all. The best solution in this case is to take it to a sink, water it, then after a few minutes turn it on its side for a minute or two to let excess water drain out. Image Saucers Beware of pots with permanently attached saucers where emptying the overflow is difficult or impossible. I learned this the hard way when my basil leaves turned black because the roots were waterlogged - the result of a permanently attached saucer. Drainage of excess water is vital for the health of plant roots. It is much better to use a pot with a detachable saucer. Be sure to empty the excess water so soils have an opportunity to dry out somewhat. Title Double potting Body Slipping a container inside a slightly larger one is called double potting. When double potting is used, the plants grow in a pot liner. Often this is a plain plastic pot. This allows you to slip the pot liner in or out of the decorative container without disturbing the plant roots. Check to be sure that the plants in the pot liner never stand in water (unless you have aquatic plants) If water accumulates in the bottom of the larger container, remove the inside pot and drain the water from the outside pot. Place gravel in the bottom of the outer pot if the decorative pot is deep enough. A little excess water can accumulate in the gravel without the plant roots having to stand in water. Double potting techniques may be used: to overcome the problem of no drainage hole in an attractive decorative pot; to quickly change seasonal displays; to combine plants with different environmental requirements; to minimize extreme soil temperature fluctuations; to keep desirable but invasive plants from spreading. Title Skip the gravel inside the bottom of individual or pot liners Body It is a myth that a layer of gravel (inside the bottom of an individual pot) beneath the soil improves container drainage. Instead of extra water draining immediately into the gravel, the water "perches" or gathers in the soil just above the gravel. The water gathers until no air space is left. Once all the available soil air space fills up, then excess water drains into the gravel below. So gravel in the bottom does little to keep soil above it from being saturated by overwatering. Damp gravel placed in a saucer underneath the pot may help by increasing the humidity in the immediate area of the plants as the water evaporates from the gravel surfaces. Self-watering pots Self-watering pots use various methods to effectively draw water from a bottom reservoir into the soil without causing the soil to become too wet. Water may be drawn up into the soil by capillary action (or wicking) through small soil columns, rope wicks, or the use of moisture sensors. Self-watering containers are especially useful for weekend cottages and people who do not have time to check water needs daily. Consistently available water is great for vegetables, and tropical houseplants. Imagine a dozen stalks of sweet corn producing ears on your patio in a self -watering container like the Earth Box™. Plants that need to dry out like thick-leaved cactus and succulents do not usually warrant the extra cost of a self-watering container. For more information, use a web search engine using the keywords "self-watering planter" or "self-watering container" or "earth box." Decorative pots and wraps without drainage holes. Sometimes you fall in love with a wonderful container that has no drainage hole. This would be the perfect time to find a pot liner to fit inside so both you and the plants are happy. In addition to decorative pots, decorative foil or plastic pot wraps are a form of double potting. The wrap keeps water from leaking out where it is not wanted. To protect the plant from becoming prone to root rot, pierce a hole in the bottom of the wrapper or foil. Then place the container on a saucer. Or, take the container to a sink, remove the wrapper, and then water. Let the water drain freely out the holes in the bottom of the pot. After the pot finishes draining, replace the wrapper. Changing seasonal displays.Double potting makes changing out seasonal plants a breeze. Fresh plants are easily rotated in and tired ones out of a large landscape container holding multiple plants. Double potting makes it possible to sink individual potted plants into the landscape (or remove them) without disturbing the roots. Combining plants with differing requirements. Plants that have different soil drainage (aeration) requirements can be combined in a landscape planter if they are in their own separate pots. This allows different watering practices for each plant. On the other hand, if there is not enough light, two sets of plants can be used. As individual plants begin to decline, rotate them back to a higher-light nursery. Replace them with healthy plants. It is less work to choose plants with similar environmental needs or ones well adapted to conditions available. However, this is not always possible. Reducing fluctuation of soil temperature. Roots are generally more susceptible to cold damage than stems and leaves. When overwintering an otherwise hardy plant, sink the potted plant into the ground. Then mulch over the soil with wood chips, soil or other material to help it survive winter. The soil temperature underground does not fluctuate like container soil which is exposed to wind and extremes of temperature. Restraining invasive plants.Some plants are attractive, but spread invasively by underground roots. They invade space that does not belong to them. For example, most people love mints for their fragrance and culinary uses, but they can spread aggressively in the garden. Sink a large pot into the ground. Then place the invasive mint in a slightly smaller pot liner. This limits the spread of the roots. College of Agricultural, Consumer & Environmental Sciences Illinois Extension 101 Mumford Hall (MC-710) 1301 W. Gregory Dr. Urbana, IL 61801 Email: extension@illinois.edu EEO myExtension Login Select Language​▼ Original text Rate this translation Your feedback will be used to help improve Google Translate Cookie Notice We use Cookies on this site to enhance your experience and improve our marketing efforts. Click on “About Cookies” to learn more. 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https://www.math.cmu.edu/~bkell/magmas.pdf
Definitions! • A binary operation on a set S is an operation that takes two elements of S as input and produces one element of S as output. • A set S is closed under an operation if whenever the inputs to the operation come from S, the output of the operation is in S too. ◦[Note that in order for an operation to qualify as a binary operation on a set under the first definition above, the set must be closed under the operation, because the definition of binary operation requires that the output of the operation be in the set. If the set isn’t even closed under the operation, then the operation does not qualify as a binary operation on the set, and none of the following definitions apply.] • A binary operation ⋆on a set S is commutative if a ⋆b = b ⋆a for all elements a and b in S. • A binary operation ⋆on a set S is associative if (a ⋆b) ⋆c = a ⋆(b ⋆c) for all elements a, b, and c in S. • Let S be a set with a binary operation ⋆. An element e in S is an identity element (or just an identity) if e ⋆a = a and a ⋆e = a for every element a in S. • Let S be a set with a binary operation ⋆and an identity element e. ◦Let a be an element in S. If there exists an element b in S such that a ⋆b = e and b ⋆a = e, then the element a is invertible, and b is an inverse of a. ◦If every element a in S is invertible, then the binary operation ⋆itself is called invertible. • A binary operation ⋆on a set S is idempotent if a ⋆a = a for every element a in S. • A set with a binary operation is called a magma. [Note that the set must be closed under the operation—otherwise the operation wouldn’t qualify as a binary operation on the set!] • There are some special names for magmas that have additional properties. ◦Semigroup: associativity. ◦Monoid: associativity and identity. ◦Group: associativity, identity, and invertibility. ◦Abelian group: associativity, identity, invertibility, and commutativity. ◦Semilattice: associativity, commutativity, and idempotence. ◦Bounded semilattice: associativity, commutativity, idempotence, and identity. MAGMA associativity SEMIGROUP identity MONOID invertibility GROUP commutativity ABELIAN GROUP commutativity and idempotence SEMILATTICE identity BOUNDED SEMILATTICE 1. The set of positive integers under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 2. The set of nonnegative integers under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 3. The set of all integers under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 4. The set of positive integers under the operation of subtraction. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 5. The set of all integers under the operation of subtraction. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 6. The set of integers under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 7. The set of rational numbers under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 8. The set of integers under the operation of division. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 9. The set of nonzero rational numbers under the operation of division. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 10. The set of positive integers under the operation of integer division: division where the remainder is thrown away. For example, under integer division, 38 ÷ 5 = 7, because the remainder of 3 is thrown away. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 11. The set of positive integers under the operation of exponentiation. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 12. The set of real numbers under the operation of exponentiation. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 13. The set of rational numbers whose denominators are 1 or 2, under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 14. The set of rational numbers whose denominators are 1, 2, or 3, under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 15. The set of real numbers in the interval [0, 1], under the operation ⋋defined by a ⋋b = a+b 2 . closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 16. The set of all real numbers under the operation ⋌defined by a ⋌b = a+2b 3 . closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 17. The set of rational numbers under the operation ⋄defined by a ⋄b = ab a+b. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 18. The set of positive rational numbers under the operation ⋄defined above. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 19. The set of rational numbers under the operation ⊞defined by a ⊞b = ab + 1. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 20. The set of positive integers under the operation ⊛defined by a ⊛b = 2ab. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 21. The set of real numbers under the operation ∨defined by a ∨b = max{a, b}. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 22. The set of real numbers under the operation ∧defined by a ∧b = min{a, b}. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 23. The set of positive integers under the operation ▲defined by a ▲b = gcd(a, b). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 24. The set of positive integers under the operation ▼defined by a ▼b = lcm(a, b). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 25. The set {0, 1, 2, 3, 4} under the operation of addition modulo 5, written ⊕5. Addition modulo 5 is done by adding the two numbers together and then taking the remainder when the sum is divided by 5. For example, 2 ⊕5 4 = 1, because 2 + 4 = 6, and the remainder when 6 is divided by 5 is 1. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 26. The set {0, 1, 2, 3, 4} under the operation of multiplication modulo 5, written ⊗5. Multiplication mod-ulo 5 is done by multiplying the two numbers together and then taking the remainder when the product is divided by 5. For example, 3 ⊗5 4 = 2, because 3 × 4 = 12, and the remainder when 12 is divided by 5 is 2. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 27. The set {1, 2, 3, 4} under the operation of multiplication modulo 5. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 28. The set {1, 2, 3, 4, 5} under the operation of multiplication modulo 6, written ⊗6. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 29. The set of integers under the operation ◁defined by a ◁b = a. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 30. The set of integers under the operation ◀defined by a ◀b = −a. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 31. The set of real numbers under the operation ¶ defined by a ¶ b = the least integer that is greater than a + b. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 32. The set of states of the U.S., under the operation ♣defined by a ♣b = Kentucky. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 33. The set of all sets of integers, under the operation of union. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 34. The set of all sets of integers, under the operation of intersection. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 35. The set of positive integers under the operation ⋉defined by a ⋉b = the number you get by writing a down b times. For example, 1702 ⋉3 = 170217021702. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 36. The set of nonnegative integers under the operation ↷defined by a ↷b = the number you get by doing “move the first digit of a to the end” b times. For example, 12345 ↷1 = 23451, 67890 ↷2 = 89067, 203 ↷1 = 32 (why?), and 149283317 ↷57 = 283317149. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 37. The set of Boolean truth values (T and F, “true” and “false”) under the operation ∧defined by T∧T = T, T ∧F = F, F ∧T = F, and F ∧F = F. (This is the Boolean “AND” operation.) closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 38. The set of Boolean truth values under the operation ∨defined by T ∨T = T, T ∨F = T, F ∨T = T, and F ∨F = F. (This is the Boolean “OR” operation.) closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 39. The set of Boolean truth values under the operation ⊕defined by T ⊕T = F, T ⊕F = T, F ⊕T = T, and F ⊕F = F. (This is the Boolean “XOR” operation.) closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 40. Rock Paper Scissors: The set {r, p, s} under the operation • defined by r • p = p and p • r = p (“paper beats rock”), p • s = s and s • p = s (“scissors beat paper”), r • s = r and s • r = r (“rock beats scissors”), r • r = r (“rock ties with rock”), p • p = p (“paper ties with paper”), and s • s = s (“scissors tie with scissors”). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 41. The set of 2 × 2 matrices of real numbers, under the operation of matrix addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 42. The set of 2 × 2 matrices of real numbers, under the operation of matrix multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 43. The set of all finite strings formed from the letters A, B, C, . . . , Z, under the operation of string concatenation. (For example, the string “ABC” concatenated with the string “WXYZ” yields the string “ABCWXYZ”.) closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 44. The set of polynomial functions in x with integer coefficients, under the operation of function compo-sition, written with the symbol ◦, and defined as follows: if f and g are two functions, then f ◦g is the function defined by f g(x)  . For example, if f is the function defined by f(x) = 7x3 + 5x −12 and g is the function defined by g(x) = 4x −1, then f ◦g is the function defined by f g(x)  , which is 7(4x −1)3 + 5(4x −1) −12 = 448x3 −336x2 + 104x −24. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 45. The set of all ordered pairs (x1, x2) where x1 is an integer and x2 is a real number, under the operation ‡ defined by (a1, a2) ‡ (b1, b2) = (a1 + b1, a2 · b2). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 46. The set of permutations of a set of five elements under the operation of composition. Such a permutation can be written as five numbers, like “35142,” in which every number from 1 to 5 appears exactly once. The permutation 35142 means that the first element in the output is the third element in the input (that’s the 3), and the second element in the output is the fifth element in the input (that’s the 5), and the third element in the output is the first element in the input (that’s the 1), and so on. For example, applying the permutation 35142 to the input ABCDE gives the output CEADB, and applying the same permutation 35142 to the input BDACE gives the output AEBCD. [Note that a permutation is an action—it’s a “verb,” not a “noun.”] The operation of composition means doing one permutation and then another. Composition is written with the symbol ◦, and it is done from right to left; so 43125◦35142 means, “First do the permutation 35142, and then do the permutation 43125 on that output.” Applying the composition of permutations 43125 ◦35142 to the input ABCDE gives the output DACEB, which is the same output as if you had applied the single permutation 41352; so 43125 ◦35142 = 41352. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice More problems! 47. The set of even integers under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 48. The set of odd integers under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 49. The set of even integers under the operation of subtraction. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 50. The set of odd integers under the operation of subtraction. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 51. The set of even integers under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 52. The set of odd integers under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 53. The set of positive even integers under the operation of exponentiation. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 54. The set of positive odd integers under the operation of exponentiation. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 55. The set of all rational numbers whose denominators are powers of 2, under the operation of addition. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 56. The set of all nonzero rational numbers whose denominators are powers of 2, under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 57. The set of real numbers under the operation ⊙defined by a ⊙b = 7ab. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 58. The set {1, 5, 7, 11} under the operation ⊗12, multiplication modulo 12. (Remember, multiplication modulo 12 is done by multiplying the two numbers together and then taking the remainder when the product is divided by 12.) closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 59. The set {0} under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 60. The set of all real numbers under the operation defined by a b = √ a2 + b2. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 61. The set of all nonnegative real numbers under the operation defined above. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 62. The set of all real numbers under the operation 3 defined by a 3 b = 3 √ a3 + b3. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 63. The set Z ∪{∞} (that is, the set of integers together with ∞) under the operation ∨defined by a ∨b = max{a, b}. Note that max{a, ∞} = ∞for all a in this set, and max{∞, b} = ∞for all b in this set. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 64. The set Z ∪{−∞, ∞} (that is, the set of integers together with −∞and ∞) under the operation ⊕ defined by a ⊕b =      0, if a = ∞and b = −∞, or a = −∞and b = ∞; ∞, if a = ∞and b ̸= −∞, or b = ∞and a ̸= −∞; −∞, if a = −∞and b ̸= ∞, or b = −∞and a ̸= ∞; a + b, if a ̸= ±∞and b ̸= ±∞. For example, 2 ⊕5 = 7, 3 ⊕∞= ∞, 7412 ⊕−∞= −∞, −∞⊕−∞= −∞, ∞⊕∞= ∞, and −∞⊕∞= 0. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 65. The set of positive integers under the operation § defined by a § b = ab + ba. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 66. The set of all nonnegative integers that can be expressed as the sum of two perfect squares (i.e., the set {0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, . . .}, because, for example, 0 = 02+02, 5 = 12+22, 20 = 22+42, and 25 = 02 + 52 = 32 + 42), under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 67. The set of nonnegative real numbers under the operation defined by a b = πa2b. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 68. The set of all positive integers under the operation of tetration, written ↑↑and defined by a ↑↑b = aa. . .a | {z } b copies of a . Note that this “power tower” is evaluated top-down: for example, 7 ↑↑3 = 777 = 7(77), not (77) 7. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 69. The set of perfect squares, {0, 1, 4, 9, 16, 25, . . .}, under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 70. The set of Fibonacci numbers under the operation ▼defined by a ▼b = lcm(a, b). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 71. The set of Fibonacci numbers under the operation ▲defined by a ▲b = gcd(a, b). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 72. The set of real numbers of the form a + b √ 5 where a and b are integers, under the operation of multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 73. The set of integers under the operation ?: defined by a ?: b =  a, if a ̸= 0; b, if a = 0. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 74. The set {a, b, c} under the operation ∗defined by the table below. ∗ a b c a a b c b b c a c c a b closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 75. The set {0, 1 2, 1} under the operation →defined by the table below. → 0 1 2 1 0 1 1 1 1 2 0 1 1 1 0 1 2 1 closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 76. The set {x, y, z} under the operation ↔defined by the table below. ↔ x y z x x z y y z y x z y x z closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 77. The set {α, β, γ, δ} under the operation ⊚defined by the table below. ⊚ α β γ δ α δ γ β α β β β β β γ γ γ γ γ δ α β γ δ closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 78. The subset {1, −1, i, −i} of the complex numbers under the operation × defined by the table below. × 1 −1 i −i 1 1 −1 i −i −1 −1 1 −i i i i −i −1 1 −i −i i 1 −1 closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 79. The set of complex numbers under the operation of complex addition, defined by (a + bi) + (c + di) = (a + c) + (b + d)i. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 80. The set of complex numbers under the operation of complex multiplication, defined by (a+bi)·(c+di) = (ac −bd) + (bc + ad)i. [Hint: There is another operation on complex numbers, a unary operation, that takes the complex number a+bi as input and produces the complex number a a2+b2 + −b a2+b2  i as output.] closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 81. The set of nonzero complex numbers under the operation of complex multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 82. The set of complex numbers a+bi such that a2 +b2 = 1, under the operation of complex multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 83. The set of all points inside the unit circle [that is, the set of all points (x, y) whose distance from the origin (0, 0) is less than 1] under the operation • defined by a • b = the midpoint of the line segment joining a and b. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 84. The set of all points outside the unit circle under the operation • defined above. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 85. The set of all nonempty, closed, bounded intervals on the real number line (i.e., intervals of the form [a, b] with a ≤b), under the operation ⊔defined by [a, b] ⊔[c, d] = [min{a, c}, max{b, d}]. For example, [−2.8, 1] ⊔[π, 7] = [−2.8, 7]. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 86. The set of polynomial functions in x with integer coefficients, whose coefficients add up to 0, under the operation of multiplication. For example, f = 3x5 −4x2 + 1 is in this set, because 3 −4 + 1 = 0, and g = −17x3 + 20x2 −5x + 2 is in this set, because −17 + 20 −5 + 2 = 0; and f · g = (3x5 −4x2 + 1) × (−17x3 + 20x2 −5x + 2) = −51x8 + 60x7 −15x6 + 74x5 −80x4 + 3x3 + 12x2 −5x + 2. [Hint: Think about evaluating one of these polynomials at x = 1.] closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 87. The set of all sets of integers under the operation of set difference, written \ and defined by A\B = the set of all elements of A that are not elements of B. For example, {−5, −2, 3, 17, 21} \ {−2, 0, 14, 17} = {−5, 3, 21}. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 88. The set of all sets of integers under the operation of symmetric difference, written △and defined by A △B = (A \ B) ∪(B \ A), where \ is the operation of set difference defined above. In other words, A △B is the set of all elements of A that are not elements of B, together with all elements of B that are not elements of A. Equivalently, A △B = (A ∪B) \ (A ∩B). For example, {−5, −2, 3, 17, 21} △ {−2, 0, 14, 17} = {−5, 0, 3, 14, 21}. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 89. The set of all sets of integers under the operation of Minkowski addition, written + and defined by A + B = the set of all numbers that you can get by adding one number in A and one number in B. For example, {−8, 1, 3} + {0, 2, 7} = {−8, −6, −1, 1, 3, 5, 8, 10}. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 90. The set of all finite strings formed from the letters A, B, C, . . . , Z, under the operation ⊣defined by a ⊣b = the longest string of letters that appears at the beginning of both a and b. For example, CATFISH ⊣CATAMARAN = CAT and FARMHOUSE ⊣FIREHOUSE = F. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 91. The set of 2 × 2 matrices a c b d such that ad −bc = 1, under the operation of matrix multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 92. The set of 2 × 2 matrices a c b d such that a + b = 1 and c + d = 1, under the operation of matrix multiplication. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 93. The set of all 8-digit strings of digits 0, 1, 2, . . . , 9 (with leading zeroes allowed) under the operation + defined just like ordinary addition, except that leading zeroes are kept in the sum, and if the sum would be a 9-digit number then only the last 8 digits are kept. For example, 02814019 + 03152944 = 05966963 and 51043819 + 72010038 = 23053857. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 94. The set of all strings, made of the digits 0, 1, 2, . . . , 9, that are one-way infinite to the left, under the operation + defined just like ordinary addition, starting at the rightmost digit and proceeding leftwards, with carries. Of course, this process of adding digits (and carrying digits to the left) will require infinitely many steps. For example, . . . 31439 + . . . 52486 = . . . 83925. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 95. The set of ordered pairs of integers under the operation ◁▷defined by (x1, y1) ◁▷(x2, y2) = (x1, y2). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 96. The set of ordered pairs of integers under the operation ▷ ◁defined by (x1, y1) ▷ ◁(x2, y2) = (y1, x2). closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 97. The set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} under the operation ⟳defined by a ⟳b = the number you get to by starting at a in the picture below and following the arrows for b steps. 1 2 3 4 5 6 7 8 9 10 closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 98. The set of nonnegative integers under the operation ∔defined by a ∔b = the number that you get by writing a and b in binary and performing binary addition without carries, and then converting back to base ten. For example, 12 in binary is 1100, and 42 in binary is 101010; adding those numbers in binary without carries, we get 1100 + 101010 100110 and 100110 in binary is 38; so 12 ∔42 = 38. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 99. The set of positive integers under the operation ! ⃝defined by the following process: Let afirst be the first digit of a, and let alast be the last digit of a. Let bfirst be the first digit of b, and let blast be the last digit of b. Let a′ be the number that you get from a by replacing every occurrence of the digit blast with the digit bfirst, and let b′ be the number that you get from b by replacing every occurrence of the digit alast with the digit afirst. Then a ! ⃝b = a′ + b′. For example, 1234 ! ⃝812443 = 1284 + 812113 = 813397. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice 100. Quaternions: The set {1, −1, i, −i, j, −j, k, −k} under the operation × defined by the table below. × 1 −1 i −i j −j k −k 1 1 −1 i −i j −j k −k −1 −1 1 −i i −j j −k k i i −i −1 1 k −k −j j −i −i i 1 −1 −k k j −j j j −j −k k −1 1 i −i −j −j j k −k 1 −1 −i i k k −k j −j −i i −1 1 −k −k k −j j i −i 1 −1 For example: i × i = −1, j × j = −1, k × k = −1, i × j = k, j × k = i, k × i = j, j × i = −k, k × j = −i, i × k = −j. closed commutative associative identity: invertible idempotent magma semigroup monoid group abelian group semilattice bounded semilattice
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https://scholar.harvard.edu/files/schwartz/files/2-diffusion_0.pdf
Matthew Schwartz Statistical Mechanics, Spring 2021 Lecture 2: Diffusion 1 Introduction If you we put a drop of red dye in water, it will slowly diffuse throughout the water. Why does this happen? How fast does it happen? What is going on microscopically? The microscopic mechanism of diffusion is very simple: the dye molecules start densely concen-trated near one point. Then they get bumped by neighboring molecules until they are spread out all over. To model this process, we can suppose that the dye molecule moves a distance between collisions and after each collision its direction is completely randomized. This approximation is called a random walk. Although the distance between collisions has some variation and the direction of scattering is somewhat correlated with the initial direction, because molecules collide billions of times per second, the law of large numbers applies to their net displacement and random walks provide an excellent approximation to real diffusion. Random walks are actually quite common. They can be used to model any stochastic process. For another example, say you're playing blackjack with a friend. You are both expert players and evenly matched. Sometimes you win, sometimes she wins. Each time you play, you bet 1 dollar. This is a 1-D random walk. Say you play N games. Although we can't say who will be winning after N games, we can predict how by much they would be winning. The 2-dimensional random walk is sometimes called the drunkard's walk. The idea is that a drunkard leaves a party late at night, takes a step in one direction, then gets totally disoriented and takes a step in another direction. How far will she get after N steps? 2 1D random walk Let's work out the blackjack problem. We'll make it a little more interesting. Say you have a probability a of winning and your opponent has a probability b = 1 −a of winning. If you play N times, the chance of you winning m of them is BN(m) = ambN −m  N m  (1) This is known as the binomial distribution. The factor  N m  =N Cm = N! m!(N −m)! (2) is known as the binomial coefficient. It is the number of ways of picking m of the games for you to have won out of the N total games.1 Binomial coefficients come up in the expansion of powers of sums. Namely (a + b)N = X m=0 N ambN −m  N m  (3) So the binomial distribution is simply the mth term in this sum. In fact, this relationship makes it easy to see that the probabilities sum to 1: X m=0 N BN(m) = X m=0 N ambN −m  N m  = (a + b)N = 1 (4) 1. To see this, first note that if we want to sort N objects, N of them can go first, then N −1 second, and so on, so there are N! permutations. If we are instead splitting into a set of size m and a set of size N −m, we don't care about the order within each set. So we have to divide by the permutations of each set, giving N! m!(N −m)!. 1 since b = 1 −a. How much can you expect to be winning after N games? This is determined by the number of games you win, namely the expected value of m: hmi = X m=0 N mBN(m) = X m=0 N mambN −m  N m  (5) Although b = 1 −a we can compute the sum on the right most easily if we allow a and b to be unrelated. Then, we note that this sum is the same as the sum in Eq. (4) if we differentiate with respect to a, then multiply by a. So we have X m=0 N mambN −m  N m  = a@a " X m=0 N ambN −m  N m # = a@a(a + b)N = Na(a + b)N −1 (6) Now that we can compute the sum for any a and b we can take the case of interest where b =1−a and get hmi = Na (7) Similarly, the standard deviation is σm = X m=0 N (m2 −hmi2)BN(m) v u u u u t = Nab p (8) As N ! 1 the binomial distribution approaches a Gaussian, by the central limit theorem. Thus knowing the mean and standard deviation, we know the whole answer: BN(m) ! 1 2πNab p exp  −(m −Na)2 2N ab  (9) We can derive this by studying the falloffof ln BN(m) at large N, but using the central limit theorem is easier. Try checking the agreement of Eqs. (1) and (9) for some numerical values. For the blackjack game, hmi = Na is the expected number of times you win. The expected number of times you lose is hmi =N. If you gain $1 when you win, or lose $1 when you lose, then hwinningsi = N(a −b) × $1 (10) and the standard deviation is (try it to test your understanding): σwinnings = 2 Nab p × $1 (11) For a fair match a=b= 1 2 and so the expected winnings are hwinningsi=0 with standard deviation σ = N p (12) That σ grows as N p is exactly what we expect for the sum of random values by the central limit theorem. The a = b = 1 2 case is sometimes called an unbiased 1D random walk. For example, if you play 100 games for $1 each and are evenly matched, then hwinningsi = 0 and σwinnings = 100 p × $1=$10. This means that after 100 games, we don't know who's winning but there is a 32% chance someone is up by at least $10. For an unbiased 1D random walk, the mean displacement is 0. In this case, the typical scale for displacement is better described by the RMS fluctuation, which reduces to the standard deviation when the mean is zero. That is, the RMS fluctuation is xrms=σ= N p . Typically, RMS fluctuations are used for quantities that average to zero (as in a unbiased random walk), but there is no hard and fast rule about when to use the mean displacement and when to use the RMS fluctuation. Just like there are many measures of uncertainty (σ, fraction within 50% of mean, width at half maximum, etc), there are many measures of discplacement (x ¯, xrms, etc.). Typically these are all similar (when nonzer) and determined by dimensional analysis, so you can just substitute one for another, often depending on which leads to the simplest looking final expression. 2 Section 2 Let us compare the binomial distribution to the Poisson distribution. The binomial distribution BN(m) is defined for discrete N and m, in contrast to the Poisson distribution Pm(t) = (λt)m m! e−λt which has discrete m but continuous t. For a binomial distribution, the smallest interval is one discrete step, with probability of occurrence a. For Poisson, we can take an arbitrarily small timestep ∆t with probability dP =λ∆t. If we identify the interval for the binomial distribution with that of the Poisson distribution by setting a = λ∆t = λ t N, and then take the limit N ! 1 we find lim N!1BN(m) = lim N!1  N m  | | | | | | | | | | | | | | | | | | | | | | | | | | | ||{z} } } } } } } } } } } } } } } } } } } } } } } } } } } } } pick mto decay  1 −λ t N N −m | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |{z} } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } N −m do not decay  λ t N m | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |{z} } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } m dceays = (λt)m m! e−λt (13) which formally recovers the Poisson distribution from the binomial distribution. Note that this requires taking N ! 1 holding λt = Na fixed, and therefore a ! 0 and b ! 1. In other words, a Poisson process is like a random walk that always goes in one direction but you don't know when the step will be taken. Although they can be related, as we have seen, you should really think of binomial and Poisson distributions as being relevant in different contexts: binomial is used when the steps are discrete and incoherent (random directions) and Poisson is used when time is continuous but the steps are coherent (counts always increase). If we are flipping coins, then a = 1 2 is fixed, and so the Poisson distribution is not relevant since it needs a = 0. For a decay process, a decay can happen at any time t and so the binomial distribution is not appropriate. After a given time t, a Poisson process can have potentially an infinite number of events. With a binomial process the time is the number of steps N, so the number of possible events is always bounded. 2.1 Random walks in 2D and 3D For the 2D case, a popular picture of the random walk is a drunkard stumbling around. In each time step she moves a distance L in some random direction. In 3D you can imagine a dye molecule diffusing in water and in each time step it bumps into something, and then gets buffeted into a different direction. For simplicity, we'll assume in the 2D and 3D cases that the distance is the same each step and the angle totally random. Where will the drunkard or molecule be after N steps? Let us say that in the jth step she moves by a displacement ~j. The vectors ~j all have length . The dot product of two vectors is ~j · ~k =2cosθjk (14) where θjk is the angle between the two steps. Since we are assuming the angle is random, then the expectation value of this dot product is zero: h~j · ~ki = 21 π Z 0 π dθ cosθ = 0 (15) Now let s ~N be the total displacement from the origin after N timesteps. s ~N = X j=1 N ~j (16) Then hs ~N 2 i = h(s ~N −1 + ~N)2i = hs ~N −1 2 i + 2hs ~N −1 · ~Ni + h~N 2 i (17) Now, the angle between ~N and s ~N −1 is totally random, so hs ~N −1·~Ni=0, and h ~N 2 i=2. So we find hs ~N 2 i = hs ~N −1 2 i +2 (18) Since this is true for any N we can compute that hs ~N 2 i = hs ~N −1 2 i + 2 = hs ~N −2 2 i + 22 = ··· = N2 (19) and the RMS distance moved is N p, just like in the 1D case. 1D random walk 3 3 Diffusion from random walks Diffusion refers to the net spreading of the distribution of molecules due to random molecular motion. Think about an individual molecule in a gas, say some CO molecule coming out of a car's exhaust. It leaves the exhaust and moves in a straight line until it hits another molecule, in which case it is buffeted essentially randomly in a different direction. As all the CO molecules are doing the same thing, on average, the net effect is a diffusion of the CO gas. We want to compute the probability distribution Pt(x) for where a CO molecule is after a time t and then use this to determine the equation of motion of the density of the gas. 3.1 Collisions in a gas It's helpful to discuss random walks for gases in terms of a set of convenient physical quantities. An important one is • τ = the collision time is the average time a molecule goes before colliding with another molecule The number of collisions in a time t is then N = t τ (20) A related quantity is • = the mean free path is the average distance a molecule goes between collisions The mean free path is related to the collision time by = v ¯τ (21) where • v ¯ = the average molecular speed, v ¯= hjv ~ji. Sometimes a more useful quantity is the root-mean-square velocity vrms = hv ~2i p . We can also use the speed of sound cs in a gas, which is of course limited by the speed by which the molecules move. All three of these, v ¯; vrms and cs are related by coefficients of order one, as we will see once we understand gases in more detail in future lectures. For example, in air at room temperature, v ¯= 467m s , vrms = 507m s and cs = 346m s . The mean free path is related to the density and size of the molecules. Treating molecules as spheres of radius R, two molecules will hit if their centers are within 2R of each other. Thus you can think of a moving molecule as having an effective cross sectional area of σ =π(2R)2. This effective cross sectional area is also called the collisional cross section. After N collisions a molecule will have swept out a volume V =Nσ. The number of molecules it hits during this sweeping is N =Vn with • n = the number density = number of molecules per unit volume We will use number density a lot in statistical mechanics. It is interchangeable with the • ρ = the mass density. as ρ=mn where m is the mass of a molecule (or the average mass of a molecule if the gas is mixed). Thus, = 1 nσ (22) Bigger molecules have bigger cross sectional areas so they will have smaller mean free paths. Since liquids are more dense than gases, generally they will have smaller mean free paths. For example, the radius of a typical atom is around the Bohr radius a0 = 0.05 nm. So an air molecule, such as N2 or O2, has a radius of around R≈2a0∼0.1nm. Thus σ ≈π(2R)2=0.14nm2 in air. Air has a density of ρ=1.3 kg m3 and an average mass of m=4.81×10−26 kg molecule, so its number density is n = ρ m = 2.6 × 1025 1 m3. Note that n−1/3 = 3.3nm so air molecules are around 3nm apart on average. The mean free path is = 1 nσ = 0.26µm. The collision time is then τ = v ¯ = 0.57ns. 4 Section 3 These are useful numbers to have in your head: in air at room temperature, molecules have velocities around v∼500 m s , are around R∼0.1nm big and n−1/3∼1nm apart. They collide around once every nanosecond (one billion times per second) after having moved around ∼100nm (one thousand molecule lengths). 3.2 Diffusion from random walks Let's now consider the probability distribution Pt(x) for where a CO molecule is after a time t. We'll start in one dimension. Treating molecular interactions as a random walk, we take a = b = 1 2 since the molecule should be equally likely to be knocked left as right. Such a random walk is unbiased. For an unbiased random walk, the mean displacement is x ¯ = 0 and therefore does not tell us much about how fast the molecules are diffusing. Instead, the RMS displacement is more useful. The RMS displacement after a time t is, from Eq. (12) xrms = N p = t τ r =v ¯t p (23) This x ∼ t p behavior is the key characteristic of a random walk. Note that this is going to be a much smaller distance than an unhindered molecule would move on average, ∆x∼vt. Knowing the mean (x ¯ = 0) and the standard deviation (σ = v ¯t p ) we can immediately write down the full probability distribution for large times (t ≫τ) using the central limit theorem: Pt(x) = 1 2πv ¯t r exp  −x2 2tv ¯  (24) Note that probability distribution satisfies the differential equation @Pt(x) @t = D@2Pt(x) @x2 (25) where D = 1 2v ¯. This is the 1D diffusion equation. You can easily check by plugging Eq. (24) into Eq. (25). We defined Pt(x) as a probability distribution for one particular CO molecule in a gas. But the same probability distribution holds for any molecule. Since there are usually an enormous number N ∼1024 of gas molecules, if each one has a probability Pt(x) of being at the the point x then number density will be simply n(x; t) = NPt(x) (26) To be precise, the number density is not exactly the same as the probability distribution since, classically, a particle is either at a particular position or not. So we should think of n(x;t)=NPt(x) as referring to the number density averaged over time. (We'll return to this averaging in the next lecture, in the context of ergodicity.) Thus we find @n(x; t) @t = D@2n(x; t) @x2 (27) In 2 or 3 dimensions, the resulting equation is the rotationally symmetric version of this: @n(x ~ ; t) @t = Dr ~ 2n(x ~ ; t) (28) This is known as the diffusion equation. It describes how substances move due to random motion. The coefficient is D = 1 2v ¯= 1 22 τ (29) This coefficient D is called the diffusion constant and the relation D = 1 2 2 τ is known as the Einstein-Smoluchowski equation. Diffusion from random walks 5 For example, in air≈0.26µm and τ ≈0.57ns so D =5.9×10−5m2/s. This means that it takes 1 day for an individual gas molecule to diffuse 1 meter. Then it takes 100 days for it to diffuse 10 meters. These numbers are characteristic of diffusion processes: diffusion in air over macroscopic distances is generally very slow, and almost always dominated by convection (see Section 4.1) and other forms of energy transport. On the other hand, on smaller length scales where convention is irrelevant like a cell, diffusion can be dominant. For example, the diffusion constant for proteins in water is around D ≈10−11m2 s . This is tiny compared to D for gases, but the typical distances proteins diffuse are also tiny, and time scales like distance squared. To diffuse across a cell of size 10−4m it takes 15 minutes. To diffuse across a cell nucleus of size 6µm it takes only 3.6 seconds. Just because the diffusion equation looks simple does not mean it has trivial consequences! For example, it is mathematically identical to the Schrödinger equation, which accounts for a great variety of interesting physics. It also also mathematically identical to the heat equation: heat conduction is a diffusive process. The diffusion equation is linear, so that if n1(x ~ ; t) and n2(x ~ ; t) are solutions, then so is their sum. In particular if we start with a bunch of particles at some positions xi, then they will diffuse independently of each other. This gives us a way to solve the diffusion equation in general. For one particle, starting at x ~ =0, the solution is given by Eq. (24). Note that at t =0, this solution really does represent a localized source. In fact, the limit as t ! 0 of this solution is one of the possible definitions of a δ-function: lim t!0  1 4πDt 3/2 exp  −x ~ 2 4Dt  = δ3(x ~) (30) where δ3(x ~) = δ(x)δ(y)δ(z). Thus Eq. (24) is a solution to the diffusion equation with boundary condition n(x ~ ; t) = δ3(x ~) at time t = 0: it describes the diffusion away from a point source. Since any function can be described as a set of points, we can construct any solution to the diffusion equation by combining the solutions as in Eq. (24). More precisely, if n0(x ~ ;0) is the number density at time 0, then the solution for all times is n(x ~ ; t) = Z d3y  1 2πv ¯t 3/2 exp  −(x ~ −y ~)2 2tv ¯  n0(y ~; 0) (31) To check this, we note that the right-hand side satisfies the diffusion equation and Eq. (30) verifies the boundary condition at t=0. Solving differential equations in this way is known as the Green's function method.2 It converts solving a difficult differential equation to doing an integral. Eq. (31) has a simple physical interpretation: the number of molecules at a point x ~ are those that have walked there randomly from any other point y ~ over the time t. 4 Fick's laws of diffusion The approach to diffusion we discussed was based on a microscopic picture of random walks of individual molecules. We can also approach diffusion from the continuum perspective. Let us continue to denote the number density by n(x ~ ; t) and let us also denote the local velocity as the vector field v ~(x ~ ; t). It can be helpful to think of n(x ~ ; t) as the density of a fluid, like water in a stream, and v ~(x; t) as its velocity at the point x ~ at time t. For simplicity, lets assume that n and v ~ are constant in the y and z directions, so they only depend on x, n = n(x; t), v ~ = (vx(x; t); 0; 0). Now, the total number of molecules between x1 and x2 can only change if particles flow in or out of that region. So @ @t Z x1 x2 dxn(x; t) = n(x1; t)vx(x1; t) | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |{z} } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } number coming in past x1 per unit time − n(x2; t)vx(x2; t) | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |{z} } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } } number going out past x2 per unit time (32) =− Z x1 x2 dx@x[n(x; t)vx(x; t)] (33) 2. In general, a Green's function satisfies OG(x; t) = δ(x)δ(t) for some differential operator O. In our case, the Green's function is G(x ~ ; t) = P (x ~ ; t)θ(t) = 1 4πDt q exp h −x ~ 2 4Dt i θ(t) which satisfies [@t −Dr ~ 2]G(x ~ ; t) = δ3(x ~)δ(t). 6 Section 4 Pulling the d dt on the left into the integral, and using that x1 and x2 are arbitrary, we get @ @tn(x; t) = −@ @xJx(x; t) (34) where Jx(x; t) = n(x; t)vx(x; t) (35) The 3D version of this equation is called the continuity equation @n @t + r ~ · J ~ = 0 (36) and J ~(x ~ ;t) is called the flux. The flux is the number density times velocity J ~(x;t)=n(x;t)×v ~(x;t). It gives the number of particles passing by a given point per unit area per unit time. Note that the velocity field and the density are in principle independent, like position and velocity are independent in classical mechanics. How does the flux related to the density during diffusion? Well, if the density is constant in position, then the net diffusion should be zero and the flux should vanish. Conversely, if the density has some spacial gradient, then there should be net flux from high density to low density. Thus, the leading order thing we could imagine is that J ~ is proportional to the concentration gradient: J ~ = −Dr ~ n (37) with D a proportionality constant. We put in the minus sign so that D would be a positive number (if @xn > 0, so the gradient increases to the right, then particles flow to the left). This is known as Fick's first law. It's a law because we didn't derive it (neither did Fick), it just seems reasonable. Note that Fick's first law is not some general property of fluxes – you can certainly have a nonzero flux at constant density, like current in a river, if there's some potential driving the flow. Ficks first law it is a statement about fluxes in diffusive systems, where there is no other source of motion other than random motion. Once we have Eq. (37) we can plug into the continuity equation, Eq. (36) to get @n(x ~ ; t) @t = Dr ~ 2n(x ~ ; t) (38) This is also known as Fick's second law. It is none other than the diffusion equation. Since Fick's second law follows from Fick's first law, in fact, we have justified Fick's first law through our analysis of random walks. Moreover, through our analysis of random walks, we have related D to properties of the gas, D = 1 2v ¯ as in Eq. (29). For example, say we have some lemmings that come out of a hole, walk left or right randomly, and maybe fall offa cliffa distance a away. Let their density be n(z), with z = 0 the hole and z = a the cliff. At a given z, over a short time δt, the number of lemmings that leave to the right is proportional to n(z) and those that come in from the right is propotional to n(z + δz). Similarly, the number that leave to the left is n(z) and those coming in from the left proportional to n(z −δz). So the net change in the number at z is going to be proportional to n(t+δt) −n(t) /[n(z +δz) −n(z)]+[−n(z) +n(z −δz)]. That is, @n @t / @2n @z2 . This is a physical way to see why diffusion only occurs if the second derivative is nonzero. In a steady state situation, @n @t = 0, the solution to the diffusion equation is n(z) = n0 + z a(n1 −n0). Then Jz = −D a (n1 −n0). This is a constant flux of lemmings coming out of the hole and falling of the cliff. Note that since Jz is constant, @zJz = 0 and the continuity equation is satisfied. To get a feel for how fast diffusion is, the diffusion constant in water for nitrogen molecules is D = 2 × 10−9 m2 s . Recalling from Eq. (22) that = 1 nσ, so D = 1 2 v ¯ nσ, bigger molecules should have smaller diffusion rates. Indeed, benzene molecules C6H6 in water have D = 1 × 10−9 m2 s . For large molecules like proteins in water the diffusion constant is even smaller D ≈10−11 m2 s . In gases, densities n are smaller so is larger and the diffusion constants are generally larger. For example, CO molecules in air at room temperature and pressure have D = 2 × 10−5 m2 s . To use the diffusion contant, we can either plug in the exponential solution, Eq. (24), or more simply use Eq. (23): xrms = 2Dt p (39) Fick's laws of diffusion 7 For example, taking a dye molecule in water with D ≈10−9m2 s , to move ∆x = 1m would take (∆x)2 2D = 31 years. So clearly diffusion is not the main mechanism by which dyes move around in water. By the way, thermal conduction is very much like diffusion. Instead of the diffusion equation, temperature satisfies the heat equation: @T(x; y; z; t) @t = r ~ 2T(x; y; z; t) (40) where is called the coefficient of thermal diffusivity. This equation describes diffusion of temper-ature, rather than diffusion of particle number. The derivation of the heat equation is identical to the derivation of Fick's second law, with conservation of energy replacing conservation of particle number. The analog of Fick's first law for thermal conduction is called Fourier's law. Fourier's law is an empirical observation that the rate of heat flow is proportional to the temperature difference. We'll return to thermal conduction when we talk about temperature and heat in future lectures. 4.1 Convection (optional) Diffusion refers to the motion of a molecule through random collisions. Think of a liquid in equilib-rium and just try to follow one molecule. Convection occurs when the system is not in equilibrium to begin with. In such situations, there can be coherent convective currents, like a hot or cold wind, that move the dye much faster than through a random walk. Or if you dropped the dye into the water with a dropper it hits with some force and has some inertia; then it takes a while for the system to equilibrate and the dye molecules are for a while moving much faster than they would if there were only diffusion. If there is some external effect causing the medium to flow with velocity v ~conv(x ~ ; t), then there will be flux even if there is no concentration gradient. We can introdude the convective flux J ~conv(x ~ ; t) = v ~conv n(x ~ ; t) (41) to describe this situation. Adding this convective flux to the diffusive flux, we get a new term: @n(x ~ ; t) @t = Dr ~ 2n(x ~ ; t) + v ~conv·rn(x ~ ; t) (42) This is called the generalized diffusion equation and describes situations where diffusion and convection are both important. Unfortunately, it is usually impossible to determine J ~conv(x ~ ; t), since when there is convection usually molecules are all moving around in different directions and it is a horribly non-linear process. Think about this next time you pour milk into your coffee – all those little eddy currents and funny shapes are convective. Good luck describing them analytically! Convection is almost always studied with numerical simulations. So for diffusion to actually be visible, a system has to be very calm – no temperature, density or chemical gradients. A place where diffusion is more important than convection is in biology. In biological systems, temperature is often very constant, convection is small, and molecules do not have to move very far. Diffusion of heat (thermal conduction) is the dominant mechanism of heat transfer in solids, for example as you heat up a pan on the stove. However, when you heat a room, convection dominates and the heat equation, Eq. (40) is not relevant. 5 Brownian motion An important application of the diffusion equation is to study Brownian motion. In 1827, a botanist named Robert Brown collected some pollen one Spring afternoon and put it in some water in a Petri dish in his lab, then went to bed. When he woke up, he found that the pollen grain had moved a significant distance. “It's alive!” he concluded. In fact, the pollen moved not because it is alive, but rather because it underwent a random walk due to the water molecules surrounding it constantly giving it little kicks. This movement is called Brownian motion, after Robert Brown. 8 Section 5 Brownian motion refers to the random walk of a large particle due to stochastic collisions with smaller particles. Although each hit from a small particle does very little, the hits add up to a macroscopically observable displacement. You can see Brownian motion easily with a microscope, where a dust particle or a bacterium will move a finite distance in a reasonable time. What is fascinating about this migration is that you cannot resolve the small molecules, like water molecules, in the microscope, so it looks like the big particle is moving by magic. Of course, it is not magic, and indeed we can deduce the existence of “invisible” molecules from Brownian motion of something visible. Einstein used this insight to measure Avogadro's number, as we will now see. The molecular collisions have another effect too – they slow down a moving particle, through a drag force. Indeed, drag, that you experience running your hand through the air or in water is a collective effect of many small molecules impeding the motion. Drag forces are macroscopic and can be measured without ever talking about molecules. For example we could rub the particle to make it electrically charged, then pull it with the electric force and measure the resistance. Or we could tie a tether to it, add a weight, and pull it with gravity. A drag force, by definition, slows down a particle, so if v ~ = 0 it should vanish. Thus, the leading effect in an expansion around v ~ = 0 of drag is that it is linear in the velocity. Calling the mass of the particle m and the external force F ~ext, the drag force we can then write m d2x ~ dt2 + µ d x ~ dt = F ~ext + F ~B (43) where x ~ is the position of the particle, µ is the drag coefficient (also called mobility), and F ~B is the force due to Brownian motion. That is, F ~B is a rapidly varying function of time, representing the small forces that the molecules give from whatever direction they come in at. Generally, when F ~ext is nonzero then F ~B can be neglected. So we turn F ~ext on to measure µ, then turn F ~ext off. Once µ is measured and F ~ext turned offwe can look at the distance the dust particle moves due to Brownian motion alone. Of course, the expected value is hx ~ i = 0, by symmetry, so we want to look at the RMS displacement xrms = hx ~ 2i p . We want to know how x ~ 2 changes with time: d dtx ~ 2 = 2 x ~ · d x ~ dt = 2 x ~ · v ~ (44) First note that Eq. (43) implies d dtv ~ = −µ mv ~ + 1 mF ~B (45) So then we have d dtx ~ · v ~ =  d dtx ~  · v ~ + x ~ ·  d dtv ~  = v ~ · v ~ −µ mx ~ · v ~ + 1 mx ~ · F ~B (46) Now let's take the average of all the quantities in this equation over very many molecular collisions. Since the force F ~B due to molecular collisions is in a random direction, on average it is totally uncorrelated to the current position of the particle, so hx ~ · F ~Bi = 0 as in Eq. (15). Then we have simply d dthx ~ · v ~ i = hv ~2i −µ mhx ~ · v ~i (47) The general solution of this equation starting at x ~ = 0 is hx ~ · v ~i = m µ hv ~2i  1 −e−µt m  (48) For late times t ≫m µ , the exponential is very small and we see that hx ~ · v ~i = m µ hv ~2i is constant in time. Then, from Eq. (44) we get d dthx ~ 2i = 2m µ hv ~2i (49) So that hx ~ 2i = 2m µ hv ~2i t (50) The RMS displacement is therefore xrms = hx ~ 2i p = vrms 2m µ r t p (51) Brownian motion 9 where vrms = hv ~2i p . Comparing to Eq. (39) we have D = 1 µhmv ~2i (52) This is known as the Einstein relation. Since µ is the macroscopic drag coefficient and xrms and t are measurable, this equation lets us determine the average kinetic energy of each molecule. Using also the total energy in a system, we can then determine the number of molecules. More precisely, we can derive an explicit formula for N A if we borrow a result from Lecture 4, that the average kinetic energy per particle is h1 2m v ~2i = 3R 2N AT with R the ideal gas constant and T the temperature. (It should not be obvious to you that the average kinetic energy is determined by the temperature, but it's true as we'll show in Lecture 4.) Then Eq. (52) implies N A = 3RT µD (53) So by measuring the temperature (with a thermometer), the drag coefficient (with an external force) and the rate of diffusion (from Brownian motion), the number of water molecules N A can be determined. This is how Albert Einstein proposed to measure Avogadro's number N A using Brownian motion in 1905. 5.1 Viscosity Consider a ball falling through a fluid. It is accelerated from a downward force due to gravity and encounters resistance, or drag, from the fluid. There are two sources of drag. The first, called inertial drag, is due to the ball bonking molecules in front of it and speeding them up. In a time ∆t the ball will sweep out a volume ∆V =πR2jv ~j∆t, thereby accelerating a mass ∆m = ρ∆V of fluid to velocity roughly v ~. The average acceleration is then a ~ = v ~ ∆t, so that the force is F ~inertial=∆ma ~ =πR2ρjv ~jv ~. This goes like the square of the velocity. This inertial drag is relevant at large velocities, but for slow objects it is always going to be subdominant to a drag foce that is linear in v. For slow objects (slow compared to typical molecular speeds), the dominant drag force is due to viscosity, called viscous drag. It is viscous drag that is important for Brownian motion, and related to random walks. Viscosity is another physical effect whose microscopic origin is in the stochastic collision of molecules. The more precise name for viscosity is dynamic shear viscosity. It measures how a fluid responds to shear forces: you push the top layer and ask how much the bottom layer moves. Intuitively, viscosity is a measure of how well a fluid flows. A shear force is applied to an area. Think of floating a block of wood on water and applying a external force F ~ext to move it parallel to the surface. The bigger the area of the block, the more force it puts on the water. The water then responds by picking up some velocity v ~, in the same direction as the force. The deeper you go into the water, the slower it will go. So we might expect ηv ~ z = = = = = = = = = = = = = = = = = = ? F ~ext A (54) with z the depth. This isn't quite right, because we don't know that the z dependence is exactly 1 z. The right way to think about this shear force is that we apply it only to the top of the water. When the top of the water moves, it pulls along the layer below that (by layer we mean layer of molecules, or just some abstract infinitesimal thickness of the fluid), and so on. So we write η@v ~ @z = F ~ext A (55) The parameter η is this equation is called the dynamic shear viscosity. Now lets return to the sphere falling through the fluid. It has a downward force due to gravity. As it moves down, it displaces the molecules, but also imparts velocity to the fluid in the direction transverse to its motion 10 Section 5 Figure 1. A ball falling through a viscous fluid has a downward force due to gravity and a drag force due to viscosity. The definition of viscosity tells us that the sphere induces a velocity of the fluid η@v ~ @z = F ~g 4πR2 (56) where 4πR2 is the surface area of the sphere. At the surface of the sphere, the fluid velocity is the same as the sphere velocity. Thus we can solve this equation to see how the fluid velocity changes with distance. The faster the sphere is falling, the faster the fluid will go. Eventually, all the energy used by gravity to accelerate the sphere will be taken up by the work done to move the water, and the sphere will stop accelerating. Thus, there will be some effective drag force. Working out all the factors (an annoyingly tedious calculation), the result is F ~drag = 6πηR v ~ (57) This is known as the Stokes drag force. This equation relates a property of the fluid (the viscosity), to the resistance experienced by an external object as it is being forced through the fluid (the viscous drag force). It is linear in the velocity of the object, compared to the inertial force which was quadratic in velocity, so at small velocities, the viscous drag force dominates. Keep in mind that viscosity is a property of the fluid itself, while drag depends on what is being dragged through the fluid. The Stokes force applies when that thing is a sphere of radius R. For a different object, like a cube, the precise form of the force would be different, but the scaling with η and v ~ and some measure of size R is universal. Indeed, it is fixed by dimensional analysis. Recall that we defined mobility as the drag coefficient in Eq. (43). So µ = 6πηR (58) This is known as the Stokes relation. Plugging it into the Einstein relation in Eq. (52) gives D = 1 6πηRhmv ~2i (59) This is known as the Einstein-Stokes relation. It relates the diffusion constant D and the viscosity, telling us that viscosity also has its origin in the microscopic random walks of the molecules in the fluid. 6 Summary In this lecture we have studied diffusion. The main concepts to understand are • Random walks: A particle/system has a fixed probability of moving in each direction. We are interested in the net motion after N steps. • If we move left with probability a and right with probability b =1 −a, the chance of taking m steps right is given by the binomial distribution: BN(m) = ambN −m  N m  where  N m  = N! m!(N −m)!. Summary 11 • The average distance moved in a random walk after N steps scales as σ ∼ N p where is the step size. This is true in any number of dimensions. • A bunch of molecules random-walking is called diffusion. Then the number of steps N is proportional to the time t so the distance moved is ∆x∼ t p. This scaling of distance / square-root of time is the characteristic feature of diffusion. • In the continuum limit, molecules are described by a number-density n(x ~ ; t). • The number density satisfies the diffusion equation @n(x ~ ; t) @t = Dr ~ 2n(x ~ ; t) when diffusion is dominant. • The exact solution to the diffusion equation with boundary conditions n(x ~ ; t) = δ3(x ~) is n(x ~ ; t) = 1 4πDt q exp h −x ~ 2 4Dt i . This confirms the scaling ∆x∼ t p . • The diffusion equation comes from a modeling of the microscopic system as undergoing random walks. An alternative classical-field approach (no particles) uses Fick's laws. These assume the conservation of the amount of stuff(the continuity equation @n @t +r ~ ·J ~ =0, where J ~ =n×v ~ is the flux) and that flux is proportional to the concentration gradient J ~ =−Dr ~ n, leads to the same diffusion equation. • The conduction of heat is described by the heat equation: @T(x ~ ; t) @t = r ~ 2T(x ~ ; t). This equation has identical form as the diffusion equation, and so also has ∆x∼ t p . • In liquids and gases, diffusion is rarely dominant unless the system is very calm. Convec-tion, coming from non-equilibrium initial conditions, is usually is more important. In solids, diffusion and the heat equation work well. • A large molecule in a reasonably calm liquid or gas can often be modeled well by a random walk/diffusion. When this applies, we say it undergoes Brownian motion. For such mole-cules, the diffusion consent is given by the Einstein relation: D = 1 µhmv ~2i with µ the drag coefficient. Measuring D and µ is one way to measure Avogadro's number: N A = 3RT µD . • The drag coefficient is also related to viscosity η by µ=6πηR. Then D = 1 6πηRhmv ~2i. This is called the Einstein-Stokes relation, The main lesson from the last section is simply that the drag force, mobility, diffusion, viscosity and random walks are all related. I don't expect you to remember all these formulas, and I certainly don't want you to memorize them. Just try to have the basic ideas straight. Viscosity is a macroscopically measurable property of a material. When a large particle moves in a viscous material, it undergoes Brownian motion. The bigger the particle, the smaller the diffusion constant, and the slower it moves. The higher the viscosity, the larger the drag force, and the slower the particle moves. Small molecules moving in a fluid also undergo random walks. For small molecules, of around the same size as the molecules in the fluid, the diffusion constant is D = 1 2v ¯ with the mean free path. In this case, it is not useful to think in terms of drag forces and viscosity since the diffusing particle has essentially no inertia. A tricky point from this lecture is that diffusion is an equilibrium phenomena. Equilibrium does not mean there is no time-dependence at all, but that the macroscopic properties of the system are static. We'll discuss equilibrium more next lecture. You should think of @n(x ~ ; t) @t = Dr ~ 2n(x ~ ; t) as describing the change in the number density of a small subset of the molecules in a bath, for example, dye in water, or of a macromolecule undergoing Brownian motion in a bath. The water in the bath doesn't change, even though individual water molecules might move. When the whole system is out of equilibrium, for example when the boundary conditions are open (gas released into a vacuum) or if you violently drop the dye in, then the diffusion equation doesn't apply. 12 Section 6
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https://education.ti.com/en/Activity/activity-detail?id=384FB053735B4C86BBF76AA6E018891C
Classroom Activities: Exploring Higher Degree Polynomials - Texas Instruments - content Education Technology Sign Up for TI Emails » Products Graphing Calculators Computer Software TI-Nspire™ Technology TI-Innovator™ Technology Data Collection TI-Nspire™ Navigator™ Scientific Calculators Financial Calculators Basic and Elementary Downloads Guidebooks - All products Guidebooks - TI-84 Plus CE Guidebooks - TI-84 Plus CE Python Guidebooks - TI-Nspire™ Technology TI-84 Plus CE Apps and Updates TI-Nspire CX Updates All software, OS and Apps Activities All Classroom Activities 84 Activity Central Math Nspired Building Concepts in Mathematics TI Codes STEM Activities Science Nspired TI-Innovator Lessons Educators Math Resources Science Resources STEM Resources Special Offers Financial Literacy Resources Computer Science Resources Professional Development Webinars Tutorials Customer Support Educator Support Programs IT Admin Support Exams Knowledge Base Contact Us Where to Buy SiteUS and Canada Activities Software, OS updates and apps Guidebooks | Texas Instruments Activities Downloads Home Download Download All Files (ZIP) ##### Lesson Files Student Activity DOC Student Activity PDF Teacher Notes PDF Exploring_Polynomials_Factors_Roots_and_Zeros_Teacher.doc TI-Nspire Activity TNS Alignment Standards Alignment Textbook Alignment Subject Area Math: Algebra II: Polynomials Author Texas Instruments Level 9-12 Activity Time 45 Minutes Device TI-Nspire™ Apps for iPad® TI-Nspire™ CX series TI-Nspire™ CX CAS/CX II CAS TI-Nspire Version 3.2 Report an Issue Exploring Higher Degree Polynomials Activity Overview Investigate graphical and algebraic representations of a polynomial function and its linear factors. Objectives Students will discover that the zeros of the linear factors are the zeros of the polynomial function. Students will discover that the real zeros of a polynomial function are the zeros of its linear factors. Students will determine the linear factors of a quadratic function. Students will connect the algebraic representation to the geometric representation. Students will see the effects of a double and/or triple root on the graph of a cubic function. Students will see the effects of the leading coefficient on a cubic function. Vocabulary Zeros Double or triple root Leading coefficient About the Lesson This lesson merges graphical and algebraic representations of a polynomial function and its linear factors. As a result, students will: Manipulate the parameters of the linear functions and observe the resulting changes in the polynomial function. 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188230
https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/0c07cbdc9c352c39eb9539b31ded90d7_MIT8_04S16_LecNotes4.pdf
Lecture 4 B. Zwiebach February 18, 2016 Contents 1 de Broglie wavelength and Galilean transformations 1 2 Phase and Group Velocities 4 3 Choosing the wavefunction for a free particle 6 1 de Broglie wavelength and Galilean transformations We have seen that to any free particle with momentum p, we can associate a plane wave, or a “matter wave”, with de Broglie wavelength λ = h/p, with p = jpj. The question is, waves of what? Well, this wave is eventually recognized as an example of what one calls the wavefunction. The wavefunction, as we will see is governed by the Schr¨ odinger equation. As we have hinted, the wavefunction gives us information about probabilities, and we will develop this idea in detail. Does the wave have directional or polarization properties like electric and magnetic fields in an electromagnetic wave? Yes, there is an analog of this, although we will not delve into it now. The analog of polarization corresponds to spin! The effects of spin are negligible in many cases (small velocities, no magnetic fields, for example) and for this reason, we just use a scalar wave, a complex number Ψ(x, t) 2 C (1.1) that depends on space and time. A couple of obvious questions come to mind. Is the wavefunction measurable? What kind of object is it? What does it describe? In trying to get intuition about this, let’s consider how different observers perceive the de Broglie wavelength of a particle, which should help us understand what kind of waves we are talking about. Recall that h p = λ = h 2π 2π = ℏk, (1.2) λ where k is the wavenumber. How would this wave behave under a change of frame? We therefore consider two frames S and S0 with the x and x0 axes aligned and with S0 moving to the right along the +x direction of S with constant velocity v. At time equal zero, the origins of the two reference frames coincide. The time and spatial coordinates of the two frames are related by a Galilean transformation, which states that x0 = x −vt, t0 = t . (1.3) Indeed time runs at the same speed in all Galilean frames and the relation between x and x0 is manifest from the arrangement shown in Fig. 1. Now assume both observers focus on a particle of mass m moving with nonrelativistic speed. Call the speed and momentum in the S frame v e and p = mv e, respectively. It follows by differentiation with 1 | | ∈ ′ ′ ′ ′ − ′ ′ Figure 1: The S0 frame moves at speed v along the x-direction of the S frame. A particle of mass m moves with speed v , f e, and thus momentum p = mv in the S frame. respect to t = t0 o the first equation in (1.3) tha e t dx0 dt0 = dx −v , (1.4) dt which means that the particle velocity v e0 in the S0 frame is given by v 0 = v −v . (1.5) Multiplying by the mass m we find the relati e on be e tween the momenta in the two frames p0 = p −mv. (1.6) The momentum p0 in the S0 frame can be appreciably different from the momentum p in the S frame. Thus the observers in S0 and in S will obtain rather different de Broglie wavelengths λ0 and λ! Indeed, h λ0 = p0 = h mv 6= λ, (1.7) p − This is very strange! As we review now, for ordinary waves that propagate in the rest frame of a medium (like sound waves or water waves) Galilean observers will find frequency changes but no change in wavelength. This is intuitively clear: to find the wavelength one need only take a picture of the wave at some given time, and both observers looking at the picture will agree on the value of the wavelength. On the other hand to measure frequency, each observers must wait some time to see a full period of the wave go through them. This will take different time for the different observers. Let us demonstrate these claims quantitatively. We begin with the statement that the phase φ = kx −ωt of such a wave is a Galilean invariant. The wave itself may be cos φ or sin φ or some combination, but the fact is that the physical value of the wave at any point and time must be agreed by the two observers. The wave is an observable. Since all the features of the wave (peaks, zeroes, etc, etc) are controlled by the phase, the two observers must agree on the value of the phase. In the S frame the phase can be written as follows φ = kx −ωt = k(x −ω k t) = 2π λ (x −V t) = 2πx λ −2πV λ t, (1.8) where V = ω ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ̸= ′ ̸ is the wave velocity. Note that the wavelength is read from the coefficient of x and ω is k minus the coefficient of t The two observers should agree on the value of φ. That is, we should have φ0 ′(x′ 0, t′ 0) = φ(x, t) (1.9) 2 where the coordinates and times are related by a Galilean transformation. Therefore 2π φ0(x0, t0) = λ (x −V t) = 2π λ (x0 + vt0 −V t0) = 2π λ x0 −2π(V −v)t0. (1.10) λ Since the right-hand side is expressed in terms of primed variables, we can read λ0 from the coefficient of x0 and ω0 as minus the coefficient of t0: λ0 = λ (1.11) 2π ω0 = λ (V −v) = 2πV λ  1 −v V  = ω  1 −v . (1.12) V This confirms that, as we claimed, for a physical wave propagating in a  medium, the wavelength is a Galilean invariant and the frequency transforms. So what does it mean that the wavelength of matter waves change under a Galilean transforma-tion? It means that the Ψ waves are not directly measurable! Their value does not correspond to a measurable quantity for which all Galilean observers must agree. Thus, the wavefunction need not be invariant under Galilean transformations: Ψ(x, t) 6= Ψ0(x0, t0) , (1.13) where (x, t) and (x0, t0) are related by Galilean transformations and thus represent the same point and time. You will figure out in Homework the correct relation between Ψ(x, t) and Ψ0(x0, t0). What is the frequency ω of the de Broglie wave for a particle with momentum p? We had p = ℏk (1.14) which fixes the wavelength in terms of the momentum. The frequency ω of the wave is determined by the relation E = ℏω , (1.15) which was also postulated by de Broglie and fixes ω in terms of the energy E of the particle. Note that for our focus on non-relativistic particles the energy E is determined by the momentum through the relation p2 E = . (1.16) 2m We can give three pieces of evidence that (1.15) is a reasonable relation. 1. If we superpose matter waves to form a wave-packet that represents the particle, the packet will move with the so called group velocity vg, which in fact coincides with the velocity of the particle. The group velocity is found by differentiation of ω with respect to k, as we will review soon: dω vg = dk = dE dp = d dp  p2 2m  = p = v . (1.17) m 2. The relation is also suggested by special relativity. The energy and the momentum components of a particle form a four-vector: E c , p  (1.18) 3 ′ ′ ′ − ′ ′ − ′ ′ − − ′ ′ ′ ′ ′ − − − ̸= ′ ′ ′ ′ ′ ′ ′ ′ ′ ̸ ′ ̸ Similarly, for waves whose phases are relativistic invariant we have another four-vector ω , k (1.19) c Setting two four-vectors equal to each other is a con  sistent choice: it would be valid in all Lorentz frames. As you can see, both de Broglie relations follow from E c , p  = ℏ ω , k c  . (1.20) 3. For photons, (1.15) is consistent with Einstein’s quanta of energy, because E = hν = ℏω. In summary we have p = ℏk, E = ℏω . (1.21) These are called the de Broglie relations, and they are valid for all particles. 2 Phase and Group Velocities To understand group velocity we form wave packets and investigate how fast they move. For this we will simply assume that ω(k) is some arbitrary function of k. Consider a superposition of plane waves ei(kx−ω(k)t) given by ψ(x, t) = Z dk Φ(k)ei(kx−ω(k)t). (2.22) We assume that the function Φ(k) is peaked around some wavenumber k = k0, as shown in Fig. 2. Figure 2: The function Φ(k) is assumed to peak around k = k0. In order to motivate the following discussion consider the case when Φ(k) not only peaks around k0 but it also is real (we will drop this assumption later). In this case the phase ϕ of the integrand comes only from the exponential: ϕ(k) = kx −ω(k)t . (2.23) We wish to understand what are the values of x and t for which the packet ψ(x, t) takes large values. We use the stationary phase principle: since only for k ∼k0 the integral over k has a chance to give a non-zero contribution, the phase factor must be stationary at k = k0. The idea is simple: if a function is multiplied by a rapidly varying phase, the integral washes out. Thus the phase must have zero derivative at k0. Applying this idea to our phase we find the derivative and set it equal to zero at k0: dϕ dk k0 = x −dω dk k0 t = 0 . (2.24) 4 ∼ This means that ψ(x, t) is appreciable when x and t are related by dω x = t dk , (2.25) k0 showing that the packet moves with group velocity dω vg = . dk (2.26) k0 Exercise. If Φ(k ) is not real write Φ(k) = jΦ(k)jeiφ(k 0 ). Find the new version of (2.25) and show that the velocity of the wave is not changed. Let us now do a more detailed calculation that confirms the above analysis and gives some extra insight. Notice first that ψ(x, 0) = dk Φ(k)eikx . (2.27) We expand ω(k) in a Taylor expansion around k Z = k0 dω ω(k) = ω(k0) + (k −k0) + O (k 2 dk −k0) . (2.28) k0 Then we find, neglecting the O((k −k )2  0 ) terms i(k k ) dω t ψ(x, t) = Z dk Φ(k) eikx e−iω(k0)te − −0 dk jk0 . (2.29) It is convenient to take out of the integral all the factors that do not depend on k: iω(k )t+ik dω t ik dω t ψ(x, t) = e − 0 0 dk jk0 Z dk Φ(k)eikxe − dk jk0 (2.30) iω(k )t+ik dω t ik x dω t = e − 0 0 dk jk dk k 0 Z dk Φ(k)e  − j 0  . Comparing with (2.27) we realize that the integral in the above expression can be written in terms of the wavefunction at zero time: −iω(k dω 0)t+ik0 dk j t dω ψ(x, t) = e k0 ψ  x − (2.31) dk t . k0  The phase factors in front of the expression are not important in trac king where the wave packet is. In particular we can take the norm of both sides of the equation to find dω jψ(x, t)j = ψ  x − t, dk 0 k0  If ψ(x, 0) peaks at some value x it is clear from the abov e equation . (2.32) 0 that jψ(x, t)j peaks for dω x − dω t = x0 ! x = x0 + dk k dk 0 sho t , (2.33) k0 wing that the peak of the packet moves with velocity vgr = dω, ev aluated at k0. dk 5 − − − − − − | − | − | − | − | − | − − | | − → | | | | 3 Choosing the wavefunction for a free particle What is the mathematical form of the wave associated with a particle a particle with energy E and momentum p? We know that ω and k are determined from E = ℏomega and p = ℏk. Let’s suppose that we want our wave to be propagating in the +x ˆ direction. All the following are examples of waves that could be candidates for the particle wavefunction. 1. sin (kx −ωt) 2. cos (kx −ωt) 3. ei(kx−ωt) = eikxe−iωt - time dependence / e−iωt 4. e−i(kx−ωt) = e−ikxeiωt - time dependence / e+iωt In the third and fourth options we have indicated that the time dependence could come with either sign. We will use superposition to decide which is the right one! We are looking for a wave-function which is non-zero for all values of x. Let’s take them one by one: 1. Starting from (1), we build a superposition in which the particle has equal probability to be found moving in the +x and the −x directions. Ψ(x, t) = sin (kx −ωt) + sin (kx + ωt) (3.1) Expanding the trigonometric functions this can be simplified to Ψ(x, t) = 2 sin(kx) cos(ωt) . (3.2) But this result is not sensible. The wave function vanishes identically for all x at some special times ωt = π 2 , 3π 2 , 5π, ... 2  (3.3) A wavefunction that is zero cannot represent a particle. 2. Constructing a wave function from (2) with a superposition of left and right going cos waves, Ψ(x, t) = cos(kx −ωt) + cos(kx + ωt) = 2 cos(kx) cos(ωt) . (3.4) This choice is no good, it also vanishes identically when ωt = π 2 , 3π, ... 2 3. Let’s try a similar superposition of exponentials from (3), with both  having the same time dependence Ψ(x, t) = ei(kx−ωt) + ei(−kx−ωt) (3.5) = (eikx + e−ikx) e−iωt (3.6) = 2 cos kx e−iωt . (3.7) This wavefunction meets our criteria! It is never zero for all values of x because e−iωt is never zero. 6 − − − − ∝ − − − − ∝ − − − − − − − − − − 4. A superposition of exponentials from (4) also meets our criteria Ψ(x, t) = e−i(kx−ωt) + e−i(−kx−ωt) (3.8) = (eikx + e−ikx) eiωt (3.9) = 2 cos kx eiωt . (3.10) This is never zero for all values of x Since both options (3) and (4) seem to work we ask: Can we use both (3) and (4) to represent a particle moving to the right (in the +x ˆ direction)? Let’s assume that we can. Then, since adding a state to itself should not change the state, we could represent the right moving particle by using the sum of (3) and (4) Ψ(x, t) = ei(kx−ωt) + e−i(kx−ωt) = 2 cos(kx −ωt) . (3.11) This, however, is the same as (2), which we already showed leads to difficulties. Therefore we must choose between (3) and (4). The choice is a matter of convention, and all physicists use the same convention. We take the free particle wavefunction to be Free particle wavefunction : Ψ(x, t) = ei(kx−ωt) , (3.12) representing a particle with p = ℏk , and E = ℏω . (3.13) In three dimensions the corresponding wavefunction would be Free particle wavefunction : Ψ(x, t) = ei(k·x−ωt) , (3.14) representing a particle with p = ℏk , and E = ℏω . (3.15) Andrew Turner and Sarah Geller transcribed Zwiebach’s handwritten notes to create the first LaTeX version of this document. 7 − − −− − − − − − − − · − MIT OpenCourseWare 8.04 Quantum Physics I Spring 2016 For information about citing these materials or our Terms of Use, visit:
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https://nickelinstitute.org/media/4675/ni_inco_279_corrosionresisthydrochloric.pdf
CORROSION RESISTANCE OF NICKEL AND NICKEL-CONTAINING ALLOYS IN HYDROCHLORIC ACID, HYDROGEN CHLORIDE AND CHLORINE (CEB-3) A PRACTICAL GUIDE TO THE USE OF NICKEL-CONTAINING ALLOYS NO 279 Distributed by NICKEL INSTITUTE Produced by INCO CORROSION RESISTANCE OF NICKEL AND NICKEL-CONTAINING ALLOYS IN HYDROCHLORIC ACID, HYDROGEN CHLORIDE AND CHLORINE (CEB-3) A PRACTICAL GUIDE TO THE USE OF NICKEL-CONTAINING ALLOYS NO 279 This handbook was first published prior to 1985 by INCO, The International Nickel Company, Inc. Today this company is part of Vale S.A. The Nickel Institute republished the handbook in 2020. Despite the age of this publication the information herein is considered to be generally valid. Material presented in the handbook has been prepared for the general information of the reader and should not be used or relied on for specific applications without first securing competent advice. The Nickel Institute, their members, staff and consultants do not represent or warrant its suitability for any general or specific use and assume no liability or responsibility of any kind in connection with the information herein. Nickel Institute communications@nickelinstitute.org www.nickelinstitute.org Page 2 Nickel-clad steel jacketed reactor used for organic chlorinations. It was built in accordance with the A.P.I. A.S.M.E. code for unfired pressure vessels and operates at a temperature of 650° F, and pressures of 25 lb. per sq. in in the body and 125 Ib per sq. in. in the jacket. Page 3 Resistance of Nickel and High Nickel Alloys to Corrosion by Hydrochloric Acid, Hydrogen Chloride and Chlorine† CONTENTS HYDROCHLORIC ACID .................................................. Page 3 CHLORINE AND HYDROGEN CHLORIDE MONEL alloy 400 and Nickel 200....................................... 4 AT HIGH TEMPERATURES ........………….. Page 11 Acid Concentration.......................................................... 5 Chlorine ..................……………………………….......... 12 Temperature of the Solution........................................... 5 Hydrogen Chloride .......………………………............... 13 Aeration and Oxidizing Agents....................................... 6 Metallurgical Considerations ...………………......... 14 Oxidizing Salts in Solution.............................................. 6 Applications ..............………………………….......... 14 Velocity of the Solution ................................................... 7 Film Formation ................................................................ 7 PLANT CORROSION TESTS ....……………………............. 15 Effect of Stress................................................................ 8 Chlorinated Solvents ...........………………………..... 15 Organic Chlorinations ..........………………………..... 19 INCONEL alloy 600 ............................................................. 8 Synthetic Resins and Rubbers .....…………….…...... 21 HASTELLOY Alloys............................................................. 8 Other Organic Chlorides .........……………………..... 22 Oxidizing Salts................................................................. 8 Pickling of Iron and Steel .....…..…………………...... 23 Nickel-Iron Alloys ................................................................. 9 Galvanizing and Soldering Fluxes ...………………... 24 Petroleum Refining ...............……………………....... 25 HYDROGEN CHLORIDE AND CHLORINE........................... 9 Inorganic Acid Chlorides ...........…………………...... 27 Pulp and Paper .................………………………....... 29 Other Applications ...............………………………..... 31 Hydrochloric Acid ICKEL and high nickel alloys are among the few metallic materials having useful resistance to hydrochloric acid solutions. Nominal compositions of the nickel alloys most commonly used in this service are shown in Table 1. Selection of the most useful and economical of these materials usually depends upon the concentration, aeration and temperature of the acid and upon such other factors as velocity, film formation, continuity of exposure, allowable metallic content of the solution and the physical properties of the material. The performance of all of the high-nickel materials in hydrochloric acid solutions has been determined by service experience and by numerous laboratory and plant corrosion tests which are summarized in the following pages. TABLE 1 Nominal Compositions of High-Nickel Alloys Commonly Used with Hydrochloric Acid Solutions Material Nickel % Copper % Iron Chromium Molybdenum Silicon Manganese Carbon Other Nickel 200°.......... 99.0• 0.1 0.15 .... .... 0.05 0.25 0.1 …. Nickel 201°.......... 99.0 0.1 0.15 .... .... 0.05 0.25 0.02 Max. …. Duranickel alloy 301. 94.0 0.05 0.35 .... .... 0.50 0.30 0.25 A14.5 Ti 0.50 Monel alloy 400°.... 67 30 1.4 .... .... 0.1 1 0.15 …. Monel alloy K-500... 66 29 0.9 .... .... 0.25 0.4 0.15 A1 .2.75 Monel alloy 506b.... 63 31 2.0 .... .... 3.0 0.9 0.1 …. Monel alloy 505b.... 65 28 2.0 .... .... 4.0 0.9 0.1 …. Inconel alloy 600°... 77 0.25 7.0 15.0 .... 0.25 0.25 0.08 ..... Ti 2.50 Inconel alloy X-750.. 73 .05 7.0 15.0 .... 0.40 0.50 0.05 Al 0.75 { Cb & Ta 0.90 Chlorimet# 2b....... 63 .... 3.0 Max. 32 1.0 1.0 0.10 …. Chlorimet# 3b....... 60 3.0 Max. 18.0 18 1.0 1.0 .07 …. NI-Resist (Type 1)b.. 13.5-17.5 5.5-7.5 Bal. 1.75-2.50 .... 1.0-2.5 1.0-1.5 '3.00 Max. .... Ni-Resist (Type 2)b.. 18.0-22.0 0.5 Max. Bal. 1.75-3.00 .... 1.0-2.5 0.8-1.5 3.00 Max. .... Ni-Resist (Type 3)b.. 28.0-32.0 0.5 Max. Bal. 2.5-3.5 . . . 1.0-2.0 .40-.80 2.75 Max. .... Hastelloy alloy Ell.. 60 .... 6 …. 32 1 1 …. Hasfelloy alloy C... 51 .... 6 17 19 1 1 .... W 5 Hastelloy alloy D b.. 85 3 .... 10 1 Al 1 Hastelloy alloy F.. 48 .... 15 22 6.5 1 1.5 0.05 .Max. .... † Previously published as Technical Bulletin T-29. INCO Registered Trademark. Haynes Stellite Company, Trademark. # Duriron Co., Inc., Trademark. a Including cobalt. b Available only in cast form. c Also available in form of clad steel. N Page 4 MONEL alloy 400 and Nickel 200 The results of laboratory corrosion tests of MONEL nickel-copper alloy 400 and Nickel 200 over a wide range of concentrations and temperatures, and a description of the test methods used have been presented by Friend and Knapp. 1 Test specimens were cut from standard cold-rolled commercial sheets. Tests were made in air-saturated and in air-free solutions to give conditions representative of maximum and minimum corrosion rates respectively. Tests in air-saturated solutions were made in the circular path corrosion testing apparatus shown in Figure 1. This apparatus is similar to that described by Fraser, Ackerman and Sands. 2 The methods of supporting the test specimens and of aerating the solutions are shown in Figure 2. Tests in air-free solutions were made by saturating the solutions with nitro-gen. These tests were made in the rotating-spindle type apparatus shown in Figure 3. This apparatus was used because it is entirely closed and permits the maintenance of a nitrogen atmosphere over the solution surface as well as in the solution itself. In most cases, the test specimens were moved 1 W. Z. Friend and B. B. Knapp, "Behavior of Nickel and High Nickel Alloys in Hydrochloric Acid and Hydrogen Chloride," Trans. Am. Inst. Chem. Engrs. 39, 731 (1943). 2 O. B. J Fraser, D. E. Ackerman and J. W. Sands, Ind. Eng. Chem. 19, 332 (1927). FIG. 2-Method for supporting test specimen and for aerating the test solution. at a velocity of 15.7 feet per minute in air-saturated tests and 21.6 feet per minute in air-free tests. The two different types of apparatus in which the tests were made could not be adjusted conveniently to give the same speed in each. Test periods were of 24 hours duration in air-saturated and 48 hours in air-free solutions except where the specific effect of duration of test on corrosion rate was being determined. The test methods proved to be under statistical control. FIG. 1-Circular-path corrosion testing apparatus. FIG. 3-Rotating-spindle type corrosion testing apparatus. Page 5 Acid Concentration In this bulletin, all acid concentrations are expressed in per cent by weight of HCI. Corrosion rates are shown in milligrams per square decimeter per day (mdd.), and in inches penetration per year (ipy.). The latter unit is based on the assumption of continuous exposure, 24 hours a day and 365 days a year, on one surface of the metal only. Conversion from mdd. to ipy. for any metal or alloy is made by the following equation where d is the density of the metal in gm per cu. cm. Figure 4 shows the effect of hydrochloric acid concentration upon the corrosion rates of MONEL nickel copper alloy 400 and nickel in air-free and air-saturated solutions at 30° C (86° F). FIG. 4-Corrosion of nickel and MONEL alloy 400 in hydrochloric acid solutions at 30° C (86° F). It will be noted from Figure 4 that the corrosion rates of MONEL alloy 400 and nickel at room temperature are approximately the same up to 10 per cent acid concentration. In this concentration range, the corrosion rates of both materials in air-free solutions are normally under 0.01 ipy. The rates for both materials are considerably increased by saturating the solutions with air. Above about 15 per cent acid concentration, the corrosion rates of MONEL alloy 400 begin to increase rather sharply. In practice, applications Of MONEL alloy 400 at room temperature will be, in general, limited to concentrations under about 10 per cent in aerated solutions and under about 20 per cent in air-free solutions. Nickel is superior to MONEL alloy 400 in resistance to hydrochloric acid concentrations above 20 per 3 A. Seidell, "Solubilities of Inorganic and Metal Organic Compounds," Third Edition, D. Van Nostrand Co. Inc., New York (1940). mdd. x (0.001437) = ipy. d cent at room temperature, possibly due to the fact that nickel chloride is less soluble than cuprous or cupric chloride in this range of acid concentrations. The solubilities of these chlorides in hydrochloric acid solutions, as calculated from the data of Seidell 3 are shown in Figure 5. The corrosion data in Figure 4 indicate that occasionally nickel may find suitable applications in acid concentrations up to 30 per cent concentration, either aerated or unaerated, at room temperature. Its performance in air-saturated solutions is of particular interest. As shown in Figure 4, the corrosion rate was a maximum of 574 mdd. at 5 per cent concentration, then fell off to 500 mdd. at 10 per cent and did not exceed this figure at any higher concentration up to and including 30 per cent. At 25 per cent HCl concentration, its corrosion rate took a sharp drop to an average of 304 mdd. Several test specimens, tested in different solutions of this concentration, checked very closely. In drawing the curve for nickel in air-saturated acid in Figure 4, it has been considered prudent to disregard this low point because in commercial practice, it may be diffi-cult to maintain exactly this concentration. The use of nickel with concentrated hydrochloric acid solutions has been very limited and there is little service data to provide confirmation of these test results. If the performance of nickel is due to the relatively low solubility of its corrosion products, caution should be used in applying these test data to service in concentrated acid where high velocities are encountered. Temperature of the Solution An increase in temperature increases the corrosion rates of both MONEL alloy 400 and nickel in hydrochloric acid solutions, although nickel is affected more in this respect than MONEL alloy 400. Corrosion rates of both materials in 5 per cent hydrochloric acid, in FIG. 5-Solubility of cupric chloride, cuprous chloride, and nickel chloride in hydrochloric acid solutions. Page 6 air-free and air-saturated solutions, at temperatures up to 90° C (194° F) are shown in Figure 6. In unaerated 5 per cent acid, MONEL alloy 400 is usefully resistant up to about 75° C (167° F) and nickel up to 55° C (130° F). Applications of both materials, but particularly of MONEL alloy 400, in lower concentrations at higher temperatures are common. In most of the processes in which hydrochloric acid is formed as a result of hydrolysis of chlorides or chlorinated solvents, acid concentrations are less than 0.5 per cent; these are being withstood satisfactorily at temperatures up to 300 to 400° F Applications of MONEL alloy 400 and nickel in air-saturated hydrochloric acid above room temperature are usually limited to concentrations under 3 or 4 per cent. MONEL nickel-copper alloy 400 is being applied successfully to handling aerated acid of 2 per cent concentration at 120° F and of 1 per cent concentration at 180° F The results of laboratory corrosion tests of MONEL alloy 400 and nickel in boiling hydrochloric acid solutions are given in Table 2. MONEL alloy 400 can be used often in boiling solutions of 1 per cent or lower concentration 'and in this application is superior to nickel. Under boiling conditions, ebullition sweeps the solution free of air so that aeration is not usually an important factor. FIG. 6-Effect of temperature on corrosion of nickel and MONEL alloy 400 in 5% hydrochloric acid. Corrosion of MONEL alloy 400 and Nickel in Boiling Hydrochloric Acid Velocity: None Aeration: None Duration of tests: 10 days TABLE 2 vice for breaking up the air stream into small bubbles and distributing them through the solution. The rate at which air may be replenished at a quiet solution-air surface is probably quite small compared to the rate at which oxygen can be removed by corrosion. Consequently, in an ordinary metal tank, open to the air and containing an unagitated solution, the air content of the solution may be only a few per cent of saturation except right at the liquid line. In estimating the probable life of equipment, it is important to have information on the degree of aeration to be expected, and on the presence of any impurities which will affect oxygen availability. Many organic compounds such as foods, fats and sugars will react with dissolved oxygen, removing it from solution. Dyeing and bleaching materials are often oxidizing or reducing. The presence of oxidizing salts in acid solutions may have a very strong effect on corro-sion. Oxidizing Suits in Solution Oxidizing salts such as cupric and ferric salts, when dissolved in significant amounts in hydrochloric acid solutions, will increase considerably the corrosiveness of the solutions toward MONEL alloy 400 and nickel. To illustrate the corrosive effect of cupric chloride, laboratory tests were made with MONEL alloy 400 in air-saturated hydrochloric acid solutions of various concentrations, at 30° C (86° F) with and without the initial addition of cupric chloride to the solutions. The duration of the tests was 24 hours. The average mdd.= mg per sq dm per day. ipy.= inch penetration per year. Aeration and Oxidizing Agents In Figures 4 and 6, corrosion rates have been shown in air-saturated and air-free solutions in order to provide information as to upper and lower limits of corrosion which conceivably might be encountered in practice and because these conditions can be readily maintained in laboratory testing. It is apparent that a high degree of aeration will considerably increase corrosion of both nickel and MONEL alloy 400 in hydrochloric acid solutions. In practice, the continued existence of completely air-saturated solutions is not common. A high degree of aeration would not ordinarily be expected unless air actually is bubbled continuously through the solution as a part of the process, or unless the solution is sprayed or poured through air, or churned up with considerable air in pumping. Air saturation by mechanical means such as by stirring is rather a slow process and requires an efficient de- Corrosion Rate Monel alloy 400 Nickel Add Concentration % HCI by wt. mdd. ipy. mdd. ipy. 0.5 178 0.029 1,875 0.304 1.0 258 0.042 4,200 0.680 5.0 1,500 0.244 35,400 5.74 Page 7 FIG. 7-Effect of copper ions on corrosion of MGNEL alloy 400 in air-saturated hydrochloric solutions at 30° C (86° F). corrosion rates obtained for both high and low copper content are shown in Figure 7. In one set of tests, no copper was added originally to the test solutions and in the other tests, 0.5 gm per liter were added. The latter amount is high enough that the copper added to the solutions by corrosion of the MONEL alloy 400 specimens was only a small percentage of the total. It will be noted that with a copper content of 0.5 gm per liter the corrosion rate remained at a fairly constant high level regardless of the acid concentration. In other tests at 80° C (176° F) when the corrosion rate of MONEL alloy 400 in air-saturated 0.5 per cent hydrochloric acid was 98 mdd. (0.016 ipy.), the addition of 0.5 per cent cupric chloride to the air-saturated solution increased the corrosion rate to 1104 mdd. (0.18 ipy.). Other oxidizing chloride salts, such as ferric chlo-ride and mercuric chloride, are also highly corrosive to nickel and MONEL alloy 400 except in very dilute solutions. In the effluent from ferric chloride coagula-tion of sewage, where the ferric chloride content of the effluent was only 0.2 to 0.3 per cent, the corrosion rate of nickel was only 55 mdd. (0.009 ipy.) and that of MONEL alloy 400, 86 mdd. (0.014 ipy.). Stannic chloride is apparently not as corrosive to nickel as some of the other oxidizing salts. In three weeks tests in a 26% solution at 24° C (75° F), the corrosion rate of nickel was only 109 mdd. (0.018 ipy.) and that of MONEL alloy 400 was 179 mdd. (0.029 ipy.). The addition of appreciable amounts of such oxi-dizing salts as chromates, dichromates, nitrates and peroxides to hydrochloric acid solutions may make them highly corrosive to nickel and MONEL alloy 400. Cuprous and ferrous chlorides in reducing or air-free hydrochloric acid solutions do not increase the corrosion rates. Velocity of the Solution The usual effect of an increase in velocity of relative motion between metal and liquid is to increase the corrosion rate. It brings fresh acid and oxygen, if present, to the corroding surface, removes spent acid, and thins the diffusion film through which soluble reacting substances and corrosion products must pass. If movement is relatively swift, it may prevent the retention of what otherwise might be protective films. Velocities of 15 to 22 feet per minute were used in many of the laboratory tests described in this bulletin because they favor reproducibility of results and represent conditions often found in plant practice. Film Formation The formation of films or accumulation of solid products of corrosion upon a metal surface often has a marked effect on corrosion rates. This is particularly true where the deposits are dense and are insoluble in the solution, thus preventing acid and oxygen from reaching the metal surface. The effect of such protective deposits is usually to reduce corrosion to an extent dependent upon the distribution and physical structure of the deposit. In the case of MONEL alloy 400 in pure hydrochloric acid solutions, there is no evidence of such a protective effect from corrosion products. With nickel, as mentioned previously, there is some evidence that the relatively low solubility of nickel chloride may give a certain amount of protection in the more concentrated cold solutions. In some plant processes in which hydrochloric acid is mixed with other materials, non-metallic films which tend to reduce corrosion may be formed some-times on a metal surface. This is exemplified by the oily films which may occur in exposure to sewage or in the processing of oily substances. The addition of protein materials such as milk albumen to dilute, aerated acid solutions is often ef-fective in reducing corrosion as is evident from the results of laboratory corrosion tests given in Table 3. Solution Average Corrosion Rate, mdd. Air-saturated, 2% HCl......................... 1875 Air-saturated, 2 °fo HCl plus 0.5 % milk albumen ................ 745 Air-saturated, 5% HCl......................... 2230 Air-saturated, 5 % HCl plus 0.5% milk albumen ................. 1115 Corrosion rates are high due to accumulation of cupric chloride in solutions from exposure of MONEL alloy 400 to air-saturated test- solutions for 7 day periods. For normal rate of MONEL alloy 400 in 24 hr tests in air-saturated solutions see Fig. 4. It is believed that the inhibiting effect is due to the film-forming characteristics of the protein, to- TABLE 3 Effect of Protein Addition an Corrosion of MGNEL alloy 400 in Hydrochloric Acid Solutions Temperature: 30° C (86° F) Duration of Tests: 7 days Velocity: 16 ft per min. Page 8 gether with its reaction with some of the oxygen present. Effect of Stress Experience with nickel and MONEL nickel-copper alloy 400 equipment under various conditions of stress has demonstrated that these materials are not subject to stress-corrosion cracking in hydrochloric acid solutions. INCONEL nickel-chromium alloy 600 which is discussed below, also is free from this form of attack. INCONEL alloy 600 While INCONEL alloy 600 possesses fairly good resistance to dilute hydrochloric acid solutions, its performance, because of its chromium content, will not usually be quite as good as that of MONEL alloy 400 or nickel. It has demonstrated good resistance to cold, aerated acid in concentrations under about 2 per cent. The results of a number of laboratory tests are given in Table 4. Caution should be used in exposing INCONEL alloy 600 to hot hydrochloric acid solutions except where the acid concentrations are very low. It has been applied successfully to such processes as the production of synthetic resins, distillation of coal tar, evaporation of gelatine, distillation of pectin, and handling of glandular compounds where small amounts of residual hydrochloric acid are present from catalytic or acid extraction processes, and particular resistance to organic materials is required. F I G. 8-Corrosion of H A S T E L L O Y alloy B in hydrochloric acid solutions. HASTELLOY alloy C has useful resistance to all concentrations of hydrochloric acid at room temperature and is used successfully up to 50° C (122° F). Near this limiting temperature, however, HASTELLOY alloy B is the better choice of material, the only exception being when oxidizing salts or traces of free chlorine are present. HASTELLOY alloy D possesses moderate resistance to hydrochloric acid, although the highest temperature for which it is recommended is about 40° C (105° F). HASTELLOY alloy F has good resistance to hydro-chloric acid of all concentrations at room tempera-ture, and may be used in dilute solutions at tempera-tures up to 65° C (150° F). Oxidizing Salts Ferric chloride, cupric chloride, and other oxidizing salts in solution, considerably increase the corrosion of HASTELLOY alloys B and D in hydrochloric acid solutions and none of these alloys is recommended for handling such solutions except where the concentrations of oxidizing salt is extremely small. HASTELLOY alloy C, however, is outstanding in its resistance to oxidizing salt solutions and, at moderate temperatures, to hydrochloric acid solutions containing appreciable amounts of oxidizing salts. HASTELLOY alloy C is resistant to ferric chloride solutions at temperatures up to about 70° C (158° F) and to cupric chloride solutions up to about 40° C (105° F). In Table 5 are given the results of a number of laboratory corrosion tests of HASTELLOY alloy C in quiet ferric chloride solutions and mixtures of hydro-chloric acid and ferric chloride. TABLE 4 Corrosion of INCONEL alloy 600 in Hydrochloric Acid Tempera- ture Corrosion Rate Arid concen- tration, % HCl by wt. °C °F Dura- tion of Test Aeration veloc- ity ft. per min. mdd. ipy. 5 30 86 20 hr H2 saturated 16.5 78 0.013 5 30 86 20 hr Air saturated 16.5 581 0.097 5.9 30 86 5 days Air saturated Quiet 267 0.045 5.9 80 176 5 days Air saturated Quiet 4,223 0.71 5 85 185 20 hr H2 saturated 16.5 9,550 1.59 5 8S 185 20 hr Air saturated 16.5 11,630 1.95 HASTELLOY Alloys For handling hydrochloric acid in concentrations and at temperatures above those indicated for the suitable performance of nickel and MONEL alloy 400, the most resistant of the high nickel alloys is the HASTELLOY alloy B. As indicated in the data 4 of Figure 8, HASTELLOY alloy B has unusually high resistance at all concentrations and temperatures, even at the boiling point, It is recommended particularly for service in hydrochloric acid solutions in the temperature range from 70 to 110° C (158 to 230° F) and for handling wet hydrogen chloride gas. The penetration rate is only 148 mdd. (0.023 ipy.) in a 20 per cent boiling solution, generally considered to be one of the most corrosive conditions. 4 “Hastellov Corrosion Resistant Atloya,” May 1957 Edition. Haynes Stefte Company, Kokomo, Indiana. Page 9 TABLE 5 Corrosion of HASTELLOY alloy C in Ferric Chloride Solutions and Mixtures of Ferric Chloride and Hydrochloric Acid Nickel-Iron Alloys The addition of nickel to iron increases its resistance to hydrochloric acid solutions at atmospheric temperature. The improvement is roughly in proportion to the nickel content where the nickel addition is above about 12 per cent. So far as nickel steels are concerned, the ones most likely to have practical application in dilute hydrochloric acid applications are the series of nickel-iron alloys containing 25 to 60 per cent nickel used for their special expansion or magnetic characteristics, as for example the INVAR or ELINVAR alloys containing about 36 per cent nickel. In Table 6 are given the results of several corrosion tests of high nickel steels in hydrochloric acid solutions at room temperature. Corrosion rates of an ELINVAR alloy in dilute solutions at several temperatures are shown in Table 7. TABLE 6 Laboratory Corrosion Tests of Nickel Steels in Quiet Hydrochloric Acid Solutions at Room Temperature 25% Ni Steel 32% Ni-2% Cr Steel_ Acid Concentration, % HCl by wt. mdd. ipy. mdd. ipy. 1.8 70 0.013 … … 5 … … 96 0.017 9 160 0.029 … … 10 … … 168 0.036 15 576 0.105 … … 17.2 ... ... 360 0.064 TABLE 7 Laboratory Corrosion Tests of ELINVAR Type Alloy in Quiet Unaerated Hydrochloric Acid Solutions Corrosion Rate Acid Concentra- tion, % HCI Room Temp. 52°C (125°F) 74°C (165°F) by wt. mdd. ipy. mdd. ipy. mdd. ipy. 1 25 0.005 48 0.009 73 0.013 2 26 0.005 50 0.009 131 0.024 4 27 0.005 54 0.010 291 0.053 8 26 0.005 206 0.038 1260 0.23 Trademark of Soc. Anon. de Commentry-Fourchambault et Decazivi11e (Aciéries d'Imphy). The nickel-iron alloys most commonly used with dilute hydrochloric acid at room temperature are the nickel and nickel-copper cast irons of the NI-RESIST cast iron series. Corrosion rates of NI-RESIST Type 1 cast iron (13.5 to 17.5 per cent nickel, 5.5 to 7.5 per cent copper, and 1.75 to 2.5 per cent chromium) in unaerated hydrochloric acid solutions at room tem-perature are given in Table 8. Corrosion of NI-RESIST Type I Cast Iron in Unaerated Hydrochloric Acid Solutions at Room Temperature TABLE 8 Corrosion Rate Ni-Resist Type I Cast Iron Plain Cast Iron Acid Concentration, % HC1 by wt. mdd. ipy. mdd. ipy. 1.8 25 0.005 4476 0.9 3.6 74 0.015 5.0 88 0.018 7440 1.5 10.0 81 0.016 6186 1.2 20.0 230 0.045 .... NI-RESIST Type 3 (containing 28 to 32 per cent nickel), because of its higher nickel content, will usually have somewhat better resistance to dilute hydrochloric acid solutions than NI-RESIST Type 1 and NI-RESIST Type 2, as indicated by plant tests, the results of which are given in the latter part of this bulletin. Both aeration and increase in temperature considerably increase the corrosion rates of the NI-RESIST alloys. For example, tests in aerated 2 per cent hydrochloric acid at 49° C (120° F) gave corrosion rates of 520 mdd. (0.10 ipy.) for NI-RESIST Type 1 and 15,000 mdd. (3.0 ipy.) for plain cast iron. Hydrogen Chloride and Chlorine All of the nickel alloys considered in this bulletin are resistant to dry chlorine and hydrogen chloride, most of them even at considerably elevated temperatures, as indicated in the following section of the bulletin. MONEL alloy 400 is a standard material for trim on chlorine cylinder and tank car valves, for orifice plates in chlorine pipe lines, and is used frequently for parts of chlorine dispensing equipment. Wet hydrogen chloride at temperatures below the dew point will usually behave about the same as concentrated hydrochloric acid as discussed in a preceding section of the bulletin. HASTELLOY alloy B is the most resistant nickel alloy. Wet chlorine at temperatures below the dew point or aqueous solutions containing considerable amounts of free chlorine are highly corrosive to all of these alloys except HASTELLOY alloy C. MONEL alloy 400, nickel and INCONEL alloy 600 frequently can be used with solutions containing 3 gm per liter or less available chlorine in discontinuous operations such as in cyclic textile bleaching with hypochlorite solutions where the bleaching cycle is followed by rinsing and acid “souring” operations in the same vessel. In Temperature Corrosion Rate Solution °C °F mdd. ipy. 5% Ferric Chloride... Room ... 2.0 0.0003 10% Ferric Chloride... Room 2.0 0.0003 10% Ferric Chloride ... 65 149 42.0 0.0068 45% Ferric Chloride... 30 86 59.0 0.0095 45% Ferric Chloride... 65 149 294.0 0.047 5% Ferric Chloride plus 1% HCl ... Room ... 2.0 0.0003 5% Ferric Chloride plus 5% HCl... Room ... 2.4 0.0004 10% Ferric Chloride plus 1 % HCl ... Room ... 2.0 0.0003 10% Ferric Chloride plus 5% HCI... Room ... 2.1 0.0003 10% Ferric Chloride plus 10% HCl.. 70 158 1272 0.21 Page 10 higher concentrations attack is likely to be severe and accompanied by pitting. Short-time tests of MONEL alloy 400 in sodium hypochlorite solutions of 3 gm per liter available chlorine have shown average rates of corrosion of about 0.001 ipy. in contact with the solutions for 8 hours daily over a one-year period in cyclic textile bleaching. Lower or higher rates have been found to apply for shorter or longer periods of daily contact with the solutions. Nickel behaves similarly to MONEL alloy 400 in weak hypochlorite solutions, but is generally inferior to MONEL alloy 400 in concentrations where the available chlorine content is over 3 gm per liter. Tests with INCONEL alloy 600 have indicated that it is somewhat more resistant than either MONEL alloy 400 or nickel to corrosion by hypochlorites, especially where the concentration of available chlorine is over 3 gm per liter. MONEL alloy 400, nickel, INCONEL alloy 600 are resistant to the very dilute hypochlorite solutions, usually containing less than 500 ppm. available chlo-rine used for sterilizing purposes. Prucha 5 reported 5 M. J. Prucha, "Corrosive Action of Washing Powders and Chemical Sterilizers on Metals used in Milk Plant Equipment," Proc. International Assn. Milk Dealers, Plant Sect. 22, 54 (1929). TABLE 10 Effect of Inhibitors on Corrosion by Sodium Hypochlorite Solutions Temperature: 40° C (104° F) Duration of tests: 16 hrs Beaker tests-No agitation MONEL alloy 400 NICKEL INCONEL alloy 600 Solution Composition gm per liter Corrosion Rate Corrosion Rate Corrosion Rate Available Chlorine Sodium Silicate Trisodium Phosphate mdd. ipy. Max. Depth Pitting, inch mdd. ipy. Max. Depth Pitting, inch mdd. ipy. Max. Depth Pitting, inch 6.5 … ... 692 0.113 .011 321 0.052 .022 69 0.012 .027 6.5 0.5 … 107 0.018 .005 64 0.010 .014 15 0.003 .022 6.5 0.5 51 0.008 .006 122 0.020 .024 15 0.003 .017 6.5 2.0 … 13 0.002 .007 9 0.001 None 6 0.001 .008 6.5 ... 2.0 21 0.003 .005 57 0.009 None 7 0.001 .012 3.3 … ... 243 0.040 .007 183 0.030 .014 29 0.005 .032 3.3 0.5 6 0.001 None 22 0.004 None 7 0.001 .008 3.3 ... 0.5 26 0.004 .003 40 0.006 None 6 0.001 .007 0.1 … ... 23 0.004 .003 26 0.004 None 12 0.002 .006 0.1 0.5 … 2 0.0003 None 3 0.0005 None 4 0.0007 .003 0.1 ... 0.5 8 0.0013 None 4 0.0006 None 4 0.0007 None the results of tests in sodium hypochlorite sterilizing solutions shown in Table 9. Inhibitors, such as sodium silicate (water-glass) or trisodium phosphate, have marked effect in reducing corrosion rates of nickel, MONEL alloy 400 and INCONEL alloy 600 in hypochlorite solutions. The inhibitive effect holds for solutions containing as mucn as 6.5 gm per liter available chlorine. Data illustrating the inhibitive properties of these salts in sodium hypochlorite solutions are given in Table 10. It is important to note that, in those solutions containing an inhibitor, corrosion is more uniform and less confined to local areas. Specimens in contact with solutions containing 6.5 gm per liter available chlorine show some susceptibility to local corrosion, even with the inhibitor present, though the tendency toward such attack has been reduced by the inhibitor. With an inhibitor present in concentrations of 3.3 gm per liter available chlorine, or less, the corrosion of MONEL alloy 400 and nickel is uniform, with local attack entirely absent. In laboratory tests, as little as 0.025 cc of sodium silicate solution (1.4 sp gr) per liter of bleaching solution has been found effective in reducing corrosion. HASTELLOY alloy C is outstanding in its resistance to strong chlorine solutions and to wet chlorine gas, in the latter case being limited to temperatures below about 40° C (105° F). Its maximum corrosion rate in wet chlorine gas at room temperature is 0.001 ipy. In water vapors at 170° C (340° F) containing 1000 ppm chlorine, HASTELLOY alloy C corroded at a rate of 0.0003 ipy. In strong bleaching solutions containing over 10% available chlorine, there has been some indication that HASTELLOY alloy C may have a TABLE 9 Corrosion of MONEL alloy 400 and Nickel in Sodium Hypochlorite Sterilizing Solutions Temperature: 25° C (77° F) Corrosion Rate 35 ppm Available Chlorine 100 ppm Available Chlorine 500 ppm Available Chlorine mdd. ipy. mdd. ipy. mdd. ipy. Monel alloy 400. Nickel. . . . . 0 0.61 0 0.0001 1.3 2.1 0.0002 0.0003 5.1 5.1 0.0008 0.0008 Page 11 Chlorine and Hydrogen Chloride at High Temperatures Developments in recent years in the chemical process industries have been in the direction of using increasingly higher temperatures in order to accelerate reactions or to accomplish reactions not feasible at lower temperatures. This trend has applied to processes involving chlorine and hydrogen chloride, as in the production of new organic or inorganic chlorides, and the treatment of titanium and zirconium ores. In many cases the successful development of the process has depended upon finding suitable materials of construction to withstand the high temperature halogen gases and salts. A most commendable corrosion study is that of Brown, DeLong and Auld 6 from whose work much of the following discussion has been abstracted. The corrosion rates of many metals and alloys in dry chlorine and dry hydrogen chloride increase relatively slowly with increase in temperature up to a critical point, which varies with the individual material. Above this point further increase in temperature rapidly accelerates attack. To some degree the corrosion rate is roughly proportional to the vapor pressure of the particular metallic chloride involved, as in the case of nickel, shown in Figure 9. However, may be inherently resistant to the formation of surface chloride films. Still others may ignite in chlorine above a certain temperature with evolution of heat, thus raising both the metal temperature and the rate of reaction. Melting points and vapor pressure data for many metal halides have been published by Quill. 7 The above authors conducted laboratory corrosion tests of a number of materials in dry chlorine and dry hydrogen chloride with test runs of 2 to 6 hours and 10 to 20 hours, using a gas flow rate of 1.3 ft per minute. As indicated in the data for INCONEL nickel-chromium alloy 600 in dry chlorine, Figure 10, the FIG. 9-Corrosion of nickel in dry chlorine and dry hydrogen chloride. the corrosion resistance may not be predicted solely by this method. Some metal chlorides melt or decompose at temperatures at which the vapor pressure is still low, and the reaction can proceed since the protective effect of the coating is lost. Other materials 6 "Corrosion by Chlorine and by Hydrogen Chloride at High Temperatures," M. H. Brown, W, B. DeLong, and J. R. Auld, Ind. do Eng. Chemistry, Vol. 39, No. 7 pp 839-844. 7 L. L. Quill, "Chemistry & Metallurgy of Miscellaneous Materials: Thermodynamics," pp 193-206. McGraw-Hill Book Co. 1950. longer test runs yielded somewhat lower corrosion rates, probably due to the effect of time on the formation of protective films. The results of two hour runs with INCONEL alloy 600 in dry hydrogen chloride are given in Figure 11. FIG. 11-Corrosion of INCONEL alloy 600 in dry hydrogen chloride gas. FIG. 10-Corrosion of INCONEL nickel-chromium alloy 600 in dry chlorine gas. Page 12 From this study the authors prepared Table 11 listing temperatures at which various corrosion rates were exceeded, and suggested upper temperature limits for continuous service. These limits are to be Hoskins Mfg. Co., Trademark. Duriron Co., Inc., Trademark. a It is emphasized that these values are based on short-time laboratory tests under controlled conditions. They should be interpreted only as being indicative of the limitations of the materials and should not be used for estimation of the service life of equipment. b Ignites at about 600° F c Ignites at 450-500° F d Ignites at 400-450° F interpreted as a rough guide of maximum temperature at which the given materials can be used without serious attack in dry chlorine or dry hydrogen chloride. Such values obviously have limitations, for one service application may call for practically no material loss, while another may permit a substantial corrosion allowance. These suggested temperature limits are believed to be conservative, however, for corrosion rates based on longer testing periods might be expected to be considerably lower for many of the materials which develop protective chloride films. In many industrial processes the chlorine or hydro-gen chloride may be neither pure nor dry. Air, moisture, or chemicals may be present to complicate the corrosion action. A limited amount of data dealing 8 Tseitlin, Kh. L. "The Corrosion of Metals in the Presence of Chlorine at Elevated Temperatures." J1. of Applied Chemistry, USSR, Vol. 27, #9, 1954. 9 Friend, W. Z. and Knapp, B. B., Trans. of Amer. Inst. Chem. Engrs., Vol. 39, 731-53 (1943). 10 Heinemann G., Garrison, F. G. as Haber, P. A., Industrial Eng. Chemistry, Vol. 38, 497-9, 1946. TABLE 11 Corrosion of Metals a with the effect of moisture or air dilution which has been reported by Brown et al, 6 Tseitlin' 8 and, Friend and Knapp, 9 will be included in the discussion to follow. Chlorine For service temperatures up to 540°C (1000° F), the most resistant alloys are nickel, INCONEL alloy 600 and the high nickel alloy, HASTELLOY alloy B. Closely following these is HASTELLOY alloy C, useable to 480° C (900° F). MONEL alloy 400 appears to be suitable up to 430° C (800° F), and the chromium-nickel stainless steels up to 345° C (650° F). Both platinum and gold form unstable chlorides in hot chlorine, with the result that their usefulness is restricted to the relatively low temperatures shown in Table 11. Copper, cast iron, and carbon steel should all be restricted to service temperatures of less than 205° C (400° F) in dry chlorine, for they all tend to ignite at somewhat higher temperatures. Tseitlin 8 has reported the ignition of copper at 345° C (650°F) at a low gas flow velocity, and at 260 to 300° C (500 to 575° F) at higher velocities. Brown et al, observed the ignition of steel at 230° C (450° F) towards the end of tests of 16 and 20 hours duration. Heinemann, Garrison and Haber 1 0 reported the ignition of steel at 250° C (484° F) in tests of 30 minutes or less. Tseitlin investigated the effect of velocity of chlorine flow and moisture addition to the gas at temperatures of 260-300° C (500-575° F). Metals and alloys studied were aluminum, carbon steel, copper, Type 347 stainless steel, and nickel. At low velocities (up to 40 ml/min) aluminum was not corroded in dry chlorine, but at the higher velocity (250 ml/min) the reaction proceeded violently so as to melt the sample. With 0.4% mois-ture present in the chlorine the corrosion of aluminum was reduced to a very low rate even at the higher velocity, due apparently to the formation of a protective oxide film. While Tseitlin reported very little corrosion of carbon steel in dry chlorine at the low velocity, ignition occurred within 15 minutes at the higher velocity. Again moisture addition greatly reduced the attack on the steel, due to formation of an oxide film. Copper corroded at a substantial rate at low velocity even when moisture was present, and ignited in the higher velocity tests. Neither velocity nor moisture had any significant effect on Type 347 stainless, which exhibited a low corrosion rate, or on nickel which did not react in Tseitlin’s tests of 6 hours duration. Brown et al, included some studies of the effect of 0.4% moisture in chlorine on the corrosion of an 18% chromium-8% nickel stainless steel. At tem-peratures below 370° C (700° F) the presence of moisture increased the corrosion rate significantly but had no effect at higher temperatures. Tseitlin’s work did not show this effect since his tests were conducted in the temperature range found by Brown where the moisture effect began to disappear. Approx. Temp. at Which Given Corrosion Rate Is Exceeded in Short Time Tests in Dry C12, °F 0.0025 in/mo 0.005 in/mo 0.01 in/mo 0.05 in/mo 0.1 in/mo Suggested Upper Temp. Limit for Continuous Service, °F Corrosion in Dry Chlorine Nickel 950 1000 1100 1200 1250 1000 Inconel alloy 600 950 1000 1050 1200 1250 1000 Hastelloy alloy B 950 1000 1100 1200 ... 1000 Hastelloy alloy C 900 1000 1050 1200 … 950 Magnesium 850 900 950 1000 1050 850 Chromel A 800 900 1000 1150 850 Monel alloy 400 750 850 900 1000 1000 800 18-8-Mo 600 650 750 850 900 650 18-8 550 600 650 750 850 600 Platinum 900 950 1000 1050 1050 500 Hastelloy alloy D 400 450 550 … … 400 Deoxidized copper 350 450 500 500 5506 400 Carbon steel 250 350 400 450 450-400 Cast iron 200 250 350 450 450-350 2S aluminum 250 300 300 350 3504 250 Gold 250 300 350 400 400 ... Silver 100 150 250 450 500 ... Corrosion in Dry Hydrogen Chloride Platinum 2300 ... ... ... ... 2200 Gold 1800 … … … … 1600 Nickel 850 950 1050 1250 1300 950 Inconel alloy 600 800 900 1000 1250 1350 900 Hastelloy alloy B 700 800 900 1200 1300 850 Hastelloy alloy C 700 800 900 1150 1250 850 Hastelloy alloy D 550 700 850 1200 … 800 18-8-Mo 700 700 900 1100 1200 800 25-12-Cb 650 750 850 1050 1150 800 18-8 650 750 850 1100 1200 750 Carbon steel 500 600 750 1050 1150 500 Ni-Resist (Type 1) 500 600 700 1000 1100 500 Monel alloy 400 450 500 650 900 1050 450 Silver 450 550 650 850 … 450 Cast iron 400 500 600 850 950 400 Durichlor 350 400 500 650 750 350 Duriron 350 400 500 650 700 350 Copper 200 300 400 600 700 200 Page 13 Pershke and Pecherkin 1 1 reported results of 15 hour laboratory tests in chlorine containing water vapor to approximately 100% of saturation. Included were a number of copper base and iron base alloys. Most of the copper base alloys showed high rates of attack at 300° C (575° F) and were not considered to have useful resistance above 200° C (390° F). Of the iron base alloys, only stainless steels of the 18-8 chromium nickel variety showed useful resistance at 300° C (575° F), none being suitable at 500° C (930° F). Hydrogen Chloride The performance of materials in dry hydrogen chloride is somewhat analogous to their behavior in dry chlorine, as can be seen in the tabulation of Brown et al, 6 Table 11. Evidence that these temperature limitations for certain materials may be on the conservative side is supported by the more recent data of Savolainen and Blanco, 1 2 Table 12. Their tests of Cast iron and carbon steel appear to be useful in hydrogen chloride to somewhat higher temperatures than in chlorine, and while their corrosion rates increase with temperature of the hydrogen chloride, ignition has not been reported at temperatures up to 760° C (1400° F). Depending upon the corrosion allowance permissible, carbon steel might be useful to temperatures of 260 to 315° C (500 to 600° F). The effect of moisture in hydrogen chloride at elevated temperatures has been reported for a few materials. Friend and Knapp, 9 whose results are shown in Table 13, found that 0.25% moisture in TABLE 13 Corrosion of Metals and Alloys in Hydrogen Chloride at High. Temperatures A. Tests at 205° C (400° F) for 20 hr TABLE 12 Laboratory Corrosion Tests in Anhydrous HCI Gas Temperature: 500° C (930° F) Test Duration: 500 hrs Moisture content of gas was about 0.25% by weight after bubbling through concentrated hydrochloric acid and before passing over the test specimens. hydrogen chloride did not significantly alter the cor-rosion of either carbon steel or nickel in tests con-ducted at 400° F and 1000° F Pershke and Pecherkin 1 1 compared several copper base and iron base alloys in wet hydrogen chloride and wet chlorine, finding lower rates to prevail in wet hydrogen chloride. Brown et al, reported that 0.2% moisture in hydrogen chloride did not appreciably alter the corrosion rate of alloys tested at temperatures above the dew point. The effect of air dilution on the corrosivity of hydrogen chloride is of interest to operators of muri-atic acid furnaces, where air leakage is difficult to exclude completely. At temperatures over 370° C (700° F) the corrosion of cast iron has been observed to increase several fold with some air present, while little or no effect was noted at temperatures of 500 hours duration in anhydrous hydrogen chloride at 500° C (930° F) resulted in corrosion rates con-siderably lower than those observed by Brown. Notable exceptions in the behavior of materials in hydrogen chloride and chlorine are those of platinum and gold, which exhibit such high resistance to hydrogen chloride attack that the chloride salts are not formed. Platinum appears to be outstanding in its resistance, and useable to temperatures of the order of 1200° C (2200° F). Nickel and the high nickel alloys such as INCONEL alloy 600 and HASTELLOY alloys B and C appear to be the most practical alloys for service to temperatures of about 540° C (1000° F), and reference to their uses in industrial applications is mentioned below. While the chromium-nickel family of stainless steels appears to have suitable corrosion resistance at temperatures up to 400 to 425° C (750 to 800° F), one case of stress corrosion cracking of Type 347 stainless steel has been observed in wet mixed gases containing HCl. That certain of the chromium-nickel-iron alloys are suitable is evident by the successful use in muriatic furnace parts of a cast alloy; having a composition of 37% Ni, 27% Cr, 3% Mo, 2% Cu, 0.25% C, balance iron. 11 Pershke, V. & Pecherkin L., Khimistroi, 6, 140-1 (1934). 12 “New Developments in Head End Methods for Preparation of Fuels for Aqueous Processing.” J. E. Savolainen & R. E. Blanco, Chemical Engineering Progress, Vol. 53, No. 2., February 1957, pp 78-81. Corrosion Rates Wet Dry Material mild. ipy. mild. ipy. Carbon Steel . . . . . 5.2 0.0009 5.4 0.0009 Nickel . . . . . . . . . . 2.0 0.0003 1.6 0.0003 Corrosion Rate Material mdd. ipy. Hastelloy alloy B............................... 39 0.0025 Nickel ............................................... 48 0.003 Inconel alloy 600............................... 51 0.003 Type 304L Stainless .......................... 200 0.011 Corrosion Rates, ipy. Weight Losses, mg/sq dm Wet Dry wet Material 4 hr 8 hr 20 hr 47 hr 20 hr 4 hr 8 hr 20 hr 47 hr Carbon Steel .. 0.17 0.21 ... ... 0.25- … … … … 0.42 Nickel ............ 0.12 0.07 0.028 ... 0.037 126 143 140 ... Monel alloy 400 ... … … ... 0.049 ... 124 165 ... 600 Inconel alloy 600 ... … … 0.034 ... ... 110 105 165 ... Hastelloy alloy B ...... 0.14 0.08 ... ... ... ... Type 316 Stainless (18% Cr, 8% Ni, 3% Mo ..... 0.17 0.14 . ... ... .. Platinum ........ Unattacked B. Tests at 538°C (1000°F) Page 14 F I G. 12-Specially designed chlorinating furnace used for extraction of precious and semi-precious metals from ores. Built of nickel and nickel-clad steel, it operates at temperatures up to 425° C (800° F). 315 to 340° C (600 to 650° F). Air dilution had little effect on the corrosion resistance of 25% Cr12% Ni (columbium containing), but did lower the temperature at which appreciable attack began on platinum and gold. Where the major consideration is resistance to corrosion at temperatures below the dew point, but where provision against high-temperature attack is also desired, the indicated order of preference for materials of construction would be platinum, HASTELLOY alloy B, MONEL alloy 400, nickel, INCONEL alloy 600, copper and carbon steel. Nickel 201 and INCONEL alloy 600 would be rated above the HASTELLOY alloys in cases where the major requirement would be resistance to attack at temperatures from 450 to 540° C (850 to 1000° F). In the synthesis of hydrogen chloride from hydro-gen and chlorine, the preferred design of combustion chamber makes use of metal construction with the temperature of the metal being controlled within proper limits by means of water jackets. In this con-nection, it appears that the danger of corrosion by condensed hydrochloric acid at some temperature near the dew point is greater than by dry hydrogen chloride at some elevated temperature. Maude 1 3 has pointed out that the allowable temperature range for any metal or alloy may be contracted at its lower end by the hygroscopic nature of the chloride of the metal concerned; for example, iron should not be used at a temperature less than 30° C (54° F) above the HCl-H 2O dew point, while copper may be used so long as the temperature is 5° C (9° F) above the dew point. In view of the hygroscopic nature of nickel chloride, it would be prudent to assume that hydro-chloric acid might condense on a nickel surface at a temperature as much above the dew point as in the case of iron, i.e. 30° C (54° F). Metallurgical Considerations In the selection of materials for high temperature chlorine or hydrogen chloride service, particularly at temperatures above 370° C (700° F), not only must the corrosion rate of the material be considered, but 1 2 A . H . M a u d e , “ A n h y d r o u s H y d r o g e n C h l o r i d e G a s ” , T r a n s . A m e r . I n s t . C h e m . E n g r s . , V o l . 3 8 , p p 8 6 5 - 8 8 4 ( 1 9 4 2 ) . the effect that the temperature may have on the mechanical properties of the material. It has been shown above that nickel and high nickel alloys have useful corrosion resistance to temperatures of the order of 540° C (1000° F). In considering the use of commercial nickel at elevated temperatures, it is necessary to consider two of its limitations. Nickel 200, having a nominal content of up to 0.10% C., is subject to embrittlement by intergranularly precipitated carbon or graphite when held at temperatures of 425 to 760° C (800 to 1400° F) for extended periods of time. Nickel 201 (low-carbon nickel), having a maximum carbon content of 0.01 to 0.02%, is not subject to such embrittlement, providing carbonaceous materials are not in contact with it. Both Nickel 200 and Nickel 201 are subject to intergranular embrittlement by sulfur compounds at temperatures above about 315° C (600° F). In applications where carbon or sulfur compounds are present, or where high temperature strength is required, INCONEL alloy 600 is frequently substituted for Nickel 200 or Nickel 201. At temperatures too high for the suitable performance of metals, use is sometimes made of fused silica, silica brick linings or graphite. Applications Nickel and INCONEL nickel-chromium alloy 600 have both been successfully used for chlorination equipment at temperatures up to 540° C (1000° F). The nickel-clad pressure vessel shown in the photograph on page two, operates at 340° C (650° F), and the chlorinating furnace equipment illustrated in Figure 12, consisting of both nickel and nickel-clad steel, operates at 425° C (800° F). Where arrangements can be made for cooling the metal surface, nickel and INCONEL alloys have been used at even higher gas temperatures. For example, chlorine at 1100° C (2000° F) is being handled in a jacketed nickel tube cooled with high velocity water; chlorine at 760° C (1400° F) is Page 15 in the combustion of chlorine and hydrogen to form HCl. INCONEL nickel-chromium alloy 600 has been em-ployed in the treatment of an organic material with HCl at 700° C (1300° F), and platinum linings are employed in the thermal cracking of a chlorinated organic forming HC1 at a temperature of 750° C (1380° F). In the Kroll process or similar processes for the production of zirconium and titanium, a first step is the chlorination of the ores to form ZrC14 or TiC14, which is subsequently purified and reduced to the metal sponge by reaction with magnesium or sodium. The chlorinating reactor, where temperature may rise to as high as 900° C (1650° F), is usually lined with silica brick. INCONEL alloy 600 or Nickel 201 are used for heating the incoming chlorine to about 400° C (750° F). ZrC14 or TiC14 vapors are condensed in an initial condenser made of nickel at 95 to 205° C (200 to 400° F), and sometimes an after condenser of steel operating at 150° C (300° F) or lower. Purification of ZrCl4 is carried out by placing the metal salt in INCONEL alloy 600 cans and heating to about 1200° F in a retort made of Type 309 (25 Cr-12 Ni) or Type 310 (25 Cr-20 Ni) stainless steel. The sublimed chloride collects on cooled coils at the top of the retort and is allowed to drop into alloyed steel reduction crucibles frequently located in the same retort. In corrosion tests of 18 hours in ZrCl4 at 845° C (1550° F), the corrosion rate of INCONEL alloy 600 was 0.0035 inches per month and that of nickel was 0.017 inches per month. Nickel 201 has been used for cylindrical retorts for the sublimation of ZrCl4 at temperatures of 430 to 540° C (800 to 1000° F). Plant Corrosion Tests When dealing with straight hydrochloric acid solu-tions or mixtures where known concentrations of acid are added, the choice of suitably resistant materials often can be based on the laboratory test data in pure solutions. Frequently it is more difficult to predict which will be the most economical material for reac-tions in which hydrochloric acid is formed by the hydrolysis of chlorides or chlorinated compounds in the presence of water, or by the presence of small amounts of water in reactions involving hydrogen chloride. In such cases it is usually advisable to make corrosion tests in plant or pilot-plant equipment under actual operating conditions. Numerous plant corrosion tests have been made in such processes and the results of some of the typical tests are presented in the following pages to show the 1 4 H. E. Searle and F. L. LaQue, “Corrosion Testing Methods,” Pro-ceedings Am. Soc. Testing Mat. 35, 249 (1935). 1 5 Am. Soc. Testing Mat. Standards, Part 3, p. 257, (1958). comparison in performance between the various nickel alloys and other materials. The testing device used in most of these tests was the spool-type specimen holder described by Searle and LaQue. 1 1 This method of testing is substantially in accord with the A.S.T.M. Recommended Practice for Conducting Plant Corrosion Tests, A 224-46 1 5 Briefly, the assembly consists of previously cleaned and weighed speci-mens, in duplicate, of the several metals to be tested, mounted on a spool-type holder, with non-metallic parts of bakelite, other organic plastics or porcelain to separate and insulate the specimens from each other, from the metal strengthening members of the device, and from the equipment in which the test is made. Details are shown in Figure 13. The completed FIG. 13-Spool-type specimen holder. assemblies are installed in the desired locations, and allowed to remain for a sufficient length of time to give reliable indications of corrosion behavior. Upon completion of a test, the assembly is removed, dis-mantled, and the specimens examined, cleaned of all adhering corrosion product, and reweighed. From the known weight losses, areas of specimens, and duration of tests, the corrosion rates are calculated. In addition, the specimens are examined for pitting or other local attack. Chlorinated Solvents Chlorinated hydrocarbon solvents are used exten-sively in dry cleaning, metal degreasing, and a wide variety of solvent extraction operations in the process industries. When free from water and at atmospheric temperature, they are not corrosive and steel equip-ment ordinarily can be used to handle them. However, in the presence of entrained water or a water layer, and, particularly, at elevated temperatures, such as are encountered in the distillation and re- Page 16 covery of the solvents, there is appreciable hydrolysis with resulting formation of dilute hydrochloric acid. The development of acidity is accelerated by light and air. Under such conditions, steel frequently is rapidly attacked. Over a period of many years of service MONEL alloy 400 and nickel have shown a high degree of resistance to corrosion under these conditions with a wide variety of chlorinated hydro-carbons. Both materials are used for distillation and recovery equipment in the manufacture and applica-tion of the solvents. Brallier 1 6 found that the rate of decomposition of carbon tetrachloride by hydrolysis in the presence of water is extremely slow provided that no metal that is attacked by the mixture is present. Carbon tetra-chloride was allowed to boil continuously for a week under a layer of water in glass apparatus without the development of any evidence of acidity. A similar test with a sample of steel in the mixture resulted in the decomposition of 1.2 per cent of the carbon tetra- Additional tests have been made by Brallier 17 by exposing specimens to carbon tetrachloride-water mixtures boiling under reflux with the results shown in Table 14. Brallier 17 made a further study of the resistance of metals and alloys to corrosion by several different chlorinated solvents and mixtures of the solvents with water, both at room temperature and boiling. Tests in boiling mixtures of solvent and water exposed to light for prolonged periods constituted exceptionally severe conditions of exposure. Pure nickel was found to be the best of the several materials tested, with MONEL alloy 400 a very good second. The results obtained are reported in Table 15. In room temperature tests, specimens were exposed to solvent and air, with water absent, and to solvent and air with water present. In both cases, 90 per cent of the area of the specimen was in contact with the solvent. TABLE 14 Effects of Metals on the Decomposition of Boiling Carbon Tetrachloride in the Presence of Water Experimental Conditions: Amount of carbon tetrachloride used: 480 to 560 gm Duration test: 7 days Metal exposure: 25% of area to liquid carbon tetrachloride 25% of area to liquid water 50% of area to mixed vapors Material Total Area, of Specimen, sq cm Loss of Weight. of Specimen, mg Weight of Chlorine as Chloride from Hydrolysis, mg Blank ............. ... .... 31 Blank ............. ... .... 30 Blank ............. ... .... 26 Steel .............. 57.5 3517 3829 Steel .............. 42.8 2117 2237 Steel .............. 42.8 4419 5022 Steel .............. 42.8 4933 5913 Nickel ............ 53.4 42 87 Nickel ............ 54.4 18 54 Nickel ............ 52.0 63 89 Nickel ............ 52.0 443 504 Monel alloy 400 ... 54.4 103 148 Monel alloy 400 ... 53.9 44 95 Monel alloy 400 ... 52.5 62 93 Monel alloy 400 ... 53.5 57 105 chloride to form hydrochloric acid. When nickel was substituted for the steel, and the test repeated, the extent of formation of hydrochloric acid was only 0.0067 per cent of the carbon tetrachloride. 1 6 P. S. Brallier, “Properties of Carbon Tetrachloride,” Laundry Age, Sept. 1, 1931, p. 88. 1 7P. S. Brallier, Private Communication (1934). 1 8P. S. Brallier, Private Communication (1934). FIG. 14-Dry cleaning system employing chlorinated solvents. All parts in contact with solvent are made of MONEL alloy 400 or nickel. In tests at the boiling point, where specimens were exposed to solvent and vapor with water absent, about 25 per cent of the area of each specimen was in contact with the solvent and 75 per cent with the vapor. In tests with water present, about 25 per cent of the area of each specimen was exposed to solvent, 25 per cent to the water layer and the remaining 50 per cent to the vapors. In the rectification of chlorinated solvents during their production, corrosive conditions are usually most severe in the still or reboiler, or in the bottom sections of distilling columns where water tends to accumulate. This is indicated by the results of plant corrosion tests in a continuous process for rectifica- Page 17 tion of crude carbon tetrachloride, given in Table 16 and in a batch process for rectification of crude trichlorethylene, given in Table 17. It will be noted from Table 16 that, in addition to MONEL alloy 400, nickel and INCONEL alloy 600 and NI-RESIST Types 1, 2 and 3 have useful resistance to the corrosive condi- tions encountered, and these materials are used where a cast material is required for valves, pumps and fittings, and in some cases for cast bubble caps and plates. Common applications of MONEL alloy 400 and nickel are in the construction of dry cleaning ma- TABLE 15 Results of Corrosion Tests in Chlorinated Solvents and Their Vapors Mixture contained 90 per cent carbon tetrachloride and 10 per cent ethylene dichloride by volume. CORROSION RATE Test 1 Test 2 Test 3 Test 4 Material mdd. ipy. mdd. ipy. mdd. ipy. mdd. ipy. Monel alloy 400........................... 1.7 0.0003 1.1 0.0002 1.7 0.0003 5.3 0.0009 Nickel ........................................ 1.3 0.0002 1.1 0.0002 1.9 0.0003 2.1 0.0003 Inconel alloy 600 ......................... 1.0 0.0002 0.9 0.0002 2.2 0.0004 3.3 0.0006 Ni-Resist (Type 3)........................ 3.0 0.0005 2.8 0.0005 4.5 0.0009 6.6 0.0013 Ni-Resist (Type 1)........................ 2.8 0.0005 1.9 0.0004 6.8 0.0013 10.7 0.0021 Mild Steel ................................... 43.8 0.0081 25.2 0.0046 x 1180 0.217 Cast Iron..................................... 36.5 0.0073 25.8 0.0052 800 0.160 640 0.128 x Specimens completely corroded away before end of test. Original thickness of specimens 0.060 inch. CORROSION RATE Tests at 25-30°C (77-86°F) Tests at Boiling Point Water Layer Present Water Layer Absent Water Layer Present Water Layer Absent Solvent mdd. ipy. mdd. ipy. mdd. ipy. mdd. ipy. a. NICKEL.............................. Carbon Tetrachloride .............. 0.12 0.00002 0.02 0.000003 11.1 0.0018 0.16 0.00003 Chloroform ............................. 0.36 0.00006 0.17 0.00003 0.73 0.0001 1.3 0.0002 Ethylene Dichloride................ 0.08 0.00001 0.04 0.000007 2.2 0.0004 0.20 0.00003 Trichlorethylene...................... 2.3 0.0004 0.94 0.00015 5.9 0.001 0.14 0.00002 Carbon Tetrachloride Ethylene Dichloride } ........ 0.02 0.000003 0.02 0.000003 1.09 0.0002 0.34 0.00006 b. MONEL alloy 400 .............. Carbon Tetrachloride .............. 0.66 0.0001 0.06 0.00001 27 0.0044 0.23 0.00004 Chloroform ............................. 0.10 0.00002 0.08 0.00001 27.3 0.0045 0.93 0.0002 Ethylene Dichloride................ 0.15 0.00002 0.06 0.00001 16.5 0.0027 0.17 0.00003 Trichlorethylene...................... 4.1 0.0007 0.45 0.00007 67.1 0.011 0.36 0.00006 Carbon Tetrachloride Ethylene Dichloride } ........ 0.12 0.00002 0.06 0.00001 6.1 0.0010 0.59 0.0001 c. MILD STEEL ..................... Carbon Tetrachloride .............. 44.3 0.0082 0.35 0.00006 874 0.16 0.30 0.00006 Chloroform ............................. 12.9 0.0024 3.7 0.0007 65.9 0.012 46.8 0.0086 Ethylene Dichloride................ 7.6 0.0014 0.30 .000006 73.4 0.0135 9.0 0.0016 Trichlorethylene...................... 7.0 0.0013 3.3 .00006 37.8 0.007 . . . . . . Carbon Tetrachloride Ethylene Dichloride } ........ 48.4 0.0089 0.04 .0000007 1079 0.20 2.5 0.0005 TABLE 16 Results of Plant Corrosion Tests in a Continuous Process for the Distillation and Rectification of Crude Carbon Tetrachloride Duration of Tests: 133 days Test 1-Suspended in vapor stream above top plate of bubble cap rectification column. Approximate analysis of vapor: Carbon Tetrachloride 99.398% Sulfur Chloride 0.600% Carbon Bisulfide 0.002% Temperature: 78° C (171° F) Test 2-Suspended from Plate No. 19 of column, which is three plates above feed plate. Temperature: 80° C. (176° F) Test 3-Suspended in vapor space immediately below bottom plate of column. Vapor approximately 98% sulfur chloride. Temperature: 125° C (257° F) Test 4-Suspended in reboiler for column, completely immersed in liquid with following approximate analysis: Sulfur Chloride 98.3% Carbon Tetrachloride 0.13% Iron as FeCl, 0.01% Temperature: 138° C (280° F) Page 18 CORROSION RATE Test 1 Test 2 Test 3 Test 4" Material mdd. ipy. mdd. ipy. mdd. ipy. mdd. ipy. Nickel ................... 8.7 0.0015 0.6 0.0001 0.7 0.0001 9.9 0.0016 Monel alloy 400 ...... 9.2 0.0014 0.7 0.0001 0.7 0.0001 5.7 0.0009 Inconel alloy 600 .... 5.6 0.0009 0.2 <0.0001 0.1 <0.0001 1.9 0.0001 Hastelloy alloy B .... 6.8 0.0011 0.5 <0.0001 0.4 <0.0001 3.8 0.0006 Hastelloy alloy C .... 0.5 <0.0001 0.04 <0.0001 0.2 <0.0001 0.3 <0.0001 Copper ................... x x x x x x x x < = Less than. x Specimens completely corroded away before end of test. Original thickness of specimens 0.031 inch. TABLE 18 Results of Corrosion Tests in Solvent Reclamation Unit of Dry Cleaning Machine Handling Carbon Tetrachloride CORROSION RATE Average Temp. Nickel MONEL alloy 400 NI-RESIST (Type 1) Point of Test Duration of Test, Days °C °F mdd. ipy. mild. ipy. mdd. ipy. Main Still Liquid........................................... 66 77 170 0.21 0.00003 0.08 0.00001 5.0 0.0011 Main Still Vapor ........................................... 66 71 160 0.11 0.00002 0.15 0.00003 1.8 0.00035 Auxiliary Condenser Tubes......................... 96 71 160 9.5 0.0015 8.9 0.0015 . . . . . . . . . Auxiliary Condenser Shell (Bottom)......... 128 66 150 1.6 0.0003 3.1 0.0005 . . . . . . . . . Sump Tank.................................................... 66 27 80 0.43 0.00007 0.69 0.0001 6.4 0.0012 Main Storage Tank ....................................... 98 ROOM 1.0 0.00016 1.4 0.0002 11.2 0.0022 Specimens were in the form of pieces bent so as to clamp over the tubes, from which they were insulated by means of a lacquer on the under side of the specimens. These Specimens were circular discs laid at the bottom of the condenser shell underneath the tubes and insulated from the shell by means of a lacquer. chines, for such parts as solvent recovery stills and condensers, washer cylinders, water separators, lint traps and solvent storage tanks. Such machines use a variety of solvents including carbon tetrachloride, trichlorethylene, perchlorethylene and solvent mixtures. From the clothes, the solvents pick up water, greases, metallic salts and other impurities which frequently tend to cause hydrolysis. The results of corrosion tests in the solvent reclamation unit of a dry cleaning machine using carbon tetrachloride are given in Table 18. The results of corrosion tests in the liquid and vapor sections of the solvent recovery still of a dry cleaning machine using perchlorethylene as solvent, are given in Table 19. For metal degreasing applications, a variety of chlorinated solvents are used including trichlorethylene, carbon tetrachloride, perchlorethylene, ethylene dichloride, propylene dichloride, ortho dichlorben- zene, and others. MONEL alloy 400 and nickel are used for degreasing tanks and accessories as well as for solvent recovery equipment. TABLE 19 Results of Corrosion Tests in Solvent Reclamation Equipmen of Dry Cleaning Unit Handling Duration of Tests: 64 days Test 1: In vapors at 127° C (260 ° F) Test 2: In liquid at 152 ° C (305 ° F) Corrosion Rate Test 1 Test 2 Material mdd. ipy. mild. ipy. Monel alloy 400........ 0.83 0.0001 1.4 0.0002 Nickel ....................... 0.64 0.0001 1.3 0.0002 Inconel alloy 600 ...... 0.11 <0.0001 0.14 <0.0001 Ni-Resist (Type 1)..... 20.1 0.0039 5.7 0.0011 Mild Steel ................. 86.4 0.016 27.2 0.005 Cast Iron ................... 56.1 0.011 18.7 0.004 < = Less than. TABLE 17 Results of Plant Corrosion Tests in a Batch Process for the Distillation and Rectification of Crude Trichlorethylene Duration of Tests: 228 days Test 1-Suspended in pot still, alternately exposed to trichlorethylene liquid plus water, and trichlorethylene vapor plus steam. Temperature: 100° C (212° F) Test 2-Suspended in vapor space above Plate No. 4 of bubble cap rectification column, exposed to vapor and entrained liquid, plus steam. Temperature: 76 to 93° C (169 to 198° F) Test 3-Suspended in vapor space above Plate No. 14 of column, exposed to vapor and entrained liquid. Temperature: 73° C (163° F) Test 4-Suspended in vapor space above top plate, No. 25, of column. Reflux enters column on this plate. Temperature: 68 to 70° C (154 to 158° F) Page 19 FIG. 15-Completely automatic unit built entirely of MONEL alloy 400 for recovery of chlorinated solvents from the solvent degreasing of metal parts. In the solvent extraction of animal and vegetable oils, several chlorinated solvents are used. A typical example of the use of MONEL alloy 400 in this service is for solvent recovery apparatus in the production of cod liver extract using ethylene dichloride as solvent. Organic Chlorinations In the field of organic chlorinations, a wide range of corrosive conditions is encountered. In some chlorinations, the chlorine as well as all the other reacting materials are charged to the reactor in dry or essentially dry condition. In others the presence of water may be desirable. In some chlorinations, the presence of even small amounts of iron is highly detrimental to the process from the standpoints of undesirable catalytic effect, or effect on color of the product. In others iron is a necessary catalyst in the reaction. Some chlorinations are carried out at room temperature and others at considerably elevated temperatures. In view of the many factors involved, consider-, able care must be used in the selection of chlorinators and other process equipment. A considerable number of organic chlorinations are in the group of those where the raw materials are dry or essentially so, and the presence of iron salts is detrimental to the process or the product. In this class of chlorinations, nickel, nickel-clad steel, MONEL alloy 400 and sometimes 1NCONEL alloy 600 are used for chlorinators. They have suitable resistance to corrosion and the metals have no detrimental catalytic effect on the process or on the color of the product. Nickel 200 or MONEL alloy 400 reaction vessels, coils and agitators are being used successfully in the chlorination of paraffin and olefine hydrocarbons, benzol, aniline and a number of other organic compounds. In some instances, steel chlorinators have been lined with Nickel 200 or MONEL alloy 400 sheet, thus greatly increasing the useful life of the equipment. Examples of the corrosion resistance of several materials in dry chlorination reactions at moderate and elevated temperatures are given in Table 20. It will be noted that as the temperature of operation is increased, the corrosivity increases, with the best corrosion resistance being provided by the high nickel alloys. Experimental work with heptane-chlorine mixtures at temperatures up to 300° C (575° F) and pressures up to 1000 lb per sq in showed that with metals and alloys containing considerable amounts of iron, ferric chloride was formed, which catalyzed the decomposition of the chloro-hydrocarbons with formation of undesirable hydrogen chloride, polychlorides, and gum. A pure nickel reaction tube functioned satisfactorily from the start, and the substitution of MONEL nickel-copper alloy 400 for other ferrous alloy parts of the apparatus proved satisfactory. FIG. 16-Nickel 200 vessel used for the chlorination of paraffin. Page 20 < - Less than. In wet chlorinations HASTELLOY alloy C can be used frequently although it is subject to the tempera-ture limitations described on page 10. The production of hexachlorethane by direct chlorination of perchlor-ethylene in the presence of water is an example of a particularly corrosive condition. In Table 21 are TABLE 21 Plant Corrosion Test in Wet Recirculated Perchlorethylene Containing 2 to 16 Per Cent Hexachlorethane and Saturated with Chlorine Temperature: Atmospheric to 25°C (77° F) Duration of test: 50 days a Specimens originally 0.031 inch thick. b Specimens originally 0.188 inch thick. given the results of a plant test in recirculated wet perchlorethylene containing 2 to 16 per cent hexachlorethane and saturated with chlorine gas at approximately 25° C (77° F). At higher temperature of operations wet chlorina-tions may be so extremely corrosive that metallic materials of construction can not be justified, and materials such as glass lined steel become the pre-ferred construction materials. An example of such a condition is shown in Table 22 in a process for the wet chlorination of ethanol at 93° C (200° F). TABLE 22 Plant Corrosion Test in Wet Chlorination of Ethanol Mixture of ethanol, chlorine, HCl, chlorinated products, and water at 48°Be. Test made in liquid phase. Temperature: 93°C (200°F) Duration of Test: 20 hrs Corrosion Rate Material mdd. ipy. Monel alloy 400 .............................. 12,700 2.07 Nickel ............................................ 3,100 .50 Inconel alloy 600 ........................... 13,100 2.22 Hastelloy alloy B............................ 3,500 .54 Hastelloy alloy C............................ 3,600 .58 Mild Steel ...................................... 25,200 4.8 Ni-Resist (Type 1) .......................... 26,400 5.2 Cast Iron ........................................ 28,500 5.7 In reactions involving wet hydrogen chloride with no free chlorine present HASTELLOY alloy B is usually the most resistant material. The above corrosion data illustrates the wide range of corrosion conditions which may prevail in chlorina-tion reactions. Prior to the selection of materials for process equipment for a new chlorination process it is advisable to conduct corrosion tests during the laboratory or pilot plant stage of the development. TABLE 20 Results of Plant Corrosion Tests in Dry Chlorination Processes Test 1-Plant corrosion test in chlorination of acetylene to make acetylene tetrachloride. Mixture of crude acetylene tetrachloride with 10 to 21 gm per liter excess chlorine and 0.15 to 1.65 gm per liter dissolved iron. Acidity as HCl, 0.7 to 6.6 gm per liter. Temperature: 42.3 to 48.9° C (Ave. 46° C) 108 to 120° F (Ave. 115° F) Duration of test: 282 days. Velocity: Approx. 0.5 ft per sec. Test 2-Plant corrosion test in chlorination of phenol. Mixture of phenol, chlorophenol, chlorine and HC1 gas, and ferric chloride. In liquid phase. Temperature: 50-60° C (123-140° F). Test duration: 19 days. Test 3-Same as Test 2. In vapor phase. Test 4-Plant corrosion test in chlorination of naphthalene. Mixture of chlorine, HCl, and naphthalene chloride. Specimens exposed in splash zone. Temperature; 165° C (330° F). Test duration: 52 days. Test 5-Plant test in chlorination of methane. Specimens exposed in bottom of chlorinator. Temperature: 230° C (446° F). Test duration: 9 days. CORROSION RATE Test 1 Test 2 Test 3 Test 4 Test s Material mdd. I ipy. mdd. ipy. mdd. ipy. mdd. ipy. mdd. ipy. Monel alloy 400 ......... 0.25 <0.0001 43 0.007 12.0 0.002 3.6 0.0006 37 0.006 Nickel........................ 0.27 <0.0001 43 0.007 6.2 0.001 2.0 0.0003 12.4 0.002 Inconel alloy 600........ 0.79 0.0001 50 0.009 5.6 0.001 6.3 0.0011 14.8 0.0025 Hastelloy alloy B........ … ... 45 0.007 13 0.002 1.5 0.0002 ... ... Hastelloy alloy C........ 3.0 0.0005 4.4 0.0007 6.2 0.001 5.8 0.001 7.5 0.0012 Ni-Resist (Type 3) ....... 3.5 0.0007 ... ... ... . . . . . . . . . 222 0.043 Ni-Resist (Type 1) ...... 5.6 0.0011 97 0.019 92 0.018 31.2 0.006 1430 0.283 1% Ni Cast Iron.......... 18.6 0.0037 … … … … … … 16000 2.92 Cast Iron .................... ... ... 215 0.043 165 • 0.033 65.5 0.013 ... … Mild Steel .................. ... ... 164 0.030 186 0.034 134 0.025 ... ... Corrosion Rate Max. Depth Pitting, Material mdd. ipy. inch Monel alloy 400.......... 442 0.072 None Nickel ........................ 216 0.035 Perforated a Inconel alloy 600........ 292 0.049 .008 Hastelloy alloy B........ 431 0.067 None Hastelloy alloy C........ 205 0.033 None Ni-Resist (Type 3) ....... 513 0.097 None Ni-Resist (Type 1) ...... 722 0.140 None Tank Steel.................. 1940 0.36 Perforated Cast Iron ................... Completely graphitized b Page 21 Synthetic Resins and Rubbers In the production of phenolic resins, hydrogen chlo-ride is used frequently as a catalyst. A common con-centration of HCl is about 0.1 per cent by weight of the charge. The raw materials used are essentially dry, and the temperatures involved are usually high enough that the hydrogen chloride and any moisture present, which tend to concentrate in the vapor phase in the top of the reactor, are in vapor form so that corrosive conditions are often not severe. However, iron salts usually have an objectionable effect in catalysis and upon the color of the resin product, and the possibility that condensed moisture will sometimes be present is such that steel often is not a satisfactory material for reactors. Nickel and nickel-clad steel are used for resin reactors and autoclaves and for condensers, vapor and condensate lines associated therewith, because of the resistance of nickel to the hydrochloric acid conditions which sometimes may occur, its beneficial effect on the color of the product, and its relatively good thermal conductivity. MONEL alloy 400 and INCONEL alloy 600 also are used for reactors in some cases. Table 23 shows the results of corrosion tests in resin reactors in four different plants. In Test 3, it will be noted that the temperature was low enough and the moisture content high enough for appreciable hydrochloric acid condensation to take place in the vapor space and that MONEL alloy 400 was more resistant than nickel. Test 4 is an example of a case where INCONEL alloy 600 might be preferred because of its resistance to the organic materials at the high temperature used. †Wrought age-hardeaable high-nickel alloy applicable where higher hardness or strength are required than are available with nickel. In the production of synthetic resins and rubbers, involving the use of chlorinated hydrocarbon monomers, polymers or reagents, nickel and MONEL alloy 400 are used in applications where processing may involve the formation of dilute hydrochloric acid solutions by hydrolysis or the accumulation of small concentrations of hydrogen chloride in water wash solutions. The following will indicate some of the products or processes in which such applications exist: Vinyl chloride and vinylidene chloride polymers and copolymers involving vinyl chloride, vinylidene chloride, ethylene dichloride and trichlorethane, Chloroprene polymers Thio-rubbers involving ethylene dichloride Chlorostyrene polymers Chlorinated rubber Chlorinated paraffins Silicone resins involving silicon tetrachloride DURANICKEL alloy 3011 dies and other parts of nickel are frequently used in the forming of shapes and fabrication of vinylidene and vinyl chloride plastics. In the production of styrene, nickel and MONEL alloy 400 are sometimes used for covers of alkylation reactors in the synthesis of ethyl benzene using aluminum chloride and hydrogen chloride as catalysts. As a general rule corrosive conditions are less severe in the vapor sections of these reactors than where the metal is in contact with the liquid catalyst complex, which may be corrosive if significant amounts of moisture are present. In this case HASTELLOY alloy B is used. TABLE 23 Plant Tests in Vapors from Phenol-Formaldehyde Resin Reactors Where Hydrogen Chloride Used as Catalyst Test 1-In hydrogen chloride vapor plus steam, in vapor line from reactor. Concentration of HCl in vapor varied from 2 to 5 per cent. Temperature: 149 to 204° C (300 to 400° F) Duration of test: 46 days Test 2-In vapor containing dilute HCl and steam, in vapor portion of reactor. Temperature: 121 to 149° C (250 to 300° F) Duration of test: 21 days Test 3- In vapor space of reactor. Hydrogen chloride used as catalyst to extent of 0.1 to 0.175 per cent by weight of charge. Temperature: 99 to 104° C (210 to 220° F) Duration of test: 57 days Test 4-In resin reactor just above HCl gas inlet. Temperature: Up to 350° C (660° F) Duration of test: 864 hours CORROSION RATE Test 1 Test 2 Test 3 Test 4 Material mdd. ipy. mdd. ipy. mdd. ipy. mdd. ipy. Monel alloy 400........... Nickel ........................ Inconel alloy 600 ......... Hastelloy alloy B ......... Mild Steel ................... Ni-Resist (Type 1) ....... Cast Iron .................... 20 10 8 … 46 ... ... 0.003. 0.002 0.001 0.009 ... ... … 10 6 5 3 … 26 2161 0.002 0.001 0.001 0.0005 … 0.005 0.43 75 567 316 ... ... ... ... 0.012 0.092 0.048 ... ... ... ... 113 63 16 15 ... 44 132 0.018 0.010 0.003 0.002 … 0.009 0.064 Page 22 FIG. 17-One of the large companies transports bulk quantities of benzyl chloride in this 2,135-gallon capacity, two-section tank truck. Tanks are constructed from solid Nickel to protect the purity-sensitive chemical from contamination by corrosive attack. Other Organic Chlorides Nickel, MONEL alloy 400, INCONEL alloy 600 and NI-RESIST equipment can be used with numerous other non-oxidizing organic chlorides in which some hydrolysis can be expected to occur at elevated temperatures. For example, MONEL alloy 400 and Nickel 200 are used in handling amyl chloride, and Nickel 200 condensers are used for condensing allyl chloride. Nickel 200 shipping drums and tank trucks are employed for transporting both benzyl chloride and benzoyl chloride. Monochlorobenzene is usually made by treating benzene with chlorine in a steel chlorinator since both constituents are available in the water free condition. However, if traces of water are present in chlorobenzenes, some hydrolysis occurs, with the result that steel or cast iron is attacked at significant rates, as shown in the plant corrosion test data, Table 24. Nickel and MONEL alloy 400 have found uses for handling chlorobenzenes containing moisture. Plant Corrosion Test in Monochlorobenzene, Containing Traces of Water and Phenol Temperature: 60-70°C (140-158°F) Test Duration: 57 Days Slight Aeration and Agitation TABLE 24 FIG. 18-Evaporator for the concentration of pectin containing a small amount of hydrochloric acid. The evaporator body is built of 15% INCONEL alloy 600-clad steel 5/16 in and 3/8 in thick. Tube sheets and tubes are of solid INCONEL alloy 600. Material Corrosion Rate mdd. ipy. Monel alloy 400 .................... 58 0.009 Nickel.................................. 51 0.008 Inconel alloy 600 .................. 42 0 007 Copper ................................. 116 0.019 Ni-Resist (Type 1) ................ 148 0.029 Cast Iron .............................. 721 0.145 Mild Steel ............................ 627 0.115 Page 23 Chloroacetic acid is frequently a difficult material to handle from a corrosion standpoint. Nickel 200 and MONEL alloy 400 have been found to be suitably resistant under some conditions. Where conditions are too severe for these materials, HASTELLOY alloy C usually will be satisfactory. Calcott, Whetzel and Whittakerl 19 report the corrosion test data given in Table 25. TABLE 25 Laboratory Corrosion Tests in Chloroacetic Acid Having Specific Gravity of 1.358 at 50° C Temperature of tests: 100° C (212° F) INCONEL alloy 600 is used for the extraction and concentration of acidified vitamins. Pickling of Iron and Steel Pickling of steel strip or steel parts, particularly where coatings of enamel, zinc, tin or plated metals are to be applied subsequently, is done often in hydrochloric acid solutions. The most common acid concentrations are from 5 to 10 per cent HCl by weight. Where pickling is done by dipping, the acid is used frequently at room temperature, but higher temperatures are used sometimes for continuous processes where it is desired to speed up the reaction. Frequently organic inhibitors are added to the acid baths. The temperature of the solution has an important bearing on the selection of metals and alloys for pickling equipment. For room temperature pickling with hydrochloric acid solutions up to 10 per cent HCl concentration, MONEL alloys and nickel have good corrosion resistance. MONEL alloy 402 is used widely for pickling equipment such as baskets, crates, racks, chains, hooks and even for solid metal pickling tanks. It is also used for bolts, nuts, stay-rods, and other parts supporting wooden pickling tanks. Pickling conditions are particularly favorable because the pickling reactions use up any oxygen that may be dissolved in the acid, and the hydrogen evolved by reaction of the acid with the steel tends to keep the solution in a reducing condition. In addition, MONEL alloy 402 crates, baskets, and chains are protected galvanically by the steel parts with which they are in contact. Consequently, corrosion rates are very low, and it is not unusual to find MONEL alloy 402 pickling crates which show no appreciable corrosion after 10 to 15 years service in room temperature pickling solutions. Solid MONEL alloy 400 pickling tanks also have shown good performance handling room temperature solutions. At one plant, where range parts are pickled Material Corrosion Rate mdd. ipy. Nickel ............................................ 142 0.023 Copper ........................................... 1620 0.262 Lead............................................... 5170 0.656 Mild Steel ...................................... 2030 0.378 Cast Iron ........................................ 1430 0.286 The results of other laboratory tests in molten 100 per cent chloroacetic acid at 70° C (158° F) and in 90 per cent chloroacetic acid solution at 23° C (73° F) are given in Table 26. Dilute hydrochloric acid conditions are encountered in the distillation and processing of some foods and other organic materials which have been extracted with hydrochloric acid. MONEL alloy 400, nickel and INCONEL alloy 600 are frequently used in such processes to provide corrosion resistance and protect the purity of the products. As examples, nickel and INCONEL alloy 600 evaporators are used with acidified gelatine, MONEL alloy 400 and INCONEL alloy 600 evaporators are used with acidified pectin, and 19 W. S. Calcott, J. C. Whetzel, and H. F. Whittaker, "Corrosion Tests and Materials of Construction for Chemical Engineering Apparatus," D. Van Nostrand Co., New York (1923), p. 118. TABLE 26 Laboratory Corrosion Tests in Chloroacetic Acid Test 1: Molten pure chloroacetic acid at 70° C (158° F). Test specimens submerged (and half-submerged in the case of nickel) in the liquid in closed Mason jars. Test 2: Saturated aqueous solution containing 90 parts of chloroacetic acid and 10 parts of water at 23° C (73° F). Test specimens submerged (and half-submerged in the case of nickel) in liquid in closed Mason jars. CORROSION RATE Test 1 Test 2 First 40 hr Next 64 hr Next 42 hr First 97.5 hr Next 44 hr Material mdd. ipy. mdd. ipy. mdd. ipy. mdd. ipy. mdd. ipy. Monel alloy 400 ... 18 0.003 30 0.005 61 0.010 5.0 0.0008 6.1 0.0010 Nickel Totally . . . . . . 111 0.018 93 0.015 111 0.018 5.6 0.0009 5.0 0.0008 Half submerged 130 0.021 111 0.018 148 0.024 22 0.0036 21 0.0034 Inconel alloy 600 . 30 0.005 137 0.023 131 0.022 5.0 0.0008 5.0 0.0008 Hastelloy alloy B . 2.2 0.00034 2.2 0.00035 8 0.0012 2.6 0.0004 4.5 0.0007 Hastelloy alloy C . 1.7 0.00028 0.2 0.00003 0 0 0.5 0.00008 0.1 0.00002 Page 24 F I G . 19-Solid M O N E L alloy 400 pickling tanks of welded construction handling cold 9 to 10 per cent hydrochloric acid. These tanks are in good condition after more than 10 years service. The pickling crates are also of welded M O N E L alloy 400 construction. prior to enamelling, solid MONEL alloy 400 tanks have been in use over 10 years handling 9 per cent hydrochloric acid and show very little evidence of attack. The relatively low copper content of MONEL alloy 400 is of importance in equipment for the pickling of steel for enamelling or coating since it avoids the development of “copper flash” on the steel surface. Some alloys are subject, under conditions likely to be encountered in tie-rod service, to a peculiar form of corrosion akin to the well-known “dezincification” of yellow brasses. The net result, in the case of tie-rods of such alloys, is the formation of a mechanically weak outer layer, increasing in thickness as service time goes on, around a correspondingly decreasing core of unaltered material. Ultimately, the cross-sectional area of the unaltered core becomes too small to sus-tain the loading involved and failure occurs. Fortunately, MONEL alloys 400 and 402 do not corrode in this manner, so that tie-rods of these materials can be depended upon to maintain their high unit strength throughout their useful life. For pickling acid concentrations higher than 10 to 15 per cent HCl at room temperature, nickel is likely to be more resistant than MONEL alloy 400, as is indicated by the corrosion data in Figure 4. In this higher concentration range, nickel can be used frequently for baskets and crates, although it is not usually recommended for solid metal pickling tanks. Reports have been received of the satisfactory performance Of MONEL alloy 400 baskets, crates, hooks and chains in acid concentrations under 10 per cent HCl at temperatures up to boiling. Its performance will usually depend upon the amount of galvanic protection received from the steel being pickled, the portion of time it is actually exposed to the acid, and upon whether an inhibitor is used. MONEL alloy 400 would be superior to nickel under such conditions, as is shown in Figure 6. In no case would it be used for a solid pickling tank in elevated temperature acid. For pickling tanks and heating coils used with hot hydrochloric acid solutions HASTELLOY alloy B is usually the most resistant wrought metallic material. MONEL alloy 400 is used for “nip” or flux rolls and for guide parts carrying steel strip from pickling tanks to flux tanks in continuous galvanizing systems. Caution should be observed in the use of MONEL alloy 400 or nickel for pickling drums, either of con-tinuous or batch types, which are not completely immersed in the acid bath. In some cases, these drums have only the bottom portion immersed to pick up the acid and in others the entire drum may be suspended above the bath and acid sprayed into it. In such in-stallations, the exposed surfaces of the drum are covered with a thin layer of acid which becomes highly aerated in its travel through the air. Furthermore, the. drums may be only partly loaded, with the load constantly shifting, so that the actual area of contact between the drum and loading may be too small to provide suitable galvanic protection. To provide satisfactory performance, the drum should be immersed completely in the pickling acid, or, if suspended partly or entirely above the acid surface, the pickling tank and drum should be enclosed with a sealed hood which will provide continuously a hydrogen atmosphere around the drum. Where this cannot be done, HASTELLOY alloy B is the preferred material of construction for drums. If MONEL alloy 400 hoods are installed over con-tinuous hydrochloric acid pickling equipment, they should be installed with the edges of the inlet end immersed in the pickling acid or in a water seal or some other arrangement should be used to keep out air and maintain a reducing atmosphere within the system. In any hood design, cognizance should be taken of the fact that hydrogen and air may form combustible mixtures. The behavior of bare MONEL alloy 400 or nickel in open type hoods above pickling tanks will depend upon their location with respect to the pickling equipment and the acid content of the fumes. They may not be satisfactory if the combination of acid concentration and aeration is too severe. In such cases, a usually satisfactory installation can be made by coating the MONEL alloy 400 or nickel with an acid-resisting paint. If breaks occur in the paint, the underlying alloy will provide suitable resistance until the breaks can be repaired. Galvanizing and Soldering Fluxes In the batch processes for galvanizing iron or steel parts it is customary practice to flux the parts after pickling and prior to dipping in the molten zinc bath. The flux bath is usually a water solution of zinc chloride, or a combination of zinc and ammonium chlorides, acidified with a small amount of hydrochloric acid. MONEL alloy 400 baskets, tanks, and heating coils have been used successfully for the fluxing operation providing the bath is not aerated. Labo- Page 25 ratory test data for the unaerated conditions are shown in Table 27. TABLE 27 Laboratory Corrosion Test in Galvanizing Flux Solution of zinc chloride and ammonium chloride. Specimens half immersed in solution. No Agitation Temperature: 15.5°C (60°F) Test duration: 60 days Corrosion Rate Material mdd. ipy. Monel alloy 400 .................. 10.9 0.002 Nickel................................ 9.9 0.002 Inconel alloy 600 ................ ' Based on area of specimen immersed. "Accelerated attack at liquid level. In continuous operation procedures employing conveyer mechanisms the flux solution may be sprayed or cascaded over the parts, the solution being recirculated through a system of tanks. Under these conditions of high aeration the flux solution becomes considerably more corrosive, as shown by the plant corrosion test data of Table 28. For such conditions from this alloy are in use for galvanizing of steel nuts and bolts. Petroleum Refining In the distillation of crude petroleum containing entrained salt water, corrosion of iron frequently is severe, due to hydrolysis of the brines to form dilute hydrochloric acid. While hydrolysis is likely to be greatest at the higher temperatures, corrosion is likely to be most severe in the top sections of distillation towers because temperatures are lowest in those sections permitting the maximum amount of acid condensation to occur. Acid formation is likely to be greatest when the brines contain considerable proportions of magnesium and calcium chlorides because these hydrolyze more readily than sodium or potassium chlorides. In a case cited by Hamlin and Turner", where the entrained brine contained 2.2 per cent Mg C1z and 6.5 per cent NaCl, an acidity equal to 8.23 grains of HCl per gallon of water was present after passage through the dephlegmators. Where petroleum contained one per cent brine, HCl was formed in the ratio of 3.39 pounds per 1000 barrels of oil. TABLE 28 Plant Corrosion Test in Galvanizing Flux 25% Zinc chloride solution, acidified to pH 4 with hydro-chloric acid. Specimens immersed in recirculation tank, in aerated solution. Temperature: 25°C (77°F) Test Duration: 56 days Severely pitted and perforated. HASTELLOY alloy C exhibits the best corrosion resistance of the high nickel alloys. In connection with batch galvanizing operations wherein MONEL nickel-copper alloy 400 or HASTELLOY alloy C baskets may be used to handle the parts through the pickling and fluxing processes, the user should be cautioned against employing the same basket for immersion in the molten zinc bath. These alloys can be intergranularly attacked by molten zinc, leading to premature failure. The usual practice has been to employ steel baskets for the zinc dipping operation. More recently the 20 % Ni, 21 % v1 -, 20 % Co, 3% Mo alloy, N-155, has been found to be resistant to the molten zinc bath, and baskets fabricated 20M. L. Hamlin and F. M. Turner, Jr., "The Chemical Resistance of Engineering Materials." The Chemical Catalog Co., Inc., New York (1923), p. 62. FIG. 20-Refinery accumulator tanks lined with 0.083 in. thick MONEL alloy 400 sheet, plug welded on 4 in. centers, to resist corrosion by naphtha containing dilute hydrochloric acid. Because of its resistance to this hydrochloric acid condition, MONEL alloy 400 is used to a considerable extent for lining the upper parts of distillation towers and for adjacent equipment such as bubble caps and trays, dephlegmators, condensers and condensate and reflux lines. Some of the new distillation towers have been constructed with MONEL alloy 400 lining for the top section, INCONEL alloy 600 lining for the middle section, and stainless steel lining for the bottom section. In Table 29 are given the results of plant corrosion tests in the upper sections of distillation towers in three different refineries handling salty crudes. The charging stock in each case had passed through water settling tanks before distillation. Corrosion Rate Material mdd. ipy. Monel alloy 400......................... 348 0.057 Nickel ....................................... 378 0.061 Inconel alloy 600 ....................... 224 .038 Hastelloy alloy B ....................... 102 .016 Hastelloy alloy C ....................... 0.4 .0001 Lead ......................................... 146 .019 Page 26 Corrosion Rate Test 1 Test 2 Test 3 Material mdd. ipy. mdd. ipy. mdd. ipy. Monel alloy 400..... 57 0.0093 18 0.003 46 0.0075 Nickel ................... 28 0.0045 37 0.006a 43 0.0069 Inconel alloy 600... 72 0.0120 100 0.017a . . . . . . Ni-Resist (Type 1) . 57 0.011 260 0.051 Mild Steel.............. 290 0.054 263 0.049 Cast Iron ............... 250 0.050 900 0.181 . . . . . . a Local attack in form of pitting to maximum depth of 0.016". Completely destroyed. Original thickness 0.031". Several refineries use chlorinated solvents in solvent extraction or dewaxing of lubricating oil. One dewaxing process employs a mixture of carbon tetrachloride and ethylene dichloride. In the recovery of the solvents by steam distillation, the formation of dilute hydrochloric acid by hydrolysis results in corrosive condition in distillation equipment and condensers which can be overcome by the use of MONEL alloy 400 or nickel. The results of corrosion tests in the condenser of the chlorinated solvents recovery unit at one refinery are given in Table 30. Isomerization processes using aluminum chloride and other metallic chlorides plus hydrogen chloride together as catalysts, provide a corrosive condition TABLE 30 Plant Corrosion Test in Condenser of Chlorinated Solvents Recovery Unit at Petroleum Refinery Temperature: 60 to 93° C (140 to 200° F) Duration of test: 99 days FIG. 21-Section. of a large petroleum fractionating tower. The top part is lined with MONEL alloy 400 sheet to withstand corrosion by dilute HCl formed by hydrolysis of brine in the petroleum. where there is a certain amount of residual water dissolved or entrained in the charging stocks. The catalyst complex is circulated usually as a viscous tarry mixture of the chlorides and hydrocarbons and where this "tar" contacts the metal surfaces it is likely to be particularly corrosive. While nickel has been used for isomerization reactors, corrosion rates of the order of 0.036 ipy in the catalyst chamber at 350° F, and 0.025 ipy in the FIG. 22-Piping, pumps, and valves for the isomerization units are made of HASTELLOY alloy B. Chloride catalyst sludge settles out from the tanks and is pumped back through HASTELLOY alloy B equipment. TABLE 29 Plant Corrosion Tests in Distillation of Crude Oils Containing Entrained Brine Where Some Hydrochloric Acid Is Formed by Hydrolysis of Brine Test 1- On top tray of bubble tower of crude topping unit handling Kansas crude. Some hydrogen sulfide also present. Cooled reflux enters top tray. Temperature: 250 to 260° F (121 to 126° C) Duration of test: 80 days Test 2- In top of bubble tower of crude topping unit handling West Texas crude. Temperature: 250° F (121° C) Duration of test: 60 days Test 3- On top tray of bubble tower of crude topping unit handling Michigan crude. Cooled reflux enters top tray. Temperature: About 200° F (93° C) Duration of test: 90 days Corrosion Rate Material mdd. ipy. Monel alloy 400 .................. 65 0.0106 Nickel ............................... 37 0.0061 Inconel alloy 600 ............... 28 0.0047 Ni-Resist (Type 1) .............. 65 0.0126 Mild Steel .......................... 304 0.0560 Cast Iron ........................... 471 0.0943 Page 27 FIG. 23-A vacuum evaporator equipped with MONEL alloy 400 tubes used for the evaporation of zinc chloride. catalyst removal column have been observed. HASTELLOY alloy B has shown considerably better corrosion resistance and is the preferred material for reactor linings. The results of plant corrosion tests in the liquid and vapor sections of a "tar" knock-out vessel are shown in Table 31. In some cases where particular care is taken to dehydrate the charging stocks, nickel may be suitable for parts of isomerization equipment. Additional information on petroleum refining ap-plications is given in the reprint, "The Selection of TABLE 31 Plant Corrosion Tests in "Tar" Knock-Out Vessel of Hydrocarbon Isomerization Unit Employing Aluminum Chloride and Hydrogen Chloride as Catalyst Temperature: 107 to 121° C (Avg. 116° C) 225 to 250° F (Avg. 240° F) Duration of tests: 44 days Corrosion Rate In Liquid In Vapor Material mdd. ipy. mdd. ipy. Monel alloy 400 ... 160 0.026 149 0.024 Nickel ................ 59 0.010 79 0.013 Inconel alloy 600 . 63 0.011 71 0.012 Hastelloy alloy B . 2.5 0.0004 5 0.0008 Ni-Resist (Type 1) 270 0.053 266 0.052 Mild Steel ........... 873 0.162 435 0.081 Cast Iron............. 425 0.085 413 0.083 Materials for Some Petroleum Refinery Applications". Inorganic Acid Chlorides Aqueous solutions of most of the inorganic acid chlorides hydrolyze to a certain extent to form dilute hydrochloric acid, particularly at temperatures associated with their evaporation. MONEL nickel-copper alloy 400, nickel, and in some cases INCONEL nickel-chromium alloy 600 have useful resistance to corrosion by many of these salt solutions. As pointed out at the beginning of this bulletin, aeration of hydrochloric acid solutions tends to increase the corrosion rate of MONEL alloy 400 and nickel. Likewise, aeration of the acid chloride salt solutions will also tend to increase the corrosion rate of MONEL alloy 400 and nickel. Listed below are several of the acid chloride salts with which MONEL alloy 400, nickel or INCONEL alloy 600 have been used, and corrosion data collected from plant and laboratory tests are given in Tables 32 to 40. Aluminum Chloride-Al Cl3 Nickel has been used for dissolving tanks, and for isomerization and Friedel Craft reactions wherein aluminum chloride is employed as a catalyst. Ammonium Chloride-NH4 Cl Nickel has been employed as tanks for the mixing of battery pastes containing ammonium chloride, and for evaporator tubes in the concentrating of ammonium chloride solutions to 40%. FIG. 24-Welded construction was used for this 40 in. diameter by 24 in. high pure nickel centrifugal basket used with an acid chloride salt. Welded to the 3/8 in. thick shell is a cast nickel hub. Page 28 Antimony Chloride-Sb Cl3 Nickel has been used for condensing of antimony chloride vapors containing moisture. Magnesium Chloride-Mg Cl2 In equipment for the evaporation and handling of hot magnesium chloride used in the production of magnesium, INCONEL nickel-chromium alloy 600 has been used satisfactorily. The iron, copper, and nickel content of the finished magnesium must be kept extremely low. INCONEL alloy 600 is used because of its high degree of corrosion resistance and because it (as well as MONEL alloy 400 and nickel) is free from stress-corrosion cracking in boiling magnesium chloride solutions. MONEL alloy 400 and NI-RESIST are used for such equipment as filters, pumps and valves. Phosphorus Oxychloride-POCI Nickel is used for both shipping drums and tanks for truck shipment of the commercial product. Nickel and nickel clad steel are employed in the processing equipment for production of tricresyl phosphate wherein phosphorus oxychloride is a reactant. Phosphorus Trichloride-P Cl3 Nickel is used for reactors, reflux columns, piping and valves in the manufacture of the product, as well as for shipping drums. Pyrosulfuryl Chloride-S205Cl Nickel is used for shipping drums. Nitrosyl Chloride-NOCI Both nickel and INCONEL alloy 600 are used for piping, and valves in the handling of the dry product following its manufacture. Nickel is standard material for shipping cylinders. Thlonyl Chloride-SO Cl2 Nickel is used for shipping drums. Thlophosphoryl Chloride-PS Cl3 Nickel is used for shipping drums. Sulfuryl Chloride-S02 Cl2 Nickel is used for shipping drums. Sulfur Monochloride-S2 Cl2 Nickel and MONEL alloy 400 are both resistant to sulfur monochloride as encountered in the process of distillation and rectification of crude carbon tetrachloride. (See Table 16, Tests 3 and 4.) Tin Tetrachloride Nickel and MONEL alloy 400 are both resistant to hot solution encountered in distillation. (See data Table 37.) Zinc Chloride In the production of vulcanized fibre, paper is treated with 70° Be zinc chloride solution which subsequently is washed from the wet fibre sheet. MONEL alloy equipment is used in the form of MONEL alloy 400-covered drying and processing rolls and in evaporators for recovering and concentrating the zinc chloride solution. Corrosion rates in the concentration of zinc chloride are shown in Table 38. The resistance of MONEL alloy 400 protects the color of the fibre board and keeps its metallic content at a minimum, so that uniformly good dialetric properties are maintained for electrical insulating uses. TABLE 32 Plant Corrosion Test in Evaporation of Ammonium Chloride From 28 to 40 Per Cent NHsCl Test spool immersed in liquor Temperature: 102º C. (216º F.) Duration of test: 762 hr Corrosion Rate Material mdd. ipy. Monel alloy 400 .............. 73 0.012 Nickel ............................ 52 0.0084 Inconel alloy 600............. 3 0.0005 Ni-Resist (Type 1) ........... 63 0.012 Cast Iron ........................ 1780 0.36 Pitted to maximum depth of 0.006 inch. TABLE 33 Plant Corrosion Test in Mixture of Arsenic Trichloride and Sulfur Monochloride with Small Amount of Moisture Arsenic trichloride content varied from 72 to 100% Sulfur monochloride varied from 0 to 28% Test spool located in top of first pass of vertical condenser subject to vapor and some condensate. Temperature: 120 to 130° C (248 to 266° F) Duration of test: 618 hours Corrosion Rate Material mdd. ipy. Monel alloy 400 ................ 23 0.0038 Nickel ............................. 8 0.0013 Inconel alloy 600 .............. 10 0.0016 Ni-Resist (Type 1) ............. 36 0.007 Mild Steel ........................ 117 0.022 Cast Iron .......................... 148 0.030 TABLE 34 Plant Corrosion Test in Evaporation of 37 Per Cent Manganous Chloride Solution Test spool half submerged at liquor level in open pan into which atmospheric drum dryer dips. Temperature: 99 to 107° C (Avg. 104° C) 210 to 225° F (Avg. 220° F) Duration of test: 19 days Corrosion Rate Material mdd. ipy. Monel alloy 400.......................... 116 0.019 Nickel........................................ 187 0.030 Inconel alloy 600 ........................ 166 0.028 Ni-Resist (Type 1)....................... 204 0.040 Mild Steel .................................. 1280 0.236 Cast Iron.................................... 1538 0.308 Page 29 TABLE 35 Laboratory Corrosion Tests in Phosphorus Pentachloride Test 1: at 76° C (169° F) for 19 days Test 2 : at 150° C (302° F) for 3 days TABLE 39 Laboratory Corrosion Test in Mixture of Zinc Chloride, Ammonium Chloride and Water Composition of solution: 66% ZnCl2 20% NH4Cl 14% Water Specimens immersed in open jars Temperature: 83 to 95° C (182 to 204° F) Duration of test: 2 hours TABLE 36 Plant Test in Mixture of Phosphoric, Hydrochloric, and Cresylic Acids with Phosphorus Oxychloride Test spool at liquid line in reaction kettle Temperature: 82° C (180° F) Duration of test: 174 hours TABLE 37 Plant Corrosion Test in Distillation of Crude Tin Tetrachloride Corrosion Rate Material mdd. ipy. Monel alloy 400 ................... Nickel ................................. Inconel alloy 600 .................. Hastelloy alloy C .................. Ni-Resist (Type 1)................. Cast Iron.............................. 20 24 42 54 118 210 0.0032 0.0039 0.0071 0.0087 0.0228 0.0420 TABLE 38 Plant Corrosion Test in Double-Effect Zinc Chloride Evaporato Test 1:-Weak liquor effect, 7.9 to 21% ZnC12 Temperature: 32 to 38° C (90 to 100° F) Vacuum: 26 to 28" Hg. Duration of test: 210 days Test 2:-Strong liquor effect, 21 to 69% ZnC12 Temperature: 110 to 116° C (230 to 240° F) Vacuum: 15 to 18" Hg Duration of test: 90 days TABLE 40 Plant Corrosion Test in Evaporator Concentrating a Mixture of Magnesium and Calcium Chloride Brines to 50 Per Cent Chlorides Temperature: Boiling under vacuum Duration of test: 26 days Pulp and Paper Wood pulp is bleached with either a hypochlorite solution (in some cases sodium chlorite), or by direct chlorination using chlorine water or chlorine gas. Bleaching is done by either single-stage or multi-stage processes. Single-stage bleaching usually is done with hypochlorite solutions, the bleaching being followed by washing on vacuum filters. In multi-stage bleaching, direct chlorination frequently is used in the first stage and hypochlorite treatment in subsequent stages. It is customary in multi-stage bleaching to wash the pulp on vacuum filters after each stage. The direct chlorination, or acid bleaching, frequently is followed by a caustic neutralizing treatment before washing. In the preparation of strong hypochlorite bleach solutions or in the bleaching processes themselves, either with hypochlorite or chlorine, corrosive conditions usually are so severe that HASTELLOY alloy C is the only highly resistant wrought metallic material. The results of corrosion tests in the preparation of strong calcium hypochlorite bleach liquor are given in Table 41. MONEL alloy 400 has satisfactory resistance in the washing of hypochlorite bleached pulps and is used for wire covers on vacuum washers and for lining the vats and repulper sections of these washers. Table 42 gives the results of corrosion tests in the filtrate from a vacuum washer handling hypochlorite bleached Corrosion Rate Material mdd. ipy. Monel alloy 400 ................. 12 0.002 Nickel ............................... 17 0.003 Inconel alloy 600................ 0.6 0.0001 Ni-Resist (Type 1................ 19 0.004 Mild Steel.......................... 410 0.076 Cast Iron ........................... 150 0.030 Corrosion Rate Test I Test 2 Material mdd. ipy. mdd. ipy. Monel alloy 400 ........ Nickel ...................... 2.6 1.3 0.0004 0.0002 . . 1.7 . . 0.0003 Corrosion Rate Material mdd. ipy. Ni-Resist (Type 1)............... 95 0.0184 Ni-Resist (Type 2)............... 116 0.0225 Cast Iron ............................ 3420 0.684 Corrosion Rate Material mdd. ipy. Monel alloy 400 .............. 459 0.074 Nickel....................... 106 0.017 Inconel alloy 600............. 588 0.099 Corrosion Rate Test I Test 2 Material mdd. ipy. mdd. ipy. Monel alloy 400 ............. 28 0.0045 98 0.016 Nickel............................ 29 0.0046 245 0.040 Test spool located in still, above steam coil and below liquid level. Temperature: 104 to 116° C (220 to 240° F) Duration of test: 432 hours Page 30 TABLE 41 Plant Test in Preparation of Calcium Hypochlorite Bleach Liquor in Paper Mill The pulp from direct chlorine bleaching usually is too corrosive for the use of MONEL alloy 400 washer covers. For this purpose, Type 316 stainless steel is usually a satisfactory material as indicated by Table 43 giving the results of tests in chlorine bleached sul- TABLE 43 Plant Corrosion Test in Chlorine Bleached Paper Stock Specimens originally 0.032 inch thick. pulp. At one mill the following services were obtained from MONEL nickel-copper alloy 400 wire cloth covers on washers and thickeners: TABLE 42 Plant Corrosion Test in Filtrate From Washer Handling Hypochlorite Bleached Soda Pulp Test specimens immersed in filtrate sump Chlorine content of filtrate, 0.009 gm per liter pH 6.0 Temperature: 21 to 25° C (70 to 78° F) Duration of test: 72 days Rate of filtrate flow: 1050 gal per min Corrosion Rate Material mdd. ipy. Monel alloy 400 .................... 0.45 <0.0001 Nickel................................... 0.26 <0.0001 Inconel alloy 600 ................... 0.69 0.0001 Hastelloy alloy C................... 0.17 <0.0001 Ni-Resist ............................... 10. 0.002 Cast Iron ............................... 20. 0.004 < = Less than. < = Less than. fite pulp containing 0.02 gm of free chlorine and 0.15 gm of hydrochloric acid per liter. In the filtrate from these washers, the chlorine content frequently is reduced to the point where MONEL alloy 400 and NI-RESIST can be used for such applications as filtrate piping. The results of corrosion tests in the filtrate from the washing of chlorine bleached pulp are given in Table 44. TABLE 44 Plant Corrosion Test in Filtrate from Washer Handling Chlorine Bleached Soda Pulp Corrosion Rate Material mdd. ipy. Monel alloy 400 .................. 9 0.0014 Nickel ................................ 11 0.0018 Inconel alloy 600................. 0.22 <0.0001 Hastelloy alloy C................. 0.13 <0.0001 Ni-Resist ............................ 39. 0.0075 Cast Iron ........................... 78. 0.016 < = Less than. In recent years the chlorine dioxide pulp bleaching process has grown considerably, to supply the demand for high brightness, high strength pulps. The generation of chlorine dioxide involves reactions wherein mixtures of methanol, sulfurous or sulfuric acid, chlorine and chlorine dioxide are present at temperatures up to 150 ° F The corrosive conditions associated with the process are rather severe for most of the corrosion resisting alloys usually employed in pulp handling, and non-metallics are employed wherever possible. A corrosion test study at one plant using chlorine dioxide bleaching has been reported by Available chlorine content approximately 40 gm per liter plus small amounts of calcium chloride, calcium chlorate, calcium carbonate, and free lime. Fresh solution made up twice a week Temperature: 16 to 38° C (60 to 100° F) Duration of test: 30 days Corrosion Rate Remarks Material mdd. ipy. Monel alloy 400. . . . 455 0.074 Perforated by pitting Nickel. . . . . . . . . . . . 958 0.155 Perforated by pitting Inconel alloy 600. . . 276 0.047 Perforated by pitting 18-8 Mo Stainless. . (Type 316). . . . . . 21 0.004 Perforated by pitting Hastelloy alloy C. . . 0.75 0.0001 . . Silicon Cast Iron. . . (14.5% Si). . . . . . 2.1 0.0004 . . Ni-Resist. . . . . . . . . 155 0.030 . . Stock contains 0.15 gm per liter of HCl and 0.02 gm per liter of free chlorine. Test spool attached to cylinder mold of pulp thickener. Duration of test: 90 days Corrosion Rate Material mdd. ipy. Monel alloy 400...................... 262 0.0425 18-8 Mo Stainless (Type 316) .. 0.07 <0.0001 Hastelloy alloy C.................... 0.03 <0.0001 Ni-Resist (Type 1) .................. 120 0.0232 Cast Iron................................ 886 0.177 Thickener handling hypochlorite bleached soda pulp 1st Monel alloy 400 cover..... Gave 26 months service 2nd Monel alloy 400 cover.... Gave 61 months service 3rd Monel alloy 400 cover .... In good condition after 21 months service Thickener handling hypochlorite bleached sulfite pulp 1st Monel alloy 400 cover ..... Gave 45 months service 2nd Monel alloy 400 cover .... In satisfactory condition after 63 months service Washer handling hypochlorite bleached soda pulp 1st Monel alloy 400 cover ..... Gave 14 months service 2nd Monel alloy 400 cover .... Gave 6½ years service In a multi-stage bleach system, MONEL alloy 400 covers on washers handling soda pulp from the caustic extraction stage and the hypochlorite stage, gave the following services: Caustic extraction stage washer - MONEL alloy 400 covers averaged 12 months. Hypochlorite stage washers (high and low den- sity)-MONEL alloy 400 covers averaged 2 years: Test specimens immersed in filtrate sump Chlorine content of filtrate, 0.011 gm per liter, pH 3.5. Temperature: 23 to 33° C (74 to 92° F) Duration of test: 72 days Rate of filtrate flow: 720 gal per min Page 31 Teeple and Adams 21 in which the corrosion test data shown in Tables 45 to 48 were obtained. One equipment manufacturer has found cast HASTELLOY alloy C to be suitable for the chlorine dioxide TABLE 47 Plant Test in Chlorine Dioxide Solution Receiver Chlorine dioxide concentration 4 to 5 grams per liter, pH 2 to 3.5. Temperature: 2°C (36° F) Test Duration: 351 hrs TABLE 45 Plant Test in Top of Chlorine Dioxide Reaction Tower Media consists of concentrated sulfuric acid, 32% sodium chlorate, methanol, and chlorine dioxide gas. Temperature: 57°C (135°F) Test Duration: 351 hours Material Corrosion rate, ipy.• Remarks Tantalum ......................... 0.0001 No attack LaBour R-55 ........................ 0.002 No pitting Durichlor.......................... 0.003 Incipient pits Chemical lead ................... 0.006 Etched Duriron............................. 0.007 Etched Hastelloy alloy C (cast). ..... 0.013 No pitting Chlorimet 3............................. 0.020 No pitting Durimet 20........................... 0.027 0.059 max. pitting b Type 316 stainless steel..... 0.15 Perforated 0.031 in. c Inconel alloy 600.................... > 0.37 Spec. corroded away d Type 430 stainless steel..... > 0.46 Spec. corroded away d Monel alloy 400................. > 0.48 Spec. corroded away d Nickel............................... > 0.48 Spec. corroded away d Type 304 stainless steel..... > 0.55 Spec. corroded away d 88/10/2 Bronze ................. > 1.8 Spec. corroded away d The LaBour Co., Inc., Trademark. > = More than. Duriron Co., Inc., Trademark. a Corrosion rate expressed in terms of inch penetration per year based upon the continuous exposure of one surface. b One of the two duplicate specimens showed no pitting on the surface but was perforated (0.189 in in 351 hr) under the spacer, i.e., in a crevice. The other duplicate specimen showed the max. pitting as indicated. c Both duplicate specimens were perforated (0.031 in in 351 hr) and there was gross subsurface corrosion up to Y4 in. back from the specimen edges. d Specimen corroded away during exposure period. Minimum corrosion rate is shown. TABLE 46 Plant Test in Chlorine Dioxide Gas Line to the Absorber 8 to 10% chlorine dioxide gas plus some condensed chlorine dioxide solution. Flow rate: 250 cu ft per min Temperature: 66°C (150°F) Test Duration: 350 hours Material Corrosion rate, ipy. Remarks Tantalum.............................. 0.0001 No attack Durichlor ......................... 0.003 No pitting Chlorimet 3............................. 0.015 No pitting Chemical lead................... 0.030 No pitting Hastelloy alloy C (cast)...... 0.046 No pitting a Durimet 20.............................. 0.19 Perforated b Type 316 stainless steel .... 0.29 Perforated c Inconel alloy 600............... 0.30 Perforated d Type 304 stainless steel .... 0.33 Perforated e Nickel .............................. 0.35 Etchhd f Monel alloy 400 ................ 0.36 Etchhd f Type 430 stainless steel .... 0.40 Perforated e LaBour R-55 ........................ 0.46 0.095 in. max. pitting Duriron ............................ 0.62 Gross attacks g 88/10/2 Bronze................. 0.68 Heavy etch a Subsurface corrosion on one of duplicate spec. b Specimen perforated (0.189 in.) during exposure period. c Specimen perforated (0.031 in.) during exposure period-two thirds of specimen missing. d Specimen perforated (0.031 in.) during exposure period-specimen looks like filigree. e Specimen perforated (0.031 in.) during exposure period-one third of specimen missing. f One half of specimen missing. g Also deep pitting. 21 H. O. Teeple & R. L. Adams, Jr. "A Corrosion Study in a Chlorine Dioxide Pulp Bleaching Plant". TAPPI Vol. 38, #1, January 1955. Material Corrosion rate, ipy. Remarks Tantalum.................... 0.0001 No attack Duriron...................... 0.0001 No pitting Durichlor ................... 0.0001 No pitting Type 316 stainless steel 0.0001 No pitting Durimet 20 ................ 0.0007 No pitting LaBour R-55............... 0.0036 No pitting Hastelloy alloy C (cast) 0.0050 No pitting Type 304 stainless steel 0.0051 Incipient pitting a Chlorimet 3 ................ 0.014 Macro etch Type 430 stainless steel 0.025 Perforated b Inconel alloy 600 ........ 0.11 Perforated c Chemical lead............. 0.21 No pitting Nickel........................ 0.32 Etched d 88/10/2 Bronze ........... 0.34 Heavy etch Monel alloy 400 .......... 0.48 Specimen almost gone a Specimen pitted 0.018 in under spacer (crevice), during exposure period. b Specimen perforated 0.031 in. in line of rolling direction during expo sure period. c Specimen perforated 0.031 in. during exposure period. d Severe intergranular corrosion. TABLE 48 Plant Test in Vent Line from Top of Bleach Tower Spent chlorine dioxide gas Temperature: 68°C (155°F) Test Duration: 338 hrs Material Corrosion rate, ipy. Remarks Tantalum .............................. 0.0001 No attack Durichlor......................... 0.0001 Incipient pitting Hastelloy alloy C (cast). .... 0.0002 No attack Durimet 20.............................. 0.0002 No attack Chlorimet 3............................. 0.0003 No attack Duriron............................ 0.0004 Incipient pitting Type 316 stainless steel.... 0.0005 Pitted a Type 304 stainless steel.... 0.0006 Pitted a Monel alloy 400................ 0.0006 Incipient pitting Nickel.............................. 0.0007 Slightly etched Inconel alloy 600 .............. 0.0007 Perforated b Type 430 stainless steel.... 0.0009 Pitted c 88/10/2 Bronze................. 0.0052 Localized etching Chemical lead .................. 0.014 Pitted 0.035 in max. d a Two severely pitted areas on each specimen, otherwise surface is good b Only one perforation on each specimen, otherwise surface is good. c Only one severely pitted area on each specimen, otherwise surface is good. d Average pit depth 0.025 in. mixer at the bottom of the pulp bleaching tower, but cast Type 317 stainless suffered a high corrosion rate. Where corrosive conditions are too severe for the use Of HASTELLOY alloy C, titanium or tantalum may be used. Other Applications Corrosion test data and service experience are available for nickel alloys in numerous other industries and processes involving hydrochloric acid or chlorine which, for lack of space, cannot be included in this bulletin. Such information relating to specific processes or corrosion problems will be furnished upon request made to INCO's Development and Re-search Department.
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31.3 Substructure of the Nucleus – College Physics chapters 1-17 Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Book Contents Navigation Contents Introduction Preface to College Physics I. 1 Introduction: The Nature of Science and Physics 1.Introduction to Science and the Realm of Physics, Physical Quantities, and Units 2.1.1 Physics: An Introduction 3.1.2 Physical Quantities and Units 4.1.3 Accuracy, Precision, and Significant Figures 5.1.4 Approximation II. 2 Kinematics 6.Introduction to One-Dimensional Kinematics 7.2.1 Displacement 8.2.2 Vectors, Scalars, and Coordinate Systems 9.2.3 Time, Velocity, and Speed 10.2.4 Acceleration 11.2.5 Motion Equations for Constant Acceleration in One Dimension 12.2.6 Problem-Solving Basics for One-Dimensional Kinematics 13.2.7 Falling Objects 14.2.8 Graphical Analysis of One-Dimensional Motion III. 3 Two-dimensional Kinematics 15.Introduction to Two-Dimensional Kinematics 16.3.1 Kinematics in Two Dimensions: An Introduction 17.3.2 Vector Addition and Subtraction: Graphical Methods 18.3.3 Vector Addition and Subtraction: Analytical Methods 19.3.4 Projectile Motion 20.3.5 Addition of Velocities IV. 4 Dynamics: Force and Newton's Laws of Motion 21.Introduction to Dynamics: Newton’s Laws of Motion 22.4.1 Development of Force Concept 23.4.2 Newton’s First Law of Motion: Inertia 24.4.3 Newton’s Second Law of Motion: Concept of a System 25.4.5 Normal, Tension, and Other Examples of Forces 26.4.4 Newton’s Third Law of Motion: Symmetry in Forces 27.4.6 Problem-Solving Strategies 28.4.7 Further Applications of Newton’s Laws of Motion 29.4.8 Extended Topic: The Four Basic Forces—An Introduction V. 5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity 30.Introduction: Further Applications of Newton’s Laws 31.5.1 Friction 32.5.2 Drag Forces 33.5.3 Elasticity: Stress and Strain VI. 6 Uniform Circular Motion and Gravitation 34.Introduction to Uniform Circular Motion and Gravitation 35.6.1 Rotation Angle and Angular Velocity 36.6.2 Centripetal Acceleration 37.6.3 Centripetal Force 38.6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force 39.6.5 Newton’s Universal Law of Gravitation 40.6.6 Satellites and Kepler’s Laws: An Argument for Simplicity VII. 7 Work, Energy, and Energy Resources 41.Introduction to Work, Energy, and Energy Resources 42.7.1 Work: The Scientific Definition 43.7.2 Kinetic Energy and the Work-Energy Theorem 44.7.3 Gravitational Potential Energy 45.7.4 Conservative Forces and Potential Energy 46.7.5 Nonconservative Forces 47.7.6 Conservation of Energy 48.7.7 Power 49.7.8 Work, Energy, and Power in Humans 50.7.9 World Energy Use VIII. 8 Linear Momentum and Collisions 51.Introduction to Linear Momentum and Collisions 52.8.1 Linear Momentum and Force 53.8.2 Impulse 54.8.3 Conservation of Momentum 55.8.4 Elastic Collisions in One Dimension 56.8.5 Inelastic Collisions in One Dimension 57.8.6 Collisions of Point Masses in Two Dimensions 58.8.7 Introduction to Rocket Propulsion IX. 9 Statics and Torque 59.Introduction to Statics and Torque 60.9.1 The First Condition for Equilibrium 61.9.2 The Second Condition for Equilibrium 62.9.3 Stability 63.9.4 Applications of Statics, Including Problem-Solving Strategies 64.9.5 Simple Machines 65.9.6 Forces and Torques in Muscles and Joints X. 10 Rotational Motion and Angular Momentum 66.Introduction to Rotational Motion and Angular Momentum 67.10.1 Angular Acceleration 68.10.2 Kinematics of Rotational Motion 69.10.3 Dynamics of Rotational Motion: Rotational Inertia 70.10.4 Rotational Kinetic Energy: Work and Energy Revisited 71.10.5 Angular Momentum and Its Conservation 72.10.6 Collisions of Extended Bodies in Two Dimensions 73.10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum XI. 11 Fluid Statistics 74.Introduction to Fluid Statics 75.11.1 What Is a Fluid? 76.11.2 Density 77.11.3 Pressure 78.11.4 Variation of Pressure with Depth in a Fluid 79.11.5 Pascal’s Principle 80.11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement 81.11.7 Archimedes’ Principle 82.11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action 83.11.9 Pressures in the Body XII. 12 Fluid Dynamics and Its Biological and Medical Applications 84.Introduction to Fluid Dynamics and Its Biological and Medical Applications 85.12.1 Flow Rate and Its Relation to Velocity 86.12.2 Bernoulli’s Equation 87.12.3 The Most General Applications of Bernoulli’s Equation 88.12.4 Viscosity and Laminar Flow; Poiseuille’s Law 89.12.5 The Onset of Turbulence 90.12.6 Motion of an Object in a Viscous Fluid 91.12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes XIII. 13 Temperature, Kinetic Theory, and the Gas Laws 92.Introduction to Temperature, Kinetic Theory, and the Gas Laws 93.13.1 Temperature 94.13.2 Thermal Expansion of Solids and Liquids 95.13.3 The Ideal Gas Law 96.13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature 97.13.5 Phase Changes 98.13.6 Humidity, Evaporation, and Boiling XIV. 14 Heat and Transfer Methods 99.Introduction to Heat and Heat Transfer Methods 100.14.1 Heat 101.14.2 Temperature Change and Heat Capacity 102.14.3 Phase Change and Latent Heat 103.14.4 Heat Transfer Methods 104.14.5 Conduction 105.14.6 Convection 106.14.7 Radiation XV. 15 Thermodynamics 107.Introduction to Thermodynamics 108.15.1 The First Law of Thermodynamics 109.15.2 The First Law of Thermodynamics and Some Simple Processes 110.15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency 111.15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated 112.15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators 113.15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy 114.15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation XVI. 16 Oscillatory Motion and Waves 115.Introduction to Oscillatory Motion and Waves 116.16.1 Hooke’s Law: Stress and Strain Revisited 117.16.2 Period and Frequency in Oscillations 118.16.3 Simple Harmonic Motion: A Special Periodic Motion 119.16.4 The Simple Pendulum 120.16.5 Energy and the Simple Harmonic Oscillator 121.16.6 Uniform Circular Motion and Simple Harmonic Motion 122.16.7 Damped Harmonic Motion 123.16.8 Forced Oscillations and Resonance 124.16.9 Waves 125.16.10 Superposition and Interference 126.16.11 Energy in Waves: Intensity XVII. 17 Physics of Hearing 127.Introduction to the Physics of Hearing 128.17.1 Sound 129.17.2 Speed of Sound, Frequency, and Wavelength 130.17.3 Sound Intensity and Sound Level 131.17.5 Sound Interference and Resonance: Standing Waves in Air Columns 132.17.4 Doppler Effect and Sonic Booms 133.17.6 Hearing 134.17.7 Ultrasound XVIII. 18 Electric Charge and Electric Field 135.Introduction to Electric Charge and Electric Field 136.18.1 Static Electricity and Charge: Conservation of Charge 137.18.2 Conductors and Insulators 138.18.3 Coulomb’s Law 139.18.4 Electric Field: Concept of a Field Revisited 140.18.5 Electric Field Lines: Multiple Charges 141.18.6 Electric Forces in Biology 142.18.7 Conductors and Electric Fields in Static Equilibrium 143.18.8 Applications of Electrostatics XIX. 19 Electric Potential and Electric Field 144.Introduction to Electric Potential and Electric Energy 145.19.1 Electric Potential Energy: Potential Difference 146.19.2 Electric Potential in a Uniform Electric Field 147.19.3 Electrical Potential Due to a Point Charge 148.19.4 Equipotential Lines 149.19.5 Capacitors and Dielectrics 150.19.6 Capacitors in Series and Parallel 151.19.7 Energy Stored in Capacitors XX. 20 Electric Current, Resistance, and Ohm's Law 152.Introduction to Electric Current, Resistance, and Ohm's Law 153.20.1 Current 154.20.2 Ohm’s Law: Resistance and Simple Circuits 155.20.3 Resistance and Resistivity 156.20.4 Electric Power and Energy 157.20.5 Alternating Current versus Direct Current 158.20.6 Electric Hazards and the Human Body 159.20.7 Nerve Conduction–Electrocardiograms XXI. 21 Circuits and DC Instruments 160.Introduction to Circuits and DC Instruments 161.21.1 Resistors in Series and Parallel 162.21.2 Electromotive Force: Terminal Voltage 163.21.3 Kirchhoff’s Rules 164.21.4 DC Voltmeters and Ammeters 165.21.5 Null Measurements 166.21.6 DC Circuits Containing Resistors and Capacitors XXII. 22 Magnetism 167.Introduction to Magnetism 168.22.1 Magnets 169.22.2 Ferromagnets and Electromagnets 170.22.3 Magnetic Fields and Magnetic Field Lines 171.22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field 172.22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications 173.22.6 The Hall Effect 174.22.7 Magnetic Force on a Current-Carrying Conductor 175.22.8 Torque on a Current Loop: Motors and Meters 176.22.9 Magnetic Fields Produced by Currents: Ampere’s Law 177.22.10 Magnetic Force between Two Parallel Conductors 178.22.11 More Applications of Magnetism XXIII. 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies 179.Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies 180.23.1 Induced Emf and Magnetic Flux 181.23.2 Faraday’s Law of Induction: Lenz’s Law 182.23.3 Motional Emf 183.23.4 Eddy Currents and Magnetic Damping 184.23.5 Electric Generators 185.23.6 Back Emf 186.23.7 Transformers 187.23.8 Electrical Safety: Systems and Devices 188.23.9 Inductance 189.23.10 RL Circuits 190.23.11 Reactance, Inductive and Capacitive 191.23.12 RLC Series AC Circuits XXIV. 24 Electromagnetic Waves 192.Introduction to Electromagnetic Waves 193.24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed 194.24.2 Production of Electromagnetic Waves 195.24.3 The Electromagnetic Spectrum 196.24.4 Energy in Electromagnetic Waves XXV. 25 Geometric Optics 197.Introduction to Geometric Optics 198.25.1 The Ray Aspect of Light 199.25.2 The Law of Reflection 200.25.3 The Law of Refraction 201.25.4 Total Internal Reflection 202.25.5 Dispersion: The Rainbow and Prisms 203.25.6 Image Formation by Lenses 204.25.7 Image Formation by Mirrors XXVI. 26 Vision and Optical Instruments 205.Introduction to Vision and Optical Instruments 206.26.1 Physics of the Eye 207.26.2 Vision Correction 208.26.3 Color and Color Vision 209.26.4 Microscopes 210.26.5 Telescopes 211.26.6 Aberrations XXVII. 27 Wave Optics 212.Introduction to Wave Optics 213.27.1 The Wave Aspect of Light: Interference 214.27.2 Huygens's Principle: Diffraction 215.27.3 Young’s Double Slit Experiment 216.27.4 Multiple Slit Diffraction 217.27.5 Single Slit Diffraction 218.27.6 Limits of Resolution: The Rayleigh Criterion 219.27.7 Thin Film Interference 220.27.8 Polarization 221.27.9 Extended Topic Microscopy Enhanced by the Wave Characteristics of Light XXVIII. 28 Special Relativity 222.Introduction to Special Relativity 223.28.1 Einstein’s Postulates 224.28.2 Simultaneity And Time Dilation 225.28.3 Length Contraction 226.28.4 Relativistic Addition of Velocities 227.28.5 Relativistic Momentum 228.28.6 Relativistic Energy XXIX. 29 Introduction to Quantum Physics 229.Introduction to Quantum Physics 230.29.1 Quantization of Energy 231.29.2 The Photoelectric Effect 232.29.3 Photon Energies and the Electromagnetic Spectrum 233.29.4 Photon Momentum 234.29.5 The Particle-Wave Duality 235.29.6 The Wave Nature of Matter 236.29.7 Probability: The Heisenberg Uncertainty Principle 237.29.8 The Particle-Wave Duality Reviewed XXX. 30 Atomic Physics 238.Introduction to Atomic Physics 239.30.1 Discovery of the Atom 240.30.2 Discovery of the Parts of the Atom: Electrons and Nuclei 241.30.3 Bohr’s Theory of the Hydrogen Atom 242.30.4 X Rays: Atomic Origins and Applications 243.30.5 Applications of Atomic Excitations and De-Excitations 244.30.6 The Wave Nature of Matter Causes Quantization 245.30.7 Patterns in Spectra Reveal More Quantization 246.30.8 Quantum Numbers and Rules 247.30.9 The Pauli Exclusion Principle XXXI. 31 Radioactivity and Nuclear Physics 248.Introduction to Radioactivity and Nuclear Physics 249.31.1 Nuclear Radioactivity 250.31.2 Radiation Detection and Detectors 251.31.3 Substructure of the Nucleus 252.31.4 Nuclear Decay and Conservation Laws 253.31.5 Half-Life and Activity 254.31.6 Binding Energy 255.31.7 Tunneling XXXII. 32 Medical Applications of Nuclear Physics 256.Introduction to Applications of Nuclear Physics 257.32.1 Medical Imaging and Diagnostics 258.32.2 Biological Effects of Ionizing Radiation 259.32.3 Therapeutic Uses of Ionizing Radiation 260.32.4 Food Irradiation 261.32.5 Fusion 262.32.6 Fission 263.32.7 Nuclear Weapons XXXIII. 33 Particle Physics 264.Introduction to Particle Physics 265.33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited 266.33.2 The Four Basic Forces 267.33.3 Accelerators Create Matter from Energy 268.33.4 Particles, Patterns, and Conservation Laws 269.33.5 Quarks: Is That All There Is? 270.33.6 GUTs: The Unification of Forces XXXIV. 34 Frontiers of Physics 271.Introduction to Frontiers of Physics 272.34.1 Cosmology and Particle Physics 273.34.2 General Relativity and Quantum Gravity 274.34.3 Superstrings 275.34.4 Dark Matter and Closure 276.34.5 Complexity and Chaos 277.34.6 High-temperature Superconductors 278.34.7 Some Questions We Know to Ask Appendix A Atomic Masses Appendix B Select Radioactive Isotopes Appendix C Useful Information Appendix D Glossary of Key Symbols and Notation College Physics chapters 1-17 31 Radioactivity and Nuclear Physics 251 31.3 Substructure of the Nucleus Substructure of the Nucleus Define and discuss the nucleus in an atom. Define atomic number. Define and discuss isotopes. Calculate the density of the nucleus. Explain nuclear force. What is inside the nucleus? Why are some nuclei stable while others decay? (See [link].) Why are there different types of decay (αα size 12{α} {}, ββ size 12{β} {} and γγ size 12{γ} {})? Why are nuclear decay energies so large? Pursuing natural questions like these has led to far more fundamental discoveries than you might imagine. Why is most of the carbon in this coal stable (a), while the uranium in the disk (b) slowly decays over billions of years? Why is cesium in this ampule (c) even less stable than the uranium, decaying in far less than 1/1,000,000 the time? What is the reason uranium and cesium undergo different types of decay (αα size 12{α} {} and ββ size 12{β} {}, respectively)? (credits: (a) Bresson Thomas, Wikimedia Commons; (b) U.S. Department of Energy; (c) Tomihahndorf, Wikimedia Commons) We have already identified protons as the particles that carry positive charge in the nuclei. However, there are actually two types of particles in the nuclei—the proton and the neutron, referred to collectively as nucleons, the constituents of nuclei. As its name implies, the neutron is a neutral particle (q=0q=0 size 12{q=0} {}) that has nearly the same mass and intrinsic spin as the proton. [link] compares the masses of protons, neutrons, and electrons. Note how close the proton and neutron masses are, but the neutron is slightly more massive once you look past the third digit. Both nucleons are much more massive than an electron. In fact, mp=1836memp=1836me size 12{m rSub { size 8{p} } =”1836″ m rSub { size 8{e} } } {} (as noted in Medical Applications of Nuclear Physics and mn=1839memn=1839me size 12{m rSub { size 8{n} } =”1839″ m rSub { size 8{e} } } {}. [link] also gives masses in terms of mass units that are more convenient than kilograms on the atomic and nuclear scale. The first of these is the unified atomic mass unit (u), defined as 1 u=1.6605×10−27kg.1 u=1.6605×10−27kg. size 12{“1 u”=1 “.” “6605”´”10″ rSup { size 8{-“27″} } ” kg”} {} This unit is defined so that a neutral carbon 12C12C atom has a mass of exactly 12 u. Masses are also expressed in units of MeV/c2MeV/c2. These units are very convenient when considering the conversion of mass into energy (and vice versa), as is so prominent in nuclear processes. Using E=mc2E=mc2 size 12{E= ital “mc” rSup { size 8{2} } } {} and units of mm size 12{m} {} in MeV/c2MeV/c2 size 12{“MeV/”c rSup { size 8{2} } } {}, we find that c2c2 size 12{c rSup { size 8{2} } } {} cancels and EE size 12{E} {} comes out conveniently in MeV. For example, if the rest mass of a proton is converted entirely into energy, then E=mc2=(938.27 MeV/c2)c2=938.27 MeV.E=mc2=(938.27 MeV/c2)c2=938.27 MeV. size 12{E= ital “mc” rSup { size 8{2} } = ( “938” “.” “27” “MeV/”c rSup { size 8{2} } ) c rSup { size 8{2} } =”938” “.” “27”” MeV”} {} It is useful to note that 1 u of mass converted to energy produces 931.5 MeV, or 1 u=931.5 MeV/c2.1 u=931.5 MeV/c2. size 12{“1 u”=”931” “.” 5″ MeV/”c rSup { size 8{2} } } {} All properties of a nucleus are determined by the number of protons and neutrons it has. A specific combination of protons and neutrons is called a nuclide and is a unique nucleus. The following notation is used to represent a particular nuclide: ZAXN,ZAXN, size 12{“” lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } } {} where the symbols AA size 12{A} {}, XX size 12{X} {}, ZZ size 12{Z} {} , and NN size 12{N} {} are defined as follows: The number of protons in a nucleus is the atomic number ZZ size 12{Z} {}, as defined in Medical Applications of Nuclear Physics. X is the symbol for the element, such as Ca for calcium. However, once ZZ size 12{Z} {} is known, the element is known; hence, ZZ size 12{Z} {} and XX are redundant. For example, Z=20Z=20 size 12{Z=”20″} {} is always calcium, and calcium always has Z=20Z=20 size 12{Z=”20″} {}. NN size 12{N} {} is the number of neutrons in a nucleus. In the notation for a nuclide, the subscript NN size 12{N} {} is usually omitted. The symbol AA size 12{A} {} is defined as the number of nucleons or the total number of protons and neutrons, A=N+Z,A=N+Z, size 12{A=N+Z} {} where AA size 12{A} {} is also called the mass number. This name for AA size 12{A} {} is logical; the mass of an atom is nearly equal to the mass of its nucleus, since electrons have so little mass. The mass of the nucleus turns out to be nearly equal to the sum of the masses of the protons and neutrons in it, which is proportional to AA size 12{A} {}. In this context, it is particularly convenient to express masses in units of u. Both protons and neutrons have masses close to 1 u, and so the mass of an atom is close to AA size 12{A} {} u. For example, in an oxygen nucleus with eight protons and eight neutrons, A=16A=16 size 12{A=”16″} {}, and its mass is 16 u. As noticed, the unified atomic mass unit is defined so that a neutral carbon atom (actually a 12C12C size 12{“” lSup { size 8{“12”} } C} {} atom) has a mass of exactly 12 uu size 12{u} {}. Carbon was chosen as the standard, partly because of its importance in organic chemistry (see Appendix A). Masses of the Proton, Neutron, and Electron| Particle | Symbol | kg | u | MeV c 2 | --- --- | Proton | p | 1.67262×10−271.67262×10−27 size 12{1 “.” “67262” times “10” rSup { size 8{ – “27”} } } {} | 1.007276 | 938.27 | | Neutron | n | 1.67493×10−271.67493×10−27 size 12{1 “.” “67493”´”10″ rSup { size 8{-“27”} } } {} | 1.008665 | 939.57 | | Electron | e | 9.1094×10−319.1094×10−31 size 12{9 “.” “1094”´”10″ rSup { size 8{-“31”} } } {} | 0.00054858 | 0.511 | Let us look at a few examples of nuclides expressed in the ZAXNZAXN notation. The nucleus of the simplest atom, hydrogen, is a single proton, or 11H11H (the zero for no neutrons is often omitted). To check this symbol, refer to the periodic table—you see that the atomic number ZZ of hydrogen is 1. Since you are given that there are no neutrons, the mass number AA is also 1. Suppose you are told that the helium nucleus or αα particle has two protons and two neutrons. You can then see that it is written 24He224He2. There is a scarce form of hydrogen found in nature called deuterium; its nucleus has one proton and one neutron and, hence, twice the mass of common hydrogen. The symbol for deuterium is, thus, 12H112H1 (sometimes DD is used, as for deuterated water D2OD2O). An even rarer—and radioactive—form of hydrogen is called tritium, since it has a single proton and two neutrons, and it is written 13H213H2. These three varieties of hydrogen have nearly identical chemistries, but the nuclei differ greatly in mass, stability, and other characteristics. Nuclei (such as those of hydrogen) having the same ZZ size 12{Z} {} and different NN size 12{N} {} s are defined to be isotopes of the same element. There is some redundancy in the symbols AA size 12{A} {}, XX size 12{X} {}, ZZ size 12{Z} {}, and NN size 12{N} {} . If the element XX size 12{X} {} is known, then ZZ size 12{Z} {} can be found in a periodic table and is always the same for a given element. If both AA size 12{A} {} and XX size 12{X} {} are known, then NN size 12{N} {} can also be determined (first find ZZ size 12{Z} {}; then, N=A−ZN=A−Z size 12{N=A-Z} {}). Thus the simpler notation for nuclides is AX,AX, size 12{“” lSup { size 8{A} } X} {} which is sufficient and is most commonly used. For example, in this simpler notation, the three isotopes of hydrogen are 1H, 2H,1H, 2H, and 3H,3H, while the αα particle is 4He4He. We read this backward, saying helium-4 for 4He4He, or uranium-238 for 238U238U. So for 238U238U, should we need to know, we can determine that Z=92Z=92 for uranium from the periodic table, and, thus, N=238−92=146N=238−92=146 size 12{N=”238″-“92″=”146”} {}. A variety of experiments indicate that a nucleus behaves something like a tightly packed ball of nucleons, as illustrated in [link]. These nucleons have large kinetic energies and, thus, move rapidly in very close contact. Nucleons can be separated by a large force, such as in a collision with another nucleus, but resist strongly being pushed closer together. The most compelling evidence that nucleons are closely packed in a nucleus is that the radius of a nucleus, rr size 12{r} {}, is found to be given approximately by r=r0A1/3,r=r0A1/3, size 12{r=r rSub { size 8{0} } A rSup { size 8{1/3} } } {} where r0=1.2 fmr0=1.2 fm and AA is the mass number of the nucleus. Note that r3∝Ar3∝A size 12{r rSup { size 8{3} } µA} {}. Since many nuclei are spherical, and the volume of a sphere is V=(4/3)πr3V=(4/3)πr3 size 12{V= ( 4/3 ) pr rSup { size 8{3} } } {}, we see that V∝AV∝A size 12{V prop A} {} —that is, the volume of a nucleus is proportional to the number of nucleons in it. This is what would happen if you pack nucleons so closely that there is no empty space between them. A model of the nucleus. Nucleons are held together by nuclear forces and resist both being pulled apart and pushed inside one another. The volume of the nucleus is the sum of the volumes of the nucleons in it, here shown in different colors to represent protons and neutrons. How Small and Dense Is a Nucleus? (a) Find the radius of an iron-56 nucleus. (b) Find its approximate density in kg/m3kg/m3, approximating the mass of 56Fe56Fe to be 56 u. Strategy and Concept (a) Finding the radius of 56Fe56Fe is a straightforward application of r=r0A1/3,r=r0A1/3, given A=56A=56. (b) To find the approximate density, we assume the nucleus is spherical (this one actually is), calculate its volume using the radius found in part (a), and then find its density from ρ=m/Vρ=m/V. Finally, we will need to convert density from units of u/fm3u/fm3 to kg/m3kg/m3. Solution (a) The radius of a nucleus is given by r=r0A1/3.r=r0A1/3. Substituting the values for r0r0 and AA yields r=(1.2 fm)(56)1/3=(1.2 fm)(3.83)=4.6 fm.r=(1.2 fm)(56)1/3=(1.2 fm)(3.83)=4.6 fm. (b) Density is defined to be ρ=m/Vρ=m/V, which for a sphere of radius rr is ρ=mV=m(4/3)πr3.ρ=mV=m(4/3)πr3. Substituting known values gives ρ=56 u(1.33)(3.14)(4.6 fm)3=0.138 u/fm3.ρ=56 u(1.33)(3.14)(4.6 fm)3=0.138 u/fm3. Converting to units of kg/m3kg/m3, we find ρ=(0.138 u/fm3)(1.66×10–27kg/u)(1 fm10–15m)=2.3×1017kg/m3.ρ=(0.138 u/fm3)(1.66×10–27kg/u)(1 fm10–15m)=2.3×1017kg/m3. Discussion (a) The radius of this medium-sized nucleus is found to be approximately 4.6 fm, and so its diameter is about 10 fm, or 10–14m10–14m. In our discussion of Rutherford’s discovery of the nucleus, we noticed that it is about 10–15m10–15m in diameter (which is for lighter nuclei), consistent with this result to an order of magnitude. The nucleus is much smaller in diameter than the typical atom, which has a diameter of the order of 10–10m10–10m. (b) The density found here is so large as to cause disbelief. It is consistent with earlier discussions we have had about the nucleus being very small and containing nearly all of the mass of the atom. Nuclear densities, such as found here, are about 2×10142×1014 times greater than that of water, which has a density of “only” 103kg/m3103kg/m3. One cubic meter of nuclear matter, such as found in a neutron star, has the same mass as a cube of water 61 km on a side. Nuclear Forces and Stability What forces hold a nucleus together? The nucleus is very small and its protons, being positive, exert tremendous repulsive forces on one another. (The Coulomb force increases as charges get closer, since it is proportional to 1/r21/r2 size 12{1/r rSup { size 8{2} } } {}, even at the tiny distances found in nuclei.) The answer is that two previously unknown forces hold the nucleus together and make it into a tightly packed ball of nucleons. These forces are called the weak and strong nuclear forces. Nuclear forces are so short ranged that they fall to zero strength when nucleons are separated by only a few fm. However, like glue, they are strongly attracted when the nucleons get close to one another. The strong nuclear force is about 100 times more attractive than the repulsive EM force, easily holding the nucleons together. Nuclear forces become extremely repulsive if the nucleons get too close, making nucleons strongly resist being pushed inside one another, something like ball bearings. The fact that nuclear forces are very strong is responsible for the very large energies emitted in nuclear decay. During decay, the forces do work, and since work is force times the distance (W=FdcosθW=Fdcosθ size 12{W= ital “Fd””cos”θ} {}), a large force can result in a large emitted energy. In fact, we know that there are two distinct nuclear forces because of the different types of nuclear decay—the strong nuclear force is responsible for αα size 12{α} {} decay, while the weak nuclear force is responsible for ββ size 12{β} {} decay. The many stable and unstable nuclei we have explored, and the hundreds we have not discussed, can be arranged in a table called the chart of the nuclides, a simplified version of which is shown in [link]. Nuclides are located on a plot of NN size 12{N} {} versus ZZ size 12{Z} {}. Examination of a detailed chart of the nuclides reveals patterns in the characteristics of nuclei, such as stability, abundance, and types of decay, analogous to but more complex than the systematics in the periodic table of the elements. Simplified chart of the nuclides, a graph of NN size 12{N} {} versus ZZ size 12{Z} {} for known nuclides. The patterns of stable and unstable nuclides reveal characteristics of the nuclear forces. The dashed line is for N=ZN=Z size 12{N=Z} {}. Numbers along diagonals are mass numbers AA size 12{A} {}. In principle, a nucleus can have any combination of protons and neutrons, but [link] shows a definite pattern for those that are stable. For low-mass nuclei, there is a strong tendency for NN size 12{N} {} and ZZ size 12{Z} {} to be nearly equal. This means that the nuclear force is more attractive when N=ZN=Z size 12{N=Z} {}. More detailed examination reveals greater stability when NN size 12{N} {} and ZZ size 12{Z} {} are even numbers—nuclear forces are more attractive when neutrons and protons are in pairs. For increasingly higher masses, there are progressively more neutrons than protons in stable nuclei. This is due to the ever-growing repulsion between protons. Since nuclear forces are short ranged, and the Coulomb force is long ranged, an excess of neutrons keeps the protons a little farther apart, reducing Coulomb repulsion. Decay modes of nuclides out of the region of stability consistently produce nuclides closer to the region of stability. There are more stable nuclei having certain numbers of protons and neutrons, called magic numbers. Magic numbers indicate a shell structure for the nucleus in which closed shells are more stable. Nuclear shell theory has been very successful in explaining nuclear energy levels, nuclear decay, and the greater stability of nuclei with closed shells. We have been producing ever-heavier transuranic elements since the early 1940s, and we have now produced the element with Z=118Z=118 size 12{Z=”118″} {}. There are theoretical predictions of an island of relative stability for nuclei with such high ZZ size 12{Z} {} s. The German-born American physicist Maria Goeppert Mayer (1906–1972) shared the 1963 Nobel Prize in physics with J. Jensen for the creation of the nuclear shell model. This successful nuclear model has nucleons filling shells analogous to electron shells in atoms. It was inspired by patterns observed in nuclear properties. (credit: Nobel Foundation via Wikimedia Commons) Section Summary Two particles, both called nucleons, are found inside nuclei. The two types of nucleons are protons and neutrons; they are very similar, except that the proton is positively charged while the neutron is neutral. Some of their characteristics are given in [link] and compared with those of the electron. A mass unit convenient to atomic and nuclear processes is the unified atomic mass unit (u), defined to be 1 u=1.6605×10−27 kg=931.46 MeV/c2.1 u=1.6605×10−27 kg=931.46 MeV/c2. A nuclide is a specific combination of protons and neutrons, denoted by ZAXNor simplyAX,ZAXNor simplyAX, size 12{“” lSup { size 8{A} } X} {} ZZ size 12{Z} {} is the number of protons or atomic number, X is the symbol for the element, NN size 12{N} {} is the number of neutrons, and AA size 12{A} {} is the mass number or the total number of protons and neutrons, A=N+Z.A=N+Z. size 12{A=N+Z} {} Nuclides having the same ZZ size 12{Z} {} but different NN size 12{N} {} are isotopes of the same element. The radius of a nucleus, rr size 12{r} {}, is approximately r=r0A1/3,r=r0A1/3, where r0=1.2 fmr0=1.2 fm. Nuclear volumes are proportional to AA size 12{A} {}. There are two nuclear forces, the weak and the strong. Systematics in nuclear stability seen on the chart of the nuclides indicate that there are shell closures in nuclei for values of ZZ size 12{Z} {} and NN size 12{N} {} equal to the magic numbers, which correspond to highly stable nuclei. Conceptual Questions The weak and strong nuclear forces are basic to the structure of matter. Why we do not experience them directly? Define and make clear distinctions between the terms neutron, nucleon, nucleus, nuclide, and neutrino. What are isotopes? Why do different isotopes of the same element have similar chemistries? Problems & Exercises Verify that a 2.3×1017kg2.3×1017kg size 12{2 “.” 3 times “10” rSup { size 8{“17”} } “kg”} {} mass of water at normal density would make a cube 60 km on a side, as claimed in [link]. (This mass at nuclear density would make a cube 1.0 m on a side.) m=ρV=ρd3⇒a=mρ1/3=2.3×1017kg1000 kg/m313=61×103 m=61 kmm=ρV=ρd3⇒a=mρ1/3=2.3×1017kg1000 kg/m313=61×103 m=61 km Find the length of a side of a cube having a mass of 1.0 kg and the density of nuclear matter, taking this to be 2.3×1017 kg/m32.3×1017 kg/m3 size 12{2 “.” 3´”10″ rSup { size 8{“17″} } ” kg/m” rSup { size 8{3} } } {}. What is the radius of an αα size 12{α} {} particle? 1.9 fm1.9 fm size 12{1 “.” 9″ fm”} {} Find the radius of a 238Pu238Pu size 12{“” lSup { size 8{“238”} } “Pu”} {} nucleus. 238Pu238Pu size 12{“” lSup { size 8{“238”} } “Pu”} {} is a manufactured nuclide that is used as a power source on some space probes. (a) Calculate the radius of 58Ni58Ni size 12{“” lSup { size 8{“58”} } “Ni”} {}, one of the most tightly bound stable nuclei. (b) What is the ratio of the radius of 58Ni58Ni size 12{“” lSup { size 8{“58”} } “Ni”} {} to that of 258Ha258Ha size 12{“” lSup { size 8{“258”} } “Ha”} {}, one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom. (a) 4.6 fm4.6 fm size 12{4 “.” “6 fm”} {} (b) 0.61 to 10.61 to 1 size 12{0 “.” “61 to 1”} {} The unified atomic mass unit is defined to be 1 u=1.6605×10−27kg1 u=1.6605×10−27kg size 12{1″ u”=1 “.” “6605”×”10″ rSup { size 8{-“27”} } “kg”} {}. Verify that this amount of mass converted to energy yields 931.5 MeV. Note that you must use four-digit or better values for cc size 12{c} {} and ∣qe∣∣qe∣ size 12{ lline q rSub { size 8{e} } rline } {}. What is the ratio of the velocity of a ββ size 12{β} {} particle to that of an αα size 12{α} {} particle, if they have the same nonrelativistic kinetic energy? 85.4 to 185.4 to 1 size 12{“85” “.” “4 to 1”} {} If a 1.50-cm-thick piece of lead can absorb 90.0% of the γγ size 12{γ} {} rays from a radioactive source, how many centimeters of lead are needed to absorb all but 0.100% of the γγ size 12{γ} {} rays? The detail observable using a probe is limited by its wavelength. Calculate the energy of a γγ size 12{γ} {}-ray photon that has a wavelength of 1×10−16m1×10−16m size 12{1 times “10” rSup { size 8{ – “16”} } m} {}, small enough to detect details about one-tenth the size of a nucleon. Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe. 12.4 GeV12.4 GeV size 12{“12” “.” “4 GeV”} {} (a) Show that if you assume the average nucleus is spherical with a radius r=r0A1/3r=r0A1/3 size 12{r=r rSub { size 8{0} } A rSup { size 8{1/3} } } {}, and with a mass of AA size 12{A} {} u, then its density is independent of AA size 12{A} {}. (b) Calculate that density in u/fm3u/fm3 size 12{“u/fm” rSup { size 8{3} } } {} and kg/m3kg/m3 size 12{“kg/m” rSup { size 8{3} } } {}, and compare your results with those found in [link] for 56Fe56Fe size 12{“” lSup { size 8{“56”} } “Fe”} {}. What is the ratio of the velocity of a 5.00-MeV ββ size 12{β} {} ray to that of an αα size 12{β} {} particle with the same kinetic energy? This should confirm that ββ size 12{β} {}s travel much faster than αα size 12{β} {}s even when relativity is taken into consideration. (See also [link].) 19.3 to 1 (a) What is the kinetic energy in MeV of a ββ size 12{β} {} ray that is traveling at 0.998c0.998c? This gives some idea of how energetic a ββ size 12{β} {} ray must be to travel at nearly the same speed as a γγ ray. (b) What is the velocity of the γγ ray relative to the ββ size 12{β} {} ray? Glossary atomic mass the total mass of the protons, neutrons, and electrons in a single atom atomic number number of protons in a nucleus chart of the nuclides a table comprising stable and unstable nuclei isotopes nuclei having the same ZZ size 12{Z} {} and different NN size 12{Ns} {}s magic numbers a number that indicates a shell structure for the nucleus in which closed shells are more stable mass number number of nucleons in a nucleus neutron a neutral particle that is found in a nucleus nucleons the particles found inside nuclei nucleus a region consisting of protons and neutrons at the center of an atom nuclide a type of atom whose nucleus has specific numbers of protons and neutrons protons the positively charged nucleons found in a nucleus radius of a nucleus the radius of a nucleus is r=r0A1/3r=r0A1/3 size 12{r=r rSub { size 8{0} } A rSup { size 8{1/3} } } {} Previous/next navigation Previous: 31.2 Radiation Detection and Detectors Next: 31.4 Nuclear Decay and Conservation Laws Back to top License College Physics chapters 1-17 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. 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What is Conditional Probability? (Video & Practice Questions) Skip to content Online Courses Study Guides Flashcards Online Courses Study Guides Flashcards Menu Conditional Probability What is Conditional Probability? (Video & Practice Questions) On this page Review of Basic Probability Conditional Probability Conditional Probability Practice Questions [x] Transcript - [x] Practice Today, we’re going to take a look at conditional probability. This type of probability is all about finding the odds of something happening given that something else has already occurred. Review of Basic Probability Let’s start with a refresher on basic probability. The probability of something happening can be expressed as the number of desired outcomes divided by the number of total possible outcomes: P(A)P(A)==Number of desired outcomes Total number of possible outcomes Number of desired outcomes Total number of possible outcomes Example #1 Here’s an example of a basic probability problem. There are 10 marbles in a bag: three are blue, three are red, and four are orange. What are the odds of pulling out a red marble? To solve this, we’d place the number of desired outcomes on the top (in this case, it’s 3 because there are three red marbles) and the total number of possible outcomes on the bottom (this is 10 since there are 10 total marbles in the bag). P(A)P(A)==3 red marbles 10 total marbles 3 red marbles 10 total marbles Seems pretty straightforward, right? Conditional Probability Well, conditional probability adds a bit of a twist to this, as the objects or people in question often have more than one possible attribute. Example #1 Let’s look at an example problem with more than one possible attribute. Out of 100 houses sold, 50 had a garage, 40 had a pool, and of these, 10 had both a garage and a pool, leaving 20 houses with neither. Given that a house had a garage, what are the odds that it also had a pool? What do we do with all this information? If we draw it as a Venn diagram, we can make more sense of it: We can now see the overlap between the houses sold with garages and the houses sold with pools, all within the larger set of houses sold. It’s easier to see now that of the 50 houses sold with a garage, 10 of them also had a pool. So the odds are 10 50 10 50, which can be expressed as the fraction1 5 1 5, or 20%. Fortunately, there’s a faster way to do this than by drawing a diagram every time you encounter a conditional probability question. Conditional Probability Formula There’s a handy formula we can use: P(B|A)=P(B and A)P(A)P(B|A)=P(B and A)P(A) We read this formula as “the probability P of event B happening given that event A has happened is equal to the probability of events B and A both happening over the probability that event A has happened.” In these types of problems, both probabilities on the right side of the equation are usually given or can be determined from a sentence or table. So in our sample problem, the P(B|A)P(B|A) is the 10 houses that were sold with both a garage and a swimming pool, out of the total 100 houses. This fraction can be written as 1 10 1 10. The P(A) is the probability of a house being sold with a garage (event A), which is 50 houses out of the 100, or 5 10 5 10. Now that our formula is complete, we can divide 1 10 1 10 by 5 10 5 10 and we get 1 5 1 5, which is equal to 20%, just as we figured out from our diagram. Example #2 Let’s try one more with numbers that aren’t quite so neat. In a small town, there are 216 men: 144 of them have brown hair and 56 of them have mustaches. 24 of the men have brown hair and mustaches. What are the odds that a man chosen at random has a mustache given that he has brown hair? Let’s plug into our formula. The probability of event A (brown hair) and event B (mustache) is the 24 men with both, divided by the 216 total men. The probability of event A is 144 divided by the 216 total men. We divide 24 216 24 216 by 144 216 144 216 using fraction division. The 216s can be canceled out, so we are left with 24 144 24 144. This fraction can be reduced to 1 6 1 6, or ≈16.67%. All right, that’s all for this review. I hope this increased your odds of understanding conditional probability. Thanks for watching, and happy studying! Conditional Probability Practice Questions Question #1: What is the formula for conditional probability? P(B|A)=P(A)P(B)P(B|A)=P(A)P(B) P(B|A)=P(B)P(A)P(B|A)=P(B)P(A) P(B|A)=P(A and B)P(A)P(B|A)=P(A and B)P(A) P(B|A)=P(B and A)P(A and B)P(B|A)=P(B and A)P(A and B) [x] Show Answer Answer: The conditional probability formula allows us to determine the probability that Event B will occur, given the knowledge that Event A has already occurred. P(B|A)P(B|A) stands for “the probability of B given A.” The probability of Event B can be determined by dividing the probability of Events A and B occurring by the probability of Event A occurring. [x] Hide Answer Question #2: The Venn diagram below shows sections for the probability of Event A occurring and the probability of Event B occurring. What does the shaded middle section represent? The shaded middle section represents P(A B)P(A B) The shaded middle section represents P(A and B)P(A and B) The shaded middle section represents P(A not occurring)P(A not occurring) The shaded middle section represents P(B not occurring)P(B not occurring) [x] Show Answer Answer: The middle section of the Venn diagram represents the overlap of Events A and B, where both events occur. [x] Hide Answer Question #3: Which Venn diagram represents the following scenario: 100 customers have purchased cars at Jason’s car dealership this year. 40 of those customers purchased cars with touch-screen GPS, and 30 customers purchased cars with snow tires. 20 customers purchased cars with both touch-screen GPS and snow tires. [x] Show Answer Answer: The far left portion of the Venn diagram represents the customers who purchased cars with touch-screen GPS but no snow tires. The far right portion of the Venn diagram represents the customers who purchased cars with snow tires but not touch-screen GPS. The middle section represents customers who purchased cars with both touch-screen GPS and snow tires. [x] Hide Answer Question #4: Consider the same group of 100 customers from Jason’s car dealership. 40 purchased GPS, 30 purchased snow tires, and 20 purchased both. If a randomly selected customer has purchased the GPS, what would the probability be that they also purchased the snow tires? 20% 30% 40% 50% [x] Show Answer Answer: The conditional probability formula can be used to solve this problem. The probability of Event A occurring (customer purchasing GPS) is 40 out of 100 customers, or 0.4. The probability of Event B occurring (customer purchasing snow tires) is 30 out of 100, or 0.3. The probability of a customer purchasing both, or Event A and B occurring, is 20 out of 100, or 0.2. We know P(A) = 0.4 and P (A and B) = 0.2 so we can plug these into the formula. P(A)=P(A and B)P(A)P(A)=P(A and B)P(A) becomes P(B|A)=0.2 0.4 P(B|A)=0.2 0.4, which simplifies to 0.5, or 50%. [x] Hide Answer Question #5: Sybil took two science tests. The probability of her passing both tests is 0.6 and the probability of her passing the first test is 0.8. What is the probability of her passing the second test if she passed the first test? 65% chance of passing the second test 75% chance of passing the second test 85% chance of passing the second test 95% chance of passing the second test [x] Show Answer Answer: The conditional probability formula can be used to solve this problem. The probability of Event A occurring (passing the first test) is 0.8. The probability of Event A and B occurring (passing both tests) is 0.6. We know P(A)=0.8 P(A)=0.8 and P(A and B)=0.6 P(A and B)=0.6 so we can plug these into the formula. P(B|A)=P(A and B)P(A)P(B|A)=P(A and B)P(A) becomes P(B|A)=0.6 0.8 P(B|A)=0.6 0.8 which simplifies to 0.75 or 75%. [x] Hide Answer Return to Probability Videos 397924 by Mometrix Test Preparation | Last Updated: August 20, 2025 On this page Review of Basic Probability Conditional Probability Conditional Probability Practice Questions Why you can trust Mometrix Raising test scores for 20 years 150 million test-takers helped Prep for over 1,500 tests 40,000 5-star reviews A+ BBB rating Who we are About Mometrix Test Preparation We believe you can perform better on your exam, so we work hard to provide you with the best study guides, practice questions, and flashcards to empower you to be your best. Learn More... 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HomeOn proofs for monotonicity of a function involving the psi and exponential functions Article Licensed Unlicensed Requires Authentication Cite this On proofs for monotonicity of a function involving the psi and exponential functions Wen-Hui LiWen-Hui Li Search for this author in: De Gruyter De Gruyter Brill| Google Scholar , Feng QiFeng Qi Search for this author in: De Gruyter De Gruyter Brill| Google Scholar and Bai-Ni GuoBai-Ni Guo Search for this author in: De Gruyter De Gruyter Brill| Google Scholar Published/Copyright:April 2, 2013 or Purchase Article 30,00 € Purchase Article £23.00 Purchase Article $42.00 Published by Become an author with De Gruyter Brill Submit ManuscriptAuthor InformationExplore this Subject From the journalVolume 33 Issue 1 MLA MLA APA Harvard Chicago Vancouver MLA APA Harvard Chicago Vancouver Li, Wen-Hui, Qi, Feng and Guo, Bai-Ni. "On proofs for monotonicity of a function involving the psi and exponential functions" , vol. 33, no. 1, 2013, pp. 45-50. Li, W., Qi, F. & Guo, B. (2013). On proofs for monotonicity of a function involving the psi and exponential functions. , 33(1), 45-50. Li, W., Qi, F. and Guo, B. (2013) On proofs for monotonicity of a function involving the psi and exponential functions. Vol. 33 (Issue 1), pp. 45-50. Li, Wen-Hui, Qi, Feng and Guo, Bai-Ni. "On proofs for monotonicity of a function involving the psi and exponential functions" 33, no. 1 (2013): 45-50. Li W, Qi F, Guo B. On proofs for monotonicity of a function involving the psi and exponential functions. 2013;33(1): 45-50. Copy Copied to clipboard BibTeXEndNoteRIS Article Abstract In the note, the authors review the past and long proofs and supply two new and short proofs for a monotonicity result that the function ψ(x)+ln(e 1/x−1) is strictly increasing on (0,∞), where ψ(x) is the psi function. Keywords: Proof; monotonicity; psi function; exponential function Correspondence address: Tianjin Polytechnic University, School of Science, Department of Mathematics, Tianjin City, 300387, Volksrepublik China, qifeng618@gmail.com; qifeng618@hotmail.com; qifeng@qq.com Published Online:2013-04-02 Published in Print:2013-03 © by Oldenbourg Wissenschaftsverlag, München, Germany You are currently not able to access this content. Not sure if you should have access? Please log in using an institutional account to see if you have access to view or download this content. Purchase Article $42.00Purchase Article $42.00 You are currently not able to access this content. Not sure if you should have access? Please log in using an institutional account to see if you have access to view or download this content. Purchase Article $42.00Purchase Article $42.00 Articles in the same Issue Articles in the same Issue Uniform spacing of zeros of orthogonal polynomials for locally doubling measures A uniqueness polynomial for equi-polar meromorphic functions The spectrum of the Hausdorff operator on a subspace of L 2(ℝ) An application of q-mathematics on absolute summability of orthogonal series On proofs for monotonicity of a function involving the psi and exponential functions On the (M,λ n) method of summability Developments of the theory of generalized functions or distributions – A vision of Paul Dirac Readers are also interested in: Capacity solutions for anisotropic variable exponent parabolic-elliptic systems with degenerate term Analysis Doubly warped product Hermitian manifold with constant holomorphic sectional curvature Analysis On spherical shells containing all the zeros of a quaternionic polynomial Analysis Next Search journal Search the content of this journal Keywords for this article Proof; monotonicity; psi function; exponential function Articles in the same Issue Uniform spacing of zeros of orthogonal polynomials for locally doubling measures A uniqueness polynomial for equi-polar meromorphic functions The spectrum of the Hausdorff operator on a subspace of L 2(ℝ) An application of q-mathematics on absolute summability of orthogonal series On proofs for monotonicity of a function involving the psi and exponential functions On the (M,λ n) method of summability Developments of the theory of generalized functions or distributions – A vision of Paul Dirac Sign up now to receive a 20% welcome discount Subscribe to our newsletter Institutional Access How does access work? 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Multiplying complex numbers in polar form | Precalculus | Khan Academy Khan Academy 9090000 subscribers 172 likes Description 16583 views Posted: 28 Jun 2021 Keep going! Check out the next lesson and practice what you’re learning: Sal shows how to multiply two complex numbers in polar form by multiplying their moduli, adding their arguments, and simplifying. 7 comments Transcript: we're given two different complex numbers here and we want to figure out what is the product pause this video and see if you can figure that out all right now let's work on this together so we know from the form that it's written here that the modulus of w sub 1 is equal to 3 and we know that the argument of w sub 1 is equal to 330 degrees and by the same line of reasoning we know that the modulus of w sub 2 is equal to 2 and that the argument of w sub 2 is going to be equal to we can see that right over here 120 degrees now when you multiply complex numbers you could view as one transforming the other we've seen this in multiple examples so let's imagine that we are transforming w2 by multiplying it by w1 so what is going to happen well let me write it here so what's the resulting modulus of w1 times w2 well we're just going to scale up w2's modulus by w1's modulus or essentially we're just going to multiply the two so this is going to be equal to 6 3 times 2 and then the argument of w sub 1 times w sub 2. if we start at w sub 2's argument 120 degrees and then we rotate it by w sub 1's argument well then you're going to add these two angles that gets you to 450 degrees so this is equal to 450 degrees which is more than a complete rotation and so if we wanted to give it an angle between 0 and 360 degrees if we just subtract 360 from that that is going to be equal to 90 degrees and so we can rewrite this here or we can rewrite the product as w sub 1 times w sub 2 is equal to its modulus 6 times cosine of its argument 90 degrees plus i times sine of its argument now we know what the cosine and sine of 90 degrees is cosine of 90 degrees is equal to 0 and sine of 90 degrees is equal to 1. so all of this simplifies quite nicely all you're left with is a 6 times i so this is equal to 6 i and we are done
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How to Use the Addition Rule for Disjoint Events | Statistics and Probability | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright How to Use the Addition Rule for Disjoint Events High School Statistics Skills Practice Mary Bechtel Instructors Mary Bechtel Mary Bechtel has taught numerous middle and high school math topics for five years. She holds a Master's in Curriculum and Instruction as well as a Bachelor's in Mathematics. She has presented at local, state, and national education conferences for over five years and has two publications on teaching mathematics. View bio Example SolutionsPractice Questions How to Use the Addition Rule for Disjoint Events Step 1: Determine if the two events are disjoint. Step 2: Determine the probability of the first event occurring, P(A). Step 3: Determine the probability of the second event occurring, P(B). Step 4: Calculate the probability of either of the events occurring by adding the probabilities P(A) and P(B). What are the Addition Rule and Disjoint Events? Addition Rule of Probability: a rule used to determine the probability of either event occurring. It is calculated by adding the probabilities of each of the two events then subtracting the probability of both events happening at the same time. Addition Rule of Probability Formula:P(A or B)=P(A)+P(B)−P(A and B) Disjoint Events: events that can not occur at the same time. Disjoint events are also called mutually exclusive events. Addition Rule of Probability of Disjoint Events Formula: a variation of the general Addition Rule of Probability formula. Since it is impossible for two disjoint events to occur at the same time, P(A and B)=0. For two disjoint events, P(A or B)=P(A)+P(B)−P(A and B)P(A or B)=P(A)+P(B)−0 P(A or B)=P(A)+P(B) Let's use these steps and definitions to work through two examples of using the addition rule for disjoint events. Examples of Using the Addition Rule for Disjoint Events Example 1 What is the probability of rolling an even number or a 1 on a fair six-sided number cube? Write your answer as a decimal rounded to the nearest hundredth. Step 1: Determine if the two events are disjoint. Since it is not possible to roll an even number and a 1 at the same time, these events are disjoint. Step 2: Determine the probability of the first event occurring, P(A). The first event is rolling an even number. There are 3 even numbers on a 6 sided number cube, so P(A)=P(rolling an even number)=3 6=0.5 Step 3: Determine the probability of the second event occurring, P(B). The second event is rolling a 1. There is a single 1 on a 6 sided number cube, so P(B)=P(rolling a 1)=1 6=0.17 Step 4: Calculate the probability of either of the events occurring by adding the probabilities P(A) and P(B). We now solve P(rolling an even number or a 1)=P(rolling an even number)+P(rolling a 1)=0.5+0.17=0.67 The probability of rolling an even number or a 1 is 0.67. Example 2 What is the probability of drawing an even number or a King from a standard deck of cards? Write your answer as a percentage rounded to the nearest whole number. Step 1: Determine if the two events are disjoint. Since it is not possible to draw an even number and a King at the same time, these events are disjoint. Step 2: Determine the probability of the first event occurring, P(A). The first event is drawing an even number. The even numbers in a deck of cards are 2, 4, 6, 8, and 10. There are four of each number. Therefore, there are 20 even-numbered cards out of 52 total cards in the deck. This means P(A)=P(drawing even number)=20 52=0.38=38% Step 3: Determine the probability of the second event occurring, P(B). The second event is drawing a King. There are 4 Kings out of 52 total cards in the deck, so P(B)=P(drawing a King)=4 52=0.08=8% Step 4: Calculate the probability of either of the events occurring by adding the probabilities P(A) and P(B). We now solve P(drawing an even number or a King)=P(drawing an even number)+P(drawing a King)=38%+8%=46% The probability of drawing an even number or a King from a standard deck of cards is 46%. Get access to thousands of practice questions and explanations! 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CheckYourMath Home Length / Distance Conversions CONVERSIONS Length / Distance Weight / Mass Temperature Fuel Economy Area Volume Time Pressure Angle Energy Frequency Power / Electricity Speed / Velocity Digital Storage Data Rates Color Value MORE Geometry Convert nm to cm - Length / Distance Conversions Online calculator to convert nanometers to centimeters (nm to cm) with formulas, examples, and tables. Our conversions provide a quick and easy way to convert between Length or Distance units. Conversion Calculator Enter your value in the conversion calculator below. Convert nm to cm TIP: If the result of your conversion is 0, try increasing the "Decimals". How to convert nm to cm: Enter a value in the nm field and click on the "Calculate cm" button. Your answer will appear in the cm field. Conversion Definitions The following is a list of definitions relating to conversions between nanometers and centimeters. What is a nanometer (nm)? A nanometer is a unit of Length or Distance in the Metric System. The symbol for nanometer is nm. There are 10,000,000 nanometers in a centimeter. The International spelling for this unit is nanometre. What is a centimeter (cm)? A centimeter is a unit of Length or Distance in the Metric System. The symbol for centimeter is cm. There are 0.0000001 centimeters in a nanometer. The International spelling for this unit is centimetre. Conversion Formula Let's take a closer look at the conversion formula so that you can do these conversions yourself with a calculator or with an old-fashioned pencil and paper. The formula to convert from nm to cm is: cm = nm ÷ 10,000,000 Conversion Example Next, let's look at an example showing the work and calculations that are involved in converting from nanometers to centimeters (nm to cm). Nanometer to Centimeter Conversion Example Task: Convert 750,000 nanometers to centimeters (show work) Formula: nm ÷ 10,000,000 = cm Calculations: 750,000 nm ÷ 10,000,000 = 0.075 cm Result: 750,000 nm is equal to 0.075 cm Conversion Table For quick reference purposes, below is a conversion table that you can use to convert from nm to cm. Nanometers to Centimeters Conversion Chart | nanometers (nm) | centimeters (cm) | --- | | 1 nm | 0.0000001 cm | | 2 nm | 0.0000002 cm | | 3 nm | 0.0000003 cm | | 4 nm | 0.0000004 cm | | 5 nm | 0.0000005 cm | | 6 nm | 0.0000006 cm | | 7 nm | 0.0000007 cm | | 8 nm | 0.0000008 cm | | 9 nm | 0.0000009 cm | | 10 nm | 0.000001 cm | | 20 nm | 0.000002 cm | | 30 nm | 0.000003 cm | | 40 nm | 0.000004 cm | | 50 nm | 0.000005 cm | | 75 nm | 0.0000075 cm | | 100 nm | 0.00001 cm | | 250 nm | 0.000025 cm | | 500 nm | 0.00005 cm | | 750 nm | 0.000075 cm | | 1,000 nm | 0.0001 cm | | 2,500 nm | 0.00025 cm | | 5,000 nm | 0.0005 cm | | 7,500 nm | 0.00075 cm | | 10,000 nm | 0.001 cm | | 25,000 nm | 0.0025 cm | | 50,000 nm | 0.005 cm | | 75,000 nm | 0.0075 cm | | 100,000 nm | 0.01 cm | | 250,000 nm | 0.025 cm | Related Units This table provides a summary of the Length or Distance units within their respective measurement systems. | Unit | Symbol | Measurement System | Description | --- --- | | inches | in or " | US Customary Units/Imperial System | 36 inches = 1 yard | | feet | ft or ' | US Customary Units/Imperial System | 1 foot = 12 inches | | yards | yd | US Customary Units/Imperial System | 1 yard = 3 feet | | miles | mi | US Customary Units/Imperial System | 1 mile = 1760 yards or 5280 feet | | picometers | pm | Metric System | 1 m = 1,000,000,000,000 pm | | nanometers | nm | Metric System | 1 m = 1,000,000,000 nm | | micrometers | μm | Metric System | 1 m = 1,000,000 μm | | millimeters | mm | Metric System | 1 m = 1,000 mm | | centimeters | cm | Metric System | 1 m = 100 cm | | decimeters | dm | Metric System | 1 m = 10 dm | | meters | m | Metric System | base unit | | decameters | dam or dkm | Metric System | 1 dam = 10 m | | hectometers | hm | Metric System | 1 hm = 100 m | | kilometers | km | Metric System | 1 km = 1,000 m | | megameters | Mm | Metric System | 1 Mm = 1,000,000 m | | gigameters | Gm | Metric System | 1 Gm = 1,000,000,000 m | | nautical miles | M or NM or nmi | Non-SI (International) | 1 nmi = 1,852 meters | | angstroms | Å | Non-SI (International) | 10,000,000,000 Å = 1 m | | rods | rod | Non-SI (International) | 320 rods = 1 mile | Note: For Length and Distance conversions, US Customary Units and the Imperial System are equivalent. Find a Conversion Looking for a conversion? Select a conversion type and the desired units. Share on: Home | About Us | Contact Us | Donate While using this site, you agree to have read and accepted our Terms of Service and Privacy Policy. Copyright © 2012-2025 CheckYourMath.com. All rights reserved.
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Lesson 4: Estimating School Store Purchases – Raymond Geddes Skip to content Spend $45.00 to get free shipping Spend $45.00 to get free shipping Spend $45.00 to get free shipping Spend $45.00 to get free shipping Spend $45.00 to get free shipping Spend $45.00 to get free shipping Spend $45.00 to get free shipping Spend $45.00 to get free shipping Menu Shapes Navigation bar Close (esc) Shop Offers $1 and Under Geddes Exclusives Best Sellers Bundles New Items Sale Items Weekly Specials Price Reductions Clearance School Supplies Writing Supplies & Accessories Pens Pencils Pencil Toppers & Grips Pencil Pouches Mechanical Pencils & Lead Refills Highlighters Erasers Pencil Sharpeners Scented School Supplies Novelty Toys & Games Stress Balls Keychains Plush 3D Printed Toys Putty, Slime, and Goo Squishy Toys Tech Gadgets Wearables Classroom Supplies Art Supplies Backpacks Back-to-School Essentials Bookmarks Desk Pets Fidget & Sensory Toys Happy Birthday Incentives & Rewards Paper Products Treasure Boxes Fan-Favorite Brands Pete the Cat Dr. Seuss™ Dog Man™ Diary of a Wimpy Kid Seasonal Valentines Easter/Spring Halloween/Fall Christmas/Winter Grinch School Stores School Kits Sold Individually Goodie Bags Contact Us Log in Create account Search Search Facebook Twitter Pinterest Instagram TikTok YouTube LinkedIn Shop Offers $1 and Under Geddes Exclusives Best Sellers Bundles New Items Sale Items Weekly Specials Price Reductions Clearance School Supplies Writing Supplies & Accessories Pens Pencils Pencil Toppers & Grips Pencil Pouches Mechanical Pencils & Lead Refills Highlighters Erasers Pencil Sharpeners Scented School Supplies Novelty Toys & Games Stress Balls Keychains Plush 3D Printed Toys Putty, Slime, and Goo Squishy Toys Tech Gadgets Wearables Classroom Supplies Art Supplies Backpacks Back-to-School Essentials Bookmarks Desk Pets Fidget & Sensory Toys Happy Birthday Incentives & Rewards Paper Products Treasure Boxes Fan-Favorite Brands Pete the Cat Dr. Seuss™ Dog Man™ Diary of a Wimpy Kid Seasonal Valentines Easter/Spring Halloween/Fall Christmas/Winter Grinch School Stores School Kits Sold Individually Goodie Bags Contact Us Log in Search our site Search Close (esc) Cart items Lesson 4: Estimating School Store Purchases Return to Lesson Plan Home Page | Download This Lesson | Next Lesson Lesson Title: Estimating School Store Purchases? Grade Band: 3-5 Lesson Length: Approximately 2 days NCTM Standard: Number and Operations Standard Learning Objectives The student will identify appropriate uses for estimation. The student will round purchase prices to estimate total purchases in a school store setting. The student will practice estimating by using rounding and mental computations to arrive at estimates. Connection to Bloom’s Taxonomy Comprehension Application Analysis Synthesis Lesson Materials Calculators Clear Jar ( or other container ) Crayons or Colored Pencils Dice Dictionary ( Printed and/or on the Web ) Happy Cap Erasers ( or any other small sized school store item appropriate for estimating quantity ) Index Cards Item Cards from the Geddes Kit List Pencils Worksheets for lesson plan 4 (see sidebar) Making Connections What is an estimate? Ask several students to use a dictionary or on-line dictionary to define the word “estimate.” Record several of those definitions on the board. In general, an estimate is an approximate calculation. Use Word Origin by Hannie in the sidebar for additional clarification. Ask students to think about times when they may use estimating outside of the classroom. Have you ever purchased something on sale? What does it mean to buy something on sale? Have students brainstorm responses to the following two questions. Some possible responses may include: To determine how long it will take to drive somewhere unfamiliar To determine how long it will take to complete a homework assignment To determine how much time will be needed to complete a project at work or home To determine how much money to bring to the grocery store to purchase a week’s worth of food To determine what time to set the alarm clock based on how long it will take to get ready in the morning To determine how many suitcases to pack for a week-long family vacation To determine how much gasoline will be needed to fill the car gas tank To determine a weekly or monthly budget for household expenses To determine how much food to prepare for a party Discuss why estimation is useful and explain to students the overall purpose of estimation. Students should understand that estimates provide approximations or reasonable answers to questions. Although the exact amount or number might not be known, an estimate helps to plan and prepare. Estimates actually help people to solve problems. The world around us is filled with unknowns. We do not and will not always know or have the exact answer to a question or problem. However, estimates provide people with a way of making reasonable, intelligent approximations when the exact answer is not easily or readily available. Exploring and Learning Fill a jar with Happy Cap Erasers or any other small-sized school store item. Show a single item to students so that they have an idea as to the item’s shape and size. Then show the class the filled jar of erasers and ask them to estimate the number of erasers or items in the jar. Record student estimations on the board. As a class, look at the list of estimates and determine if there was a wide range of estimates or were the numbers fairly close to one another. Discuss how estimates are derived by asking students to share their approaches, thought processes, and methods for developing their estimates. Often estimates are developed based on prior knowledge, rough calculations, or simply an educated guess. Explain that estimating is very useful when making purchases. It is a helpful way to calculate quickly the total amount to be purchased and to ensure you have enough money for your purchases. Use the Estimating Store Purchases - Guided Practice (PDF) as a transparency to help guide students through the process of estimation. Explain to students that making estimates in a store may involve rounding, mental calculations, or using paper and pencil to do quick calculations. It is often better to estimate a little higher to ensure that you have enough money to afford the purchase. Present students with the following scenario: RG and Hannie are purchasing for their Raymond Geddes Elementary School Store. Today in math class, they learned about estimation and want to practice what they have learned for the school store. RG and Hannie will help estimate the total price of two different purchases. With this information, the students will know whether or not they have enough money to buy what they want to sell at the school store. Sniffer will tabulate the exact total price using a calculator. Can you help RG, Hannie, and Sniffer estimate the students’ purchases? 7. To help complete the scenario, pair students together and provide them with a copy of Estimate Store Purchases Worksheet (PDF). Explain and list on the board or as a transparency the following instructions: Use rounding to generate an estimate for each item. Use a calculator to determine the actual total price. Does each student have enough money to make his/her purchases? As a class, review the results and determine if Natalie and Matt have enough money to make their purchases? Student estimations should determine that Natalie does not have enough money to purchase all the items that she wanted to buy. Review with students Natalie’s 2 alternatives found on the Estimate Store Purchases Worksheet Key (PDF). The scenario involving Matt provides a different look at estimation. Some students will overestimate Matt’s total purchase price when rounding upward. Estimating by rounding upward showed that Matt could not afford all his items. However the actual price for all of his items was less than the $75.00 he had to spend. Rounding each item’s price upward yielded an overestimate. Review the upward estimation on the Estimate Store Purchases Worksheet (PDF) with students. Explain that when you are making purchases, it is often better to overestimate than to underestimate the total purchase price. Provide students with additional practice at estimating purchase totals by creating item cards using index cards and the Geddes Kit List (PDF). You can create up to 20 item cards, each containing the item name and corresponding purchase price. For example, one card will state “Retro Pencil $8.64/Display”. Fold the cards in half and place them in a container. Call two students to the front of the classroom to lead the activity. One student will roll a die and the other will draw an item card from the container. Divide the remainder of the class into teams of 3-4 students. All students will participate by estimating (using mental computation only) a total purchase price within their assigned teams. Remind students that you do not want the actual total price, but an estimate. To estimate in this activity means to round the item price first and then to multiply that price by the quantity rolled on the die. Ask each team to jot down their estimate. Students will share their estimates aloud after each card is drawn and die rolled. Did everyone arrive at the same estimate? If not, discuss any differences in creating the estimate. Extended Learning and Practice Open up your own school store using the Geddes School Store Kit and our School Stores: An Operating Manual. Operating a school store is an excellent opportunity for inquiry based/hands-on learning that is both fun and rewarding. Create an instant real life business in your classroom by using the merchandise in the Geddes Kit and following the school store operations manual. A school store will allow opportunity for raising money, provides a service to students, and creates an atmosphere of responsibility while building valuable team working skills. This lesson demonstrated how customers use estimation prior to making purchases. Identify several other ways estimation can be used within a school store. Some possibilities include: estimating and forecasting future sales based on prior actual sales data, estimating the amount of shelf space needed to display store items, or simply estimating how much help will be needed to operate the store. Challenge students by using two dice to perform mental computations for each item card created in step 10 of the Exploring and Learning component. Explain to students that estimation is very helpful when the quantities increase and you can begin to group by tens. Have students think about what it means to overestimate versus underestimate. Ask which do you think is better, and why. Provide one or two examples to support your answer. Visit (U.S. Department of Education Archives) for additional estimation activities that can be completed at the grocery store. Visit Figure This! Math Challenges for Families. The Challenge Index contains estimation challenges. This website was developed by the National Council of Teachers of Mathematics. Assessment The lesson objectives can be assessed by evaluating the Estimate Store Purchases Worksheet (PDF) with the Estimate Store Purchases Answer Key (PDF). Use the Assessment of Student Progress (PDF) to assess students’ overall abilities to meet the lessons learning objectives which include rounding purchase prices, estimating individual item totals, and estimating overall purchase amounts. Closure Provide each student with an index card and have them answer the following questions on one side of the index card: What are two new things that you have learned? What else would you like to learn about this topic? On the back side of the index card, instruct the students to draw a picture of something they learned about during this lesson. The index cards can be hole punched and held together with a simple shower curtain ring. Learning Model Making Connections Exploring and Learning Extended Learning and Practice Assessment Closure Lesson Worksheets Geddes Kit List Estimating Store Purchases Guided Practice Estimate Store Purchases Worksheet Estimate Store Purchases Key Assessment of Student Progress Download all Worksheet PDFs Download all worksheets & the lesson. Teaching Strategies Brainstorming Guided Practice Paired Learning See teaching strategies for all lessons. Literary Connection Great Estimations by Bruce Goldstone is a fun book that teaches children to make intelligent estimations of grouped items. Children are encouraged to not count individual items on a page, but to estimate the items by grouping and clump counting. Estimation skills increase with practice and this book provides page after page of colored photos of grouped items such as bugs, candy, animals or people in a swimming pool. Goldstone has written several educational books and is well known for writing math word problems. In fact, he estimates that he has written more than 30,000 word problems, more than 12,000 pages in different textbooks, and over 2,000,000 individual words! See book recommendations for all lessons. Word Origins Estimation is from the Latin word aestimationem which means a valuation, and from the word aestimare which means to value. Used as a noun, estimation can be a document or a calculation that provides an approximate value of something such as an item like a car or house. It tells you how much something is worth or how much it cost. The damage to the car is estimated at five hundred dollars. I estimate the jar of jelly beans to weigh about 3 pounds. It can also be used to judge character or provide opinion. It is my estimation that our football team will win the championship this year. In my estimation the class will do well on the math test. See word origins for all lessons. Vocabulary List Estimate: To give an approximate calculation. Round: To give a close approximation of a number, a form of estimating. See vocabulary words for all lessons. Free Shipping We offer free shipping on purchases over $45 Top-notch Support Contact us to get in touch with our support team. 888-431-1722 Call us any time Monday to Friday, 8:00am through 4:00pm EST Secure Payments Your payment information is processed securely GEDDES is your wholesale school supply store! Schools earn Extra Credit Rewards on every purchase and enjoy 30-Day Billing. Businesses save every day, and all customers enjoy our 100% Satisfaction Guarantee with 100 years of experience. Facebook Twitter Pinterest Instagram TikTok YouTube LinkedIn About Raymond Geddes Contact Us Catalog Library Frequently Asked Questions Raymond Geddes Wholesale Satisfaction Guarantee Why Shop With Us? 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https://www.youtube.com/watch?v=lx0_n30CgAI
Constructing a REGULAR POLYGON: GeoGebra Beginner Exercise 1 Tim Brzezinski 13700 subscribers 47 likes Description 29626 views Posted: 24 Apr 2018 Screencast displays how to use the REGULAR POLYGON tool to construct various regular polygons in GeoGebra. Try it yourself here: 3 comments Transcript:
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https://www.enchantedlearning.com/math/followthearrows/
Follow-the-Arrows Math Puzzles - EnchantedLearning.com Discover more Math Puzzle Puzzles Enchanted Learning Enchanted Learning Center Mathematics Advertisement. EnchantedLearning.com is a user-supported site. As a bonus, site members have access to a banner-ad-free version of the site, with print-friendly pages. Click here to learn more. (Already a member? Click here.) 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https://www.reddit.com/r/Mcat/comments/1990nnf/since_when_does_3d_orbital_fill_before_4s_have_i/
Since when does 3d orbital fill before 4s? Have I been delulu? / UWhirlpool confusing question : r/Mcat Skip to main contentSince when does 3d orbital fill before 4s? Have I been delulu? / UWhirlpool confusing question : r/Mcat Open menu Open navigationGo to Reddit Home r/Mcat A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to Mcat r/Mcat•2 yr. ago medd27 Since when does 3d orbital fill before 4s? Have I been delulu? / UWhirlpool confusing question I am going to try to reword the question and answers so this does not get removed. Question states that Fe is ionized to Fe2+. When this happens, electrons are lost from the 4s orbital. Based on this, do electrons for iron fill the 3d orbital or the 4s orbital first? A. Electrons fill the 3d orbital first because according to Aufbau principle, electrons fill the lower energy orbitals first, and electrons are lost from the higher energy levels. B. 4s fills first because 3d orbitals are higher in energy than 4s orbitals. I picked B but the answer is A. I am confused!! I thought 4s always fills first then 3d because 4s is lower energy than 3d. I do know that when removing electrons, we remove 4s first. Am i confused? Do I have this all mixed up?? Plus iron is d-block so I thought electrons are lost from the d-orbitals for d-block elements?? Read more Share Related Answers Section Related Answers Electron filling order 3d vs 4s orbitals Why d orbital energy can be lower than 4s Electron configuration of zinc and copper Top MCAT study schedules for busy students Effective MCAT CARS practice techniques New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of January 17, 2024 Reddit reReddit: Top posts of January 2024 Reddit reReddit: Top posts of 2024 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
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https://digitalcommons.latech.edu/cgi/viewcontent.cgi?article=1001&context=mathematics-senior-capstone-papers
Published Time: Tue, 13 Oct 2020 18:43:36 GMT Pisano Periods: A Comparison Study Katherine L. Willrich May 6, 2019 Project Advisor: John R. Doyle, Ph.D. Abstract The Pisano period, denoted π(n), is the period during which the Fibonacci sequence repeats after reducing the original sequence modulo n. More generally, one can similarly define Pisano periods for any linear recurrence sequence; in this paper we compare the Pisano periods of certain linear recurrence sequences with the Pisano periods of the Fibonacci sequence. We first construct recurrence sequences, defining the initial values as integers from 2 to 1000 and second values as 1. This paper discusses how the constructed sequences are related to the matrix M = [1 11 0 ] reduced modulo n. We offer a proof to show that the order of M is equal to the Pisano period of the Fibonacci sequence reduced modulo n. Further, we provide data showing that there are few discrepancies between the order of M and the Pisano periods of the constructed sequences reduced modulo n, for n from 2 to 1000. Finally, we detail progress made in the analysis of the comparison between the Pisano periods of the Fibonacci and Lucas sequences. Keywords: Pisano period, second order linear recurrence sequence, Fibonacci sequence 1 Introduction and Motivation One of the most studied sequences in mathematics, now known as the Fibonacci sequence, has been found written on ancient Sanskrit tablets, according to Wall . Since it was introduced to the Western world by the Italian mathematician Fibonacci, other mathemati-cians have followed suit, discovering the properties of other linear recurrence sequences. 1Pisano Periods 2Even given the breadth of research, there are still lingering questions regarding linear recur-rence sequences, their properties, and how they are interrelated. This research investigates some properties of second order linear recurrence sequences, of which the definition is found in [2, 7]. Definition 1. Let R be a ring, and let B, C ∈ R be constants such that B, C 6 = 0 . A second order linear recurrence sequence over R is a sequence {Ak}∞ k=0 of elements of R for which the following is true: Ak = Ba k−1 + Ca k−2, k ∈ Z+, k ≥ 2. (1) Each second order linear recurrence sequence is determined by two initial values, called seed values; altering one or more of these seed values alters the entire sequence. This research considers sequences that each have a second seed value of 1 and for which B = C = 1. Therefore, the notation herein identifies sequences based solely on their initial values, with the understanding that their second seed values are 1. We denote linear sequences Ak(i), where k is the sequence index and i is the initial seed value. The Fibonacci sequence is a second order linear recurrence sequence with initial seed value 0 and second seed value 1. This sequence is denoted Fk = Ak(0), and it is defined as follows: Fk+2 = Fk+1 + Fk, k ∈ Z+, k ≥ 2. (2) Thus, the Fibonacci sequence is defined by adding two prior values to find the next value in the sequence. The first few values are given below: 0, 1, 1, 2, 3, 5, 8, 13 , 21 , 34 , 55 , 89 , . . . . (3) Every value of the Fibonacci sequence can be generated by the matrix M = [1 11 0 ] ; (4) more precisely M k = [1 11 0 ]k = [Fk+1 Fk Fk Fk−1 ] by . For example, M 2 = [1 11 0 ]2 = [2 11 1 ] = [F3 F2 F2 F1 ] . The Pisano period of the Fibonacci sequence modulo n , denoted π0(n), is the period in which the Fibonacci sequence repeats after reducing the original sequence modulo n. Currently, there is no known general formula for the Pisano periods of the Fibonacci sequence by [3, 5, 10, 11]. An interesting question to pose, however, is how the Pisano periods of other second order linear recurrence sequences compare to those of the Fibonacci sequence; changing only the initial value of the Fibonacci sequence, we compare the resulting Pisano periods in the same modulus. Noting that M is an invertible, square matrix with det( M ) = −1, we consider the following definition by : Pisano Periods 3 Definition 2. Let n ∈ Z+. Let A be a m × m matrix with integer entries, and let A be reduced mod n. Suppose there exists k ∈ Z+ such that Ak = Im mod n, where Im is the m × m identity matrix. The order of A is the smallest such k. Proposition 1. If a matrix’s determinant is ±1, the order exists. In particular, for every positive integer n, there exists a positive integer k such that M k ≡ I2 mod n. Proposition 1 is a consequence of Lagrange’s Theorem from group theory. Because the recurrence matrix has a determinant of −1, we are ensured that the sequences related to M k are finite and cyclic. These sequences reduced mod n produce strings of repeating sequence values. Thus, we are ensured that strings of repeating sequence values occur at a minimum of j, where j is the order of M mod n.We consider the cyclic nature of the sequences reduced modulo n and ask whether it is possible for the sequence to repeat with greater frequency than the order of M . One may assume that because changing one seed value alters the entire sequence, the resulting Pisano periods would be changed dramatically. Surprisingly, there are few discrepancies between the Pisano periods of the Fibonacci sequence reduced modulo n and other second order linear recurrence sequences that differ by only the initial seed value. By comparing the Pisano periods of the Fibonacci sequence with those of the Lucas sequence, Ak(2), we conjecture that the discrepancies exist only at moduli n where the 2 × 1 seed value matrix is an eigenvector of M modulo n. 2 Background and Related Studies 2.1 Closed Form of the Fibonacci Sequence As noted in the general definition (1), linear recurrence sequences can be written as a com-bination of terms with constant coefficients. The Fibonacci sequence, (2), is no exception. There is, however, another form for the sequence. The following appears in [1, 2, 4]. Theorem 1. Let F0 = 0 and F1 = 1 . Let Fk be defined recursively in accordance with (2). Then, for all k ≥ 0, Fk = 1 √5(φk − ¯φk) (5) where φ = 1 + √52 , ¯φ = 1 − √52 . Somer and Robinson state in [8,10] that the closed forms of other linear recurrence sequences can also be found using Binet’s formulas for recursion, in conjunction with the seed values of the sequence. Pisano Periods 4 2.2 Eigenvectors and eigenvalues of the recurrence matrix M The eigenvectors for the recurrence matrix M given in (4) are determined by the usual method: det( M − λI ) = 0 ⇔ ∣∣∣∣1 − λ 11 −λ ∣∣∣∣ = 0 ⇔ (1 − λ)( −λ) − 1 = 0 ⇔ λ2 − λ − 1 = 0 . By the quadratic formula, the roots of the polynomial are φ and ¯φ. These are, therefore, the eigenvalues of the recurrence matrix M that generates the Fibonacci sequence . Substituting first λ = φ, then λ = ¯φ into M − λI , we solve for eigenvectors in the following way: [1 − φ 11 −φ ] [ 1 x ] = [00 ] ⇔ [1 − φ + x 1 − φx ] = [00 ] ⇒ 1 − φx = 0 ⇒ x = 1 φ ⇒ x = 21 + √5. Multiplying by the conjugate of the denominator, we see that x = 2(1 − √5) (1 + √5)(1 − √5) = 2(1 − √5) −4= 1 − √5 −2= − ¯φ. Therefore, our first eigenvector is v1 =  11 − √5 −2  = [ 1 − ¯φ ] , which has an eigenvalue of φ . Likewise, v2 =  11 + √5 −2  = [ 1 −φ ] has an eigenvalue of ¯φ.Pisano Periods 5 2.3 Pisano periods of the Fibonacci sequence: Chinese Remainder Theorem The following definition is found in [2, 10, 11]. Definition 3. Let i ∈ Z and let Ak(i) be a second order linear recurrence sequence. When Ak(i) is reduced mod n, the Pisano period, denoted πi(n), is the period with which the sequence repeats. This research considers only second order linear recurrence sequences with second seed values of 1; we therefore use notation that denotes the initial seed value and the modulus, Ak(i, n )where k is the index, i is the initial value, and n is the modulus. For instance, when the Fibonacci sequence is reduced modulo 2, denoted as Fk(0 , 2), the following is produced: Fk(0) = 0 , 1, 1, 2, 3, 5, 8, 13 , 21 , . . . ⇒ Fk(0 , 2) = 0 , 1, 1, 0, 1, 1, 0, 1, 1, . . . After the reduction, a repeating string of the Fibonacci sequence 0,1,1 is evident. Because the string length is 3, the Pisano period of the Fibonacci sequence reduced modulo 2 is 3. That is, π0(2) = 3. Currently, there is no general formula for the Pisano periods of the Fibonacci sequence or any other second order linear recurrence sequence. Wall and Somer detail that the Chinese Remainder Theorem is applicable for the determi-nation of Pisano periods [10, 11]. Theorem 2 (Chinese Remainder theorem for Pisano periods) . If m and n are co-prime, then πi(mn ) is equal to the least common multiple of πi(m) and πi(n). Therefore, Pisano periods of linear recurrence sequences can be determined by first decom-posing the modulus into prime powers, then finding the least common multiple of those powers’ Pisano periods. Interestingly, the Pisano periods of prime powers of the Fibonacci sequence do not follow a general trend (see ). 2.4 Anomalous primes 2 and 5 When studying the Pisano periods of the Fibonacci sequence, the Chinese Remainder The-orem tells us that determining the periods of the prime factors of the modulus is important. However, there are two primes that are anomalous with regard to Pisano periods. First, we consider the closed form of the Fibonacci sequence from (5). We see that the form has a factor of 1 √5. Likewise, the closed form contains a 2 in the denominators of both φ and ¯φ.This suggests that calculating the Pisano periods for prime factors of moduli 2 and 5 may produce anomalous results. Pisano Periods 6 3 Methods This section describes the methods used to investigate the properties of linear recurrence sequences reduced modulo n. In particular, it details how the sequences are constructed, how the are related to the recurrence matrix M , and how their properties are compared in this research. 3.1 Constructing recurrence sequences This research focuses on comparing the Pisano periods of the Fibonacci sequence with those of linear recurrence sequences with one dissimilar seed value; the second seed value for each sequence will be 1. Therefore, we include in our study sequences that begin with integers i ∈ { 2, 3, . . . , 1000 }. The sequences are defined as follows: Ak+2 (i) = Ak+1 (i) + Ak(i), k ∈ Z+. For instance, the Lucas sequence, with initial value of 2, is denoted Ak(2), with the under-standing that the second seed value is 1. 3.2 Relating constructed sequences to the recurrence matrix M As detailed in (4), the matrix M generates the Fibonacci sequence when it is raised to powers k. When M k is multiplied by a 2 × 1 matrix that contains only the seed values of a linear recurrence sequence, the entire sequence can be generated. In this research, the initial values range from 2 to 1000, and the second seed value of each constructed sequence is 1. Therefore, for the linear recurrence sequences considered in this research, the following holds: M k [1 i ] = [Ak+1 (i) Ak(i) ] , where A0(i) = i. Let us consider, for instance, the Lucas sequence with initial value 2: Ak(2) = M k [12 ] . To find the next two sequence values, we set k = 2 and write: M 2 [12 ] = [1 11 0 ]2 [12 ] = [43 ] = [A3(2) A2(2) ] . Thus, the recurrence matrix M is related to all linear recurrence sequences that we will study. Pisano Periods 7 3.3 Proof: the order of M is equal to the Pisano period of the Fibonacci sequence We show that the order of recurrence matrix M is equal to the Pisano period of the Fibonacci sequence, thereby eliminating the need to raise matrices to powers and reduce by a modulus. Theorem 3. The order of M reduced modulo n is equal to the Pisano period of the Fibonacci sequence reduced modulo n.Proof. Let j be order of M ; that is, let j be the minimal positive integer with M j = [1 00 1 ] ,and let l be the Pisano period of the Fibonacci sequence reduced modulo n.Then, M l+1 = [Fl+2 Fl+1 Fl+1 Fl ] = [F2 F1 F1 F0 ] = [1 00 1 ] ≡ M mod n. ⇒ M l+1 ≡ M mod n ⇒ M l+1 M −1 ≡ M M −1 mod n ⇒ M l ∼= I2 mod n Therefore, l ≥ j.Now consider M j+k = M j · M k = I2 · M k = M k. Therefore, [Fj+k+2 Fj+k+1 lFj+k+1 Fj+k ] = [Fk+2 Fk+1 Fk+1 Fk ] , so Fj+k = Fk. Therefore, l ≤ j Since we have both j ≤ l and l ≤ j, by anti-symmetry, we conclude that j = l. Hence, the order of M reduced modulo n is equal to the Pisano period of the Fibonacci sequence reduced modulo n. 4 Data Based on Theorem 3, we can compare the Pisano period of the Fibonacci sequence to the Pisano period of a constructed sequence to determine whether the sequence repeats at an index k which is less than the order j of the matrix. To make this comparison, we calculate the Pisano periods of the constructed sequences with initial values between 2 and 1000 reduced by moduli between 2 and 1000. All necessary calculations are made via Sage. Code is first written to generate the Fibonacci sequence and its Pisano periods reduced modulo n, 2 ≤ n ≤ 1000. Then, additional code Pisano Periods 8Table 1: Sample Data Set Initial Value i Domain Anomalous Moduli n Ratio: πi(n)/ π0(n) i = 5 x + 2 0 ≤ x ≤ 199 5n, 1 ≤ n ≤ 200 5 i = 3 + 11 x 0 ≤ x ≤ 90 11, 22 2 i = 4 + 19 x 0 ≤ x ≤ 52 19, 38 2 i = 8 + 71 x 0 ≤ x ≤ 13 71, 142 2 i = 11 + 131 x 0 ≤ x ≤ 7 131, 262 2 i = 13 + 181 x 0 ≤ x ≤ 5 181, 362 2 i = 15 + 239 x 0 ≤ x ≤ 4 239, 478 2 i = 16 + 271x 0 ≤ x ≤ 3 271, 542 2 i = 18 + 31 x 0 ≤ x ≤ 31 31, 62 2 i = 19 + 379 x 0 ≤ x ≤ 2 379, 758 2 i = 23 + 29 x 0 ≤ x ≤ 33 29, 58 2 i = 24 + 599 x 0 ≤ x ≤ 1 599 2 i = 23 551 2 i = 25 649 2 i = 36 121, 242 2 i = 42 361, 722 2 i = 80 209, 418, 341, 682 2generates the Pisano periods of other second order linear recurrence sequences with initial values i ∈ Z, 2 ≤ i ≤ 1000, reduced modulo n, 2 ≤ n ≤ 1000. Finally, we use Sage to generate the ratios of the Pisano periods of the Fibonacci sequence to those of the other linear recurrence sequences in the same moduli. The code used to generate the data is detailed in Appendix A. The data that was generated with this code was cross-referenced by hand and with reference tables in . Table 1 details a sample data set that is representative of the data collected in this research. The first column indicates the initial value i of the sequence described. The second column, called “Anomalous Moduli n,” lists moduli for which the sequence Ak(i) has a different Pisano period than the Pisano period of the Fibonacci sequence for the same modulus. The third column displays the following ratio: Pisano period for Ak(i) mod n divided by the Pisano period of the Fibonacci sequence mod n.The data collected can be divided into three classifications:(initial values and moduli that produce a ratio of 5), (initial values and prime moduli that produce a ratio of 2), and (initial values and composite moduli that produce a ratio of 2). Pisano Periods 9 5 Results Overview The sample data above illustrates that when there are discrepancies between the Pisano period of the Fibonacci sequence and the Pisano period of the constructed sequences, the ratio between the two is 5 or 2. Upon examining the data generated in Sage, we see that every initial value i ∈ { 2, 7, 12 , 17 , 22 , 27 , . . . } has anomalous moduli 5 n, for all 1 ≤ n ≤ 200. We also observe that moduli that have a ratio of 2 as compared to the Pisano periods of the Fibonacci sequence tend to appear in pairs and fall into one of two categories: n ending in 1 (hence 2 n ending in 2), or n ending in 9 (hence 2 n ending in 8). 5.1 Analysis of ratios: Fibonacci to Lucas Pisano periods To better understand the circumstances that produce anomalous moduli, we compare the resulting Pisano periods of the Lucas sequence to those of the Fibonacci sequence. First, we note that the Lucas sequence, Ak(2), has anomalous moduli at 5 n. The ratio at these mod-uli is 5; this implies that the sequence repeats 5 times as often as the order of M demands. Therefore, the matrix [12 ] must become an eigenvector of the matrix M k mod 5 n, for some k ≥ 1 such that M k 6 = I2. To determine at which moduli this is possible, we consider: M [12 ] = [1 11 0 ] [ 12 ] = [31 ] Since the eigenvalues of M are φ and ¯φ, to be an eigenvector for M mod n, [31 ] ≡ φ [12 ] ≡ [ φ 2φ ] mod n or [31 ] ≡ ¯φ [12 ] ≡ [ ¯φ 2 ¯φ ] mod n. It suffices to consider when 1 ≡ 2φ mod n 1 ≡ 2φ mod n ⇒ 1 ≡ 2( √5 + 1 2 ) mod n This expression is true only when √5 is 0 when reduced mod 5. Therefore, the Lucas sequence’s seed value matrix becomes an eigenvector with eigenvalue 1 at n = 5. 5.2 Analysis of ratio 2: Quadratic Reciprocity As noted earlier, the moduli with ratio 2 tend to occur in pairs, with n ending in 1 or 9 and 2n ending in 2 or 8. When we examine the circumstances necessary for this to be true in Pisano Periods 10 moduli p, where p is a prime number, we see that in anomalous moduli, [1 i ] is an eigenvector for M mod p. This means that i ≡ − φ mod p or i ≡ − ¯φ mod p. Based on the eigenvalues for M , this would imply that √5 exists mod p. The following appears in : Definition 4 (Legendre symbol) . Let p be a prime and a be an integer not divisible by p.The Legendre symbol is given by: (ap ) = { 1 if a is a quadratic residue modulo p −1 if a is a nonresidue modulo p where a quadratic residue modulo p is a nonzero number that is congruent to a square modulo p. We use the Law of Quadratic Reciprocity to determine in which moduli p 5 is a quadratic residue, since 5 is a quadratic residue mod p if and only if 5 is a square mod p. In Legendre symbol notation, we write: (5 p ) = 1 (6) In , Silverman defines the Law of Quadratic Reciprocity as follows: Theorem 4 (Law of Quadratic Reciprocity) . Let p and q be distinct odd primes. (−1 p ) = { 1 if p ≡ 1 mod 4 −1 if p ≡ 3 mod 4 (2 p ) = { 1 if p ≡ 1, 7 mod 8 −1 if p ≡ 3, 5 mod 8 ( qp ) = (pq ) if p ≡ 1 mod 4 or q ≡ 1 mod 4 − (pq ) if p ≡ 3 mod 4 and q ≡ 3 mod 4 By the Law of Quadratic Reciprocity, we can compute (6) in the following manner: (5 p ) = 1 ⇔ (p 5 ) = 1 . Therefore, the number 5 has a square root if and only if p = 2 , p = 5, or p ≡ 1, 4 mod 5. Thus, in these moduli, we will find discrepancies between the Pisano period of a constructed sequence and the Pisano period of the Fibonacci sequence. 6 Conclusion and future studies Second order linear recurrence sequences modulo n are finite and cyclic, but it is possible for the sequences to repeat with more frequency than the order guarantees. Because the order Pisano Periods 11 of the recurrence matrix M reduced modulo n is equal to the Pisano period of the Fibonacci sequence reduced modulo n, comparing the Pisano periods of constructed sequences with those of the Fibonacci sequence in the same moduli allows us to determine the moduli and initial values for which this is possible. Surprisingly, the results of this project show that altering the initial seed value of the Fibonacci sequence produces few instances of change in the Pisano periods, modulo n. Within the search parameters, the ratio between the two is always 1, 2, or 5. When the ratio is 1, the sequence does not repeat more frequently than the order of the associated recurrence matrix M mod n. We consider the anomalous moduli that produce a ratio of 5 or 2. When the ratio is 5, this occurs at moduli 5 n. The comparison study between the Lucas sequence and the Fibonacci sequence suggests that this occurs when the seed value matrix becomes an eigenvector with eigenvalue 1, forcing the sequence to repeat 5 times as often as the order demands. The eigenvalues of the Fibonacci sequence allows us to determine that this is possible when n = 5. When the ratios between the Pisano periods of the Fibonacci sequence and the constructed sequence for the same modulus are 2, they appear in pairs, n and 2 n. This suggests that the seed value matrix for the sequence becomes an eigenvector for M mod n. This is only possible for integers n such that 5 is a square. We can use the Law of Quadratic Reciprocity to determine that this is possible for prime values of n. That is, a ratio of 2 will be produced for prime moduli n, if n is equivalent to 1 mod 5 or 4 mod 5. In the future, studies should be conducted to compare the closed forms of second order linear recurrence sequences to that sequence’s anomalous moduli, in an effort to generalize the occurrence of anomalies. For moduli beyond the scope of this research, studies need to be conducted to determine if anomalous moduli with ratio 2 always appear in pairs. A Sage Code The following Sage code was used to collect data for second order linear recurrence sequences with initial seed values from 2 to 1000 and second seed values of 1, reduced modulo n for 2 ≤ n ≤ 1000. The code first generates the Pisano periods of the Fibonacci sequence, then the Pisano periods of the other linear recurrence sequences. Lastly, the code produces a ratio of the Pisano period for the sequence to the Pisano period for the Fibonacci sequence in the same modulus. start=0 list = []; for n in range(2,1000): k=0 a=start b=1 while s[:k]!=s[k:]or k<2: k=len(s)//2; a,b=b,a+b Pisano Periods 12 list+=[k]; for startx in range(2,1000): listx = []; for n in range (2,1000): sx=[] k=list[n-2] kx=0 ax=startx bx=1 while sx[:kx]!=sx[kx:]or kx<2: sx+=[ax%n]; kx=len(sx)//2; ax,bx=bx,ax+bx listx+=[kx]; print s,n,kx/k; References Mireille Bousquet-M´ elou and Marko Petkovˇ sek. Linear recurrences with constant coef-ficients: the multivariate case. Discrete Mathematics , 225(1-3):51–75, 2000. Graham Everest, Alfred Jacobus Van Der Poorten, Igor E Shparlinski, Thomas Ward, et al. Recurrence sequences , volume 104 of Mathematical Surveys and Monographs .American Mathematical Society Providence, RI, 2003. Sergio Falcon and ´Angel Plaza. k-Fibonacci sequences modulo m. Chaos, Solitons & Fractals , 41(1):497–504, 2009. Charles M Fiduccia. An efficient formula for linear recurrences. SIAM Journal on computing , 14(1):106–112, 1985. Mulatu Lemma, Samera Mulatu, and Krubel Seyoum. The fascinating and mysterious Fibonacci numbers. Advances and Applications in Mathematical Sciences , 13(5):219, 2014. Tam´ as Lengyel. The order of the Fibonacci and Lucas numbers. Fibonacci Quart ,33(3):234–239, 1995. Richard J. Mathar. A table of Pisano period lengths. mathar/public/mathar20111026.pdf , 2017. DW Robinson. A note on linear recurrent sequences modulo m. The American Mathe-matical Monthly , 73(6):619–621, 1966. Joseph H Silverman. A friendly introduction to number theory . Prentice-Hall, 2006. Pisano Periods 13 Lawrence Somer. Possible periods of primary Fibonacci-like sequences with respect to a fixed odd prime. The Fibonacci Quarterly , 20(4):311–333, 1982. DD Wall. Fibonacci series modulo m. The American Mathematical Monthly , 67(6):525– 532, 1960.
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https://pmc.ncbi.nlm.nih.gov/articles/PMC7083054/
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Search in PMC Search in PubMed View in NLM Catalog Add to search Roles of steroid hormones in oviductal function Brooke E Barton Brooke E Barton 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 Find articles by Brooke E Barton 1,#, Gerardo G Herrera Gerardo G Herrera 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 Find articles by Gerardo G Herrera 1,#, Prashanth Anamthathmakula Prashanth Anamthathmakula 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 2 Center for Reproductive Biology, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 Find articles by Prashanth Anamthathmakula 1,2, Jenna K Rock Jenna K Rock 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 Find articles by Jenna K Rock 1, Anna M Willie Anna M Willie 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 Find articles by Anna M Willie 1, Emily A Harris Emily A Harris 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 2 Center for Reproductive Biology, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 Find articles by Emily A Harris 1,2, Ken-Ichi Takemaru Ken-Ichi Takemaru 3 Department of Pharmacological Sciences, Stony Brook University, Stony Brook, New York, 11794 Find articles by Ken-Ichi Takemaru 3, Wipawee Winuthayanon Wipawee Winuthayanon 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 2 Center for Reproductive Biology, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 Find articles by Wipawee Winuthayanon 1,2, Author information Article notes Copyright and License information 1 School of Molecular Biosciences, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 2 Center for Reproductive Biology, College of Veterinary Medicine, Washington State University, Pullman, WA 99164 3 Department of Pharmacological Sciences, Stony Brook University, Stony Brook, New York, 11794 Corresponding author: Wipawee Winuthayanon, Biotechnology/Life Sciences Building, 1770 NE Stadium Way, Pullman, WA 99164, phone: 509-335-8296, w.winuthayanon@wsu.edu Equal contributions, co-first authors Author contribution statement: BEB, GGH, PA, JKR, AMW, and EAH wrote and edited the paper. GGH, AMW and EAH performed histological staining, immunostaining and analyzed data. KIT provided human oviduct samples, wrote and edited the paper. WW conceived the study, analyzed data, wrote and edited the paper. Collection date 2020 Mar 1. PMC Copyright notice PMCID: PMC7083054 NIHMSID: NIHMS1560609 PMID: 32040278 Abstract The oviduct (known as the fallopian tube in humans) is the site for fertilization and pre-implantation embryo development. Female steroid hormones, estrogen and progesterone, are known to modulate the morphology and function of cells in the oviduct. In this review, we focus on the actions of estrogen and progesterone on secretory, ciliated, and muscle cell functions and morphologies during fertilization, pre-implantation embryo development, and embryo transport in humans, laboratory rodents and farm animals. We review some aspects of oviductal anatomy and histology and discuss current assisted reproductive technologies (ARTs) that bypass the oviduct and their effects on embryo quality. Lastly, we review the causes of alterations in secretory, ciliated, and muscle cell functions that could result in embryo transport defects. Keywords: cilia, ectopic pregnancy, estrogen, fallopian tube, muscle, oviduct, progesterone, secretory, steroids Gross anatomy and cell physiology of the oviduct The oviduct is a tubular organ that connects the ovary and the uterus. In humans, the oviduct is referred to as the fallopian tube. We will use the term “the oviduct” hereafter for continuity. In humans, the oviduct is curved, whereas in some mammals, especially in rodents, is coiled. The oviduct is the site of fertilization, pre-implantation embryo development and is made up of five main regions: the fimbria, the infundibulum, the ampulla, the isthmus, and the uterotubal junction (Fig. 1). The fimbria, a fringe of finger-like structures projecting from the infundibulum, is responsible for egg “pick-up” into the oviduct after ovulation. The fimbria and infundibulum are adjacent to the ampulla, where fertilization occurs. The isthmus and uterotubal junction (UTJ) are the proximal regions of the oviduct. The UTJ then exits into the uterine cavity. The isthmus functions as a sperm reservoir and is also thought to limit polyspermy by allowing only a gradual entry of sperm into the ampulla (reviewed by Suarez 2002). Figure 1: Open in a new tab Gross and histological morphologies of human and mouse oviducts. A. Human oviduct. Top panel: a drawing for gross anatomy of the oviduct. Purple and orange layers represent myosalpinx and endosalpinx layers, respectively. Middle and bottom left panels: Hematoxylin and Eosin (H&E) staining of the ampulla and the isthmus regions of the oviduct demonstrating three tissue layers: endosalpinx, myosalpinx, and mesosalpinx. Middle and bottom right panels: Ampulla and isthmus at higher magnification. B. Mouse oviduct. Top panel: gross anatomy of the female reproductive tract including ovaries, oviducts, and the uterus. Middle and bottom left panels: the cross section of the ovary and the whole oviduct indicating three different regions; the infundibulum, the ampulla, and the isthmus. Middle and bottom right panels: Ampulla and isthmus at higher magnification. Ciliated epithelial cells; CC. Non-ciliated (secretory) epithelial cells; non-CC. The primary regions of the oviduct are joined by distinct segments. First, the ampullary-isthmic junction (AIJ) separates the ampulla from the isthmus. At the AIJ, there is a transition from tall and branching mucosal folds in the ampulla to shorter, simpler folds in the isthmus. The smooth muscle transitions from a thin layer in the ampulla to a thick layer in the isthmus. Lastly, the UTJ (also known as the intramural portion of the oviduct in humans) joins the isthmus to the uterus. In numerous species (such as rats, mice, pigs and cows), the UTJ facilitates the regulation of sperm entering the oviduct (reviewed by Holt and Fazeli 2010). Recent findings using transgenic mice with sperm expressing fluorescent proteins showed that the majority of the sperm accumulates around the opening of the UTJ 15 minutes after coitus (Muro, et al. 2016). Additionally, there are at least 13 proteins expressed on sperm that are required for passage through the UTJ (reviewed by Okabe 2015). For example, proteins in a disintegrin and metalloproteinase (ADAM) family (necessary for cell migration, cell adhesion, and cell interactions) are required for sperm migration to the oviduct. Specifically, ADAM1B and ADAM2 dimerize and form fertilin. The presence of fertilin in the endoplasmic reticulum of spermatids then subsequently leads to a localization of ADAM3 on the mature sperm surface. Male mice with a global deletion of Adam3 are completely sterile due to an impairment of sperm transport through the UTJ (reviewed by Fujihara, et al. 2018). However, the precise mechanism of ADAM3-mediated sperm migration is still unclear. These data suggest that normal sperm motility is an important factor, but protein-protein interactions between sperm surface and the UTJ are also essential for successful sperm entry into the oviduct. The oviduct is made up of three structural components (Fig. 1A). The outside layer is the mesosalpinx, which surrounds the myosalpinx and the inner mucosal endosalpinx. Mesosalpinx The mesosalpinx, a part of the broad ligament, anchors the oviduct to the body wall. The outermost layer of the oviduct is the serosa (Fig. 1A). The mesosalpinx also supplies the oviduct, and the uterine and ovarian arteries, with the vascular system through the serous membrane and into the muscle layer underneath (Hunter 1988). These veins and arteries allow for circulation throughout the organ, which is partly responsible for the generation and maintenance of tubal fluid via transudation. The mesosalpinx also connects the oviduct to the lymphatic system. The interconnection of vascular, nervous, and lymphatic systems allows for the oviduct to receive all necessary nutrients as well as to eliminate waste products into lymphatic drainage. Myosalpinx The myosalpinx is the muscular layer of the oviduct, consisting of an outer longitudinal and an inner circular layer of smooth muscle. The use of scanning electron microscopy after tissue maceration showed that the myosalpinx contains complex networks of smooth muscle cells (SMC) (reviewed by Muglia and Motta 2001). Muglia and Motta demonstrated that the histoarchitecture of the myosalpinx in the oviduct varies between regions and among species. For the UTJ, the muscular layer can be classified as either a barrier-like or a sphincter-like structure depending on the species (Fig. 2A-B). Barrier-like structures are characterized by robust musculature, rich in densely packed SMC fibers (in rats and pigs). Sphincter-like structures are classified by geometrically organized independent muscle fibers, called sphincter-like type a (in rabbits and sheep). Sphincter-like type b structures show loosely interwoven plexiforms, which are characterized by uneven distributions and alignments of muscle fibers (in cows and humans). In the isthmus, bundles of SMC also orient in an intermingled fashion which gives rise to a plexiform musculature (Fig. 2C-D). In humans, the myosalpinx in the ampulla is comprised of bundles of inner and outer SMC arranged into clockwise and counter-clockwise spiral fibers that also intermingle into a plexiform. Figure 2: Open in a new tab Scanning electron microscopy (SEM) images of the myosalpinx and the inner endosalpinx. A-D. SEM images of the myosalpinx in the uterotubal junction (UTJ) and isthmic regions after tissue maceration techniques (adapted from (Muglia and Motta 2001) with permission with original scale bars shown). A. Barrier-like salpinx in rat UTJ, bar = 25 μm. B. Sphincter-like type a salpinx in ewe UTJ, bar = 5 μm. C. Plexiform arrangement of myosalpinx in the isthmus in C. human (bar = 100 μm) and D. rabbit (bar = 50 μM). E-F. SEM images of inner endosalpinx at the ampulla region including ciliated and non-ciliated (or secretory) epithelial cells from E. human oviduct [×7,810 magnification, modified with permission from (Seki, et al. 1978)] and F. mouse oviduct [×9,000 magnification, modified with permission from (Dirksen and Satir 1972)]. Motile cilia are present at the apical membrane of ciliated cells and microvilli are present on the apical surface of secretory cells. Muscle contractions in the oviduct facilitate sperm transport to the fertilization site and oocyte transport to the uterus. In mice, a specialized network of pacemaker cells called interstitial cells of Cajal generate slow electric waves that underlie myosalpinx contractions (Dixon, et al. 2010). In humans, similar cells have been identified and are referred to as interstitial cells of Cajal-like cells (Popescu, et al. 2005). The myosalpinx contracts and relaxes depending on different levels of steroid hormones, which will be discussed in a later section. Muscle contractions create a back-and-forth motion of the oocytes/embryos within the oviduct before arrival to the uterus. In rodents, peristaltic contractions of the myosalpinx result in movement of tubal fluid that is crucial for sperm transport from the isthmus to the ampulla (Hino and Yanagimachi 2019). As shown in Fig. 1A, the myosalpinx is thicker in the isthmus than in the ampulla or infundibulum. Endosalpinx The endosalpinx is the inner mucosal layer of the oviduct and is comprised of epithelial cells and lamina propria mucosa (Castro, et al. 2019). The lamina propria mucosa is a network of connective tissues containing fibroblasts (or stromal cells) and other mononuclear cell populations (Pauerstein and Eddy 1979). In mammals, the oviductal epithelium is made-up of secretory (or non-ciliated) and ciliated epithelial cells (Fig. 2E-F). Secretory epithelial cells are characterized by the presence of microvilli on the apical surface. In contrast, ciliated epithelial cells have motile cilia at the apical surface. Functions of epithelial and muscle cells of the oviduct The fine-tuning of tubal fluid flow aids in the fertilization of eggs and congruent transport of embryos from the oviduct into the uterus. Approximate duration of embryo transport varies between species. In pigs, embryos do not enter the uterus until 2 days post fertilization. In other species, this embryo transport takes longer, with embryos exiting the oviduct closer to 3 days in mice and cows, and ~3.5 days in humans (reviewed by Croxatto 2002). Recently developed optical coherence microscopy for in vivo imaging in mice showed that zygotes are located in the ampulla at 0.5 days post-coitus (dpc) (Moore, et al. 2018). Furthermore, at 1.5 dpc, 2-cell embryos are in the isthmus and at day 3, embryos are generally located in the UTJ. Secretory cells Secretory cells are abundant in the proximal region of the oviduct, specifically throughout the isthmus (Fig. 3). Recent findings using lineage tracing showed that some secretory cells in mouse neonatal oviducts act as epithelial progenitors, giving rise and differentiating into both secretory and ciliated epithelial cells (Ghosh, et al. 2017). Using in vivo ligation in mice, it was found that secretory cells in the isthmus continuously secrete fluid into the oviductal lumen (Hino and Yanagimachi 2019), in addition to the transudation mechanism previously described. Tubal fluid is comprised of ions, energy substrates (such as glucose, pyruvate and lactate), amino acids, prostaglandins (PGs), steroid hormones (E 2 and P 4) as well as various proteins (such as albumin, glycoproteins and lipoproteins) and growth factors [including epidermal growth factor (EGF), fibroblast growth factor (FGF), transforming growth factor (TGF), insulin-like growth factor (IGF), and IGF-binding protein (IGFBP)] (reviewed by Aguilar and Reyley 2005). There are several secreted proteins in tubal fluid that contribute to embryo development and transport (reviewed by Coy and Yanagimachi 2015). Figure 3: Open in a new tab Ciliated epithelial cells (CC) are more abundant in the distal part of the oviduct (including infundibulum and ampulla) whereas the secretory epithelial cells (SC) are more prominent in the proximal part including the isthmus. Muscle cell (MC) layer is also thickest in the proximal compared to the distal region of the oviduct (as also seen in the histological analysis in Fig. 1). Lamina propria (LP) represents fibroblasts underneath epithelial cells in the mucosal folds. Zygote, 2-cell, 4- to 8-cell, morula, and blastocyst stage embryo develops within the oviducts before the transport into the uterine cavity. One factor from secretory cells is oviductal glycoprotein 1 (OVGP1, encoded by Muc9 gene). OVGP1 is involved in sperm capacitation, sperm-oocyte binding, oocyte penetration, modification of the zona pellucida (ZP), and regulation of polyspermy (Avilés, et al. 2010). In pigs and cows, incubation of oviductal fluid-containing OVGP1 significantly reduces sperm-ZP binding and subsequently prevents polyspermy (Coy, et al. 2008), potentially through increased resistance of proteolytic digestion of ZP and decreased affinity for sperm binding due to modified ZP proteins. Recent studies showed that the presence of recombinant porcine OVGP1 improves fertilization rates (Algarra, et al. 2016) but does not affect developmental rates of bovine embryos (Algarra, et al. 2018). In mice, OVGP1 is localized to perivitelline space but does not bind to the ZP (Kapur and Johnson 1986). However, in a subsequent study, OVGP1 was shown to be associated with the ZP of the ovulated oocyte and facilitates sperm-ZP adhesion (Lyng and Shur 2009). Nevertheless, global deletion of Ovgp1 in mice has no fertility defect (Araki, et al. 2003). Moreover, Ovgp1 is a pseudogene in rats and horses (Avilés, et al. 2010). These findings suggest that the functional significance of OVGP1 in female fertility varies among species. Another important protein is fetuin B, a liver-derived plasma protein that is also expressed locally in the oviduct in response to estrogen signaling (Winuthayanon, et al. 2015). In mice, fetuin B is essential for fertilization, as global deletion of Fetub causes female infertility (Dietzel, et al. 2013). Therefore, it is likely that secretory cells of the oviduct facilitate fertilization and pre-implantation embryo development by secreting necessary factors into the oviductal lumen. Ciliated cells Ciliated cells are most abundant in the infundibulum and ampulla (Figs. 1 and 3). There are two proposed functions of oviductal cilia in reproduction: 1) to facilitate movement of tubal fluid and embryos toward the uterus and 2) to create a rheotaxis signal for sperm to reach the ampulla (Miki and Clapham 2013). However, recent in vivo studies by Hino and Yanagimachi showed that rheotaxis does not seem to play a role in sperm movement in mice (Hino and Yanagimachi 2019). Therefore, the function of unidirectional flow generated by ciliated epithelial cells is still under controversy. Nevertheless, a genetic study in mice showed that cilia of appropriate length are essential for oviductal function as a global loss of Kif19 gene (kinesin family member 19) has significantly longer cilia in the oviduct compared to controls (Niwa, et al. 2012). These elongated cilia in Kif19 −/− females led to an ineffective fluid flow causing an obstruction of cell debris and an occluded lumen, resulting in complete sterility. In contrast, conditional deletion of Cep164, encoding a key ciliary protein and central regulator of primary ciliogenesis, in ciliated cells caused a reduction in both length and number of multicilia compared to control mice (Siller, et al. 2017). Surprisingly, these female mice lacking Cep164 in ciliated cells are fertile, indicating that if the oviductal lumen is not occluded, eggs and embryos are able to transport to the uterus and that shortening cilia has no overt effect on fertility in mice. In mice, ciliary beat frequency (CBF) cycles at 7-10 Hertz (Hz) at 0.5 dpc and decreases significantly to ~2.5-5.0 Hz at 2.5 dpc (Wang, et al. 2015, Li, et al. 2017). It is likely that this reduction in CBF at 2.5 dpc decelerates the luminal fluid flow and helps retain the embryo within the oviduct to prevent the premature arrival of the embryos into the uterus before 3.5 dpc. Muscle cells The layer of smooth muscle cells is thicker in the isthmus than in the ampulla (Figs. 1 and 3). Contraction of smooth muscle cells results in an increased velocity of fluid flow. As mentioned above, muscle contraction is a major contributor to adovarian tubal fluid flow, supporting sperm transport from the isthmus to the ampulla (Hino and Yanagimachi 2019). It is also postulated that muscle contraction facilitates the embryo transport process, in addition to the function of cilia. Studies in rats and rabbits showed that in the absence of smooth muscle contraction after treatment with isoproterenol, a β-adrenergic agonist, ciliated cells were able to make up most of the tubal flow and egg transport was not affected (Halbert, et al. 1976, Halbert, et al. 1989). However, a study in mice showed that disruption of cannabinoid receptor (Cnr1−/−) resulted in impaired embryo transport due to an alteration of smooth muscle contraction and relaxation (Wang, et al. 2004). In mouse oviducts, one of the Ca 2+-activated Cl− channels protein called anoctamin-1, encoded by Tmem16a gene, in the interstitial cells of Cajal in the myosalpinx, is responsible for the generation of spontaneous slow electrical wave and may be involved in egg transport (Dixon, et al. 2012). Dixon et al. showed that the oviduct from Tmem16a−/− females completely lack their slow wave activity. However, whether this slow wave generated by interstitial cells of Cajal is required for female fertility remains unclear as Tmem16a−/− mice die before puberty. Overall, oviductal muscle contraction does plays a crucial role in female reproduction but appears to vary among species and therefore must be the subject of future investigations as the mechanisms are widely unknown. Roles of estrogen and progesterone in oviductal function Estrogen (E 2) and progesterone (P 4) are female steroid hormones that act through nuclear and membrane receptors. E 2 is produced by granulosa cells in the ovary and exhibits its activity through estrogen receptor α and β (encoded by Esr1 and Esr2 genes) (Hewitt, et al. 2016). P 4 is secreted from peri-ovulatory granulosa and luteal cells and acts through progesterone receptors (encoded by Pgr gene) (Conneely, et al. 2003, Mulac-Jericevic, et al. 2003). In mice and rats, ESR1 is expressed in epithelial, stromal, and muscle cells of the oviduct (Okada, et al. 2003, Winuthayanon, et al. 2015) (Fig. 4). Comparatively, immunohistochemical analysis of PGR showed higher expression in stromal and muscle cells compared to epithelial cells in the infundibulum and ampulla, respectively, while PGR expression is observed in all three cell types in the isthmus of mouse oviducts (Fig. 4). In addition to classical nuclear receptors, P 4 also functions through membrane progesterone receptors (mPRs) and their expression is tissue-specific (Kowalik, et al. 2013). Nutu et al. showed that mPRβ protein is detected in the cilia whereas the mPRγ is localized at the apical plasma membrane of ciliated cells in mouse oviducts and human oviducts (Nutu, et al. 2009). Figure 4: Open in a new tab Expression of estrogen and progesterone receptor (ESR1 and PGR) proteins in the mouse oviduct at the infundibulum, the ampulla, and the isthmus. Oviducts were randomly collected from female mice at different stages of the estrus cycle. E; Epithelium, S; Stroma, M; Muscle cells. Classical studies from the 1950s-1990s in several species have described that both secretory and ciliated cells change their morphology in response to changes in levels of E 2 and P 4 (reviewed by Abe 1996). As shown in rhesus monkeys (Brenner 1969), epithelial cells of the oviduct in the fimbria are cuboidal at day 2 of the menstrual cycle (Fig. 5). At days 5-6 (follicular phase, high levels of circulating E 2), epithelial cell height drastically increases and ciliation is observed. At day 15 (early luteal phase, E 2 level declines while P 4 level rises), the epithelial cells regress and atrophy (decreased cell height and underwent partial de-ciliation as ciliary apparatus pinched off and shed into the lumen). At the end of the menstrual cycle, in which both E 2 and P 4 are at basal levels (day 27), the epithelial cells return to cuboidal morphology and cilia are rarely present. Unlike the fimbria and ampulla, ciliated and secretory cells in the isthmus do not drastically regress during luteal phase (Steffl, et al. 2008). Whether or not these ciliated cells undergo apoptotic cell death and renew each cycle is still debatable and appears to vary among species (Steffl, et al. 2008). As mentioned above, secretory cells in neonatal mouse oviducts could give rise to the ciliated cell population, driven by Wnt/β-catenin signaling pathway, due to its pluripotency during development (Ghosh, et al. 2017). However, it is unlikely that these terminally differentiated de-ciliated epithelial cells could become secretory cells during luteal phase. Figure 5: Open in a new tab Images from light microscopy of epithelial cell layer of oviducts collected from the fimbria region at different days of the menstrual cycle in Rhesus monkeys. Days 2, 5-6, 15, and 27 represent days after menstruation. All images are taken at ×1200 magnification (adapted from (Brenner 1969) with permission). In the oviduct, both E 2 and P 4 have also been shown to modulate sperm function in the female reproductive tract. In humans, P 4, secreted from cumulus cells, was shown to be a chemoattractant for sperm (Oren-Benaroya, et al. 2008). The presence of P 4 increased intracellular Ca 2+ in sperm flagella through the potentiation of CatSper, a pH-dependent Ca 2+ channel in human sperm (Lishko, et al. 2011). The effect of P 4 on sperm motility appears to be mediated by mPRα (Tan, et al. 2019), not classical PGRs (Lishko, et al. 2011). These data are supported by the finding that global deletion of Pgr (Pgr−/−) has no effect on male fertility in mice (Lydon, et al. 1995). Inhibition of mPR activity also significantly decreases acrosome reactions in human sperm (Sabeur, et al. 1996). These findings indicate that P 4 acts as a chemotactic cue and stimulates the capacitation-related events in human sperm. In contrast, the mechanism and effects of E 2 on sperm capacitation and motility are less defined. ESR1 and ESR2 are expressed in the head and midpiece of human sperm, respectively (Solakidi, et al. 2005). In mice, treatment of sperm with E 2 significantly increases capacitation and acrosome reactions (Adeoya-Osiguwa, et al. 2003). Additionally, studies using a global Esr1-knockout mouse model (Esr1−/−) showed a severe reduction in sperm motility and sperm count, leading to an inability to fertilize eggs in vitro (Eddy, et al. 1996). However, the defect in Esr1−/− mice appears to be due to a disruption of spermatogenesis and seminiferous tubule development. Steroid action in secretory epithelial cells Tubal fluid is not only produced by transudation of fluid from capillaries, as mentioned above, but also by the secretory cells that line the oviduct. Oviducts are filled with this tubal fluid that not only nourishes and protects the egg and embryo but also maintains sperm motility, viability and storage and aids in embryo transport to the implantation site. In rats and pigs, expression of aquaporins (water channels encoded by Aqp genes; including Aqp5, Aqp8, and Aqp9, on the apical plasma membrane along the oviduct) is regulated by both E 2 and P 4 (Branes, et al. 2005, Skowronski, et al. 2011). In mice the expression of Aqp5, specific to secretory cells, is highest during estrus when E 2 levels peak (Nah, et al. 2017). As such, these aquaporins may contribute to the regulation of fluid within the oviductal lumen. Accordingly, Hino and Yanagimachi showed that, at estrus, the isthmus region produced approximately 2.2 μl of fluid per hour in mice (Hino and Yanagimachi 2019). In cows, sheep and rabbits, oviductal fluid volume drastically increases after E 2 treatment or at estrus (McDonald and Bellve 1969, Roberts, et al. 1975, Gott, et al. 1988). Therefore, E 2 is a positive regulator for secretory function in the oviduct. In addition to fluid production, E 2 also increases oviductal protein production. Some of these secreted proteins, called embryotrophic factors, are postulated to be involved in embryo development (Aviles, et al. 2010). These factors include OVGP1 (Buhi, et al. 1992, Chen, et al. 2013), demilune cell and parotid protein (DCPP) (Lee, et al. 2006), insulin-like growth factor binding proteins (IGFBP) (Lai, et al. 1996), among others (Bauersachs, et al. 2003). In addition to positive regulation of embryo development, our group also found that E 2 through ESR1 in oviductal epithelial cells is necessary for in vivo embryo protection, as deletion of Esr1 in these cells leads to embryo death prior to the two-cell stage (Winuthayanon, et al. 2015). This defect is mainly due to excess protease activity in the oviductal lumen, resulting in a disruption of embryonic plasma membranes. P 4 has the opposite effect of E 2. When P 4 levels are high, such as during the luteal phase, aquaporin expression and fluid secretions decrease (McDonald and Bellve 1969, Skowronski, et al. 2011). In rabbits, oviductal fluid production is attenuated when P 4 is administered alone or co-administered with E 2 (Bishop 1956), suggesting that P 4 opposes E 2-induced fluid secretion. It was also found that rabbits treated with P 4 have higher protein content in the oviductal fluid when compared to non-treated controls, suggesting that P 4 increases protein concentration and alters the viscosity of the oviductal fluid (Hamner and Fox 1968). In primary porcine oviductal epithelial cells, treatment with P 4 is correlated with a decrease in OVGP1 expression (Chen, et al. 2013). A combination of decreasing fluid production and increasing protein content characterizes the negative regulatory effects of P 4 on oviductal fluid. Therefore, E 2 and P 4 oppositely affect the production of tubal fluid and work in a balanced unison throughout the estrous/menstrual cycle to maintain appropriate fluid levels and protein content. Steroid actions in ciliated epithelial cells The function of ciliated epithelial cells is crucial for recruiting the released egg to the ampulla. Ciliated cells, in addition to the muscle cells, create tubal fluid flow in the oviduct. As shown in Fig. 5, the length of cilia in rhesus monkey is increased in the presence of E 2 and decreased by P 4 (Brenner 1969). Normal cilia formation in female mice with a global loss of Esr1 (Esr1−/−) suggests that ESR1 is not required for ciliogenesis in the oviduct (Okada, et al. 2004). However, E 2 induces the differentiation of ciliated epithelial cells in rats by neonatal day 5, leading to the conclusion that E 2 facilitates but is not required for cilia development (Okada, et al. 2004). Additionally, Okada et al. showed that inhibition of ESR activity using ESR antagonist, ICI182,780, hinders ciliary differentiation in neonatal rat oviducts. Our recent studies showed that loss of ESR1 in oviductal epithelial cells (Esr1 d/d) causes an increase in ciliary length and decrease in CBF compared to control littermates, resulting in complete embryo retention in the oviduct and failed implantation (Li, et al. 2017). Although ESR2 is exclusively expressed in the cilia of oviductal ciliated epithelial cells, a global deletion of Esr2 (Esr2−/−) does not affect embryo transport (Li, et al. 2017). These findings indicate that E 2 signaling through ESR1 is crucial for the regulation of length and function of cilia in mouse oviduct and that signaling through ESR2 may be secondary. P 4 has the opposite effect of E 2 in regard to ciliation. High circulating level of P 4 causes de-ciliation and decreases CBF (Brenner 1969, Mahmood, et al. 1998). In mice, cows and humans, treatment with P 4 causes a rapid decrease of CBF in the oviduct (Mahmood, et al. 1998, Wessel, et al. 2004, Bylander, et al. 2010). However, global deletion of Pgr in mice has no effect on ciliogenesis as cilia remain present in the oviductal epithelial cells of Pgr−/− females (Akison, et al. 2014). Treatment with RU486, a PGR antagonist, suppresses P 4-mediated actions on CBF (Bylander, et al. 2013) suggesting a requirement of classical PGR-regulation for ciliated cell activity in the oviduct. Transient receptor potential cation channel subfamily V member 4 (or TRPV4) was shown to regulate Ca 2+ intracellular influx and the beating of ciliated epithelial cells (Lorenzo, et al. 2008). In hamsters, activation of TRPV4 caused an increase in oviductal CBF (Andrade, et al. 2005). Expression of TRPV4 protein in both human airways and mammary gland epithelial cells, as well as in vascular smooth muscle cells, is suppressed by P 4 treatment (Jung, et al. 2009). As such, the action of P 4 on decreasing CBF may also result from a suppression of TRPV4 expression in ciliated epithelial cells in addition to classical steroid hormone regulation of oviductal CBF. Steroid action in muscle cells Both ESR1 and PGR proteins are detected in the muscle cell layers of the oviduct in rodents (Okada, et al. 2003, Winuthayanon, et al. 2015) as shown in Fig. 4. In rats, E 2 increases contraction of primary smooth muscle cells through a non-genomic mechanism via the induction of inositol trisphosphate production (Reuquén, et al. 2015). In cows, rabbits and rats, E 2 and P 4 are involved in the production of PGs (especially PGE 2 and PGF 2α) and endothelins (EDNs) in oviductal epithelial cells, which are crucial for oviductal muscle contractility (Spilman 1974, Rosselli, et al. 1994a, Rosselli, et al. 1994b, Wijayagunawardane, et al. 1999, Al-Alem, et al. 2007, Parada-Bustamante, et al. 2012). In rats, E 2 upregulates the expression of cyclooxygenase-2 (COX2; enzyme responsible for biosynthesis of PG precursor) in oviductal epithelial cells, leading to an increase in the production of PGs in the whole oviductal tissue (Pérez Martínez, et al. 2006). Therefore, it is possible that oviductal epithelial cells are a source of PGs production in rats that subsequently acts on muscle cells. However, direct evidence to support this speculation has not been evaluated. In cows, treatment with E 2 increased the expression of PG synthetases and prostanoid receptors (EP2, EP4, and FP) within the smooth muscle cells of the oviduct (Huang, et al. 2015, Huang, et al. 2018). When bound to PGs, EP2 and FP increase contraction and EP4 decreases contraction (Huang, et al. 2015), therefore resulting in an increased rate of oviductal contraction. Overall, E 2 acts through PGs and EDNs to increase tubal contractility. P 4, on the other hand, decreases muscle contractility in the oviduct to induce relaxation. In cows, EDN1 mRNA in epithelial cells and the production of PGE 2 and PGF 2α are lowest during the luteal phase, at which circulating P 4 is elevated (Wijayagunawardane, et al. 2001, Priyadarsana, et al. 2004). Corresponding to levels of EDN1, PGE 2 and PGF 2α, tubal contraction is also significantly lower during the luteal phase compared to the follicular or post-ovulatory phases (Wijayagunawardane, et al. 2001). In summary, a fine-tuned balance of E 2 and P 4 action plays a crucial role in secretory and ciliated epithelial cells as well as muscle cells in the oviduct during sperm transport, fertilization, embryo transport, and embryo development. Collectively, E 2 and P 4 have direct and opposing effects on ciliated, secretory, and muscle cell types to prepare the microenvironment in the oviduct for successful establishment of early pregnancy (summarized in Fig. 6). Figure 6: Open in a new tab Actions of estrogen and progesterone in the oviductal cells in the oviducts. Estrogen and progesterone exhibit different effects on secretory, ciliated, and muscle cells in the oviduct. Estrogen- and progesterone-mediated signals are represented with green and red lines, respectively. The overall summation of estrogen and progesterone action is to increase and decrease the embryo transport rate, respectively, within the oviduct. CBF; ciliary beat frequency. Defective oviductal function Potential negative effects of assisted reproductive technologies (ARTs) when the oviduct is bypassed Assisted Reproductive Technologies (ARTs) are fertility treatments that include artificial insemination, ovarian stimulation, gamete intra-fallopian transfer (GIFT), zygote intra-fallopian transfer (ZIFT), intrauterine insemination, in vitro fertilization (IVF) and embryo transfer (ET). GIFT and ZIFT procedures allow gametes to fertilize or embryos to develop inside the oviduct. IVF procedures, however, require fertilization to occur in a culture dish. Fertilized eggs are developed in culture to cleavage or blastocyst stages, when the embryos are then transferred into the uterine cavity. IVF includes a specialized procedure in which sperm are injected directly into the oocyte’s cytoplasm, called intracytoplasmic sperm injection (ICSI). Eggs fertilized by ICSI are also developed to blastocyst stage. As such, we will refer to “IVF” for both IVF and ICSI techniques. As a result, IVF bypasses the entire oviduct, whereas GIFT and ZIFT utilize the oviduct for fertilization and/or pre-implantation embryos to develop in vivo. In the United States, more than 99% of all ARTs performed are IVF procedures (Sunderam, et al. 2017) compared to 65% in Europe (De Geyter, et al. 2018). The Developmental Origins of Health and Disease (DoHaD) hypothesis argues that the environment with which a fetus interacts in utero can have lasting effects on the health of the offspring (reviewed by Wadhwa, et al. 2009), yet it is less conclusive concerning the effect of the oviductal environment on the embryo prior to uterine implantation. Because human embryos undergo genome activation (i.e., activation of transcription) during day 3 of development (at 4-8 cell stage) while traveling through the oviduct (Braude, et al. 1988), it is crucial that the embryo is subjected to the optimal microenvironment within the oviduct to support proper embryonic genome activation. As GIFT and ZIFT are less common in comparison to IVF (Sunderam, et al. 2017), it is difficult to accurately compare the incidence of diseases in babies-conceived and developed in the presence (GIFT and ZIFT) or the absence (IVF and ICSI) of oviducts. Nevertheless, ARTs have been shown to alter the epigenome of resulting offspring in mice and humans as well as to increase the frequency of epigenetic disorders such as Angelman or Beckwith-Wiedemann syndromes (reviewed by Uyar and Seli 2014). Early studies showed that co-culturing gametes or fertilized eggs with epithelial cells improved both fertilization rates and embryo quality (Bongso, et al. 1989, White, et al. 1989). In mice and cows, introduction of oviductal fluid extracellular vesicles (oEVs) have been shown to improve embryo development, cryoresistance (Almiñana, et al. 2017, Lopera-Vasquez, et al. 2017), and increase ARTs birth rates in mice (Qu, et al. 2019). In cows, oEVs secreted from oviductal epithelial cells in vivo (obtained by flushing oviducts) compared to oEVs obtained from in vitro primary oviductal epithelial cell culture presented different proteins, among which notably, OVGP1 was present only in oEVs from in vivo origin (Almiñana, et al. 2017). Furthermore, oEVs produced in vivo were found to be under the hormonal regulatory effects of the estrous cycle (Alminana, et al. 2018). Lopera-Vasquez et al. also demonstrated that addition of oEVs collected from the isthmus improved the developmental competence and quality of bovine embryos (Lopera-Vasquez, et al. 2017). In addition to oEVs, optimal culture media for the development of embryos to the blastocyst stage in IVF settings have been developed. There are two approaches, including one-step (monophasic) media and sequential media tailored for metabolically changing embryos during different stages. However, a recent meta-analysis showed that the superiority of sequential over monophasic media for several pregnancy outcomes was still inconclusive (Sfontouris, et al. 2016), suggesting a need for further studies. Efforts have also been made to improve embryo quality in ARTs by providing a simulated oviductal environment for fertilization and early embryo development ex vivo. Ferraz et al. developed a 3-dimentional microfluidic model of bovine oviduct epithelial cells, called an oviduct-on-a-chip, which has been shown to produce bovine embryos with epigenetic patterning more similar to that of in vivo derived embryos compared to embryos derived using current optimized IVF protocols (Ferraz, et al. 2017). Recently, pig blastocysts produced in vitro in the presence of natural reproductive fluids (follicular, oviductal and uterine fluids) showed higher quality in terms of cell number and hatching ability, with gene expression and methylation patterns closer to naturally conceived embryos than blastocysts produced in non-supplemented ARTs-liquids (Canovas, et al. 2017). Similarly, bovine blastocysts cultured in the presence of oviductal and uterine fluids supported embryo development and improved blastocyst cryosurvival, DNA methylation and antioxidant activity (Hamdi, et al. 2018). Ectopic pregnancy If embryo transport in the oviduct does not properly occur (i.e., if the embryo travels too slowly due to improper regulation of ciliary beating, muscle contractions, or fluid flow), the embryo may implant itself outside of the uterus, known as an ectopic pregnancy. Implantation of embryos in the oviduct (known as a tubal pregnancy) accounts for 90% of all ectopic pregnancies; the remaining 10% of ectopic pregnancies occur in the abdominal cavity, interstitial, ovary, and cervix (Panelli, et al. 2015). A recent retrospective population-based longitudinal cohort study showed that women with ectopic first pregnancies had an increased risk of adverse birth outcomes (such as preterm birth, low birth weight and placental abruption) during subsequent intrauterine pregnancies. Furthermore, 10% of these women had subsequent ectopic pregnancies (Chouinard, et al. 2019). Ectopic pregnancy accounts for 5% of maternal deaths in developed countries and less than 1% in developing countries (Khan, et al. 2006). This is mainly due to ruptures at the site of implantation leading to internal bleeding. Many factors may contribute to increasing the likelihood of ectopic pregnancies, including mal-developed oviducts (infection-induced or anatomical defect), ARTs procedures (due to increased uterine peristalsis in the cervix-to-fundus direction following embryo transfer), as well as improper regulation of oviductal mechanisms (Shaw, et al. 2010, Perkins, et al. 2015, Rombauts, et al. 2015). Interestingly, tubal pregnancies appear to be restricted to primates and this could be due to anatomical differences at the utero-tubal junction, known as the intramural segment in humans. Compared to laboratory and farm animals, there is no clear-cut distinction between the endometrium and endosalpinx in primates, leading to mixing of oviductal and uterine fluids and environments which could result in tubal pregnancies (reviewed by Corpa 2006). Disruption of estrogen signaling from exposure to endocrine disruptors such as bisphenol A (BPA), used for manufacturing plastics, has been shown to increase the incidence of embryo retention in mouse oviducts (Xiao, et al. 2011). Some plastics used for ARTs have been found to contain BPA (Gatimel, et al. 2016). Although the levels are low in these products, BPA could negatively impact embryo quality and implantation (Ehrlich, et al. 2012). As discussed previously, oviductal microenvironments play a role in transporting and nurturing developing embryos before implantation through fluid secretions, ciliary beating, and tubal muscle contractions. Ji et al. showed that expression of AQP9 is decreased in oviduct tissues collected from patients who had ectopic pregnancies (Ji, et al. 2013). In addition to AQP, the presence of MUC1, a glycoprotein expressed in epithelial cells of the upper reproductive tract, inhibits embryo implantation by acting as an anti-adhesive molecule (Aplin, et al. 2001). MUC1 expression on the apical membrane of uterine epithelial cells is lost shortly before embryo attachment (Aplin, et al. 2001). Aplin et al. also showed that MUC1 protein was detected on the epithelial surface of ampulla in both humans and monkeys, suggesting that the presence of MUC1 may provide anti-adhesive activity against embryo attachment in the oviduct preventing ectopic pregnancy. Accordingly, ectopic pregnancy is associated with low MUC1 expression (Al-Azemi, et al. 2009). These data together demonstrate that secretory cells are necessary for normal embryo transport in the oviduct. In addition to the secretory cells, ciliated epithelial cells in the oviduct also facilitate proper embryo transport. In rats, levonorgestrel (a synthetic P 4 analog used in plan B contraceptive pills) decreases CBF in the oviduct (Zhao, et al. 2015). In humans, increased circulating levels of levonorgestrel have also been associated with ectopic pregnancies (Sheffer-Mimouni, et al. 2003, Graner, et al. 2019). Recently Li et al. demonstrated that levonorgestrel decreases expression of TRPV4 channels in human fallopian tubal epithelial cell line OE-E6/E7 (Li, et al. 2019). Cultured primary ciliated epithelial cells of human oviducts treated with levonorgestrel showed a significant reduction of CBF in a TRPV4-dependent manner (Li, et al. 2019). In fact, oviducts from patients with ectopic pregnancies have significantly lower levels of TRPV4 protein expression (Li, et al. 2019). In addition to the effect of high levels of P 4, cigarette smoking also increases risk of ectopic pregnancies (1.6-3.5 times higher in smokers vs. non-smokers) (Saraiya, et al. 1998). Exposure to cigarette smoke significantly decreases CBF of ciliated epithelial cells in hamster oviducts (Knoll, et al. 1995). Inhalation of cigarette smoke not only affects ciliary function, but it also slows muscle contractions in hamster oviducts (DiCarlantonio and Talbot 1999). Together these findings suggest that excess exposure to P 4 (or synthetic P 4) as well as cigarette smoking can cause a disruption in ciliated epithelial and muscle cell functions in the oviduct and could result in ectopic pregnancies. Conclusion: Regulation of the mechanical processes in the oviduct including the physiological function of secretory, ciliated, and muscle cells is crucial for maintaining proper embryo quality and transport within the oviduct, along with supporting various sperm functions. Alterations in steroid hormone signaling can result in a disruption of normal embryo development and can also lead to ectopic pregnancies or infertility. Acknowledgements: The authors thank Lana Lim for an initial contribution of the manuscript and acknowledge the Stony Brook University Medicine (SBUM) BioBank, a Department of Pathology and Cancer Center Core Facility at the Stony Brook University School of Medicine for human oviduct specimens. Funding: This work was supported by grants from the Eunice Kennedy Shriver National Institute of Child Health & Human Development of the National Institutes of Health (award number R01HD097087) and the start-up fund from Washington State University, College of Veterinary Medicine to WW and the National Heart, Lung, and Blood Institute (award number R01HL139643) to KIT. Footnotes Declaration of interest: There is no conflict of interest that could be perceived as prejudicing the impartiality of the research reported. References: Abe H 1996. 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[DOI] [PMC free article] [PubMed] [Google Scholar] ACTIONS PDF (2.2 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Gross anatomy and cell physiology of the oviduct Functions of epithelial and muscle cells of the oviduct Roles of estrogen and progesterone in oviductal function Defective oviductal function Conclusion: Acknowledgements: Footnotes References: Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
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https://physics.stackexchange.com/questions/639020/accelerated-system-container-tension-to-be-calculated
homework and exercises - Accelerated system + container . Tension to be calculated - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Accelerated system + container . Tension to be calculated Ask Question Asked 4 years, 4 months ago Modified4 years, 4 months ago Viewed 195 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. So there is this question which is as follows the tension in a string holding a solid block below the surface of a liquid is T T, as shown. It is given that the density of the solid is more than that of the liquid. Find the tension in the string if the system accelerates upwards with acceleration a a. This is the figure provided: The answer was T T(1+a g)(1+a g) I'm not sure why there is a plus sign there. My reasoning was that I imagined a real situation where this would happen. Due to the upward acceleration, the block would naturally have a tendency to move towards the base of the container which can be verified by the direction of the pseudo force. So naturally the tension in string should decrease and there should be a minus sign. If the system accelerates upwards at very high values of 'a' , then the tension should be zero, right ? homework-and-exercises newtonian-mechanics acceleration density buoyancy Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited May 26, 2021 at 4:37 Qmechanic♦ 222k 52 52 gold badges 636 636 silver badges 2.6k 2.6k bronze badges asked May 26, 2021 at 3:29 Dusty_WandererDusty_Wanderer 186 2 2 silver badges 12 12 bronze badges 2 Why is the block buoyant if it is denser than the liquid?Adrian Howard –Adrian Howard 2021-05-26 04:14:49 +00:00 Commented May 26, 2021 at 4:14 oops my bad. its the other way round. the block is less dense Dusty_Wanderer –Dusty_Wanderer 2021-05-26 04:31:34 +00:00 Commented May 26, 2021 at 4:31 Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. In this question drawing a free body is of great help. In the first case when there was no acceleration of the block upward we come with equation T+mg=d v g here m is the mass of the body d is the density of the liquid v is the volume of the immersed body In the second case we see that new tension equation would be T'+ mg+ma(pseudo force will cause acc on the block downwards)= d v(g+a) An important point here is that the buoyant force will also increase as we have to always take the "effective value". so that just plainly gives the desired eqn T'=T(1+a/g) on solving the two equations. So, concluding it you were right about the fact that the block tends to move downwards as confirmed by pseudo force but at the same time there was an increase in the buoyant force which can compensate the effect and hence the tension actually increases. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 26, 2021 at 5:59 Mandava KarthikeyaMandava Karthikeya 36 3 3 bronze badges Add a comment| This answer is useful 1 Save this answer. Show activity on this post. There are actually two opposing forces in play here. As we accelerate upwards there is a downwards pseudo force which tries to decrease the tension but the buoyant force also increases due to acceleration and it tries to push the block upwards thus increasing the tension. Both effects are increased in the same proportion that is by the same factor (1+a g)(1+a g), so the net result is that the tension is increased by the same factor (because the laws of motion are linear) and thus you get that result. As this logic says, even for very high values of ′a′ the tension should not be zero. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 26, 2021 at 4:08 Manas ChoudharyManas Choudhary 143 6 6 bronze badges Add a comment| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions homework-and-exercises newtonian-mechanics acceleration density buoyancy See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 2Why does centre of mass of ice-container system shift in absence of any net external force? 1When is tension positive? 2Closed container buoyant force 1Tension in a string moving in elevator 3Work done by tension on a system-generalisation 2Can buoyancy be explained in terms of kinetic and potential energy exchanges between the buoyant object submerged in a fluid? 0Why there is no change to the height of a cube immersed in a fluid when the container is kept in a lift going against gravity? 0Why is Tension = Friction here? 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188245
https://www.echemi.com/community/how-would-i-answer-find-the-molar-concentration-of-acidic-acid-in-5-m-v-vinegar_mjart22040910803_405.html
How would I answer find the molar concentration of acidic acid in 5% m/V vinegar? - ECHEMI  Community  Sign in | Join free Home>Community>How would I answer find the molar concentration of acidic acid in 5% m/V vinegar? Upvote 14 Downvote Chemical reactions Acids Chemistry Posted by Abdullah Al Mahmud Arafat How would I answer find the molar concentration of acidic acid in 5% m/V vinegar? Vinegar does not contain sulfuric acid. Vinegar is a natural product of fermentation and aging from many different sources, including wine. The alcohol is converted to acetic acid. You can buy many different flavors of vinegar in the supermarket. You will also see white vinegar. This is vinegar which has been distilled to remove nearly all of the colored compounds and flavor components in natural vinegar. It is mostly water and acetic acid. Dinky DaoFollowFollowing Vinegar does not contain sulfuric acid. Vinegar is a natural product of fermentation and aging from many different sources, including wine. The alcohol is converted to acetic acid. You can buy many different flavors of vinegar in the supermarket. You will also see white vinegar. This is vinegar which has been distilled to remove nearly all of the colored compounds and flavor components in natural vinegar. It is mostly water and acetic acid. More Upvote VOTE Downvote It varies a lot depending on the source of the raw materials but usually about 5% to 8% Benjamin van AsFollowFollowing It varies a lot depending on the source of the raw materials but usually about 5% to 8% More Upvote VOTE Downvote Vinegar, being a commercial product, can have variances in the amount of particular ingredients it contains, and also differs from region to region and season to season, but its main component is natural acetic acid, and large amounts of water to dilute the same and other ingredients. Charles ZellstinesFollowFollowing Vinegar, being a commercial product, can have variances in the amount of particular ingredients it contains, and also differs from region to region and season to season, but its main component is natural acetic acid, and large amounts of water to dilute the same and other ingredients. More Upvote VOTE Downvote MW = 60 g/mol acid 5%m/V 5 g———in 100 mL solution X————-in 1000 mL solution X = 50 g 50 /60 = 0.83 mol M = 0.83 mol/L Chandrakala ChaulagainFollowFollowing MW = 60 g/mol acid 5%m/V 5 g———in 100 mL solution X————-in 1000 mL solution X = 50 g 50 /60 = 0.83 mol M = 0.83 mol/L More Upvote VOTE Downvote The key here is that this is a conversion like others you do. The 5% number means a percentage or parts per hundred parts. But you need to know if it is mass percent [indicated with the symbol (m/m)%] or mass volume [indicated with the symbol (m/v)%] percent. If not given, assume (m/v)% for vinegar, as that's what is used on vinegar bottles from the store. So, the 5% them means: 5 g acetic acid per 100 mL of solution. Now, you need the density of the solution. It is in the units: grams of solution / volume of solution. So suppose they want the molarity of 5% acetic acid. You need a volume of solution to start. So assume you have 1000 mL of solution. You are trying to get to moles acetic acid per liter of solution, the units of molarity of acetic acid. So you have: 1000 mL solution Density from Google: 1.02 g solution/ mL solution, only needed for (m/m)%, and unless told otherwise it's easier to assume (m/v)%. Percentage: 5 g acetic acid per 100 mL solution Molar mass: 60.06 g acetic acid/ mole acetic acid. Volume conversion: 1000 mL/ 1L Now, you want to end up with moles acetic acid per liter solution. Start the 5%: (5 g acetic acid/ 100 mL solution) x (1000 mL solution)= 0.5 g acetic acid. (50. g acetic acid) x (1 mole acetic acid/ 60.06 g acetic acid) = 0.833 moles acetic acid. Now, 0.833 moles acetic acid / 1L solution = 0.833 Molar Now, if you have 5% (m/m) then you'll need the density. (1000 mL solution) x (1.02 g solution/ mL solution) = 1020 g solution (1020 g solution) x (5 g acetic acid / 100 g solution) = 51. g acetic acid (51. g acetic acid) x (1 mole acetic acid/ 60.06 g acetic acid) = 0.849 moles acetic acid (0.849 moles acetic acid/ 1 L solution) = 0.849 Molar Croren RspsFollowFollowing The key here is that this is a conversion like others you do. The 5% number means a percentage or parts per hundred parts. But you need to know if it is mass percent [indicated with the symbol (m/m)%] or mass volume [indicated with the symbol (m/v)%] percent. If not given, assume (m/v)% for vinegar, as that's what is used on vinegar bottles from the store. So, the 5% them means: 5 g acetic acid per 100 mL of solution. Now, you need the density of the solution. It is in the units: grams of solution / volume of solution. So suppose they want the molarity of 5% acetic acid. You need a volume of solution to start. So assume you have 1000 mL of solution. You are trying to get to moles acetic acid per liter of solution, the units of molarity of acetic acid. So you have: 1000 mL solution Density from Google: 1.02 g solution/ mL solution, only needed for (m/m)%, and unless told otherwise it's easier to assume (m/v)%. Percentage: 5 g acetic acid per 100 mL solution Molar mass: 60.06 g acetic acid/ mole acetic acid. Volume conversion: 1000 mL/ 1L Now, you want to end up with moles acetic acid per liter solution. Start the 5%: (5 g acetic acid/ 100 mL solution) x (1000 mL solution)= 0.5 g acetic acid. (50. g acetic acid) x (1 mole acetic acid/ 60.06 g acetic acid) = 0.833 moles acetic acid. Now, 0.833 moles acetic acid / 1L solution = 0.833 Molar Now, if you have 5% (m/m) then you'll need the density. (1000 mL solution) x (1.02 g solution/ mL solution) = 1020 g solution (1020 g solution) x (5 g acetic acid / 100 g solution) = 51. g acetic acid (51. g acetic acid) x (1 mole acetic acid/ 60.06 g acetic acid) = 0.849 moles acetic acid (0.849 moles acetic acid/ 1 L solution) = 0.849 Molar More Upvote VOTE Downvote The acid found in vinegar is acetic acid, one of the simplest forms of carboxylic acids. It is acid because the - COOH group at one of its ends is able to release the H+ ion making it an acid, although a very weak one. Banti Singh (???? ????)FollowFollowing The acid found in vinegar is acetic acid, one of the simplest forms of carboxylic acids. It is acid because the - COOH group at one of its ends is able to release the H+ ion making it an acid, although a very weak one. More Upvote VOTE Downvote You can dilute it with water, or you can neutralize the acid by adding a base such as baking soda. Note that if you neutralize acetic acid to sodium acetate, the result is not the same as diluting the acid with water. You can get to the same acidity level either way, but the flavor would be different. Alexandra MorozFollowFollowing You can dilute it with water, or you can neutralize the acid by adding a base such as baking soda. Note that if you neutralize acetic acid to sodium acetate, the result is not the same as diluting the acid with water. You can get to the same acidity level either way, but the flavor would be different. More Upvote VOTE Downvote (200 x 0.1)+ (10 x V)= (200+ V) x1 So 20 + 10V = 200 + V So (10V-V)= (200- 20) So 9 V= 180 V= 180/9= 20 Therefore on adding 20- ml of 10-M HCl, to 200- ml of 0.1-M HCl, we will get 220- ml of 1M HCl. Brian McNellisFollowFollowing (200 x 0.1)+ (10 x V)= (200+ V) x1 So 20 + 10V = 200 + V So (10V-V)= (200- 20) So 9 V= 180 V= 180/9= 20 Therefore on adding 20- ml of 10-M HCl, to 200- ml of 0.1-M HCl, we will get 220- ml of 1M HCl. More Upvote VOTE Downvote You have to understanf what a 5%m/v solution of acetic acid is. This is 5 g acetic acid dissolved in 100 mL of solution. Now you want to calculate the molarity of this solution Molarity means : moles of solute dissolved in 1.0 L solution How do you calculate moles of solute? Moles = mass / molar mass Molar mass CH3COOH = 60.0 g/mol Mol CH3COOH in 5 g = 5 g / 60 g/mol = 0.083 mol This is dissoolved in 100 mL solution = 0.1 L solution Molarity = 0.083 mol / 0.1 L Molarity = 0.83 mol /L Omar HassanFollowFollowing You have to understanf what a 5%m/v solution of acetic acid is. This is 5 g acetic acid dissolved in 100 mL of solution. Now you want to calculate the molarity of this solution Molarity means : moles of solute dissolved in 1.0 L solution How do you calculate moles of solute? Moles = mass / molar mass Molar mass CH3COOH = 60.0 g/mol Mol CH3COOH in 5 g = 5 g / 60 g/mol = 0.083 mol This is dissoolved in 100 mL solution = 0.1 L solution Molarity = 0.083 mol / 0.1 L Molarity = 0.83 mol /L More Upvote VOTE Downvote Weak acid Dan ChadwickFollowFollowing Weak acid More Upvote VOTE Downvote Fluoroantimonic acid Charles GreenFollowFollowing Fluoroantimonic acid More Upvote VOTE Downvote Related Posts How many ml are in 1 glass of water? 26 Upvotes · 2 Comments What is the best solution for cleaning a meth pipe? 15 Upvotes · 3 Comments What is dicyanin, and what is it used for? 14 Upvotes · 3 Comments How do you prepare 0.5M of an HCl solution? 14 Upvotes · 8 Comments Is it safe to use expired salt packets from neti pots for nasal irrigation? 19 Upvotes · 2 Comments How do I prepare a 1M KOH solution? 16 Upvotes · 8 Comments How do I make a 1N H2SO4 solution? 15 Upvotes · 10 Comments Wholesale About ECHEMI Encyclopedia Contact Us Local Mall Partners Market Price & Insight Marketing Kit Trade Data ECHEMI Group Join Us News Sitemap For Buyers For Suppliers Copyright@Qingdao ECHEMI Digital Technology Co., Ltd. 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https://www.merckmanuals.com/home/lung-and-airway-disorders/bronchiectasis-and-atelectasis/atelectasis
honeypot link IN THIS TOPIC Causes Symptoms Diagnosis Treatment Prevention More Information OTHER TOPICS IN THIS CHAPTER Bronchiectasis Atelectasis Atelectasis By Trevor Steinbach, MD, University of Colorado Reviewed/Revised Modified Apr 2025 v1141047 VIEW PROFESSIONAL VERSION Atelectasis is a condition in which all or part of a lung becomes airless and collapses. Causes | Symptoms | Diagnosis | Treatment | Prevention | More Information | Topic Resources The Trachea and Bronchi Blockage of the bronchial tubes is a common cause of atelectasis. Shortness of breath can develop if oxygen levels are low or pneumonia occurs. Chest x-ray or computed tomography is used to confirm the diagnosis. Treatment may involve making sure deep breathing occurs, relieving airway blockages, or both. The main function of the lungs is to absorb oxygen into the bloodstream from the atmosphere and to expel carbon dioxide from the blood into the exhaled breath (gas exchange—see figure Gas Exchange Between Alveolar Spaces and Capillaries ). For gas exchange to occur, the small air sacs within the lungs (alveoli) must remain open and filled with air. Alveoli are kept open by the elastic structure of the lung and by a liquid lining called surfactant. Surfactant counters the natural tendency of the alveoli to close (collapse). Periodic deep breaths, which people take unconsciously, and coughing also help keep alveoli open. Coughing expels any mucus or other secretions that could block the airways leading to the alveoli. If the alveoli are closed for any reason, they cannot participate in gas exchange. The more alveoli that are closed, the less gas exchange occurs. Accordingly, atelectasis can decrease the level of oxygen in the blood. The body compensates for a small amount of atelectasis by closing off (constricting) the blood vessels in the affected area. This constriction redirects blood flow to alveoli that are open so that gas exchange can continue. The Trachea and Bronchi video Causes of Atelectasis Common causes of atelectasis usually involve one of the following: Blockage of one of the tubes (bronchi) that branch off from the trachea (windpipe) and lead to the lung tissue Conditions that decrease deep breathing or suppress a person’s ability to cough The blockage may be caused by something inside the bronchus, such as a plug of mucus, a tumor, or an inhaled foreign object (such as a pill, a piece of food, or a toy). Alternatively, the bronchus may be blocked by something pressing from the outside, such as a tumor or an enlarged lymph node. Blockage from the outside can also occur if the pleural space (the space outside of the lung but inside of the chest) contains a large amount of fluid ( pleural effusion ) or air ( pneumothorax ). When a bronchus or a smaller airway (bronchiole) becomes blocked, the air in the alveoli beyond the blockage is absorbed into the bloodstream, causing the alveoli to shrink and collapse. The area of collapsed lung may become infected because bacteria and white blood cells can build up behind (to the inside of) the blockage. Infection is particularly likely if atelectasis persists for several days or more. If atelectasis persists for months, the lung may not easily re-expand. Any condition that decreases deep breathing or suppresses a person’s ability to cough can cause or contribute to atelectasis. Large doses of opioids or sedatives can decrease deep breathing. Atelectasis is common after general anesthesia, which temporarily suppresses a person’s cough and drive to breathe. Atelectasis is particularly common after chest or abdominal surgery because the effects of receiving general anesthesia may be added to the pain of deep breathing, so people take only shallow breaths. Chest or abdominal pain due to other causes (for example, injury or pneumonia ) also makes taking a deep breath painful. Certain neurologic conditions, immobility, and chest deformities can limit chest movement and thus decrease deep breathing, as can abdominal swelling. People who have obesity are also at greater risk of developing atelectasis. Did You Know... | | | Taking deep breaths after surgery can help prevent atelectasis. People who smoke can decrease their risk of atelectasis after surgery by stopping smoking, ideally 6 to 8 weeks before the surgery. | Symptoms of Atelectasis Atelectasis itself does not cause any symptoms except sometimes shortness of breath. The presence and severity of shortness of breath depend on how rapidly atelectasis develops and how much of the lung is affected. If atelectasis involves a limited portion of the lung or develops slowly, symptoms may be mild or not even noticed. If a large number of alveoli are affected, particularly if atelectasis occurs rapidly, shortness of breath may be severe. The heart rate and breathing rate may increase, and sometimes the person may look bluish (a condition called cyanosis ) because oxygen levels in the blood are low. This appearance may be darker brown or black in people with dark skin and easier to identify in the eyes (conjunctiva), lips, and nail beds. Symptoms may also reflect the disorder that caused atelectasis (for example, chest pain due to an injury) or a disorder that results from atelectasis (for example, chest pain with deep breathing, due to pneumonia ). Diagnosis of Atelectasis Chest x-ray Computed tomography of the chest Doctors suspect atelectasis based on a person’s symptoms, the physical examination findings, and the setting (for example, after surgery, chest injury, or use of certain drugs) in which the symptoms occurred. A chest x-ray or CT scan that shows the airless area confirms the diagnosis. Sometimes additional testing like bronchoscopy (inserting a viewing tube into the bronchus)may be done to find the cause or remove a blockage. Treatment of Atelectasis Deep breathing and coughing Relief of airway blockages by suctioning or bronchoscopy Treatment of atelectasis may involve making sure deep breathing occurs, relieving airway blockages, or both. Sometimes blockages can be relieved when a patient’s airway is suctioned by a health care practitioner. A blockage that cannot be removed by suctioning may require removal by bronchoscopy . Sometimes other methods are necessary. For example, if a tumor is blocking an airway, the blockage can sometimes be relieved by surgery, radiation therapy, chemotherapy, or laser treatment. If mucus is plugging the airways, doctors sometimes give drugs to try to thin the mucus or open the airways. Treatment of complications Symptoms and complications of atelectasis may require treatment. People may require: Supplemental oxygen Antibiotics, if bacterial infection is suspected Rarely, insertion of a breathing tube (endotracheal intubation) and mechanical ventilation Prevention of Atelectasis Atelectasis may be prevented by making sure deep breathing occurs. Whenever possible, conditions that cause shallow breathing for long periods should be treated. People who smoke can decrease their risk of atelectasis after surgery by stopping smoking , ideally 6 to 8 weeks before surgery. All people who have surgery should be encouraged to breathe deeply, cough regularly, and move about as soon as possible after the operation. The use of devices to encourage voluntary deep breathing, called incentive spirometry , and certain exercises, including changing position to increase the drainage of lung mucus and other secretions, may also help prevent atelectasis. More Information The following English-language resource may be useful. Please note that The Manual is not responsible for the content of this resource. National Heart. Lung, and Blood Institute: Atelectasis : Discussion of diagnosis, including screening, as well as prevention and treatment Test your Knowledge Take a quiz on this topic This Site Uses Cookies and Your Privacy Choice Is Important to Us We suggest you choose Customize my Settings to make your individualized choices. Accept Cookies means that you are choosing to accept third-party Cookies and that you understand this choice. See our Privacy Policy
188247
https://www.onemathematicalcat.org/algebra_book/online_problems/at_least_most.htm
HomeAbout MeMain Table of ContentsTestimonialsContact Practice with the Phrases ‘at least’ and ‘at most’ The audio controller is at the bottom of the web page. (Look down!) This gives easy access while reading. The yellow highlighting follows my voice! Depending on your connection speed, the audio may take some time to load. (It will display 0:00 until it's ready.) Of course I'm biased—but I think it's worth the wait! (Click for cat book) The sentence ‘x is at least 5’ means that the least x is allowed to be is 5. It can be 5, or any number greater than 5. So, the phrase ‘x is at least 5’ means ‘x≥5’. The sentence ‘x is at most 10’ means that the most x is allowed to be is 10. It can be 10, or any number less than 10. So, the phrase ‘x is at most 10’ means ‘x≤10’. More generally: | | | | --- | sentence | meaning of sentence | equivalent sentence | | x is at least k | the least x is allowed to be is k; it can be k or any number greater than k | x≥k | | x is at most k | the most x is allowed to be is k; it can be k or any number less than k | x≤k | Note from Dr. Burns (the website creator) Welcome—so glad you're here! My sister and I have finished our house-clean-out adventure! Here it is on Zillow (30 Mt Laurel Way, Monterey, MA). Here are more photos (with captions). Want to say hello? Sign my guestbook! Example Suppose that x takes on values between −3 and 5. Then: x is at least −3 x is at most 5 x is greater than or equal to −3 x is less than or equal to 5 Master the ideas from this section by practicing below: When you're done practicing, move on to: Identifying Inequalities with Variables as True or False Practice
188248
https://abelprize.no/sites/default/files/2021-04/Abel%20prize%202012%20Endre%20Szemeredi%20Arne%20Sletsj%C3%B8e%20Number%20theory%20eng.pdf
Abel Prize Laureate 2012 Arithmetic progressions An arithmetic progression is a sequence of natural numbers of constant difference. The set of natural numbers contains a lot of arithmetic progressions, but if we consider smaller proper subsets of the natural numbers, it is not obvious that there are any arith­ metic progressions. A general problem is to decide how small a randomly chosen subset can be, and still contain arithmetic progressions. Endre Szemerédi In primary school children learn by heart the two times table, 2,4,6,…, the three times table, 3,6,9,12,15,…, and so on. The mathematical term for a times table is arithmetic progression. The concept of arithmetic progression also includes finite sequences like 10,13,16,19. This is an arith­ metic progression of length 4, of (constant) differ­ ence 3 and initial value 10. The two times table is an arithmetic progression of infinite length, of difference 2 and initial value 2. An arithmetic pro­ gression is determined by its length, difference and initial value. If you are asked to write up the arith­ metic progression of length 5, difference 7 and ini­ tial value 23, you should answer 23,30,37,44,51. Notice that 4,5,6,7,8,9 is also an arithmetic pro­ gression, the difference in this case is 1. The Dutch mathematician Pierre Joseph Henry Baudet formulated in 1921 the following conjec­ ture: If one divides the natural numbers 1,2,3,.. ad infinitum into a random number of boxes, then there is nevertheless always at least one box which contains an arithmetic progression of arbitrary length. Baudet died shortly after, at the age of 30. The conjecture was proved in 1927 by another Dutch mathematician, Bartel Leendert van der Waerden. A strengthening of van der Waerdens result was conjectured by Pál Erdős and Pál Turán in 1936. They believed that the reason for the existence of Colouring By colouring the natural numbers using only two colours, say red and blue, it is easy to see that 9 consequent numbers e.g. 1,2,...,9, are needed to ensure that there is an arithmetic progression of length 3. Why? Let us try to prove the opposite, so suppose the sequence 1,2,...,9 does not contain arith­ metic progressions of length 3. For this rea­ son, number 1, 5 and 9 cannot be all equally coloured. So assume first that 1 and 5 are red, and 9 is blue. Since 1 and 5 are red, 3 has to be blue. But 9 is also blue, so 6 has to be red. Now 5 and 6 are red, forcing 4 and 7 to be blue. Number 8 has to be red since 7 and 9 are blue, and number 2 has to be red since 3 and 4 are blue. But then 2, 5 and 8 are all red, which is a contradic­ tion. The case where 1 and 9 are red and 5 is blue is treated similarly. On the other hand, the sequence of length 8, given by RBRB­ BRBR has no arithmetic progressions of length 3. Thus 9 is a sharp bound for this property. This means that the socalled van der Waerden number W(2,3)=9. In 1953, Klaus Friedrich Roth proved that any subset of the integers with positive upper density contains an arithmetic progression of length 3. In 1969, Endre Szemerédi proved that the subset must contain an arithmetic progression of length 4, and then in 1975 proved that any subset with positive upper density must contain arithmetic progressions of arbitrary length, known as Szemerédis theorem. Erdős formulated in 1973 a stronger version of the Erdős-Turán conjecture: Let A be a subset of the natural numbers such that the sum of their recipro­ cals exceeds any natural number. Then A must have arithmetic progressions of arbitrary length. Erdős offered a prize of US$ 3000 for a proof of this con­ jecture at the time. The problem is currently worth US$ 5000. One can prove that the sets of natural numbers of positive upper density necessarily have divergent reciprocal sums. Thus Erdős´ conjecture implies Szemerédis theorem. It is also known that the set of primes have divergent reciprocals, first proved by Leonhard Euler in 1737. The theorem of Ben Green and Terence Tao from 2004 about the existence of arithmetic progressions of arbitrary length in the set of prime numbers is a special case of this conjecture. The results concerning existence of arithmetic progressions are based on the interplay between size, randomness and structure. The bigger the sets are, the more likely it is that they have arithmetic progressions. Szemerédi states that positive upper density is a sufficient condition. For even smaller sets, of zero density, some additional structure is needed. The set of prime numbers has zero density, but Green and Tao show that this set nevertheless has some structural similarities to the natural num­ bers, enough to prove the existence of arithmetic progressions, based on Szemerédi`s theorem. Upper density Example 1. Let A be the set of even num­ bers. For a given N the set of even num­ bers between 1 and N has cardinality N/2 if N is an even number and (N-1)/2 if N is odd. Thus the fraction we are looking for is 1/2, which is the upper density for the even numbers. Example 2. Now let A be some finite set, say all natural numbers from 1 to 100. For N less than 100, the fraction is 1, but for N greater than 100 the fraction will decrease and eventually tend to zero. So the upper density is 0. Example 3. We consider the set of powers of 10, A={10,100,1000,…}. If we compare this set to the set {1,2,3,…,10k} for some natural number k, it is easy to see that the fraction will be k/10k. As k grows, this frac­ tion will tend to 0, and again the upper den­ sity is 0. arithmetic progressions is that some colour occu­ pies a set of natural numbers of strict positive up­ per density. For a subset A of the natural numbers, the upper density is defined as follows: For each natural number N we intersect the set A with the set {1,2,…,N}, count the number of integers in the intersection and divide by N. This rational number between 0 and 1 measures the size of A compared to all integers between 1 and N. We do this for increasing numbers N. If the fraction for huge N never exceeds a certain number, we say that this number is an upper bound for the fractions. The smallest possible upper bound for huge N is called the upper density for A. Notice that there is also a concept of lower den­ sity, given as the greatest lower bound for the frac­ tion, when N is a huge number.
188249
https://www.albert.io/blog/denatured-enzyme-ap-biology-review/
Published Time: 2025-03-26T15:46:57+00:00 Denatured Enzyme: AP® Biology Review | Albert Resources Skip to content Subjects Classes Pricing Schedule a Demo Resources Help Center Albert Academy Try Albert School Success Free Resources Subjects Classes Pricing Schedule a Demo Resources Help Center Albert Academy Try Albert School Success Free Resources Log In Sign Up ➜ AP® Biology RESOURCES PRACTICE FREE RESPONSE ASSESSMENTS FAQ RESOURCES PRACTICE FREE RESPONSE ASSESSMENTS FAQ Denatured Enzyme: AP® Biology Review The Albert Team Last Updated On: March 26, 2025 What We Review Toggle - [x] Introduction What Is a Denatured Enzyme? What Happens to an Enzyme When It Is Denatured? Environmental Factors Leading to Denaturation Factors Affecting Enzyme Denaturation Reversibility of Enzyme Denaturation Practical Applications and Examples Practice Problems Conclusion Sharpen Your Skills for AP® Biology Need help preparing for your AP® Biology exam? Introduction Enzymes are proteins that catalyze biochemical reactions, speeding up critical processes in cells. In AP® Biology, understanding how enzymes function—and especially how they can become denatured—is essential for grasping how various biological systems work. This article explains how enzymes become denatured, what causes denaturation, and why it matters for students preparing for the AP® Biology exam. Start Practicing AP® Biology on Albert now! What Is a Denatured Enzyme? Enzymes, like most proteins, have a complex three-dimensional structure that is crucial for their function. When an enzyme denatures, its three-dimensional structure unravels, causing it to lose its functional shape. Without this proper shape, the enzyme’s active site—which binds to substrate molecules—no longer fits the substrate correctly, leading to a loss of catalytic activity. What Happens to an Enzyme When It Is Denatured? Disruption of Structure: During denaturation, the breakdown of hydrogen bonds, ionic bonds, and disulfide bridges disrupts the enzyme’s folded shape. Loss of Activity: Because the enzyme can no longer bind to its substrate effectively, its catalytic activity declines significantly or disappears altogether. Environmental Factors Leading to Denaturation Temperature: Extremely high temperatures can break the delicate bonds holding the enzyme’s shape, leading to denaturation. pH: Each enzyme functions optimally at a particular pH. Deviations from this pH range can alter the charge interactions within the protein, disrupting its structure. Factors Affecting Enzyme Denaturation Enzymes generally function within narrow temperature and pH ranges. Even slight shifts can impact activity. Emma,CC BY-SA 4.0, via Wikimedia Commons Temperature: Optimal Range: Most enzymes in humans work best at around 37°C (98.6°F). Above the Optimal Range: High temperatures increase molecular motion, potentially tearing apart the enzyme’s structure. Below the Optimal Range: Although low temperatures slow down reaction rates, they usually do not permanently denature an enzyme; however, extremely low or high temperature fluctuations can sometimes cause irreversible damage. pH and Hydrogen Ion Concentration: The pH Equation: pH = –log [H+]. Hydrogen Ion Impact: Changes in [H+] concentration can alter the chemical bonds within the enzyme, especially in the active site. Too many or too few hydrogen ions can disrupt ionic and hydrogen bonds, leading to denaturation. Ready to boost your AP® Biology scores? Explore our plans and pricing here! Reversibility of Enzyme Denaturation In some cases, mild denaturation can be reversed if conditions return to normal—often called renaturation. This typically happens if the changes in temperature or pH are not too severe or prolonged, allowing the enzyme to refold into its functional shape. However, many instances of denaturation are permanent, especially when the polypeptide chain becomes irreversibly tangled or fragmented. Practical Applications and Examples Biological Systems: In our digestive tract, enzymes like pepsin and trypsin have specific pH ranges. If abnormal conditions denature these enzymes, they can impair digestion. Industrial and Lab Settings: Brewers and manufacturers often monitor temperature and pH to maintain optimal enzyme activity for processes like fermentation. Even slight deviations can ruin a batch of product. Practice Problems Describe one way high temperature causes enzyme denaturation. How does pH alteration impact an enzyme’s active site structure? If an enzyme is denatured at a temperature slightly above its optimal range, under what conditions might it regain activity? Calculate the pH of a solution if the [H+] is 1 × 10⁻⁵ M. Provide an example of how a denatured enzyme might affect a specific metabolic pathway in the human body. Conclusion Enzyme function is tightly linked to its structure. When denaturation occurs, it often loses or reduces its ability to catalyze reactions. This concept is central in AP® Biology because it illustrates the delicate balance that organisms must maintain to ensure proper enzyme activity. By thoroughly understanding what happens when an enzyme is denatured, students can master a key principle underlying many biological processes. By exploring these topics and practicing with relevant questions, AP® Biology students can enhance their understanding of enzymes and the critical concept of denaturation, setting a solid foundation for success on the exam. Sharpen Your Skills for AP® Biology Are you preparing for the AP® Biology test? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now! Ready to boost your AP® Biology scores? Explore our plans and pricing here! Need help preparing for your AP® Biology exam? Albert has hundreds of AP® Biology practice questions, free response, and full-length practice tests to try out. Start practicing AP® Biology on Albert now! Interested in a school license?​ Bring Albert to your school and empower all teachers with the world's best question bank for: ➜ SAT® & ACT® ➜ AP® ➜ ELA, Math, Science, & Social Studies ➜ State assessments Options for teachers, schools, and districts. EXPLORE OPTIONS Popular Posts AP® Score Calculators Simulate how different MCQ and FRQ scores translate into AP® scores AP® Review Guides The ultimate review guides for AP® subjects to help you plan and structure your prep. 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188250
https://www.aae.org/specialty/wp-content/uploads/sites/2/2018/05/COL042Spring2018CBCTinDiagnosis.pdf
ENDODONTICS: Colleagues for Excellence The Impact of Cone Beam Computed Tomography in Endodontics: A New Era in Diagnosis and Treatment Planning Spring 2018 Published for the Dental Professional Community by the aae.org/colleagues ENDODONTICS: Colleagues for Excellence 2 Clinical examination and diagnostic imaging are essential components of preoperative diagnosis and treatment planning in endodontics. Thorough clinical examination must be performed before considering any radiographic examination. Accurate diagnostic imaging allows the clinician to better visualize the area in question, and supports the clinical diagnosis, treatment plan and outcome assessment. Conventional two-dimensional (2-D) radiographs continue to be the most popular method of imaging today. However, the diagnostic potential of periapical radiographs (PA) is limited. Information may be difficult to interpret, especially when the anatomy and background pattern is complex. These PAs have inherent limitations due to the compressed three-dimensional (3-D) structures in a 2-D image. In addition, interpreting the film-based radiograph or digital image is a somewhat subjective process. Goldman et al. (1) showed that the agreement among six examiners was only 47% when evaluating healing of periapical lesions using 2-D periapical radiographs. In a follow-up study, Goldman et al. (2) also reported that when examiners evaluated the same films at two different times, they only had 19%–80% agreement with their previous interpretations. A recent study evaluated interobserver and intraobserver reliability in detecting periradicular radiolucencies by using a digital radiograph system. Agreement among all six observers for all images was less than 25%, and agreement for five of six observers was approximately 50% (3). New radiographic imaging systems recently have become available for use in dentistry. Among these new imaging technologies is cone beam computed tomography (CBCT). In 2000, the U.S. Food and Drug Administration approved the first CBCT unit for dental use in the United States. CBCT systems are available in different fields of view (FOV). CBCT with limited FOV, which typically is used for endodontic diagnosis, ranges in diameter from 40 –100 mm. Full FOV CBCT, typically used for orthodontics or facial structure imaging ranges from 100 –200 mm. The voxel size generally is smaller for the limited version (0.1– 0.2 mm vs. 0.3– 0.4 mm), thus offering higher resolution and greater utility for endodontic applications. The limited FOV is the most ideal for use in endodontic diagnosis and treatment planning because it is capable of providing images with sufficient spatial resolution at a low radiation dose. The advent of CBCT can overcome challenges in image interpretation by enabling the clinician to visualize the dentition and the relationship of anatomic structures in three dimensions. CBCT units reconstruct the projection data to provide inter-relational images in three orthogonal planes (axial, sagittal and coronal). This newsletter will review the utilization of CBCT in endodontic diagnosis and management of periapical pathology, diagnosis of pain, cracked teeth and vertical root fractures, nonsurgical and surgical cases, inflammatory resorptive defects and traumatic injuries. In May 2015 an updated joint position statement of the American Association of Endodontists and the American Academy of Oral and Maxillofacial Radiology was published. The intent of the updated statement is to provide scientifically based guidance to clinicians regarding the use of CBCT in endodontic treatment and reflect new developments since the 2010 statement (4). The updated statement addresses the potential applications and recommendations for use of limited FOV CBCT in different phases of treatment, several of which will be reflected in this newsletter. Endodontic Diagnosis and Detection of Periapical Lesions Endodontic diagnosis is dependent upon evaluation of the patient’s chief complaint, medical and dental history, and clinical and radiographic examination. Based on recommendation two of the CBCT position statement, limited FOV CBCT should be considered the imaging modality of choice for diagnosis in patients who present with contradictory or nonspecific clinical signs and symptoms associated with untreated or previously endodontically treated teeth. CBCT imaging has the ability to detect periapical pathology prior to it being apparent on PAs (5) (Figure 1). Clinical studies demonstrate that periapical radiolucency was detected in 20% of cases using radiographs, compared to 48% using CBCT (6). Ex vivo studies in which simulated periapical lesions were created showed similar findings (7-8). Diagnosis of pain is a challenging process for the clinician prior to and after treatment. In challenging diagnostic pain cases, the clinical and radiographic evaluation of the patient may be inconclusive. The ability to determine the etiology of the pain can be attributed to the limitations in both clinical vitality testing and intraoral radiographs to detect the source of the pain. Fig. 1. A patient was referred for evaluation and treatment of tooth #30. Patient presented with pain to percussion and no response to cold testing. Periodontal probing depths were all less than 3mm. Root canal treatment was indic ated based on 2-D radiographic findings and clinical tests. 1A is a periapical radiograph of tooth #30. 1B is a CBCT sagittal view of tooth #30. A periapical low-density area involving both mesial and distal roots as well as the furcation was detected. 3 The Impact of Cone Beam Computed Tomography in Endodontics: A New Era in Diagnosis and Treatment Planning Persistent pain following root canal treatment also presents a diagnostic challenge to the clinician. Atypical odontalgia (AO) is an example of persistent dentoalveolar pain (9). The diagnostic yield of intraoral radiographs and CBCT was evaluated in the differentiation between patients presenting with apical periodontitis and suspected AO without the evidence of periapical bone destruction. CBCT imaging detected 17% more teeth with apical bone loss (apical periodontitis) than intraoral radiographs (10). Preoperative Anatomy Assessment The success of endodontic treatment depends on the identification of all root canal systems so that they can be treated. Recommendation three of the position statement recommends CBCT for initial treatment of teeth with the potential for extra canals and suspected complex morphology, such as mandibular anterior teeth, and maxillary and mandibular premolars and molars, and dental anomalies (Figure 2). The superior ability of CBCT to accurately explore tooth anatomy and identify the prevalence of a second mesiobuccal canal (MB2) in maxillary molars when compared to the gold standard (clinical and histologic sectioning) has been well documented (11-12). CBCT showed higher mean values of specificity and sensitivity when compared to intraoral radiographic assessments in the detection of the MB2 canal (13). Endodontic Diagnosis and Detection of Cracked Teeth and Vertical Root Fractures 2-D radiographs are of limited value for the diagnosis of VRFs and usually only provide indirect evidence of the presence of a VRF. Recommendation six suggests CBCT imaging if clinical examination and 2-D intraoral radiography are inconclusive in the detection of vertical root fracture. Cracked teeth represent a diagnostic and restorative dilemma for both clinicians and patients because of their complicated and vague symptoms and unpredictable prognosis. Treatment plans for cracked teeth depend on the extent and location of the cracks and the severity of the symptoms. In cases of marginal ridge cracks, early detection by CBCT is not possible. Over time, a mesial or distal pattern of bone loss may develop that is indicative of the extent of the longitudinal crack. Several studies have demonstrated the validity of utilizing CBCT to detect vertical root fractures (VRFs). In a comparative study, the sensitivity and specificity of CBCT and PAs in detecting VRF were evaluated. The sensitivity and specificity were 79.4% and 92.5% respectively for CBCT and 37.1% and 95% respectively for PAs. The same study reported that the specificity of CBCT was reduced in the presence of root canal filling material (14). Higher sensitivity and specificity were observed in a clinical study where the definitive diagnosis of VRF was confirmed at the time of surgery to validate CBCT findings, with sensitivity being 88% and specificity 75% (15). Several case series studies have concluded that CBCT is a useful tool for the diagnosis of VRF. In vivo and laboratory studies (16, 17) evaluating CBCT in the detection of VRFs agreed that sensitivity, specificity and accuracy of CBCT generally was higher and more reproducible when compared to traditional radiographs (Figure 3). However, these results should be interpreted with caution because detection of VRFs is dependent on the size of the fracture, presence of radiographic artifacts caused by obturation materials and posts and the spatial resolution of the CBCT. A recent study evaluated the diagnostic ability of a CBCT scan to assess longitudinal root fractures in prosthetically treated teeth (18). The presence of gutta-percha or cast-gold posts reduced the overall sensitivity and specificity. This was attributed to star-shaped streak artifacts that mimic fracture lines in axial views. Another significant problem which can affect the quality and accuracy of CBCT images is the scatter and beam hardening caused by high density neighboring structures, such as enamel, metal posts and restorations. If a substance Fig. 2. An example of using CBCT imaging for identification of complex tooth anatomy in a mandibular first molar. CBCT axial and coronal views (top) demonstrating the presence of a middle mesial canal (arrow). One-year recall CBCT axial and coronal views (bottom) demonstrating the obturation of the complex mesial canal anatomy. Fig. 3. Example of the sensitivity of CBCT in detecting vertical root fractures. Periapical radiograph of tooth #30 (A). 3-D reconstruction demonstrating the mid-root buccal defect (B) (arrow). Line corresponds to mid-root buccal plate fenestration. Coronal view demonstrating the mid-root buccal bone loss (C) (arrow). Surgical degranulation of the defect confirming the vertical root fracture in the mesial root (D) (arrow). ENDODONTICS: Colleagues for Excellence 4 that causes this scattering and beam hardening is close to or associated with the tooth being assessed, the overall sensitivity and specificity is dramatically reduced. A thorough dental history, noting classic clinical and radiographic signs and symptoms such as pain, swelling, presence of a sinus tract and/or presence of an isolated deep periodontal pocket can be helpful hints to suggest the presence of a VRF. Radiographically, a combination of periapical and lateral root radiolucency ‘‘halo’’ appearance is valuable information indicating the possible presence of VRFs. Several of the previously mentioned clinical and radiographic elements have to align to establish a presumptive diagnosis of VRF (19); however, dye examination, usually requiring surgical exposure, is still the gold standard for diagnosis of VRF. Nonsurgical and Surgical Treatment Planning Diagnostic information directly influences treatment planning and clinical decisions. Accurate data leads to better treatment decisions and potentially more predictable outcomes (21). Recommendations seven and eight from the AAE/AAOMR position statement identify a variety of circumstances where CBCT imaging may be appropriate in nonsurgical and surgical treatment planning. Several studies evaluated the impact of CBCT on diagnosis, clinical decision making and treatment planning (22-24). Ee et al evaluated and compared endodontic treatment planning with CBCT and periapical radiography (22). Thirty endodontic cases completed in a private endodontic practice were randomly selected to be included in the study. Each case was required to have a preoperative digital PA and a CBCT scan. Three Board-certified endodontists reviewed the 30 preoperative PAs. Two weeks later, the CBCT studies were reviewed in random order by the same evaluators. The evaluators were asked to select a preliminary diagnosis and treatment plan based solely on interpretation of the periapical and CBCT images. Diagnosis and treatment planning choices were then compared to determine if there was a change from the PA to the CBCT scan. Under the conditions of the previous study, CBCT was a more accurate imaging modality for diagnosis of endodontic pathosis when compared to diagnosis using only PAs. An accurate diagnosis was reached in 36.6% to 40% of the cases when using PAs compared to an accurate diagnosis in 76.6% to 83.3% of the cases when using CBCT. The previous study also demonstrated that the treatment plan may be directly influenced by information gained from CBCT studies as the examiners altered their treatment plan after viewing the CBCT scan in 62.2% of the cases overall (range from 56.6% to 66.7%). This high number indicates that CBCT had a significant influence on the examiners’ treatment plan. Figures 4A and 4B are examples of the utilization of CBCT imaging in nonsurgical treatment planning. The use of CBCT has been recommended for treatment planning of endodontic surgery (25-27). CBCT visualization of the true extent of periapical lesions and their proximity to important vital structures and anatomical landmarks is superior to that of PAs. The use of CBCT has enabled the clinician to evaluate the true extent of periapical lesions and their spatial relationship to important anatomical landmarks and vital structures (Figure 5). Fig. 4A. 3-D rendering of the MB root of tooth #14 in a coronal view (top) and axial view (bottom) demonstrating the untreated MB2. Fig. 4B. Coronal view of two-year recall demonstrating complete healing of the pathology related to MB root after nonsurgical retreatment (top). 3-D rendering axial view demonstrating the obturated MB1 and MB2 (bottom). Fig. 5A. 3-D rendering demonstrating the relation of the inferior alveolar nerve (arrow) to the periapical defect and the apex of tooth #29. Fig. 5B. A cropped 3-D coronal view rendering demonstrating the mental foramen, bundle and IAN in relation to the base of the periapical defect (arrow). Note the apical root resorption. 5 The Impact of Cone Beam Computed Tomography in Endodontics: A New Era in Diagnosis and Treatment Planning Endodontic Diagnosis and Detection of Inflammatory Resorptive Defects Diagnosis and detection of root resorption is often challenging due to the quiescent nature of onset and varying clinical presentation. Recommendation 12 suggests CBCT imaging in the localization and differentiation of external and internal resorptive defects and the determination of appropriate treatment and prognosis (Figure 6). Definitive diagnosis and treatment planning ultimately is dependent on the radiographic representation of the extent and progression of the disease. 2-D imaging offers a limited diagnostic potential when compared to 3-D imaging (28-29). A 3-D classification for external cervical resorption (ECR) recently published (30) takes into account the ECR lesion height, circumferential spread and proximity to the root canal, thus classifying ECR in three dimensions. This novel and clinically relevant 3-D classification should allow for more effective and accurate communication of ECR lesions between clinicians. It also will allow for an objective assessment of the effect of the ECR on treatment outcomes. Endodontic Diagnosis and Detection of Traumatic Dental Injuries Traumatic dental injuries (TDIs) present a clinical challenge with regard to diagnosis, treatment planning and prognosis. Radiographic assessment is important to identify the location, type and severity of TDIs. According to the 2012 International Association of Dental Traumatology guidelines (31), a series of PAs from different angulations and an occlusal film are recommended for evaluation of TDIs. Unfortunately, 2-D imaging has limitations in the diagnosis and detection of TDI due to projection geometry, magnification, superimposition of anatomic structures, distortion and processing errors. The AAE/AAOMR position statement recommends CBCT for the diagnosis and management of a variety of traumatic dental injuries, absent other maxillofacial or soft tissue injury that may require other advanced imaging modalities. In the diagnosis of horizontal root fractures (HRFs) utilizing 2-D imaging, the fracture line will be evident radiographically only if the central beam is directed within four degrees of the fracture line (32). 2-D imaging has limited accuracy in identifying the location, severity and extent of HRF and could lead to misdiagnosis, improper treatment and an unfavorable outcome. Because of the limitations of intraoral radiography, CBCT is suggested as the preferred imaging modality for diagnosis of HRF (33). CBCT overcomes several of the limitations of 2-D imaging by providing a considerable amount of 3-D information about the nature and extent of the HRF. The significant difference in the nature of HRF when assessed with 2-D radiographs compared to CBCT has been reported (33). Limited FOV CBCT imaging seems to be generally advantageous in the diagnosis, assessment of prognosis, treatment planning and treatment follow up of HRF cases (Figure 7). In pediatric patients with TDI, it should be noted that children are at greater risk than adults from a given dose of radiation because they are inherently more radiosensitive. Fig. 6. Example of the utilization of CBCT imaging in diagnosis and treatment planning of inflammatory resorptive defects. Periapical radiograph of tooth #21 (A). The patient was referred for evaluation and treatment of an internal resorptive defect. Coronal view of tooth #21 demonstrating the external/internal resorptive defect (B). 3-D reconstruction of the resorptive defect (B). Note in B and C the perforated root on the buccal aspect (white arrow) rendering tooth #21 nonrestorable. The treatment plan changed after reviewing the CBCT. Fig. 7. A 15-year-old patient with history of trauma to the anterior maxilla was referred for consultation for teeth #8 and 9. Tooth #8 had a grade I mobility and tooth #9 had a grade II+ mobility. Periapical radiograph (A) showed mid-root horizontal root fractures teeth #8 and #9. Periodontal probing depths were WNL. Marginal and attached gingiva demonstrated normal color and architecture. Sagittal views of teeth #8 and 9, respectively (B, C). Note the oblique nature of the root fractures and bone fill between the coronal and apical segments in C. Since the patient was asymptomatic and CBCT revealed no periradicular pathosis, a palatal splint was suggested to address the mobility but no endodontic intervention was recommended at the time. ENDODONTICS: Colleagues for Excellence 6 Summary Several studies evaluated the use of CBCT in endodontics (35-41) and found that its use overcomes many of the limitations of periapical radiography. The increased diagnostic information provided by the CBCT study should result in more accurate diagnosis and improved decision making for the management of complex endodontic problems. It is a desirable addition to the endodontist’s armamentarium. The effective radiation dose to patients when using CBCT is higher than conventional 2-D radiography and the benefit to the patient must therefore outweigh any potential risks of the additional radiation exposure, especially in children. Radiation dose should be kept as low as reasonably achievable (42-43). The value of CBCT for endodontic diagnosis and treatment planning should be determined on an individual basis to assure that the benefit: risk assessment supports the use of CBCT. Dentists should utilize the referral to an endodontist as part of the dental team as endodontists use the latest diagnostic imaging, specialty training and techniques to ensure the patient benefits from their diagnostic and treatment planning expertise. References 1. Goldman M, Pearson AH, Darzenta N. Endodontic success: who’s reading the radiograph? Oral Surg Oral Med Oral Pathol 1972;33:432–7. 2. Goldman M, Pearson AH, Darzenta N. Reliability of radiographic interpretation. Oral Surg Oral Med Oral Pathol 1974;38:287–93. 3. Tewary S, Luzzo J, Hartwell G. Endodontic radiography: Who is reading the digital radiograph? J Endod 2011;37:919-21. 4. Joint Position Statement of the American Association of Endodontists and the American Academy of Oral and Maxillofacial Radiology on the Use of Cone Beam Computed Tomography in Endodontics 2015 Update. 5. De Paula-Silva FW, Wu MK, Leonardo MR, da Silva LA, Wesselink PR. Accuracy of periapical radiography and cone-beam computed tomography scans in diagnosing apical periodontitis using histopathological findings as a gold standard. J Endod 2009;35:1009-12. 6. Patel S, Wilson R, Dawood A, Mannocci F. The detection of periapical pathosis using periapical radiography and cone beam computed tomography – part 1: preoperative status. Int Endod J 2012;8:702-10. 7. Sogur E, Grondahl H, Bakst G, Mert A. Does a combination of two radiographs increase accuracy in detecting acid-induced periapical lesions and does it approach the accuracy of cone-beam computed tomography scanning. J Endod 2012;38:131-6. 8. Patel S, Dawood A, Mannocci F, Wilson R, Pitt Ford T. Detection of periapical bone defects in human jaws using cone beam computed tomography and intraoral radiography. Int Endod J 2009;42:507-15. 9. Nixdorf D, Moana-Filho E. Persistent dento-alveolar pain disorder (PDAP): Working towards a better understanding. Rev Pain 2011;5:18-27. 10. Pigg M, List T, Petersson K, Lindh C, Petersson A. Diagnostic yield of conventional radiographic and cone-beam computed tomographic images in patients with atypical odontalgia. Int Endod J 2011;44:1092-1101 11. Blattner TC, Goerge N, Lee CC, Kumar V, Yelton CGJ. Efficacy of CBCT as a modality to accurately identify the presence of second mesiobuccal canals in maxillary first and second molars: a pilot study. J Endod 2012;36:867-70. 12. Michetti J, Maret D, Mallet J-P, Diemer F. Validation of cone beam computed tomography as a tool to explore root canal anatomy. J Endod 2010;36:1187-90. 13. Vizzotto MB, Silveira PF, Arús NA, Montagner F, Gomes BP, Da Silveira HE. CBCT for the assessment of second mesiobuccal (MB2) canals in maxillary molar teeth: effect of voxel size and presence of root filling. Int Endod J 2013;46:870-6. 14. Hassan B, Metska ME, Ozok AR, van der Stelt P, Wesselink PR. Detection of vertical root fractures in endodontically treated teeth by a cone beam computed tomography scan. J Endod 2009;35:719-22. 15. Edlund M, Nair MK, Nair UP. Detection of vertical root fractures by using cone-beam computed tomography: a clinical study. J Endod 2011;37:768-72. 16. Metska ME, Aartman IH, Wesselink PR, Özok AR. Detection of vertical root fracture in vivo in endodontically treated teeth by cone-beam computed tomography scans. J Endod 2012;38:1344-7. 17. Brady E, Mannocci F, Wilson R, Brown J, Patel S. A comparison of CBCT and periapical radiography for the detection of vertical root fractures in non-endodontically treated teeth. Int Endod J 2014;477:35-46. 18. Melo SLS, Bortoluzzi EA, Abreu M, Correa LR, Correa M. Diagnostic ability of a cone-beam computed tomography scan to assess longitudinal root fractures in prosthetically treated teeth. J Endod 2010;36:1879-82. 7 The Impact of Cone Beam Computed Tomography in Endodontics: A New Era in Diagnosis and Treatment Planning 19. Tsesis I., Rosen E., Tamse A., Taschieri S, Kfir A. Diagnosis of vertical root fractures in endodontically treated teeth based on clinical and radiographic indices: a systematic review. J Endod 2010;36:1455-58. 20. Fayad MI, Ashkenaz PJ, Johnson BR. Different representations of vertical root fractures detected by cone-beam volumetric tomography: a case series report. J Endod 2012;10:1435-42. 21. Liang YH, Li G, Wesselink PR, Wu MK. Endodontic outcome predictors identified with periapical radiographs and cone-beam computed tomography scans. J Endod 2011;37:326-31. 22. Ee J, Fayad MI, Johnson BR. Comparison of endodontic diagnosis and treatment planning decisions using cone-beam volumetric tomography versus periapical radiography. J Endod 2014;40:910-6. 23. Rodríguez G, Abella F, Sindreu F, Patel S, Roig M. Influence of Cone-beam Computed Tomography in Clinical Decision Making among Specialist. J Endod 2017;43:194–9. 24. Rodríguez G, Patel S, Sindreu F, Roig M, Abella F. Influence of Cone-beam Computed Tomography on Endodontic Retreatment Strategies among General Dental Practitioners and Endodontists. J Endod 2017;43:1433-7. 25. Venskutonis T, Plotino G, Tocci L, Gambarini G, Maminskas J, Juodzbalys G. Periapical and endodontic status scale based on periapical bone lesions and endodontic treatment quality evaluation using cone-beam computed tomography. J Endod 2015;41:190-6. 26. Bornstein MM , Lauber R, Pedram Sendi P, and Arx T. Comparison of periapical radiography and limited cone-beam computed tomography in mandibular molars for analysis of anatomical landmarks before apical surgery. J Endod 2011;37:151-7. 27. Low KM, Dula K, Bürgin W, Arx T. Comparison of periapical radiography and limited cone-beam tomography in posterior maxillary teeth referred for apical surgery. J Endod 2008;34:557-62. 28. Estrela C, Bueno MR, De Alencar AH, Mattar R, Valladares Neto J, Azevedo BC, De Araújo Estrela CR. Method to evaluate inflammatory root resorption by using cone beam computed tomography. J Endod 2009;35:1491-7. 29. Durack C, Patel S, Davies J, Wilson R, Mannocci F. Diagnostic accuracy of small volume cone beam computed tomography and intraoral periapical radiography for the detection of simulated external inflammatory root resorption. Int Endod J 2011;44:136-47. 30. Patel S, Foschi F, Mannocci F, Patel K. External cervical resorption: a three dimensional classification. Int Endod J 2018;51:206-214. 31. Diangelis AJ, Andreasen JO, Ebeleseder KA, et al. International Association of Dental Traumatology guidelines for the management of traumatic dental injuries: 1—fractures and luxations of permanent teeth. Dent Traumatol 2012;28:2–12. 32. Rud J, and Omnell KA. Root fracture due to corrosion. Diagnostic aspect. Scand Journal of Dental Research 1970;78:397-403. 33. Contemporary management of horizontal root fractures to the permanent dentition: diagnosis--radiologic assessment to include cone-beam computed tomography. J Endod 2013 Mar;39:S20-5. 34. Iikubo M, Kobayashi K, Mishima A, et al. Accuracy of intraoral radiography, multidetector helical CT, and limited cone-beam CT for the detection of horizontal tooth root fracture. Oral Surg Oral Med Oral Pathol Oral Radiol Endod 2009;108:e70-4. 35. Patel S, Durack C, Abella F, Shemesh H, Roig M, Lemberg K. Cone beam computed tomography in endodontics : a review. Int Endod J 2015;48:3-15. 36. Cotton TP, Geisler TM, Holden DT, et al. Endodontic applications of cone-beam volumetric tomography. J Endod 2007;33:1121-32. 37. Patel S, Dawood A, Whaites E, Pitt Ford T. New dimensions in endodontic imaging: part 1. Conventional and alternative radiographic systems. Int Endod J 2009;42:447-62. 38. Patel, S.: New dimensions in endodontic imaging: part 2. Cone beam computed tomography. Int Endod J 2009;42:463-75. 39. Scarfe WC, Farman AG, Sukovic P. Clinical applications of cone-beam computed tomography in dental practice. J Can Dent Assoc 2006;72:75-80. 40. Nair MK, Nair UP. Digital and advanced imaging in endodontics: a review. J Endod 2007;33:1-6. 41. Setzer FC, Hinckley N, Kohli MR, Karabucak B. A survey of cone-beam computed tomographic use among endodontic practitioners in the United States. J Endod 2017;43:699-704 42. Farman AG, Farman TT. A comparison of 18 different x-ray detectors currently used in dentistry. Oral Surg Oral Med Oral Pathol Oral Radiol Endod 2005;99:66-71. 43. Farman AG. ALARA still applies. Oral Surg Oral Med Oral Pathol Oral Radiol Endod 2005;100:395-7. © 2018 American Association of Endodontists (AAE), All Rights Reserved Information in this newsletter is designed to aid dentists. Practitioners must use their best professional judgment, taking into account the needs of each individual patient when making diagnosis/treatment plans. The AAE neither expressly nor implicitly warrants against any negative results associated with the application of this information. If you would like more information, consult your endodontic colleague or contact the AAE. Did you enjoy this issue of Colleagues? Are there topics you would like to cover in the future? We want to hear from you! Send your comments and questions to the American Association of Endodontists at the address below, and visit the Colleagues online archive at aae.org/colleagues for back issues of the newsletter. 211 E. Chicago Ave., Suite 1100 Chicago, IL 60611-2691 Phone: 800-872-3636 (U.S., Canada, Mexico) or 312-266-7255 Fax: 866-451-9020 (U.S., Canada, Mexico) or 312-266-9867 Email: info@aae.org facebook.com/endodontists @SavingYourTeeth youtube.com/rootcanalspecialists www.aae.org Endodontic Case Study This feature in Colleagues for Excellence highlights endodontic treatment that demonstrates the benefits of treatment planning and partnership with an endodontist to improve patient outcomes. A 14-year-old male was referred for consultation for tooth #31. The panoramic radiograph revealed a radiolucency related to tooth #31 (Figure 1-A). The tooth responded positively to pulp sensitivity tests and periodontal probings were WNL except an 8 mm pocket on the mid-buccal surface. The patient was previously seen by two other dental specialists who both recommended extraction of #31. Figure 1-B (a) Periapical radiograph of tooth #31. (b) Coronal view showing the buccal location of the periapical radiolucency in relation to tooth #31. Notice the apical third of the root is surrounded by bone which explains the positive response to cold and EPT. The lines on the PA correspond to the axial section views in C and D, which demonstrate the origin of the lesion as tooth #32. 3-D reconstruction (e) demonstrates the buccal location of the lesion and explains the isolated deep periodontal probing on the buccal aspect of tooth #31. Tooth #32 was extracted and the lesion was submitted for biopsy. The biopsy result was a dentigerous cyst. Panoramic image is a one-year recall demonstrating complete bony remodeling around #31 (Figure 1-C). Contributed by Dr. Mohamed I. Fayad The AAE wishes to thank Dr. Mohamed I. Fayad for authoring this issue of the newsletter, as well the following article reviewers: Drs. Scott L. Doyle, Steven J. Katz, Keith V. Krell, Garry L. Myers and Jaclyn F. Rivera. Exclusive Online Bonus Materials: The Impact of Cone Beam Computed Tomography in Endodontics This issue of the Colleagues newsletter is available online at aae.org/colleagues with the following bonus materials: • Full-Text Article: Patel K, Mannocci F, Patel S. The Assessment and Management of External Cervical Resorption with Periapical Radiographs and Cone-beam Computed tomography: a clinical study. J Endod 2016;42:1435-40. • AAE/AAOMR Joint Position Statement: Use of Cone Beam Computed Tomography in Endodontics. 2015 Update. • Full-Text Article: Ee J, Fayad MI, Johnson BR. Comparison of endodontic diagnosis and treatment planning decisions using cone-beam volumetric tomography versus periapical radiography. J Endod 2014;40:910-6. Fig. 1-C. Fig. 1-B. Fig. 1-A.
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https://hc.edu/center-for-christianity-in-business/2014/06/13/the-ten-commandments-in-the-world-of-business-thou-shalt-not-covet/
The Tenth Commandment (10th Commandment) in the World of Business Skip to main contentSkip to sidebarSkip to footer ×Search Search ApplyVisitRequest Info Academics Majors & ProgramsAcademic CalendarColleges & SchoolsRecords & TranscriptsMoody LibraryAcademic Services Admissions & Aid Undergraduate AdmissionsGraduate AdmissionsOnline AdmissionsTuition & AidFinancial Aid Life at HCU Student LifeSpiritual LifeHousing & DiningStudent ServicesStudent Success About About HCUVision & MissionLeadershipThe CampusNews, Events, Publications Huskies AthleticsAlumniGivingMyHCU Student Links My Application Status Access Student Systems Payment Portal Academic Schedule Program Catalog University Store Campus Map & Parking Community Links News, Events & Publications Facility Rentals Commencement Social Media Channels Personnel Directory Phone Directory Job Opportunities ApplyVisitRequest Info AthleticsAlumniGivingMyHCU Academics Majors & ProgramsAcademic CalendarColleges & SchoolsRecords & TranscriptsMoody LibraryAcademic Services Admissions & Aid Undergraduate AdmissionsGraduate AdmissionsOnline AdmissionsTuition & AidFinancial Aid Life at HCU Student LifeSpiritual LifeHousing & DiningStudent ServicesStudent Success About About HCUVision & MissionLeadershipThe CampusNews, Events, Publications Search and Sidebar Menu The Tenth Commandment in the World of Business Center for Christianity in Business About About CCB Our Leadership CCB Scholarships Contact Donate Publications The Christian Business Review Articles Bookstore Media Christian Business Leader Podcast Webinars Prior CCB Luncheons and Seminars Videos Events All Events Mentoring Conference Networking Luncheon and Seminar Professional Management Seminars CCB Articles Posted on June 13, 2014 June 13, 2025 SHARE: See also: A Biblical View of Wealth By Dr. Richard J. Martinez Thou Shalt Not Covet Alas As the equipment and workers left the premises, I was pleased to see the upgrades my new neighbor recently made to his house, yard, and landscaping. Very nice trees, all new windows, paint, gutters, new fence, etc. I am certainly excited about any property improvements my neighbors make. Who wouldn’t be? But it didn’t take long for my admiration to turn into something less positive. My yard could use some new trees. And I would love to install a circular driveway in front of my house, like many of my neighbors have. My fence is rotting in places. I wouldn’t say that I became “pouty” over my neighbor’s improvements, but I moved from appreciation to…what? Envy? Maybe, maybe not. Coveting? Well, what is coveting, anyway? Browse Related Degrees Bachelor Degrees Business Majors Biblical Studies Christian Ministry Theological Studies Graduate Degrees Business Master’s Christian Leadership Theological Studies Browse All Degrees » The Journey Here I begin a series of commentaries around the Ten Commandments, and how each may be understood in the business and marketplace context. For reasons that will become apparent, I will discuss these critical elements of the Christian life in reverse order; that is, beginning with the Tenth Commandment. Moreover, while a full theological exploration of each of these commandments could approach book length, I will limit my discussion as best as I can. Coveting Exodus 20:17: “You shall not covet your neighbor’s house;you shall not covet your neighbor’s wife, or his male servant, or his female servant, or his ox, or his donkey, or anything that is your neighbor’s.” So, to clarify terms, the coveting that is forbidden by God in the tenth commandment involves allowing a legitimate appreciation or desire for things that others possess (or that I do not possess) to be corrupted into an illegitimate obsession – one that often leads to a multitude of other sins. That is, it is natural and good to have an appreciation for that which is valuable and beautiful. I appreciate the improvements my neighbor has made to his house. I can celebrate his improved lot in life (pun intended). But it is entirely possible for someone in my position – the neighbor, the observer, the unimproved lot – to move from contentment about my own possessions to jealousy, hatred, theft, destruction. Now, my neighbor really has nothing to worry about from me, but admiration, appreciation, and contentment can quickly become corrupted in any of us when we see what others have acquired. And, of course, covetousness can be measured by degrees. I might not burn my neighbor’s trees, graffiti his windows, or steal his gutters, but I might become too quick to judge him harshly, or gossip about his materialism, or avoid him altogether, out of jealousy or spite. C.S. Lewis takes note of the particular frustration that God must experience as He sees our discontent with His provision: He has provided a rich, beautiful world for people to live in. He has given them intelligence to show them how it can be used, and conscience to show them how it ought to be used. He has contrived that the things they need for their biological life (food, drink, rest, sleep, exercise) should be positively delightful to them. And, having done all this, He sees all His plans spoiled…by the crookedness of the people themselves. All the things He has given them to be happy with they turn into occasions for quarreling, and jealousy, and excess and hoarding, and tomfoolery. Marketplace Coveting As we consider the particular temptations toward coveting in the business context, we begin by simplifying the meaning. We are speaking generally of wanting or desiring what we do not have, such that it leads us to discount our appreciation of what we do have, or to compromise our biblical principles in order to obtain the object of our desire. I may covet many things in the marketplace: The job my former classmate just landed The promotion my colleague deserved and was awarded The attractive woman with whom I have been working on a work project Status symbols such as parking spaces, or office locations In fact, the things that are most likely to tempt us to covet will change across time in our business lives. We can learn much from Maslow’s familiar “Hierarchy of Needs” here. Maslow suggested that humans are motivated to satisfy different types of needs in different stages of life. He specified the following (from lower-order needs to higher-order needs): Survival and physiological needs (e.g. food, shelter) Safety and Security needs (e.g. neighborhood, job) Relationship and belongingness needs (e.g. spouse, friends, memberships) Esteem needs (e.g. awards, advancement, status symbols) Self-actualization needs (e.g. fulfillment, legacy, creativity) Early Challenges Those of us in the earliest stages of a business calling are most likely seeking things that ensure fulfillment of “lower-level” needs. We desire a job or position that ensures that rudimentary bills, such as rent and student loans, can be paid. We seek a job in an industry or firm that is established and stable. These are all normal and understandable desires. In this quest, however, we don’t always have… well, enough. As soon as my budgeted bills are covered by income, I wonder why I can’t afford a little more, a little fun. Surely I deserve to be doing what “everybody else” is doing, right? Surely I should be in my own house, not this crummy apartment. Someone with my potential should be in a car more fitting of my (desired) social class. And soon the thankfulness (even thrill) I once had for my job is gone. We know that people in all business positions constantly compare their efforts and rewards with others, making determinations of “equity.” Soon I am making note of the (likely undeserving) peers around me who have better jobs, more money, and “nicer things”, even though I am likely smarter, harder-working, and more deserving. I have moved from contentment with the manner in which God has provided for my needs to coveting what does not belong to me. At this point we violate the tenth commandment. Worse yet, it is this early discontent that leads many young people to incur the financial indebtedness that has become the enabler of their covetousness. Those of us pursuing service in the business world are especially vulnerable to this early covetousness. Many of the rewards of business success are the exact type of material rewards that most appeal to people seeking material security. It is very easy to fall into coveting in the materialistic, money-minded world of business. Constant Challenges Interestingly enough, people at all stages of business life are susceptible to the coveting that is related to our relationship and membership needs. The fact is we are working with and around people (mostly), and much of what we do is stressful. We seek many forms of comfort through our work relationships. We form bonds of camaraderie with co-workers and team members. We build social communities at work that facilitate productivity and creativity. These are all positive outcomes of the organizational experience. But there is a dark side of marketplace life that arises from coveting. I will mention two here: the sexual coveting that comes from proximity, and the social coveting that comes from the “Inner Ring”. There are many wonderful, fulfilling, and legitimate relationships that form in the context of business life. But it is no great revelation for me to mention that humans in the business world often engage in sexually-motivated activities with co-workers and colleagues. Proximity and the pressures of business life create interpersonal sparks that are a constant temptation for most people. This creates the modern-day scenario that God has in mind when He commands, “…you shall not covet your neighbor’s wife…”. I am speaking, of course, specifically of illicit relationships, adulterous affairs, and the coveting that leads to exploitation, such as sexual harassment and the creation of a sexually hostile work environment. Modern business life is susceptible to these temptations, especially when we allow ourselves to “go there,” that is, to covet sexual relationships that are outside of marriage. Enough said. The second relationship challenge is related to what management researchers have identified as in-groups and out-groups. According to the Leader-Member Exchange model of organizations, managers and other leaders create for themselves an “in-group” of employees who are most-trusted, first-consulted, and who have access to inside information not shared with others (those in the out-group). This is, of course, not problematic, as such arrangements facilitate decision-making and progress. C.S. Lewis refers to this group as the Inner Ring, and he notes that it exists in all organizations, not just businesses. The great concern we ought to have, according to Lewis, is that our human nature is to covet membership in the inner-ring: I believe that in all men’s lives at certain periods, and in many men’s lives at all periods…one of the most dominant elements is the desire to be inside the local Ring and the terror of being left outside… A thing may be morally neutral and yet the desire for that thing may be dangerous… This desire [for membership in the Inner Ring] is one of the great permanent mainsprings of human action… Of all passions the passion for the Inner Ring is most skillful in making a man who is not yet a very bad man do very bad things. We know, in the business community, that our bosses, our organizations, our industries have Inner Rings. We are tempted to covet entrance into these Rings. It is the coveting that leads us to compromise our principles, to demonstrate our loyalties in problematic ways, to leave behind our former friends who are keeping us from being as “relevant” as we wish to be. As Lewis notes, we are at our best when we simply find ourselves in the Rings that matter, from having been diligent and productive in our work, and faithful to our friends and employers. Later Challenges The greater marketplace temptations for those in later stages of business careers, or who have “risen beyond” the needs at the lower levels of Maslow’s Hierarchy, are those pertaining to power, esteem, and status. It is one thing to earn and receive authority, responsibility, power, and the titles that define them. It is another thing to look across the hall, or across the street, to titles and powers we have not yet attained, and then to begin the maneuvering we deem necessary to obtain what is not ours. At higher levels of management, it is even possible to project our coveting into the organizational context. Since I know that executive compensation is positively correlated with organizational size, I may covet the acquisition that will drastically increase my firm’s market position. What won’t I do to make that happen? Ambition is a funny thing. We likely agree that one should be ambitious to do good (e.g. Titus 2:14; 1 Pet. 3:13), and to be productive, and to grow in life and in skills and in accomplishments. But we also know that ambition can come to control us (e.g. James 4:13-17). One needs only to watch a few episodes of “American Greed” (a CNBC network series: to have a sense of the destructive pattern. There is a “WOW!” factor upon seeing the story be repeated again and again – “white collar” and business criminals are typically swayed through overwhelming temptation to have what others have, to be what others have become, to live a life that is beyond their reach through patient, legitimate means. Bluntly, the stupidity displayed across the real-life stories in the series is a consequence of coveting, and a testimony to the goodness of the tenth commandment. Having gotten a view of the “high life” from colleagues, customers, or otherwise, many people in many business activities lie, cheat, and steal (literally) to realize the new desires that have been awoken within them. Their common refrain is that it is not simply the money or the toys that compels them, but it is the status, the esteem, the access to ever-better company, that most tempts them. They covet a new, better life. It is not just the neighbor’s house, or his wife, or his servants or possessions that we covet. It is all of it. The whole thing. We want it all. And we want it now. Like it or not, the higher we rise in the business world, the greater is the temptation to covet, and the greater is our access to the resources that may make all of our covetous dreams come true. Thou Shalt Not Covet Placed at the end of the Commandments given to Moses on Sinai, it is tempting to view covetousness as a relatively minor transgression. Earlier I noted that covetousness is problematic, in one sense, because it might lead to the commission of other sins. But let’s be clear: coveting is a grievous sin in its own right, and the tenth commandment is specifically named as one of the ten transgressions we are commanded to avoid. How then shall we be obedient to the tenth commandment? Three Things that will Limit our Coveting Cultivate a spirit and attitude of Contentedness – It is not easy to see past the trappings of materialism in a material world. As my human nature feels constrained by the limits of physical life, it is not unusual to feel as though I must maximize the material experience. YOLO, right! And yet, my God has promised to provide for my needs (Matthew 6:25-34). This includes everything at every level of Maslow’s Hierarchy. God provides for my physiological needs, as noted in Matt. 6. He is my Rock, my Security (e.g. Heb. 13:6). He provides the friends and relations and love that I need. In terms of esteem, He assures me that I have great value in His eyes. And, in the end, I am made whole in Him. As John the Baptist said, He must increase, and I must decrease (John 3:29-30). I am a new creation in Christ (2 Cor. 5:17). If I cannot be thrilled and contented and completed by the joy of knowing God and doing His will, then I do not know Him very well, and I must continue to seek Him in His Word. Believe me when I say that I am not there yet, like many of you, I imagine. Cultivate a habit of Prayer for control and perspective – As a follower of Jesus Christ, I know that I am not here (in this world) simply for me. Frankly, I am already saved and redeemed by the blood of Christ, and my entire purpose for living and working in the marketplace is to serve God and perform His will. That is, I am here to be His hands and feet. Where is the place for coveting from that perspective? And yet, I am indeed still human. I lose perspective. I get caught up in the here and now of this world. It is by God’s grace alone that I can escape the clutches and bondage of the materialism and greed that leads to coveting. I must be in the habit of praying every day (and sometimes more) for God’s wisdom and discernment. I must pray continuously for the Holy Spirit’s help in controlling my impulses and shaping where my mind goes. I must pray that I can “be anxious for nothing”. I must pray something like this: Heavenly Father, please help me to realize who You are, and who I am in You. Help me to remember your purpose and will for my life. Allow me to be and remain thankful and full of joy for this day that You have given me here, and for your provisions as I work out Your calling on my life. For each blessing you shower upon me and my family, Lord, I praise you. Father, I submit to the leading of Your Holy Spirit as I face temptations on this day, that each victory over temptation will be for your glory. Amen Cultivate a system of Accountability – My wife and I work to hold one another accountable when it comes to coveting and discontentedness. Each of us needs to develop honest, realistic, ongoing conversations with someone who knows us well, and who can snap us back to reality when that covetous gaze comes over us. Since coveting is essentially a longing for that which we do not have and do not need, I need to be held accountable by my wife, who is able to remind me of how blessed we are. She is faithful to remind me (for example) that we do not NEED a circular driveway (sigh), and we would be better off using our resources to eliminate debt or support specific missions projects. Once we have been launched into the world of business, the pressure is high to BE something special, to HAVE something beautiful, to MOVE into ever-higher circles of influence and prestige. These things in and of themselves are not bad, but they can become the objects of our obsessions instead of the means of glorifying God. We must surround ourselves with partners and communities that can keep us grounded as God elevates our ministries’ impacts. We need to be reminded that the glittering ring that is within our grasp is not always gold, and even if it is, grasping it may come at a cost. Should I take the promotion and the elevated salary that comes with it, even if my church brothers point out that it may exact a cost on my family life? That it may impact my ability to teach or preach or engage in other ministries? Maybe the answer is yes, I should take this step, sensing the Lord’s leading in this way. But I need to hear these arguments, I need to have these conversations. I must have a system of accountability. In the end, we will be tempted go beyond admiration into the realm of coveting. We will fail to be restrained from coveting at times. But we are clear: coveting is sinful, it is forbidden in the tenth commandment, and it represents unique challenges for those of us in the world of business. Let us pray, hold one another accountable, and be ever-thankful and contended with the table that has been prepared for us by a loving God, Who meets all of our needs abundantly. ABOUT THE AUTHOR: Dr. Rick Martinez is Chair of the Department of Management, Marketing, and Business at Houston Christian University. He holds a Ph.D. in Management from (THE) Texas A&M University, and is blessed to be involved teaching and preaching ministries at his church, First Colony Bible Chapel in Sugar Land, TX. Rick, by God’s grace, is married to Peggy, and they have three awesome children, Mariah, Tyler, and Sierra. Please contact atrjmartinez@HC.edu. C.S. Lewis, (1970). “The Trouble with ‘X’…” In Walter Hooper (Ed.), God in the Dock, (p. 152). Grand Rapids, MI: W. B. Eerdmans Publishing. Or, at least tend to focus on lower-level needs before being concerned with higher-level needs. This is reflected in pursuit of lower-level needs at earlier stages of life and career; Maslow’s model is much-critiqued, but has a certain intuitive appeal; Maslow, A.H. (1943). “A Theory of Human Motivation,” Psychological Review, pp. 370-396. Equity Theory is presented in Adams, J. (1965). “Inequality in Social Exchange,” in L. Berkowitz (Ed.) Advances in Experimental Social Psychology (New York: Academic Press). Graen, G. (1976). “Role-making processes within complex organizations”. In M. D. Dunnette (Ed.), Handbook of industrial and Organizational Psychology(pp. 1201-1245). Chicago: Rand McNally CCB About About CCB Our Leadership CCB Scholarships Contact Donate Publications The Christian Business Review Articles Bookstore Media Christian Business Leader Podcast Webinars Prior CCB Luncheons and Seminars Videos Events All Events Mentoring Conference Networking Luncheon and Seminar Professional Management Seminars Newsletter Signup to receive access to all past issues of the Christian Business Review Journal plus two new "Christianity in Business" articles each month. indicates required Email Address First Name Last Name Browse article archives Browse article archives See also… Majors and Degree Programs Student Life Houston Christian University Online (281) 649-3000 7502 Fondren Rd Houston, TX 77074 Instagram Facebook X (Twitter) TikTok LinkedIn YouTube Get Started Majors and Programs Tuition & Aid Admissions Visit HCU Student Life Online Learning The Graduate School Resources Academic Calendar Moody Library University Store Colleges and Schools Campus Map & Parking Police and Emergency Job Opportunities Contact Social Media Channels News, Events & Publications Request Information Phone Directory Personnel Directory Facility Rentals Office of Student Accounts Nondiscrimination Notice Title IX Policies Privacy Accreditation © 2025 Houston Christian University
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https://artofproblemsolving.com/wiki/index.php/Alternating_sum?srsltid=AfmBOorCHiRVQcRU9fZFlCYSI4YrmKVjWXRP6H_L2AXG632i0dLWk0VR
Art of Problem Solving Alternating sum - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Alternating sum Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Alternating sum An alternating sum is a series of real numbers in which the terms alternate sign. For example, the alternating harmonic series is . Alternating sums also arise in other cases. For instance, the divisibility rule for 11 is to take the alternating sum of the digits of the integer in question and check if the result is divisble by 11. Given an infinite alternating sum, , with , if corresponding sequence approaches a limit of [0|zero]] monotonically then the series converges. Error estimation Suppose that an infinite alternating sum satisfies the the above test for convergence. Then letting equal and the -term partial sum equal , the Alternating Series Error Bound states that The value of the error term must also have the opposite sign as , the last term of the partial series. Examples of infinite alternating sums This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
188253
https://www.colorado.edu/cas/outreach/curriculum-units/middle-school/japanese-haiku-and-haiga-one-week-unit-plan
Center for Asian Studies College of Arts and Sciences Japanese Haiku and Haiga: A One Week Unit Plan Japanese Haiku and Haiga: A One Week Unit Plan By: John Keveanos Norte Vista High School Riverside, California Monday: Day 1 Lesson: An Introduction to Japanese Haiku Poetry; the Rules of Haiku. Theme of the Lesson: Introduce the Japanese Haiku poem and the rules that define the art form. California State Standards: Reading 2.2 - Analyze the way in which clarity of meaning is affected by the patterns of organization, hierarchical structures, repletion of the main ideas, syntax, and word choice in the text. Literary Response and Analysis 3.1 - Analyze the characteristics of subgenres (e.g., satire, parody, allegory, pastoral) that are used in poetry, prose, plays, novels, short stories, essays, and other basic genres. Literary Response and Analysis 3.2 - Analyze the way in which the theme or meaning of a selection represents a view comment on life, using textual evidence to support the claim. Literary Response and Analysis 3.3 - Analyze the ways in which irony, tone, mood, the author's style, and the "sound" of language achieve specific rhetorical or aesthetic purposes or both Literary Response and Analysis 3.3 - Analyze the ways in which poets use imagery, personification, figures of speech, and sounds to evoke readers' emotions. ELD Standard: Listening and Speaking Cluster 2: Listen attentively to the stories and information and identify important details and concepts by using verbal and nonverbal reponses. Objectives: By the end of this lesson, the student will... identify a Japanese Haiku poem. recognize the various rules of traditional Haiku poetry analyze and differentiate the elements of Haiku poetry Materials: "Traditional Japanese Kabuki Music": A collection from numerous sources; Photographs taken from the "Okunohoshimiche" Tour 2009; Writing and Enjoying Haiku: A Hands-On Guide by Jane Reichhold (Kodansha International Ltd., Tokyo, Japan, 2002.) The students will receive a handout that will provide them with "Six Basic Rules of Haiku" found on pages 49-52. Also, "A Checklist for Revising Haiku" and "Believe It or Not- More Rules" found on Pages 73-39. The Language of Literature: World Literature (McDougal Littell, Evanston. IL, 2002). - The students will be examining Haiku poetry on pages 541-549 of the textbook. Prior Knowledge: By the time we reach this unit on Haiku poetry, the students will have spent two weeks examining Ancient Japanese Literature. They will have read and analyzed short selections from The Pillow Book (Sei Shonagon); The Tale of Genji (Lady Murasaki Shikibu); Zen Teachings and Parables (Muso soseki); The Deserted Crone (a Noh Drama by Zeami Motokiyo) and an assortment of Tanka Poetry (Ono Komachi, Lady Ise, Ki Tsurayuki, and Saigyo). On the Friday before this unit this students were given the handouts on Haiku poetry rules taken from the Reichhold text and asked to read and study them. Anticipatory Set: As the students enter the classroom, they will hear the sounds of "Traditional Japanese Kabuki Music" being played. They will see assorted photographs of the natural beauty of Japan that the teacher took on the Okunohoshimichi Study Tour 2009. Introductory Activity: A KWL Graphic Organizer: What do you know about Japan, Japanese, Culture, and Japanese Poetry? Lesson Body: What the teacher does: The teacher allows the students ten minutes to listen to the music; view images of rural Japan; and jot down their ideas about Japan, its culture and poetry. The teacher will then ask the students to share the ideas aloud. What the student does: The students share their knowledge of Japan, Japanese Culture, and Japanese Poetry. What the teacher does: Then, the teacher directs the students attention to the photos projecten on the screen by the LCS. The teacher describes how Japan is a virtual garden and the importance of the preservation of nature to the Japanese. The teacher informs the students that Japanese Haiku Poetry is a literary expression of their love of nature; that is the central subject of the art form. The teacher explains how the haiku poet seeks to discover a "Haiku Moment" which is an inspiration from an observation of nature. (Also known as an "Ah-ha" moment.) The teacher asks the students to take out their Haiku Handout and take a few moments to review Reichhold's Haiku rules starting with the six that the author has identified as the most important: "1. Write in three lines that are short, long, short without counting syllables.; 2. Make sure the haiku has a fragment and a phrase.; 3. Have some element of nature.; 4. Use words in the present tense.; 5. Avoid capital letters or punctuation.; 6. Avoid rhymes." (Note: The teacher should stress that a traditional haiku poem contains seventeen syllables that are distributed over three lines of verse: five in the first and third lines with seven in line two. Also that the author's reference to an "element of nature" comes in the form of a seasonal word or "Kigo"). What the teacher does: The teacher presents the students (on the LCS or the whiteboard) with five Haiku poems for analysis: The teacher also encourages the students to take out a piece of paper so that they can take Cornell Notes during their analysis of Haiku poetry. after morning snow onion shoots rise in the garden like little signposts -Matsuo Basho The teacher then leads the students in an analysis both the form of the Haiku poem. What the student does: The student analyzes the poem using the handout. Responses may include: "The syllables are wrong" (The teacher may point out that this poem had been translated from Japanese and the translation may not be able to recreate the rule regarding syllables without ruining the poet's intent). "'Spring' is the Kigo" (Actually, "withered" and "warm" are also Kigo words). "The poem is in the present tense". (The teacher agrees). "There is a phrases and a fragment" (The fragment is line one). The students also agrees that: "There are no capital letters"; "There is only one article - 'the' garden"; "There is no mention of humanity". (Not in a classic, traditional Haiku, the teacher points out.) What the teacher does: The teacher asks the students about hte content of the poem: "What do you think is the theme of the poem?" (The students decide that it is about the changing seasons). "Is this a proper subject for a Haiku?", the teacher asks. (The students agree it is due to its consideration of nature.) The teacher then asks them to one again review their handout to see how well the Haiku lives up to the rules of a traditional Haiku. What the student does: The students may make observations regarding whether or not the poem..." ...is about ordinary things (written) in an ordinary way using ordinary language:". (It is); "Eliminates all the possible uses of gerunds ("ing" endings on verbs). (It does not); "Attempt to have levels of meaning in the haiku. On the surface it is a set of simple images; underneath, a philosophy or lesson of life". (It deals with the passage of time which was a major theme of the previously studies Tanka Poetry). What the teacher does: The teacher presents a new Haiku poem: white blossoms of the pear and a woman in moonlight reading a letter -Yosa Buson The teacher then asks them to once again review their handout to see how well the Haiku lives up to the rules of a traditional Haiku. What the student does: The students go about their analysis of both form and content. Comments may include: "Again, the syllables are wrong". (A problem of translation). "The word 'and' in line two ruins the fragment and phase" (Perhaps this is also an imperfection of the translation). "'Blossoms' is the Kigo". (Yes, a Spring word). "A human is depicted in this poem" (Yes, this is a rule that is frequently broken by authentic Haiku poets and seems to create a subgenre of Haiku poetry). What the teacher does: The teacher presents a new Haiku poem: The pheasant cries as if it has just noticed the mountain. -Kobayashi Issa What the teacher does: The teacher then asks them to once again review their handout to see how well the Haiku lives up to the rules of a traditional Haiku. What the student does: Again the students point out how the poem is not using symbols properly; lack a phrase and fragment; uses punctuation and a capital letter; and seems to lack a Kigo. (The teacher gets a chance to reiterate the problems of translation from Japanese to English; the capricious disregard of the assumed "rules" that many poets employ in their Hiakus; and that "pheasant" and "mountain" could be seasonal word for Autumn). Lesson Conclusion/ Re-teaching: What the teacher does: The teacher asks the students to use their Cornell Notes in order to extend their knowledge of information about Japan, its culture and its Haiku poetry to the KWL Graphic Organizer that they started at the beginning of the period. What the student does: The students apply their Cornell Notes to their KWL Graphic Organizer. Homework/ Reinforcement: What the teacher does: The teacher assigns the students to write two Haiku poems that adhere to the guidelines that Jane Reichhold has suggested. Reiterate that all Haiku poetry does not necessarily have to conform to all these rules (as we saw quite often in this lesson) but let us start by seeing how close we can come to pure Haiku poetry. Tuesday: Day 2 Lesson: Experiencing ginko - an inspiration walk around campus in order to collect ideas to improve the students' Haiku poetry skills; Peer editing of student Haiku poems. Theme of the Lesson: Finding "Haiku Moments" for the creation of poetry and taking a closer look at nature in order to improve the students' use of imagery. California State Standards: Literary Response and Analysis 3.4 - Analyze ways in which poet's use imagery, personification, figures of speech, and sounds to evoke readers emotions. Writing Strategies 1.5 - Use language in natural, fresh, and vivid ways to establish a specific tone. Research and Technology 1.6 - Develop presentations by using clear research questions and creative critical research strategies (e.g., field studies, oral histories, interviews, experiments, electronic sources). ELD Standards: Writing Strategies and Applications Cluster 1 ES: Write responses to selected literature that exhibits an understand of the text, using detailed (Note: Here this will applied to a peer editing exercise). Objectives: By the end of this lesson, the student will... describe nature using imagery with greater eloquence. apply this imagery to the creation of Haiku poetry. critique and peer edit their Haiku poetry homework. Materials: New selections from the teacher's "Traditional Japanese Kabuki Music" CD. A trio of miniature Banzai Trees. Assorted photographs of the natural beauty of Japan that the teacher took on the Okunohoshimichi Study Tour 2009. Prior Knowledge: What the students have learned from the handout; yesterday's critique of haiku poetry; their attempt to write two traditional Haikus and homework. Anticipatory Set: As the students enter the classroom, they will hear the sounds of "Traditional Japanese Kabuki Music" being played. They will see assorted photographs of the Okunhosimichi Hiking Trail in the Naruko area. This breathtaking nature walk over hill and dale was part of Matsuo Basho's original trek. The trio of potted miniature Bonsai Trees are placed prominently on a table before the LCS projection screen. Lesson Body: What the teacher does: The teacher introduces the students to the Japanese concept of "Ginko" - a nature walk intended to inspire the creation of Haiku poetry. The teacher (in this case, me) talks about where the incredible photos were taken: on a Ginko-like hike through the Japan's northern interior, near Naruko. The teacher will explain that Haiku poet Matsuo Basho had walked this path 320 years ago. The teacher will state that Basho, (whom we will study in greater depth tomorrow) seemed to make all of scenic Japan a "Ginko". The teacher then tells the students that they will today go on a "Ginko" by strolling around the campus of our high school. The teacher asks the students to get a notebook so they can jot down ideas. What the student does: The students grab a notebook and a pen and assemble outside the classroom. What the teacher does: The teacher explains that although Norte Vista High School in Riverside, California may not be the Garden of Eden, it still provides us with some glimpses of natural beauty. The teacher reiterates that the purpose of the walk is to discovery and describe images of natural beauty. Examples should include floral, foul, insects, weather, sounds and aromas. Anything that will create greater imagery in your Haiku. Then, the teacher leads the student on a stroll around campus, wishing them well in their quest for “Haiku Moments”. What the student does: The student will find numerous “Haiku Moments” (even at Norte Vista High School). The albatross birds that swarm around and squawk near the Performing Arts Center; the Ivy-covered walls with its changing leaves; the majestic, snow- capped mountains seen from the Quad which gives the school its name. What the teacher does: The teacher encourages the students to point out their observations and to avoid using clichéd adjectives or trite descriptions. For instance, that is not just a tree: What kind of a tree is it? Or, the bird is singing, but what would be a more precise word to let the reader hear the sound? What the teacher does: The teacher leads the students back to class. Once back inside the classroom, the teacher points out the potted miniature Bonzai Trees which also feature little ceramic figurines of poets, workers and herons and other birds. These are meant for people to find serenity and inspiration anywhere. The teacher asks the students to write a sentence that describes the scene in one of the Bonzai pots, perhaps employing some of the descriptive language found on the Ginko. What the student does: The students write their descriptions. Conclusion/ Re-teaching: What the teacher does: The teacher asks that some of the students read their work aloud before it is collected. (A few students do).Then the teacher asks the students to take out the two Haiku poems that they had written as homework. The teacher asks the students to split into pairs to analyze and critique each others poems. Is the poem faithful to the rules of traditional Haiku? But now, after our Ginko, how might the poem be improved? How could it be made more vivid? The teacher gives the students the rest of the period to make changes before they are required to turn in their work. What the student does: The students work in pairs peer editing their poems to be turned in at the end of the period. Homework/ Reinforcement: What the teacher does: The teacher assigns the students two new Haiku poems that must employ at least two examples of imagery. One of the poems must conform to Reichhold’s rules. The other may take liberties (e.g., a human presence, a syllable out of place, etc.) What the student does: The students write down the assignment and turn in their peer edited and rewritten poems before going to their next class. Wednesday: Day 3 Lesson: Poems by Matsuo Basho; Introduction to Haiga. Theme of the Lesson: An introduction to Haiku’s most revered poet and to a subgenre of Haiku that merges art illustration and poetry. California State Standards: Literary Response and Analysis 3.2- Analyze the way in which the theme or meaning of a selection represents a view or comment on life, using textual evidence to support the claim. Response and Analysis 3.3- Analyze the ways in which irony, tone, mood, the author’s style, and the “sound” of language achieve specific rhetorical or aesthetic purposes or both. Writing Application 2.1- Write fictional, autobiographical, or biographical narratives... c. Make effective use of descriptions of appearance, images, shifting perspectives, and sensory details. (Note: There are no California Standards that require students to write poetry, but I feel this one comes the closest to Haiku). ELD Standards: Listening and Speaking Cluster 2: Listen attentively to the stories and information and identify important details and concepts by using verbal and nonverbal responses. Listening and Speaking Cluster 8: Respond to messages by asking questions, challenging statements, or offering examples that affirm the messages. Cluster 9 Prepare and deliver short oral presentations. Objectives: By the end of this lesson, the student will... ...recognize the contribution of Matsuo Basho to the art of Haiku poetry. ...appraise and critique some of Basho’s poems. ...write and illustrate Haiga poetry. Materials: An assortment of different recordings of the Japanese classic “Cherry Blossoms”; Assorted photographs of the natural beauty of Japan that the teacher took on the Okunohoshimichi Study Tour 2009, especially all the different statues of Basho that are found all along the route of the Okunohosimichi trail; white typing paper; a box of art supplies that includes crayons, markers, and colored pencils and pens. Prior Knowledge: What the students continue to learn from the handout; yesterday’s Ginko and peer edit of their Haikus. Anticipatory Set: As the students enter the classroom, they will hear the sounds of the Japanese musical class, “Cherry Blossom” , played on different instruments with different interpretations. They will see assorted photographs of the natural beauty of Japan that the teacher took on the Okunohoshimichi Study Tour 2009, especially all the different statues of Basho that are found all along the route of the Okunohosimichi trail; A handout on the life of Matsuo Basho found on Wikipedia. The Language of Literature: World Literature (McDougal Littell), Evanston. IL, 2002).-The students will be examining Basho’s Haiku poetry on pages 542 of the textbook Introductory Activity: The students read over the handout about the life of Matsuo Basho. Lesson Body: What the teacher does: The teacher reinforces the handout information by talking about the Okunohosimichi Study Tour of Japan 2009. I, in this case, would stress the importance of Basho’s art and legacy and how the Japanese people and Press reacted to our endeavors. The teacher would then ask the students to open their McDougal Littell The Language of Literature: World Literature textbooks to page 542: Matsuo Basho Poems- Translated by Sam Hamill. The teacher then refers to three of Basho’s poems that have been written on the white board and selects three pairs of students to come up and critique the poem applying the Reichhold rules as well as making their own analysis of the poem. The teacher then leads the class in a reading of the three poems. They are: Pitifully- under a great soldier’s empty helmet, a cricket sings Withered winter grass- waves of warm spring air shimmering just above Sick on my journey, only my dreams will wander these desolate moors What the student does: The students study the poems while those selected to critique the poems come to the whiteboard and begin to circle words and underline phrases; identifying the Kigo; and writing down other observations in a visual diagram of their critique. What the teacher does: The teacher asks the first pair of students to present their critique of the poem to the class. What the student does: The students might start with a formal critique (e.g., the poem is close to the proper syllable count; identification of fragment and phrase; Kigo; present tense; the translator’s use of a capital letter and a hyphen). What the teacher does: The teacher asks the students to discuss if they think these liberties are helpful to the translation. What the student does: The students’ answers to this question might lead to some consensus on how closely a poet must stick to Reichhold’s guidelines. The students at the white board continue their critique by discussing the poem’s content. They may point out that the Haiku (as Reichhold suggests) “(a)ttempts to have levels of meaning...On the surface it is a set of simple images; underneath a philosophy or lesson of life” (Reichhold, p.77) They might regard the poem as a profound statement regarding the futility of war. Reichhold states (on page 76) that the Haiku should “(s)ave the ‘punch’ for the end line”. The singing cricket serves this purpose. They might call attention to Reichhold’s suggestion that the Haiku “(u)se only images from nature with no mention of humanity”. Yet this Haiku makes a statement about humanity without the direct presence of an individual human. (Note: The process of students critiquing the next to poems follow a similar pattern to the one previously described). What the teacher does: The teacher might make reference to the third poem, sometimes called Basho’s “Death Poem” and its relevance considering that Basho would himself die a short time later, at 46. The Rest of the Lesson Body: What the teacher does: The teacher provides a definition of Haiga as a Haiku that is accompanied by art work which complements the Haiku. Reichhold describes a Haiga as “(i)llustrations that break up the reader’s headlong rush for the next haiku”. The teacher asks the students to choose a partner (but not the same one from the peer editing session yesterday). The teacher then tells them that their task will be to write an illustrate two Haiga poems- one traditional that follows Reichhold’s rules and one more freeform. The teacher then makes the art supplies available to the students and asks them to begin working. What the student does: The students work on creating two Haiku and begin to create illustrations that will illustrate them. What the teacher does: Perhaps the teacher will play soft Japanese music to underscore the students’ endeavor. Also, the teacher will walk around the classroom providing suggestions, answering questions, and making sure that all the students are on task. Conclusion/ Re-teaching: The Haiga-making task at hand provides the students with reinforcement of all lessons learned in the unit thus far. Homework/ Reinforcement: What the teacher does: The teacher tells the students that they need to continue to work on their Haigas. They will be presented to the class on Friday as part of a Gallery Walk Exhibition. What the student does: The students assign each other tasks regarding the Haiga project. Then, they return the art supplies and prepare to leave. Thursday: Day 4 Lesson: Poems by Yosa Buson; Creating Haiga poetry. Theme of the Lesson: An introduction to one of Haiku’s most revered poet and a dynamic lesson in merging illustration and poetry. California State Standards: Literary Response and Analysis 3.2- Analyze the way in which the theme or meaning of a selection represents a view or comment on life, using textual evidence to support the claim. Response and Analysis 3.3- Analyze the ways in which irony, tone, mood, the author’s style, and the “sound” of language achieve specific rhetorical or aesthetic purposes or both. Writing Application 2.1- Write fictional, autobiographical, or biographical narratives... c. Make effective use of descriptions of appearance, images, shifting perspectives, and sensory details. ELD Standards: Listening and Speaking Cluster 2: Listen attentively to the stories and information and identify important details and concepts by using verbal and nonverbal responses. Listening and Speaking Cluster 8: Respond to messages by asking questions, challenging statements, or offering examples that affirm the messages. Cluster 9 Prepare and deliver short oral presentations. Objectives: By the end of this lesson, the student will... ...recognize the contribution of Yusa Buson to the art of Haiku poetry. ...appraise and critique some of Buson’s poems. ...write and illustrate Haiga poetry. Materials: A compilation CD of famous Japanese Pop Hits entitled “Japan- The Greatest Hits Ever”.; Assorted photographs of the natural beauty of Japan that the teacher took on the Okunohoshimichi Study Tour 2009; white typing paper; a box of art supplies that includes crayons, markers, and colored pencils and pens; Bamboo place mats to be used as a matting for the Haigas; String to hang the Haigas; A handout on the life of Yusa Buson found on Wikapedia; The Language of Literature: World Literature (McDougal Littell, Evanston. IL, 2002).-The students will be examining Buson’s Haiku poetry on page 544 of the textbook. Prior Knowledge: What the students continue to learn from the handout; yesterday’s experience reading Basho’s poetry and creating Haiga. Anticipatory Set: As the students enter the classroom, they will hear the sounds of the Japanese Pop Hits from the compilation CD entitled “Japan- The Greatest Hits”. They will see assorted photographs of the natural beauty of Japan that the teacher took on the Okunohoshimichi Study Tour 2009, especially the different monuments upon which Basho’s poetry is inscribed, which stresses the importance of Haiku poetry to the Japanese people; A handout on the life of Yusa Buson found on Wikipedia. Introductory Activity: The students read over the handout about the life of Yusa Buson. Lesson Body: What the teacher does: The teacher should stress the importance of Buson’s art and legacy. The teacher then asks the students to open their The Language of Literature: World Literature (McDougal Littell) textbooks to page 543: Yusa Buson Poems- Translated by Robert Haas. The teacher then refers to three of Buson’s poems that have been written on the white board and selects three pairs of students to come up and critique the poem applying the Reichhold rules as well as making their own analysis of the poem. The teacher then leads the class in a reading of the three poems. They are: Coolness- the sound of the bell as it leaves the bell The old man cutting barley- bent like a sickle It cried three times, the deer, then silence What the student does: The students study the poems while those selected to critique the poems come to the whiteboard and begin to circle words and underline phrases; identifying the Kigo, etc.; and writing down other observations in a visual diagram of their critique. What the teacher does: The teacher asks the first pair of students to present their critique of the poem to the class. What the student does: The students might start with a formal critique (e.g., the poem is close to the proper syllable count; identification of fragment and phrase; Kigo; present tense; the translator’s use of a capital letter and a hyphen). Note: The rest of this part of the Lesson Body analyses the Haiku poems as explained in the lesson plans described on Days 1 and 3. The Rest of the Lesson Body: What the teacher does: The teacher then makes the art supplies available to the students and asks them to take out their Haiga projects and continue working. What the student does: The students work on finishing their two haikus and the creation of illustrations that will adorn them. What the teacher does: Perhaps the teacher will play soft Japanese music to underscore the students endeavor. Also, the teacher will walk around the classroom providing suggestions, answering questions, commenting on their work, and making sure that all the students are on task. The teacher provides the students with bamboo place mats, paste, tape, string, tacks, and everything else the students will need to matte their work and hang it on the walls of the classroom. Conclusion/ Re-teaching: The haiga-making task at hand and the hanging of their work, provides the students with reinforcement of all lessons learned in the unit thus far. Homework/ Reinforcement: What the teacher does: The teacher advises the students that they need to copy their work and prepare to present their poem aloud. They will be presented to the class tomorrow as part of a Gallery Walk Exhibition. What the student does: The students copy their work. (Each student is to present one of their poems). The student arrange with their partners who will present which poem then prepare to leave. Friday: Day 5 Lesson: Poems by Kobayashi Issa; Introduction to Haiga Gallery Walk and Presentations. Theme of the Lesson: An introduction to another of Haiku’s most revered poet and poetry presentations and Gallery Walk. California State Standards: Literary Response and Analysis 3.2- Analyze the way in which the theme or meaning of a selection represents a view or comment on life, using textual evidence to support the claim. Response and Analysis 3.3- Analyze the ways in which irony, tone, mood, the author’s style, and the “sound” of language achieve specific rhetorical or aesthetic purposes or both. Writing Application 2.1- Write fictional, autobiographical, or biographical narratives... c. Make effective use of descriptions of appearance, images, shifting perspectives, and sensory details. (Note: There are no California Standards that require students to write poetry, but I feel this one comes the closest to Haiku).Listening and Speaking Strategies 2.5- Recite poems selections from speeches, or dramatic soliloquies with attention to performance details to achieve clarity, force, and aesthetic effect and to demonstrate an understanding of the meaning (e.g., Hamlet’s soliloquy “To Be or Not to Be”). ELD Standards: Listening and Speaking Cluster 3: Speak clearly and comprehensively by using standard English grammatical forms, sounds, intonation, pitch, and modulation.: Reading Fluency and Systematic Vocabulary Development Cluster 7: Read aloud with appropriate pacing, intonation, and expression increasingly complex narrative and expository texts.; Reading Comprehension Cluster 2: Orally identify the features of simple excerpts of public documents by using key words or phrases.; Writing Strategies and Applications Cluster 1 ES: Write responses to selected literature that exhibit understanding of the text, using detailed sentences and transitions. Listening and Speaking Cluster 9: Prepare and deliver short oral presentations. Objectives: By the end of this lesson, the student will... ...recognize the contribution of Kobayashi Issa to the art of Haiku poetry. ...appraise and critique some of Issa’s poems. ...describe and recite Haiga poetry. Materials: An assortment of different recordings of the Japanese classics both modern and traditional; The Language of Literature: World Literature (McDougal Littell, Evanston. IL, 2002).-The students will be examining Issa’s Haiku poetry on page 544 of the textbook; A handout on the life of Kobayashi Issa found on Wikipedia. The Language of Literature: World Literature (McDougal Littell, Evanston. IL, 2002).-The students will be examining Issa’s Haiku poetry on page 544 of the textbook. Prior Knowledge: What the students continue to learn from the handout and all other activities this week. Anticipatory Set: As the students enter the classroom, they will hear the sounds of the Japanese music- traditional and modern selections. They are encouraged to examine all their Haigas that grace the walls of the classroom. Also, the teacher distributes a copy of the Wikipedia piece on Kobayashi Issa and encourages them to read it. Introductory Activity: The students read over the handout about the life of Kobayasi Issa. Lesson Body: What the teacher does: The teacher should stress the importance of Issa’s art and legacy and his contribution to the art of Haiku . The teacher then asks the students to open their The Language of Literature: World Literature: (McDougal Littell) textbooks to page 544: Kobayasi Issa Poems- Translated by Robert Haas. The teacher then refers to three of Issa’s poems that have been written on the white board and selects three pairs of students to come up and critique the poem applying the Reichhold rules as well as making their own analysis of the poem. The teacher then leads the class in a reading of the three poems. They are: A huge frog and I, staring at each other, neither of us moves. Asked how old he was, the boy in the new kimono stretched out all five fingers. In a dream my daughter lifts a melon to her soft cheek. What the student does: The students study the poems while those selected to critique the poems come to the whiteboard and begin to circle words and underline phrases; identifying the Kigo; and writing down other observations in a visual diagram of their critique. What the teacher does: The teacher asks the first pair of students to present their critique of the poem to the class. What the student does: The students might start with a formal critique (e.g., the poem is close to the proper syllable count; identification of fragment and phrase; Kigo; present tense; the translator’s use of a capital letter and a hyphen). Note: The rest of this part of the Lesson Body analyses the Haiku poems as explained in the lesson plans described on Days 1 and 3. The Rest of the Lesson Body: What the teacher does: The teacher then directs half of the pairs of students to stand by their pair of Haiga while the other half in small groups will do a Gallery Walk. The presenters will recite their poems and comment on their meaning and inspiration, etc. The Gallery walkers must make notes on what they believe to be the most perfect Haiga and explain why. What the student does: The first group of presenters and Gallery Walkers begin. What the teacher does: Perhaps the teacher will play soft Japanese music to underscore the students endeavor. Also, the teacher will walk around the classroom observing the quality of the students Oral Presentations. and making sure that all the students are on task. What the teacher does: After the first group of presenters has displayed their Haiga, then the teacher will ask the students to change places and the presenters will now become Gallery Walkers and the Gallery Walkers now become presenters. Conclusion/ Re-teaching: What the teacher does: Asks the students to finish choosing what they felt were the best haiga and why. What the student does: The students go over their notes and make their selections and why their thought their choices of top haiga were superior. Homework/ Reinforcement: What the teacher does: The teacher congratulates the student on a unit well done and assigns no homework. A Note on Unit Assessment: There are a number of ways that assessment was used. The three most important criteria that I, as the teacher, would use. They are: Effort- A strong engagement with the lesson. Creativity- The joy of participation and/ or and outlet of personal expression. Excellence- A striving and achievement of a mastery in understanding and recreation of the art form. Of course, the poetic assignments and Oral Presentations are great factors in grading. Included is a test that covers the entire Japanese Unit that I administer> World Literature: English IV UnitTest 3: Capturing the Moment Ancient Japanese LiteratureMr. Keveanos, Room 602 Name:____________________________ Period:_____ Japanese Literature truly began with the invention of Kana, a writing system based on Chinese characters. This occurred sometime between 784-1185, during the reign of which imperial dynasty: A) Heian B) Kamakura C) Ashikaga D) Tokugawa Sei Shonagon wrote “The Pillow Book” which is considered the first example of Japanese Literature. The book describes imperial court life and its genre is that of a: A) Novel B) Diary C) Novella D) Collection of Short Stories. “Lists” are among the many selections that Sei includes in the book. Which is NOT one of the “Lists” she includes: A) Embarrassing Things B) Hateful Things C) Humorous Things D) Graceful Things The book is entitled “The Pillow Box” because they were private thoughts that were written in the bedroom and stashed away in the drawers of the wooden pillows on which the nobility slept. (Circle one) True False “Zen Teachings and Parables” are writings meant to illustrate the teachings of which religion: A) Confucianism B) Buddhism C) Hinduism D) Taoism “Zen Teachings and Parables” are a form of what type of writing. ___________________________________________________ Zen Masters and their disciples aspire to achieve a state of spiritual enlightenment known as: ___________________________________________________ In “Contamination of Virtue”, Muso Soseki describes the virtue of doing good for the sake of good, not to do good to seek a reward. He writes: “Contamination and Noncontamination refer to the state of mind of the doer, not to the good deed itself”. This teaching is in the form of a: ___________________________________________________ “Publishing the Sutras” tells the story of Tetsugen, a devotee of Zen. In it, the writer claims that “(t)he Japanese tell their children that Tetsugen made three sets of sutras, and that the first two invisable sets surpass even the last”. This is said because the Tetsugen gave the money to the Buddhist priests to build a golden temple where everyone who forever have a glorious place to pray. (Circle one) True False 10. The parable “Right & Wrong”, Bankei, a Zen master, backs a thief among his students, telling them that they know right from wrong but “(w)ho will teach him if I do not?” Hearing this, the thief did which of the following; A) Confessed B) Cried C) Laughed D) Escaped 11. “The Deserted Crone” by Zeami Mutokiyo is an example of Kabuki Drama. (Circle One) Yes Noh 12. This type of Drama has religious roots and is said to lead to spiritual enlightenment. Name the religion that this dramatic form is based upon. 13 &14. There are two major reasons why this type of Japanese Drama is similar to Greek drama. They are: 1)____________________________________________ 2)____________________________________________ The play’s title, “The Deserted Crone”, refers to an old woman who was left to die on Mount Tonagawa. (Circle one) True False “Tanka” Poetry actually means_____________ ______________ 17. Tanka Poetry consists of 31 syllables divided among five lines. (Circle one) True False Which of the following is NOT a subject of Tanka Poetry: A) The significance of love B) The necessity for war C) The beauty of nature D) The passage of time 19. Haiku poetry possesses three qualities that are greatly valued in Japanese Art. These are: A)Precision, delicacy, and drama B)Precision, economy, and power C)Precision, economy and delicacy D)Precision, delicacy, and grace 20-23. A Haiku consists of ____lines of poetry. Line one has_____ syllables; Line two has_______ syllables and Line three has _____ syllables. Haikus also contain a “Kigo” which is a Along with their rhyming structure, Haikus are known for their: A) Humor B) Imagery C) Figurative Language D) Drama South, Southeast and West Asia Outreach Program (SSEWA) Program for Teaching East Asia Curriculum Units Elementary Middle School High School The American Association of Teachers of Japanese (AATJ)
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https://www.youtube.com/watch?v=DqpSmtSfbvk
Kinematic Equations for Constant Acceleration This “edukate” Guy 1780 subscribers 77 likes Description 10852 views Posted: 29 Aug 2012 This is an instructional video that illustrates and discusses the derivation of the kinematic equations for particles (objects) moving in one dimension under constant acceleration. It is important to understand where these powerful expressions came from. 6 comments Transcript: Introduction [Music] welcome to another physics video today we talk about deriving the four kinematic equations for constant acceleration uh if you have not watched the video on constant acceleration please watch that before watching this one in the video on constant acceleration we did derive the four kinematic equations but it was kind of mixed with teaching constant acceleration as well and it's so important to learn how to derive equations and expressions by yourself that I really want to touch on this topic a little bit more um for instance imagine that you're getting ready to take your test you've been studying really hard uh and one of the problems on the test is asking for the final position of some object under constant acceleration and darn it you forgot the formula you forgot the expression you for you forgot it so what are you going to do how are you going to figure it out so understanding how the four equations came to be and understanding where they come from guess what if you did forget on your test you can now duplicate it and it's extremely simple so let's go ahead and get started in the constant Excel acceleration video it's stated that the instantaneous acceleration is equal to the average acceleration and that's because the acceleration is constant so since this is true we can use the average acceleration which is the one without the limit sign we could use this to uh derive some interesting Expressions some powerful expressions for objects under constant acceleration in I should say one dimension all right so what is the average acceleration well acceleration is equal to the change in velocity over the change in time that's the average acceleration all right so what deriving the four kinematic equations is all about is basically algebra gymnastics you're looking at your expression and you're trying to find other ways you can write it so there's there's a couple things here already the change in velocity that can be Rewritten because we know change in velocity is nothing more than its final velocity minus in its initial velocity and change in time is nothing more than its final time minus its initial time which for this video we will just call T since the initial time will be zero all right so imagine that you see a problem and it's asking you for the final velocity of an object under constant an acceleration and it gives you the acceleration it gives you the initial velocity and it gives you the time well it might be a little easier to use a formula that's solved already for the final velocity so all you have to do is plug in your information so where does that come from well you substitute you do your algebra gymnastics and you substitute these these uh equal equal expressions in for your original so you get acceleration is equal to final velocity minus initial velocity over T time all right now if a problem is asking you for your final velocity well let's go ahead and solve for it let's see if we can get an expression that is in terms that is uh geared at solving for the final velocity so you multiply each side by time right this is just simple algebra all right so this leaves you with final velocity minus initial velocity is equal to the acceleration times the time and we want to solve for final velocity so we add initial velocity to each side and the final velocity is equal to the initial velocity plus the acceleration time time keep in mind super important keep in mind that this is for constant acceleration only this does not apply if the Accel acceleration is not constant all right because then we you know you'd still have the limit sign in front of everything and it'd be all it'd be a huge mess this is only for constant acceleration but if you look in your physics textbook or anywhere else you'll see that usually this is the first powerful kinematic equation for constant Final Position acceleration all right so you finished that problem you found your you found your final you F you found your final velocity and everything you're on to your next problem on your test and suddenly it's asking you about the final position of an object knowing the initial position and the uh knowing the average velocity and knowing the time uh and it asks you for the final position and you're like darn it I don't remember that formula I don't remember that expression well there is a very very easy way of deriving the the another um kinematic equation and that's realizing since constant acceleration varies linearly in time meaning it's constant um the average velocity is actually going to be equal to the initial velocity plus the final velocity divid by two it's just going to it simply will be the arithmetic means all right and we also know that average velocity is equal to your change in X over your change in time your displacement over time which we know to be your final position minus your initial position over over T all right so combine now if we were trying to make an expression for final position then we should probably combine these two uh these two ideas here and solve for our final position so plugging this in for our average velocity we get the initial velocity plus the final velocity over 2 is equal to the final position minus the initial position over T now it's simply once again algebra gymnastics multiply each side by T and you get2 of the initial velocity plus the final velocity times the time right equals the final position minus the initial position now if we add the initial position to each side we actually solve for the final position which is the initial position plus 1/2 of the initial velocity plus the final velocity times the time so here is the second main K uh kinematic equation for constant acceleration and once again it was derived from very simple expressions do not try to memorize these practice deriving Final Velocity [Music] these all right so you finished that last problem you're moving on to the next one it's asking you for the final position of another object this time though it doesn't give you the final velocity which you notice in your second kinematic equation here that that's kind of a requirement for this particular equation so and and on top of that you forgot the third kinematic equation so what do you do well you basically do some more olyp um algebraic gymnastics for your final velocity here you could actually plug in your first your first kinematic equation right into where it has the V there let's go ahead and do that and see if we can get something without final velocity cuz notice if we plug in this first expression we no longer will have a final velocity in our second kinematic equation which is the goal so our final position is equal to our initial position plus 12 of the initial velocity plus the initial velocity plus a cuz we just substituted this in instead of our final velocity right and all of that times T simplifying get final position is equal to the initial position plus 12 of the initial velocity time time + 12 of the initial velocity time time + 12 of the acceleration time squared now we combine like terms which are these two we get the final position is equal to the initial position plus the initial velocity time time plus 1 12 of the acceleration time time squar so it took me less than probably 1 minute to derive that one yet it's powerful it gives you a relationship for the final position that doesn't even include its final velocity so you now have another tool in your tool belt to solve constant acceleration problems and you don't even have to remember the tool you just have to remember how to do some algebra gymnastics all right right and the fourth kinematic equation is a little bit more difficult than the other two that we have derived in the sense that it requires a little bit more algebra but basically for this one the idea is we want to get uh an equation for the final velocity but one that does not have time within it so what we're going to do is we're going to use equation one we're going to solve for time plug it into equation two solve for final velocity so let's begin and um final velocity is equal to the initial velocity plus acceleration time time so we're going to solve for T here uh acceleration time time is equal to the final velocity minus the initial velocity time is equal to the final velocity minus the initial velocity over acceleration all I did there was divided each side by a so this is what we're going to use to plug into the second kinematic equation in order to um uh get rid of the time variable within our uh expression so we have the final position is equal to the initial position plus 12 of the initial velocity plus the final velocity times time which is the final velocity minus the initial velocity over acceleration so simplifying this we have the final position is equal to the initial position plus now if you multiply these together you're actually going to get V final position squared minus initial position squar and go ahead and verify that using the uh distributive property but I'm going to run out of space if I do it here and then all of that is over 2 a all right now from here remember we're looking for an equation for velocity final velocity that doesn't have time within it and if you look here there is no time symbol in it all that's left to do is to solve for the final velocity first subtract your initial position from each side which gets rid of this one here and multiply both sides by 2 a but I'll go ahead and do it one step at a time here so I'm going to move this to the left hand side v^2 final position squar minus initial position squar over 2 a is equal to the final position is the initial position all right now for this next part we're going to multiply each side by 2 a in order to cancel this denominator here this gives us v^ squar final position squared minus initial position squared equals 2 time acceleration time the quantity of the displacement basically just the displac placement and all right so all we have to do from here is add the initial velocity to each side and we get the final velocity squared is equal to the initial velocity squared plus 2 a the final position minus the initial position all right and this is the fourth kinematic equation [Music] here is the list of the four chematic equations you can use for constant acceleration these are only for constant acceleration thanks for watching the video and if you have a test coming up with a a timed test make sure that you do try to memorize these to some extent but your first priority should always be to master the derivation for it instead of just memorizing the formula without knowing the derivation
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https://www.pathologyoutlines.com/topic/transfusionmedrbcexchange.html
Pathology Outlines - Red blood cell exchange Menu Chapters By Subspecialty Autopsy & forensics Bone, joints & soft tissue Bone & joints Soft tissue Breast Clinical pathology Chemistry, toxicology & UA Coagulation Hematology & immune disorders Lab admin & management Microbiology & infectious diseases Transfusion medicine Cytopathology Dermatopathology Skin melanocytic tumor Skin nonmelanocytic tumor Skin nontumor GI / liver Anus & perianal Appendix Colon Esophagus Gallbladder & extrahep bile ducts Liver & intrahepatic bile ducts Pancreas Small intestine & ampulla Stomach GU / adrenal Adrenal gland & paraganglia Bladder & urothelial tract Kidney tumor Penis & scrotum Prostate gland & seminal vesicles Testis & paratestis Gyn Cervix Fallopian tubes & broad ligament Ovary Placenta Pleura & peritoneum Uterus Vulva & vagina Head & neck Ear Larynx, hypopharynx & trachea Mandible & maxilla Nasal cavity & nasopharynx Oral cavity & oropharynx Salivary glands Thyroid & parathyroid Hematopathology Bone marrow neoplastic Bone marrow nonneoplastic Hematology & immune disorders Lymph nodes & spleen, nonlymphoma Lymphoma & related disorders Informatics / digital pathology Medical renal Neuropathology CNS & pituitary tumors CNS nontumor Eye Muscle & peripheral nerve nontumor Staging Stains & CD markers / molecular Molecular markers Stains & CD markers Thoracic Heart & vascular pathology Lung Mediastinum Pleura & peritoneum WHO classification of tumors Jobs Fellowships Conferences / Webinars Authors Editorial Board Authors Author Information Author Instructions Directory Books Question Bank Other Links About Us Advertise Blog Cases CCN CME (online) Comment Contact Us FAQ Games Grants Newsletters Statistics Testimonials Videos Textbook Autopsy & forensics Bone, joints & soft tissue Bone & joints Soft tissue Breast Clinical pathology Chemistry, toxicology & UA Coagulation Hematology & immune disorders Lab admin & management Microbiology & infectious diseases Transfusion medicine Cytopathology Dermatopathology Skin melanocytic tumor Skin nonmelanocytic tumor Skin nontumor GI / liver Anus & perianal Appendix Colon Esophagus Gallbladder & extrahep bile ducts Liver & intrahepatic bile ducts Pancreas Small intestine & ampulla Stomach GU / adrenal Adrenal gland & paraganglia Bladder & urothelial tract Kidney tumor Penis & scrotum Prostate gland & seminal vesicles Testis & paratestis Gyn Cervix Fallopian tubes & broad ligament Ovary Placenta Pleura & peritoneum Uterus Vulva & vagina Head & neck Ear Larynx, hypopharynx & trachea Mandible & maxilla Nasal cavity & nasopharynx Oral cavity & oropharynx Salivary glands Thyroid & parathyroid Hematopathology Bone marrow neoplastic Bone marrow nonneoplastic Hematology & immune disorders Lymph nodes & spleen, nonlymphoma Lymphoma & related disorders Informatics / digital pathology Medical renal Neuropathology CNS & pituitary tumors CNS nontumor Eye Muscle & peripheral nerve nontumor Staging Stains & CD markers / molecular Molecular markers Stains & CD markers Thoracic Heart & vascular pathology Lung Mediastinum Pleura & peritoneum WHO classification of tumors Jobs Fellowships Conferences Directory Books Cases Advertise × The Directory now requires login - if you have a profile, click the "Forgot your password?" link to activate it. Authors Editorial Board Authors Author Information Author Instructions Author Awards Blog Question Bank CCN CME Comment Contact Us Games Grants Newsletters Statistics Testimonials Home>Transfusion medicine> Red blood cell exchange Transfusion medicine Therapeutic apheresis Red blood cell exchange Authors:Timothy Carll, M.D., Geoffrey Wool, M.D., Ph.D. Editorial Board Member:Kyle Annen, D.O. Deputy Editor-in-Chief:Patricia Tsang, M.D., M.B.A. Last author update: 14 February 2022 Last staff update: 14 February 2022 Copyright: 2002-2025, PathologyOutlines.com, Inc. PubMed Search: Red blood cell exchange[TI] OR erythrocytapheresis "free full text"[SB] Page views in 2024: 1,645 Page views in 2025 to date: 736 Table of Contents Definition / general | Essential features | Terminology | American Society for Apheresis (ASFA) guidelines | Indications for red blood cell exchange | Indications for erythrocytapheresis | Replacement fluid | Exchange volume | Simple exchange, depletion and depletion / exchange | Vascular access | Anticoagulation | Circuit priming | Adverse events | Sample assessment & plan | Additional references | Practice question #1 | Practice answer #1 | Practice question #2 | Practice answer #2 | Practice question #3 | Practice answer #3 Cite this page: Carll T, Wool G. Red blood cell exchange. PathologyOutlines.com website. Accessed September 28th, 2025. Definition / general Automated technique that separates red blood cells (RBCs) from whole blood and replaces with replacement fluid When replacement fluid is donor packed red blood cells (pRBCs), the procedure is referred to as red blood cell exchange (RBCX) When replacement fluid is crystalloid or colloid solution, the technique is referred to as erythrocytapheresis (Greek: aphaíresis, "taking away") Essential features Therapeutic technique directed at correcting disease driven by congenital or acquired abnormality of red blood cells Sickle cell disease (SCD) is the most common indication; American Society for Apheresis (ASFA) category I indications for RBCX in SCD include treatment of acute stroke and chronic prevention of stroke Other indications include parasitemias, such as babesiosis and malaria Red blood cell exchange requires donor pRBCs; having valid antibody screen testing on the patient is crucial and blood should be antigenically matched in sickle cell patients Hereditary hemochromatosis (HH) and polycythemia vera (PV) are ASFA category I indications for erythrocytapheresis, with modest efficacy advantages over simple therapeutic phlebotomy Terminology Total blood volume (TBV): Estimated from patient height, weight and sex Preferred method is Nadler formula Red blood cell volume (mass) (RBCV): Calculated by multiplying total blood volume and hematocrit (Hct) Hemoglobin S as a percentage of blood hemoglobin (HbS%): Marker of disease severity in sickle cell disease best measured by high performance liquid chromatography (HPLC) Fraction of cells remaining (FCR): Fraction of residual patient red blood cells following exchange, a marker of procedure performance 1.0 would indicate no replacement of patient cells, 0.0 would reflect complete replacement Whole blood to anticoagulant ratio (WB:AC): Inversely reflects degree of anticoagulation in extracorporeal circuit Ratios of 10:1 to 15:1 are commonly used for un-anticoagulated patients Inlet rate: Refers to volume of whole blood drawn per unit time from patient Directly affects procedure runtime American Society for Apheresis (ASFA) guidelines Published triennially (every 3 years) Categories published based on best evidence; organized into 4: Category I: disorders for which apheresis is accepted as first line therapy Category II: disorders for which apheresis is accepted as second line therapy Category III: optimum role of apheresis therapy is not established; decision making should be individualized Category IV: disorders in which published evidence demonstrates or suggests apheresis to be ineffective or harmful Reference: J Clin Apher 2019;34:171 Indications for red blood cell exchange Sickle cell disease: Congenital hemoglobinopathy caused by beta globin mutation Disease can be caused by homozygous HbS or compound heterozygotes (HbSC, HbS beta thalassemia) In HbSS patients with acute exacerbation, goal of RBCX is generally to reduce HbS to 30% or lower Homozygotic patients with no HbS% data available can generally be assumed to have 90 - 100% HbS unless transfused within past 100 days In patients with HbSC disease or other symptomatic compound heterozygous sickling disorder, goal of RBCX is generally to increase HbA to 70% or higher (i.e. the combined fraction of [HbS + HbX] is 30% or less) Acute indications Acute stroke: category I Acute chest syndrome, severe: category II Other complications (vaso-occlusive crisis, dactylitis, priapism, sequestration): category III In patients with prior stroke or other severe sequelae of SCD or to prevent complication of pregnancy / surgery, goal of RBCX is to maintain HbS < 30 - 50%; HbS% data is generally necessary for procedure planning Note: simple transfusion of pRBCs may be indicated as a temporizing measure while vascular access is obtained and blood is procured Chronic indications Chronic stroke prophylaxis (primary or secondary): category I Pregnancy: category II Recurrent veno-occlusive crises: category II Preoperative management: category III Parasitemias: Severe infection by intraerythrocytic parasites, typically Plasmodium or Babesia Diagnosis requires laboratory evidence of hemolysis and blood smear examination demonstrating organisms RBCX is generally adjunctive to antiparasitic therapy when > 10% parasitemia with severe symptoms 2 times RBC volume exchange is standard Babesiosis: category II Malaria: category III ABO incompatibility: Prevention / amelioration of hemolysis in major ABO incompatibility may be achievable through RBCX Major ABO incompatible red cell transfusion: Rationale is to remove incompatible transfused cells before total hemolysis occurs In practice, cells hemolyze rapidly and practice is not recommended Note: plasma exchange could be considered for removal of excess free hemoglobin to treat moderate to severe nephrotoxicity caused by hemolysis of incompatible RBCs Red cell alloimmunization: Usually to prevent immunization to RhD by RhD negative women in setting of accidental or emergency exposure to RhD positive cells or massive hemorrhage by RhD positive pregnancy Category III Minor ABO incompatible hematopoietic stem cell transplantation: To prevent hemolysis of recipient RBCs by donor isohemagglutinins Effectively accelerates patient ABO conversion Category III Reference: J Clin Apher 2019;34:171 Indications for erythrocytapheresis Hereditary hemochromatosis: Treat iron overload by removal of RBC mass, goal ferritin < 50 ng/mL Erythrocytapheresis can more rapidly correct hyperferritinemia than therapeutic phlebotomy but incurs a higher procedure cost Category I Polycythemia vera: Treat polycythemia by removal of RBC mass, goal hematocrit ≤ 45% Erythrocytapheresis can more rapidly correct polycythemia than therapeutic phlebotomy and may be preferred in hemodynamically unstable patients Category I Reference: J Clin Apher 2019;34:171 Replacement fluid For RBCX, obtain type and screen and procure donor pRBCs for replacement fluid, if needed Patients with sickle cell disease require prophylactic antigen matching per National Heart, Lung and Blood Institute (NHLBI) guidelines Procuring antigen negative compatible blood may cause delay ABO, RH group and K recommended for sickle cell patients which no prior history of antibody formation Extended matching (ABO, RH group, K, Duffy, Kidd and Ss systems) recommended for patients with a history of antibody formation Reference: JAMA 2014;312:1033 Exchange volume Removal of pathogenic substances in apheresis follows decay equation: Y/Yo = e-X Y = residual pathogenic substance (note: fraction of cells remaining = Y/Yo) Yo = initial pathogenic substance X = number of volumes exchanged (in RBCX, red blood cell volume equivalents) Determine the required number of volumes to exchange and convert into donor pRBC volume Recall that most pRBC units are stored with additive solution and have a hematocrit of approximately 60% Prediction of RBCX efficiency is optimized by accurate knowledge of pRBC unit hematocrit Example: Patient with 5 L total blood volume and 30% hematocrit has 90% HbS RBCX is requested with an end goal of 30% Goal fraction of cells remaining is (30/90) = 33% Solving for X indicates that ~1.1 times the RBC volume should be exchanged Patient's RBC volume is (5,000 mL x 0.3) = 1,500 mL 1,650 mL of red cell mass should be exchanged or (1,650/0.60) = 2,750 mL of pRBCs Exchange should be performed with 9 - 10 units of pRBCs Automated calculator mobile app is available for free from TerumoBCT Note that exchange becomes less efficient with higher volumes; most acute procedures tend to exchange about 1.0 times RBC volume Note: end procedure patient hematocrit can be programmed; increasing patient hematocrit in this fashion requires further pRBCS but may improve symptoms and reduce reticulocytosis Keep in mind that increasing end hematocrit will increase patient iron load and serum ferritin End hematocrit should generally be ≤ 33% (and never higher than 36%) to avoid hyperviscosity in SCD; end hematocrit of 30% is typical Reference: Transfusion 2018;58:1965 Simple exchange, depletion and depletion / exchange In simple exchange, donor pRBCs are transfused to patient in an isovolemic fashion to offset removal of diseased RBCs In depletion, crystalloid or colloid replacement is infused instead of donor pRBCs; this process is standard in erythrocytapheresis for polycythemia vera or hereditary hemochromatosis Hybrid depletion / exchange can be performed, wherein a brief depletion phase is followed by exchange with donor pRBCs This marginally improves procedure efficiency and may have implications on iron homeostasis (contested) Reference: Transfusion 2018;58:1965 Vascular access In adults, choice of vascular access should be able to support inlet rate of 30 mL/min or ideally up to and above 60 mL/min Generally, procedure is performed continuously with 2 points of intravenous access Peripheral IVs: Least invasive Should be considered in patients with adequate veins for peripheral access before resorting to central lines Central venous catheters: Femoral access confers higher infection rate but fewer placement complications (e.g. pneumothorax) than internal jugular or subclavian access and does not require imaging to confirm placement Nontunneled lines are appropriate for short term inpatient exchange in patients who fail peripheral access Tunneled lines (Permcath) can be used for intermediate duration of procedures on outpatient basis Peripherally inserted central catheters (PICCs) generally cannot sustain required inlet rates and are inappropriate for apheresis procedures Ports: Appropriate for patients requiring chronic exchange on outpatient basis Note required inlet rates; standard infusion ports are not adequate Vortex (AngioDynamics) or PowerFlow (Becton-Dickinson) ports are recommended Reference: Transfusion 2018;58:569 Anticoagulation Majority of apheresis procedures use citrate to anticoagulate extracorporeal circuit Patients who are systemically anticoagulated (e.g. high dose heparin therapy) may require less or even no anticoagulation Citrate prevents clotting by binding calcium that is required for secondary hemostasis Default whole blood to anticoagulant ratio is 12:1; increasing the ratio decreases the anticoagulation dose to patient and slightly increases speed of procedure at risk of extracorporeal clot formation Reference: Transfus Apher Sci 2019;58:132 Circuit priming Extracorporeal circuit must be purged of air; saline is used for initial purge as part of standard kit operation When exchanging patients with low total blood volume (infants or small children) or at risk of severe intraprocedural anemia, circuit may then be primed with pRBCs (1 unit is sufficient) pRBC prime if extracorporeal volume (ECV) exceeds 15% of patient total blood volume Spectra Optia therapeutic kit ECV is 185 mL plus blood warmer tubing volume or other circuit dead volume Rinseback refers to end of procedure purging of circuit back to patient If pRBCs were used for prime, rinseback may be modified, as rinseback would result in net transfusion of the unit used for priming Adverse events Use of a blood warmer is recommended for prevention of hypothermia Citrate toxicity: Most frequent adverse effect, related to hypocalcemia Often limited to paresthesias or numbness Severe complications include nausea / vomiting, muscle spasms or cardiac arrhythmias Severe citrate toxicity in red blood cell exchange is rare; oral calcium repletion is generally sufficient to resolve mild symptoms Transfusion reactions: Apheresis provider should evaluate for acute reactions to donor pRBCs, including allergic reaction, transfusion related acute lung injury (TRALI) and acute hemolysis Apheresis and other patient providers should be alert to delayed transfusion reactions that occur 24 hours to 28 days later, such as delayed hemolysis or transfusion transmitted infection Line and procedural related complications: Improper line placement or pneumothorax following internal jugular or subclavian line placement must be ruled out with plain film chest Xray Hypotension during apheresis is common due to fluid shifts but is a known complication in setting of ACE inhibitor use due to contact activation of bradykinin in extracorporeal circuit Line infection and bacteremia may present as fever during procedure and should be prevented by proper line care Reference: Transfusion 1999;39:282 Sample assessment & plan Assessment and plan: John Doe is a 25 year old man with history of sickle cell disease (HbSS) who is admitted for leg and back pain consistent with veno-occlusive crisis. He is normally treated with Hydrea as an outpatient. His baseline hemoglobin is 10 g/dL and his current hemoglobin is also 10 g/dL. He has not been transfused this admission. 2 days into admission, he developed shortness of breath and an oxygen requirement (saturating 86% on room air). Chest plain film imaging shows bilateral opacities concerning for acute chest syndrome. Apheresis service is consulted for urgent red cell exchange by the inpatient Heme / Onc consulting team. A central line is currently being placed. Severe acute chest syndrome is a category II (grade 1C) indication for red cell exchange according to 2019 ASFA guidelines (J Clin Apher 2019;34:171). Emergent exchange is indicated for acute chest syndrome with severe features (SaO2 < 90%). The apheresis service will perform inpatient red cell exchange to reduce his HbS to < 30% and to effect an acute change in his respiratory condition with the following parameters: Height: 69 inches / 175 centimeters Weight: 165 pounds / 75 kilograms Hematocrit: 30% per most recent CBC TBV: ~5,000 mL RBCV: ~1,500 mL Current HbS%: currently unknown (no Hb eval sent), assume 90% Goal HbS%: < 30% Goal FCR: 33.3% Type & screen drawn on admission is negative. All past antibody screens are negative. Patient is group A positive. STAT order placed with blood supplier for 10 units group A pRBC as below. Primary team to verify appropriate line placement and vascular access. Nurse to perform simple red cell exchange with 10 units of antigen negative, crossmatch compatible pRBCs, negative for C, E, K antigens and negative for HbS (SickleDex). Set start and end hematocrit both to 30%. Set WB:AC ratio to 12. Nurse to monitor for signs and symptoms of hypocalcemia. 2 g PO calcium carbonate ordered PRN. Page blood bank physician on call for any issues or patient reactions. Postprocedure: Patient tolerated red cell exchange well. During the visit, he was comfortable and did not complain of itching, numbness or paresthesias. No adverse reactions were noted by nursing. Per Optia instrument, the final FCR was 29%; therefore, residual HbS% is expected to be approximately 26%. Primary team: Please draw hemoglobin electrophoresis to assess HbS%. This patient has been seen by and discussed with the attending blood bank physician. Their written attestation is to follow. Additional references McLeod: Apheresis - Principles and Practice, 3rd Edition, 2010, Fung: Technical Manual, 19th Edition, 2017, Shaz: Transfusion Medicine and Hemostasis - Clinical and Laboratory Aspects, 3rd Edition, 2019 Practice question #1 Which of the following conditions is considered an American Society for Apheresis (ASFA) category I indication for red cell exchange? Acute chest syndrome in sickle cell disease Babesiosis Prevention of hemolysis in ABO major incompatible packed red blood cell transfusion Stroke in sickle cell disease Thrombotic thrombocytopenic purpura Practice answer #1 D. Stroke in sickle cell disease. In sickle cell disease, acute treatment or prevention of stroke are both category I ASFA indications for red blood cell exchange. Thrombotic thrombocytopenic purpura is a category I indication for therapeutic plasma exchange (TPE). Severe acute chest syndrome is a category II indication for red blood cell exchange. Prevention of hemolysis in major incompatible transfusion is unlisted in the 2019 ASFA guidelines but is not recommended. Babesiosis is a category II indication for red blood cell exchange. Comment Here Reference: Red blood cell exchange Practice question #2 A red cell exchange is requested for a 50 kg, 165 cm man with sickle cell disease with acute stroke. His hematocrit is 27% and his HbS is 90%. Because of antigen matching requirements, only 4 units of packed red blood cells (300 mL volume each, at 60% hematocrit) are immediately available. What is the expected postprocedural HbS if all 4 units are used and end hematocrit is left at 27%? 27% 30% 45% 50% 70% Practice answer #2 C. 45%. The available donor red cell mass to exchange is (300 mL x 0.60 x 4) = 720 mL. The patient's total blood volume is ~3,900 mL (via Nadler formula). At a hematocrit of 27%, his red blood cell volume is ~1,050 mL. Using all 4 units would amount to a (720/1,050) = 0.69 red blood cell volume exchange, with a resulting fraction of cells remaining of e-0.69 = 50%. If the patient's starting HbS is 90%, the expected postexchange HbS should be 45%. Comment Here Reference: Red blood cell exchange Practice question #3 Which of the following vascular access methods is inappropriate for red cell exchange? Arteriovenous fistula Existing extracorporeal circuits (e.g. extracorporeal membrane oxygenation) Mahurkar catheter Nonpower single lumen port Quinton catheter Practice answer #3 D. Nonpower single lumen port. Most single lumen nonpower injectable ports, such as most standard chemo infusion ports, cannot support the pressure and inlet rates needed for apheresis procedures. Additionally, 2 access sites are necessary for continuous apheresis. Quinton and Mahurkar catheters are both central venous catheters (CVCs) generally used for hemodialysis and are adequate for apheresis. Special attention should be given to the catheter diameter and length to ensure flow rates of up to 60 mL/min or greater can be supported. For reference, typical inlet rates in hemodialysis may exceed 120 mL/min. Arteriovenous fistulas are acceptable for access but require training for safe access. Extracorporeal circuits are also acceptable for apheresis procedures but the additional extracorporeal volume should be factored into consideration and coordination with extracorporeal membrane oxygenation team is necessary. Comment Here Reference: Red blood cell exchange Back to top Home>Transfusion medicine> Red blood cell exchange Sign up for our Email Newsletters Comment © Copyright PathologyOutlines.com, Inc. Click here for information on linking to our website or using our content or images. 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Sets and Functions - SAS Skip to Main Content You are impersonating . Stop Impersonating SAS Standards Search Standards View Standards Vertical Viewer English Language Development Standards (2020) Download Standards Download PSSA Anchors and Eligible Content Download Keystone Anchors Download PA Core Implementation Download PASA DLM Essential Elements Conceptual Vertical Alignment Viewer Assessments Assessment Center Project Based Assessment Keystone Exams Classroom Diagnostic Tools PSSA Analyzing Root Cause Assessment Literacy Early Learning: Pre-Kindergarten to Grade 3 PASA Online Assessment Resources Curriculum Framework Search Curriculum Framework Download Curriculum Framework Curriculum Maps PA Standards Instructional Frameworks: ELA PA Standards Instructional Frameworks: Math PA Standards Instructional Frameworks: Personal Finance Early Learning: Pre-Kindergarten to Grade 3 Instruction Educator Effectiveness Instructional Toolkits Teacher of the Year Pennsylvania Learns Early Learning: Pre-Kindergarten to Grade 3 PA Roadmap: Focus on Effective Instruction Pennsylvania High School Graduation Requirements Seal of Biliteracy Toolkit STEELS Hub Educator Professional Development Resource PDE English Learner (EL) Portal PA Disability Inclusive Curriculum Pilot Program Pennsylvania Inspired Leadership (PIL) Materials & Resources Search My Lessons Voluntary Model Curriculum (sample unit and lesson plans) Learning Progressions Pennsylvania Literacy Early Learning: Pre-Kindergarten to Grade 3 POWER Library Career Readiness TDA Toolkit Career Ready Toolkit Organ and Tissue Donation Awareness Toolkit Act 35 Civics Toolkit PBS Toolkit Content Collections CPR Toolkit Holocaust, Genocide, and Human Rights Toolkit Information and Media Literacy Toolkit September 11 Resource Toolkit Personal Finance Toolkit STEM & CS in PA USA's 250th Toolkit Safe & Supportive Schools Safe & Supportive Schools Menu Standards Assessments Curriculum Framework Instruction Materials & Resources Safe & Supportive Schools Quick Links 's SAS Tools Login Username Password Log In RegisterForgot Password? Profile ePortfolio My Assessments My Curriculum Maps My Lessons My Library Communities PD Center Workshop Evaluation Logout Quick Links Share Suggestion Subject Please include a subject for your suggestion. Suggestion Please enter information about your suggestion. Please check the "I'm not a robot" checkbox. Send Close Sets and Functions Lesson Plan Sets and Functions Options Printer Friendly Version Email Grade Levels Related Academic Standards2.11.A1.B Describe rates of change as modeled by linear equations. 2.8.A1.C Identify and represent patterns algebraically and/or graphically. 2.8.A1.D Demonstrate an understanding and apply properties of functions (domain, range) and characteristics of linear functions. CC.2.2.HS.C.1 Use the concept and notation of functions to interpret and apply them in terms of their context. CC.2.2.HS.C.2 Graph and analyze functions and use their properties to make connections between the different representations. CC.2.2.HS.C.3 Write functions or sequences that model relationships between two quantities. CC.2.2.HS.C.4 Interpret the effects transformations have on functions and find the inverses of functions. CC.2.2.HS.C.5 Construct and compare linear, quadratic and exponential models to solve problems. CC.2.2.HS.C.6 Interpret functions in terms of the situation they model. CC.2.4.HS.B.1 Summarize, represent, and interpret data on a single count or measurement variable. CC.2.4.HS.B.2 Summarize, represent, and interpret data on two categorical and quantitative variables. CC.2.1.HS.F.3 Apply quantitative reasoning to choose and Interpret units and scales in formulas, graphs and data displays. CC.2.1.HS.F.4 Use units as a way to understand problems and to guide the solution of multi-step problems. CC.2.2.8.B.2 Understand the connections between proportional relationships, lines, and linear equations. CC.2.2.8.C.1 Define, evaluate, and compare functions. CC.2.2.8.C.2 Use concepts of functions to model relationships between quantities. Assessment AnchorsA1.2.1 Functions A1.2.2 Coordinate Geometry Eligible ContentA1.2.1.1.1 Analyze a set of data for the existence of a pattern and represent the pattern algebraically and/or graphically. A1.2.1.1.2 Determine if a relation is a function given a set of points or a graph. A1.2.1.1.3 Identify the domain or range of a relation (may be presented as ordered pairs, a graph, or a table). A1.2.1.2.1 Create, interpret and/or use the equation, graph or table of a linear function. A1.2.1.2.2 Translate from one representation of a linear function to another (graph, table and equation). A1.2.2.1.1 Identify, describe and/or use constant rates of change. Big Ideas Bivariate data can be modeled with mathematical functions that approximate the data well and help us make predictions based on the data. Families of functions exhibit properties and behaviors that can be recognized across representations. Functions can be transformed, combined, and composed to create new functions in mathematical and real world situations. Mathematical functions are relationships that assign each member of one set (domain) to a unique member of another set (range), and the relationship is recognizable across representations. Numbers, measures, expressions, equations, and inequalities can represent mathematical situations and structures in many equivalent forms. Relations and functions are mathematical relationships that can be represented and analyzed using words, tables, graphs, and equations. There are some mathematical relationships that are always true and these relationships are used as the rules of arithmetic and algebra and are useful for writing equivalent forms of expressions and solving equations and inequalities. Concepts Algebraic properties and processes Analysis of one and two variable (univariate and bivariate) data Functions and multiple representations Linear relationships: Equation and inequalities in one and two variables Linear system of equations and inequalities Polynomial functions and equations Competencies Extend algebraic properties and processes to quadratic, exponential, and polynomial expressions and equations and to matrices, and apply them to solve real world problems. Represent functions (linear and non-linear) in multiple ways, including tables, algebraic rules, graphs, and contextual situations and make connections among these representations. Choose the appropriate functional representation to model a real world situation and solve problems relating to that situation. Write, solve, and interpret systems of two linear equations and inequalities using graphing and algebraic techniques. Write, solve, graph, and interpret linear equations and inequalities to model relationships between quantities. Objectives In this unit, students will investigate notations via sets and functions. Students will: [IS.3 - Struggling Learners] analyze patterns. examine multiple representations/notations of sets and functions, and draw connections between the two. examine components and characteristics of sets and functions. learn notation and use and interpret them in context. Essential Questions What notations are generally accepted throughout mathematics? What role do notations play in the realm of mathematical understanding? How are mathematical notations used as part of the problem-solving process? Vocabulary [IS.1 - Struggling Learners][IS.2 - Struggling Learners] Counting Numbers: Any of the positive whole numbers, 1, 2, 3,…. Disjoint Set: Two sets are disjoint if there is no point which belongs to each of the sets, i.e., if the intersection of the sets is the null set. Domain: For a function, the set of all values which the independent variable may take on. Element: A single component found within a set. Empty Set: A set without any elements, denoted with { } or Ø. Finite Set: A set with a definitive number of elements listed, i.e., A = {1, 2, 3}. Function: A relation in which each input element is mapped to a unique output element. A function describes a set and points to elements in the set. Infinite Set: A set that is not finite, one whose members cannot be enumerated, i.e., F = {all real numbers}. Integer: Any of the positive or negative whole numbers, including 0, i.e., 0, ±1, ±2, ±3,… Intersection: The common ground for two sets; the elements common to sets A and B; the elements found in Set A AND Set B. Interval Notation: A notation that uses the endpoints of the set to describe the elements, i.e., [4, ∞) represents the set of all real numbers greater than or equal to four. Irrational:A real number not expressible as an integer or quotient of integers. Linear Function: A function of degree one. Mapping: Two sets related in such a way that to each element of set A there corresponds a unique element f(x) of a space B; then there is said to be a mapping or map f of the set A in the set B, and the point f(x) is said to be the image of the point x. Natural Numbers: Any of the positive integers, 1, 2, 3,…. Null Set: The set which is empty, has no members. Proper Subset: A subset, in which the sets are not equal; for example, the factors of 6 {1, 2, 3, 6} are a proper subset of the factors of 18 {1, 2, 3, 6, 18}. Range:For a function, the set of values the function may take on. Rational:An algebraic expression which involves no variable in an irreducible radical or under a fractional exponent; a number that can be expressed as an integer or as a quotient of integers. Relation: A subset of a set associated with another set. Roster Notation: A notation where the elements of each set are simply listed, i.e., Set A= {1, 2, 3,…}. Set: A group of elements. Set Builder Notation: A notation where the set is described with symbols, formally, i.e., A = is the set of all x, such that x is an element of the natural numbers. Subset: A set embedded in another set. Union: The combining of two sets, such that all elements of both sets are included in the combined set; elements found in Set A OR Set B. Universal Set: The set that contains all other sets, including the set itself. Venn Diagram: A graphic organizer that shows the relationships between sets by encircling combinations of individual elements. Whole Number: Any of the positive integers, 1, 2, 3,…. Duration 120–150 minutes/3–4 class periods [IS.4 - All Students] Prerequisite Skills Prerequisite Skills haven't been entered into the lesson plan. Materials copies of Set Relationships handout (M-A1-2-1_Set Relationships.docx) copies of Set Examples handout (M-A1-2-1_Set Examples.docx) copies of the Relation and Function sheet (M-A1-2-1_Relation and Function and KEY.docx). Related Unit and Lesson Plans Notation History of Numbers and Symbols Patterns and Sequences (enrichment lesson) Related Materials & Resources The possible inclusion of commercial websites below is not an implied endorsement of their products, which are not free, and are not required for this lesson plan. Formative Assessment View [IS.12 - All Students] Observe/evaluate class discussion by listening for students’ capacity to discriminate between infinite and finite sets; offer examples of null sets; and explain the relationship between sets, subsets, and proper subsets. Evaluate student performance on: set illustrations. Venn diagram illustrations (by checking for accuracy of inclusion and exclusion of set elements). creation of universal set and subsets (by noting their appropriateness). determination of functions and supporting reasons (by checking the difference between relations that are and are not functions). creation of specific relations. Evaluate the materials for the mathematics-journal article for clarity and accuracy. Suggested Instructional Supports View Scaffolding, Active Engagement, Modeling, Explicit Instruction W:The lesson is presented in a manner that connects sets and functions, while examining components of each, using various notations/representations. Students are offered opportunities to explore notations in a very open-ended way via class discussion and participation in discovery-oriented activities, both independently and as part of a group. H:The connection of sets to the real number set and subsets will lead students towards realizing the applicability and presence of sets in mathematics. The presentation of a variety of representations, forms, and notations will reveal the multitudinous ways that sets can be represented. E:The lesson is divided into two parts. Part 1 focuses on the concept of a set, notations and representations used when discussing sets, types of sets, and operations on sets. Part 2 discusses the concepts of relations and functions and connects these concepts to the idea of a set. Multiple notations and representations are investigated. R:Students are offered many opportunities to reflect, revisit, revise, and rethink ideas via the open-ended activities intertwined throughout. Classroom discussion provides extra opportunities for students to voice and debate ideas. E:Due to the scaffolding and modeling used in the lesson, students have the opportunity to self-evaluate their own understanding within the structure of the lesson. Engagement in both independent and group activities allows for self-evaluation. T:The premise of a presentation and study of many forms of notations is inherently geared towards multiple learning styles. O:The lesson moves from abstract presentation to concrete ideas and generalizations. IS.1 - Struggling Learners The number of vocabulary terms listed may be too extensive for struggling learners. Consider using only the terms that they will need to apply to this lesson. IS.2 - Struggling Learners Consider using the following methods with regard to vocabulary for struggling learners: 1. Define vocabulary using student friendly terms. Provide both examples and non-examples. 2. Review vocabulary before each lesson. 3. Provide opportunities throughout the lesson for students to apply the vocabulary they have learned. 4. Use graphic organizers such as the Frayer Model, Verbal Visual Word Association, Concept Circles, etc. IS.3 - Struggling Learners Consider providing struggling learners with written examples of the patterns to analyze, the sets of functions and how to make connections between the two, and the characteristics and components of the functions. IS.4 - All Students Consider pre-teaching the concepts critical to the lesson. Use formative assessments throughout the lesson to determine the level of student understanding. Use follow-up reinforcement as necessary. IS.5 - Struggling Learners Consider providing struggling students with written examples of these when defining them. IS.6 - All Students Consider using student friendly terms when defining this to students along with examples both on the board and on paper. IS.7 - Struggling Learners Consider providing struggling students with written examples so that they can follow along as you explain. You might also consider providing non-examples to increase their understanding. IS.8 - Struggling Learners Consider providing struggling learners with a more visually friendly graphic organizer. IS.9 - Struggling Learners Consider putting this in writing for the struggling students. IS.10 - Struggling Learners Struggling students may have difficulty in reading the description/meaning of this table. Consider putting an example in the description/meaning and limiting it to one representation. IS.11 - Struggling Learners Consider allowing struggling learners to provide other forms of representation. IS.12 - All Students Consider using multiple forms of formative assessments such as questioning. Instructional Procedures View Part 1 Tell students,“Today we are going to discuss sets,elements, and functions.A set is a group of elements. An element is simply a component found within a set. A function can be used to describe a set and point to the elements in the set. We will talk more about functions in a bit.” [IS.5 - Struggling Learners] Show the following box on the board or an overhead projector: “Now, let’s think about our real number system.Does anyone know the two disjointed number sets in our real number system?”[IS.6 - All Students](Allow all responses, guiding the answer to rational and irrational numbers. In summarizing responses, note that rationals can be expressed as numerator divided by denominator, while irrationals cannot.) “We have the set of rational numbers and the set of irrational numbers. Within the rational number set, we have particular embedded number sets. One set embedded in another set is called a subset. In this case, since the sets are not equal, we have what is called a proper subset.” “Which numbers does the set of rational numbers contain?”(_integers, whole numbers, counting numbers_) “Which sets of numbers are subsets of the rational numbers?”_(integers, whole numbers, natural numbers)_ “The rational numbers include all of the typical numbers we think about, including natural numbers (counting numbers), whole numbers, and integers.” “Notice that the rational numbers also include other numbers, such as those that terminate or repeat. All rational numbers can be written in the form.” “We can also write each of these numbers as sets. Shown below is the roster method, whereby we simply list elements of each set.” Roster Method Natural Numbers{1, 2, 3,…} Whole Numbers{1, 2, 3,…} Integers{…−2, −1, 0, 1, 2,…} Ask, “Can we illustrate the sets in another way?”(Yes; we often express some of these mathematical ideas in a special type of notation.)“Yes, first let’s draw a Venn Diagram to represent these three subsets of the real number system.” “Next, let’s write the number sets, using something called set builder notation.[IS.7 - Struggling Learners] With set builder notation, you describe the set with symbols, using a formal notation. For example, if we were to indicate the sets of natural numbers, whole numbers, and integers, respectively, using set builder notation, we would write: A =the set of all x, such that x is an element of the Natural Numbers. B =the set of all x, such that x is an element of the Whole Numbers. C =the set of all x, such that x is an element of the Integers. Note that Z is used, rather than I so it is distinct from the I notation for the irrational numbers.” “Notice that we denote each set with a letter. The letters A, B, and C serve to name the sets. By doing so, we can refer to them as Set A, Set B, and Set C. Thus far, we have seen sets with and without names. An example of a set without a name is {3, 4, 5,…}. We know this is a set. It just does not have a name. Normally, we do wish to name the set, especially when we are looking at and comparing more than one set. For example, here are three sets: {3, 4, 5,… } {1, 2,… } ” “We must identify which set we are speaking about. Later on, when we look at relations, and specifically, functions of sets, it’s important to identify the set.” “Also, notice how each notation for the number sets is put into words. We could write each set simply using words. A = {the set of natural numbers greater than two} B = {the set of whole numbers} C = {the suits of a deck of cards}” “We can go even further. If we wanted to show that one particular number is an element of the set of integers, we could write.” “What if we wanted to identify only a part of the set of real numbers? Remember, the real numbers include both rational and irrational numbers.” “What if we are only concerned with those real numbers greater than or equal to 4? We can represent this set, using the roster method, set-builder notation, and interval notation, as well as with several other representations.” Summary of Notations for Real Numbers Greater Than or Equal to 4 [IS.8 - Struggling Learners] NotationRepresentation of SetAny Specific Notes? Set-Builder Notation A = (read, “the set of all real numbers, x, such that x is greater than or equal to 4”)Since we didn’t specify that x was an element of a specific subset, we realize we are working with the set of real numbers. Interval NotationSince we are including the number 4, we use brackets. A number that is not included is indicated with a parenthesis. Important: Infinity is never closed with a bracket in order to show the difference between to two limits of the interval. Graphic Words The set of all real numbers, greater than or equal to 4. Other? Interval Notation Tell students,“Interval notation is a new one that we haven’t explored yet. Interval notation uses the endpoints of the set to describe the elements. Again, we can talk about specific elements by saying . We are saying that 7 ‘is an element of’ the set A.” Activity 1 “It is time to practice using multiple representations of various sets. Consider these sets: 1)The four most recent presidents of the United States 2)Integers less than −2 and greater than −9 3)The four seasons of the year 4)All polygons with four or fewer sides 5)Rational numbers greater than ” (Answers: 1. Barack Obama, George W. Bush, Bill Clinton, George H. W. Bush; –8, –7, –6, –5, –4, –3; 3. Winter, Spring, Summer, Autumn; 4. Quadrilateral, triangle; 5. {n|n>}) “The information you have just learned is useful in illustrating sets generally for a variety of uses. You might have noticed key differences in some of the sets. We have what are called finite sets and infinite sets. A finite set has a definitive number of elements listed. For example, B = {1, 2, 3} is an example of a finite set. An infinite set is a set that is not finite. The set can either be a countable infinite set, as with the set of all natural numbers, or an uncountable infinite set, as with the set of all real numbers. D = {1, 2, 3,… } and F = {all real numbers} are examples of infinite sets.” “The focus of this lesson is on different types of number notations. Thus, it is important that you recognize and have facility working with all different representations of sets. We will now look at some of the common conventions of sets, using an example set. Suppose we are interested in the set of all whole numbers less than 8. We can represent this set in the following ways.” Notations for the Set of All Whole Numbers Less Than 8 Name of RepresentationWhat It Looks Like Words The set of all whole numbers less than 8. Words with set name Let A be the set of all whole numbers less than 8. Words with set name and brackets A = {The set of all whole numbers less than 8}. Graphic Number line Roster notation without name{0, 1, 2, 3, 4, 5, 6, 7} Roster notation with name A = {0, 1, 2, 3, 4, 5, 6, 7} Set-builder notation without name Set-builder notation with name Interval notation not suitable Activity 2 Tell the class, “Choose one numeric and one non-numeric set to represent. For each set, include all of the representative forms we have discussed, as well as any others you can think of.”Allow students time to work on each set in pairs or as a small group.Encourage as many different ideas from students in the discussion and ask them to evaluate each other’s suggestions. Then ask: “So, what is the most commonly/widely accepted set notation, i.e., most prevalent notation?” “What is the most common convention used to talk about a set?” “We’ve looked at several different versions already, but is there one or more that are ‘just better’ than the others?” “What is our standard notation?” “Are the others just loose forms of the formal convention? Are they accepted?” “If we were to place a few agreed-upon set identification conventions in a file for future students to view, what would they be?” Tell students that the four most widely-accepted notations of sets are: [IS.9 - Struggling Learners] Words with set name and brackets (A = {The set of all whole numbers less than 8}) Roster notation with name (A = {0, 1, 2, 3, 4, 5, 6, 7}) Set-builder notation with name (Image 21) Interval notation ([0,8)) “Notice that the most widely accepted notations are formal and include the name of the set. Now, let’s look at specific types of sets and relationships between sets. There are two types of sets we have not discussed, at least not directly. These sets are the empty set and the universal set. The empty set, denoted with either empty brackets, { }, or Ø, is simply a set without any elements. This might seem to be a contradiction. An example should clarify.” Ask students to offer examples of empty sets. Lead the discussion to evaluate their examples. “Now, what if we discuss a universal set? We actually already have. The real number system is an example of a universal set. A universal set is the set that contains all other sets, including itself. Since the real number system includes the sets of rational numbers and set of irrational numbers, as well as all other real numbers, it is indeed the universal set.” “A universal set can relate to any encompassing set, however. Let’s look at the Venn Diagram below.” Note: A Venn diagram is a graphic organizer that shows the relationship between the sets. Explain that a Venn diagram is a very general way of representing a universal set and its subsets. Activity 3 Tell students, “Using any sets that you wish, draw a Venn diagram to represent the universal set and any subsets.” “Many times, we want to show the relationships between sets and/or to compare sets. We now need to talk about ways to represent relationships between sets. First, let’s think of what these representations could be. How could you relate two sets?” “Well, we could look at the combination of the elements of both sets. We could look at the common elements. We could look at the elements contained in one, but not contained in the other. How could we display these relationships?” “We could actually use any of the methods we’ve previously looked at.” Put the following statement on the board or an overhead projector: Let U be the set of integers from −4 to 8. Let A = {−1, 0, 3, 4} and B = {2, 3, 6, −3} “Let’s use Venn diagrams to represent the relationships we’re looking at, discover the meaning of the symbolism, and represent the solutions in a variety of ways. Consider the table below.”[IS.10 - Struggling Learners] Distribute copies of the Set Relationships handout for student reference (M-A1-2-1_Set Relationships.docx). RelationshipDescription/MeaningVenn DiagramOther Representations This is read, “A union B,” which means “the elements found in Set A OR Set B.” This is read, “A intersect B,” which means “the elements common to Sets A and B,” or “the elements found in Set A AND Set B.” This is read, “A complement,” which means “those elements not found in Set A.” We are looking at the complement of A relative to B, not to the universal set. This is read, “B complement,” which means “those elements not found in Set B.” We are looking at the complement of B relative to A, not to the universal set. This is read, “A minus B,” which means “the elements found in set A minus the elements found in Set B”; also known as the complement of B relative to A. This is read, “B minus A,” which means “the elements found in set B minus the elements found in Set A”; also known as the complement of A relative to B. Note:Complement has the same idea as “negation.” For example p and ~p represent “p” and “not p” respectively. Activity 4 Divide students into groups of three or four. Say to the groups: “Create a universal set and at least two subsets. Choose the most effective representation of each relationship shown above. Provide a justification for why that representation makes the most sense and should be used in this case.” “Create a short PowerPoint presentation,[IS.11 - Struggling Learners] describing when various representations for union, intersection, complement, and difference are desirable. What seems to be the most commonly accepted convention for indicating set relationships?” Part 2 “Now that you have a firm understanding of the idea of sets, let’s use a function to create a set.” “Before we discuss functions, we need to discuss the idea of relations. All functions are relations, but not all relations are functions. This fact is very important to note.” “Does anyone know what a relation is? Apart from mathematics, what is a relation?”(Students may provide examples of husband, wife; child, mother; friends; grandparents, etc.) “Now, let’s think about some relations. In particular, mapping. A mapping relationship is one where every element in the first set matches to one or more elements in the second set.”Distribute copies of the Set Examples handout (M-A1-2-1_Set Examples.docx), which contains these examples: Example: Pig is mapped to Sheep. These two elements are paired. Similarly, cat is not mapped. A B −4 0 2 8 5 9 2 3 6 1 R = {(2,9), (3,4), (3,0), (8,2)} “Looking at the examples, what definition would you give for a relation? A relation is simply a subset of a set associated with another set. A relation is denoted by a capital R.” “A relation does not restrict the ability of an input value to be mapped to more than one output value. In other words, a relation would allow an element in Set A to be mapped to more than one element in Set B.” “A function behaves in a different way. In fact, functions are used to create sets using a rule given by the function. A function is a relation in which each element in the domain is mapped to one and only one element in the range. In other words, each input value (or value from Set A) is mapped to only one output value (or value from Set B). It is important that you realize input values are the same as domain values.” “Can anyone give an example of a function?”Allow students time to provide many different examples. “What if we had a cow mapped to a giraffe, a giraffe mapped to an elephant, and a cow mapped to an elephant? Would that relation constitute a function? No, because the cow is mapped to more than one animal. Let’s look at the relations from the list of animals above.”(M-A1-2-1_Set Relationships.docx). Set Examples RepresentationRelation?Function?Supporting Reasons Yes Horse is paired with two elements in second set. Yes Each element in domain is paired with one and only one element in range. A B −4 0 2 8 5 9 2 3 6 1 Yes 2 is paired with 8 and 3. Set A = {(2,9), (3,4), (3,0), (8,2)}Yes 3 is paired with two elements in range. Activity 5 Distribute copies of Set Examples (M-A1-2-1_Set Examples.docx). Students should fill in the columns for “Function?” and “Supporting Reasons” in the table. The purpose of the activity is to allow students to discover the idea behind functions. Say,“Now, let’s use a function to create a set. For example, suppose we have the function. Suppose our set is the set of output values. Let’s name the set Set G. Read this as g of x, not g times x.” “What elements would Set G contain?”Give students time to articulate their thoughts. “Set G would contain −5, −3, −1, 1, and 3. Are there other elements? How many? Can you list them all? Count them? This set is an uncountable infinite set because we could insert any input value for x and get a different output value. The input values (or domain) could be any real number, since we don’t have a stipulation here.” “Did you notice that no input value is mapped to more than one output value? If not, let’s look at a couple of representations: a) a table and b) a graph.” a) x g(x) −2−5 −1−3 0−1 1 1 2 3 b) Ask students to explain why the graph represents a function. Be sure they include a description of the appearance of the graph in their explanation. “Did anyone notice the type of function shown above? Are there different types of functions? How are they different? How can you tell?”Allow some time for discussion. “What is special about a linear function? Can this component be recognized via an equation, table, graph, or words?”Invite discussion, related to rate of change and linearity and the multiple representations thereof. Activity 6 “Let’s examine some relations from Set A to Set B, and determine whether or not they represent functions. Suppose we have Sets A and B. Set A = {2, 5, 6} and Set B = {3, −1, 9, 0}.”Distribute copies of the Relation and Function sheet (M-A1-2-1_Relation and Function and KEY.docx). Ask students to fill in the “Function?” and “Reasons” columns in the table below. RelationFunction?Reasons Yes Each domain is paired with exactly one range. Yes Each domain is paired with exactly one range. No The domain of 2 is paired with 9 and 0, two values. Yes Each domain is paired with exactly one range. Activity 7 Divide students into groups of three or four. “Using any manipulatives you like, create four relations, two of which are only relations and two of which are both relations and functions. Be prepared to present the relations and discuss reasons supporting your representations.” To review the lesson, tell students,“Your assignment is to write an article for a leading mathematics journal. Describe key representations, related to sets and functions. Discuss any similarities, differences, and deliver a consensus, related to varying representations and ‘best’ notation. Support your ideas and provide at least three illustrations.” Extension: Ask students to create specific types of functions, including linear and nonlinear, using a variety of representations. The creation of nonlinear functions with manipulatives would be especially beneficial. Students could illustrate the numbers/objects that go in each set according to the function. You may suggest to them what some nonlinear functions look like. For example, _y_ = _x_ 2, _y_ = _x_ 3. Related Instructional Videos Note: Video playback may not work on all devices. 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https://tajseer.files.wordpress.com/2014/06/re-nfa-dfa.pdf
Compiler design – RE&DFA&NFA Page 1 of 9 From Regular Expression to NFA to DFA 1. Regular expression Consider the regular expression r = (a|b)abb, that matches {abb, aabb, babb, aaabb, bbabb, ababb, aababb,……} To construct a NFA from this, use Thompson’s construction. This method constructs a regular expression from its components using ε-transitions. The ε transitions act as “glue or mortar” for the subcomponent NFA’s. An ε-transition adds nothing since concatenation with the empty string leaves a regular expression unchanged (concatenation with ε is the identity operation). Step 1. Parse the regular expression into its subexpressions involving alphabet symbols a and b and ε: ε, a, b, a|b, (), ab, abb These describe a. a regular expression for single characters ε, a, b b. alternation between a and b representing the union of the sets: L(a) U L(b) c. Kleene star () d. concatenation of a and b: ab, and also abb Subexpressions of these kinds have their own Nondeterministic Finite Automata from which the overall NFA is constructed. Each component NFA has its own start and end accepting states. A Nondeterministic Finite Automata (NFA) has a transition diagram with possibly more than one edge for a symbol (character of the alphabet) that has a start state and an accepting state. The NFA definitely provides an accepting state for the symbol. It is proven (Kleene’s Theorem) that RE and FA are equivalent language definition methods. Based on this theoretical result practical algorithms have been developed enabling us actually to construct FA’s from RE’s and simulate the FA with a computer program using Transition Tables. In following this progression an NFA is constructed first from a regular expression, then the NFA is reconstructed to a DFA, and finally a Transition Table is built. The Thompson’s Construction Algorithm is one of the algorithms that can be used to build a Nondeterministic Finite Automaton (NFA) from RE, and Subset construction Algorithm can be applied to convert the NFA into a Deterministic Finite Automaton (DFA). The last step is to generate a transition table. We need a finite state machine that is a deterministic finite automaton (DFA) so that each state has one unique edge for an input alphabet element. So that for code generation there is no ambiguity. But a nondeterministic finite automaton (NFA) with more than one edge for an input alphabet element is easier to construct using a general algorithm - Thompson’s construction. Then following a standard procedure, we convert the NFA to a DFA for coding. Compiler design – RE&DFA&NFA Page 2 of 9 Take these NFA’s in turn: a. the NFA’s for single character regular expressions ε, a, b start ε start a 2 3 start b 4 5 b. the NFA for the union of a and b: a|b is constructed from the individual NFA’s using the ε NFA as “glue”. Remove the individual accepting states and replace with the overall accepting state a ε 2 3 ε start 1 6 b ε 4 5 ε c. Kleene star on (a|b). The NFA accepts ε in addition to (a|b) ε a ε 2 3 ε start ε ε 0 1 6 7 b ε 4 5 ε ε Compiler design – RE&DFA&NFA Page 3 of 9 d. concatenate with abb ε a ε 2 3 ε start ε ε a b b 0 1 6 7 8 9 10 ε 4 5 ε b ε This is the complete NFA. It describes the regular expression (a|b)abb. The problem is that it is not suitable as the basis of a DFA transition table since there are multiple ε edges leaving many states (0, 1, 6). Converting the NFA into a DFA A Deterministic Finite Automaton (DFA) has at most one edge from each state for a given symbol and is a suitable basis for a transition table. We need to eliminate the ε-transitions by subset construction. Definitions Consider a single state s. Consider a set of states T Operation Description ε-closure(s) Set of NFA states reachable from NFA state s on ε-transitions alone ε-closure(T) Set of NFA states reachable from set of states T on ε-transitions alone move(T,a) Set of states to which there is a transition on input symbol a from some NFA state in T We have as input the set of N states. We generate as output a set of D states in a DFA. Theoretically an NFA with n states can generate a DFA with 2n states. Start the Conversion 1. Begin with the start state 0 and calculate ε-closure(0). a. the set of states reachable by ε-transitions which includes 0 itself is { 0,1,2,4,7}. This defines a new state A in the DFA A = {0,1,2,4,7} Compiler design – RE&DFA&NFA Page 4 of 9 2. We must now find the states that A connects to. There are two symbols in the language (a, b) so in the DFA we expect only two edges: from A on a and from A on b. Call these states B and C: a B start A b C We find B and C in the following way: Find the state B that has an edge on a from A a. start with A{0,1,2,4,7}. Find which states in A have states reachable by a transitions. This set is called move(A,a) The set is {3,8}: move(A,a) = {3,8} b. now do an ε-closure on move(A,a). Find all the states in move(A,a) which are reachable with ε-transitions. We have 3 and 8 to consider. Starting with 3 we can get to 3 and 6 and from 6 to 1 and 7, and from 1 to 2 and 4. Starting with 8 we can get to 8 only. So the complete set is {1,2,3,4,6,7,8}. So ε-closure(move(A,a)) = B = {1,2,3,4,6,7,8} This defines the new state B that has an edge on a from A Find the state C that has an edge on b from A c. start with A{0,1,2,4,7}. Find which states in A have states reachable by b transitions. This set is called move(A,b) The set is {5}: move(A,b) = {5} d. now do an ε-closure on move(A,b). Find all the states in move(A,b) which are reachable with ε-transitions. We have only state 5 to consider. From 5 we can get to 5, 6, 7, 1, 2, 4. So the complete set is {1,2,4,5,6,7}. So ε-closure(move(A,a)) = C = {1,2,4,5,6,7} This defines the new state C that has an edge on b from A a B A={0,1,2,4,7} start B={1,2,3,4,6,7,8} A C={1,2,4,5,6,7} b C Now that we have B and C we can move on to find the states that have a and b transitions from B and C. Compiler design – RE&DFA&NFA Page 5 of 9 Find the state that has an edge on a from B e. start with B{1,2,3,4,6,7,8}. Find which states in B have states reachable by a transitions. This set is called move(B,a) The set is {3,8}: move(B,a) = {3,8} f. now do an ε-closure on move(B,a). Find all the states in move(B,a) which are reachable with ε-transitions. We have 3 and 8 to consider. Starting with 3 we can get to 3 and 6 and from 6 to 1 and 7, and from 1 to 2 and 4. Starting with 8 we can get to 8 only. So the complete set is {1,2,3,4,6,7,8}. So ε-closure(move(A,a)) = {1,2,3,4,6,7,8} which is the same as the state B itself. In other words, we have a repeating edge to B: a a B A={0,1,2,4,7} start B={1,2,3,4,6,7,8} A C={1,2,4,5,6,7} b C Find the state D that has an edge on b from B g. start with B{1,2,3,4,6,7,8}. Find which states in B have states reachable by b transitions. This set is called move(B,b) The set is {5,9}: move(B,b) = {5,9} h. now do an ε-closure on move(B,b). Find all the states in move(B,b) which are reachable with ε-transitions. From 5 we can get to 5, 6, 7, 1, 2, 4. From 9 we get to 9 itself. So the complete set is {1,2,4,5,6,7,9}}. So ε-closure(move(B,a)) = D = {1,2,4,5,6,7,9} This defines the new state D that has an edge on b from B A={0,1,2,4,7}, B={1,2,3,4,6,7,8}, C={1,2,4,5,6,7}, D{1,2,4,5,6,7,9} a b a B D start A b C Compiler design – RE&DFA&NFA Page 6 of 9 Find the state that has an edge on a from D i. start with D{1,2,4,5,6,7,9}. Find which states in D have states reachable by a transitions. This set is called move(D,a) The set is {3,8}: move(D,a) = {3,8} j. now do an ε-closure on move(D,a). Find all the states in move(B,a) which are reachable with ε-transitions. We have 3 and 8 to consider. Starting with 3 we can get to 3 and 6 and from 6 to 1 and 7, and from 1 to 2 and 4. Starting with 8 we can get to 8 only. So the complete set is {1,2,3,4,6,7,8}. So ε-closure(move(D,a)) = {1,2,3,4,6,7,8} =B This is a return edge to B: A={0,1,2,4,7}, B={1,2,3,4,6,7,8}, C={1,2,4,5,6,7}, D{1,2,4,5,6,7,9} a b a B D start a A b C Find the state E that has an edge on b from D k. start with D{1,2,4,5,6,7,9}. Find which states in D have states reachable by b transitions. This set is called move(B,b) The set is {5,10}: move(D,b) = {5,10} l. now do an ε-closure on move(D,b). Find all the states in move(D,b) which are reachable with ε-transitions. From 5 we can get to 5, 6, 7, 1, 2, 4. From 10 we get to 10 itself. So the complete set is {1,2,4,5,6,7,10}}. So ε-closure(move(D,b) = E = {1,2,4,5,6,7,10} This defines the new state E that has an edge on b from D. Since it contains an accepting state, it is also an accepting state. A={0,1,2,4,7}, B={1,2,3,4,6,7,8}, C={1,2,4,5,6,7}, D={1,2,4,5,6,7,9}, E={1,2,4,5,6,7,10} a b b a B D E start a A b C Compiler design – RE&DFA&NFA Page 7 of 9 We should now examine state C Find the state that has an edge on a from C m. start with C{1,2,4,5,6,7}. Find which states in C have states reachable by a transitions. This set is called move(C,a) The set is {3,8}: move(C,a) = {3,8} we have seen this before. It’s the state B A={0,1,2,4,7}, B={1,2,3,4,6,7,8}, C={1,2,4,5,6,7}, D={1,2,4,5,6,7,9}, E={1,2,4,5,6,7,10} a b b a B D E start a A a b C Find the state that has an edge on b from C n. start with C{1,2,4,5,6,7}. Find which states in C have states reachable by b transitions. This set is called move(C,b) The set is {5}: move(C,b) = {5} o. now do an ε-closure on move(C,b). Find all the states in move(C,b) which are reachable with ε-transitions. From 5 we can get to 5,6,7,1,2,4. which is C itself So ε-closure(move(C,b)) = C This defines a loop on C a b b a B D E start a A a b C b Finally we need to look at E. Although this is an accepting state, the regular expression allows us to repeat adding in more a’s and b’s as long as we return to the accepting E state finally. So Compiler design – RE&DFA&NFA Page 8 of 9 Find the state that has an edge on a from E p. start with E{1,2,4,5,6,7,10}. Find which states in E have states reachable by a transitions. This set is called move(E,a) The set is {3,8}: move(E,a) = {3,8} We saw this before, it’s B. So a a b b a B D E start a A a b C b Find the state that has an edge on b from E q. start with E{1,2,4,5,6,7,10}. Find which states in E have states reachable by b transitions. This set is called move(E,b) The set is {5}: move(A,b) = {5} We’ve seen this before. It’s C. Finally a a b b a B D E start a A a b C b b That’s it ! There is only one edge from each state for a given input character. It’s a DFA. Disregard the fact that each of these states is actually a group of NFA states. We can regard them as single states in the DFA. In fact it also requires other as an edge beyond E leading to the ultimate accepting state. Also the DFA is not yet optimized (there can be less states). Compiler design – RE&DFA&NFA Page 9 of 9 However, we can make the transition table so far. Here it is: State Input a Input b A B C B B D C B C D B E E B C NFA vs. DFA (Time-Space Tradeoffs The table below summarizes the worst-case for determining whether an input string x belongs to the language denoted by a regular expression r using recognizers constructed from NFA and DFA. Automaton Space Time NFA O(|r |) O( |r | x | x |) DFA O(2 |r| ) O (| x |) | r | is the length of r, and | x | is the length of x. Example: For the regular expression (a | b)a(a | b)…(a | b) , where there are n-1 (a | b)’s at the end There is no DFA with fewer than 2n states.
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https://puzzling.stackexchange.com/questions/71268/class-seating-arrangement
logical deduction - Class Seating Arrangement - Puzzling Stack Exchange Join Puzzling By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Class Seating Arrangement Ask Question Asked 7 years ago Modified7 years ago Viewed 243 times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. There are 25 25 students with distinct heights in a class. The seats are arranged in the class like a square array (5 5 x 5 5) and students are seated such a way that each person will be taller than both the student in front of and left to her/him. In how many different ways can this operation be done? Hint: If this question was asked for 9 9 students with 3 3 x 3 3 square array matrix, then the answer would be 42 42. logical-deduction combinatorics Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked Sep 2, 2018 at 9:36 OrayOray 33.6k 6 6 gold badges 68 68 silver badges 233 233 bronze badges 7 2 The answer is always 42xhienne –xhienne 2018-09-02 11:57:09 +00:00 Commented Sep 2, 2018 at 11:57 I don't think so. Do that in a 2x2 matrix. Then there are only 24 ways to seat the students and these still include combinations that conflict the given requirement. (Edit: oh i saw the link only now... sorry :-)puck –puck 2018-09-02 13:49:22 +00:00 Commented Sep 2, 2018 at 13:49 @puck he is making a joke :)Oray –Oray 2018-09-02 13:50:31 +00:00 Commented Sep 2, 2018 at 13:50 1 Hey, you seem to love these kinds of math puzzles. I think you might like this site proposal (not accessible via Google Chrome, but Safari works). And if you want, you can join the related chatroom.Mr Pie –Mr Pie 2018-09-02 16:38:52 +00:00 Commented Sep 2, 2018 at 16:38 1 @user477343 this kind of puzzle is like finding the way how to do it, not actually solve it manually in my opinion, and learn as well while searching ;)Oray –Oray 2018-09-02 16:45:35 +00:00 Commented Sep 2, 2018 at 16:45 |Show 2 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. The answer is 701149020 701149020 Because We may rank the students by height {1,2,3,⋯,25}{1,2,3,⋯,25}. Since each student must be taller than any to their left (in their row) and any in front of ("above") them (in their column) then any such square is a standard Young tableau of shape (5,5,5,5,5)(5,5,5,5,5) (by definition). The number of standard Young tableau of shape λ λ, d λ d λ, is given by the hook-length formula, d λ=n!∏h λ(i,j)d λ=n!∏h λ(i,j) ...where the hook length, h λ(i,j)h λ(i,j), is the number of cells which are either in row i i but not to the left of j j or in column j j but not above i i. As such d(5,5,5,5,5)=25⋅24⋅23⋅22⋅21⋅20⋅19⋅18⋅17⋅16⋅15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 9⋅8⋅7⋅6⋅5⋅8⋅7⋅6⋅5⋅4⋅7⋅6⋅5⋅4⋅3⋅6⋅5⋅4⋅3⋅2⋅5⋅4⋅3⋅2⋅1 d(5,5,5,5,5)=25⋅24⋅23⋅22⋅21⋅20⋅19⋅18⋅17⋅16⋅15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 9⋅8⋅7⋅6⋅5⋅8⋅7⋅6⋅5⋅4⋅7⋅6⋅5⋅4⋅3⋅6⋅5⋅4⋅3⋅2⋅5⋅4⋅3⋅2⋅1 d(5,5,5,5,5)=25⋅24⋅23⋅22⋅21⋅20⋅19⋅18⋅17⋅16⋅15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1(5⋅5)⋅(8⋅6⋅7⋅6⋅5)⋅(6⋅3)⋅(4⋅4)⋅(5⋅3)⋅(7⋅2)⋅(4⋅3)⋅9⋅8⋅7⋅6⋅5⋅4⋅2⋅1 d(5,5,5,5,5)=25⋅24⋅23⋅22⋅21⋅20⋅19⋅18⋅17⋅16⋅15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1(5⋅5)⋅(8⋅6⋅7⋅6⋅5)⋅(6⋅3)⋅(4⋅4)⋅(5⋅3)⋅(7⋅2)⋅(4⋅3)⋅9⋅8⋅7⋅6⋅5⋅4⋅2⋅1 d(5,5,5,5,5)=25⋅24⋅23⋅22⋅21⋅20⋅19⋅18⋅17⋅16⋅15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 25⋅(8⋅6⋅7⋅6⋅5)⋅18⋅16⋅15⋅14⋅12⋅9⋅8⋅7⋅6⋅5⋅4⋅2⋅1 d(5,5,5,5,5)=25⋅24⋅23⋅22⋅21⋅20⋅19⋅18⋅17⋅16⋅15⋅14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 25⋅(8⋅6⋅7⋅6⋅5)⋅18⋅16⋅15⋅14⋅12⋅9⋅8⋅7⋅6⋅5⋅4⋅2⋅1 d(5,5,5,5,5)=24⋅23⋅22⋅21⋅20⋅19⋅17⋅13⋅11⋅10⋅3 8⋅6⋅7⋅6⋅5 d(5,5,5,5,5)=24⋅23⋅22⋅21⋅20⋅19⋅17⋅13⋅11⋅10⋅3 8⋅6⋅7⋅6⋅5 d(5,5,5,5,5)=24⋅23⋅22⋅21⋅20⋅19⋅17⋅13⋅11⋅10⋅3(8⋅6 2)⋅(7⋅6 2)⋅(5⋅2⋅2)d(5,5,5,5,5)=24⋅23⋅22⋅21⋅20⋅19⋅17⋅13⋅11⋅10⋅3(8⋅6 2)⋅(7⋅6 2)⋅(5⋅2⋅2) d(5,5,5,5,5)=24⋅23⋅22⋅21⋅20⋅19⋅17⋅13⋅11⋅10⋅3 24⋅21⋅20 d(5,5,5,5,5)=24⋅23⋅22⋅21⋅20⋅19⋅17⋅13⋅11⋅10⋅3 24⋅21⋅20 d(5,5,5,5,5)=23⋅22⋅19⋅17⋅13⋅11⋅10⋅3 d(5,5,5,5,5)=23⋅22⋅19⋅17⋅13⋅11⋅10⋅3 d(5,5,5,5,5)=701149020 d(5,5,5,5,5)=701149020 Note that with 9 9 students this would be: d(3,3,3)=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 5⋅4⋅3⋅4⋅3⋅2⋅3⋅2⋅1 d(3,3,3)=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 5⋅4⋅3⋅4⋅3⋅2⋅3⋅2⋅1 d(3,3,3)=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1(3⋅3)⋅(4⋅2)⋅5⋅4⋅3⋅2⋅1 d(3,3,3)=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1(3⋅3)⋅(4⋅2)⋅5⋅4⋅3⋅2⋅1 d(3,3,3)=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 9⋅8⋅5⋅4⋅3⋅2⋅1 d(3,3,3)=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 9⋅8⋅5⋅4⋅3⋅2⋅1 d(3,3,3)=7⋅6=42 d(3,3,3)=7⋅6=42 In general for squares of side n n these are the n n th n n-dimensional Catalan numbers with an entry at OEIS A039622 ...and for rectangles of sides m,n m,n the result has an entry at OEIS A060854 Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 2, 2018 at 15:29 answered Sep 2, 2018 at 14:42 Jonathan AllanJonathan Allan 21.4k 2 2 gold badges 60 60 silver badges 111 111 bronze badges 1 1 very good explanation! u even explained th general formula thanks.Oray –Oray 2018-09-02 16:46:32 +00:00 Commented Sep 2, 2018 at 16:46 Add a comment| Your Answer Thanks for contributing an answer to Puzzling Stack Exchange! 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https://www.r-bloggers.com/2014/09/bayesian-first-aid-poisson-test/
Bayesian First Aid: Poisson Test | R-bloggers MENU Home About RSS add your blog! Learn R R jobs► Submit a new job (it’s free) Browse latest jobs (also free) Contact us R-bloggers ========== R news and tutorials contributed by hundreds of R bloggers ---------------------------------------------------------- Home About RSS add your blog! Learn R R jobs Submit a new job (it’s free) Browse latest jobs (also free) Contact us Bayesian First Aid: Poisson Test Posted on September 4, 2014 by Rasmus Bååth in R bloggers | 0 Comments [This article was first published on Publishable Stuff, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here) Want to share your content on R-bloggers?click here if you have a blog, or here if you don't. ShareTweet As the normal distribution is sort of the default choice when modeling continuous data (but not necessarily the best choice), the Poisson distribution is the default when modeling counts of events. Indeed, when all you know is the number of events during a certain period it is hard to think of any other distribution, whether you are modeling the number of deaths in the Prussian army due to horse kicks or the numer of goals scored in a football game. Like the t.test t.test in R there is also a poisson.test poisson.test that takes one or two samples of counts and spits out a p-value. But what if you have some counts, but don’t significantly feel like testing a null hypothesis? Stay tuned! Bayesian First Aid is an attempt at implementing reasonable Bayesian alternatives to the classical hypothesis tests in R. For the rationale behind Bayesian First Aid see the original announcement. The development of Bayesian First Aid can be followed on GitHub. Bayesian First Aid is a work in progress and I’m grateful for any suggestion on how to improve it! The Model The original poisson.test poisson.test function that comes with R is rather limited, and that makes it fairly simple to construct the Bayesian alternative. However, at first sight poisson.test poisson.test may look more limited than it actually is. The one sample version just takes one count of events $x$ and the number of periods $T$ during which the number of events were counted. If your ice cream truck sold 14 ice creams during one day you would call the function like poisson.test(x = 14, T = 1) poisson.test(x = 14, T = 1). This seems limited, what if you have a number of counts, say you sell ice cream during a whole week, what to do then? The trick here is that you can add up the counts and the number of time periods and this will be perfectly fine. The code below will still give you an estimate for the underlying rate of ice cream sales per day: ice_cream_sales = c(14, 16, 9, 18, 10, 6, 13) poisson.test(x = sum(ice_cream_sales), T = length(ice_cream_sales)) ice_cream_sales = c(14, 16, 9, 18, 10, 6, 13) poisson.test(x = sum(ice_cream_sales), T = length(ice_cream_sales)) ice_cream_sales = c(14, 16, 9, 18, 10, 6, 13) poisson.test(x = sum(ice_cream_sales), T = length(ice_cream_sales)) Note that this only works if the counts are well modeled by the same Poisson distribution. If the ice cream sales are much higher on the weekends, adding up the counts might not be a good idea. poisson.test poisson.test is also limited in that it can only handle two counts; you can compare the performance of your ice cream truck with just one competitor’s, no more. As the Bayesian alternative accepts the same input as poisson.test poisson.test it inherits some of it’s limitations (but it can easily be extended, read on!). The model for the Bayesian First Aid alternative to the one sample possion.test possion.test is: Here $x$ is again the count of events, $T$ is the number of periods, and $lambda$ is the parameter of interest, the underlying rate at which the events occur. In the two sample case the one sample model is just separately fitted to each sample. As $x$ is assumed to be Poisson distributed, all that is required to turn this into a fully Bayesian model is a prior on $lambda$. In the the literature there are two common recommendations for an objective prior for the rate of a Poisson distribution. The first one is $lambda propto 1 / lambda$ which is the same as $log(lambda) propto text{const}$ and is proposed, for example, by Villegas (1977). While it can be argued that this prior is as non-informative as possible, it is problematic in that it will result in an improper posterior when the number of events is zero ($x = 0$). I feel that seeing zero events should tell the model something and, at least, not cause it to blow up. The second proposal is Jeffreys prior $lambda propto 1 / sqrt{lambda}$, (as proposed by the great BUGS Book) which has a slight positive bias compared to the former prior but handles counts of zero just fine. The difference between these two priors is very small and will only matter when you have very few counts. Therefore Bayesian First Aid alternative to poisson.test poisson.test uses Jeffreys prior. So if the model uses Jeffreys prior, what is the $lambda sim text{Gamma}(0.5, 0.00001)$ doing in the model definition? Well, the computational framework underlying Bayesian First Aid is JAGS and in JAGS you build your model using probability distributions. The Jeffreys prior is not a proper probability distribution but it turns out that it can be reasonably well approximated by ${Gamma}(0.5, epsilon)$ with $epsilon rightarrow 0$ (in the same way as $lambda propto 1 / lambda$ can be approximated by ${Gamma}(epsilon, epsilon)$ with $epsilon rightarrow 0$). The bayes.poisson.test bayes.poisson.test Function The bayes.poisson.test bayes.poisson.test function accepts the same arguments as the original poisson.test poisson.test function, you can give it one or two counts of events. If you just ran poisson.test(x=14, T=1) poisson.test(x=14, T=1), prepending bayes. bayes. runs the Bayesian First Aid alternative and prints out a summary of the model result (like bayes.poisson.test(x=14, T=1) bayes.poisson.test(x=14, T=1)). By saving the output, for example, like fit <- bayes.poisson.test(x=14, T=1) fit <- bayes.poisson.test(x=14, T=1) you can inspect it further using plot(fit) plot(fit), summary(fit) summary(fit) and diagnostics(fit) diagnostics(fit). To demonstrate the use of bayes.poisson.test bayes.poisson.test I will use data from Boice and Monson (1977) on the number of women diagnosed with breast cancer in one group of 1,047 tuberculosis patients that had received on average 102 X-ray exams and one group of 717 tuberculosis patients whose treatment had not required a large number of X-ray exams. Here is the full data set: Here WY stand for woman-years (as if woman-years would be different from man-years, or person-years…). While the data is from a relatively old article we are going to replicate a more recent reanalysis of that data from the article Testing the Ratio of Two Poisson Rates by Gu et al. (2008). They tested the alternative hypothesis that the rate of breast cancer per person-year would be 1.5 times greater in the group that was X-rayed compared to the control group. They tested it like this: no_cancer_cases <- c(41, 15) person-centuries rather than person-years to get the estimated rate on a more interpretable scale. person_centuries <- c(28.011, 19.025) poisson.test(no_cancer_cases, person_centuries, r = 1.5, alternative = "greater") no_cancer_cases <- c(41, 15) # person-centuries rather than person-years to get the estimated rate # on a more interpretable scale. person_centuries <- c(28.011, 19.025) poisson.test(no_cancer_cases, person_centuries, r = 1.5, alternative = "greater") no_cancer_cases <- c(41, 15) person-centuries rather than person-years to get the estimated rate on a more interpretable scale. person_centuries <- c(28.011, 19.025) poisson.test(no_cancer_cases, person_centuries, r = 1.5, alternative = "greater") ## ## Comparison of Poisson rates ## ## data: no_cancer_cases time base: person_centuries ## count1 = 41, expected count1 = 41.61, p-value = 0.291 ## alternative hypothesis: true rate ratio is greater than 1.5 ## 95 percent confidence interval: ## 1.098 Inf ## sample estimates: ## rate ratio ## 1.856 and concluded that “There is not enough evidence that the incidence rate of breast cancer in the X-ray fluoroscopy group is 1.5 times to the incidence rate of breast cancer in control group”. It is oh-so-easy to interpret this as that there is no evidence that the incidence rate is more than 1.5 times higher, but this is wrong and the Bayesian First Aid alternative makes this clear: library(BayesianFirstAid) bayes.poisson.test(no_cancer_cases, person_centuries, r = 1.5, alternative = "greater") library(BayesianFirstAid) bayes.poisson.test(no_cancer_cases, person_centuries, r = 1.5, alternative = "greater") library(BayesianFirstAid) bayes.poisson.test(no_cancer_cases, person_centuries, r = 1.5, alternative = "greater") ## Warning: The argument ‘alternative’ is ignored by bayes.poisson.test ## ## Bayesian Fist Aid poisson test – two sample ## ## number of events: 41 and 15, time periods: 28.011 and 19.025 ## ## Estimates [95% credible interval] ## Group 1 rate: 1.5 [1.1, 1.9] ## Group 2 rate: 0.80 [0.43, 1.2] ## Rate ratio (Group 1 rate / Group 2 rate): ## 1.8 [1.1, 3.4] ## ## The event rate of group 1 is more than 1.5 times that of group 2 by a probability ## of 0.754 and less than 1.5 times that of group 2 by a probability of 0.246 . The warning here is nothing to worry about, there is no need to specify what alternative is tested and bayes.poisson.test bayes.poisson.test just tells you that. So sure, the evidence is far from conclusive, but given the data and the model there is a 75% probability that the incidence rate is more than 1.5 times higher in the X-rayed group. That is, rather than just saying that there is not enough evidence we have quantified how much evidence there is, and the evidence actually slightly favors the alternative hypothesis. This is also easily seen in the default plot of bayes.poisson.test bayes.poisson.test: plot( bayes.poisson.test(no_cancer_cases, person_centuries, r = 1.5)) plot( bayes.poisson.test(no_cancer_cases, person_centuries, r = 1.5) ) plot( bayes.poisson.test(no_cancer_cases, person_centuries, r = 1.5) ) Comparing More than Two Groups Back to the ice cream truck, say that you sold 14 ice creams in one day and your competitors Karl and Anna sold 22 and 7 ice creams, respectively. How would you estimate and compare the underlying rates of sold ice creams of these three trucks when bayes.poisson.test bayes.poisson.test only accepts counts from two groups? When you want to go off the beaten path the model.code model.code function is your friend as it takes the result from Bayesian First Aid method and returns R and JAGS code that replicates the analysis you just ran. In this case start by running the model with two counts and then print out the model code: fit <- bayes.poisson.test(x = c(14, 22), T = c(1, 1)) model.code(fit) fit <- bayes.poisson.test(x = c(14, 22), T = c(1, 1)) model.code(fit) fit <- bayes.poisson.test(x = c(14, 22), T = c(1, 1)) model.code(fit) ### Model code for the Bayesian First Aid two sample Poisson test ### require(rjags) # Setting up the data x <- c(14, 22) t <- c(1, 1) # The model string written in the JAGS language model_string <- "model { for(group_i in 1:2) { x[group_i] ~ dpois(lambda[group_i] t[group_i]) lambda[group_i] ~ dgamma(0.5, 0.00001) x_pred[group_i] ~ dpois(lambda[group_i] t[group_i]) } rate_diff <- lambda - lambda rate_ratio <- lambda / lambda }" # Running the model model <- jags.model(textConnection(model_string), data = list(x = x, t = t), n.chains = 3) samples <- coda.samples(model, c("lambda", "x_pred", "rate_diff", "rate_ratio"), n.iter=5000) # Inspecting the posterior plot(samples) summary(samples) Just copy-n-paste this code directly into an R script and make the following changes: Add the data for Karl’s ice cream truck: x <- c(14, 22) x <- c(14, 22) → x <- c(14, 22, 7) x <- c(14, 22, 7) t <- c(1, 1) t <- c(1, 1) → t <- c(1, 1, 1) t <- c(1, 1, 1) Change the JAGS code so that you iterate over three groups instead of two: for(group_i in 1:2){ for(group_i in 1:2) { → for(group_i in 1:3){ for(group_i in 1:3) { And that’s it! Now we can run the model script and take a look at the estimated rates of ice cream sales for the three trucks. plot(samples) plot(samples) plot(samples) If you want to compare many groups you should perhaps consider using a hierarchical Poisson model. (Pro tip: John K. Kruschke’s Doing Bayesian Data Analysis has a great chapter on hierarchical Poisson models.) References Boice, J. D., & Monson, R. R. (1977). Breast cancer in women after repeated fluoroscopic examinations of the chest. Journal of the National Cancer Institute, 59(3), 823-832. link to article (unfortunatel behind paywall) Gu, K., Ng, H. K. T., Tang, M. L., & Schucany, W. R. (2008). Testing the ratio of two poisson rates. Biometrical Journal, 50(2), 283-298. doi: 10.1002/bimj.200710403, pdf Villegas, C. (1977). On the representation of ignorance. Journal of the American Statistical Association, 72(359), 651-654. doi: 10.2307/2286233 Lunn, D., Jackson, C., Best, N., Thomas, A., & Spiegelhalter, D. (2012). The BUGS book: a practical introduction to Bayesian analysis. CRC Press. pdf of chapter 5 on Prior distributions Related ShareTweet To leave a comment for the author, please follow the link and comment on their blog: Publishable Stuff. R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job. Want to share your content on R-bloggers?click here if you have a blog, or here if you don't. ← Previous post Next post → Most viewed posts (weekly) Which data science skills are important ($50,000 increase in salary in 6-months) PCA vs Autoencoders for Dimensionality Reduction Better Sentiment Analysis with sentiment.ai Self-documenting plots in ggplot2 5 Ways to Subset a Data Frame in R How to write the first for loop in R Date Formats in R Sponsors Our ads respect your privacy. Read our Privacy Policy page to learn more. Contact us if you wish to place your banner here. 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188260
https://d.cosx.org/d/102862-102862
线性回归中有截距与无截距的差别是什么？ - COS论坛 | 统计之都 | 统计与数据科学论坛 主站 新手须知 简体中文 简体中文 English 注册 登录 统计学生物统计 线性回归中有截距与无截距的差别是什么？ 登录以回复 登录以回复 1 / 5 条 最早内容 1 / 5 条十一月 2010 0 条未读 最新回复 foxmouse 2010年11月3日发布 #1 2010年11月3日星期三 15点24分 一组数，进行一元回归分析，得出有截距和无截距两个方程 y=1.1950x y=1.1261x+11.87 P值与R平方都很好，现在不知道怎么对比两个方程那个更好？高人能否指点一二？谢谢 y x 124 99 134 108 153 126 186 158 184 153 206 177 196 163 196 163 215 176 228 190 230 191 236 202 240 199 236 202 回复 12 天 后 DavidLung 2010年11月15日发布 #2 2010年11月15日星期一 15点42分 1.根据研究的问题，看截距是否有现实意义。 2.最初的技术方法就是看残差呗，然后深入些，就是模型检验云云。 回复 agri521 2010年11月15日发布 #3 2010年11月15日星期一 18点32分 线性回归中的截距 回复 yihui 2010年11月15日发布 #4 2010年11月15日星期一 21点54分 回复 第3楼 的 agri521：我感觉你引用的英文网页中的解释好像不对，无截距的回归并非让ybar为0，只是x=0的时候y必须为0而已（严格来说是E(Y)=0）。另外你说的“无截距线性回归的R2和F值总是高于有截距线性回归”这话也不能肯定，无截距和有截距的R2并没有直接关系。我模拟了一下，确实无截距的R2会偏大，但并非严格大。具体原因我得想想。 回复 微微 2010年11月16日发布 #5 2010年11月16日星期二 05点48分 在推导TSS=RSS+ESS时，用了sum(y-y(hat))=0的假设，这个假设在无截距回归的情况下是不成立的，可能出现RSS>TSS。 x y 2.38 51.11 2.44 50.63 2.7 51.82 2.98 52.97 3.32 54.47 3.12 53.33 2.14 49.9 2.86 51.99 3.5 55.81 3.2 52.93 2.78 52.87 2.7 52.36 2.36 51.38 2.42 50.87 2.62 51.02 2.8 51.29 2.92 52.73 3.04 52.81 3.26 53.59 2.3 49.77 在Eviews中，计算拟合优度用的并非ESS/TSS，而是1-RSS/TSS，用这组数据作无截距回归，得到的R2为-12.9。其他统计软件如果使用ESS/TSS计算拟合优度，那么得到的R2可能大于1。 回复 说点什么吧... 正在加载… 加载论坛时出错，请强制刷新页面重试。
188261
https://dialnet.unirioja.es/descarga/articulo/8648906.pdf
H Revista Integración Escuela de Matemáticas Universidad Industrial de Santander Vol. 40, No. 1, 2022, pág. 1–23 Some notes about power residues modulo prime Yuki Kiriu a, Diego A. Mejía b , a Shizuoka Salesio High School, Shizuoka, Japan. b Shizuoka University, Faculty of Science, Creative Science Course (Mathematics), Shizuoka, Japan. Abstract. Let q be a prime. We classify the odd primes p ̸= q such that the equation x2 ≡q (mod p) has a solution, concretely, we find a subgroup L4q of the multiplicative group U4q of integers relatively prime with 4q (modulo 4q) such that x2 ≡q (mod p) has a solution iffp ≡c (mod 4q) for some c ∈L4q. Moreover, L4q is the only subgroup of U4q of half order containing −1. Considering the ring Z[ √ 2], for any odd prime p it is known that the equation x2 ≡2 (mod p) has a solution iffthe equation x2 −2y2 = p has a solution in the integers. We ask whether this can be extended in the context of Z[ n √ 2] with n ≥2, namely: for any prime p ≡1 (mod n), is it true that xn ≡2 (mod p) has a solution iffthe equation D2 n(x0, . . . , xn−1) = p has a solution in the integers? Here D2 n(¯ x) represents the norm of the field extension Q( n √ 2) of Q. We solve some weak versions of this problem, where equality with p is replaced by 0 (mod p) (divisible by p), and the “norm" Dr n(¯ x) is considered for any r ∈Z in the place of 2. Keywords: Power residues modulo prime, quadratic residues, Legendre sym-bol, norms of field extensions, irreducible polynomials. MSC2010: 11A15, 11C20, 11R04. Sobre residuos de potencias módulo primo Resumen. Sea q un número primo. Clasificamos los primos impares p ̸= q tal que la ecuación x2 ≡q (m´ od p) tiene solución, concretamente, hay un subgrupo L4q del grupo multiplicativo U4q de los enteros primos relativos con 4q (módulo 4q) tal que x2 ≡q (m´ od p) tiene solución si y solo si p ≡c (m´ od 4q) para algún c ∈L4q. Aún más, L4q es el único subgrupo de U4q con la mitad del orden que contiene a −1. En conexión con el anillo Z[ √ 2], para cualquier primo impar p se sabe que la ecuación x2 ≡2 (m´ od p) tiene solución si y solo si x2 −2y2 = p tiene so-lución en los enteros. Nos preguntamos si esta situación se puede extender al 0 Email: auki_sal@yahoo.co.jp; bdiego.mejia@shizuoka.ac.jp. Received: 18 March 2021, Accepted: 29 September 2021. To cite this article: Y. Kiriu and D.A. Mejía, Some notes about power residues modulo prime, Rev. Integr. temas mat., 40 (2022), No. 1, 1-23. doi: 10.18273/revint.v40n1-2022001 1 2 Yuki Kiriu & Diego A. Mejía contexto de Z[ n √ 2] con n ≥2, a saber: para cualquier primo p ≡1 (m´ od n), ¿la ecuación xn ≡2 (m´ od p) tiene solución si y solo si D2 n(x0, . . . , xn−1) = p tiene solución en los enteros? Aquí D2 n(¯ x) representa la norma de Q( n √ 2) co-mo extensión del campo Q. Solucionamos algunas versiones débiles de este problema, donde igualdad con p se reemplaza por 0 (m´ od p) (divisible por p), y la “norma” Dr n(¯ x) se considera para cualquier r ∈Z en lugar de 2. Palabras clave: Residuos de potencias módulo primo, residuos cuadráticos, símbolo de Legendre, normas de extensiones de campos, polinomios irreduci-bles. 1. Introduction In this work, we prove several properties and present problems related with quadratic residues and its generalization to n-th power residues modulo prime, all in the framework of elementary number theory. Before entering into the subject, we first fix some basic notations. Notation 1.1. In the following, m > 1 is an integer and q is a prime. (1) Fq denotes the field of integers modulo q, which is the prime field of order q, and F× q denotes its associated multiplicative group. (2) More generally, Um denotes the multiplicative group of integers modulo m that are relatively prime with m. Note that Uq = F× q . (3) Let G be a group with identity element 1G. For any r ∈G, the order of r in G, which we denote by OG(r), is the smallest positive integer n satisfying rn = 1G in case it exists, otherwise OG(r) is infinite. When G = Um, for r ∈Um we abbreviate Om(r) := OUm(r), which is the smallest positive integer n such that rn ≡1 (mod m) (which always exists because Um is finite). We can of course extend this notion for any r ∈Z that is relatively prime with m, so Om(r) = Om(r0) where r0 is the residue obtained after dividing r by m. (4) The number of elements of a set A is denoted by #A. When G is a group, #G is also called the order of G. When G is a finite group and r ∈G, OG(r) divides #G. Therefore, since #Um = φ(m) where φ denotes Euler’s phi function, Om(r) | φ(m) for any integer r relatively prime with m. In particular, if q does not divide r then Oq(r) | φ(q) = q −1. (5) Let r ∈Z be relatively prime with m. Since Om(r) | φ(m), there is a unique (positive) integer nm(r) satisfying Om(r)nm(r) = φ(m). Therefore, due to the definition of Om(r), nm(r) is the largest n | φ(m) such that r φ(m) n ≡1 (mod m). The notion of nm(r) is not standard, but it will be very useful in the context of power residues modulo prime, as well as in characterizations of Om(r). Euler’s criterion for quadratic residues modulo prime can be easily generalized to power residues as follows (see e.g. [5, Thm. 3.11], [8, Thm. 1.29] and [3, Prop. 4.2.1]). [Revista Integración Some notes about power residues modulo prime 3 Theorem 1.2 (Generalized Euler’s criterion). Let r ∈Z, p a prime not dividing r and let n be a positive integer. Then the equation xn ≡r (mod p) has a solution iff r p−1 gcd(p−1,n) ≡1 (mod p). Even more, if the equation xn ≡r (mod p) has a solution then it has gcd(p−1, n)-many incongruent solutions modulo p in total. As a consequence, Corollary 1.3. Let r ∈Z and p a prime not dividing r. Then np(r) is the largest n | p−1 such that r has an n-th root modulo p. Moreover, the following statements are equivalent for any positive integer n: (i) xn ≡r (mod p) has a solution. (ii) r p−1 gcd(p−1,n) ≡1 (mod p). (iii) gcd(p −1, n) | np(r). Proof. The equivalence (i) ⇔(ii) is Theorem 1.2; the equivalence (ii) ⇔(iii) can be seen from the definition of np(r) (see Notation 1.1(5)). □ ✓ ✓ ✓ In this view, np(r) plays a very important role in relation with power residues modulo p. The main results of this paper are divided in two parts, the first about quadratic reci-procity, and the second about power reciprocity modulo prime. Main results 1: On quadratic residues Fix r ∈Z. When p is an odd prime not dividing r (i.e. gcd(p, r) = 1), whether r is a quadratic residue modulo p is determined by the Legendre symbol, which is defined by r p  =  1 if the equation x2 ≡r (mod p) has a solution, −1 otherwise. (1.4) In the case r = 2, the problem of whether 2 is a quadratic residue modulo an odd prime is already solved. Theorem 1.5 (See e.g. [1, Thm. 9.6]). If p is an odd prime then  2 p  = 1 iffp ≡±1 (mod 8). We ask about similar characterizations for any integer r. Problem 1.6. Let r ∈Z. Is there a positive integer m(r) and a set L(r) ⊆Um(r) such that, for any prime p not dividing r,  r p  = 1 iffthe residue of p modulo m(r) is in L(r)? If so, can L(r) be characterized in some way? Vol. 40, No. 1, 2022] 4 Yuki Kiriu & Diego A. Mejía The answer to the first question should not be difficult due to the quadratic reciprocity law, but the characterization of L(r) is more interesting for settling the general problem. In fact, due to the property ab p  = a p   b p  , (1.7) the interesting case of Problem 1.6 is when r is a prime. In this case, we proved the following main result: Theorem A (Theorem 3.5). Let q be a prime. Then (a) There is only one subgroup of U4q with order #U4q 2 containing −1. This subgroup is denoted by L4q. (b) For any prime p ̸= q,  q p  = 1 iffthe residue of p modulo 4q is in L4q. This theorem becomes a tool to calculate  r p  for any r ∈Z relatively prime with p. This is presented in Theorem 3.6 (and at the end of Section 3). In the case of composite r, due to Equation (1.7) an extension of Theorem A is reasonable when r is square free. In this case we can find a subgroup L4r of U4r containing −1 as in (b), but in general this group is not unique as in (a). Details are presented in Theorem 3.7 and in the discussion that follows it. Main results 2: On power residues We aim to generalize the following result to power residues. Theorem 1.8 (See e.g. [2, Thm. 256] and ). Let p be an odd prime. Then the following statements are equivalent. (i) The equation x2 ≡2 (mod p) has a solution. (ii) The equation x2 −2y2 = p has an integer solution. This is related to the characterization of irreducible elements of the ring Z[ √ 2]: an odd prime p in Z is still a prime in Z[ √ 2] iffthe equation x2 −2y2 = p does not have integer solutions (see [2, Thm. 256]). Recall that x2 −2y2 is the norm of x + y √ 2 in the field extension Q( √ 2) of Q. For any n ≥2, denote by D2 n(x0, . . . , xn−1) the norm of x0 + x1 n √ 2 + . . . xn−1 n √ 2n−1 in the field extension Q( n √ 2) of Q. This norm is defined (even in a more general context) in Section 4, but we just state here that D2 n(x0, . . . , xn−1) is an integer when x0, . . . , xn−1 ∈ Z. So we ask whether Theorem 1.8 can be generalized in the following sense. Problem 1.9. Let n > 2 and p a prime such that p ≡1 (mod n). Are the following statements equivalent? (1) The equation xn ≡2 (mod p) has a solution. [Revista Integración Some notes about power residues modulo prime 5 (2) The equation D2 n(x0, . . . , xn−1) = p has an integer solution. The solution of this problem seems to rely on tools in algebraic number theory that would go beyond elementary number theory. In these terms, we managed to solve weaker versions of the problem, where in some of them (2) is replaced by D2 n(x0, . . . , xn−1) ≡0 (mod p). The trivial solution of this equation is x0 = . . . = xn−1 = 0, so we aim for non-trivial solutions. On the other hand, our results deal with any integer r in place of 2, so we used a general version Dr n(x0, . . . , xn−1) of the norm (which is defined in detail in Section 4). Theorem B (Theorem 5.1). Let p be a prime, r ∈Z, n ∈Z+ and r0 ∈Fp such that r ≡r0 (mod p). (a) The polynomial xn −r0 is irreducible in Fp[x] iffthe equation Dr n(x0, . . . , xn−1) ≡0 (mod p) does not have a non-trivial solution in the integers. (b) If n ≥2 and the equation xn ≡r (mod p) has a solution, then Dr n(x0, . . . , xn−1) ≡0 (mod p) has a non-trivial solution in Zn satisfying −p 1 n < xi < p 1 n for all 0 ≤i < n. The proof of Theorem B(b) is inspired in the proof of Theorem 1.8 presented in the post . As a consequence, we obtain the following equivalence when n is a prime. Corollary (Corollary 5.2). Let p and q be primes, r ∈Z. Then the following statements are equivalent: (i) xq ≡r (mod p) has a solution. (ii) Dr q(x0, . . . , xn−1) ≡0 (mod p) has a non-trivial solution. We can also conclude some weakening of the implication (2) ⇒(1) of Problem 1.9, which yields the real implication when n is a prime. Theorem C (Theorem 5.3). Assume that p is a prime, n ≥2, r ∈Fp and r0 ∈Fp such that r ≡r0 (mod p). If the polynomial xn −r0 is irreducible in Fp[x] then Dr n(¯ x) = p does not have a solution in the integers. In particular, (2) ⇒(1) of Problem 1.9 is true when n is a prime. We also present a simple proof of Theorem 1.8 using Theorem B (see Theorem 5.4), where 2 can also be replaced by r ∈{−2, −1}. This shortens the proof in a little bit. We remark that “xn −r is irreducible in Fp[x]” is stronger than “xn ≡r (mod p) does not have a solution”. For instance, if p ∈{7, 17, 23, 31, 41, 47, 71}, the equation x2 ≡2 (mod p) has a solution, but xp−1 ≡2 (mod p) does not have one. On the other hand, if a0 is a solution of x2 −2 = 0 in Fp then, in Fp[x]: xp−1 −2 = x2( p−1 2 ) −a2 0 = (x p−1 2 −a0)(x p−1 2 + a0). This means that xp−1 −2 is reducible in Fp[x]. More details about irreducibility of xn −r are presented in Section 4. We do not have any counter-example for Problem 1.9 even when xn −2 is reducible in Fp[x]. Vol. 40, No. 1, 2022] 6 Yuki Kiriu & Diego A. Mejía Indirect motivation The motivation of this work is related with the study of Mersenne primes, although we do not present explicit results about them. A Mersenne number is an integer of the form 2n −1 with n ∈Z+ (positive integer), and a Mersenne prime is a primer number of this form. It is well known that, whenever 2n −1 is a prime, n must be a prime. Another curious fact is that, whenever 2n −1 is a Mersenne prime, there is only one (odd) prime p such that Op(2) | n, that is, such that 2n ≡1 (mod p). Even more, since n must be prime, n = Op(2). The converse situation is interesting: if n is a prime and there is only one prime p such that Op(2) | n, then 2n −1 = pe for some e ∈Z+. Hence, when e = 1, 2n −1 is a Mersenne prime; but if e > 1 then p is a Wieferich prime, i.e., a prime number p satisfying 2p−1 ≡1 (mod p2). Recall that so far only two Wieferich primes are known, namely 1093 and 3511, and Silverman proved under the abc-conjecture that there are infinitely many non-Wieferich primes . The previous observation indicates that understanding Op(2) would lead to a better understanding of Mersenne primes and would trigger possible characterizations. On the other hand, since Op(2) is associated with np(2), according to Corollary 1.3 we can discover a lot about np(r) in general by studying power residues modulo p. Concerning Op(r) for some fixed integer r > 1, the pattern of the sequence of Op(r) for prime p relatively prime with r seems to be very erratic , but On(r) in general can be determined in terms of Op(r) for prime p | n, see Theorems 2.1–2.3. In particular, Ope(r) is deeply related with Wieferich primes (in base r). A more detail discussion is presented in Section 2. Structure of the paper Section 2. We discuss some simple aspects related with Om(r) and np(r). In particular, we show expressions of Om(r) for composite m, and a method to obtain n-th roots of 1 modulo a prime p, in particular np(r)-th roots of 1. The contents of this section are known and unrelated with the main results, but we present them in accordance with the “indirect motivation” above. Section 3. This is dedicated to the proof of Theorem A and to further discussions about groups associated with quadratic reciprocity. Section 4. We present some preliminaries in algebra that are going to be required in the proof of the main results about power residues modulo prime. Section 5. We prove our main results about power residues modulo prime, in particular Theorems B and C. Section 6. We discuss research related to this work. 2. Multiplicative order We first show how the multiplicative order modulo composite numbers can be calculated. [Revista Integración Some notes about power residues modulo prime 7 Theorem 2.1 (See e.g. [5, §3.2, Thm. 3.6]). Let p be an odd prime and r ∈Z, r ̸= ±1 relatively prime with p. Assume that e0 is the maximum integer such that Ope0 (r) = Op(r). Then, for any e ≥1, Ope(r) =  Op(r) when e ≤e0, pe−e0Op(r) otherwise. The previous result has a deep connection with Wieferich primes. In fact, an odd prime p is a Wieferich prime in base r if p ∤r and Op2(r) = Op(r).1 Very few of these numbers are known for each r > 1. The following is a version of Theorem 2.1 for p = 2. The proof is almost the same, so we omit it. Theorem 2.2. Assume r ∈Z is odd, r ̸= ±1. If e0 ≥2 is the maximum integer such that O2e0 (r) = O4(r) then, for any e ≥2, O2e(r) =  O4(r) when e ≤e0, 2e−e0O4(r) otherwise. Now we look at the case when m > 1 is composite but not a prime power, so we assume that it has prime factorization m = Qs i=1 pei i (s ≥2). Theorem 2.3. When gcd(r, m)=1, Om(r) = lcm(Ope1 1 (r), Ope2 2 (r), . . . , Opes s (r)). Proof. Let us suppose b := lcm(Ope1 1 (r), Ope2 2 (r), . . . , Opes s (r)). We need to prove the following. 1. rb ≡1 (mod m). For any i ≤s we know that r Opei i (r) ≡1 (mod pei i ) and Op ei i (r) | b, so rb ≡1 (mod pei i ), i.e. pei i | rb −1. Since pei i and pej j are relatively prime when i ̸= j, we conclude that m | rb −1. 2. b is the minimal number satisfying the equation rx ≡1 (mod p) Assume rx ≡1 (mod m). This implies rx ≡1 (mod pei i ) for any i ≤s, so Op ei i (r) | x. Therefore b | x, so by (1) b is the minimum we claim. □ ✓ ✓ ✓ Notice that, by the Chinese remainder theorem, the map Zm →Ls i=1 Zp ei i that sends a to the tuple (a1, . . . , as) of residues modulo pei i is a ring isomorphism, and when restricted to Um it gives a group isomorphism onto ⊕s i=1Up ei i . So the previous result can be seen as a particular case of the following fact: if G = Lk i=1 Gi is a direct sum of groups of finite order and ¯ a = (a1, . . . , ak) ∈G, then OG(¯ a) = lcm(OG1(a1), . . . , OGk(ak)). (A similar proof works.) As a consequence, we obtain the following modular equation using Euler’s phi function. 1The standard definition is rp−1 ≡1 (mod p2), which is equivalent thanks to Theorem 2.1: If Op2(r) ̸= Op(r) then Op2(r) = pOp(r), which does not divide p −1. Vol. 40, No. 1, 2022] 8 Yuki Kiriu & Diego A. Mejía Corollary 2.4. If gcd(r, m) = 1 and c = φ(m) gcd(φ(pe1 1 ), φ(pe2 2 ), . . . , φ(pes s )) then rc ≡1 (mod m). Proof. Since lcm(a1, a2, . . . , am) · gcd(a1, a2, . . . , am) | a1a2 · · · am, by Theorem 2.3 we can prove that Om(r) | lcm(φ(p1 e1), φ(p2 e2), . . . , φ(ps es)) and lcm(φ(p1 e1), φ(p2 e2), . . . , φ(ps es)) | c. The theorem follows immediately. □ ✓ ✓ ✓ The previous result can be generalized as well in the context of direct sums of groups: if ¯ a ∈G and c = #G gcd(#G1,...,#Gk) then ¯ ac = 1G, i.e. OG(¯ a) | c. From here until the end of this section, we assume that p is a prime and gcd(r, p) = 1. We look at the effect of the power of Op(r) in F× p , namely, properties of kOp(r) for k ∈Fp. In fact, these properties come from more general results. First, we show that {kOp(r) : k ∈F× p } gives the full set of np(r)-th roots of 1 modulo p, which can be generalized as follows. Theorem 2.5. Let n ≥1 be an integer. Then all the n-th roots of unity can be obtained from the set A :=  a p−1 gcd(n,p−1) : a ∈F× p Moreover, if rp is a primitive root of p then the set above coincides modulo p with B := n rp ℓ p−1 gcd(n,p−1) : 0 ≤ℓ< gcd(n, p −1) o , and their members are pairwise incongruent modulo p. Proof. We define m(n) := p−1 gcd(n,p−1) and b := rpm(n). For any a ∈F× p , if a ≡rpk (mod p) then am(n) ≡rpkm(n) (mod p). If we put k = d · gcd(n, p −1) + ℓfor some d ∈Z and 0 ≤ℓ< gcd(n, p −1), then km(n) = d(p −1) + ℓm(n). So we get am(n) ≡ rpm(n)ℓ≡bℓ (mod p). This shows A ⊆B (modulo p). The converse inclusion is trivial. By Theorem 1.2, the equation xn ≡1 (mod p) has exactly gcd(n, p −1)-many solutions in Fp. On the other hand, since Op(b) = gcd(n, p −1), it is clear that bℓn ≡1 (mod p) for all 0 ≤ℓ< gcd(n, p −1), and that the bℓare pairwise incongruent modulo p. This shows that B is the complete set of n-th roots of unity. □ ✓ ✓ ✓ Corollary 2.6. The set of solutions for the equation xnp(r) ≡1 (mod p) (i.e. the set of np(r)-th roots of unity modulo p) is n aOp(r) : a ∈F× p o = n rp ℓOp(r) : 0 ≤ℓ< np(r) o (modulo p). [Revista Integración Some notes about power residues modulo prime 9 Recall the following properties of roots of unity modulo p. Lemma 2.7. Let n ≥1 and assume that a is an n-th root of 1 modulo p. Then: (a) If a ≡1 (mod p) then n−1 X i=0 ai ≡n (mod p). (b) If a ̸≡1 (mod p) then n−1 X i=0 ai ≡0 (mod p). Proof. Property (a) is trivial; since (a −1) n−1 X i=0 ai = an −1 ≡0 (mod p), it is clear that a ̸≡1 (mod p) implies (b). □ ✓ ✓ ✓ As a consequence, we can show the behaviour of the sum of kOp(r) for 1 ≤k ≤p −1, or even more generally: Theorem 2.8 (See e.g. [8, Pg. 67]). Let n ∈Z+ . Then: (a) p −1 | n ⇔ p−1 X k=1 kn ≡p −1 (mod p). (b) p −1 ∤n ⇔ p−1 X k=1 kn ≡0 (mod p). Proof. Fix a primitive root rp of p, and for each 1 ≤k < p choose ek < p −1 such that rpek ≡k (mod p). We have the following: p−1 X k=1 kn ≡ p−1 X k=1 (rp ek)n ≡ p−1 X k=1 rp nek ≡ p−2 X i=0 rp ni (mod p). Note that any member of F× p is a (p −1)-th root of 1, so we can apply Lemma 2.7 to conclude: (a) if rpn ≡1 (mod p) then p−2 X i=0 rp ni ≡p −1 (mod p); (b) if rpn ̸≡1 (mod p) then p−2 X i=0 rp ni ≡0 (mod p). It is easy to verify that rpn ≡1 (mod p) is equivalent to p−1 | n, so the result follows. □ ✓ ✓ ✓ Vol. 40, No. 1, 2022] 10 Yuki Kiriu & Diego A. Mejía Corollary 2.9. Let r ∈Z such that gcd(r, p) = 1. Then: (a) Op(r) = p −1 ⇔ p−1 X k=1 kOp(r) ≡p −1 (mod p). (b) Op(r) ̸= p −1 ⇔ p−1 X k=1 kOp(r) ≡0 (mod p). 3. Groups associated with quadratic residues This section is dedicated to the proof of Theorem A. Recall the Legendre symbol  r p  as presented in Equation (1.4). It is known that the map F× p →U4, r 7→  r p  is a group homomorphism, where U4 = {1, −1} as a multiplicative group,2 so L∗ p := ( a ∈F× p : a p  = 1 ) (3.1) is a subgroup of F× p of order p−1 2 (half of the order of F× p ). We look at the following converse situation: given an integer r, characterize the odd primes p relatively prime with r such that  r p  = 1. This is associated with np(r) in the following sense. Lemma 3.2. Let p be an odd prime, r ∈Z such that gcd(r, p) = 1. Then the following statements are equivalent: (i)  r p  = 1. (ii) x2 ≡r (mod p) has a solution. (iii) r p−1 2 ≡1 (mod p). (iv) np(r) is even. Proof. The equivalence (i) ⇔(ii) follows from the definition of Lagrange’s symbol. The others are a direct consequence of Corollary 1.3 (applied to n = 2). □ ✓ ✓ ✓ First, we look at the case when r = q is a prime. If q = 2 we have the following situation. Theorem 3.3. If p is an odd prime then the following statements are equivalent. (i)  2 p  = 1. (ii) p ≡±1 (mod 8). 2This is isomorphic to the additive group Z2. [Revista Integración Some notes about power residues modulo prime 11 (iii) 2 p−1 2 ≡1 (mod p). (iv) np(2) is even. Proof. (i) ⇔(ii) is known, see Theorem 1.5. The rest follows by Lemma 3.2. □ ✓ ✓ ✓ We aim to generalize Theorem 3.3 for any r in the place of 2, concretely, to find a condition like in (ii) that characterizes  r p  for any odd prime p relatively prime with r. An observation about the case r = 2: Denote L8 := {1, −1} as a subgroup of U8. Note that this is the only subgroup of U8 of order 2 (half of the order of U8) that contains −1. Theorem 3.3 says that  2 p  = 1 iffp ≡c (mod 8) for some c ∈L8, which validates Theorem A for r = 2. Assume that r = q is an odd prime. If p ̸= q is an odd prime then, by the quadratic reciprocity law: q p  = (−1) q−1 2 p−1 2 p q  . (3.4) We start assuming q ≡−1 (mod 4),3 in which case q p  = (−1) p−1 2 p q  . Therefore,  q p  = 1 iffone of the following cases hold: (i) p ≡1 (mod 4) and p ≡a (mod q) for some a ∈L∗ q (see Equation (3.1)), or (ii) p ≡−1 (mod 4) and p ≡b (mod q) for some b ∈Uq ∖L∗ q. For any odd prime q0: by the Chinese remainder theorem, the map Fq0 : Z4q0 →Z4 ⊕Fq0 that sends any x to the pair (x0, x1) of remainders modulo 4 and q0 respectively, is a ring isomorphism. When this map is restricted to U4q0 it becomes a group isomorphism onto U4 ⊕F× q0. Coming back to our argument, using the previous terminology we conclude that  q p  = 1 iffp ≡c (mod 4q) for some c ∈U4q0 such that c satisfies one of the following conditions: (⋆)q 1: Fq(c) = (1, a) for some a ∈L∗ q (by (i)), or (⋆)q 2: Fq(c) = (−1, b) for some b ∈Uq ∖L∗ q (by (ii)). Let L4q be the set of c ∈U4q satisfying either (⋆)q 1 or (⋆)q 2. Since L′ (4,q) := {(e, a) ∈U4 ⊕Uq : either e = 1 and a ∈L∗ q, or e ̸= 1 and a / ∈L∗ q} 3Although the easy case is q ≡1 (mod 4), we decided to start with the other case for convenience of the presentation. Vol. 40, No. 1, 2022] 12 Yuki Kiriu & Diego A. Mejía is a subgroup of U4 ⊕Uq and L4q is the inverse image under Fq of this subgroup, we conclude that L4q is a subgroup of U4q. Moreover, L4q has order q −1, which is half of the order of U4q, and −1 ∈L4q: Since L∗ q has order q−1 2 , it is clear that the order of L′ (4,q) is double, that is, q −1, and this is the order of L4q; note that Fq(−1) = (−1, −1) and −1 / ∈L∗ q because q ≡−1 (mod 4), so it satisfies (⋆)q 2 and we get −1 ∈L4q. We turn to the case when q ≡1 (mod 4). By Equation (3.4) we obtain that  q p  =  p q  , so  q p  = 1 iffp ≡a (mod q) for some a ∈L∗ q. Using the ring isomorphism Fq introduced before, define L4q := {c ∈U4q : Fq(c) = (e, a) for some e ∈U4 and a ∈L∗ q}. Since this is the inverse image under Fq of U4 ⊕L∗ q and this is a subgroup of U4 ⊕Uq of size q −1, we conclude that L4q is a subgroup of U4q of order q −1 (half of the order of U4q). Even more, −1 ∈L4q because Fq(−1) = (−1, −1) and, since q ≡1 (mod 4), −1 ∈L∗ q. The previous argument is then summarized in the following result, which generalizes Theorem 3.3 and concludes the proof of Theorem A. Theorem 3.5. Let q ̸= p be prime numbers with p odd. Then  q p  = 1 iffp ≡c (mod 4q) for some c ∈L4q. Moreover, L4q is the unique subgroup of U4q with order q −1 (half of the order of U4q) that contains −1. Proof. According to the previous discussion, it remains to show that, whenever q is an odd prime, L4q is the unique subgroup of U4q as in the statement. So let G be a subgroup of U4q of order q −1 with −1 ∈G. This indicates that (−1) := {1, −1} is a subgroup of G, so when taking quotients U4q/G ∼ = (U4q/(−1))/(G/(−1)). Note that U4q/(−1) ∼ = U2q and G/(−1) is a subgroup of U4q/(−1) of order q−1 2 . So it is enough to show that U2q contains only one subgroup of order q−1 2 . By the Chinese remainder theorem, U2q is isomorphic to U2 ⊕F× q , which is isomorphic to F× q itself. Since F× q is a cyclic group, it only contains one subgroup of order q−1 2 , which concludes the proof. □ ✓ ✓ ✓ Now we turn to the more general case r ∈Z+. If r is a square then trivially  r p  = 1 for any odd prime p relatively prime with r; if r = Qs i=1 qei i is the prime factorization of r and r is not a square, and p is an odd prime relatively prime with r, then by (1.7): r p  = s Y i=1 qi p ei = Y i∈S qi p  = Q i∈S qi p  [Revista Integración Some notes about power residues modulo prime 13 where S := {i : ei is odd}. Therefore, the general case reduces to when r is square free, that is, it has its prime factorization of the form q1 · · · qm (when all prime powers are 1). Since r p  = m Y i=1 qi p  we obtain that  r p  = 1 iffthe number of elements of the set  i :  qi p  = −1 is even. We can express this in terms of the groups L4q thanks to Theorem 3.5. Theorem 3.6. Let r ∈Z+. (a) If r is a square then  r p  = 1 for any odd prime p with gcd(p, r) = 1. (b) Assume that r is not a square and r = Qs i=1 qei i is its prime factorization. If S := {i : ei is odd} then, for any odd prime p with gcd(p, r) = 1,  r p  = 1 iffthe number of elements of the set {i ∈S : p ≡b (mod qi) for some b ∈U4qi ∖L4qi} is even. We develop the case r = q1 · · · qm (prime factorization) a bit more. Consider the ring homomorphism F ′ r : Z →Lm i=1 Z4qi that sends x to the tuple (x1, . . . , xm) where x ≡xi (mod 4qi) for any i. Although the kernel of this map is (4r)Z, the image is not everything: as a consequence of the Chinese remainder theorem (for non-coprime moduli),4 F ′ r[Z] = ( (x1, . . . , xm) ∈ m M i=1 Z4qi : xi ≡xj (mod 4) for all i, j ) . Therefore, the map Fr : Z4r →F ′ r[Z] defined by Fr(a) = F ′ r(a), is a ring isomorphism. If we restrict this map to U4r, we get a group isomorphism onto U ′ (4,r) := F ′ r[Z] ∩ m M i=1 U4qi = ( (x1, . . . , xm) ∈ m M i=1 U4qi : xi ≡xj (mod 4) for all i, j ) . According to (b), define L′ (4,r) := {(x1, . . . , xm) ∈U ′ (4,r) : the number of elements of the set {i : xi ∈U4qi ∖L4qi} is even}. And let L4r = {x ∈U4r : Fr(x) ∈L′ (4,r)}. Therefore, for any odd prime p with gcd(p, r) = 1,  r p  = 1 iffp ≡c (mod 4r) for some c ∈L4r. 4This holds even when some qi is 2. Recall that the Chinese remainder theorem (for non-coprime moduli) states that a system of congruences x ≡ai (mod ni) (1 ≤i ≤m) has a solution iffai ≡aj (mod gcd(ni, nj)) for all i, j, and the solution (if it exists) is unique modulo lcm(n1, . . . , nm) (this is a generalization of [5, §2.4, Thm. 2.9] that can be easily proved by induction). Vol. 40, No. 1, 2022] 14 Yuki Kiriu & Diego A. Mejía It is easy to check that L′ (4,r) is a subgroup of U ′ (4,r) of half order, so L4r is a subgroup of U4r of half order. Moreover, −1 ∈L4r because {i : −1 ∈U4qi ∖L4qi} is empty by Theorem 3.5 (so it has zero elements). To summarize: Theorem 3.7. Let r ∈Z+ with prime factorization r = q1 · · · qm. Then there is a subgroup L4r of U4r of half order, containing −1, such that for any odd prime p with gcd(p, r) = 1,  r p  = 1 iffp ≡c (mod 4r) for some c ∈L4r. However, it may be that L4r is not the only subgroup of U4r of half order containing −1. For example, consider r = 15: L60 = {±1, ±7, ±11, ±17}, but {±1, ±11, ±19, ±29} is another subgroup of U60 of half order containing −1. To finish this section, we consider negative integers. If r ∈Z+ and p is an odd prime with gcd(r, p) = 1 then −r p  = −1 p  r p  . Since  −1 p  = 1 iffp ≡1 (mod 4),  −r p  can be easily calculated by Theorem 3.6. 4. Preliminaries about modules and fields Throughout this section, we fix an arbitrary integral domain R, r ∈R and a natural number n. We first discuss the ring quotient Rr n := R[x]/(xn −r). It is very common to look at this ring quotient when R is a field and xn −r is irreducible in R[x], in which case Rr n is a field. But in this work we also want to look at the situation when xn −r is reducible in R[x], in which case Rr n is not an integral domain. In any case: Lemma 4.1. The ring Rr n is a free R-module with basis {1, u, . . . , un−1} where u := x (mod (xn −r)), even more Rr n is an R-algebra. Proof. Recall that R[x] satisfies the division algorithm with monic polynomials: for any f(x), g(x) ∈R[x], if g(x) is of the form xm + am−1xm−1 + . . . + a0 (m = 0 is allowed, in which case g(x) = 1) then there are unique q(x), t(x) ∈R[x] such that f(x) = q(x)g(x)+ t(x) and t(x) has degree smaller than g(x). Now, if 0 ̸= f(x) ∈R[x] has degree smaller than n then, by applying the previous division algorithm to g(x) = xn −r, we obtain that f(x) = q(x)g(x) + t(x) for unique q(x) and t(x), the latter with degree smaller than n. Hence q(x) = 0: if q(x) ̸= 0 has degree m ≥0, then q(x)g(x), and thus f(x), have degree n+m, which contradicts that f(x) has degree smaller than n. Therefore t(x) = f(x) ̸= 0, meaning that f(x) is not a multiple of xn −r (otherwise, t(x) = 0 by the division algorithm with monic polynomials). Let R′ be the R-submodule of R[x] generated by {1, x, . . . , xn−1}, which is a free R-module. The previous paragraph shows that the surjective R-module homomorphism R′ →Rr n that sends each xi to ui has kernel equal to the zero ring, so it is an R-module isomorphism. This shows that Rr n is a free R-module with basis {1, u, . . . , un−1}. It is clear that Rr n is an R-algebra. □ ✓ ✓ ✓ [Revista Integración Some notes about power residues modulo prime 15 If xn −r is reducible in R[x] then Rr n is not an integral domain, but it is an integral domain when R is a unique factorization domain and xn −r is irreducible in R[x]. In general, Rr n can be expressed as a ring of matrices Mr n(R) such that the determinant works as the norm of the elements of the ring. Definition 4.2. (1) For ¯ x = (x0, . . . , xn−1) ∈Rn define M r n(¯ x) :=        x0 rxn−1 rxn−2 . . . rx2 rx1 x1 x0 rxn−1 . . . rx3 rx2 . . . . . . . . . ... . . . . . . xn−2 xn−3 xn−4 . . . x0 rxn−1 xn−1 xn−2 xn−3 . . . x1 x0        and denote its determinant by Dr n(¯ x). (2) If z ∈ Rr n we denote M r n(z) := M r n(¯ x) and Dr n(z) := Dr n(¯ x) where ¯ x = (x0, . . . , xn−1) ∈Rn is the unique tuple such that z = Pn−1 i=0 xiui. (3) Define Mr n(R) := {M r n(¯ x) : ¯ x ∈Rn}. When R is understood from the context we just write Mr n. These matrices actually describe the shift endomorphisms in Rr n: Lemma 4.3. If z ∈Rr n then the matrix M r n(z) characterizes the endomorphism Rr n →Rr n given by w 7→zw. Concretely, M r n(z) is the unique matrix with the following property: if w = Pn−1 i=0 xiui for some ¯ x ∈Rn, then zw = Pn−1 i=0 yiui where ¯ y = M r n(z)¯ x. As a consequence Mr n is a subring of the ring of n × n matrices with entries in R, even more, Mr n is commutative and so it is an R-algebra. In fact, it characterizes Rr n. Lemma 4.4. The function M r n : Rr n →Mr n is an R-algebra isomorphism, and the map Dr n : Rr n →R satisfies Dr n(zz′) = Dr n(z)Dr n(z′) for any z, z′ ∈Rr n. The function Dr n has the role of a norm for Rr n. In fact, when F is a field and xn −r is irreducible in F[x], F r n is a field and Dr n is its norm as an F-extension. We list the exact form of some few Dr n(¯ x) with ¯ x ∈Rn: Dr 2(¯ x) =x0 2 −x1 2r; Dr 3(¯ x) =x0 3 + x1 3r + x2 3r2 −3x0x1x2r; Dr 4(¯ x) =x0 4 −x1 4r + 4x0x1 2x2r −2x0 2x2 2r −4x0 2x1x3r + x2 4r2 −4x1x2 2x3r2+ 2x1 2x3 2r2 + 4x0x2x3 2r2 −x3 4r3. We can also talk about conjugates in Rr n. In field extensions like Q(i) and Q( √ 2), the conjugate ¯ z of some element z satisfies that z¯ z is the norm of z. In the general case we can look at the matrix characterization: for any matrix A of dimensions n × n (with entries in R), A · adj(A) = |A|In where In is the identity matrix of dimensions n × n, adj(A) is the adjugate of A and |A| is the determinant of A. Since the determinant acts as Vol. 40, No. 1, 2022] 16 Yuki Kiriu & Diego A. Mejía a norm, then adj(A) works as the (analog of the) conjugate of A. Recall that the matrix A is invertible if there is some unique matrix A−1 of dimensions n × n, with entries in R, such that AA−1 = A−1A = In. Recall that A is invertible iff|A| is a unit in R, in which case A−1 = |A|−1adj(A). In Mr n(R) we obtain: Lemma 4.5. If A ∈Mr n(R) then adj(A) ∈Mr n(R). In particular, if A ∈Mr n(R) is invertible (as a matrix) then A−1 ∈Mr n(R). Proof. An analog of the Caley-Hamilton Theorem indicates that (−1)n−1adj(A) = An−1 + cn−1An−2 + · · · + c1In where cn−1, . . . , c0 ∈R and λn + cn−1λn−1 + · · · + c0 is the characteristic polynomial of A. If A ∈Mr n then (−1)n−1adj(A) ∈Mr n by the expression above, so adj(A) ∈Mr n. In particular, when A is invertible, A−1 = |A|−1adj(A) ∈Mr n. We also present an elementary proof in the case when A ∈Mr n(R) is invertible as a matrix with entries in F, where F is the field of fractions of R. Choose z ∈R such that A = M r n(z). Since A is invertible, by Lemma 4.3 the map w 7→zw is an automorphism on F r n, so there is some z′ ∈F such that zz′ = 1, hence w 7→z′w is the inverse of the previous map. Therefore A−1 = M r n(z′) ∈Mr n(F), which implies that adj(A) = |A|A−1 ∈Mr n(F). But adj(A) is a matrix with entries in R, so adj(A) ∈Mr n(R). □ ✓ ✓ ✓ Now that we know a bit more about the structure of Rr n, we now look at sufficient and necessary conditions for the polynomial xn −r to be irreducible. Lemma 4.6. If xn −r is irreducible in R[x] then: whenever q | n is prime, xq −r = 0 does not have a solution in R. Proof. Assume that q | n is prime and xq −r = 0 has a solution v in R, that is, vq = r in R. Then, in R[x], xn −r = xq n q −vq = (x n q −v)(x n q (q−1) + . . . + vq−1), so xn −r is reducible. □ ✓ ✓ ✓ We will prove the converse in some cases of interest by using the following result. From now on, fix a field F and r ∈F. Theorem 4.7 (See [4, Ch. VI §9]). The polynomial xn −r is irreducible in F[x] iffthe following two conditions hold. (i) If q | n is prime then the equation xq −r = 0 does not have a solution in F. (ii) If 4 | n then the equation 4x4 + r = 0 does not have a solution in F. Proof. The cited reference states and proves that (i) and (ii) implies that xn −r is irreducible in F[x]. The converse implication is true for any ring R and it is easy to prove. Assume that r ∈R. Lemma 4.6 shows that xn −r irreducible in R[x] implies (i). To show that (ii) is also implied we prove that, whenever 4 | n and 4u4 + r = 0 for some u ∈R, xn −r is reducible in R[x]. Since n = 4k for some k ≥1, we get xn −r = (xk)4 + 4u4 = ((x2)k −2uxk + 2u2)((x2)k + 2uxk + 2u2). □ ✓ ✓ ✓ [Revista Integración Some notes about power residues modulo prime 17 Corollary 4.8. Let q be a prime and let F be a field. Then xq −r = 0 does not have a solution in F iffxq −r is irreducible in F[x]. Condition (ii) can be suppressed when we look at fields of prime characteristic. Theorem 4.9. Let p be a prime and assume that 4 ∤n or 4 | p −1 or p = 2. If F has characteristic p then xn −r is irreducible in F[x] iff, for any prime q | n, xq −r = 0 does not have a solution in F. Proof. We showed one direction in Lemma 4.6. To see the converse, assume that, for any prime q | n, xq −r = 0 does not have a solution in F, which means that (i) of Theorem 4.7 is valid. By using the same theorem, it is enough to show that (ii) holds, that is, the equation 4x4 + r = 0 does not have a solution in F when 4 | n. Assume that 4 | n, so either 4 | p −1 or p = 2 by hypothesis. In the case 4 | p −1 assume towards a contradiction that 4x4 + r = 0 has a solution x0 ∈F. So −r = 4x4 0 = (2x2 0)2. Let y0 := 2x2 0, so y2 0 = −r. On the other hand, by properties of the Legendre symbol, −1 p  = (−1) p−1 2 = 1 (because 4 | p −1), which means that −1 ≡z2 0 (mod p) for some z0 ∈Fp. Hence, r = (−r)(−1) = (y0z0)2, that is, the equation x2 −r = 0 has a solution in F, but this is not true by hypothesis: since 2 is prime and 2 | n, x2 −r = 0 does not have a solution in F. In the case p = 2 we have 4x4 + r = r. If 4x4 + r = 0 has a solution in F then r = 0, but 4 | n so the hypothesis says that the equation x2 = 0 does not have a solution in F, which is absurd. □ ✓ ✓ ✓ Corollary 4.10. Let p be a prime and assume that n | p −1. If F has characteristic p then xn −r is irreducible in F[x] iff, for any prime q | n, xq −r = 0 does not have a solution in F. Proof. Immediate by Theorem 4.9 because 4 | n implies 4 | p −1 when p is odd. □ ✓ ✓ ✓ In some cases, we can also characterize irreducibility of xn −r in Q[x]. Theorem 4.11. Let n be a natural number. If r ∈Q and r > 0 then xn −r is irreducible in Q[x] iffxq −r = 0 does not have a solution in Q for any prime q | n. Proof. This is a direct consequence of Theorem 4.7 since condition (ii) there is always satisfied. □ ✓ ✓ ✓ The previous result actually applies to any ordered field. To finish this section, we show that irreducible in Fp[x] is stronger than irreducible in Q[x] when r ∈Z. Corollary 4.12. Let p be a prime, r ∈Z and n ∈Z+. If r ≡r0 (mod p) and xn −r0 is irreducible in Fp[x] then xn −r is irreducible in Q[x]. Vol. 40, No. 1, 2022] 18 Yuki Kiriu & Diego A. Mejía Proof. Assume that xn −r0 is irreducible in Fp[x]. We first prove that xq −r = 0 does not have a solution in Q for any prime q | n. Using Lemma 4.6 with R = Fp, we know that xq −r0 = 0 does not have a solution in Fp for any prime q | n, which implies that the equation xq −r = 0 does not have a solution in Z, so neither in Q: if a, b ∈Z are relative prime, b > 0, and a b q −r = 0, then aq = rbq, which implies that b = 1 (if b > 1 then r = 0, so a = 0 and, since gcd(a, b) = 1, b = 1, contradiction), thus xq −r has a solution in Z. In the case r > 0 the result follows by Theorem 4.11; in the case n ∤4, the result follows by Theorem 4.7; and when r = 0, we must have n = 1 (because we assumed xn −r0 irreducible in Fp[x]) and then xn −r = x is irreducible in Q. So it remains to consider the case when r < 0 and n | 4. Here it remains to show that (ii) of Theorem 4.7 holds for F = Q. Towards a contradiction, assume that 4a4 + r = 0 for some a ∈Q. Since r ∈Z and a4 = −r 4 , we must have that a ∈Z. Therefore, modulo p we get that 4x4 + r0 = 0 has a solution in Fp, but this contradicts (ii) of Theorem 4.7 for xn −r0 in Fp[x]. □ ✓ ✓ ✓ 5. Power residues In this section we show the main results concerning power residues. We start with Theorem B. Theorem 5.1. Let p be a prime, n ∈Z+, r ∈Z and let r0 ∈Fp such that r ≡r0 (mod p). (a) The polynomial xn −r0 is irreducible in Fp[x] iffthe equation Dr n(x0, . . . , xn−1) ≡0 (mod p) does not have a non-trivial solution in the integers. (b) If xn−r is reducible in Q[x] then Dr n(¯ x) = 0 has a non-trivial solution in the integers. (c) If n ≥2 and the equation xn ≡r (mod p) has a solution, then Dr n(x0, . . . , xn−1) ≡0 (mod p) has a non-trivial solution in the integers. Even more, this solution satisfies −p 1 n < xi < p 1 n for all 0 ≤i < n. Proof. Set F := Fp. We first show (a). Assume that xn −r0 is irreducible in F[x]. Then F r0 n = F(u) is a field extension of F with u := n √r0, which is isomorphic to Mr0 n (F) by Lemma 4.4. Let ¯ x = (x0, . . . , xn−1) ̸= (0, . . . , 0) with xi ∈Fp (0 ≤i < n), and set A := M r0 n (¯ x). By Lemma 4.5 A−1 ∈Mr0 n , so Dr0 n (¯ x) ̸= 0 in Fp, that is, Dr n(¯ x) ̸≡0 (mod p). For the converse, assume that xn −r0 is reducible in F[x]. Then F r0 n is not an integral domain, so there are non-zero z, w ∈F r0 n such that zw = 0. Then, by Lemma 4.4, Dr n(z)Dr n(w) ≡0 (mod p), so either Dr n(z) ≡0 (mod p) or Dr n(w) ≡0 (mod p). To see (b): if xn −r is reducible in Q[x] then there are non-zero z, w ∈Qr n such that zw = 0. Even more, we can find non-zero vectors ¯ x, ¯ y ∈Zn such that z′w′ = 0 where z′ = Pn−1 i=0 xiui and w′ = Pn−1 i=0 yiui (here u determines the basis of Qr n as a Q-vector space). Therefore Dr n(¯ x)Dr n(¯ y) = 0, so Dr n(¯ x) = 0 or Dr n(¯ y) = 0. Now we show (c). Assume that xn ≡r (mod p) has a solution t, that is, tn ≡r (mod p). [Revista Integración Some notes about power residues modulo prime 19 Consider the set S := {x ∈Z : 0 ≤x < p 1 n } and let Sn := {(x0, . . . , xn−1) : xi ∈S (0 ≤i < n)}. Note that Sn has more than p elements (because n ≥2). Now define the function f : Sn →Fp by f(x0, . . . , xn−1) ≡x0 + x1t + · · · + xn−1tn−1 (mod p). Since Fp has p many elements, Sn has more elements than Fp, so by the pigeonhole princi-ple there are two (m0, . . . , mn−1) ̸= (m′ 0, . . . , m′ n−1) in Sn such that f(m0, . . . , mn−1) = f(m′ 0, . . . , m′ n−1). For 0 ≤i < n let ai := m′ i −mi, so f(a0, . . . , an−1) ≡f(m′ 0, . . . , m′ n−1) −f(m0, . . . , mn−1) ≡0 (mod p), ¯ a := (a0, . . . , an−1) ̸= (0, . . . , 0) and −p 1 n < ai < p 1 n , We show that ¯ a is as desired. We proceed in a similar way as in the proof of (a) first assuming that xn −r is irreducible in Q[x]. Then K := Qr n = Q(v) is a field extension of Q with v = n √r, and it is isomorphic to Mr n(Q) by Lemma 4.4. Set A := M r n(¯ a). Since this matrix is not zero, it is invertible, so A−1 ∈Mr n(Q), and even more B := adj(A) ∈Mr n(Z) by Lemma 4.5. So choose ¯ y ∈Zn such that B = M r n(¯ y). Since K is Q[x]/(q(x)) with q(x) := xn −r, we have that A = M r n(g(x) (mod (q(x)))) and B = M r n(h(x) (mod (q(x)))) where g(x) := a0 + a1x + · · · + an−1xn−1, h(x) := y0 + y1x + · · · + yn−1xn−1. Since AB = |A|In, we get that xn −r divides g(x)h(x) −|A| in Q[x], and actually in Z[x] because both polynomials have coefficients in Z and xn −r is monic. Then g(x)h(x) = j(x)q(x) + |A| for some j(x) ∈Z[x]. To finish the proof, note that g(t)h(t)−|A| = (tn −r)j(t) ≡0 (mod p), so g(t)h(t) ≡|A| (mod p). On the other hand, we know that g(t) ≡f(a0, . . . , an−1) ≡0 (mod p) so |A| ≡0 (mod p), that is, Dr n(a0, . . . , an−1) ≡0 (mod p). For the general proof of (c) we work in F r n, which is isomorphic to Mr n(F). Again set A := M r n(¯ a) which is in M r n(F), so B := adj(A) ∈Mr n(F) by Lemma 4.5. Like above, since AB = |A|In we have two polynomials g(x), h(x) ∈F[x], which g(x) as above, such that xn −r divides g(x)h(x) −|A|, so g(x)h(x) = j(x)q(x) + |A| for some j(x) ∈F[x]. Exactly as in the last part of the previous argument, we conclude that Dr n(¯ a) ≡0 (mod p). □ ✓ ✓ ✓ Thanks to the results in Section 4, the previous result takes a simple form when n is a prime. Corollary 5.2. Let p and q be primes. Then the equation xq ≡r (mod p) has a solution iffthe equation Dr q(x0, . . . , xn−1) ≡0 (mod p) has a non-trivial solution. Vol. 40, No. 1, 2022] 20 Yuki Kiriu & Diego A. Mejía Proof. The direction from left to right follows from Theorem 5.1(c). For the converse, if the equation xq ≡r (mod p) does not have a solution then the polynomial xq −r0 is irreducible in Fp[x] by Corollary 4.8 where r0 ∈Fp is the residue of r modulo p, so Dr q(x0, . . . , xn−1) ≡0 (mod p) does not have a non-trivial solution by Theorem 5.1(a). □ ✓ ✓ ✓ The next result is Theorem C, which is a weakening of (2) ⇒(1) of Problem 1.9. This actually checks this implication when n is a prime (for any r ∈Z). Theorem 5.3. Assume that p is a prime, r ∈Z, r ≡r0 (mod p) with r0 ∈Fp and n ≥2. If the polynomial xn −r0 is irreducible in Fp[x] then Dr n(x0, . . . , xn−1) = p does not have a solution in the integers. In particular, if q is a prime and xq ≡r (mod p) does not have a solution then Dr q(x0, . . . , xq−1) = p does not have a solution in the integers. Proof. By Theorem 5.1, if xn −r0 is irreducible in Fp[x] then Dr n(x0, . . . , xn−1) ≡0 (mod p) does not have a non-trivial solution. Thus, if Dr n(x0, . . . , xn−1) = p has a solution a0, . . . , an−1 ∈Z, then every ai must be a multiple of p. But this implies that Dr n(a0, . . . , an−1) is a multiple of pn, so it cannot be equal to p because n ≥2. □ ✓ ✓ ✓ We can use Theorem 5.1 to solve Problem 1.9 for n = 2, i.e., Theorem 1.8. In fact, this is valid for −1 and −2 in the place of 2, which yield well known results. Theorem 5.4. Let r ∈{−2, −1, 2}. If p is a prime then the equation x2 ≡r (mod p) has a solution iffthe equation Dr 2(x0, x1) = p has a solution in the integers. Proof. One implication follows by Theorem 5.3 because 2 is prime. So we show that, whenever x2 ≡r (mod p) has a solution, the equation Dr 2(x0, x1) = p has a solution in the integers, for r ∈{−2, −1, 2}. By Corollary 5.2, the equation Dr 2(x0, x1) ≡0 (mod p) has a non-trivial solution (a, b). Hence p divides Dr 2(a, b) = a2 −b2r. According to Theorem 5.1(c), we can find a and b between −p 1 2 and p 1 2 . Case r = 2. We claim that −2p < a2 −2b2 < p. Two cases: if a2 ≥2b2 then 0 ≤ a2 −2b2 ≤a2 < p; if a2 < 2b2 then −2p < −2b2 ≤a2 −2b2 < 0, so the claim follows. Now, since −2p < D2 2(a, b) = a2 −2b2 < p and p | D2 2(a, b), we must have that D2 2(a, b) = −p (it can not be zero because p must not divide both a and b). Note that D2 2(1, 1) = 12 −2 · 12 = −1, so p = a 2b b a · 1 2 1 1 = a + 2b 2(a + b) a + b a + 2b Hence x0 := a + 2b and x1 = a + b form an integer solution of D2 2(x0, x1) = p. Case r = −1. It is clear that 0 < a2 + b2 < 2p, so a2 + b2 = p. Case r = −2. Note that 0 < a2 + 2b2 < 3p, so either a2 + 2b2 = p or a2 + 2b2 = 2p. In the first case we are done; in the second case a must be even, so a = 2a0 for some a0 ∈Z, and 2p = a2 + 2b2 = 4a2 0 + 2b2, hence D−2 2 (b, a0) = p. □ ✓ ✓ ✓ [Revista Integración Some notes about power residues modulo prime 21 6. Discussions Problem 1.9 cannot be generalized by simply replacing 2 by any r ∈Z. For n = 2, it is known it is fine for r ∈{−2, −1, 2} as shown in Theorem 5.4, but other values of r are problematic. For example, 3y2 + p is never a square when p ≡3 (mod 4) (because it is 3 or 2 modulo 4), so D3 2(x, y) = p does not have a solution for those p. However, there are primes p ≡3 (mod 4) such that x2 ≡3 (mod p) has a solution, for example, p = 11. In this case, it could be conjectured that the equation D3 2(x, y) = p has a solution iffx2 ≡3 (mod p) has a solution and p ≡1 (mod 4). This motivates: Problem 6.1. For n ≥2 (particularly n = 2) and r ∈Z (or just free of n-powers), what are suitable necessary and sufficient conditions for a prime p to get that Dr n(¯ x) = p has a solution in the integers? As discussed in the introduction, the solution of Problem 1.9 should be related to the characterization of primes (or irreducible) elements in Z[ n √ 2], which looks very complex for general values of n. In the post it is hinted that Problem 1.9 is true for n = 3 by looking at Z[ 3 √ 2] with tools that we did not deal with in this paper. Some results of Section 5 can be generalized when xn −r is replaced by any monic polynomial in Z[x]. If R is an integral domain and q(x) ∈R[x] is a monic polynomial of degree n > 0, the theory in the first part of Section 4 can be generalized in the context of Rq(x) := R[x]/(q(x)): (I) Rq(x) is a free R-module (and an R-algebra) with basis {1, u, . . . , un−1} where u := x (mod (q(x))) (II) For any z ∈Rq(x) there is a unique matrix Mq(x)(z) that characterizes the endo-morphism Rq(x) →Rq(x), w 7→zw as in Lemma 4.3. (III) Set Mq(x) := Mq(x)(R) = {Mq(x)(z) : z ∈Rq(x)}. The function Mq(x) : Rq(x) → Mq(x) is an R-algebra isomorphism. (IV) For any z ∈Rq(x) set Dq(x)(z) := |Mq(x)(z)|. Then, for any z, z′ ∈Rq(x), Dq(x)(zz′) = Dq(x)(z)Dq(x)(z′). When ¯ x = (x0, . . . , xn1) ∈R, denote Dq(x)(¯ x) := Dq(x)(z) where z = Pn−1 i=0 xiui ∈ Rq(x). (V) If A ∈Mq(x)(R) then adj(A) ∈Mq(x)(R). Using this theory, we obtain the following results (with similar proofs as in Section 5). Theorem 6.2. Let p be a prime, q(x) ∈Z[x] a monic polynomial of degree n > 0, and let q0(x) ∈Fp[x] be the polynomial resulting from q(x) by changing its coefficients by their residues modulo p. Then: (1) q0(x) is irreducible in Fp[x] iffthe equation Dq(x)(x0, . . . , xn−1) ≡0 (mod p) does not have a non-trivial solution in the integers. Vol. 40, No. 1, 2022] 22 Yuki Kiriu & Diego A. Mejía (2) If q(x) is reducible in Q[x] then the equation Dq(x)(¯ x) = 0 has a non-trivial solution in the integers. (3) If n ≥2 and the equation q0(x) ≡0 (mod p) has a solution then the equation Dq(x)(x0, . . . , xn−1) ≡0 (mod p) has a non-trivial solution in the integers with −p 1 n < xi < p 1 n for any i. (4) If n ≥2 and q0(x) is irreducible in Fp[x] then the equation Dq(x)(¯ x) = p does not have a solution in the integers. As a digression, the equation D2 3(x0, x1, x2) = p motivates the following. Problem 6.3. Assume that a, b, c ∈{1, 2, 3} and that p is a prime. Does the equation xa + 2yb + 4zc = p have a solution in the integers? Primes p where a solution was not found (a, b, c) with FindInstance among the first 1000 primes (2, 3, 3) 2069, 5303, 6101 (3, 2, 3) 2207, 2383 (3, 3, 2) 2039, 2083, 3371, 4027, 6143, 6997, 7699 (3, 3, 3) 4079, 4091, 6449, 7507 Table 1. Instances among the first 1000 primes where a solution of xa + 2yb + 4zc = p was not found in Wolfram Mathematica with the command FindInstance, in the case when at least two of a, b, c are equal to 3. First four primes p where a (a, b, c) solution was not found with FindInstance (2, 2, 3) 22691, 25903, 27191, 27241 (2, 3, 2) 37571, 39191, 41263, 44357 (3, 2, 2) 24907, 51043, 51637, 53717 Table 2. First four prime p instances where a solution of xa + 2yb + 4zc = p was not found in Wolfram Mathematica with the command FindInstance, in the case when only one of a, b, c is equal to 3. For any p ∈Z (not necessarily prime): it is easy to find a solution when either a, b or c is equal to 1; and the case a = b = c = 2 has a positive answer, as mentioned in [1, §13.3, Prob. 8(a)]. So this leaves the case 2 ≤min{a, b, c} ≤max{a, b, c} = 3. By running computations in Wolfram Mathematica with the command FindInstance (see below), a solution was not found for some primes in all the subcases (but this is not a proof that the solution does not exist). FindInstance[x^a+2y^b+4z^c==p,{x,y,z},Integers] [Revista Integración Some notes about power residues modulo prime 23 See details in Tables 1 and 2: in Table 1 we look at the case when at least two of a, b, c are equal to 3, where solutions were not found for some primes below 10000; in Table 2 we look at the case when only one of a, b, c is equal to 3, where solutions were not found for some primes beyond 20000. Acknowledgements This work was supported by: Future Scientists School at Shizuoka University, Global Science Campus supported by the Japan Science and Technology Agency (both authors); Grant-in-Aid for Early Career Scientists 18K13448, Japan Society for the Promotion of Science (second author). We would like to thank the anonymous referee for carefully reading the paper and for pointing out mistakes and unclear parts, which helped to improve the presentation. References Burton D.M., Elementary Number Theory, McGraw Hill Education (India) Pvt Ltd, 7th Indian ed., New Delhi, 2012. Hardy G.H. and Wright E.M., An introduction to the theory of numbers, Oxford University Press, 6th ed., Oxford, 2008. Ireland K. and Rosen M., A classical introduction to modern number theory from series Graduate Texts in Mathematics, Springer-Verlag, 2nd ed., vol. 84, New York, 1990. doi: 10.1007/978-1-4757-2103-4 Lang S., Algebra from series Graduate Texts in Mathematics, Springer-Verlag, 3rd ed., vol. 211, New York, 2002. doi: 10.1007/978-1-4613-0041-0 Nathanson M.B., Elementary Methods in Number Theory from Graduate Texts in Mathe-matics, Springer-Verlag, 1st ed., vol. 195, New York, 2000. doi: 10.1007/b98870 Pomerance C., “The multiplicative order mod n, on average”, Quebec/Maine number theory conference at Laval University, [cited on 18 march, 2021]. Silverman J.H., “Wieferich’s criterion and the abc-conjecture”, J. Number Theory, 30 (1988), No. 2, 226-237. doi: 10.1016/0022-314X(88)90019-4 Takagi T., Elementary Number Theory Lectures, Kyoritsu Shuppan, 2nd ed., Tokyo, 1971. “What is known about primes of the form x2 −2y2?”, MathOverflow. [cited on 18 march, 2021]. “What about Z[ n √ 2]?”, Mathematics StackExchange. [cited on 18 march, 2021]. Vol. 40, No. 1, 2022]
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https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1201/Unit_2._Periodic_Properties_of_the_Elements/2.07%3A_Magnetic_Properties_of_Atoms_and_Ions
Skip to main content 2.7: Magnetic Properties of Atoms and Ions Last updated : May 8, 2021 Save as PDF 2.6: Slater's Rules 2.8: Sizes of Atoms and Ions Page ID : 36523 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives To understand the difference between paramagnetism and diamagnetism The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic forms: including paramagnetism, and diamagnetism, ferromagnetism, and anti-ferromagnetism. Paramagnetism Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. Hund's Rule states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can spin in either direction, they display magnetic moments in any direction. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, O2 is a good example of paramagnetism (described via molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet: Figure 2.7.1: As shown in the video, molecular oxygen (O2 is paramagnetic and is attracted to the magnet. Incontrast, Molecular nitrogen, N2, however, has no unpaired electrons and it is diamagnetic (this concept is discussed below); it is therefore unaffected by the magnet. There are some exceptions to the paramagnetism rule; these concern some transition metals, in which the unpaired electron is not in a d-orbital. Examples of these metals include Sc3+, Ti4+, Zn2+, and Cu+. These metals are the not defined as paramagnetic: they are considered diamagnetic because all d-electrons are paired. Paramagnetic compounds sometimes display bulk magnetic properties due to the clustering of the metal atoms. This phenomenon is known as ferromagnetism, but this property is not discussed here. Diamagnetism Diamagnetic substances are characterized by paired electrons—except in the previously-discussed case of transition metals, there are no unpaired electrons. According to the Pauli Exclusion Principle which states that no two identical electrons may take up the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly repelled by a magnetic field as demonstrated with the pyrolytic carbon sheet in Figure 2.7.2. Figure 2.7.2: Levitating pyrolytic carbon: A small (~6mm) piece of pyrolytic graphite levitating over a permanent neodymium magnet array (5mm cubes on a piece of steel). Note that the poles of the magnets are aligned vertically and alternate (two with north facing up, and two with south facing up, diagonally). from Wikipedia. How to tell if a substance is paramagnetic or diamagnetic The magnetic form of a substance can be determined by examining its electron configuration: if it shows unpaired electrons, then the substance is paramagnetic; if all electrons are paired, the substance is diamagnetic. This process can be broken into four steps: Find the electron configuration Draw the valence orbitals Look for unpaired electrons Determine whether the substance is paramagnetic or diamagnetic Example 2.7.1: Chlorine Atoms Step 1: Find the electron configuration For Cl atoms, the electron configuration is 3s23p5 Step 2: Draw the valence orbitals Ignore the core electrons and focus on the valence electrons only. Step 3: Look for unpaired electrons There is one unpaired electron. Step 4: Determine whether the substance is paramagnetic or diamagnetic Since there is an unpaired electron, Cl atoms are paramagnetic (albeit, weakly). Example 2.7.2: Zinc Atoms Step 1: Find the electron configuration For Zn atoms, the electron configuration is 4s23d10 Step 2: Draw the valence orbitals Step 3: Look for unpaired electrons There are no unpaired electrons. Step 4: Determine whether the substance is paramagnetic or diamagnetic Because there are no unpaired electrons, Zn atoms are diamagnetic. Exercise 2.7.1 How many unpaired electrons are found in oxygen atoms ? How many unpaired electrons are found in bromine atoms? Indicate whether boron atoms are paramagnetic or diamagnetic. Indicate whether F- ions are paramagnetic or diamagnetic. Indicate whether Fe2+ ions are paramagnetic or diamagnetic. Answer (a): : The O atom has 2s22p4 as the electron configuration. Therefore, O has 2 unpaired electrons. Answer (b): : The Br atom has 4s23d104p5 as the electron configuration. Therefore, Br has 1 unpaired electron. Answer (c): : The B atom has 2s22p1 as the electron configuration. Because it has one unpaired electron, it is paramagnetic. Answer (d): : The F- ion has 2s22p6 has the electron configuration. Because it has no unpaired electrons, it is diamagnetic. Answer (e): : The Fe2+ ion has 3d6 has the electron configuration. Because it has 4 unpaired electrons, it is paramagnetic. References Pettrucci, Ralph H. General Chemistry: Principles and Modern Applications. 9th. Upper Saddle River: Pearson Prentice Hall, 2007 Sherman, Alan, Sharon J. Sherman, and Leonard Russikoff. Basic Concepts of Chemistry Fifth Edition. Boston, MA: Houghton Mifflin Company, 1992. Print. Outside Links superphysics.netfirms.com/ pp_magnetism.html www.transtutors.com/chemistry...-elements.aspx 2.6: Slater's Rules 2.8: Sizes of Atoms and Ions
188263
https://artofproblemsolving.com/wiki/index.php/Parity?srsltid=AfmBOoo41xBuYPpTRDE9qpbaspGadDL0vPaEqLt4UKobSrQhghSfpqdn
Art of Problem Solving Parity - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Parity Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Parity Parity refers to whether a number is even or odd. While this may seem highly basic, checking the parity of numbers is often an useful tactic for solving problems, especially with proof by contradictions and casework. This concept begins with integers. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without remainder; an odd number is an integer that is not evenly divisible by 2. (The old-fashioned term "evenly divisible" is now almost always shortened to "divisible".) A formal definition of an even number is that it is an integer of the form n = 2k, where k is an integer; it can then be shown that an odd number is an integer of the form n = 2k + 1. This only applies to integers, not fractions or decimals. Contents 1 Problems 2 Solution 1 3 Solution 2 3.1 Introductory 3.2 Intermediate 3.2.1 Problem 3.2.2 Solution 3.2.3 Problem 2 3.2.4 Solution Problems Example from 1997 AJHSME: Problem: Ten balls numbered to are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is Solution 1 For the sum of the two numbers removed to be even, they must be of the same parity. There are five even values and five odd values. No matter what Jack chooses, the number of numbers with the same parity is four. There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack's is . Solution 2 We find that it is only possible for the sum to be even if the numbers added are both even or odd. We will get an odd number when we add an even and odd. We can use complementary counting to help solve the problem. There are a total of possibilities since Jack can choose numbers and Jill can pick . There are possibilities for the two numbers to be different since Jack can pick any of the numbers and Jill has to pick from numbers in the set with a different parity than the one that Jack picks. So the probability that the sum will be odd is . Subtracting this by one gets the answer (edited by qkddud) Introductory Many AMC 8 problems fit this category, help us out by putting problems here! Intermediate 2000 AIME II Problem 2: Problem A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola ? Solution Note that and have the same parities, so both must be even. We first give a factor of to both and . We have left. Since there are factors of , and since both and can be negative, this gives us lattice points. 2008 AIME I Problem 3 Problem 2 There exist unique positive integers and that satisfy the equation . Find . Solution Completing the square, . Thus by difference of squares. Since is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since , the factors must be and . Since , we have and ; the latter equation implies that . Indeed, by solving, we find is the unique solution. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
188264
https://memodiction.com/word/Meander/
Meander Memodiction What's MemoDiction CategoriesGREGRE_SET_1TOEFLEnglishGRE_SET_2 Search 🌙☀️ ✕ Meander Verb Definition: To follow a winding and intricate course; to wander aimlessly or casually without a fixed direction. Synonyms: Wander, Roam, Ramble Antonyms: Advance, Progress, Navigate Examples: We decided to meander through the park and explore its hidden paths. The river meandered lazily through the valley, creating beautiful curves and bends. Mnemonics: Imagine yourself taking a leisurely walk without a set path, allowing yourself to meander through different areas. Etymology: Origin: The word "meander" originates from the Greek word "maiandros," referring to the Meander River in Asia Minor, which was known for its winding course. Historical Usage: The word has been used in English since the 16th century to describe both physical and metaphorical winding journeys or courses. Related Idioms: Idiom: "Meandering thoughts" Explanation: Describing thoughts that wander or lack a focused direction. Misconceptions: Misconception: Confusing "meander" with "wander" or "ramble." Explanation: While all three words involve a sense of aimlessly moving, "meander" specifically implies a winding and intricate course. Share: TwitterFacebookLinkedIn Email me Memodiction • 2025 •memodiction.com Powered by BinaryDecimal × Help Us Improve MemoDiction! 🎯 We're working hard to make your vocabulary learning experience even better Your feedback is invaluable! Share your thoughts and help us build features that truly matter to you. 📝 Take Quick Survey (2 min) ⏰ Remind me later 🔒 Your responses help us prioritize improvements
188265
https://www.reddit.com/r/lumetrium_definer/comments/1jfq46u/realworld_pronunciation_examples_from_youglish_in/
Real-world pronunciation examples from YouGlish in Definer popup dictionary extension for language learning : r/lumetrium_definer Skip to main contentReal-world pronunciation examples from YouGlish in Definer popup dictionary extension for language learning : r/lumetrium_definer Open menu Open navigationGo to Reddit Home r/lumetrium_definer A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to lumetrium_definer r/lumetrium_definer r/lumetrium_definer This community is devoted to the discussion of browser extension "Definer - Popup Translator and Dictionary". How to use it, tips and tricks, interesting ideas and opinions, news and updates. 280 Members Online •6 mo. ago DeLaRoka Real-world pronunciation examples from YouGlish in Definer popup dictionary extension for language learning Tutorial YouGlish is a video pronunciation dictionary that helps you improve your pronunciation and listening skills in foreign languages. It searches YouTube for videos where native speakers naturally use your target words, playing just the relevant fragments so you can hear authentic pronunciations in real contexts. How to Use YouGlish in Definer Select a word or phrase on a page in your browser Wait for Definer's bubble to pop up Click the YouGlish icon in the sidebar You'll instantly see results for your selected text: Sorry, something went wrong when loading this video. View in app Demo Customizing YouGlish Source To adjust your settings, go to: Options → Sources → YouGlish → Settings. Language By default, Definer uses whatever language you've selected in the bubble's language picker (the "Match selected" option). Want a specific language instead? Simply choose it from the dropdown. Dialect / Accent For many languages, you can select specific regional accents or dialects – perfect when you need to master pronunciation for a particular area. Scenarios Scenarios are automated sequences of actions that YouGlish performs after loading a video. Definer currently offers two scenarios: Default and Triple Repeat & Next. Default Play videos normally without any automatic actions. Triple Repeat & Next Automatically repeats each segment three times before moving to the next video. Perfect for when you're trying to nail down a challenging pronunciation. Sorry, something went wrong when loading this video. View in app Have a great idea for a new scenario? Share it in the comments! Autoplay You can toggle whether videos start automatically: Enabled: Videos begin playing as soon as YouGlish loads Disabled(default): You'll need to click play to start Sorry, something went wrong when loading this video. View in app Example of autoplay DISABLED. Click the play button to start the video. Daily Search Limits YouGlish limits free accounts to 20 searches per day. After that, you'll need to wait until tomorrow for more searches. To increase or remove this limit, you'll need aYouGlish Subscription. Note that Definer doesn't control this limit – it's set by YouGlish. Searches only count toward your daily limit when you specifically open the YouGlish source in Definer. Read more Share Related Answers Section Related Answers Uses of popup translators in daily browsing Integrating AI in dictionary extensions User interface design for language apps Offline capabilities for popup translators Extending Definer with custom dictionaries New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
188266
https://pt.scribd.com/document/605263653/Blood-Anatomy-Physiology-Handouts
Blood Anatomy Physiology Handouts | PDF | Blood Type | White Blood Cell Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 166 views 6 pages Blood Anatomy Physiology Handouts Blood has several functions: 1. It transports gases, nutrients, waste, hormones, and more between tissues and organs. 2. It helps form clots to prevent blood loss from injuries. 3. It prote… Full description Uploaded by Kids Jang AI-enhanced title and description Go to previous items Go to next items Download Save Save Blood-Anatomy-Physiology-Handouts For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Blood-Anatomy-Physiology-Handouts For Later You are on page 1/ 6 Search Fullscreen LOOD ANATOMY & PHYSIOLOGY Functions of the Blood Blood is unique; it is the only fluid tissue in the body. 1. Carrier of gases, nutrients, and waste products. Oxygen enters blood in the lungs and is transported to cells. Carbon dioxide, produced by cells, is transported in the blood to the lungs, from which it is expelled. Ingested nutrients, ions, and water are carried by th e blood from the digestive tract to cells, and the waste products of the cells are moved to the kidneys for elimination. 2. Clot formation. Clotting proteins help stem blood loss when a blood vessel is injured. Transport of processed molecules. Most substances are produced in one part of the body and transported in the blood to another part. 4. Protection against foreign substances. Antibodies help protect the body from pathogens. 5. Transport of regulatory molecules. Various hormones and enzymes that regulate body processes are carried from one part of the body to another within the blood. 6. Maintenance of body temperature. Warm blood is transported from the inside to the surface of the body, where heat is released from the blood. 7. pH and osmosis regulation. Albumin is also an important blood buffer and contributes to the osmotic pressure of blood, which acts to keep water in the blood stream. Components of Blood Essentially, blood is a complex connective tissue in which living blood cells, the formed elements, are suspended. Physical Characteristics and Volume: Blood is a sticky, opaque fluid with a characteristic metallic taste.  Color. Depend ing on the amount of oxygen it is carrying, the color of blood v aries from scarlet red (oxygen-rich) to a dull red (oxygen-poor).  Weight. Blood is heavier than water and about f ive times thicker, or more viscous, largely because of its formed elements.  pH. Blood is slightly alkaline, with a pH between 7.35 and 7.45 .  Temperature. Its temperature ( 38 degrees Celsius , or 100.4 degrees Fahrenhei t) is always slightly higher than body temperature. PLASMA Plasma, which is approximately 90 percent water, is the liquid part of the blood.  Dissolved substances. Examples of dissolved substances include nutr ients, salts (electrolytes), respiratory gases, hormones, plasma proteins, and various wastes and products of cell metabolism.  Plasma proteins. Plasma proteins are the most abundant solutes in plasma; except for antibodies and protein-based hormones, most plasma proteins are made by the liver.  Composition. The composition of plasma varies continuously as cells remove or add substances to the blood; assuming a healthy diet, however, the composition of plasma is kept relatively constant by various homeostatic mechanisms of the body. adDownload to read ad-free FORMED ELEMENTS If you observe a stained smear of human blood under a light microscope, you will see disc-shaped red blood cells, a variety of gaudily stained spherical white blood cells, and some scattered platelets that look like debris. ERYTHROCYTES Erythrocytes, or red blood cells, function primarily to ferry oxygen in blood to all cells of the body.  Anucleate. RBCs differ from other blood cells because they are anucleate, that is, they lack a nucleus; they also contain a very f ew organelles.  Hemoglobin. Hemoglobin, an iron bearing protein, transports the bulk of oxygen that is carried in the blood.  Microscopic appearance. Erythrocytes are small, flexible cells shaped like biconcave discs- flattened discs with depressed centers on both sides; they look like miniature doughnuts when viewed with a microscope.  Number of RBCs. There are normally about 5 million cells per cubic millimeter of blood; RBCs outnumber WBCs by about 1000 to 1 and are t he major factor contributing to blood viscosity.  Normal blood. Clinically, normal blood contains 12-18 grams of hemoglobin per 100 milliliters (ml); the hemoglobin content is slightly higher in men (13-18 g/dl) than in women (12-16 g/dl). LEUKOCYTES Although leukocy tes, or white blood cells, are far less num erous than red blood ce lls, they are crucial to body defense against disease.  Number of WBCs. On average, there are 4,000 to 11,000 WBC/mm3 , and they account for less than 1 percent of total body volume.  Body defense. Leukocytes form a protective, movable army that helps defend the body against damage by bacteria, viruses, parasites, and tumor cells.  Diapedesis. W hite blood cells are able to slip into and out of t he blood vessels- a process called diapedesis. adDownload to read ad-free  Positive chemotaxis. In addition, W BCs can locate areas of tissue damage and infection in the body by responding to certain chemicals that diffuse from the damaged cells; this capability is called positive chemotaxis.  Ameboid motion. Once they have “caught the scent”, the WBCs move through the tissue spaces by ameboid motion (they form flowing cytoplasmic extensions that help move them along).  Leukocytosis. A total WBC count above 11, 000 cells/mm3 is referred to as leukocyto sis.  Leukopenia. The opposite condition, leukopenia, is an abnormally low W BC count. Granulocytes. Granulocytes are granule-containing WBCs; they have lobed nuclei, which typically consist of several rounded nuclear areas connected by thin strands of nuclear material, and includes neutrophils , eosinophils , and basophils .  Neutrophils. Neutrophil are the most numerous of the WBCs; they have a multilobed granules and very fine granules that respond to acidic and basic stains; neutrophils are avid phagocyte s at sites of acute infection, and are particularly partial to bacteria and fungi.  Eosinophils. Eosinophi ls have blue red nucleus that resembles an old-fashioned telephone receiver and sport coarse, lysosome-like, brick-red cytoplasmic granules; their number increases rapidly during allergies and infections by parasitic worms or entering via the skin.  Basophils. Basophils, the rarest of the W BCs, contain large, histamine-containing granules that stain dark blue; histamine is an inflammatory chemical that makes blood vessels leaky and attracts other WBCs to the inflammatory site. Agranulocytes. The second group of W BCs, the agranulocy tes, lack visible cytoplasmic granules; their nuclei are closer to the norm- that is, they are spherical; they are spherical, oval, or kidney-shaped; and they include lymphocytes and monocytes .  Lymphocytes. Lymphocytes have a large, dark purple nucleus that occupies most of the cell volume; they tend to take up residence in lymphatic tissues, where they play an important role in the immune response.  Monocytes. Monocytes are the largest of the WBCs; when they migrate into the tissues, they transform into macrophages with huge appetites; macrophage s are very important in fighting chronic infections. PLATELETS Platelets are not cells in the strict sense; they are fragments of bizarre multinucleate cells called megakaryocytes , which pinch off thousands of anucleate platelet “pieces” that quickly seal themselves off from surrounding fluids; platelets are needed for the clotting process that occurs in plasma when blood vessels are ruptured or broken. Hematopoiesis Blood cell formation, or hematopoiesis, occurs in red bone marrow, or myeloid tissue.  Hemocystoblast. All the formed elements arise from a common type of stem cell, the hematocystoblast.  Descendants of hemocystoblasts. The hemocystobla st forms two types of descendants- the lymphoid stem cell , which produces lymphocytes, and the myeloid stem cell , which can produce all other classes of formed elements. F orma tion of R ed B loo d C el ls – Because they are anucleate, RBCs are unable to synthesize proteins, grow, or divide.  Life span. As they age, RBCs become more rigid and begin to fragment, or fall apart, in 100 to 120 days .  Lost RBCs. Lost cells are replaced more or less continuously by the division of hemocysto blasts in the red bone marrow. adDownload to read ad-free Immature RBCs. Developing RBCs div ide many times and then begi n synthesiz ing huge amounts of hemoglobin. Reticulocyte. Suddenly, when enough hem oglobin has been accumulated, the nucleus and most organelles are ejected and the cell collapses inward; the result is the young RBC, called a reticulocyte because it still contains some rough endoplasmic reticulum (ER). Mature erythrocytes. Within 2 days of release, they have rejected the remaining ER and have become fully functioning erythrocytes; the entire developmen tal process from hemocystoblast to mature RBC takes 3 to 5 days .  Erythropoietin. The rate of erythrocyte production is contr olled by a hormone called erythropoetin; normally a small amount of erythropoeitin circulates in the blood at all times, and red blood cells are formed at a fairly constant rate.  Control of RBC production. An important point to remember is that it is not the relative number of RBCS in the blood that controls RBC production; control is based on their ability to transport enough oxygen to meet the body’s demands. F orma tion of W hite B loo d C el ls a nd P la te le ts – Like erythrocyte production, the formation of leukocytes and platelets is stimulated by hormones.  Colony stimulating factors and interleukins. These colony stimulating factors and interleukins not only prompt red bone marrow t o turn out leukocyte s, but also marshal up an army of WBCs to ward off attacks by enhancing the ability of mature leukocytes to protect the body.  Thrombopoeitin. The hormone thrombopoeitin accelerates the production of platelets, but little is known about how that process is regulated. Hemostasis The multistep process of hemostasis begins when a blood vessel is da maged and connective tissue in the vessel wall is exposed to blood.  Vascular spasms occur. The immediate response to blood vessel injury is vasoconstriction, which causes that blood vessel to go into spasms; the spasms narrow the blood vessel, decreasing blood loss until clotting can occur.  Platelet plug forms. Injury to the lining of vessels expos es collage fibers; platelets adhere to the damaged site and platelet plug forms.  Coagulation events occur. At the same time, the injured tissues are releasing tissue factor (TF) , a substance that plays an important role in clotting; PF3 , a phospholipid that coats the surfaces of the platelets, interacts with TF, vitamin K, and other blood clotting adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Haematology No ratings yet Haematology 81 pages Circulation No ratings yet Circulation 62 pages Heamatological Disorders No ratings yet Heamatological Disorders 42 pages Blood No ratings yet Blood 37 pages U3.3.1 HAP Body Fluids and Blood No ratings yet U3.3.1 HAP Body Fluids and Blood 70 pages Blood No ratings yet Blood 14 pages Cardiovascular System Overview No ratings yet Cardiovascular System Overview 199 pages Blood Nursing No ratings yet Blood Nursing 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188267
https://math.libretexts.org/Courses/Mount_Royal_University/Calculus_for_Scientists_I/3%3A_Derivatives/3.0%3A_Tangent_lines_and_Rates_of_change
change in distancechange in time="rise''run=average velocity.(3.0.1) t=2 t=2 t=2 t=3 f(3)−f(2)3−2=f(3)−f(2)1=−80 ft/s,(3.0.2) [2,2.5] f(2.5)−f(2)2.5−2=f(2.5)−f(2)0.5=−72 ft/s.(3.0.3) th [2,2.1] f(2.1)−f(2)2.1−2=f(2.1)−f(2)0.1=−65.6 ft/s.(3.0.4) th [2,2.01] f(2.01)−f(2)2.01−2=f(2.01)−f(2)0.01=−64.16 ft/s.(3.0.5) [2,2+h] h f(2+h)−f(2)h(3.0.6) h h=0 0/0 h −64 limh→0f(2+h)−f(2)h=limh→0−16(2+h)2+150−(−16(2)2+150)h=limh→0−64h−16h2h=limh→0−64−16h=−64. h f (2,f(2)) (2+h,f(2+h)) h=1 f f f h→0 (2,f(2)) −64 (2,86) f f(x) with h=1 f at x=2. Skip to main content 3.0: Tangent lines and Rates of change Last updated : Sep 26, 2024 Save as PDF 3: Derivatives 3.0E: Exercises Page ID : 10287 This page is a draft and is under active development. ( \newcommand{\kernel}{\mathrm{null}\,}) Thing out Loud A common amusement park ride lifts riders to a height and then allows them to freefall a certain distance before safely stopping them. Suppose such a ride drops riders from a height of 150 feet. Students of physics may recall that the height (in feet) of the riders, tt seconds after freefall (and ignoring air resistance, etc.), can be accurately modelled by f(t)=−16t2+150f(t)=−16t2+150. Using this formula, it is easy to verify that, without intervention, the riders will hit the ground at t=2.5√1.5≈3.06t=2.51.5−−−√≈3.06 seconds. Suppose the designers of the ride decide to begin slowing the riders' fall after 2 seconds (corresponding to a height of 86 ft.). How fast will the riders be travelling at that time? Answer : We have been given a position function, but we want to compute a velocity at a specific point in time, i.e., we want an instantaneous velocity. We do not currently know how to calculate this. However, we do know from common experience how to calculate an average velocity. (If we travel 60 miles in 2 hours, we know we had an average velocity of 30 mph.) We looked at this concept in Section 1.1 when we introduced the difference quotient. We have change in distancechange in time="rise''run=average velocity. We can approximate the instantaneous velocity at t=2 by considering the average velocity over some time period containing t=2. We will get a good approximation if we make the time interval small. (This fact is commonly used. For instance, high-speed cameras are used to track fast-moving objects. Distances are measured over a fixed number of frames to generate an accurate approximation of the velocity.) Consider the interval from t=2 to t=3 (just before the riders hit the ground). On that interval, the average velocity is f(3)−f(2)3−2=f(3)−f(2)1=−80 ft/s, where the minus sign indicates that the riders are moving down. By narrowing the interval we consider, we will likely get a better approximation of the instantaneous velocity. On [2,2.5] we have f(2.5)−f(2)2.5−2=f(2.5)−f(2)0.5=−72 ft/s. We can do this for smaller and smaller intervals of time. For instance, over a time span of 1/10th of a second, i.e., on [2,2.1], we have f(2.1)−f(2)2.1−2=f(2.1)−f(2)0.1=−65.6 ft/s. Over a time span of 1/100th of a second, on [2,2.01], the average velocity is f(2.01)−f(2)2.01−2=f(2.01)−f(2)0.01=−64.16 ft/s. What we are really computing is the average velocity on the interval [2,2+h] for small values of h. That is, we are computing f(2+h)−f(2)hwhere h is small. What we really want is for h=0, but this, of course, returns the familiar "0/0'' indeterminate form. So we employ a limit, as we did in Section 1.1. We can approximate the value of this limit numerically with small values of h as seen in Figure 2.1. It looks as though the velocity is approaching −64 ft/s. Computing the limit directly gives limh→0f(2+h)−f(2)h=limh→0−16(2+h)2+150−(−16(2)2+150)h=limh→0−64h−16h2h=limh→0−64−16h=−64. Figure 2.1: Approximating the instantaneous velocity with average velocities over a small time period h. Graphically, we can view the average velocities we computed numerically as the slopes of secant lines on the graph of f going through the points (2,f(2)) and (2+h,f(2+h)). In Figure 2.2, the secant line corresponding to h=1 is shown in three contexts. Figure 2.2(a) shows a "zoomed out'' version of f with its secant line. In (b), we zoom in around the points of intersection between f and the secant line. Notice how well this secant line approximates f between those two points -- it is a common practice to approximate functions with straight lines. As h→0, these secant lines approach the tangent line, a line that goes through the point (2,f(2)) with the special slope of −64. In parts (c) and (d) of Figure 2.2, we zoom in around the point (2,86). In (c) we see the secant line, which approximates f well, but not as well the tangent line shown in (d). Figure 2.2: Parts (a), (b) and (c) show the secant line to f(x) with h=1, zoomed in different amounts. Part (d) shows the tangent line to f at x=2. Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have established the foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals. Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton (1643–1727) and the German Gottfried Leibniz (1646–1716), whose images appear in Figure. When we credit Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the first to understand the relationship between the derivative and the integral. Both mathematicians benefited from the work of predecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears to have been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although it seems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notation that we commonly use today. App:Average and Instantaneous Rate of Change Tangent Lines We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of a secant line to a function at a point (a,f(a))(a,f(a)) to estimate the rate of change, or the rate at which one variable changes in relation to another variable. We can obtain the slope of the secant by choosing a value of x near a and drawing a line through the points (a,f(a))(a,f(a)) and (x,f(x))(x,f(x)), as shown in Figure. The slope of this line is given by an equation in the form of a difference quotient: msec=f(x)−f(a)x−a msec=f(x)−f(a)x−a(3.0.7) We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing xx with a+ha+h, where hh is a value close to a. We can then calculate the slope of the line through the points (a,f(a))(a,f(a)) and (a+h,f(a+h))(a+h,f(a+h)). In this case, we find the secant line has a slope given by the following difference quotient with increment hh: msec=f(a+h)−f(a)a+h−a=f(a+h)−f(a)h msec=f(a+h)−f(a)a+h−a=f(a+h)−f(a)h(3.0.8) Definition: (Newton's) difference quotient Let ff be a function defined on an interval I containing a. If x≠a is in I, then Q=f(x)−f(a)x−a is a difference quotient. Also, if h≠0 is chosen so that a+h is in I, then Q=f(a+h)−f(a)h is a difference quotient with increment h. These two expressions for calculating the slope of a secant line are illustrated in Figure. We will see that each of these two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The primary consideration in our choice usually depends on ease of calculation. In Figure(a) we see that, as the values of x approach a, the slopes of the secant lines provide better estimates of the rate of change of the function at a. Furthermore, the secant lines themselves approach the tangent line to the function at a, which represents the limit of the secant lines. Similarly, Figure(b) shows that as the values of h get closer to 0, the secant lines also approach the tangent line. The slope of the tangent line at a is the rate of change of the function at a, as shown in Figure(c). In Figure we show the graph of f(x)=√x and its tangent line at (1,1) in a series of tighter intervals about x=1. As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a good approximation to the values of the function for choices of x close to 1. In fact, the graph of f(x) itself appears to be locally linear in the immediate vicinity of x=1. Formally we may define the tangent line to the graph of a function as follows. Definition: tangent line Let f(x) be a function defined in an open interval containing a. The tangent line to f(x) at a is the line passing through the point (a,f(a)) having slope mtan=limx→af(x)−f(a)x−a provided this limit exists. Equivalently, we may define the tangent line to f(x) at a to be the line passing through the point (a,f(a)) having slope mtan=limh→0f(a+h)−f(a)h provided this limit exists. Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find equations of tangent lines. Finding Tangent line Example 3.0.1: Finding a Tangent Line Find the equation of the line tangent to the graph of f(x)=x2 at x=3. Solution First find the slope of the tangent line. In this example, use Equation. mtan=limx→3f(x)−f(3)x−3 Apply the definition. =limx→3x2−9x−3 Substitute f(x)=x2 and f(3)=9 =limx→3(x−3)(x+3)x−3=limx→3(x+3)=6 Factor the numerator to evaluate the limit. Next, find a point on the tangent line. Since the line is tangent to the graph of f(x) at x=3, it passes through the point (3,f(3)). We have f(3)=9, so the tangent line passes through the point (3,9). Using the point-slope equation of the line with the slope m=6 and the point (3,9), we obtain the line y−9=6(x−3). Simplifying, we have y=6x−9. The graph of f(x)=x2 and its tangent line at 3 are shown in Figure. Example 3.0.2: The Slope of a Tangent Line Revisited Use Equation to find the slope of the line tangent to the graph of f(x)=x2 at x=3. Solution The steps are very similar to Example. See Equation for the definition. mtan=limh→0f(3+h)−f(3)h Apply the definition. =limh→0(3+h)2−9h Substitute f(3+h)=(3+h)2 and f(3)=9. =limh→09+6h+h2−9h Expand and simplify to evaluate the limit. =limh→0h(6+h)h=limh→0(6+h)=6 We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged. Example 3.0.c: Finding the Equation of a Tangent Line Find the equation of the line tangent to the graph of f(x)=1/x at x=2. Solution We can use Equation, but as we have seen, the results are the same if we use Equation. mtan=limx→2f(x)−f(2)x−2 Apply the definition. =limx→21x−12x−2 Substitute f(x)=1x and f(2)=12 =limx→21x−12x−2⋅2x2x Multiply numerator and denominator by 2x to simplify fractions. =limx→2(2−x)(x−2)(2x) Simplify. =limx→2−12x Simplify using 2−xx−2=−1,for x≠2. =−14 Evaluate the limit. We now know that the slope of the tangent line is −14. To find the equation of the tangent line, we also need a point on the line. We know that f(2)=12. Since the tangent line passes through the point (2,12) we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation y=−14x+1. The graphs of f(x)=1x and y=−14x+1 are shown in Figure. Exercise 3.0.1 Find the slope of the line tangent to the graph of f(x)=√x at x=4. Hint : Use either Equation or Equation. Multiply the numerator and the denominator by a conjugate. Answer : \frac{1}{4} The Derivative of a Function at a Point The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation. Definition Let f(x) be a function defined in an open interval containing a. The derivative of the function f(x) at a, denoted by f′(a), is defined by f′(a)=limx→af(x)−f(a)x−a provided this limit exists. Alternatively, we may also define the derivative of f(x) at a as f′(a)=limh→0f(a+h)−f(a)h. Velocities and Rates of Change Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if s(t) is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval [a,t] if t>a or [t,a] if t<a is given by the difference quotient vave=s(t)−s(a)t−a. As the values of t approach a, the values of vave approach the value we call the instantaneous velocity at a. That is, instantaneous velocity at a, denoted v(a), is given by v(a)=s′(a)=limt→as(t)−s(a)t−a. To better understand the relationship between average velocity and instantaneous velocity, see Figure. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t=a whose position at time t is given by the function s(t). The slope of the secant line (shown in green) is the average velocity of the object over the time interval [a,t]. Example 3.0.8: Estimating Velocity A lead weight on a spring is oscillating up and down. Its position at time t with respect to a fixed horizontal line is given by s(t)=sint (Figure). Use a table of values to estimate v(0). Check the estimate by using Equation. Solution We can estimate the instantaneous velocity at t=0 by computing a table of average velocities using values of t approaching 0, as shown in Table. | t | sint−sin0t−0=sintt | --- | | −0.1 | 0.998334166 | | −0.01 | 0.9999833333 | | −0.001 | 0.999999833 | | 0.001 | 0.999999833 | | 0.01 | 0.9999833333 | | 0.1 | 0.998334166 | Average velocities using values of t approaching 0 From the table we see that the average velocity over the time interval [−0.1,0] is 0.998334166, the average velocity over the time interval [−0.01,0] is 0.9999833333, and so forth. Using this table of values, it appears that a good estimate is v(0)=1. By using Equation, we can see that v(0)=s′(0)=limt→0sint−sin0t−0=limt→0sintt=1. Thus, in fact, v(0)=1. Exercise 3.0.5 A rock is dropped from a height of 64 feet. Its height above ground at time t seconds later is given by s(t)=−16t2+64,0≤t≤2. Find its instantaneous velocity 1 second after it is dropped, using Equation. Hint : v(t)=s′(t). Follow the earlier examples of the derivative using Equation. Answer : −32 ft/s As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function. Definition: instantaneous rate of change Theinstantaneous rate of change of a function f(x) at a value a is its derivative f′(a). Example 3.0.9: Chapter Opener: Estimating Rate of Change of Velocity Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51 seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches 229.9 mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant? Solution: First observe that 60 mph = 88 ft/s, 100 mph ≈146.67 ft/s, 200 mph ≈293.33 ft/s, and 229.9 mph ≈337.19 ft/s. We can summarize the information in a table. | | | --- | | t | v(t) | | 0 | 0 | | 3.05 | 88 | | 5.88 | 147.67 | | 14.51 | 293.33 | | 19.96 | 337.19 | v(t) at different values of t Now compute the average acceleration of the car in feet per second on intervals of the form [t,19.96] as t approaches 19.96, as shown in the following table. | | | --- | | t | v(t)−v(19.96)t−19.96=v(t)−337.19t−19.96 | | 0.0 | 16.89 | | 3.05 | 14.74 | | 5.88 | 13.46 | | 14.51 | 8.05 | Average acceleration The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s). Example 3.0.10: Rate of Change of Temperature A homeowner sets the thermostat so that the temperature in the house begins to drop from 70°F at 9 p.m., reaches a low of 60° during the night, and rises back to 70° by 7 a.m. the next morning. Suppose that the temperature in the house is given by T(t)=0.4t2−4t+70 for 0≤t≤10, where t is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight. Solution Since midnight is 3 hours past 9 p.m., we want to compute T′(3). Refer to Equation. T′(3)=limt→3T(t)−T(3)t−3 Apply the definition. =limt→30.4t2−4t+70−61.6t−3 Substitute T(t)=0.4t2−4t+70 and T(3)=61.6. =limt→30.4t2−4t+8.4t−3 Simplify. =limt→30.4(t−3)(t−7)t−3 =limt→30.4(t−3)(t−7)t−3 =limt→30.4(t−7) Cancel. =−1.6 Evaluate the limit. The instantaneous rate of change of the temperature at midnight is −1.6°F per hour. Example 3.0.11: Rate of Change of Profit A toy company can sell x electronic gaming systems at a price of p=−0.01x+400 dollars per gaming system. The cost of manufacturing x systems is given by C(x)=100x+10,000 dollars. Find the rate of change of profit when 10,000 games are produced. Should the toy company increase or decrease production? Solution The profit P(x) earned by producing x gaming systems is R(x)−C(x), where R(x) is the revenue obtained from the sale of x games. Since the company can sell x games at p=−0.01x+400 per game, R(x)=xp=x(−0.01x+400)=−0.01x2+400x. Consequently, P(x)=−0.01x2+300x−10,000. Therefore, evaluating the rate of change of profit gives P′(10000)=limx→10000P(x)−P(10000)x−10000 =limx→10000−0.01x2+300x−10000−1990000x−10000 =limx→10000−0.01x2+300x−2000000x−10000 =100. Since the rate of change of profit P′(10,000)>0 and P(10,000)>0, the company should increase production. Exercise 3.0.6 A coffee shop determines that the daily profit on scones obtained by charging s dollars per scone is P(s)=−20s2+150s−10. The coffee shop currently charges $3.25 per scone. Find P′(3.25), the rate of change of profit when the price is $3.25 and decide whether or not the coffee shop should consider raising or lowering its prices on scones. Answer : Use Example for a guide. Answer : P′(3.25)=20>0; raise prices Key Concepts The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it by finding the limit of the difference quotient or the difference quotient with increment h. The derivative of a function f(x) at a value a is found using either of the definitions for the slope of the tangent line. Velocity is the rate of change of position. As such, the velocity v(t) at time t is the derivative of the position s(t) at time t. Average velocity is given by vave=s(t)−s(a)t−a. Instantaneous velocity is given by v(a)=s′(a)=limt→as(t)−s(a)t−a. Key Equations (Newton's) Difference quotient Q=f(x)−f(a)x−a Difference quotient with increment h Q=f(a+h)−f(a)a+h−a=f(a+h)−f(a)h Slope of tangent line mtan=limx→af(x)−f(a)x−a mtan=limh→0f(a+h)−f(a)h Derivative of f(x) at a f′(a)=limx→af(x)−f(a)x−a f′(a)=limh→0f(a+h)−f(a)h Average velocity vave=s(t)−s(a)t−a Instantaneous velocity v(a)=s′(a)=limt→as(t)−s(a)t−a Glossary derivative : the slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, is the derivative difference quotient : of a function f(x) at a is given by f(a+h)−f(a)h or f(x)−f(a)x−a differentiation : the process of taking a derivative instantaneous rate of change : the rate of change of a function at any point along the function a, also called f′(a), or the derivative of the function at a Source Calculus Applets using GeoGebra by Marc Renault is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Contributors and Attributions Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 3: Derivatives 3.0E: Exercises
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http://www.math.com/students/worksheet/algebra_sp.htm
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https://en.wikipedia.org/wiki/Eisenstein%27s_criterion
Jump to content Eisenstein's criterion Čeština Deutsch Español Français Galego 한국어 Italiano עברית Nederlands 日本語 Polski Português Romnă Русский Српски / srpski Svenska Українська Tiếng Việt 中文 Edit links From Wikipedia, the free encyclopedia Sufficient condition for polynomial irreducibility In mathematics, Eisenstein's criterion gives a sufficient condition for a polynomial with integer coefficients to be irreducible over the rational numbers – that is, for it to not be factorizable into the product of non-constant polynomials with rational coefficients. This criterion is not applicable to all polynomials with integer coefficients that are irreducible over the rational numbers, but it does allow in certain important cases for irreducibility to be proved with very little effort. It may apply either directly or after transformation of the original polynomial. This criterion is named after Gotthold Eisenstein. In the early 20th century, it was also known as the Schönemann–Eisenstein theorem because Theodor Schönemann was the first to publish it. Criterion [edit] Suppose we have the following polynomial with integer coefficients: If there exists a prime number p such that the following three conditions all apply: p divides each ai for 0 ≤ i < n, p does not divide an, and p2 does not divide a0, then Q is irreducible over the rational numbers. It will also be irreducible over the integers, unless all its coefficients have a nontrivial factor in common (in which case Q as integer polynomial will have some prime number, necessarily distinct from p, as an irreducible factor). The latter possibility can be avoided by first making Q primitive, by dividing it by the greatest common divisor of its coefficients (the content of Q). This division does not change whether Q is reducible or not over the rational numbers (see Primitive part–content factorization for details), and will not invalidate the hypotheses of the criterion for p (on the contrary it could make the criterion hold for some prime, even if it did not before the division). Examples [edit] Eisenstein's criterion may apply either directly (i.e., using the original polynomial) or after transformation of the original polynomial. Direct (without transformation) [edit] Consider the polynomial Q(x) = 3x4 + 15x2 + 10. In order for Eisenstein's criterion to apply for a prime number p it must divide both non-leading coefficients 15 and 10, which means only p = 5 could work, and indeed it does since 5 does not divide the leading coefficient 3, and its square 25 does not divide the constant coefficient 10. One may therefore conclude that Q is irreducible over Q (and, since it is primitive, over Z as well). Note that since Q is of degree 4, this conclusion could not have been established by only checking that Q has no rational roots (which eliminates possible factors of degree 1), since a decomposition into two quadratic factors could also be possible. Indirect (after transformation) [edit] Often Eisenstein's criterion does not apply for any prime number. It may however be that it applies (for some prime number) to the polynomial obtained after substitution (for some integer a) of x + a for x. The fact that the polynomial after substitution is irreducible then allows concluding that the original polynomial is as well. This procedure is known as applying a shift. For example consider H = x2 + x + 2, in which the coefficient 1 of x is not divisible by any prime, Eisenstein's criterion does not apply to H. But if one substitutes x for x + 3 in H, one obtains the polynomial x2 + 7x + 14, which satisfies Eisenstein's criterion for the prime number 7. Since the substitution is an automorphism of the ring Q[x], the fact that we obtain an irreducible polynomial after substitution implies that we had an irreducible polynomial originally. In this particular example it would have been simpler to argue that H (being monic of degree 2) could only be reducible if it had an integer root, which it obviously does not; however the general principle of trying substitutions in order to make Eisenstein's criterion apply is a useful way to broaden its scope. Another possibility to transform a polynomial so as to satisfy the criterion, which may be combined with applying a shift, is reversing the order of its coefficients, provided its constant term is nonzero (without which it would be divisible by x anyway). This is so because such polynomials are reducible in R[x] if and only if they are reducible in R[x, x−1] (for any integral domain R), and in that ring the substitution of x−1 for x reverses the order of the coefficients (in a manner symmetric about the constant coefficient, but a following shift in the exponent amounts to multiplication by a unit). As an example 2x5 − 4x2 − 3 satisfies the criterion for p = 2 after reversing its coefficients, and (being primitive) is therefore irreducible in Z[x]. Cyclotomic polynomials [edit] An important class of polynomials whose irreducibility can be established using Eisenstein's criterion is that of the cyclotomic polynomials for prime numbers p. Such a polynomial is obtained by dividing the polynomial xp − 1 by the linear factor x − 1, corresponding to its obvious root 1 (which is its only rational root if p > 2): Here, as in the earlier example of H, the coefficients 1 prevent Eisenstein's criterion from applying directly. However the polynomial will satisfy the criterion for p after substitution of x + 1 for x: this gives all of whose non-leading coefficients are divisible by p by properties of binomial coefficients, and whose constant coefficient is equal to p, and therefore not divisible by p2. An alternative way to arrive at this conclusion is to use the identity (a + b)p = ap + bp which is valid in characteristic p (and which is based on the same properties of binomial coefficients, and gives rise to the Frobenius endomorphism), to compute the reduction modulo p of the quotient of polynomials: which means that the non-leading coefficients of the quotient are all divisible by p; the remaining verification that the constant term of the quotient is p can be done by substituting 1 (instead of x + 1) for x into the expanded form xp−1 + ... + x + 1. History [edit] Theodor Schönemann was the first to publish a version of the criterion, in 1846 in Crelle's Journal, which reads in translation That (x − a)n + pF(x) will be irreducible to the modulus p2 when F(x) to the modulus p does not contain a factor x−a. This formulation already incorporates a shift to a in place of 0; the condition on F(x) means that F(a) is not divisible by p, and so pF(a) is divisible by p but not by p2. As stated it is not entirely correct in that it makes no assumptions on the degree of the polynomial F(x), so that the polynomial considered need not be of the degree n that its expression suggests; the example x2 + p(x3 + 1) ≡ (x2 + p)(px + 1) mod p2, shows the conclusion is not valid without such hypothesis. Assuming that the degree of F(x) does not exceed n, the criterion is correct however, and somewhat stronger than the formulation given above, since if (x − a)n + pF(x) is irreducible modulo p2, it certainly cannot decompose in Z[x] into non-constant factors. Subsequently Eisenstein published a somewhat different version in 1850, also in Crelle's Journal. This version reads in translation When in a polynomial F(x) in x of arbitrary degree the coefficient of the highest term is 1, and all following coefficients are whole (real, complex) numbers, into which a certain (real resp. complex) prime number m divides, and when furthermore the last coefficient is equal to εm, where ε denotes a number not divisible by m: then it is impossible to bring F(x) into the form where μ, ν ≥ 1, μ + ν = deg(F(x)), and all a and b are whole (real resp. complex) numbers; the equation F(x) = 0 is therefore irreducible. Here "whole real numbers" are ordinary integers and "whole complex numbers" are Gaussian integers; one should similarly interpret "real and complex prime numbers". The application for which Eisenstein developed his criterion was establishing the irreducibility of certain polynomials with coefficients in the Gaussian integers that arise in the study of the division of the lemniscate into pieces of equal arc-length. Remarkably Schönemann and Eisenstein, once having formulated their respective criteria for irreducibility, both immediately apply it to give an elementary proof of the irreducibility of the cyclotomic polynomials for prime numbers, a result that Gauss had obtained in his Disquisitiones Arithmeticae with a much more complicated proof. In fact, Eisenstein adds in a footnote that the only proof for this irreducibility known to him, other than that of Gauss, is one given by Kronecker in 1845. This shows that he was unaware of the two different proofs of this statement that Schönemann had given in his 1846 article, where the second proof was based on the above-mentioned criterion. This is all the more surprising given the fact that two pages further, Eisenstein actually refers (for a different matter) to the first part of Schönemann's article. In a note ("Notiz") that appeared in the following issue of the Journal, Schönemann points this out to Eisenstein, and indicates that the latter's method is not essentially different from the one he used in the second proof. Basic proof [edit] To prove the validity of the criterion, suppose Q satisfies the criterion for the prime number p, but that it is nevertheless reducible in Q[x], from which we wish to obtain a contradiction. From Gauss' lemma it follows that Q is reducible in Z[x] as well, and in fact can be written as the product Q = GH of two non-constant polynomials G, H (in case Q is not primitive, one applies the lemma to the primitive polynomial Q/c (where the integer c is the content of Q) to obtain a decomposition for it, and multiplies c into one of the factors to obtain a decomposition for Q). Now reduce Q = GH modulo p to obtain a decomposition in (Z/pZ)[x]. But by hypothesis this reduction for Q leaves its leading term, of the form axn for a non-zero constant a ∈ Z/pZ, as the only nonzero term. But then necessarily the reductions modulo p of G and H also make all non-leading terms vanish (and cannot make their leading terms vanish), since no other decompositions of axn are possible in (Z/pZ)[x], which is a unique factorization domain. In particular the constant terms of G and H vanish in the reduction, so they are divisible by p, but then the constant term of Q, which is their product, is divisible by p2, contrary to the hypothesis, and one has a contradiction. A second proof of Eisenstein's criterion also starts with the assumption that the polynomial Q(x) is reducible. It is shown that this assumption entails a contradiction. The assumption that is reducible means that there are polynomials Such that The coefficient a0 of the polynomial Q(x) can be divided by the prime p but not by p2. Since a0 = c0d0, it is possible to divide c0 or d0 by p, but not both. One may without loss of generality proceed with a coefficient c0 that can be divided by p and with a coefficient d0 that cannot be divided by p. By the assumption, does not divide . Because an = cr ds, neither cr nor ds can be divided by p. Thus, if is the -th coefficient of the reducible polynomial , then (possibly with in case ) wherein cannot be divided by , because neither nor can be divided by . We will prove that are all divisible by p. As is also divisible by p (by hypothesis of the criterion), this implies that is divisible by p, a contradiction proving the criterion. It is possible to divide by , because can be divided by . By initial assumption, it is possible to divide the coefficient a1 of the polynomial Q(x) by p. Since and since d0 is not a multiple of p it must be possible to divide c1 by p. Analogously, by induction, is a multiple of for all , which finishes the proof. Advanced explanation [edit] Applying the theory of the Newton polygon for the p-adic number field, for an Eisenstein polynomial, we are supposed to take the lower convex envelope of the points (0, 1), (1, v1), (2, v2), ..., (n − 1, vn−1), (n, 0), where vi is the p-adic valuation of ai (i.e. the highest power of p dividing it). Now the data we are given on the vi for 0 < i < n, namely that they are at least one, is just what we need to conclude that the lower convex envelope is exactly the single line segment from (0, 1) to (n, 0), the slope being −1/n. This tells us that each root of Q has p-adic valuation 1/n and hence that Q is irreducible over the p-adic field (since, for instance, no product of any proper subset of the roots has integer valuation); and a fortiori over the rational number field. This argument is much more complicated than the direct argument by reduction mod p. It does however allow one to see, in terms of algebraic number theory, how frequently Eisenstein's criterion might apply, after some change of variable; and so limit severely the possible choices of p with respect to which the polynomial could have an Eisenstein translate (that is, become Eisenstein after an additive change of variables as in the case of the p-th cyclotomic polynomial). In fact only primes p ramifying in the extension of Q generated by a root of Q have any chance of working. These can be found in terms of the discriminant of Q. For example, in the case x2 + x + 2 given above, the discriminant is −7 so that 7 is the only prime that has a chance of making it satisfy the criterion. Modulo 7, it becomes (x − 3)2— a repeated root is inevitable, since the discriminant is 0 mod 7. Therefore the variable shift is actually something predictable. Again, for the cyclotomic polynomial, it becomes (x − 1)p−1 mod p; the discriminant can be shown to be (up to sign) pp−2, by linear algebra methods. More precisely, only totally ramified primes have a chance of being Eisenstein primes for the polynomial. (In quadratic fields, ramification is always total, so the distinction is not seen in the quadratic case like x2 + x + 2 above.) In fact, Eisenstein polynomials are directly linked to totally ramified primes, as follows: if a field extension of the rationals is generated by the root of a polynomial that is Eisenstein at p then p is totally ramified in the extension, and conversely if p is totally ramified in a number field then the field is generated by the root of an Eisenstein polynomial at p. Generalization [edit] Generalized criterion [edit] Given an integral domain D, let be an element of D[x], the polynomial ring with coefficients in D. Suppose there exists a prime ideal p of D such that ai ∈ p for each i ≠ n, an ∉ p, and a0 ∉ p2, where p2 is the ideal product of p with itself. Then Q cannot be written as a product of two non-constant polynomials in D[x]. If in addition Q is primitive (i.e., it has no non-trivial constant divisors), then it is irreducible in D[x]. If D is a unique factorization domain with field of fractions F, then by Gauss's lemma Q is irreducible in F[x], whether or not it is primitive (since constant factors are invertible in F[x]); in this case a possible choice of prime ideal is the principal ideal generated by any irreducible element of D. The latter statement gives original theorem for D = Z or (in Eisenstein's formulation) for D = Z[i]. Proof [edit] The proof of this generalization is similar to the one for the original statement, considering the reduction of the coefficients modulo p; the essential point is that a single-term polynomial over the integral domain D/p cannot decompose as a product in which at least one of the factors has more than one term (because in such a product there can be no cancellation in the coefficient either of the highest or the lowest possible degree). Example [edit] After Z, one of the basic examples of an integral domain is the polynomial ring D = k[u] in the variable u over the field k. In this case, the principal ideal generated by u is a prime ideal. Eisenstein's criterion can then be used to prove the irreducibility of a polynomial such as Q(x) = x3 + ux + u in D[x]. Indeed, u does not divide a3, u2 does not divide a0, and u divides a0, a1 and a2. This shows that this polynomial satisfies the hypotheses of the generalization of Eisenstein's criterion for the prime ideal p = (u) since, for a principal ideal (u), being an element of (u) is equivalent to being divisible by u. See also [edit] Cohn's irreducibility criterion Perron's irreducibility criterion Notes [edit] ^ a b Cox (2011). ^ Dorwart (1935). ^ Schönemann (1846), p. 100. ^ Eisenstein (1850), p. 166. ^ Schönemann (1850), p. 188. ^ Cassels & Fröhlich (1967), pp. 22–23, "Local fields". References [edit] Cassels, J. W. S.; Fröhlich, Albert, eds. (1967), Algebraic number theory. Cox, David A. (2011), "Why Eisenstein proved the Eisenstein criterion and why Schönemann discovered it first", American Mathematical Monthly, 118 (1): 3–31, CiteSeerX 10.1.1.398.3440, doi:10.4169/amer.math.monthly.118.01.003, S2CID 15978494. Dorwart, H. L. (1935), "Irreducibility of polynomials", American Mathematical Monthly, 42 (6): 369–381, doi:10.2307/2301357, JSTOR 2301357. Eisenstein, Gotthold (1850), "Über die Irreductibilität und einige andere Eigenschaften der Gleichung, von welcher die Theilung der ganzen Lemniscate abhängt", Journal für die reine und angewandte Mathematik, 1850 (39): 160–179, doi:10.1515/crll.1850.39.160, S2CID 122322672. Garling, D.J.H. (1986), A Course in Galois Theory, Cambridge University Press, ISBN 978-0-521-31249-3. "Algebraic equation", Encyclopedia of Mathematics, EMS Press, 2001 . Schönemann, Theodor (1846), "Von denjenigen Moduln, welche Potenzen von Primzahlen sind", Journal für die reine und angewandte Mathematik, 1846 (32): 93–118, doi:10.1515/crll.1846.32.93, S2CID 120510090. Schönemann, Theodor (1850), "Über einige von Herrn Dr. Eisenstein aufgestellte Lehrsätze, irreductible Congruenzen betreffend (S.182 Bd. 39 dieses Journals)", Journal für die reine und angewandte Mathematik, 1850 (40): 185–188, doi:10.1515/crll.1850.40.185, S2CID 199547075. Retrieved from " Categories: Polynomials Field (mathematics) Hidden categories: Articles with short description Short description is different from Wikidata Articles containing proofs
188271
https://byjus.com/maths/cube-root-of-2/
The cube root of 2, denoted as 3√2, is the value that gives the original number when multiplied by itself thrice. This definition is applicable to all the cube roots of natural numbers. Now, since 2 is not a perfect cube, therefore we cannot find it using the prime factorisation method. | | | Cube root of 2, 3√2 = 1.2599 | We will try to find the value of 3√2, without using a calculator and using the approximation method. Also read: Calculation of Cube Root of 2 As we already know, 2 is not a perfect cube, therefore we cannot use the general methods, to find its cube root. Let us find it by using approximation or we can also say, an estimation method. Assume, 3√2 = X So, here X should be equal to a number, which after getting multiplied by itself three times, gives the result as 2. But if we see, 2 is nearly equal to the cube of 1, i.e. 13 = 1 So let us just say, X = p × p × q Take p =1 and q = 2 X = 1 × 1 × 2 = 2 This is possible only when the average of the three factors (1,1,2) is roughly equal to the cube root of 2. Hence, 3√2 = (1+1+2)/3 = 4/3 = 1.3 Again let us say p = 1.3 and q = 1.18 So, 1.3×1.3×1.18 = 1.99 ≈ 2 Taking the average of three factors, 1.3,1.3 and 1.18, we get; 3√2 = (1.3+1.3+1.18)/3 = 3.78/3 = 1.26 Hence, 3√2 = 1.26 Therefore, we get the value of the cube root of 2 equal to 1.26, which is approximately equal to its actual value, i.e.,1.2599210. Cube Root of Non-Perfect Cubes To find the cube root of perfect cubes is easy, but difficult to find for the non-perfect cubes. For example, the cube root of 27 is equal to 3. Since, 27 = 3 x 3 x 3 = 33 Therefore, 3√27 = 3√(33) = 3 But for non-perfect cubes we cannot find their cubic roots this easily. Thus, here is a table for the cube roots from 1 to 20 consisting of only non-perfect cubes. | | | --- | | Number | Cube Root (3√) | | 2 | 1.260 | | 3 | 1.442 | | 4 | 1.587 | | 5 | 1.710 | | 6 | 1.817 | | 7 | 1.913 | | 9 | 2.080 | | 10 | 2.154 | | 11 | 2.224 | | 12 | 2.289 | | 13 | 2.351 | | 14 | 2.410 | | 15 | 2.466 | | 16 | 2.520 | | 17 | 2.571 | | 18 | 2.621 | | 19 | 2.668 | | 20 | 2.714 | From the table, we can see there are only two numbers eliminated from the table between 1 and 20, which are 1 and 8. Apart from that, all the natural numbers are non-perfect cubes. Video Lesson on Finding Cube Roots Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked Request OTP on Voice Call Website Post My Comment Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
188272
https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/Introductory_Biology_(CK-12)/11%3A_Invertebrates/11.04%3A_Sponges
Skip to main content 11.4: Sponges Last updated : Mar 6, 2021 Save as PDF 11.3: Invertebrate Classification 11.5: Cnidarians Page ID : 6691 ( \newcommand{\kernel}{\mathrm{null}\,}) So what exactly is a sponge? Here we have a giant barrel sponge. How can something that looks like that be considered an animal? Where's the head? Where are the legs? Where's the mouth? Sponges Invertebrates are animals without a backbone. They are the most numerous animals on Earth. Most invertebrates are insects. However, simpler invertebrates evolved before insects. Some, like the sponges you will read about in this concept, have existed virtually unchanged for hundreds of millions of years. Their continued existence is evidence that they are well adapted for their habitats. They also evolved some of the most important traits that are found in almost all animals today. Without the traits that evolved in sponges and other simple invertebrates, you would not exist. Sponges are aquatic invertebrates that make up the phylum Porifera. The word "porifera" means pore-bearing. The phylum is aptly named. As you can see from Figure below, a sponge has a porous body. There are at least 5,000 living species of sponges. Almost all of them inhabit the ocean, living mainly on coral reefs or the ocean floor. Sponge on a Coral Reef. This orange sponge is covered with pores. Can you predict the function of the pores? Structure and Function of Sponges Sponges come in a variety of shapes and sizes. For example, they may be shaped like tubes, fans, cones, or just blobs. They range in diameter from about a centimeter (0.4 inches) to over a meter (3.3 feet). Many species live in colonies that may be quite large. Adult sponges are sessile. This means they are unable to move from place to place. Root-like projections anchor them to solid surfaces such as rocks and reefs. Sponges have an internal skeleton that gives them support and protection. An internal skeleton is called an endoskeleton. A sponge endoskeleton consists of short, sharp rods called spicules (see Figure below). Spicules are made of silica, calcium carbonate, or spongin, a tough protein. They grow from specialized cells in the body of the sponge. Sponge Anatomy. A sponge lacks tissues and organs, but it has several types of specialized cells. Sponges are filter feeders. They pump water into their body through their pores. The water flows through a large central cavity called the spongocoel (see Figure above). As the water flows by, specialized collar cells (which are also known as choanocytes) filter out food particles such as bacteria. Collar cells have tiny hairs that trap the particles. They also have a flagellum that whips the water and keeps it moving. Once the food is trapped, the collar cells digest it (see Figure below). Cells called amebocytes also help digest the food. They distribute the nutrients to the rest of the body as well. Finally, the water flows back out of the body through an opening called the osculum. As water flows through the sponge, oxygen diffuses from the water to the sponge’s cells. The cells also expel wastes into the water for removal through the osculum. Collar Cell. The collar cells of sponges trap and digest food. Sponge Reproduction Sponges reproduce both asexually and sexually. Asexual reproduction occurs by budding.Figure below shows the sponge life cycle when sexual reproduction is involved. Adult sponges produce eggs and sperm. In many species, the same individuals produce both. However, they don’t produce eggs and sperm at the same time. As a result, self-fertilization is unlikely to occur. What is an advantage of avoiding self-fertilization? The sponge life cycle includes sexual reproduction. Sponges may also reproduce asexually. Sperm are released into the surrounding water through the osculum. If they enter a female sponge through a pore, they may be trapped by collar cells. Trapped sperm are delivered to eggs inside the female body, where fertilization takes place. The resulting zygote develops into a larva. Unlike the adult, the larva is motile. It is covered with cilia that propel it through the water. As the larva grows, it becomes more similar to an adult sponge and loses its ability to swim. Ecology of Sponges Sponges that live on coral reefs have symbiotic relationships with other reef species. They provide shelter for algae, shrimp, and crabs. In return, they get nutrients from the metabolism of the organisms they shelter. Sponges are a source of food for many species of fish. Because sponges are sessile, they cannot flee from predators. Their sharp spicules provide some defense. They also produce toxins that may poison predators that try to eat them. Summary Sponges are aquatic invertebrates. They make up the phylum Porifera. Sponges have specialized cells and an endoskeleton. Sponges lack tissues and body symmetry. Adult sponges are sessile filter feeders. Sponge larvae have cilia for swimming. Review Define sessile. Name an invertebrate with a sessile adult stage. Describe the skeleton of a sponge. Sponges have specialized cells called collar cells. Describe how collar cells are specialized for the functions they serve. Create a diagram of an adult sponge body plan that shows how sponges obtain food. 11.3: Invertebrate Classification 11.5: Cnidarians
188273
https://www.osmosis.org/video/Lung_volumes_and_capacities
Fall in love with Osmosis at 20% off! Save now until September 30 at 11:59 PM PT. Learn more Osmosis video - Lung volumes and capacities Back to full view 00:00 / 11:31 More Videos 07:11 Airflow, pressure, and resistance 07:52 Alveolar gas equation 11:10 Breathing cycle and regulation 09:55 Diffusion-limited and perfusion-limited gas exchange 19:15 Gas exchange in the lungs, blood and tissues Video Summary of Lung volumes and capacities Lung volumes refer to the volume of air in the lungs, measured at various phases of the respiratory cycle. Major lung volumes include the tidal volume, inspiratory reserve volume, expiratory reserve volume, and the residual volume. The tidal volume is the amount of air inhaled or exhaled during a single breath. The inspiratory reserve volume is the additional air that can be inhaled after a tidal volume has been breathed in. The expiratory reserve volume is the additional air that can be exhaled after a tidal volume has been exhaled. And the residual volume is the amount of air left in the lungs after you've finished exhaling. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see ourCookie Policy Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. You may also be able to exercise your privacy choices as described in our Privacy Policy Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Targeting Cookies These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. If you do not allow these cookies, you will experience less targeted advertising.
188274
https://courses.wccnet.edu/~palay/math160r/prob_binomial.htm
Probability: Binomial Distribution Return to Topics page The Binomial distribution refers to a whole class of discrete distributions. The Binomial distribution occurs in cases where We have a fixed number of trials (i.e., attempts) are done (run n trials). There are only 2 possible outcomes of each trial: one is called a success, the other called a failure. The probability of success is the same for each trial. We designate the probability of success on any one trial as p. This means that the probability of a failure is (1-p). The trials are independent; knowing the outcome of one trial tells you nothing about the outcome of the next trial. We are looking at a random variable, X, that is the number of successes in those n trials. For example, if we start with a coin that has a 56% probability of coming up heads when we spin the coin. We will spin it 5 times (there are 5 trials). On each spin the probability of getting heads is 0.56. The spins are independent events; the coin does not change its behavior based on its history of spins. We want to know the probability that we will get each of the following: 5 heads, 4heads and 1 tail, 3 heads and 2 tails, 2 heads and 3 tails, 1 head and 4 tails, and 5 tails. This situation where we have n=5 and p=0.56, is one instance of a binomial distribution. If we change the value of n and/or we change the value of p we get a different instance of a binomial distribution. We could use a tree diagram to compute the various probabilities. Alternatively, we can just look at the six possible results (getting 5, 4, 3, 2, 1, or 0 heads). The probability of getting 5 heads is just p5. The probability of getting 5 tails is just (1-p)5 The probability of getting the sequence HHHHT is just p4(1-p). However, this is but one of five ways to get 4 heads and 1 tail, namely, HHHHT, HHHTH, HHTHH, HTHHH, and THHHH. Each will have the same probability. Therefore, the probability of getting 4 heads and 1 tail is 5p4(1-p). In that expression, the 5 represented the number of combinations of 5 things taken 4 at a time. We recall that we represent that as 5C4. Using that we could write our probability as 5C4p4(1-p)1. | | | Why do we look at this as combinations? Really, we have five spots to fill, one for each spin of the coin. If we call the spots a, b, c, d, and e, then the question becomes which of the five slots get the 4 heads? There are 5 combinations of the five slots taken 4 at a time, namely, abcd, abce, abde, acde, and bcde. If for each of those combinations you place a head (H) in the named slot, and then fill the one empty slot with a tail (T), then you will get the five choices given above, in order. We will do this again for the next situation. | One way to get 3 heads and 2 tails is HHHTT. The probability of getting that sequence is p3(1-p)2. However, HHHTT is but one of the 10 combinations of 5 things taken 3 at a time, namely, HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT, THHTH, THTHH, and TTHHH. | | | Again, we have five spots to fill, one for each spin of the coin. If we call the spots a, b, c, d, and e, then the question becomes which of the five slots get the 3 heads? There are 10 combinations of the five slots taken 3 at a time, namely, abc, abd, abe, acd, ace, ade, bcd, bce, bde, and cde. If for each of those combinations you place a head (H) in the named slot, and then fill the two empty slots with a tail (T), then you will get the ten choices given above, in order. | With 10 different ways to get 3 heads and 2 tails the probability of getting that result is 10p3(1-p)2 or, using the combinations format, 5C2p3(1-p)2. We can see the pattern develop. For 5 trials, our probability of getting k successes, which we should write as P(X=k) but which we often write as P(k), is given by P(X=k) = 5Ckpk(1-p)5-k In fact, could make this even more general and say for n trials the probability of getting k successes, which we should write as P(X=k) but which we often write as P(k), is given by P(X=k) = nCkpk(1-p)n-k For p=0.4 and n=5 we get the following probabilities: P( X = 0) = 0.0778 P( X = 1) = 0.2592 P( X = 2) = 0.3456 P( X = 3) = 0.2304 P( X = 4) = 0.0768 P( X = 5) = 0.0102 This little list of values is nice to have, but it is but one instance of a binomial distribution. It gives the value for the particular case where n=5 and p=0.40. What about other cases, where is some other value and/or p is some other probability. For years people have created tables that give binomial probabilities for lots of different values of n and many "common" values of p. I have a web version of tables like this at Binomial Distribution Tables. A portion of the table for n=5 is given in Figure 1 with the column for p=0.40 highlighted. Figure 1 The values in the highlighted column of the table are just the values that we listed above. But the rest of that table gives the binomial distribution values for other values of p that are multiples of 0.05. Other tables on the web page provide the same kind of information for different values of n, the number of trials. What are we supposed to do if we are using the tables and we want n=5 but p=0.37? If all we have are the tables we could do the basic computations ourselves, or we could interpolate between values in the table. Thus, when n=5, p=0.37, we use the table shown in Figure 1 and note that the P(X = 3) will be between P(X = 3) for p=0.35 and P(X = 3) for p=0.40, that is between 0.1811 and 0.2304. The difference between those values is 0.2304 - 0.1811 = 0.0493. Our desired p value is 0.37 and that is 2/5 of the way between 0.35 and 0.40. Therefore, we want to go 2/5 of the 0.0493 along the way from 0.1811 toward 0.2304. But 2/5 of 0.0493 is 0.01972 and 0.1811+0.01972 = 0.20082 which we would round off to 0.2008 because we should not be adding any significant digits. Thus, using the table our best guess is that for p=0.37 we have P(X = 3)=0.2008. (As we will see below, a more accurate answer is 0.2010418, but our interpolated value of 0.2008 is not that far off from there.) Although it is nice to be able to find P(X = k) in the tables for a given n and p, we often want to know things like P(X ≤ k). For example, staying with our case of n=5 and p=0.40, to find P(X ≤ 3) we would just add the values in the cells of the table for P(X = 0), P(X = 1), P(X = 2), and P(X = 3). To help in this we mimic the list above to produce P( X ≤ 0) = 0.0778 P( X ≤ 1) = 0.3370 P( X ≤ 2) = 0.6826 P( X ≤ 3) = 0.9130 P( X ≤ 4) = 0.9898 P( X ≤ 5) = 1.0000 The web page Binomial Distribution (Cumulative) Tables has tables arranged using this scheme. A portion of the table for n=5 is given in Figure 2, again with the column for p=0.40 highlighted. Figure 2 Before leaving these tables, we should note that we could reconstruct values in Figure 1 from those in Figure 2. For example, still with n=5 and p=0.40, P(X = 3) = P(X ≤ 3) - P(X ≤ 2), or, from the table in Figure 2, P(X = 3) = 0.9130 - 0.6826 = 0.2304, the value that we had in Figure 1. Using the web, and the programming that we can put into web pages, it is now possible to create a different table for binomial probabilities. Starting at Set Up Binomial Table we can produce specialized tables for any desired number of trials and any specified probability of success. The printed version of either of the first two web pages discussed above would require 13 sheets of paper. Even with that we only get tables for the values of n from 2 to 30, and only the columns of values for p as the 19 multiples of 0.05. Prior to advanced hand-held calculators, doing the computations by hand was pretty much out of the question. Certainly a language such as R should give us some way to compute these values directly! The pbinom(k,n,p) function in R computes P(X ≤ k) for n trials with p the probability of a success. Notice that this computes the probability of getting k or fewer successes in n trials, not the probability of getting exactly k successes. Therefore, the command pbinom(3,5,0.40) should produce the same value that we found in Figure 2 for the cumulative probability of getting 3 or fewer successes out of 5 trials, each with a 0.40 probability of success. The console record of that command is given in Figure 3. Figure 3 This value, when rounded to 4 places, is the same as we had in Figure 2. If we want to find P(X = 3) using R then we will have to employ the equation P(X = 3) = P(X ≤ 3) - P(X ≤ 2). Figure 4 shows computing the two values, just so that we can confirm them, and then the difference in the values to get the desired P(X = 3). Figure 4 That last value, 0.2304, corresponds to the same value in Figure 1. With the pbinom() function we are not limited to the multiples of 0.05 for the values of the probability for success. We can get a direct answer for our earlier question of P(X = 3) for n=5 and p=0.37 by using the command pbinom(3,5,0.37)-pbinom(2,5,0.37) as eventually shown in Figure 5. Figure 5 That is the source of the "more correct" answer given above when we got an approximate answer from the table by interpolating values. This business of having to do the subtraction to find the probability of getting an exact number of successes can be a bit tiring. Perhaps we should look at some way to just get the answer. We can do this by letting R do all of the work via a new function that we will build. The code for that function is: ``` pbinomeq <- function( k, n, p ) { x<- pbinom(k,n,p) if( k>0) { y <- pbinom(k-1, n, p) x <- x - y } return( x ) } ``` The console image of defining the function is given in Figure 6. Figure 6 And two examples of using the new function, using values that we already computed, are given in Figure 7. Figure 7 As noted before, it is nice to get a confirmation of earlier work! At this point we really have all that we need for computing binomial probabilities. However, there are many ways to ask the question. Consider the following. If the probability of success in a binomial distribution is 0.42 and we do 34 trials, what is the probability of getting less than or equal to 11 successes? less than 11 successes? exactly 11 successes? more than 11 successes? equal to or more than 11 successes? no fewer than 11 successes? no more than 11 successes? something other than 11 successes? between 11 and 18 successes, including 11 and 18? between 11 and 18 successes, excluding 11 and 18? less than 11 or more than 18 successes? less than or equal to 11 or more than or equal to 18 successes? Let us go through each of those. less than or equal to 11 successes: This is the meaning of the standard call to the pbinom() function. Thus the command pbinom(11,34,0.42) will give us the answer 0.1672396. less than 11 successes: This is equivalent to finding the probability of getting 10 or fewer successes. Thus the command is pbinom(10,34,0.42). The answer is 0.09292053. exactly 11 successes: We can do this as the probability of getting 11 or less minus the probability of getting 10 or less. To do that we could use pbinom(11,34,0.42)-pbinom(10,34,0.42) t get the answer 0.07431911. Alternatively, if we have loaded our function pbinomeq() then we can use it as pbinomeq(11,34,0.42) to get the same 0.07431911. more than 11 successes: This is the complement of having 11 or fewer successes. Thus, we are really looking at 1-P(X≤11) which becomes 1-pbinom(11,34,0.42) which gives us 0.8327604. Alternatively, we could look at this as the probability of getting 12 or more successes. One might think that the command pbinom(12,34,0.42,lower.tail=FALSE) would compute this. However, the documentation for R states that for the pbinom() function lower.tail logical; if TRUE (default), probabilities are P[X≤x], otherwise, P[X>x]. Therefore, if we specify lower.tail=FALSE we will not be including the first value since we are then looking at a "greater than" situation. If we choose to use the lower.tail=FALSE option we need to start at a value below 12. Therefore the command we want is pbinom(11,34,0.42,lower.tail=FALSE) which yields the same result, 0.8327604. equal to or more than 11 successes: Now we can compute not getting 0 through 10 successes via 1-pbinom(10,34,0.42) to get 0.9070795, or we could use the lower.tail=FALSE approach , remembering to adjust the first argument to the function and use the command pbinom(10,34,0.42,lower.tail=FALSE) to get 0.9070795. no fewer than 11 successes? "No fewer than 11" means "11 or more". We did this before as 1-pbinom(10,34,0.42) or pbinom(10,34,0.42,lower.tail=FALSE) to get 0.9070795. no more than 11 successes: "No more than 11" means "11 or fewer" which we did as pbinom(11,34,0.42) which gave us 0.1672396. something other than 11 successes: This is the complement of having exactly 11 successes. We could play with the pbinom() function but it s easier to just use the pbinomeq() function as 1-pbinomeq(11,34,0.42) to get the answer 0.9256809. between 11 and 18 successes, including 11 and 18? Here we find the probability of getting 18 or fewer and then subtract the probability of getting less than 11. We can do this via the command pbinom(18,34,0.42)-pbinom(10,34,0.42) to get the result 0.8349292. between 11 and 18 successes, excluding 11 and 18: This is similar to the previous case except that we no longer want to include 18 and 11. We just need to adjust the command as pbinom(17,34,0.42)-pbinom(11,34,0.42) to get the result 0.7008324. less than 11 or more than 18 successes: This is just the complement of the probability of having between 11 and 18 successes (including the 11 and the 18) so we can code this as 1-(pbinom(18,34,0.42)-pbinom(10,34,0.42)) which gives the answer 0.1650708. less than or equal to 11 or more than or equal to 18 successes: This is just a slight change from the previous problem in that we now include 11 and 18 in the answer, not in the part being excluded. The command has to become 1-(pbinom(17,34,0.42)-pbinom(11,34,0.42)) which gives the answer 0.2991676. Now we want to look at the mean, variance, and standard deviation of the binomial random variable. In doing this we need to point out that these are parameters of the distribution, not statistics drawn from a sample distribution. That is, if we have a binomial random variable X for the case where n=5 and p=0.40, then there is a mean for this, called μX, a variance, called σX2, and a standard deviation, called σX, for that distribution. If we have a sample from that distribution then we expect that the sample statistics will be close to, though not the same as, the distribution parameters. Before we look at the parameter values let us look at a sample, or at many samples since the sample presented in Figure 8. There we have a display of 1500 experiments where each experiment involves 5 trials of a spin of a coin that is weighted so that it shows up as heads 40% of the time. For each experiment we record, and display in Figure 8 the number of successes, i.e., "heads", that we get in the 5 trials. Figure 8: changes with each reload | | | 1 1 1 3 3 2 4 4 1 2 2 3 3 3 1 3 3 2 4 1 3 1 2 2 4 1 3 1 3 1 1 2 2 1 2 0 1 2 1 2 1 1 2 3 3 1 0 0 4 2 3 3 2 3 2 2 4 3 3 2 1 3 0 1 3 4 1 0 0 4 1 2 0 1 1 3 0 2 1 3 2 2 2 2 1 1 1 1 3 4 1 0 1 3 3 1 3 2 1 2 3 3 1 2 1 2 4 2 2 3 1 2 2 2 2 3 2 3 2 2 2 2 3 0 5 3 4 1 1 4 3 3 2 4 3 3 0 0 3 1 2 2 1 2 5 2 3 3 1 2 2 2 4 2 1 2 3 1 1 3 1 2 1 2 2 3 3 2 2 1 2 1 2 2 1 2 1 2 3 3 2 4 0 1 3 3 1 3 2 3 3 3 2 1 1 2 5 3 3 3 2 0 1 1 1 2 3 2 2 1 2 3 1 2 3 2 2 3 4 2 2 1 2 4 0 2 2 2 3 1 2 3 4 3 1 4 3 4 2 1 1 4 0 3 1 3 3 3 2 3 1 2 2 2 1 3 1 1 2 4 2 3 3 3 2 3 2 3 2 0 1 4 2 2 0 0 1 3 3 1 2 2 1 2 2 3 5 2 4 3 1 2 3 4 2 5 1 2 1 2 2 2 5 1 1 0 3 1 3 2 3 1 2 4 1 2 3 1 0 2 3 2 1 3 3 3 3 2 1 2 2 1 3 3 2 1 0 1 3 1 2 3 4 2 4 2 3 2 2 4 1 4 0 3 2 1 2 4 2 3 1 4 1 3 1 3 1 1 1 2 1 1 4 1 2 1 3 1 2 1 0 0 1 3 4 2 2 1 1 1 1 4 0 1 3 1 3 3 1 3 3 1 4 2 3 3 2 1 2 3 1 3 3 3 2 2 3 1 2 2 2 2 1 4 2 2 2 2 3 3 2 1 2 2 2 1 1 4 2 0 2 2 3 2 1 2 1 0 2 3 2 3 2 2 2 2 1 3 2 0 5 2 1 1 1 1 1 1 0 2 3 2 3 0 3 3 2 0 1 2 2 3 3 2 3 0 2 2 3 2 2 3 2 1 2 3 1 1 1 2 2 2 3 3 3 3 3 2 2 2 2 3 0 2 1 1 1 1 3 2 0 1 1 4 2 1 2 2 4 3 2 2 0 2 2 3 3 2 2 1 2 3 0 1 1 2 2 1 1 3 3 2 0 4 1 2 3 2 1 1 2 0 3 1 2 3 0 3 1 2 2 3 2 0 3 3 1 3 0 3 3 1 0 1 2 1 2 4 3 3 1 1 0 2 3 2 1 2 0 1 1 3 2 2 2 2 1 4 2 1 2 2 2 2 1 4 1 0 2 2 1 4 1 1 2 2 3 3 0 0 2 1 2 0 2 4 1 2 4 2 2 3 3 3 1 0 3 3 1 2 3 1 2 3 1 0 3 0 2 1 3 3 2 3 2 2 2 4 3 3 3 1 3 2 3 3 1 0 1 2 1 4 1 1 3 1 2 2 2 4 1 2 2 1 2 2 2 3 2 1 3 3 0 2 1 1 4 3 3 1 3 2 3 2 3 1 3 0 1 3 2 2 2 3 2 2 2 2 2 3 2 1 1 1 3 3 2 2 2 3 1 3 0 3 2 2 2 3 1 3 2 3 2 1 4 2 2 3 1 2 2 1 3 2 1 2 3 1 1 3 1 2 1 0 3 1 2 1 2 3 1 0 2 3 1 1 1 3 3 4 2 3 1 1 2 3 3 2 0 1 2 2 2 1 4 1 2 2 2 4 1 4 3 2 0 1 1 3 3 1 3 2 5 2 1 3 2 2 4 1 1 1 2 0 2 2 3 0 2 3 0 2 2 2 2 3 1 2 3 0 2 1 3 2 3 1 2 1 2 2 2 2 4 1 2 4 2 5 1 1 0 2 3 4 2 1 4 3 1 1 3 2 1 4 1 4 3 2 3 4 1 2 1 2 4 3 2 0 1 5 2 2 0 1 0 2 0 2 3 1 3 1 4 1 2 3 2 1 3 2 4 2 1 3 2 3 3 2 3 2 4 3 3 1 4 0 2 2 2 2 2 4 0 2 3 4 3 3 1 4 1 3 1 2 3 3 2 1 2 2 2 3 1 4 3 0 1 3 2 1 3 2 2 2 2 3 0 1 2 1 1 2 3 1 2 2 3 1 2 1 1 2 2 3 3 5 2 4 2 2 3 4 3 2 1 2 2 0 1 1 1 2 2 2 1 2 3 2 2 2 3 1 3 1 2 1 3 1 1 1 2 2 3 2 1 2 1 0 1 2 3 0 3 3 2 1 1 2 2 1 1 2 2 3 0 0 2 3 2 3 3 3 3 1 0 3 1 3 0 1 1 1 0 2 4 1 3 1 2 2 2 2 1 1 1 0 3 1 2 3 4 4 2 2 3 3 2 1 3 0 1 2 3 2 0 2 3 1 0 2 2 0 2 2 3 4 2 2 0 1 0 3 3 0 4 2 1 5 1 2 1 3 0 1 1 2 3 4 2 2 3 3 2 3 1 2 3 2 2 3 3 2 2 1 1 0 4 2 2 2 2 2 1 2 2 3 1 2 1 1 1 2 0 2 2 2 2 3 1 4 1 1 3 3 2 3 1 1 3 0 1 3 2 3 3 1 3 3 2 2 2 3 3 2 2 2 3 2 4 3 2 2 2 3 1 3 3 2 3 1 0 0 2 3 1 2 4 1 1 2 1 1 1 1 1 2 0 1 1 2 2 0 4 4 1 1 1 3 3 2 3 0 3 1 3 3 2 2 2 3 0 2 3 2 3 2 3 2 1 3 2 1 3 0 1 1 2 1 0 3 3 0 1 1 2 2 3 1 3 2 2 1 2 1 2 3 2 2 3 0 2 3 2 1 2 1 2 3 2 2 3 2 3 1 1 2 2 3 2 4 2 1 1 4 1 2 1 2 3 0 0 1 2 0 3 2 5 1 1 1 3 1 2 3 2 2 4 3 0 1 3 2 1 5 1 2 3 2 1 1 2 2 0 2 2 1 3 2 1 0 3 3 2 2 4 0 2 3 4 2 3 3 3 0 1 3 2 0 1 3 2 0 4 3 1 2 3 1 1 2 2 4 1 2 2 2 3 3 1 2 1 2 2 0 0 3 2 2 1 1 0 1 2 3 2 1 2 2 2 3 3 4 3 3 1 2 2 3 0 1 1 2 1 3 2 2 2 0 1 3 5 0 3 2 1 2 1 3 2 3 2 1 2 3 2 2 3 1 2 2 1 2 1 1 2 0 1 2 2 2 4 1 2 0 2 3 4 0 3 1 1 3 3 1 1 2 3 1 2 3 | Some statistics to note are mean = 1.985333 variance = 1.175235 standard deviation = 1.084083 Those values change with each reload of this web page because the data in Figure 8 represents 1500 experiments. However, we have equations that give us the mean, variance, and standard deviation for the ideal binary distribution with n=5 and p=0.40. Those equations are: mean: μX = np = 50.40 = 2 variance: σX2 = np(1-p) = 5(0.40)(1-0.40)= 1.2 standard deviation: σX = sqrt(np(1-p)) = sqrt(5(0.40)(1-0.40)) = sqrt(1.2) ≅ 1.095445 You can reload this page many times and the statistics from the actual experiments should always be close to the parameters just given for the ideal distribution. These distribution parameters will come in handy later, after we have learned of the normal distribution and in those cases where we are using a binomial distribution but one with more trials than we can handle with the tables and more trials than we can handle with the software. In those cases, these parameters can be used to get a normal approximation to the binomial distribution. At this point in the course these parameters just remain interesting values, although we do recall that the mean is the expected value. Return to Topics page ©Roger M. Palay Saline, MI 48176 December, 2015 The modification date for this document is 02/26/2022 10:02:25
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logarithm with negative base and negative input blackpenredpen 1380000 subscribers 7723 likes Description 229321 views Posted: 27 Mar 2019 What if we have both a negative base and a negative input in a logarithm! People often say we cannot have a negative number inside of a logarithm. Most of the time log(negative) gives us an imaginary number. But since (-2)^3=-8, so what do you think the answer to log base -2 of -8? Check out defining natural logarithm of a negative number 👉 Subscribe for more math for fun videos 👉 💪 Support this channel, 🛍 Shop math t-shirt & hoodies: (10% off with the code "WELCOME10") 🛍 I use these markers: 408 comments Transcript: let's take a look at the base being negative 2. whoa okay as we all know there are people out there now they will tell you we cannot have negative numbers instead of logarithms because for example if you take a look of log base 2 of negative 8 to see why this right here doesn't have any answer we can put this to be k and then we can change this to an exponential version namely you can just write down 2 to the k power and you make this equal to the inside which is negative eight but the problem is that the left-hand side is exponential meaning this right here is always positive but right here on the right-hand side this right here is negative eight it's a contradiction this right here has no solution therefore it seems like this right here didn't make sense but let's take a look of this one what if we have log and then the inside is still negative eight but this time let's take a look at the base being negative 2. whoa in this case are we going to get the answer for it yes take a look make this equal to k just like hobbit earlier and don't forget this time the base is negative d2 let's give this a parenthesis so when you change this to the exponential function be sure you write down parentheses negative 2 raised to the k's power and then you make this part equal to negative 8. and now can you guys help me give me a k value to make this work thank you the answer is 3 yes so as you can see k equals 3 that will make this work in another word it seems like this right here is just equal to 3 so i'll just put down log base negative 2 of negative 8. we have a totally legitimate answer it's just 3. huh very nice isn't it very nice isn't it and whoa wow so now before we continue i just want to give a shout out to shiriha because he's the one who sent me this right here on twitter right and now i'm going to explain why do people say we cannot have negative numbers instead of logarithms the truth is this right here is actually okay but we just have to go to a complex wolf words and the deal is that right here we should have said this and that has no real valued solutions because we can actually have complex solutions for this and for that in fact we have infinite amount of complex value solutions for this and that that's very cool and i do have a video on this already be sure you guys go check it out by this video i will just write down the result from the video and i'll work out this and that for you guys in the complex world anyway let me make a note right here for you guys first of all we'll be talking about the natural log and then i really wanted inside to be negative let me just write down negative x and i will emphasize x will have to be positive so that's what you do negative x this right here has to be strictly negative and by the way natural log of zero log of zero it's not defined so don't even try this right here is i say ln of negative one times x and you can write that as ln of negative one plus ln of x ln of x if x is positive that's the inner set that's the real number part so we write that down first this right here is equal to natural log of x and then we add ln of negative one and again watch my previous video you end up with the following 2m plus 1 times pi i this i is what brings us to the complex world thanks to the ln of negative 1. this right here is the ln of negative 1 part and then this m right here is just any integer so now let's talk about how we can deal with this right here and i'm just going to put this one right here as number one and here we go we have log base two of negative eight and you see that we have a little trouble already because in order for us to use this we have to have natural log it's okay because we can just use the change of base formula right here and we can rewrite this as natural log of the inside which is negative 8 and then divide this by natural log of the base which is 2 like this on the top here this is where we have to use this formula and of course 8 is the x so we are going to get ln of 8 that's the first part right here and then let me just write down all the values so i just put down 2m plus 1 times pi i for the other part so write down plus 2 m plus 1 and this is just an odd multiple of pi i right and then we have to divide this by ln of 2 like this and that's pretty much the idea so you have the complex solutions for that and of course we can actually simplify this a little bit because for example 8 is the same as saying 2 to the third power and just as usual we can bring the power to the front so we can look at this part as 3 natural log of 2 and then of course we have this part and we just put on plus 2 m plus 1 pi i and we can of course divide this by natural log of 2 and then divide this by natural log of 2. and of course this and now will be canceled it so we actually end up with 3 plus and perhaps i'll just put down the real number part first which is this right here 2 m plus 1 times pi over natural log of 2 and then let's put the i on the side like that to make it look like a complex number it's a complex number of whole idea right a plus bi form so this right here is the solutions for that and the reason i said this is the solutions because m can be any integer if you let m to be zero then you're just talking about pi i over l one two you are just talking about the principal value it's kind of like saying the first answer for that that's pretty much the idea so this right here is it now let's take a look at that one how can we work that out in the complex world huh anyway write it down as natural log of negative 8 over natural log of negative 2 like this and like that the truth is you see something really fun because right here you can actually get infinitely many answers as well when you are in the complex world which is really cool first of all on the top i will have to follow this so let me just put on a fraction bar in black first and this right here we first get ln of 8 and then plus this part plus 2 m plus 1 pi i like this right and then for the bottom here we will also write down ln of 2 and then plus this part as well but here's the issue on the top here you have a set of solutions and this is infinitely many many solutions for that use m because m is just any integers right on the bottom here you cannot use m again technically you can because maybe this and that are equal to each other but in general it's not true so we will have to use let's say 2 we use k already let's use n plus 1 pi i so again the m and n they're just integers so here i'll just indicate m and r integers okay on the top here i know this is the same as that this is done in the sense that this is all the answers for that but if you want to see the three this is what we do we will let m to be 1 and in the meantime i will pick n to be 0. and again this is ok because m and n they are just integers and now as you can see if you put one right here you get the three and you have that three as well so let's see on the top it looks like we have let me just thread it down right here three l and two and then we add this is going to give you 3 because m is 1 3 pi i like this and then on the bottom here we have ln2 and because n is 0 that means we just have pi so we can just put down plus pi i and guess what yes on the top we can factor out the three and we just have ln2 plus pi i which is the same as ln2 plus pi on the bottom so of course you can cancel cancel and in the end you will end up with a nice number 3 which is of course this 3 huh so this 3 is just one of the answers among all the answers from here and i think this is so cool so next time when you hear people say we cannot put negative numbers instead of logarithms the reason is just because they want to stay in the real world but if you're willing to go to a complex world you can do a lot more
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PHZ7427:Solid State II ©Peter Hirschfeld 2021 July 12, 2023 Contents 1 Introduction 5 1.1 Goals in this course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Statistical mechanics of free Fermi gas . . . . . . . . . . . . . . . . . . 5 1.2.1 T = 0 Fermi sea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2.2 T > 0 Free energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2.3 Avg. fermion number. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.2.4 Fermi gas at low T. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.2.5 Classical limit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3 Second quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.3.1 Symmetry of many-particle wavefunctions . . . . . . . . . . . . . . . . 11 1.3.2 Field operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.3.3 2nd-quantized Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.3.4 Schrödinger, Heisenberg, interaction representations . . . . . . . . 16 1.4 Phonons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1.4.1 Review of simple harmonic oscillator quantization . . . . . . . . . . 17 1.4.2 1D harmonic chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 1.4.3 Debye Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.4.4 Anharmonicity & its consequences . . . . . . . . . . . . . . . . . . . . . . 21 2 Electron-electron interactions 24 2.1 Mean field theory (Hartree-Fock) . . . . . . . . . . . . . . . . . . . . . . 25 2.1.1 Validity of Hartree-Fock theory . . . . . . . . . . . . . . . . . . . . . . . . 27 2.1.2 Problem with Hartree-Fock theory . . . . . . . . . . . . . . . . . . . . . . 28 2.2 Screening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.2.1 Elementary treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.2.2 Kubo formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.2.3 Correlation functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.2.4 Dielectric constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.2.5 Lindhard function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 1 2.2.6 Thomas-Fermi theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2.2.7 Friedel oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2.2.8 Plasmons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2.3 Fermi liquid theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.3.1 Particle-hole excitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 2.3.2 Quasiparticles and quasiholes in interacting system . . . . . . . . . 43 2.3.3 Energy of quasiparticles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 2.3.4 Residual quasiparticle interactions . . . . . . . . . . . . . . . . . . . . . . 46 2.3.5 Local energy of a quasiparticle . . . . . . . . . . . . . . . . . . . . . . . . . 49 2.3.6 Thermodynamic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 2.3.7 Quasiparticle relaxation time and transport properties. . . . . . . 51 2.3.8 Effective mass m∗of quasiparticles . . . . . . . . . . . . . . . . . . . . . . 53 3 Quantum Magnetism 55 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.1.1 Atomic magnetic Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.1.2 Curie Law-free spins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 3.1.3 Magnetic interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.2 Ising model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.2.1 Phase transition/critical point . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.2.2 1D solution by transfer matrix . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.2.3 Ferromagnetic domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.3 Ferromagnetic magnons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.3.1 Holstein-Primakoff transformation . . . . . . . . . . . . . . . . . . . . . . 65 3.3.2 Linear spin wave theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 3.3.3 Dynamical Susceptibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 3.4 Quantum antiferromagnet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.4.1 Antiferromagnetic magnons . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 3.4.2 Quantum fluctuations in the ground state . . . . . . . . . . . . . . . . 74 3.4.3 Nonlinear spin wave theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.4.4 Frustrated models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.5 1D & 2D Heisenberg magnets . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.5.1 Mermin-Wagner theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.5.2 1D: Bethe solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.5.3 2D: Brief summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.6 Itinerant magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.7 Stoner model for magnetism in metals . . . . . . . . . . . . . . . . . . 77 3.7.1 Moment formation in itinerant systems . . . . . . . . . . . . . . . . . . 80 3.7.2 RKKY Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 3.7.3 Kondo model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 2 4 Electron-phonon interaction 86 4.1 Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 4.1.1 Derivation of e−-ph coupling . . . . . . . . . . . . . . . . . . . . . . . . . . 87 4.1.2 Jellium model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 4.1.3 Screening. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 4.2 Polarons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.2.1 Polaron in 1D. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.3 Bloch resistivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 4.4 Effective e−−e−interaction . . . . . . . . . . . . . . . . . . . . . . . . . . 93 5 Superconductivity 94 5.1 Phenomenology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 5.2 Electron-phonon interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 5.3 Cooper problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 5.4 Pair condensate & BCS Wavefctn. . . . . . . . . . . . . . . . . . . . . . 100 5.5 BCS Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 5.5.1 BCS wave function, gauge invariance, and number conservation.102 5.5.2 Is the BCS order parameter general? . . . . . . . . . . . . . . . . . . . . 103 5.6 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 5.6.1 Bogoliubov transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 5.6.2 Density of states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 5.6.3 Critical temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 5.6.4 Specific heat. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 5.7 Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 5.7.1 Linear response to vector potential . . . . . . . . . . . . . . . . . . . . . . 108 5.7.2 Meissner Effect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 5.7.3 Dynamical conductivity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 5.8 GL Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 5.9 Type I and Type II superconductivity . . . . . . . . . . . . . . . . . . 119 5.10 Vortex Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 5.11 Properties of Single Vortex. Lower critical field Hc1 . . . . . 127 5.12 Josephson Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 6 Topological States of Matter 135 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 6.2 Berry phase, connection, curvature and all that . . . . . . . . . 136 6.3 Berry curvature of Bloch Bands . . . . . . . . . . . . . . . . . . . . . . . . 138 6.4 Integer Quantum Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . 141 6.5 Graphene . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 6.6 Quantum spin Hall effect and Kane-Mele model . . . . . . . . . 149 6.6.1 Sz = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 3 6.6.2 Sz = −1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 6.7 Edge states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 6.8 Topological character of new insulating phase . . . . . . . . . . . 156 6.9 Kitaev model for 1D topological SC . . . . . . . . . . . . . . . . . . . . 157 ©P.J. Hirschfeld 2021 4 1 Introduction 1.1 Goals in this course These are my hopes for the course. Let me know if you feel the course is not fulfilling them for you. Be free with your criticism & comments! Thanks. • Teach basics of collective phenomena in electron systems • Make frequent reference to real experiment and data • Use 2nd quantized notation without field-theoretical techniques • Get all students reading basic CM journals • Allow students to practice presenting a talk • Allow students to bootstrap own research if possible 1.2 Statistical mechanics of free Fermi gas 1.2.1 T = 0 Fermi sea Start with simple model of electrons in metal, neglecting e−−e−interactions. Hamiltonian is ˆ H = − X j ℏ2∇2 j 2m , j = 1, . . . N particles (1) Eigenstates of each −(ℏ2∇2/2m) are just plane waves eik·r labelled by k, with ki = 2πni/Li in box with periodic B.C. Recall electrons are fermions, which means we can only put one in each single-particle state. Including spin we can put two particles (υ ↓) in each k-state. At zero temperature the ground state of N-electron system is then formed by adding particles until we run out of electrons. Energy is εk = ℏ2k2/2m, so start with two in lowest state k = 0, then add two to next states, with kx or ky or kz = 2π/L, etc. as shown. Energy of highest particle called “Fermi energy" εF, magnitude of corresponding wave vector called kF. Typical Fermi energy for metal εF ≃1eV ≃104K. At T = 0 only states with k < kF occupied (Fermi “sea" or Fermi sphere), so we can write density of electrons as 2# occupied states/Volume (2 is for spin): n = 2 L3 kF X k=0 ≃2 Z k 0 Free energy. Reminder: partition function for free fermions in grnd conical ensemble is Z = Tr e−β( ˆ H−µ ˆ N) (6) = X n1,n2...nk∞ ⟨n1, n2...n∞|e−β( ˆ H−µ ˆ N)|n1, n2...n∞⟩ (7) = X n1,n2...n∞ ⟨n1, n2...n∞|e−β(P i[εini−µni])|n1, n2...n∞⟩ (8) where i labels single-fermion state, e.g. i = k, σ, and ni runs from 0 to 1 for fermions. Since many-fermion state in occ. no. representation is simple product: |n1, n2...n∞⟩= |n1⟩|n2⟩...|n∞⟩, can factorize: Z = X n1 e−β[ε1n1−µn1] ! · · · X n∞ e−β[ε∞n∞−µn∞] ! , (9) 6 so Z = Π∞ i=0 1 + e−β(εi−µ) (10) Since the free energy (grand canonical or thermodynamic potential) is Ω= −kBT log Z, with independent variables (T, V, µ) we get Ω= −kBT P∞ i=1 log 1 + e−β(εi−µ) (11) which has the differential dΩ = −SdT −PdV −Ndµ (12) S = − ∂Ω ∂T  V,µ P = − ∂Ω ∂V  T,µ N = − ∂Ω ∂µ  T,V . (13) 1.2.3 Avg. fermion number. We may want to take statistical averages of quantum operators, for which we need the statistical operator ˆ ρ = Z−1e−β( ˆ H−µ ˆ N). Then any operator ˆ O has an expectation value ⟨ˆ O⟩= Tr(ˆ ρ ˆ O). For example, avg. no. of particles ⟨ˆ N⟩ = Tr(ρ ˆ N) (14) = Tr(e−β( ˆ H−µ ˆ N) ˆ N) Tr(e−β( ˆ H−µ ˆ N)) (15) Now note this is related to the derivative of Ωwrt chem. potential µ: ∂Ω ∂µ = −kBT ∂log Z ∂µ = −kBT Z ∂Z ∂µ (16) = −Tr(ρ ˆ N) = −⟨ˆ N⟩ (17) and using Eq. 11, we see ⟨ˆ N⟩= ∞ X i=1 1 1 + eβ(εi−µ) ≡ ∞ X i=1 n0 i (18) where ni 0 is the avg. number of fermions in a single-particle state i in equilibrium at tem-perature T. If we recall i was a shorthand for k, σ, but εk doesn’t depend on σ, we get Fermi-Dirac distribution function n0 kσ = 1 1 + eβ(εk−µ) (19) 7 1.2.4 Fermi gas at low T. Since the Fermi energy of metals is so high (∼104K), it’s important to understand the limit kBT ≪εF, where the Fermi gas is nearly degenerate, and make sure the classical limit kBT ≫εF comes out right too. Let’s calculate, for example, the entropy and specific heat, which can be obtained from the thermodynamic potential Ωvia the general thermodynamic relations S = − ∂Ω ∂T  V,µ ; CV = T ∂S ∂T  V,µ (20) From (11) and (18), and including spin, we have Ω = −2kBT X k log 1 + e−β(εk−µ) = 2kBTL3 Z dεN(ε) log 1 + e−β(ε−µ) ⇒ cV ≡ CV L3 = 2 1 kBT Z ∞ 0 dεN(ε) −∂f ∂ε  (ε −µ)2 = 2 1 kBT Z ∞ −µ dξ N(ξ) −∂f ∂ξ  ξ2 (21) where I introduced the density of k-states for one spin N(ε) = L−3X kδ(ε −εk). The Fermi function is f(ε) = 1/(1 + exp β(ε −µ)), & I defined shifted energy variable ξ = ε −µ. In general, the degenerate limit is characterized by k sums which decay rapidly for energies far from the Fermi surface, so the game is to assume the density of states varies slowly on a scale of the thermal energy, and replace N(ε) by N(εF) ≡N0. This type of Sommerfeld expansion1 assumes the density of states is a smoothly varying fctn., i.e. the thermodynamic limit V →∞has been taken (otherwise N(ε) is too spiky!). For a parabolic band, εk = ℏ2k2/(2m) in 3D, the delta-fctn. can be evaluated to find2 N(ε) = 3 2 n εF  ε εF 1/2 θ(ε). (24) This can be expanded around the Fermi level:3 1If you are integrating a smooth function of ε multiplied by the Fermi function derivative −∂f/∂ε, the derivative restricts the range of integration to a region of width kBT around the Fermi surface. If you are integrating something times f(ε) itself, it’s convenient to do an integration by parts. The result is (see e.g. Ashcroft & Mermin appendix C) Z ∞ −∞ dεH(ε)f(ε) = Z µ −∞ dεH(ε) + ∞ X n=1 an(kBT)2n d2n−1 dε2n−1 H(ε)|ε=µ (22) where an = 2 −1/22(n−1) ζ(2n) (ζ is Riemann ζ fctn., ζ(2) = π2/6, ζ(4) = π4/90, etc.). 2Here’s one way to get this: N(ε) = L−3 X k δ(ε −εk) → Z d3k (2π)3 | dε dk |−1δ(k − √ 2mε ℏ ) = Z k2dk 2π2  m ℏ2k  δ(k − √ 2mε ℏ ) = 3 2 n εF  ε εF 1/2 (23) 8 Figure 2: Density of states for parabolic spectrum, ε = k2/2m N(ξ) = N(0) + N ′(0)ξ + 1 2N ′′(0)ξ2 + . . . (26) (In a horrible misuse of notation, N(0), N(εF), and N0 all mean the density of states at the Fermi level). The leading order term in the low-T specific heat is therefore found directly by scaling out the factors of T in Eq. (21): cV ≃2k2 B 1 T N0 Z ∞ −∞ dξ −∂f ∂ξ  ξ2 = 2k2 BTN0 Z ∞ −∞ dx −∂f ∂x  x2 | {z } (27) π2/3 So cV ≃2π2 3 N0k2 BT + O(T 3). (28) This is the famous linear in temperature specific heat of a free Fermi gas. 4 1.2.5 Classical limit. I won’t calculate the classical limit. All the standard results for a Boltzman statistics gas, e.g. cV (T ≫εF) = (3/2)NkB follow immediately from noticing that the Fermi function 3When does the validity of the expansion break down? When the approximation that the density of states is a smooth function does, i.e. when the thermal energy kBT is comparable to the splitting between states at the Fermi level, δεk|εF ≃ℏ2kF δk m ≃εF δk kF ≃εF a L , (25) where a is the lattice spacing and L is the box size. At T = 1K, requiring kBT ∼δε, and taking εF /kB ≃104K says that systems (boxes) of size less than 1µm will “show mesoscopic" effects, i.e. results from Sommerfeld-type expansions are no longer valid. 4Note in (28), I extended the lower limit −µ of the integral in Eq. (21) to −∞since it can be shown that the chemical potential is very close to εF at low T. Since we are interested in temperatures kBT ≪εF , and the range in the integral is only several kBT at most, this introduces neglible error. Why: At T=0, the Fermi function n0 k →step function θ(µ −εk), so we know µ(T = 0) must just be the Fermi energy εF = 9 reduces to the Boltzman distribution, f(ε) →e−β(ε−µ), T →∞. (29) (You will need to convince yourself that the classical result µ/(kBT) →−∞is recovered to make this argument.) 1.3 Second quantization The idea behind the term "second quantization" arises from the fact that in the early days of quantum mechanics, forces between particles were treated classically. Momentum, position and other observables were represented by operators which do not in general commute with each other. Particle number is assumed to be quantized as one of the tenets of the theory, e.g. Einstein’s early work on blackbody radiation. At some point it was also realized that forces between particles are also quantized be-cause they are mediated by the exchange of other particles. In Schrödinger’s treatment of the H-atom the force is just the classical static Coulomb force, but a more complete treatment includes the interaction of the H-atom and its constituents with the radiation field, which must itself be quantized (“photons"). This quantization of the fields mediating the interac-tions between matter particles was referred to as “second" quantization. In the meantime, a second-quantized description has been developed in which both “matter fields" and “force fields" are described by second-quantized field operators. In fact, modern condensed matter physics usually does go backwards and describe particles interacting via classical Coulomb forces again,5 but these particles are described by field operators. Once the calculational rules are absorbed, calculating with the 2nd-quantized formalism is easier for most people than doing 1st-quantized calculations. They must, of course, yield the same answer, as they are rigorously equivalent. I will not prove this equivalence in class, as it is exceedingly te-dious, but merely motivate it for you below. I urge you to read the proof once for your own ℏ2(3π2n)2/3/2m. n = N L3 = 2L−3 ∞ X k n0 k = 2 Z dεN(ε)f(ε) ≃ Z µ −∞ dεN(ε) + π2 6 (kBT)2N′(ε)|ε=µ (continued on next page) ≃ Z εF −∞ dεN(ε) | {z } +(µ −εF )N(εF ) + π2 6 (kBT)2N′(ε)|ε=µ n ⇒ µ ≃εF −π2 6 (kBT)2 N′(εF ) N(εF ) Since N′/N is typically of order 1/ε2 F , corrections are small. 5Q: when does this approximation break down? 10 edification, however.6 1.3.1 Symmetry of many-particle wavefunctions Quantum mechanics allows for the possibility of indistinguishable particles, and nature seems to have taken advantage of this as a way to construct things. No electron can be distinguished from another electron, except by saying where it is, what quantum state it is in, etc. Internal quantum-mechanical consistency requires that when we write down a many-identical-particle state, we make that state noncommittal as to which particle is in which single-particle state. For example, we say that we have electron 1 and electron 2, and we put them in states a and b respectively, but exchange symmetry requires (since electrons are fermions) that a satisfactory wave-function has the form Ψ(r1, r2) = A[(ψa(r2)ψb(r1) −ψa(r1)ψb(r2)]. (30) If we have N particles, the wavefunctions must be either symmetric or antisymmetric under exchange:7 ΨB(r1 . . . ri . . . rj . . . rN) = ΨB(r1 . . . rj . . . ri . . . rN) Bosons (31) ΨF(r1 . . . ri . . . rj . . . rN) = −ΨF(r1 . . . rj . . . ri . . . rN) Fermions (32) Given a set of single-particle wave functions ϕEi(r), where Ei is a quantum number, e.g. energy (N.B. it can be any set of quantum numbers!) we can easily construct wave fctns. which satisfy statistics, e.g. Ψ B F n1...n∞(r1, . . . rN) = n1!n2! · · · n∞! N! 1/2 X P∈{E1,E2...EN} (±1) sgnP P ΠN i=1 ϕEi(ri), (33) where the factor in parentheses is the number of ways to arrange N objects in boxes, with n1 in the first box, ... Remarks on Eq. (33): 6See, e.g. Fetter & Wallecka, Quantum Theory of Many-Particle Systems 7This is related to the spin-statistics theorem first formulated by Fierz and Pauli. The proof requires the PCT theorem, proved by Schwinger, Lüders and Pauli, which says that PCT (parity-charge conjugation-time reversal) is a good symmetry for a system described by a Lorentz-invariant field theory). Recently, a generalization of Bose & Fermi statistics to particles called has been intensely discussed. Under exchange an anyon wavefunction behaves as ΨA(r1 . . . ri . . . rj . . . rN) = eiθΨA(r1 . . . rj . . . ri . . . rN) for some 0 ≤θ ≤2π. 11 • sum all permutations of the Ei’s in product ϕE1(r1)ϕE2(r2) . . . ϕEN(rN).8 • # distinct Ei’s occuring may be less than N, some missing because of multiple occupa-tion in boson case. Example: Fig. 2. Possible state of 3 noninteracting Bose particles ΨB 20100...0(r1, r2, r3) = = 1 √ 3{ϕE0(r1)ϕE0(r2)ϕE2(r3) + +ϕE2(r1)ϕE0(r2)ϕE0(r3) + +ϕE0(r1)ϕE2(r2)ϕE0(r3)} • Completely antisymmetric Fermionic wavefunction called Slater determinant: Ψn1...n∞(r1, . . . rN) =  1 N! 1/2 ϕEmin(r1) . . . ϕEmax(r1) . . . . . . ϕEmin(rN) . . . ϕEmax(rN) (34) where there are N eigenvalues which occur between Emin and Emax, inclusive, corre-sponding to N occupied states. 1.3.2 Field operators 2nd quantization is alternative way of describing many body states. We describe a particle by a field operator ˆ ψ(r) = X i aiϕEi(r) (35) where i runs over the quantum numbers associated with the set of eigenstates ϕ, ai is a “coefficient" which is an operator (I’m going to neglect the hats (ˆ ) which normally denote an operator for the a’s and a†’s), and ϕEi is a “1st-quantized" wavefunction, i.e. a normal Schrödinger wavefunction of the type we have used up to now, such that (for example) HϕEi = EiϕEi. Now we impose commutation relations h ˆ ψ(r), ˆ ψ†(r′) i ± = δ(r −r′) (36) h ˆ ψ(r), ˆ ψ(r′) i ± = h ˆ ψ†(r), ˆ ψ†(r′) i ± = 0, (37) 8You might think the physical thing to do would be to sum all permutations of the particle labels. This is correct, but actually makes things harder since one can double count if particles are degenerate (see example of 3 bosons below.) The normalization factor is chosen with the sum over all permutation of the Ei’s in mind. 12 which implies [ai, a† j]± = δij ; [ai, aj]± = [a† i, a† j]± = 0. (38) The upper sign is for fermions and the lower for bosons in Eqs. (37) and (38). Now construct many-body states from vacuum (no particles) |0⟩. a called annihilation op-erator, a† creation operator (see below). Examples & comments (all properties follow from commutation relations): Bosons: • one particle in state i: a† i|0⟩≡|1⟩i • annihilate vacuum ai|0⟩= 0 • (Bosons)9 a† ia† i|0⟩≡|2⟩i • (Bosons) two in i and one in j: a† ia† ia† j|0⟩≡|2⟩i|1⟩j • a† iai ≡ˆ ni is number operator for state i. Proof: (bosons) (a† iai)|n⟩i = (a† iai)(a† i)n|0⟩ = a† i(1 + a† iai)(a† i)n−1|0⟩= |n⟩i + (a† i)2ai(a† i)n−1|0⟩ = 2|n⟩i + (a† i)3ai(a† i)n−2|0⟩· · · = n|n⟩i Similarly show (bosons):10 • a† i|n⟩i = (n + 1)1/2|n + 1⟩i • ai|n⟩i = n1/2|n −1⟩i • many-particle state  1 n1!n2! . . . n∞! 1/2 (a† 1)n1(a† 2)n2 · · · |0⟩≡|n1, n2 . . . n∞⟩ (39) ⋆occupation numbers specify state completely, exchange symmetry included due to commutation relations! (normalization factor left for problem set) Fermions 9Analogous state for fermions is zero, by commutation relations-check! 10By now it should be clear that the algebra for the bosonic case is identical to the algebra of simple harmonic oscillator ladder operators. 13 • Anticommutation relations complicate matters, in particular note a2 = (a†)2 = 0 ⇒ni = 1 0 (Pauli principle) (40) • so a†|0⟩= |1⟩ a|1⟩= |0⟩ a†|1⟩= 0 a|0⟩= 0 (41) • many-particle state (a† 1)n1(a† 2)n2 · · · |0⟩≡|n1, n2 . . . n∞⟩ (42) ⋆note normalization factor is 1 here. • action of creation & annilation operators (suppose ns = 1): as| . . . ns . . . ⟩= (−1)n1+n2+...ns−1(a† 1)n1 · · · asa† s · · · (a† ∞)n∞|0⟩ = (−1)n1+n2+...ns−1(a† 1)n1 · · · (1 −a† sas |{z}) · · · (a† ∞)n∞|0⟩ =0 = (−1)n1+n2+...ns−1| . . . ns −1 . . . ⟩ also as| . . . 0 . . . ⟩= 0 (43) and similarly a† s| . . . ns . . . ⟩=    (−1)n1+n2+...ns−1| . . . ns + 1 . . . ⟩ ns = 0 0 ns = 1 (44) 1.3.3 2nd-quantized Hamiltonian ⋆Point: Now “it can be shown"11 that state vector |Ψ(t)⟩= X n1,n2...n∞ f(n1, n2 . . . n∞, t)|n1, n2 . . . n∞⟩ (45) satisfies the Schrödinger equation (our old friend) iℏ∂ ∂t|Ψ(t)⟩= ˆ H|Ψ(t)⟩ (46) 11Normally I hate skipping proofs. However, as mentioned above, this one is so tedious I just can’t resist. The bravehearted can look, e.g. in chapter 1 of Fetter and Wallecka. 14 if we take the "2nd-quantized" form of ˆ H ˆ H = X ij a† i⟨i|T|j⟩aj + 1 2 X ijkℓ a† ia† j⟨ij|V |kℓ⟩aℓak (47) = Z d3r ˆ ψ†(r)T(r, ∇r) ˆ ψ(r) (48) +1 2 Z Z d3r d3r′ ˆ ψ†(r) ˆ ψ†(r′)V (r, r′) ˆ ψ(r′) ˆ ψ(r) (49) where the 1st quantized Hamiltonian was H = T + V . Here i indexes a complete set of single-particle states. Translationally invariant system It may be more satisfying if we can at least verify that this formalism “works" for a special case, e.g. translationally invariant systems. For such a system the momentum k is a good quantum number, so choose single-particle plane wave states ϕi(r) = ϕkσ(r) = L−3/2eik·ruσ, (50) where uσ is a spinor like u↓= 0 1  , etc. 1st quantized T is −∇2/(2m),12 so 2nd-quantized kinetic energy is ˆ T ≡ X ij a† i⟨i|T|j⟩aj = X kσ  k2 2m  a† kσakσ . (51) Since we showed a† kσakσ is just number operator which counts # of particles in state kσ, this clearly just adds up the kinetic energy of all the occupied states, as it should. For general two-particle interaction V , let’s assume V (r, r′) = V (r −r′) only, as must be true if we have transl. invariance. In terms of the Fourier transform V (q) = 1 L3 Z d3r eiq·r V (r) (52) we have (steps left as exercise) ˆ V = 1 2 X k,k′,q σ,σ′ a† kσa† k′+qσ′V (q)ak′σak+qσ (53) Note that if you draw a picture showing k′ and k + q being destroyed (disappearing), and k and k′ + q being created, you realize that the notation describes a scattering process with two incoming particles and two outgoing ones, where momentum is conserved; in fact, it is easy to see that momentum q is transferred from one particle to another. This is called a Feynman diagram. 12I’ll set ℏ= 1 from here on out, unless required for an honest physical calculation. 15 Figure 3: Feynman diagram for 2-body interaction showing momentum conservation. 1.3.4 Schrödinger, Heisenberg, interaction representations Here we just give some definitions & reminders on different (equivalent) representations of quantum mechanics. In different reps., time dependence is associated with states (Schrödinger), operators (Heisenberg), or combination (interaction). • Schrödinger picture state | ˆ ψS(t)⟩, operators ˆ OS ̸= ˆ OS(t) i ∂ ∂t| ˆ ψS(t)⟩ = ˆ H| ˆ ψS(t)⟩ has formal solution | ˆ ψS(t)⟩= e−i ˆ H(t−t0)| ˆ ψS(t0)⟩ (54) Note ˆ H hermitian ⇒time evolution operator ˆ U ≡ei ˆ H(t−t0) is unitary. • Interaction picture (useful for pert. thy.) ˆ H = ˆ H0 + ˆ H′ (where ˆ H0 usually soluble) Def. in terms of Schr. picture: | ˆ ψI(t)⟩ = ei ˆ H0t| ˆ ψS(t)⟩ ˆ OI(t) = ei ˆ H0t ˆ OSe−i ˆ H0t ⇒i ∂ ∂t| ˆ ψI(t)⟩ = ˆ H′(t)| ˆ ψI(t)⟩ with ˆ H′(t) = ei ˆ H0t ˆ H′e−i ˆ H0t Remarks: 16 – states and ops. t-dependent in interaction picture, but time dependence of operators very simple, e.g. ˆ H0 = X k εka† kak i ∂ ∂takI(t) = ei ˆ H0t[ak, ˆ H0]e−i ˆ H0t = εkakI(t) ⇒akI(t) = ake−iεkt – Time evolution operator determines state at time t: | ˆ ψI(t)⟩= ˆ U(t, t0)| ˆ ψI(t0)⟩ From Schrödinger picture we find ˆ U(t, t0) = ei ˆ H0te−i ˆ H(t−t0)e−i ˆ H0t0 (55) (Note ([ ˆ H, ˆ H0] ̸= 0!) • Heisenberg picture state | ˆ ψH⟩ t-independent, operators ˆ OH(t) = ei ˆ Ht ˆ OSe−i ˆ Ht so operators evolve according to Heisenberg eqn. of motion i ∂ ∂t ˆ OH(t) = [ ˆ OH(t), H] (56) ⋆Note–compare three reps. at t = 0: | ˆ ψH⟩ = | ˆ ψS(0)⟩ = | ˆ ψI(0)⟩ (57) ˆ OS = ˆ OH(0) = ˆ OI(0) (58) 1.4 Phonons 1.4.1 Review of simple harmonic oscillator quantization I will simply write down some results for the standard SHO quantization from elementary QM using “ladder operators". We consider the Hamiltonian H = p2 2M + K 2 q2 (59) 17 and extract the relevant dimensions by putting ω2 = K M ξ = q Mω ℏ 1/2 −i ∂ ∂ξ = p(ℏMω)−1/2 (60) so H = ℏω 2  −∂2 ∂ξ2 + ξ2  . (61) We recall soln. goes like e−ξ2/2Hn(ξ), where Hn are Hermite polynomials, and that eigenval-ues are En = ℏω(n + 1/2) (62) Define ladder operators a, a† as a = 1 √ 2  ξ + ∂ ∂ξ  (63) a† = 1 √ 2  ξ −∂ ∂ξ  (64) Ladder ops. obey commutation relations (check) [a, a†] = 1 ; [a, a] = 0 ; [a†, a†] = 0 (65) & then H may be written (check) H = ℏω  a†a + 1 2  . (66) a, a† connect eigenstates of different quantum nos. n, as |n⟩= (a†)n (n!)1/2|0⟩, (67) where |0⟩is state which obeys a|0⟩= 0. Operating on |n⟩with a† may be shown with use of commutation relations to give a†|n⟩= (n + 1)1/2|n + 1⟩ ; a|n⟩= n1/2|n −1⟩ (68) so with these defs. the ladder operators for SHO are seen to be identical to boson creation and annihilation operators defined above in Sec. 1.3.2. 1.4.2 1D harmonic chain If we now consider N atoms on a linear chain, each attached to its neighbor with a “spring" of spring constant K as shown in figure. First let’s consider the problem classically. The 18 x x i i+1 K K K x x l l+1 eq eq Figure 4: Linear chain with spring couplings K. Dynamical variables are qi ≡xi −xeq i Hamiltonian is H = X ℓ p2 ℓ 2m + K 2 (qℓ−qℓ+1)2, (69) where the qℓ’s are the displacements from atomic equilibrium positions. Now Hamilton’s eqns. (or Newton’s 2nd law!) yield −M ¨ qj = Mω2qj = K(2qj −qj−1 −qj+1). (70) A standing sinusoidal wave qj = A cos(kaj) satisfies this equation if the eigenfrequencies have the form ω2 k = K M 2(1 −cos ka), (71) where if a is the lattice constant, k = 2π/λ. Note that for small k, ωk is linear, ωk ≃ (K/M)1/2ka.13 This is the classical calculation of the normal modes of oscillation on a 1D chain. To quan-tize the theory, let’s impose canonical commutation relations on the position and momentum of the ℓth and jth atoms: [qℓ, pj] = iℏδℓm (72) and construct collective variables which describe the modes themselves (recall k is wave vector, ℓis position) : qℓ = 1 N 1/2 X k eikaℓQk ; Qk = 1 N 1/2 X ℓ e−ikaℓqℓ pℓ = 1 N 1/2 X k e−ikaℓpk ; Pk = 1 N 1/2 X ℓ eikaℓpℓ, (73) 13Note since k = 2πn/(Na), the ωk are independent of system size 19 which leads to canonical commutation relations in wave vector space: [Qk, Pk′] = 1 N X ℓ,m e−ikaleik′am[qℓ, pm] = iℏ N X ℓ e−ial(k−k′) = iℏδk,k′. (74) Let’s now express the Hamiltonian (69) in terms of the new variables. We have, with a little algebra and Eq. (71), X ℓ p2 ℓ = X k PkP−k (75) K 2 X ℓ (qℓ−qℓ−1)2 = K 2 X k QkQ−k(2 −eika −e−ika) = M 2 X k ω2 kQkQ−k so H = 1 2M X k pkp−k + M 2 X k ω2 kQkQ−k . (76) Note that the energy is now expressed as the sum of kinetic + potential energy of each mode k, and there is no more explicit reference to the motion of the atomic constituents. To second quantize the system, we write down creation and annihilation operators for each mode k. Define ak = Mωk 2ℏ 1/2  Qk + i Mωk P−k  (77) a† k = Mωk 2ℏ 1/2  Q−k − i Mωk Pk  (78) which can be shown, just as in the single SHO case, to obey commutation relations h ak, a† k′ i = δkk′ (79) [ak, ak′] = 0 (80) h a† k, a† k′ i = 0 (81) and the Hamiltonian expressed simply as X k ℏωk  a† kak + 1 2  (82) which we could have probably guessed if we had realized that since the normal modes don’t interact, we have simply the energy of all allowed harmonic oscillators. Note this is a quantum Hamiltonian, but the energy scale in H is the classical mode frequency ωk. 1.4.3 Debye Model Let us imagine using the Hamiltonian (82) as a starting point to calculate the specific heat of a (3D) solid due to phonons. We take the Debye model for the dispersion to simplify the 20 calculation, ωk =    ck k < kD 0 k > kD (83) where the Debye wave vector kD = (6π2n)1/3 is obtained by replacing the first Brillouin zone of the solid by a sphere of radius kD which contains N wave vectors, with N the number of ions in the crystal. The average value of the Hamiltonian is U = ⟨H⟩= 3 X k ℏωk  ⟨a† kak⟩+ 1 2  = 3 X k ℏωk  1 eβℏωk −1 + 1 2  (84) since the average number of phonons in state k is simply the expectation value of a boson number operator ⟨a† kak⟩≡Tr(ρa† kak) = b(ℏωk), (85) where b(x) = (exp(βx) −1)−1 is the free Bose distribution function. The factors of 3 come from the 3 independent phonon polarizations, which we consider to be degenerate here. Taking one derivative wrt temperature, the spec. heat per unit volume is14 cV = ∂u ∂T n = 3 ∂ ∂T X k ℏck eβℏck −1 = 3 ∂ ∂T ℏc 2π2 Z kD 0 dk k3 eβℏck −1 ≃ ∂ ∂T 3(kBT)4 2π2(ℏc)3 Z ∞ 0 x3 ex −1 | {z } = ∂ ∂T π2 10 (kBT)4 (ℏc)3 = 2π2 5 kb kBT ℏc 3 (86) π4/15 So far we have done nothing which couldn’t have been done easily by ordinary 1st-quantized methods. I have reviewed some Solid State I material here by way of introduction to problems of interacting particles to which you have not been seriously exposed thus far in the condensed matter grad sequence. The second quantization method becomes manifestly useful for the analysis of interacting systems. I will now sketch the formulation (not the solution) of the problem in the case of the phonon-phonon interaction in the anharmonic crystal. 1.4.4 Anharmonicity & its consequences As you will recall from Solid State I, most thermodynamic properties of insulators, as well as neutron scattering experiments on most materials, can be explained entirely in terms of the harmonic approximation for the ions in the crystal, i.e. by assuming a Hamiltonian of the form (82). There are some problems with the harmonic theory. First, at higher temperatures atoms tend to explore the anharmonic parts of the crystal potential more 14Recall du = TdS −pdV and ∂u ∂T V = T ∂S ∂T V 21 and more, so deviations from predictions of the equilibrium theory increase. The thermal expansion of solids in the harmonic approximation is rigorously zero.15 Secondly, some important qualitative aspects of transport cannot be understood: for example, the harmonic theory predicts infinite thermal transport of insulators! (See A&M Ch. 25 for a qualitative discussion of these failures). The obvious way to go beyond the harmonic approximation is to take into account higher-order corrections to the real-space crystal potential systematically, expanding16 U = 1 2! X ℓm D(2)(ℓ, m)qℓqm + 1 3! X ℓmn D(3)(ℓ, m, n)qℓqmqn + . . . , (88) where D(n)(ℓ, m, . . . n) = ∂nU ∂qℓ∂qm . . . ∂qn ui=0 (89) are the so-called dynamical matrices.17 Using Eqs. (73,78) we find qℓ= 1 √ N  ℏ 2mω 1/2 X k Qkeikaℓ= X k (ak + a† −k)eikaℓ (90) Note that the product of 3 displacements can be written qℓqmqn = 1 (N)3/2 X k1k2k3 ei(k1aℓ+k2am+k3an)Qk1Qk2Qk3 (91) so the cubic term may be written H3 = X k1k2k3 V (3)(k1k2k3)Qk1Qk2Qk3 (92) with V (3)(k1k2k3) = X ℓmn D(3)(ℓ, m, n) exp[i(k1ℓ+ k2m + k3n)] (93) Note now that the indices ℓ, m, n run over all unit cells of the crystal lattice. Since crystal potential itself is periodic, the momenta k1, k2, and k3 in the sum are not really independent. In fact, if we shift all the sums in (93) by a lattice vector j, we would have 15This follows from the independence of the phonon energies in harmonic approx. of the system volume. (see above) Since pressure depends on temperature only through the volume derivative of the mode freqs. (see A & M p. 490),  ∂V ∂T  p =  ∂p ∂T  V  ∂p ∂V  T = 0 (87) 16I have dropped polarization indices everywhere in this discussion, so one must be careful to go back and put them in for a 2- or 3D crystal. 17Recall that the theory with only harmonic and cubic terms is actually formally unstable, since arbitrarily large displacements can lower the energy by an arbitrary amount. If the cubic terms are treated perturbatively, sensible answers normally result. It is usually better to include quartic terms as shown in figure below, however. 22 V (3)(k1k2k3) = X ℓmn D(3)(ℓ+ j, m + j, n + j)ei(k1aℓ+k2am+k3an)ei(k1+k2+k3)aj = X ℓmn D(3)(ℓ, m, n)ei(k1aℓ+k2am+k3an)ei(k1+k2+k3)aj (94) where in the last step I used the fact that the crystal potential U in every lattice cell is equivalent. Now sum both sides over j and divide by N to find V (3)(k1k2k3) = X ℓmn D(3)(ℓ, m, n) exp[i(k1ℓ+ k2m + k3n)]∆(k1 + k2 + k3), (95) where ∆(k) = 1 N X j eikaj = δk,G (96) and G is any reciprocal lattice vector. Return now to (92). We have ascertained that V (3)(ℓ, m, n) is zero unless k1+k2+k3 = G, i.e. crystal momentum is conserved. If we expand (92), we will get products of 3 creation or annihilation operators with coefficients V (3). The values of these coefficients depend on the elastic properties of the solid, and are unimportant for us here. The momenta of the three operators must be such that momentum is conserved up to a reciprocal lattice vector, e.g. if we consider the term ak1a† −k2a† −k3 we have a contribution only from k1 + k2 + k3 = G.18 Note this term should be thought of as corresponding to a physical process wherein a phonon with momentum k1 is destroyed and two phonons with momenta −k2 and −k3 are created. It can be drawn "diagrammatically" à la Feynman (as the 1st of 2 3rd-order processes in the figure below). 18As usual, processes with G = 0 are called normal processes, and those with finite G are called Umklapp processes. 23 k p q k p q k p q q' k p q q' q' k p q Figure 5: Diagrams representing phonon-phonon collision processes allowed by energy and momentum con-servation in 3rd and 4th order. Questions: • How does energy conservation enter? What is importance of processes involving de-struction or creation of 3 phonons? • If one does perturbation theory around harmonic solution, does cubic term contribute to thermal averages? • Can we calculate thermal expansion with cubic Hamiltonian? 2 Electron-electron interactions The electronic structure theory of metals, developed in the 1930’s by Bloch, Bethe, Wilson and others, assumes that electron-electron interactions can be neglected, and that solid-state physics consists of computing and filling the electronic bands based on knowldege of crystal symmetry and atomic valence. To a remarkably large extent, this works. In simple compounds, whether a system is an insulator or a metal can be determined reliably by determining the band filling in a noninteracting calculation. Band gaps are sometimes difficult to calculate quantitatively, but inclusion of simple renormalizations to 1-electron band structure known as Hartree-Fock corrections, equivalent to calculating the average energy shift of a single electron in the presence of an average density determined by all other electrons (“mean field theory"), almost always suffices to fix this problem. There are two reasons why we now focus our attention on e−−e−interactions, and why almost all problems of modern condensed matter physics as it relates to metallic systems focus on this topic: 1. Why does the theory work so well in simple systems? This is far from clear. In a good metal the average interelectron distance is of the order of or smaller than the range of the 24 interaction, e.g. the screening length ℓscr ∼(c/e2m)1/2ρ−1/6, where ρ is the density, of order 1nm for typical parameters. One might therefore expect that interactions should strongly modify the picture of free electrons commonly used to describe metals. 2. More complicated systems exhibit dramatic deviations from the predictions of band theory. I refer now not simply to large quantitative errors in the positions of 1-electron bands, but to qualitative discrepancies in the behavior of the materials. The most inter-esting modern example is the class of compounds known as the transition metal oxides, including the cuprate materials which give rise to high-temperature superconductivity (HTS).19 2.1 Mean field theory (Hartree-Fock) Let’s begin with the second-quantized form of the electronic Hamiltonian with 2-body inter-actions. The treatment will be similar to that of Kittel (Ch. 5), but for comparison I will focus on a translationally invariant system and do the calculation in momentum space. In this case as we have shown the Hamiltonian is20 ˆ H = ˆ T + ˆ V , (97) ˆ T = = X kσ  k2 2m  c† kσckσ (98) ˆ V = 1 2V X k,k′,q σ,σ′ c† kσc† k′+qσ′V (q)ck′σ′ck+qσ. (99) The 2-body interaction ˆ V contains 4 Fermi operators c and is therefore not exactly soluble. The goal is to write down an effective 2-body Hamiltonian which takes into account the 19The parent compounds of the HTS (e.g. La2CuO4 and YBa2Cu3O4 are without exception antiferromagnetic and insulating. Historically, this discovery drove much of the fundamental interest in these materials, since standard electronic structure calculations predicted they should be paramagnetic metals. This can easily be seen by simple valence counting arguments. The nominal valences in e.g. La2CuO4 are La3+, O2−, and Cu2+ The La and O ions are in closed shell configurations, whereas the Cu is in an [Ar]3d9 state, i.e. a single d-hole since there are 10 d electrons in a filled shell. The 1 hole/unit cell would then suggest a 1/2 filled band and therefore a metallic state. The crystal field splittings in the planar Cu environment give the 3dx2−y2 state being lowest in energy (see later), and since the n.n.’s are the O′s, it would seem likely that the lowest O crystal field state in the planar environment, the O 3p, will hybridize with it. Sophisticated LDA calculations confirm this general picture that the dominant band at the Fermi level is a 1/2-filled planar Cu dx2−y2- O 3p band. Instead, this compound is found experimentally to be an electrical insulator and an antiferromagnet with a Neel temperature of 300K! This strongly suggests that the band calculations are missing an important element of the physics of the high-Tc materials. In the transition metals and TMO’s band theory is notoriously bad at calculating band gaps in insulators, because of the presence of strong local Coulomb interactions which lead to electronic correlations neglected in the theory. It is therefore reasonable to assume that a similar mechanism may be at work here: extremely strong Coulomb interactions in the Cu −O planes are actually opening up a gap in the 1/2-filled band at the Fermi level and simultaneously creating a magnetic state. This phenomenon has been studied in simpler systems since the 60s and is known as the Mott-Hubbard transition. 20For the moment, we ignore the ionic degrees of freedom and treat only the electrons. To maintain charge neutrality, we therefore assume the electrons move in a neutralizing positive background (“jellium model"). 25 average effects of the interactions. We therefore replace the 4-Fermi interaction with a sum of all possible 2-body terms, c† 1c† 2c3c4 ≃ −⟨c† 1c3⟩c† 2c4 −⟨c† 2c4⟩c† 1c3 + ⟨c† 1c4⟩c† 2c3 + ⟨c† 2c3⟩c† 1c4, (100) where the + and −signs are dictated by insisting that one factor of -1 occur for each commutation of two fermion operators required to achieve the given ordering. This can be thought of as “mean field terms", in the spirit of Pierre Weiss, who replaced the magnetic interaction term Si · Sj in a ferromagnet by ⟨Si⟩Sj = ⟨S⟩Sj ≡−HeffSj, i.e. he replaced the field Si by its homogeneous mean value S, and was left with a term equivalent to a 1-body term corresponding to a spin in an external field which was soluble. The mean field ⟨S⟩in the Weiss theory is the instantaneous average magnetization of all the other spins except the spin Sj, and here we attempt the same thing, albeit somewhat more formally. The “mean field" ⟨c† kσck′σ′⟩= ⟨c† kσckσ⟩δkk′δσσ′ ≡nkσδkk′δσσ′ (101) is the average number of particles nkσ in the state kσ, which will be weighted with the 2-body interaction V (q) to give the average interaction due to all other particles (see below).21 With these arguments in mind, we use the approximate form (100) and replace the inter-action ˆ V in (99) by ˆ VHF = 1 2 X kk′q σσ′ V (q) h −⟨c† kσck′σ′⟩c† k′+qσ′ck+qσ −⟨c† k′+qσ′ck+qσ⟩c† kσck′σ′+ +⟨c† kσck+qσ⟩c† k′+qσ′ck′σ′ + ⟨c† k′+qσ′ck′σ′⟩c† kσck+qσ i = − X kq σ V (q)⟨c† kσckσ⟩c† k+qσck+qσ + V (0) X kk′ σσ′ ⟨c† kσckσ⟩c† k′σ′ck′σ′ = X kσ − X q nk−qσV (q) + nV (0) ! c† kσckσ, (102) where the total density n is defined to be n = P kσ nkσ. Since this is now a 1-body term of the form P kσ ΣHF(k)a† kσakσ, it is clear the full Hartree-Fock Hamiltonian may be written in terms of a k-dependent energy shift: ˆ HHF = X kσ ℏ2k2 2m + ΣHF(k)  c† kσckσ, (103) ΣHF(k) = − X q nk−qσV (q) | {z } + nV (0) | {z } (104) Fock Hartree (105) 21Note we have not allowed for mean fields of the form ⟨c†c†⟩or ⟨aa⟩. These averages vanish in a normal metal due to number conservation, but will be retained in the theory of superconductivity. 26 Note the Hartree or direct Coulomb term, which represents the average interaction energy of the electron kσ with all the other electrons in the system, is merely a constant, and as such it can be absorbed into a chemical potential. In fact it is seen to be divergent if V (q) represents the Coulomb interaction 4πe2/q2, but this divergence must cancel exactly with the constant arising from the sum of the self-energy of the positive background and the interaction energy of the electron gas with that background.22 The Fock, or exchange term23 is a momentum-dependent shift. 2.1.1 Validity of Hartree-Fock theory Crucial question: when is such an approximation a good one for an interacting system? The answer depends on the range of the interaction. The HF approximation becomes asymptot-ically exact for certain quantities in the limit of high density Fermi systems if interactions are suffiently long-range! This is counterintuitive, for it seems obvious that if the particles are further apart on the average, they will interact less strongly and less often, so a mean field theory like Hartree-Fock theory should work well. This is true if the interactions are short-ranged, such that for suffiently low densities the particles spend no time within an interaction length. The interaction is typically characterized by a strength V and a range a, and if the interparticle spacing r0 ≫a, the particles don’t feel the potential much and the ground state energy, for example, of such a gas can be expanded at T = 0 in powers of a/r0. If a →0, the interaction dispapears and the the only characteristic energy left at T = 0 is the zero-point energy, i.e. the energy scale obtained by confining the particle to a cage of size the interparticle spacing r0, i.e. ℏ2/(2mr2 0). This is just the Fermi energy in the case of a Fermi system. Thus we might expect HF theory to be quite good for a dilute gas of 3He, since the 3He-3He interaction is certainly short-ranged; unfortunately, liquid 3He has a ≃r0, so corrections to HF are always important. What about for a metal? We can do a simple argument to convince ourselves that HF isn’t applicable for low density systems with Coulomb interactions. Let’s rewrite the basic 2-body Hamiltonian in dimensionless variables, 22See e.g., Kittel 23The origin of the term exchange is most easily seen in many-body perturbation theory, where the Fock term is seen to arise from a scattering process where the particle in state kσ changes places with a particle from the medium in an intermediate state. 27 ˆ H = X kσ ℏ2k2 2m  c† kσckσ + 1 2V X k,k′,q σ,σ′ c† kσc† k′+qσ′ 4πe2 q2  ck′σ′ck+qσ (106) = e2 2a0r2 s    X ¯ kσ ¯ k2c† ¯ kσc¯ kσ + 3rs N X ¯ k,¯ k′,¯ q σ,σ′ c† ¯ kσc† ¯ k′+¯ qσ′  1 ¯ q2  c¯ k′σ′c¯ k+¯ qσ   , (107) where I’ve used the system volume V = N(4/3)πr3 0, Bohr radius a0 = ℏ2/me2, defined rs ≡r0/a0, and scaled all momenta as k = ¯ kr−1 0 , etc. Applying dimensional analysis, we expect the two terms P ¯ k2c†c and P c†c†(1/¯ q2)cc to be of order 1, so that it is clear that the interaction term becomes negligible in the limit rs →0 (high density). On the other hand, it’s also clear that had the 1/q2 factor not been present in the interaction, it would have scaled as 1/rs instead, and become negligible in the low density limit. One can do a more careful, formal perturbation analysis for a simple quantity like the ground state energy ⟨H⟩of the electron gas, which is discussed in many-body physics texts. The result is E0 = EHF 0 + Ecorr (108) with EHF 0 the ground state energy calculated in the independent particle approximation with the Hartree-Fock energy shifts, and Ecorr the correlation energy, defined to be the part of the g.s. energy not captured by HF. As the density increases, so does εF, which represents the average kinetic energy per electron; this increase is faster than the increase of the correlation energy, as K.E. = 3 5εF = 2.21 r2 s Ryd (109) P.E.|HF = −0.916 rs Ryd (110) Ecorr = (0.0622 log rs −0.096 + . . . ) Ryd (111) so it is clear the correlation term is less singular than the HF term in the limit rs →0. At higher densities, particles are effectively independent.24 2.1.2 Problem with Hartree-Fock theory Although we argued that the Hartree-Fock approximation becomes a better approximation in the limit of high density for electrons interacting via the Coulomb interaction, it never becomes exact, as one can see by examining the Fourier transform of the Hartree-Fock energy shift. The Hartree term itself (being k independent) is absorbed into a redefinition of the chemical potential, so the shift is (T=0): 24For a more detailed discus of these terms, see Fetter & Wallecka, Quantum Theory of Many-Particle Systems 28 δεk = ± 1 L3 X |k′|<kF 4πe2 |k −k′|2 (112) = 2e2kF π F(k/kF), (113) where F(x) is a function which has a log divergence in slope at x = 1, i.e. at the Fermi level.25 This means while the energy shift might be small compared to the Fermi energy the velocity of an electron is ∂εk/∂k|kF , which contains a term which is infinite. This problem can be traced back to the long-range nature of the Coulomb force. Two electrons at large distances r −r′ don’t really feel the full 1/|r −r′|2 but a “screened" version due to the presence of the intervening medium, i.e. The electron gas rearranges itself to cancel out the long-range part of V . 2.2 Screening 2.2.1 Elementary treatment To see how this works, let’s first examine the problem of a single fixed (i.e. infinite mass) charge placed inside an electron gas. We are interested in how the electron gas responds to this perturbation (recall in the HF approximation the perturbation would simply be the interaction of the charge with the uniform gas). But since the electrons are mobile and negatively charged, they will tend to cluster around the test charge if its charge is positive, and be repelled (i.e. have a lower density) around the test charge if it is negative. In addition, Fermi statistics lead to long-range oscillations of the electron charge density far from the test charge. This is not entirely a “Gedanken experiment". Impurities in solids may have different valences from the host, and thus acquire a localized charge, although the entire solid is still neutral. As an example of some relevance to modern condensed matter physics, consider 25The integral is straightforward & useful: 1 L3 X k′ ′ 1 (k −k′)2 = Z kF 0 k2dk 4π2 Z 1 −1 dx 1 k2 + k′2 −2kk′x = 1 8π2k Z kF 0 dk′ k′ log k + k′ k −k′ (114) = 1 8π2 k2 F −k2 k log kF + k kF −k + 2kF ! (115) so with z = k/kF and (113) we have F(z) = 1 −z2 4z log 1 + z 1 −z + 1 2 (116) 29 what happens when you put an extra Oxygen atom in the CuO2 plane of a cuprate high-Tc superconducting material. Although the crystal structure of these materials is complex, they all contain CuO2 planes as shown in the figures. The interstitial oxygen will capture two electrons from the valence band, changing it from a 2s22p4 to a 2s22p6 electronic configura-tion. The impurity will then have a net charge of two electrons. For r near the impurity, the Cu O O Cu Cu O O O O Cu Cu Cu Cu Cu Cu O O O O O O Cu O O Cu Cu O O O O Cu Cu Cu Cu Cu Cu O O O O O O a) b) O 2e Figure 6: a) The CuO2 plane of cuprate superconductor. b) Interstitial charged O impurity. electronic charge density will be reduced by the Coulomb interaction. The electronic number density fluctuation is26 n(r) ≈ Z εF −eδϕ(r) N(ω + eδϕ(r))dω ≈ Z εF +eδϕ(r) 0 N(ω′)dω′ (117) Note that this equation involves the double-spin density of states, which we will use through-out this section. On the other hand, in the bulk far away from the impurity, δϕ(rbulk) = 0, -e δφ N( ) ω N( ) ω ω ω εF εF around impurity bulk Figure 7: Local repulsion of charge due to shift of electronic density of states near impurity 26A note on signs: the electrostatic potential δϕ(r) < 0 due to the excess negative charge at the impurity. You might worry about this: the total change in ϕ should include the self-consistently determined positive charge screening the impurity. We have fudged this a bit, and the correct separation between induced and external quantities will become clear later in the discussion. Note we take e > 0 always, so the electron has charge −e. 30 so n(rbulk) ≈ Z εF 0 N(ω)dω (118) or δn(r) ≈ Z εF +eδϕ(r) 0 N(ω)dω − Z εF 0 N(ω)dω (119) Let us assume a weak potential |eδϕ| ≪εF; therefore δn(r) ≈N(εF) [εF + eδϕ −εF] = +eδϕN0 . (120) We can solve for the change in the electrostatic potential by solving Poisson’s equation. ∇2δϕ = −4πδρ = 4πeδn = 4πe2N0δϕ . (121) Define a new length k−1 TF, the Thomas-Fermi screening length, by k2 TF = 4πe2N0, so that P’s eqn. is ∇2δϕ = r−2∂rr2∂rδϕ = k2 TFδϕ, which has the solution δϕ(r) = C exp(−kTFr)/r. Now C may be determined by letting N(εF) = 0, so the medium in which the charge is embedded becomes vacuum. Then the potential of the charge has to approach q/r, so C = q, i.e. in the electron gas δϕ(r) = qe−kT F r r , (122) where q = −2e for our example. Let’s estimate some numbers, by using the free electron gas. Use kF = (3π2n)1/3, a0 = ℏ2/(me2) = 0.53 ◦ A, and N(εF) = mkF/(ℏ2π2). Then we find k−2 TF = a0π 4(3π2n)1/3 ≈ a0 4n1/3 k−1 TF ≈ 1 2  n a3 0 −1/6 (123) In Cu, for which n ≈1023 cm−3 (and since a0 = 0.53 ◦ A) k−1 TF ≈1 2 (1023)−1/6 (0.5 × 10−8)−1/2 ≈0.5 × 10−8 cm = 0.5 ◦ A (124) Thus, if we add a charged impurity to Cu metal, the effect of the impurity’s ionic potential is exponentially screened away for distances r > 1 2 ◦ A. The screening length in a semiconductor can clearly be considerably longer because n is much smaller, but because of the 1/6 power which appears in, even with n = 10−10, k−1 TF is only about 150 times larger or 75 ◦ A. What happens in the quantum-mechanical problem of an electron moving in the potential created by the impurity when the screening length gets long? As shown in the figure, the potential then correspondingly deepens, and from elementary quantum mechanics we expect it to be able to bind more electrons, i.e bound states “pop out" of the continuum into orbits bound to the impurity site. 31 r kTF -1 =0.5A r kTF -1 =5A e-kTFr r -e-kTFr r -extended bound Figure 8: Screened impurity potentials. As the density decreases, the screening length increases, so that states which were extended bound. In a poor metal (e.g., YBCO), in which the valence state is just barely unbound, decreasing the number of carriers will increase the screening length, since k−1 TF ∼n−1/6 . (125) This will extend the range of the potential, causing it to trap or bind more states–making the one free valence state bound. Of course, this has only a small effect O(1/N) effect on the bulk electrical properties of the material. Now imagine that instead of a single impurity, we have a concentrated system of such ions, and suppose that we decrease n (e.g. in Si-based semiconductors, this is done by adding acceptor dopants, such as B, Al, Ga, etc.). This will in turn, increase the screening length, causing some states that were free to become bound, eventually possibly causing an abrupt transition from a metal to an insulator. This process is believed to explain the metal-insulator transition in some amorphous semiconductors. 2.2.2 Kubo formula Many simple measurements on bulk statistical systems can be described by applying a small external probe field of some type to the system at t = 0, and asking how the system responds. If the field is small enough, the response is proportional to the external perturbation, and the proportionality constant, a linear response coefficient, is given always in terms of a correlation function of the system in the equilibrium ensemble without the perturbation. In Table 1 I list some standard linear response coefficients for condensed matter systems. 32 System Perturbation Response Coefficient metal electric field E current j conductivity σ temp. gradient ∇T heat current jQ thermal cond. κ point charge q density fluct. δn density correlation func-tion χc magnetic field B magnetization M susceptibility χs The general theory of linear response is embodied in the Kubo formula.27 Suppose a system is described by a time-independent Hamiltonian H, and a perturbation ˆ H′(t) is turned on at time t = 0. The idea is to express the subsequent time-dependence of expectation values of physical observables in terms of matrix elements of the perturbation in the unperturbed ensemble. The unperturbed ensemble with ˆ H, or ˆ H −µ ˆ N, may involve interactions, but we assume we know the exact energy eigenstates and eigenvalues. All observables can be calcu-lated by knowing the time evolution of the statistical operator ρ(t), given by the Heisenberg equation i∂ˆ ρ ∂t = [ ˆ H + ˆ H′, ˆ ρ] (126) First one defines a canonical transformation28: ˜ ρ(t) = ˆ S(t)ˆ ρ(t) ˆ S†; S(t) ≡ei ˆ Ht (127) from which one can see by substituting into (126) that i∂˜ ρ ∂t = [ ˜ H′, ˜ ρ] (128) with ˜ H′ = ˆ S ˆ H′ ˆ S†. This has the formal solution ˜ ρ(t) = ˜ ρ(0) −i Z t 0 [ ˜ H′, ˜ ρ]dt′. (129) The initial condition is ˜ ρ(0) = ˆ S(0)ˆ ρ(0) ˆ S(0)† = ˆ ρ(0), and we can iterate the rhs by substi-tuting ˆ ρ(0) for ˜ ρ(t′) in the integral, etc.29 Since we are interested only in leading order, we write ˜ ρ(t) ≃ˆ ρ(0) −i Z t 0 [ ˜ H′, ˆ ρ(0)]dt′. (130) If we now multiply on the left by ˆ S† and on the right by ˆ S, we obtain the leading order in ˆ H′ change in the original ρ: ˆ ρ(t) ≃ˆ ρ(0) −i ˆ S†(t) Z t 0 [ ˜ H′, ˆ ρ(0)]dt′  ˆ S(t), (131) 27R. Kubo, J. Phys. Soc. Jpn. 12, 570 (1957). 28The idea is to remove the “trivial" time-evolution of ˆ ρ on ˆ H, to isolate the dependence on ˆ H′. 29Note ˆ ρ(0) is the initial but still possibly interacting density matrix (statistical operator). I reserve the symbol ˆ ρ0 for the noninteracting analog. Note further that ˆ ρ(0) contains only ˆ H, not ˆ H′, since we assumed perturbation was turned on at t = 0. 33 which can now be used to construct averages in the usual way. The expectation value of an operator ˆ O at nonzero T and at some time t is ⟨ˆ O⟩ ≡ Tr(ˆ ρ(t) ˆ O) (132) ≃ Tr  ˆ ρ(0) ˆ O  −i Z t 0 dt′Tr  S†(t)[ ˜ H′, ˆ ρ(0)] ˆ S(t) ˆ O  = Tr  ˆ ρ(0) ˆ O  + i Z t 0 dt′Tr  ˆ ρ(0)[ ˜ H′, ˜ O(t)]  , (133) where in the last step the cyclic property of the trace and the definition of the Heisenberg operator ˜ O = ˆ S ˆ O ˆ S† was used. The change in the expectation value due to the perturbation is then δ⟨ˆ O⟩= i Z t 0 dt′Tr  ˆ ρ(0)[ ˜ H′, ˜ O(t)]  (134) Since typically ˆ H′ is proportional to some external (c-number) field, we can write ˆ H′ as ˆ H′(t) = Z d3r ˆ B(r)ϕ(r, t) (so ˜ H′(t) = Z d3r ˜ B(r, t)ϕ(r, t)). More often you will see the Kubo formula expressed as the change in the expectation value of the operator ˆ A in response to an applied weak field ˆ B, δ⟨ˆ A(1)⟩ = i Z t 0 Z Tr  ˆ ρ(0)[ ˜ B(1′), ˜ A(1)]  ϕ(1′)d1′ (135) ≡ Z ∞ 0 GAB(1, 1′)ϕ(1′)d1′, (136) with the notation 1 ≡r, t, σ, 1′ = r′, t′, σ′, R d1′ ≡P σ′ R d3r′dt′, etc., and we defined GAB(1, 1′) = GAB(r, t; r′, t′) ≡−iTr  ˆ ρ(0)[ ˜ A(r, t), ˜ B(r′, t′)]  θ(t −t′) .30 Remember that the operators ˜ A and ˜ B are just the Heisenberg t-dependent versions of the Schrödinger operators ˆ A, ˆ B. 2.2.3 Correlation functions As a concrete example, consider the change in the density δn(r, t) of a system in response to a local change in its density perhaps at a different place and time δn(r′, t′), where ˆ n(r) = ⟨ˆ ψ†(r) ˆ ψ(r)⟩, and ⟨ˆ ψ†(r) ˆ ψ(r)⟩ˆ H′=0 = n0. It’s usually convenient to set n0 = 0, i.e. we assume an overall neutral system. The linear response function is then χ(r, t; r′, t′) = −i⟨[ˆ n(r, t), ˆ n(r′, t′)]⟩ˆ H′=0θ(t −t′), (137) where the exectation value is taken in the unperturbed system. The function Gnn measures the average amplitude in the unperturbed ensemble a particle is to be found at r′, t′ and subsequently at r, t. If the particles are statistically independent, χ factorizes, ⟨nn′⟩∼ ⟨n⟩⟨n′⟩= 0 (no correlations), so it has the interpretation of a correlation function. 30The object Tr  ˆ ρ(0)[ ˆ AH(1), ˆ BH(1′)]  which appears on the right-hand side of (134) is related to something called a retarded 2-particle Green’s function, which I have promised not to oppress you with. 34 2.2.4 Dielectric constant The simplest way to analyze this problem conceptually is to ask what is the potential a long distance from a test charge placed into the interacting electron gas medium. This potential will be the sum of the bare potential produced by the test charge in a vacuum, plus the induced potential due to the polarization of the medium. Let’s use what we learned about linear response in the last section to calculate the electric potential in the system ϕ(r, t) due to an external potential ϕext(r′, t′) (you may think of this as the potential due to a test charge if you like). From electrodynamics we recall these two are supposed to be proportional, and that the proportionality constant is just the dielectric constant ϵ. By analogy we write, in Fourier transformed language, ϕext(q, ω) = ϵ(q, ω)ϕ(q, ω) (138) It is the dielectric constant of the medium (which is directly related to the conductivity) which we now wish to calculate. The total potential in the material is the sum of the external potential and the induced potential, ϕ = ϕind + ϕext, (139) but in fact we can calculate the induced potential ϕind by • first finding the induced change in density δρind due to the external potential, using linear response theory; then • using Poisson’s equation to calculate the induced electric potential. The additional term in the Hamiltonian in the presence of the “test charge" is ˆ H′ = e Z d3r ˆ n(r)ϕext(r, t) (140) where ˆ n is the 2nd-quantized density operator ˆ ψ† ˆ ψ. The induced change in charge density δρ = e⟨ˆ n⟩is then just given by the Kubo formula as above,31 δρind(1) = −ie2 Z t 0 Z d3r′Tr (ˆ ρ(0)[ˆ nH(1), ˆ nH(1′)]) ϕext(1′)d1′ (141) ≡ e2 Z ∞ 0 χ(1, 1′)ϕext(1′)d1′ (142) where the density-density correlation function is32 χ(1, 1′) = −iTr (ˆ ρ(0)[ˆ nH(1), ˆ nH(1′)]) θ(t −t′). (143) 31Don’t confuse charge density ρ with statistical operator ˆ ρ. 32Instead of G(2) nn, it’s more conventional to use the notation χnn or just χ. Note also the notation ˆ nH, which is the same as ˜ n in previous section for time evolution with Hamiltonian H. 35 For a translationally invariant system in Fourier transform rep., the convolution (142) reduces to δρind(q, ω) = e2χ(q, ω)ϕext(q, ω). (144) Now using Poisson’s equation q2ϕind = 4πδρind, and Eq. (139), we find33 ϕind(q, ω) = 4πe2 q2 χ(q, ω)ϕext(q, ω) ≡V (q)χ(q, ω)ϕext (145) so ϵ(q, ω) ≡ ϕext(q, ω) ϕind(q, ω) + ϕext(q, ω) = 1 1 + V (q)χ(q, ω). (146) 2.2.5 Lindhard function We’ve found the exact structure of the dielectric response,34 but not really solved it because we still don’t know the density-density correlation function χ. For the interacting system this can be a very difficult problem, but it is not too hard to find it for the free fermion gas. We will need • The definitions of Fourier transforms χ(q, ω) = Z d3(r −r′) Z d(t −t′)eiq·(r−r′)e−iω(t−t′)χ(1, 1′) c† k(t) = Z d3reik·r ˆ ψ†(1) ck(t) = Z d3re−ik·r ˆ ψ(1) (149) • the integral expression for the theta function θ(t −t′) = − Z dω′ 2πi e−iω′(t−t′) ω′ + iη+ , (150) where η+ is a positive infinitesimal.35 • the solutions of the Heisenberg equation of motion i∂t ˆ OH(t) = [ ˆ OH(t), H] for c† kσ(t) and ckσ(t) for free fermions (check): c† kσ(t) = c† kσeiϵkt (151) ckσ(t) = ckσe−iϵkt (152) 33Compare Eq. (121) 34Warning! Compare the treatment here with e.g., A& M pp. 337-340, and you will see that they arrive at the apparently different result ϵ(q, ω) = 1 −V (q)χ(q, ω), (147) which is also claimed to be exact. This is because χ is defined to be the response to the total potential ρind(q, ω) = χ|A&M(q, ω)ϕ(q, ω) (148) which is very difficult to calculate since ϕ is self-consistently determined. 35The sign of the infinitesimal is important. It determines in which complex half-plane the contour can be completed to evaluate the integral by Cauchy’s theorem. Check it! 36 • The commutator (check!) [c† k1ck2, c† k3ck4] = δk2k3c† k1ck4 −δk1k4c† k3ck2 (153) • The expectation values in the noninteracting ensemble of the number operator at finite temperature T are ⟨ˆ nk⟩= f(ϵk), (154) where f is the Fermi function. Then (suppressing spin indices)36 ⟨[nH(1), nH[1′]⟩= Tr(ˆ ρ0[nH(1), nH[1′]) = X k1...k4 ei(k1−k2)·rei(k3−k4)·r′⟨[c† k1(t)ck2(t), c† k3(t′)ck4(t′)]⟩ = X k1...k4 ei(k1−k2)·rei(k3−k4)·r′ei([ϵk3−ϵk4]t′+[ϵk1−ϵk2]t)⟨[c† k1ck2, c† k3ck4]⟩ = X k1k2 ei(k1·[r−r′]+k2·[r′−r])ei(ϵk2−ϵk1)(t′−t)⟨ˆ nk1⟩ − X k3k4 ei(k4·[r−r′]+k3·[r′−r])ei(ϵk3−ϵk4)(t′−t)⟨ˆ nk3⟩ (155) so that (using (143) and (149), and putting back the spin sum) χ0(q, ω) = X kσ f(ϵk) −f(ϵk+q) ω −(ϵk+q −ϵk) + iη . (156) This is the so-called Lindhard function, the charge susceptibility of the free Fermi gas. Note that in the free gas the major contributions to the integral (156) come from contributions where ϵk+q −ϵk −ω = 0. These correspond to 2-particle “particle-hole excitations" where the Fermi sea is given a little energy ω which promotes a particle from below the Fermi surface (say k) to above (say k + q). In the homework you will investigate when such excitations are possible within this model. The use of the free gas susceptibility χ0 for χ in Eq. (146) is a rather poor approximation, as it completely ignores the effect of the long-range Coulomb interaction on the intermedi-ate states. The self-consistent field method, or random phase approximation, replaces the induced charge without interactions with the screened charge: χ0 ≡ρind ϕext V =0 →ρind|V =0/ϵ ϕext = χ0 ϵ . (157) 36The following fermionic identity is also useful–check it! ⟨c† 1c2c† 3c4⟩= n1n3(δ12δ34 −δ14δ23 + δ14δ23δ12) + n1(δ14δ23 −δ14δ23δ12) 37 Solving (146) for ϵ gives ϵ(q, ω) ≃1 −V (q)χ0(q, ω) (RPA/SCF) (158) This is sometimes called the Lindhard approx. to the dielectric function. 2.2.6 Thomas-Fermi theory Let us consider the static limit of the theory we currently have. This will allow us to answer the original question we posed: if we insert a charge somewhere into a metal, what is the net electrostatic potential felt at some distance r away by a second charge? Taking the ω →0 limit in (156), we find χ0(q, 0) = X kσ f(εk+q) −f(εk) εk+q −εk + iη+ (159) First note that if we further take the limit q →0 we find (here N0 is double-spin dos!) χ0 → −2 X k −∂f ∂εk ≃−N0 Z dξk −∂f ∂ξk ≃−N0 Z dξkδ(ξk) = −N0. z }| { T ≪ϵF (160) The dielectric constant in the long-wavelength (i.e. small q) limit may thus be written ϵ(q, 0) ≃1 + 4πe2 q2 N0 ≡1 + k2 TF q2 , (161) where kTF is the same Thomas-Fermi wave vector k2 TF ≡4πe2N0 we discussed earlier. Now we go back: if we had placed a charge e in a vacuum, it would have produced a potential ϕext(r) = e/r, or ϕext(q) = 4πe/q2 everywhere in space. From (138) we see that the actual potential in the medium is ϕ(q) = ϕext(q)/ϵ(q) = 4πe/q2 1 + k2 TF/q2 = 4πe k2 TF + q2, (162) i.e. the Thomas-Fermi wavevector cuts off the divergence of the bare long-range Coulomb interaction. In real space ϕ(r) = X q eiq·rϕ(q) = e re−kT F r , (163) so the long-range interaction is indeed replaced by an exponentially cut off short-range one, with screening length scale k−1 TF. 2.2.7 Friedel oscillations The Thomas-Fermi theory is very intuitive and describes realistically how the singularities of the long-range Coulomb interaction are cut off by screening. It suffices to eliminate the 38 anomalies encountered in Hartree-Fock theory of transport and density of states near the Fermi level. But because it is a long-wavelength approximation, it cannot describe adequately the response of the electron gas to a short-range perturbation. Our initial example of a point charge placed in an electron gas was just such a case. To get an accurate description of the screened impurity potential, the finite-q dependence of the charge susceptibility must be accounted for. This calculation involves simply replacing our Thomas-Fermi dielectric function εTF = 1 + k2 TF/q2 with the expression involving the full Lindhard function εRPA = 1 −V (q)χ0(q, ω = 0) = 1 + k2 TF q2 F( q 2kF ), (164) where F is the same function as in (116). Then we repeat the same calculation for the impurity potential ϕ(r) = X q eiq·r ϕext(q) 1 + k2 T F q2 F( q 2kF ) ≃ x2 (2 + x2)2 cos 2kFr r3 , (165) where x = kTF/(2kF). The main interesting feature of this result is of course the fact that the potential no longer has the simple Yukawa form with exponential decay, but in fact an oscillatory 1/r3 behavior. The last step in (165) is a bit long,37 but I have attempted to give you a flavor of it in the homework. One of the main conclusions of the exercise there is that the potential is very sensitive to the details of the charge susceptibility at 2kF, which the Thomas-Fermi theory clearly gets wrong. Ultimately Friedel oscillations (and RKKY oscillations, their counterpart for magnetic spins) originate from the sharpness of the Fermi surface. 2.2.8 Plasmons An examination of the dielectric function of the solid, e.g. Eq. (146), shows that the dielectric function can be zero, meaning the system responds very strongly to a small perturbation (see Eq. (138). These are the excitation energies of the electron gas. In the random phase approximation (158) this means there are excitations when V (q)χ0(q, ω) = 1 (166) Consider first a finite size box where the energy levels ϵk are discrete. This means χ0(q, ω) is a sum of terms containing simple poles at the particle-hole excitation energies ωq(k) ≡ ϵk+q −ϵk. χ0 will change sign and diverge as ω passes through these energies, therefore there will be a solution of (166) for each excitation energy. It is simplest to focus on the limit at 37See e.g. Fetter & Wallecka 39 T = 0, where χ0(q, ω) →− X kkF  1 ω −(ξk+q −ξk) + iη − 1 ω −(ξk −ξk+q) + iη  (167) In other words, the poles in the complex ω plane are particle-hole excitation energies ±ωk(q) ≡ ξk+q −ξk, where |k + q| is always above Fermi surface and k always below.38 The situation is depicted in figure 9. At T = 0, there is a maximum frequency above which there are no Figure 9: Eigenfrequencies of electron gas in RPA. more particle-hole excitations possible for given fixed q, i.e. the largest excitation energy occurs when k = kF and k, q are parallel, so that for a parabolic band, ϵk = k2/(2m), ωmax = kFq m + q2 2m. (168) 38To arrive at this result, note from (156) χ0(q, ω) = X k f(ξk) −f(ξk+q) ω −(ξk+q −ξk) + iη = = X k f(ξk)(1 −f(ξk+q)) −f(ξk+q)(1 −f(ξk)) ω −(ξk+q −ξk) + iη = = X k  f(ξk)(1 −f(ξk+q)) ω −(ξk+q −ξk) + iη −f(ξk+q)(1 −f(ξk)) ω −(ξk+q −ξk) + iη  = X k  f(ξk)(1 −f(ξk+q)) ω −(ξk+q −ξk) + iη −f(ξk)(1 −f(ξk−q)) ω −(ξk −ξk−q) + iη  = X k  f(ξk)(1 −f(ξk+q)) ω −(ξk+q −ξk) + iη −f(ξ−k)(1 −f(ξ−k−q)) ω −(ξ−k −ξ−k−q) + iη  = X k  f(ξk)(1 −f(ξk+q)) ω −(ξk+q −ξk) + iη −f(ξk)(1 −f(ξk+q)) ω −(ξk −ξk+q) + iη  = X k f(ξk)(1 −f(ξk+q))  1 ω −(ξk+q −ξk) + iη − 1 ω −(ξk −ξk+q) + iη  lim T →0 = X kkF  1 ω −(ξk+q −ξk) + iη − 1 ω −(ξk −ξk+q) + iη  40 There is another possible root of ϵ, as also shown in Fig. (9). In the limit ω > ωmax, we may expand (167) for small q, and find χ0(q, ω) ≃ 2q2 mω2 X k<kF " 1 + 2k · q mω + 3 k · q mω 2 + . . . ≃ k3 F 3π2 q2 mω2 " 1 + 3 5 kFq mω 2 + . . . . (169) Now we look for a solution to ReV χ0 = 1 and find to leading order39 ω = ωpl 1 + O(q/kF)2 + . . .  , (170) with ωpl = 4πne2 m 1/2 (171) which is of course the plasma frequency you found by elementary methods last semester. In contrast to the particle-hole excitations, which involve exciting only 1 particle from its position in the Fermi sea, a plasma oscillation perforce involves motions of all particles and is therefore referred to as a collective mode of the system. Plasmons have been observed indirectly in metals, primarily in thin metallic films. ω q plasmon particle-hole continuum ωpl Figure 10: Excitations (poles of ϵ(q, ω)−1) at T = 0. 2.3 Fermi liquid theory In a good metal the average interelectron distance is of the order of or smaller than the range of the interaction, e.g. the screening length k−1 TF ≈0.5 (n/a3 0)−1/6, where n is the density, of order 0.1-1nm for typical parameters. One might therefore expect that interactions should 39Note this is a result for 3 dimensions. It is a useful exercise to obtain the analagous result for 2D, which has ω ∼q1/2, i.e a gapless mode. 41 strongly modify the picture of free electrons commonly used to describe metals. It turns out, however, that for T ≪TF the important degrees of freedom are mostly frozen out, and only a fraction T/TF of all electrons near the Fermi surface participate in interaction processes. This leads to the possibility of a mapping between the free Fermi gas and the interacting Fermi system. IF the elementary excitations of the many-body system can be obtained from the single-particle states of the free electron gas by turning on the interaction adiabatically (and if the ground state has the full symmetry of the Hamiltonian), the distribution function nkσ of the interacting system would be the same as that of a free gas, and the particle and hole excitations of the gas would evolve into “quasiparticle" and “quasihole" states. εk b) n k 0 k kF Fermi gas Fermi liquid a) εk 0 c) c k k kF c k < > Figure 11: a) Evolution of quasiparticle states; b) T = 0 Quasiparticle distribution function n0 k; c) T = 0 true electron distribution function ⟨c† kck⟩. 2.3.1 Particle-hole excitations To understand what this really means, let’s go back and review the definitions of the terms particles and holes. Let’s take N free Fermions, mass m in a volume L3, where we know that the eigenstates are Slater determinants of N different single particle states, each given by ψk(r) = 1 √ L3eik·r. (172) Let’s stick to T = 0 for the moment, and recall that the occupation of each state is given by nk = θ(kF −k), where kF is the radius of the Fermi sea. Now let’s add one “test" particle to the lowest available state k = kF. We then have for the chemical potential at T = 0, µ = E0(N + 1) −E0(N) = ∂E0 ∂N = k2 F 2m . (173) Particles and holes are excitations of the filled Fermi sea. Excitations of the gas consist of taking a certain number of particles and moving them to the other side of the Fermi surface, yielding particles above and an equal number of holes below the Fermi surface. These 42 excitations can be described in terms of the changes in the particle occupation numbers δnp = np −n0 p: δnk =    δk,k′ for a particle k′ > kF −δk,k′ for a hole k′ < kF . (174) At nonzero temperatures, a thermal fluctuation will create changes in occupation numbers E N(E) EF δn k =-1 particle ex. hole ex. n δ =1 k' Figure 12: Particle and hole excitations of the Fermi gas. δnp ∼1 only for excitations of energy within kBT of εF. Note that the total energy of the free gas is completely specified as a functional of the occupation numbers: E −E0 = X k k2 2m(nk −n0 k) = X k k2 2mδnk . (175) Now take the gas and put it in contact with a particle reservoir. The appropriate ther-modynamic potential is then the Gibbs free energy, which for T = 0 is F = E −µN, and F −F0 = X k  k2 2m −µ  δnk ≡ X k ξkδnk . (176) The free energy of one particle, with momentum k and occupation δnk′ = δk,k′ is ξk and it corresponds to an excitation outside the Fermi surface. The free energy of a hole δnk = −δk,k′ is −ξk, which corresponds to an excitation below the Fermi surface. For both species then the free energy of an excitation is |ξk| , which is always positive; ie., the Fermi gas is stable with respect to excitations as it must be. 2.3.2 Quasiparticles and quasiholes in interacting system Consider a system with interacting particles an average spacing a apart, so that the typical interaction energy is e2 a e−a/rT F . As discussed above, we now suppose that we can create this system by turning on the interactions adiabatically starting from the free gas in a time t, such that the system evolves while staying always in its ground state. 43 If a state of the ideal system is characterized by the distribution n0 k, then the state of the interacting system will evolve adiabatically from n0 k to nk. In fact if the system remains in its ground state, then n0 k = nk, i.e. the occupation of each state labelled by k won’t change (although the energy will). This is an assumption, and we will be studying some counterexamples to this ansatz (superconductivity and magnetism). But in general Landau expected this to hold if the symmetry of the interacting system was the same as the noninteracting one, and this appears to be the case for simple metals. Let’s follow what happens to a particle of momentum k added to the top of the non-interacting Fermi gas when we slowly turn the interaction knob. As U increases, two things k k k t=0 U=0 t=t U= e2 a e -kTFa turn on interactions Figure 13: Evolution of particle in state k on top of Fermi sea as interaction is turned on. Particle becomes “dressed" by interactions with the rest of the system which changes the effective mass but not the momentum of the excitation (“quasiparticle"). happen. The k state of the original particle is no longer an eigenstate of H, so it decays into the new excitations of the system, which according to our ansatz are also single-particle-like. Nearby particles also decay into the original state; this mixing has the effect of “dressing" the original free particle with a “cloud" of excitations of the surrounding system. However since momentum is conserved, the net excitation (called a quasiparticle)) can still be labelled with original momentum k. In the same way, holes of momentum k below the Fermi surface evolve into quasiholes. Occasionally an analogy from classical mechanics is useful: we think of a bullet being fired into a large container filled with cotton. It will drag some cotton with it, and create a “particle+cloud" with increased mass but the same momentum. It is too much to hope that the interactions between quasiparticles will vanish altogether. The N-particle ground state wave-function may still be Slater-determinant like, but this doesn’t necessarily imply that the quasiparticles are infinitely long-lived (exact) eigenstates of the system. So if each quasiparticle state k has lifetime τk, we have to worry a bit about 44 the adiabatic switching on procedure if τk < t. If so, the quasiparticle will decay before we finish switching, and we won’t be able to go reversibly back and forth. If we decrease t too much so again τk ≫t, then the switching on of U may not be adiabatic, i.e. we may knock the system into a state other than its ground state. It was Landau’s great insight to recognize that these objections are overcome for quasiparticle states sufficiently close to the Fermi energy. In this case the Pauli principle strongly restricts the decay channels for a quasiparticle, and one can argue (see below) that the lifetime becomes infinite at the Fermi energy itself. Therefore for any time t there should be a narrow range of energies near the Fermi surface for which the τk is long enough to make the whole construction work. 2.3.3 Energy of quasiparticles. As in the free gas, excitations will be characterized only by the deviation of the occupation number from the ground state occupation n0 k.40 δnk = nk −n0 k (177) (I hide the spin index for now). At suffiently low temperatures δnk ∼1 only for those excitations near the Fermi surface where the condition τk ≫t is satisfied. Therefore we should be able to formulate a theory which depends only on δnk, not on nk or n0 k, which are not well defined for k far from the Fermi surface. nk and n0 k should never enter independently and are not physically relevant. For the noninteracting system the dependence of the energy (T = 0) E on δnk is very simple: E −E0 = X k k2 2mδnk . (178) In the interacting case E[nk] is more complicated generally, but if we take δnk to be small (just a few excitations above the ground state) then we may expand: E[nk] = Eo + X k ϵkδnk + O(δn2 k) , (179) where we now define ϵk = δE/δnk. If δnk = δk,k′, then E ≈E0 + ϵk′; i.e., the energy of the quasiparticle with momentum k′ is ϵk′. As discussed, we will only need ϵk near the Fermi surface where δnk is O(1). So we may expand ϵk around the Fermi level, ϵk ≈µ + (k −kF) · ∇k ϵk|kF (180) 40Note we will now take over the notation n0 k which in the previous discussion meant the distribution function of the noninteracting system. Now it means the ground state of the interacting system. 45 where ∇kϵk = vk, the group velocity of the quasiparticle. We no longer know the exact form of ϵk, but some exact relations must be obeyed because our system obeys certain symmetries. If we explicitly put the spin-dependence back in, we have symmetries ϵk,σ = ϵ−k,−σ time-reversal (181) ϵk,σ = ϵ−k,σ parity (182) So unless we explicitly break T-symmetry by e.g. applying a magnetic field, we have ϵk,σ = ϵ−k,σ = ϵk,−σ. Furthermore, as usual ϵk has to respect the point and space group symmetries of the system, e.g. for an isotropic system ϵk depends only upon the magnitude of k, |k|, so k and vk = ∇ϵk(|k|) = ˆ k(dϵk/dk) are parallel.41 Define (m∗)−1 as the constant of proportionality at the Fermi surface42 vkF = kF/m∗ (183) Then for the interacting system at the Fermi surface Ninteracting(εF) = m∗kF π2 , (184) the same form as for the noninteracting system. Note the m∗(usually but not always > m) accounts for the dressing of the quasiparticle particle (bullet + cotton!). In other words, as far as the density of levels at the Fermi energy is concerned, the only thing which changes is the effective mass of the fermionic single-particle excitations. 2.3.4 Residual quasiparticle interactions Let’s go back to the Gibbs free energy of our system of quasiparticles characterized in some generalized excited state by distribution function δnk: F −F0 = E −E0 −µ(N −N0) . (185) where again F0, E0 etc. characterize the ground state. Thus, N −N0 = X p δnk , (186) and since E −E0 ≈ X k ϵkδnk + O(δn2) , (187) we get F −F0 ≈ X k (ϵk −µ) δnk = X k ξkδnk + O(δn2) . (188) 41For a cubic system ϵk is invariant under 90◦rotations, etc. 42Note kF is not changed by interactions (Luttinger’s theorem for isotropic system) 46 As shown in Fig. 14, we will be interested in excitations of the system which deform the Fermi surface by an amount proportional to δ. For example, suppose that along the positive kx axis, our excited system has δnk = 1 for some states k above kF, whereas along the negative kx axis some states have δnk = −1. The approximation to the free energy above kx ky Figure 14: Small deformations of the Fermi sphere, proportional to δ. will remain valid provided δ ≡1 N X k |δnk| ≪1 . (189) If we stick with our original criterion of using only states near the Fermi surface to construct our distortion, the ξk for those states with δnk ̸= 0 will also be of order δ. Thus, X k ξkδnk ∼O(δ2) , (190) so in order to keep terms of order δ2 consistently we must also add the next term in the Taylor series expansion of the energy : F −F0 = X k ξkδnk + 1 2 X k,k′ fk,k′δnkδnk′ + O(δ3) (191) where fk,k′ = δE δnkδnk′ (192) The term proportional to fk,k′, was proposed by L.D. Landau. It describes the residual interactions of the quasiparticle excitations which remain even after they are “dressed". A glance at the dimensions of the new term shows that fk,k′ ∼1/V . But if fk,k′ is an interaction between quasiparticles, each of which (in plane-wave like state) is delocalized over the whole 47 volume V , so the probability that two will interact is roughly ∼k−3 TF/V . Therefore fk,k′ ∼k−3 TF/(N0V ) (193) We can proceed as earlier with the quasiparticle energy ϵk and ask how this unknown interaction function reflects the underlying symmetries of the physical system. To this end we reintroduce the hidden spin degrees of freedom. First in the absence of an external magnetic field, the system should be invariant under time-reversal, so fkσ,k′σ′ = f−k−σ,−k′−σ′ , (194) and, in a system with reflection symmetry (parity) fkσ,k′σ′ = f−kσ,−k′σ′ . (195) Then since this implies fkσ,k′σ′ = fk−σ,k′−σ′ . (196) it must be that f depends only upon the relative orientations of the spins σ and σ′, so there are only two independent components for given k, k′, e.g. fk↑,k′↑and fk↑,k′↓. It is more conventional to decompose f into spin-symmetric and antisymmetric parts: f a k,k′ = 1 2 (fk↑,k′↑−fk↑,k′↓) f s k,k′ = 1 2 (fk↑,k′↑+ fk↑,k′↓) . (197) f a k,k′ may be interpreted as an exchange interaction, or fkσ,k′σ′ = f s k,k′ + σσ′f a k,k′. (198) where σ and σ′ are the spins of the quasiparticles with momentum k and k′, respectively. The next important assumption entering Fermi liquid theory is that the quasiparticle interactions are short-ranged. This means that fkk′ is essentially constant as a function of |⃗ k|, |⃗ k′|, and so depends only on the directions ˆ k, ˆ k′, f s,a kk′ ≃f(ˆ k, ˆ k′): fk,k′ ≃fk,k′|ϵk=ϵk′=µ (199) or equivalently, only on the angle θ between them! f α k,k′ = ∞ X l=0 f α l Pl(cos θ) . (200) Conventionally these f parameters are expressed in terms of dimensionless quantities. N0f α l = m∗kF π2 f α l = F α l . (201) We expect that the Landau parameters F s,a ℓ will be negligible for sufficiently large ℓ, so we have now achieved our goal of subsuming the complicated effects of interactions in a few phenomenological parameters. 48 2.3.5 Local energy of a quasiparticle Now consider an interacting system with a certain distribution of excited quasiparticles δnk′. Let’s add another particle, say of momentum k and see how the energy changes. The distribution function goes from (δn′ k →δn′ k + δk,k′). From Eq. 191 the free energy of the additional quasiparticle is ˜ ξk = ξk + X k′ fk′,kδnk′ , (202) (since fk′,k = fk,k′). The second term on the right corresponds to the average interaction of a quasiparticle with the other quasiparticles in the system.43 Substituting (200) and (201) in (202) gives the shift in energy of a quasiparticle due to the Landau interaction δξkσ ≡˜ ξk −ξk = 1 N0 [F s 0 δn + σF a 0 δns + 1 k2 F F s 1⃗ k · ⃗ g + ...], (203) where δn = P kσ δnkσ is the density shift, δns = P kσ σδnkσ is the spin density shift, and ⃗ g = P kσ ⃗ kδnkσ is the momentum density. The first two terms can be thought of as molecular field couplings representing the average effect of density and magnetization fluctuations in the surrounding medium on the quasiparticle energy. The third term is the famous “backflow" term which describes the effect of the quasiparticle streaming through the medium to which it couples. The flow pattern is of dipolar form, as depicted in Figure 15. Note that the Landau parameter F s 1 and m∗are not completely independent, but one has (in the Galilean invariant liquid case ONLY) m ∗/m = 1 + F s 1 /3. In a crystal this relation does not hold.44 k Figure 15: Schematic of backflow around quasiparticle of momentum k Before leaving this section, let me just mention that the importance of the local quasipar-ticle energy ˜ ξ or rather the energy shift δ˜ ξ is that when one writes down how external forces 43The structure may remind you of Hartree-Fock theory, to which it is similar, but keep in mind f is the residual interaction between quasiparticles, not electrons 44For an extensive discussion of this and many other aspects of FL theory I’ve glossed over, see G. Baym and C. Pethick, Landau Fermi-Liquid Theory : Concepts and Applications, Wiley 1991 49 couple to the system to determine, e.g. transport properties, it is the gradient of ˜ ξk which is the force a qp of momentum k experiences due to spatial inhomogeneities in the system. There is a "Landau molecular field" proportional to −∇˜ δξk = −∇ X k′ fk′kδnk′(r) (204) which is responsible, e.g. for spin waves in spin polarized dilute gases. Eq. (204) is a simplified version of Landau’s famous kinetic equation for a Fermi liquid which is the analog of the Boltzman equation for a weakly interacting gas. 2.3.6 Thermodynamic properties The expression (203) for the quasiparticle energies and the fact that the equilibrium quasi-particle distribution function is just the Fermi function enables immediate calculations of thermodynamic observables in the Fermi liquid. Because the states of the gas and liquid are in one to one correspondence, for example, one immediately knows that the expression for the entropy of the liquid is the same as that for a free Fermi gas,45 S = − X kσ [nkσ log nkσ + (1 −nkσ) log(1 −nkσ)]. (205) Note now that the distribution function nkσ in local equilibrium must be a Fermi function evaluated at energy ϵkσ given by the bare energy plus the expression (202) in general. How-ever, we are interested in calculating a quantity in global equilibrium, in which case the shifts δn, δns,⃗ g, ... vanish. In this case the calculation reduces to the usual one for a free Fermi gas. For temperatures T ≪TF we replace the sum P kσ by N0 R dξ . The temperature can then be scaled out, and the remaining integral performed, yielding the entropy density at low temperatures, s = 1 3π2N0T, (206) and the specific heat at constant volume is therefore C = T(∂s/∂T)V = (m∗kF/3)T, where in the last step we have used N0 = m∗kF/π2. The result for the liquid is identical to that for the gas, with the exception of the replacement of m by m∗. Similarly, we may derive an expression for the spin susceptibility of the Fermi liquid by constructing the magnetization δns as above, and noting that the shift in the distribution function is due to the Landau shift in the quasiparticle energy (molecular field) as well as the external field ⃗ H: 45Recall the form of the Fermi distribution function for the free gas is obtained purely from statistical considerations, not from any knowledge of the energies. Similarly the Fermi surface derivative ∂nkσ/∂ϵkσ is a delta function as in the noninteracting case since there is a discontinuity at kF . 50 δnkσ = ∂nkσ ∂ϵkσ (δϵkσ −µ0σH). (207) Using δns ≡P kσ σδnkσ, and noting that since σ∂n/∂ϵ is isotropic and odd in σ, only the ℓ= 0, spin-antisymmetric part of δϵkσ contributes to the sum, we find for the spin susceptibility χ = µ0(∂ns/∂H) χ = µ2 0N0 (1 + F a 0 ) = m∗/m 1 + F a 0 χ0, (208) where χ0 is the susceptibility of the free gas. Remember N0 here is the renormalized DOS including m∗, not m. 2.3.7 Quasiparticle relaxation time and transport properties. Calculations of transport properties require introducing the quasiparticle lifetime. Recall the quasiparticle states are not eigenstates of the many-body Hamiltonian, due to the weak residual interactions among them. We therefore expect a quasiparticle with momentum ⃗ k to decay except when it sits exactly at the Fermi surface and when T = 0, when it can easily be seen there is no phase space for scattering which conserves energy and momentum. But relaxing these conditions (⃗ k not on FS, T ̸= 0) allows phase space for scattering and hence a finite lifetime. Since the interactions among quasiparticles are assumed weak, we may use a golden rule type formula for the scattering rate" ky kx 2 4 1 3 V Figure 16: 2-quasiparticle collision: states before (1,2) and after (3,4) Let’s consider a quasiparticle with energy ξ1 = ϵ1 −µ above the Fermi level. A scattering process takes it to final state ξ3, and to conserve energy and momentum a quasiparticle with energy ξ2 must be scattered to final state ξ4. I’ve assumed the interaction V is independent 51 of momentum, frequency, etc. because the most important effects we are looking for are purely statistical, unless V is very singular.46 The "golden rule" then gives 1 τ1 = 2π Z dξ2N(ξ2) Z dξ3N(ξ3) Z dξ4N(ξ4)|V |2f(ξ2)(1 −f(ξ3))(1 −f(ξ4))δ(ξ1 + ξ2 −ξ3 −ξ4) ≃ ≃ 2πN 3 0 |V 2| Z ξ1 0 dξ3 Z ξ1−ξ3 0 dξ4 = πN 3 0 |V |2ξ2 1, where the delta function conserves energy and the Fermi factors ensure that there is a particle in state 2 and that states 3 and 4 are empty beforehand. For simplicity I evaluated the Fermi functions at T=0, giving the final simple result. Note I have neglected spin for the moment, but this is not essential.47 Note now that for N0V ≪1 (weak residual interactions) and N0 ∼1/εF, it is clear that for ξ ≪εF the scattering satisfies the condition 1 τ1 ≪ξ. (209) This means that the quasiparticle states are well-defined near the Fermi surface, since their damping is small compared to the energy of the state. ( As expected, the relaxation rate ex-actly at the Fermi surface vanishes identically.) We therefore expect that Fermi liquid theory calculations in terms of a weakly interacting quasiparticle gas will be valid for frequencies and temperatures such that ω, T ≪εF. Two further comments about the scattering rate are of potential importance. First, I give without proof the full result at finite temperatures:48 1 τ(ξ) = π|V |2 64ε3 F [(πT)2 + ξ2], (211) so that at finite T one finds that 1/τ varies as T 2 or ξ2, whichever is larger. Certain simple transport properties in a Fermi liquid reflect this directly. For example, the resistivity due to electron-electron scattering in a metal goes as T 2 at the lowest temperatures; it is usually 46In the case of the Coulomb interaction, which is very singular, screening takes care of this problem. See section 2.2.6 47What I have done which is not really Kosher, however, is to neglect the constraints placed on the scattering by momentum conservation. It turns out in three dimensions this is not terribly important, and one obtains the correct answer. It is a useful exercise to convince yourself, for example, that in 2D the relaxation rate goes like ξ2 log ξ. 48Our simple calculation has to be modified to account for momentum conservation, expressed as k1 −k3 = k4 −k2 . (210) Since ξ1 and ξ2 are confined to a narrow shell around the Fermi surface, so too are ξ3 and ξ4. This can be seen in Fig. 16, where the requirement that k1 −k3 = k4 −k2 limits the allowed states for particles 3 and 4. Taking k1 fixed, allowed momenta for 2 and 3 are obtained by rotating k1 −k3 = k4 −k2; this rotation is constrained, however, by the fact that particles cannot scatter into occupied states. This restriction on the final states reduces the scattering probability by a factor of ξ1/εF , but is the same restriction contained in the calculation above. Thus momentum conservation doesn’t affect our T = 0 estimate above, except to modify the prefactor. If these hand-waving arguments don’t satisfy you (they shouldn’t), see AGD or Fetter/Wallecka. 52 masked by other, larger sources of scattering in ordinary metals, however.49 The second remark is that if one were to find, for some microscopic model, that the relaxation rate was larger than the quasiparticle energy, the quasiparticles would be overdamped and the entire concept would be useless. One would then speak of an incoherent or "non-Fermi liquid" system. If, on the other hand, one found that 1/τ were to scale precisely with max(ξ, T) (instead of max(ξ2, T 2)), the quasiparticles would be in a certain sense critically damped, i.e. not quite well defined, but not completely washed out either. This is what is frequently referred to as a "marginal Fermi liquid".[MFL] 2.3.8 Effective mass m∗of quasiparticles Here I give a quick derivation of the relation between the Landau parameter F1s which appears in the current and the effective mass m∗. The net momentum of the volume V of quasiparticles is Pqp = 2V X k knk, net quasiparticle momentum (212) also the total momentum of the Fermi liquid. Since the number of particles equals the number of quasiparticles, however, the quasiparticle and particle currents must also be equal Jqp = Jp = 2V X k vknk net quasiparticle and particle current (213) or, since the momentum is just the particle mass times this current Pp = 2V m X k vknk net quasiparticle and particle momentum (214) where vp = ∇k˜ ϵk, is the velocity of the quasiparticle. So X k knk = m X k ∇k˜ ϵknk (215) Let’s consider an arbitrary variation of nk, and remember that ˜ ϵk depends on nk, so that δ˜ ϵk = 2V X k fk,k′δnk′ . (216) For Eq. 215, this means that X k kδnk = m X k ∇k˜ ϵkδnk (217) +m2V X k X k′ ∇k (fk,k′δnk′) nk , 49Of course this is not quite right. The electron-electron interactions in a solid conserve crysal momentum, and thus the conductivity of a pure crystal would be infinite at any temperature unless one explicitly accounts for Umklapp processes. Thus it is not true for the electrical conductivity that σ ∝τ, with τ the 1-particle lifetime due to electron-electron interactions. However the Umklapp processes do also provide a transport rate varying as T 2, so one often hears the not quite correct remark that the scattering rate in a Fermi liquid varies as T 2, therefore so does the resistivity ρ. 53 or integrating by parts (and renaming k →k′ in the last part), we get X k k mδnk = X k ∇k˜ ϵkδnk (218) −2V Z X k,k′ δnkfk,k′∇k′nk′ , The usual variational argument says that since δnk is arbitrary, we must have that the integrands are equal k m = ∇k˜ ϵk −2V X k′ fk,k′∇k′nk′ (219) At T = 0 ∇k′nk′ = −ˆ k′δ(k′ −kF). The integral may now be evaluated by using the system isotropy in k space, and choosing k parallel ˆ z. As we mostly concerned with Fermi surface properties we take k = kF, and let θ be the angle between k (or the z-axis) and k′, and finally note that on the Fermi surface |∇k˜ ϵk|k=kF = vF = kF/m∗. Thus, kF m = kF m∗+ 2V Z k′2dkdΩ (2πℏ)3 fk,k′ˆ k · ˆ k′δ(k′ −kF) (220) However, since both k and k′ are restricted to the Fermi surface ˆ k′ = cos θ, and evaluating the integral over k, we get 1 m = 1 m∗+ 2V kF Z dΩ (2πℏ)3fk,k′ cos θ , (221) If we now sum over (hidden) spins only the symmetric part of f appears 1 m = 1 m∗+ 4πV kF (2πℏ)3 Z d (cos θ) f s(θ) cos θ , (222) Now use decomposition (200), orthonormality of Legendre polynomials, and def. of Landau parameters: (223) f α(θ) = X l f α l Pl(cos θ) , (224) Z 1 −1 dxPn(x)Pm(x)dx = 2 2n + 1δnm (225) V N0f α l = V m∗kF π2ℏ3 f α l = F α l , (226) we find that 1 m = 1 m∗+ F s 1 3m∗, (227) or m∗/m = 1 + F s 1 /3. So effective mass and F1s aren’t independent, at least in a Galilean invariant system where (213) holds. 54 3 Quantum Magnetism The main purpose of this section is to introduce you to ordered magnetic states in solids and their “spin wave-like" elementary excitations. Magnetism is an enormous field, and reviewing it entirely is beyond the scope of this course. 3.1 Introduction 3.1.1 Atomic magnetic Hamiltonian The simplest magnetic systems to consider are insulators where electron-electron interactions are weak. If this is the case, the magnetic response of the solid to an applied field is given by the sum of the susceptibilities of the individual atoms. The magnetic susceptibility is defined by the the 2nd derivative of the free energy,50 χ = −∂2F ∂H2. (228) We would like to know if one can understand (on the basis of an understanding of atomic structure) why some systems (e.g. some elements which are insulators) are paramagnetic (χ > 0) and some diamagnetic (χ < 0). The one-body Hamiltonian for the motion of the electrons in the nuclear Coulomb potential in the presence of the field is Hatom = 1 2m X i  pi + e cA(ri) 2 + X i V (ri) + g0µBH · S, (229) where P i Si is the total spin operator, µB ≡e/(2mc) is the Bohr magneton, g0 ≃2 is the gyromagnetic ratio, and A = −1 2r × H is the vector potential corresponding to the applied uniform field, assumed to point in the ˆ z direction. Expanding the kinetic energy, Hatom may now be expressed51 in terms of the orbital magnetic moment operator L = P i ri × pi as Hatom = X i p2 i 2m + X i V (ri) + δH, (230) δH = µB(L + g0S) · H + e2 8mc2H2 X i (x2 i + y2 i ). (231) Given a set of exact eigenstates for the atomic Hamiltonian in zero field |n⟩(ignore degen-eracies for simplicity), standard perturbation theory in δH gives δEn = −H · ⟨n|⃗ µ|n⟩+ X n̸=n′ |⟨n|H · ⃗ µ|n′⟩|2 En −En′ + e2 8mc2H2⟨n| X i (x2 i + y2 i )|n⟩, (232) 50In this section I choose units such that the system volume V = 1. 51To see this, it is useful to recall that it’s ok to commute r and p here since, in the cross product, pz acts only on x, y, etc. 55 where ⃗ µ = −µB(L + g0S). It is easy to see that the first term dominates and is of order the cyclotron frequency ωc ≡eH/(mc) unless it vanishes for symmetry reasons. The third term is of order ωc/(e2/a0) smaller, because typical electron orbits are confined to atomic sizes a0, and is important in insulators only when the state |n⟩has zero magnetic moment (L = S = 0). Since the coefficient of H2 is manifestly positive, the susceptibility52 in the µ = 0 ground state |0⟩is χ = −∂2δE0/∂H2, which is clearly < 0, i.e. diamagnetic.53 In most cases, however, the atomic shells are partially filled, and the ground state is deter-mined by minimizing the atomic energy together with the intraatomic Coulomb interaction needed to break certain degeneracies, i.e. by “Hund’s rules".54 Once this ground state (in particlular the S, L, and J quantum numbers is known, the atomic susceptibility can be calculated. The simplest case is again the J = 0 case, where the first term in (232) again vanishes. Then the second term and third term compete, and can result in either a dia-magnetic or paramagnetic susceptibility. The 2nd term is called the Van Vleck term in the energy. It is paramagnetic since its contribution to χ is χV V = −∂2E0|2nd term ∂H2 = 2µ2 B X n |⟨0|⃗ µ|n⟩|2 En −E0 > 0. (233) 3.1.2 Curie Law-free spins In the more usual case of J ̸= 0, the ground state is 2J + 1 degenerate, and we have to be more careful about defining the susceptibility. The free energy as T →0 can no longer be replaced by E0 as we did above. We have to diagonalize the perturbation δH in the degenerate subspace of 2J + 1 states with the same J but different Jz. Applying a magnetic field breaks this degeneracy, so we have a small statistical calculation to do. The energies of the “spin" in a field are given by H = −⃗ µ · H, (234) 52If the ground state is nondegenerate, we can replace F = E −TS in the definition of the susceptibility by the ground state energy E0. 53This weak diamagnetism in insulators with filled shells is called Larmor diamagnetism 54See A&M or any serious quantum mechanics book. I’m not going to lecture on this but ask you about it on homework. 56 and since ⃗ µ = −γJ within the subspace of definite J2,55 the 2J +1 degeneracy which existed at H = 0 is completely broken. The free energy is56 F = −T log Z = −T log J X Jz=−J eβγHJz = −T log eβγH(J+1/2) −e−βγH(J+1/2) eβγH/2 −e−βγH/2  , (235) so the magnetization of the free spins is M = −∂F ∂H = γJB(βγJH), (236) where B(x) is the Brillouin function B(x) = 2J + 1 2J coth 2J + 1 2J x −1 2J coth 1 2J x. (237) Note I defined γ = µBg. We were particularly interested in the H →0 case, so as to compare with ions with filled shells or J = 0; in this case one expands for T >> γH, i.e. coth x ∼1/x + x/3 + . . . , B(x) ≃(J + 1)x/(3J) to find the susceptibility χ = −∂2F ∂H2 = γ2J(J + 1) 3T , (238) i.e. a Curie law for the high-temperature susceptibility. A 1/T susceptibility at high T is generally taken as evidence for free paramagnetic spins, and the size of the moment given by µ2 = γ2J(J + 1). 3.1.3 Magnetic interactions The most important interactions between magnetic moments in an insulator are electrostatic and inherently quantum mechanical in nature. From a classical perspective, one might expect 55This is not obvious at first sight. The magnetic moment operator ⃗ µ ∝ˆ L + g0ˆ S is not proportional to the total angular momentum operator ˆ J = ˆ L + ˆ S. However its matrix elements within the subspace of definite L, S, J are proportional, due to the Wigner-Eckart theorem. The theorem is quite general, which is of course why none of us remember it. But in it’s simpler, more useful form, it says that the matrix elements of any vector operator are proportional to those of the total angular momentum operator J. You need to remember that this holds in a basis where atomic states are written |JLSJz⟩, and that "proportional" just means ⟨JLSJz|⃗ V |JLSJ′ z⟩= g(JLS)⟨JLSJz| ⃗ J|JLSJ′ z⟩, where ⃗ V is some vector operator. In other words, within the subspace of definite J, L, S but for arbitrary Jz, all the matrix elements ⟨Jz|⃗ V |J′ z⟩are = g⟨Jz| ⃗ J|J′ z⟩, with the same proportionality const. g, called the Landé g-factor. Obviously ⃗ µ = L + g0S is a vector operator since L and S are angular momenta, so we can apply the theorem. If J = 0 but L, S are not (what we called case II in class) it is not obvious at first sight that term 1 in Eq. (232) ∝⟨0|µz|0⟩vanishes since our basis states are not eigenstates of Lz and Sz. But using the WE theorem, we can see that it must vanish. This means for the case with ground state J = 0, L, S ̸= 0, terms 2 and 3 in Eq. (232) indeed compete. Examine the matrix element in the numerator of 2 (van Vleck) in Eq. (232): while the ground state ⟨0| is still ⟨J = 0LSJz = 0| on the left, the sum over the excited states of the atom |n′⟩on the right includes other J’s besides zero! Thus the WE theorem does not apply. [for a more complete discussion, see e.g. Ashcroft and Mermin p. 654.] 56Geometric series with finite number of terms: PN n=0 xn = 1−xN 1−x . 57 Figure 17: Dimensionless Brillouin fctn. two such moments to interact via the classical dipolar force, giving a potential energy of configuration of two moments ⃗ µ1 and ⃗ µ2 separated by a distance r of U = ⃗ µ1 · ⃗ µ2 −3(⃗ µ1 · ˆ r)(⃗ µ2 · ˆ r) r3 (239) Putting in typical atomic moments of order µB = eℏ/mc and distances of order a Bohr radius r ∼a0 = ℏ2/me2, we find U ≃10−4eV , which is very small compared to typical atomic energies of order eV . Quantum-mechanical exchange is almost aways a much larger effect, and the dipolar interactions are therefore normally neglected in the discussions of magnetic interactions in solids. Exchange arises because, e.g., two quantum mechanical spins 1/2 (in isolation) can be in either a spin triplet (total S = 1), or singlet (total S = 0). The spatial part of the two-particle wavefunctions is antisymmetric for triplet and symmetric for singlet, respectively. Since the particles carrying the spins are also charged, there will be a large energetic difference between the two spin configurations due to the different size of the Coulomb matrix elements (the “amount of time the two particles spend close to each other") in the two cases.57 In terms of hypothetical atomic wave functions ψa(r) and ψb(r) for the two particles, the singlet and triplet combinations are ψ 0 1(r1, r2) = ψa(r1)ψb(r2) ± ψa(r2)ψb(r1), 57Imagine moving 2 H-atoms together starting from infinite separation. Initially the 3 S = 1 states and 1 S = 0 states must be degenerate. As the particles start to interact via the Coulomb force at very long range, there will be a splitting between singlet and triplet. 58 so the singlet-triplet splitting is approximately −J ≡E0 −E1 = ⟨0|H|0⟩−⟨1|H|1⟩ ≃ 2 Z d3r1d3r2ψ∗ a(r1)ψ∗ b(r2)V (r1, r2)ψa(r2)ψb(r1), (240) where V represents the Coulomb interactions between the particles (and possible other par-ticles in the problem).58 We’d now like to write down a simple Hamiltonian for the spins which contains the physics of this exchange splitting. This was done first around 1939 by Heisenberg, who suggested H2−spin = Jˆ S1 · ˆ S2 (242) You can easily calculate that the energy of the triplet state in this Hamiltonian is J/4, and that of the singlet state −3J/4. So the splitting is indeed J. Note that the sign of J in the H2 case is positive, meaning the S = 0 state is favored; the interaction is then said to be antiferromagnetic, meaning it favors antialigning the spins with each other.59 The so-called Heitler-London model of exchange just reviewed presented works reasonably well for well-separated molecules, but for N atoms in a real solid, magnetic interactions are much more complicated, and it is in general not sufficient to restrict one’s consideration to the 4-state subspace (singlet ⊕3 components of triplet) to calculate the effective ex-change. In many cases, particularly when the magnetic ions are reasonably well-separated, it nevertheless turns out to be ok to simply extend the 2-spin form (242) to the entire lattice: H = J X iδ ˆ Si · ˆ Si+δ (243) where J is the exchange constant, i runs over sites and δ runs over nearest neighbors.60 This is the so-called Heisenberg model. The sign of J can be either antiferromagnetic (J > 0 in this convention), or ferromagnetic (J < 0). This may lead, at sufficiently low temperature, 58For example, in the H molecule, V (r1, r2) = e2 |r1 −r2| + e2 |R1 −R2| − e2 |r1 −R1| − e2 |r2 −R2| , (241) where R1 and R2 are the sites of the protons. Note the argument above would suggest naively that the triplet state should be the ground state in the well-separated ion limit, because the Coulomb interaction is minimized in the spatially antisymmetric case. However the true ground state of the H2 molecule is the Heitler-London singlet state ψs(r1, r2) ≃ψa(r1)ψb(r2)+ψa(r2)ψb(r1). In the real H2 molecule the singlet energy is lowered by interactions of the electrons in the chemical bond with the protons–the triplet state is not even a bound state! 59Historically the sign convention for J was the opposite; J > 0 was usually taken to be ferromagnetic, i.e. the Hamiltonian was defined with another minus sign. I use the more popular convention H = J P Si · Sj. Be careful! 60One has to be a bit careful about the counting. J is defined conventionally such that there is one term in the Hamiltonian for each bond between two sites. Therefore if i runs over all sites, one should have δ only run over, e.g. for the simple cubic lattice, +ˆ x, +ˆ y, and +ˆ z. If it ran over all nearest neighbors, bonds would be double-counted and we would have to multiply by 1/2. 59 to a quantum ordered state with ferromagnetic or antiferromagnetic-type order. Or, it may not. A good deal depends on the type and dimensionality of the lattice, as we discuss below. Figure 18: a) Ferromagnetic ordering; b) antiferromagnetic ordering. Although oversimplified, the Heisenberg model is still very difficult to solve. Fortunately, a good deal has been learned about it, and once one has put in the work it turns out to describe magnetic ordering and magnetic correlations rather well for a wide class of solids, provided one is willing to treat J as a phenomenological parameter to be determined from a fit to experiment. The simplest thing one can do to take interactions between spins into account is to ask, “What is the average exchange felt by a given spin due to all the other spins?" This is the essence of the molecular field theory or mean field theory due to Weiss. Split off from the Hamiltonian all terms connecting any spins to a specific spin ˆ Si. For the nearest-neighbor exchange model we are considering, this means just look at the nearest neighbors. This part of the Hamiltonian is δHi = ˆ Si · " J X δ ˆ Si+δ # −⃗ µi · H, (244) where we have included the coupling of the spin in question to the external field as well. We see that the 1-spin Hamiltonian looks like the Hamiltonian for a spin in an effective field, δHi = −⃗ µi · Heff, (245) Heff = H − J gµB X δ ˆ Si+δ, (246) where the magnetic moment ⃗ µ ≡γS = gµBS. Note this “effective field" as currently defined is a complicated operator, depending on the neighboring spin operators ˆ Si+δ. The mean field 60 theory replaces this quantity with its thermal average Heff = H − J gµB X δ ⟨ˆ Si+δ⟩ (247) = H − zJ (gµB)2M, (248) where the magnetization M = ⟨µ⟩and z is the number of nearest neighbors again. But since we have an effective one-body Hamiltonian, thermal averages are supposed to be computed just as in the noninteracting system, cf. (236), but in the ensemble with effective magnetic field. Therefore the magnetization is M = γSB(βγSHeff) = γSB  βγS[H −zJ γ2 M]  . (249) This is now a nonlinear equation for M, which we can solve for any H and T. It should describe a ferromagnet with finite spontaneous (H →0) magnetization below a critical temperature Tc if J < 0. So to search for Tc, set H = 0 and expand B(x) for small x (we are looking for a second order transition, where M →0 continuously): M = −γSB  zJ γTc M  ≃−γS S + 1 3S zJ γTc M (250) ⇒ Tc = S(S + 1) 3 z(−J), (251) So the critical temperature in this mean field theory (unlike the BCS mean field theory!) is of order the fundamental interaction energy |J|. We expect this value to be an upper bound to the true critical temperature, which will be supressed by spin fluctuations about the mean field used in the calculation. Below Tc, we can calculate how the magnetization varies near the transition by expanding the Brillouin fctn. to one higher power in x. The result is M ∼(T −Tc)1/2. (252) Note this exponent 1/2 is characteristic of the disappearance of the order parameter near the transition of any mean field theory (Landau). 61 0.2 0.4 0.6 0.8 1 1.2 1.4 T 0.1 0.2 0.3 0.4 0.5 M Figure 19: Plot of Mathematica solution to eqn. (249) for M vs. T using −J = g = 1; z = 4; S = 1/2. Tc=1 for this choice. Upper curve: H = 0.1; lower curve: H = 0. 3.2 Ising model The Ising model61 consists of a set of spins si with z-components only localized on lattice sites i interacting via nearest-neighbor exchange J < 0: H = J X i,j∈n.n. SiSj −2µBH X i Si. (253) Note it is an inherently classical model, since all spin commutators vanish [Si, Sj] = 0. Its historical importance consisted not so much in its applicability to real ferromagnetic systems as in the role its solutions, particularly the analytical solution of the 2D model published by Onsager in 1944, played in elucidating the nature of phase transitions. Onsager’s solution demonstrated clearly that all approximation methods and series expansions heretofore used to attack the ferromagnetic transition failed in the critical regime, and thereby highlighted the fundamental nature of the problem of critical phenomena, not solved (by Wilson and others) until in the early 70’s. 3.2.1 Phase transition/critical point We will be interested in properties of the model (253) at low and at high temperatures. What we will find is that there is a temperature Tc below which the system magnetizes spontaneously, just as in mean field theory, but that its true value is in general smaller than that given by mean field theory due to the importance of thermal fluctuations. Qualitatively, the phase diagram looks like this: Below Tc, the system magnetizes spontaneously even for field H →0. Instead of investi-gating the Onsager solution in detail, I will rely on the Monte Carlo simulation of the model 61The “Ising model" was developed as a model for ferromagnetism by Wilhelm Lenz and his student Ernst Ising in the early ‘20’s. 62 > < Figure 20: Field-magnetization curves for three cases. M0 is spontaneous magnetization in ferromagnetic phase T < Tc. at The idea is as follows. We would like to minimize F = T log Tr exp −βH for a given temperature and applied field. Finding the configuration of Ising spins which does so is a complicated task, but we can imagine starting the system at high temperatures, where all configurations are equally likely, and cooling to the desired temperature T.62 Along the way, we allow the system to relax by “sweeping" through all spins in the finite size lattice, and deciding in the next Monte Carlo “time" step whether the spin will be up or down. Up and down are weighted by a Boltzman probability factor p(Si = ±1/2) = e±µHeff/T e−µHeff/T + eµHeff/T , (254) where Heff i is the effective field defined in (246). The simulation picks a spin Si in the next time step randomly, but weighted with these up and down probabilities. A single “sweep" (time step) consists of L × L such attempts to flip spins, where L is the size of the square sample. Periodic boundary conditions are assumed, and the spin configuration is displayed, with one color for up and one for down. Here are some remarks on Ising critical phenomena, some of which you can check yourself with the simulation: • At high temperatures one recovers the expected Curie law χ ∼1/T 62This procedure is called simulated annealing 63 • The susceptibility diverges at a critical temperature below the mean field value.63 Near, but not too near, the transition χ has the Curie-Weiss form χ ∼(T −Tc)−1. • With very careful application of the simulation, one should obtain Onsager’s result that very near the transition ("critical regime") M ∼(Tc −T)β, (255) with β = 1/8. The susceptibility actually varies as χ ∼|T −Tc|−γ, (256) with γ = 7/4. Other physical quantities also diverge near the transition, e.g. the specific heat varies as |T −Tc|−α, with α = 0 (log divergence). • There is no real singularity in any physical quantity so long as the system size remains finite. • The critical exponents α, β, γ... get closer to their mean field values (β = 1/2, α =, γ =,... ) as the number of nearest neighbors in the lattice increases, or if the dimen-sionality of the system increases. • The mean square size of themal magnetization fluctuations gets very large close to the transition (“Critical opalescence", so named for the increased scattering of light near the liquid-solid critical point) . • Magnetization relaxation gets very long near the transition (“Critical slowing down"). • In 1D there is no finite temperature phase transition, although mean field theory predicts one. This is an example of the Mermin-Wagner theorem, which states that for short range interactions and an order parameter which obeys a continuous symmetry, there is no long range order in 1D, and none in 2D if T > 0. Note the Ising model escapes this theorem because it has only a discrete (so-called Z2) symmetry: the magnetization can be ±S on each site. 63This is given as a homework problem. Note the value of J used in the simulation is 1/4 that defined here, since S’s are ±1. 64 3.2.2 1D solution by transfer matrix 3.2.3 Ferromagnetic domains 3.3 Ferromagnetic magnons Let’s consider the simplest example of an insulating ferromagnet, described by the ferromag-netic Heisenberg Hamiltonian H = J X iδ ˆ Si · ˆ Si+δ −2µBH0 X i ˆ Siz, (257) where J < 0 is the ferromagnetic exchange constant, i runs over sites and δ runs over nearest neighbors, and H0 is the magnetic field pointing in the ˆ z direction. It is clear that the system can minimize its energy by having all the spins S align along the ˆ z direction at T = 0; i.e. the quantum ground state is identical to the classical ground state. Finding the elementary excitations of the quantum many-body system is not so easy, however, due to the fact that the spin operators do not commute. 3.3.1 Holstein-Primakoff transformation One can attempt to transform the spin problem to a more standard many-body interacting problem by replacing the spins with boson creation and annihilation operators. This can be done exactly by the Holstein-Primakoff transformation64 ˆ S+ i = ˆ Six + i ˆ Siy = (2S)1/2 1 −a† iai 2S !1/2 ai (258) ˆ S− i = ˆ Six −i ˆ Siy = (2S)1/2a† i 1 −a† iai 2S !1/2 . (259) Verify for yourselves that these definitions S± i give the correct commutation relations [ ˆ Sx, ˆ Sy] = i ˆ Sz if the bosonic commutation relations [a, a†] = 1 are obeyed on a given lattice site. Note also that the operators which commute with the Hamiltonian are ˆ S2 and ˆ Sz as usual, so we can classify all states in terms of their eigenvalues S(S + 1) and Sz. To complete the algebra we need a representation for ˆ Sz, which can be obtained by using the identity (on a given site i) ˆ S2 z = S(S + 1) −1 2  ˆ S+ ˆ S−+ ˆ S−ˆ S+ . (260) Using (259) and some tedious applications of the bosonic commutation relations, we find ˆ Sz = S −a†a. (261) 64T. Holstein and H. Primakoff, Phys. Rev. 58, 1098 (1940). 65 Now since the system is periodic, we are looking for excitations which can be characterized by a well-defined momentum (crystal momentum) k, so we define the Fourier transformed variables ai = 1 N 1/2 X k e−ik·xi bk ; a† i = 1 N 1/2 X k eik·xi b† k, (262) where as usual the F.T. variables also satisfy the bosonic relations [bk, bk′] = δkk′, etc. Looking forward a bit, we will see that the operators b† k and bk create and destroy a magnon or spin-wave excitation of the ferromagnet. These turn out to be excitations where the spins locally deviate only a small amount from their ground state values (∥ˆ z) as the “spin wave" passes by. This suggests a posteriori that an expansion in the spin deviations a† iai (see (261)) may converge quickly. Holstein and Primakoff therefore suggested expanding the nasty square roots in (259), giving ˆ S+ i ≃ (2S)1/2 " ai − a† iaiai 4S ! + . . . = 2S N 1/2 "X k e−ik·Ribk − 1 4SN X k,k′,k′′ ei(k−k′−k′′)·Rib† kbk′bk′′ + . . . , (263) ˆ S− i ≃ (2S)1/2 " a† i − a† ia† iai 4S ! . . . = 2S N 1/2 "X k eik·Rib† k − 1 4SN X k,k′,k′′ ei(k+k′−k′′)·Rib† kb† k′bk′′ + . . . , (264) ˆ Siz = S −a† iai = S −1 N X kk′ ei(k−k′)·Rib† kbk′. (265) Note that the expansion is formally an expansion in 1/S, so we might expect it to converge rapidly in the case of a large-spin system.65 The last equation is exact, not approximate, and it is useful to note that the total spin of the system along the magnetic field is Sz,tot = X i Sz = NS − X k b† kbk, (266) consistent with our picture of magnons tipping the spin away from its T = 0 equilibrium direction along the applied field. 3.3.2 Linear spin wave theory The idea now is to keep initially only the bilinear terms in the magnon operators, leaving us with a soluble Hamiltonian, hoping that we can then treat the 4th-order and higher 65For spin-1/2, the case of greatest interest, however, it is far from obvious that this uncontrolled approximation makes any sense, despite the words we have said about spin deviations being small. Why should they be? Yet empirically linear spin wave theory works very well in 3D, and surprisingly well in 2D. 66 terms perturbatively.66 Simply substitute (263)-(265) into (257), and collect the terms first proportional to S2, S, 1, 1/S, etc. We find H = 1 2JNzS2 −2µBH0S + Hmagnon 0 + O(1), (267) where z is the number of nearest neighbors (e.g. 6 for simple cubic lattice), and Hmagnon 0 = JS N X iδkk′ h e−i(k−k′)·Rieik′·δbkb† k′ + ei(k−k′)·Rie−ik′·δb† kbk′ −ei(k−k′)·Rib† kbk′ −e−i(k−k′)·(Ri+δ)b† kbk′ i + 2µBH0 N X ikk′ ei(k−k′)·Rib† kbk′ = JzS X k h γkbkb† k + γ−kb† kbk −2b† kbk i + 2µBH0 X k b† kbk = X k [−2JzS(1 −γk) + 2µBH0] b† kbk, (268) where γk = 1 z X δ eik·δ (269) is the magnon dispersion function, which in this approximation depends only on the positions of the nearest neighbor spins. Note in the last step of (268), I assumed γk = γ−k, which is true for lattices with inversion symmetry. For example, for the simple cubic lattice in 3D with lattice constant a, γk = (cos kxa + cos kya + cos kza)/3, clearly an even function of k. Under these assumptions, the magnon part of the Hamiltonian is remarkably simple, and can be written like a harmonic oscillator or phonon-type Hamiltonian, Hmagnon 0 = P k nkωk, where nk = b† kbk is the number of magnons in state k, and ωk = −2JSz(1 −γk) + 2µBH0 (270) is the magnon dispersion. The most important magnons will be those with momenta close to the center of the Brillouin zone, k ∼0, so we need to examine the small-k dispersion function. For a Bravais lattice, like simple cubic, this expansion gives 1 −γk ≃k2,67 i.e. the magnon dispersion vanishes as k →0. For more complicated lattices, there will be solutions with ωk →const. There is always a “gapless mode" ωk →0 as well, however, since the existence of such a mode is guaranteed by the Goldstone theorem.68 The figure shows a simple 1D schematic of a spin wave with wavelength λ = 2π/k corresponding to about 10 lattice sites. The picture is supposed to convey the fact that the spin deviations from the ordered state are small, and vary slightly from site to site. Quantum mechanically, the wave function for the spin wave state contains at each site a small amplitude for the spin to be 66physically these “nonlinear spin wave" terms represent the interactions of magnons, and resemble closely terms representing interactions of phonons in anharmonic lattice theory 67Check for simple cubic! 68For every spontaneously broken continuous symmetry of the Hamiltonian there is a ωk→0 = 0 mode. 67 Figure 21: Real space picture of spin deviations in magnon. Top: ordered ground state in a state with definite Sx and/or Sy. This can be seen by inverting Eq. (264) to leading order, & noting that the spin wave creation operator b† k lowers the spin with ˆ S−= ˆ Sx −i ˆ Sy with phase e−ik·Ri and amplitude ∼1/S at each site i. 3.3.3 Dynamical Susceptibility Experimental interlude The simple spin wave calculations described above (and below) are uncontrolled for spin-1/2 systems, and it would be nice to know to what extent one can trust them. In recent years, numerical work (exact diagonalization and quantum Monte Carlo) techniques have shown, as noted, that linear spin wave calculations compare suprisingly well with such “exact" results for the Heisenberg model. But we still need to know if there are any physical systems whose effective spin Hamiltonian can really be described by Heisenberg models. In addition, keep in mind that the utility of spin wave theory was recognized long before such numerical calculations were available, mainly through comparison with experiments on simple magnets. The most useful probe of magnetic structure is slow neutron scattering, a technique developed in the 40’s by Brockhouse and Schull (Nobel prize 1994). This section is a brief discussion of how one can use neutron scattering techniques to determine the dispersion and lifetimes of magnetic excitations in solids.69 Neutrons scatter from solids primarily due to the nuclear strong force, which leads to non-magnetic neutron-ion scattering and allows structural determinations very similar to x-ray diffraction analysis. In addition, slow neutrons traversing a crystal can emit or absorb phonons, so the inelastic neutron cross-section is also a sensitive measure of the dispersion 69A complete discussion is found in Lovesey, Theory of Neutron Scattering from Condensed Matter, Oxford 1984, V. 2 68 of the collective modes of the ionic system.70 There is also a force on the neutron due to the interaction of its small magnetic dipole moment with the spin magnetic moment of the electrons in the solid. There are therefore additional contributions to the peaks in the elastic neutron scattering intensity (at the Bragg angles) corresponding to magnetic scattering if the solid has long-range magnetic order; they can be distinguished from the nonmagnetic scattering because the additional spectral weight is strongly temperature dependent and disappears above the critical temperature, or through application of an external magnetic field. Furthermore, in analogy to the phonon case, inelastic neutron scattering experiments on ferromagnets yield peaks corresponding to processes where the neutron absorbs or emits a spin wave excitation. Thus the dispersion relation for the magnons can be mapped out.71 I will not go through the derivation of the inelastic scattering cross section, which can be found in Lovesey’s book. It is similar to the elementary derivation given by Ashcroft & Mermin in Appendix N for phonons. The result is  d2σ dΩdω  inel = a2 0 k′ k ng 2F(q) o2 e−2W(q) (1 + b(ω)) × −N π(gµB)2 X αβ (δαβ −ˆ qαˆ qβ) Im χαβ(q, −ω), (271) where a0 is the Bohr radius, k and k′ are initial and final wave vector, q = k−k′, F(q) atomic form factor, e−2W(q) the Debye-Waller factor, and b(ω) the Bose distribution function, N the number of unit cells, and ω is the energy change of the neuton, k2/(2m) −k′2/(2m). The physics we are interested in is contained in the imaginary part of the dynamic susceptibility χ(q, ω). For ω < 0, this measures the energy loss by neutrons as they slow down while emitting spin waves in the solid; for ω > 0 the neutrons are picking up energy from thermally excited spin waves. To see the effect of spin excitations within linear spin wave theory, one calculates the transverse spin susceptibility χR +−(Ri −Rj, t) ≡−Tr  ˆ ρ[ˆ S+ i (t), ˆ S− j ]  θ(t) (272) and then its Fourier transform wrt momentum q and frequency ω. I won’t do this calculation explicitly, but leave it as an exercise. You express the S operators in terms of the bk’s, whose time dependence is exactly known since the approximate Hamiltonian is quadratic. At the end, after Fourier transforming, one recovers 70cf. Ashcroft & Mermin ch. 24 71Even in systems without long range magnetic order, neutron experiments provide important information on the correlation length and lifetime of spin fluctuations. In strongly correlated systems (e.g. magnets just above their critical temperature, or itinerant magnets close to a magnetic transition as in Sec. xxx), these can be thought of as collective modes with finite lifetime and decay length. This means the correlation function ⟨ˆ Sα i (t)ˆ Sα j ⟩is not const. as t, |Ri −Rj| →∞, but may fall off very slowly, as “power laws" t−β, |Ri −Rj|−γ. 69 χ+−(q, ω) = (2S N )  1 ω + ωq + i0+  . (273) Again, as for the Fermi gas, we see that the collective modes of the system are reflected as the poles in the appropriate response function. The final cross section is now proportional to Im χ+−(q, −ω) ∼δ(ω + ωq), (274) i.e. there is a peak when a magnon is absorbed (neutron’s energy k′2/(2m) is larger than initial energy k2/(2m) ⇒ω ≡k2/(2m) −k′2/(2m) < 0.). There is another similar contribution proportional to δ(ω −ωq) (emission) coming from χ−+. Thus the dispersion ωq can be mapped out by careful measurement. The one-magnon lines are of course broadened by Figure 22: Neutron scattering data on ferromagnet. I searched a bit but couldn’t come up with any more modern data than this. This is Fig. 1, Ch. 4 of Kittel, Magnon dispersions in magnetite from inelastic neutron scattering by Brockhouse (Nobel Prize 1994) and Watanabe. magnon-magnon interactions, and by finite temperatures (Debye-Waller factor). There are also multimagnon absorption and emission processes which contribute in higher order. Finally, note that I proposed to calculate χ+−, the transverse magnetic susceptibility in Eq. 272. What about the longitudinal susceptibilty χzz, defined in the analogous way but for z-components ⟨[Sz, Sz]⟩? Because the ordered spins point in the z direction, the mode for this channel is gapped, not gapless, with energy of order J! 3.4 Quantum antiferromagnet Antiferromagnetic systems are generally approached by analogy with ferromagnetic systems, assuming that the system can be divided up into two or more sublattices, i.e. infinite inter-70 penetrating subsets of the lattice whose union is the entire lattice. Classically, it is frequently clear that if we choose all spins on a given judiciously chosen sublattice to be aligned with one another, we must achieve a minimum in the energy. For example, for the classical AF Heisenberg model H = J P iδ Si · Si+δ with J > 0 on a square lattice, choosing the A-B sublattices in the figure and making all spins up on one and down on another allows each bond to achieve its lowest energy of −JS2. This state, with alternating up and down spins, is referred to as the classical Neél state. Similarly, it may be clear to you that on the triangular lattice the classical lowest energy configuration is achieved when spins are placed at 120◦ with respect to one another on the sublattices A, B, C. However, quantum magnetic systems Figure 23: Possible choice of sublattices for antiferromagnet are not quite so simple. Consider the magnetization MA on a given sublattice (say the A sites in the figure) of the square lattice; alternatively one can define the staggered magnetization as Ms = P i(−1)i⟨ˆ Si⟩(Note (−1)i means +1 on the A sites and −1 on the B sites.) Either construct can be used as the order parameter for an antiferromagnet on a bipartite lattice. In the classical Neél state, these is simply MA = NS/2 and Ms = NS, respectively, i.e. the sublattice or staggered magnetization are saturated. In the wave function for the ground state of a quantum Heisenberg antiferromagnet, however, there is some amplitude for spins to be flipped on a a given sublattice, due to the fact that for a given bond the system can lower its energy by taking advantage of the ˆ Sx ˆ S′ x + ˆ Sy ˆ S′ y terms. This effect can be seen already by examining the two-spin 1/2 case for the ferromagnet and antiferromagnet. For the ferro-magnet, the classical energy is −|J|S2 = −|J|/4, but the quantum energy in the total spin 1 state is also −|J|/4. For the antiferromagnet, the classical energy is −JS2 = −J/4, but the energy in the total spin 0 quantum mechanical state is −3J/4. So quantum fluctuations– 71 which inevitably depress the magnetization on a given sublattice–lower the energy in the antiferromagnetic case. This can be illustrated in a very simple calculation of magnons in the antiferromagnet following our previous discussion in section 7.2. 3.4.1 Antiferromagnetic magnons We will follow the same procedure for the ferromagnet on each sublattice A and B, defining ˆ SA+ i = ˆ SA ix + i ˆ SA iy = (2S)1/2 1 −A† iAi 2S !1/2 Ai (275) ˆ SA− i = ˆ SA ix −i ˆ SA iy = (2S)1/2A† i 1 −A† iAi 2S !1/2 (276) ˆ SB+ i = ˆ SB ix + i ˆ SB iy = (2S)1/2 1 −B† i Bi 2S !1/2 Bi (277) ˆ SB− i = ˆ SB ix −i ˆ SB iy = (2S)1/2B† i 1 −B† i Bi 2S !1/2 (278) SA iz = S −A† iAi (279) −SB iz = S −B† i Bi, (280) i.e. we assume that in the absence of any quantum fluctuations spins on sublattice A are up and those on B are down. Otherwise the formalism on each sublattice is identical to what we did for the ferromagnet. We introduce on each sublattice creation & annihilation operators for spin waves with momentum k: ak = 1 N 1/2 X i∈A Aieik·Ri; a† k = 1 N 1/2 X i∈A A† ie−ik·Ri (281) bk = 1 N 1/2 X i∈B Bieik·Ri; b† k = 1 N 1/2 X i∈B B† i e−ik·Ri. (282) In principle k takes values only in the 1st magnetic Brillouin zone, or half-zone, since the periodicity of the sublattices is twice that of the underlying lattice. The spin operators on a 72 given site are then expanded as ˆ SA+ i ≃ 2S N 1/2 "X k e−ik·Riak + . . . , (283) ˆ SB+ i ≃ 2S N 1/2 "X k e−ik·Ribk + . . . , (284) ˆ SA− i ≃ 2S N 1/2 "X k eik·Ria† k + . . . , (285) ˆ SB− i ≃ 2S N 1/2 "X k eik·Rib† k + . . . , (286) ˆ SA iz = S −1 N X kk′ ei(k−k′)·Ria† kak′ (287) ˆ SB iz = −S + 1 N X kk′ ei(k−k′)·Ria† kak′. (288) The expansion of the Heisenberg Hamiltonian in terms of these variables is now (compare (267)) H = −NzJS2 + Hmagnon 0 + O(1), (289) Hmagnon 0 = JzS X k h γk(a† kb† k + akbk) + (a† kak + b† kbk) i (290) Unlike the ferromagnetic case, merely expressing the Hamiltonian to bilinear order in the magnon variables does not diagonalize it immediately. We can however perform a canonical transformation72 to bring the Hamiltonian into diagonal form (check!): αk = ukak −vkb† k ; α† k = uka† k −vkbk (291) βk = ukbk −vka† k ; β† k = ukb† k −vkak, (292) where the coefficients uk, vk must be chosen such that u2 k −v2 k = 1. One such choice is uk = cosh θk and vk = sinh θk. For each k, choose the angle θk such that the anomalous terms like α† kβ† k vanish. One then finds the solution tanh 2θk = −γk, (293) and Hmagnon 0 = −NEJ + X k ωk(α† kαk + β† kβk + 1), (294) where ω2 k = E2 J(1 −γ2 k), (295) 72The definition of a canonical transformation, I remind you, is one which will give canonical commutation relations for the transformed fields. This is important because it ensures that we can interpret the Hamiltonian represented in terms of the new fields as a (now diagonal) fermion Hamiltonian, read off the energies, etc. 73 Figure 24: Integrated intensity of (100) Bragg peak vs. temperature for LaCuO4, with TN = 195K. (After Shirane et al. 1987) and EJ = JzS. Whereas in the ferromagnetic case we had ωk ∼(1 −γk) ∼k2, it is noteworthy that in the antiferromagnetic case the result ωk ∼(1 −γk)1/2 ∼k gives a linear magnon dispersion at long wavelengths (a center of symmetry of the crystal must be assumed to draw this conclusion). Note further that for each k there are two degenerate modes in Hmagnon 0 . 3.4.2 Quantum fluctuations in the ground state You may wonder that there is a constant term in (294), since Hmagnon was the part of the Hamiltonian which gave the excitations above the ground state in the ferromagnetic case. Here at T = 0 the expectation values (in 3D at least, see below) ⟨α†α⟩and ⟨β†β⟩vanish – there are no AF magnons in the ground state. But this leaves a constant correction coming from Hmagnon of −NEJ +P k ωk. The spin wave theory yields an decrease of the ground state energy relative to the classical value −NJzS2, but an increase over the quantum ferromagnet result of −N|J|S(S + 1) due to the zero-point (constant) term in (294).73 The ground-state energy is then E0 ≃−NzJS2 −NzJS + X k ωk. (296) The result is conventionally expressed in terms of a constant β defined by E0 ≡−NJzS  S + β z  , (297) 73Recall the classical Neél state, which does not contain such fluctuations, is not an eigenstate of the quantum Heisenberg Hamiltonian. 74 Figure 25: a) Inelastic neutron scattering intensity vs. energy transfer at 296K near zone boundary Q = (1, k, 0.5) for oriented LaCuO4 crystals with Neél temperature 260K; b) spin-wave dispersion ωq vs. q∥, where q∥is in-plane wavevector. (after Hayden et al. 1991) and β/z = N −1 X k  1 − q 1 −γ2 k  = 0.097 3D 0.158 2D . (298) (I am checking these–they don’t agree with literature I have) Quantum fluctuations have the further effect of preventing the staggered magnetization from achieving its full saturated value of S, as in the classical Neél state, as shown first by Anderson.74 Let us consider the average z-component of spin in equilibrium at temperature T, averaging only over spins on sublattice A of a D-dimensional hypercubic lattice. From (287), we have ⟨ˆ SA z ⟩= S −N −1 P k⟨A† kAk⟩within linear spin wave theory. Inverting the transformation (398), we can express the A’s in terms of the α’s and β’s, whose averages we can easily calculate. Note the 0th order in 1/S gives the classical result, ⟨ˆ SA z ⟩, and the deviation is the spin wave reduction of the sublattice moment δMA N = ⟨ˆ SA z ⟩−S = −1 N X k ⟨a† kak⟩ = −1 N X k ⟨(ukα† k + vkβk)(ukαk + vkβ† k)⟩ = −1 N X k u2 k⟨α† kαk⟩+ v2 k⟨β† kβk⟩+ v2 k, (299) where N is the number of spins on sublattice B. We have neglected cross terms like ⟨α† kβ† k⟩ because the α and β are independent quanta by construction. However the diagonal averages ⟨α† kαk⟩and ⟨β† kβk⟩are the expectation values for the number operators of independent bosons 74P.W. Anderson, Phys. Rev. 86, 694 (1952). 75 with dispersion ωk in equiblibrium, and so can be replaced (within linear spin wave theory) by nk ≡⟨β† kβk⟩= ⟨α† kαk⟩= b(ωk), (300) where b is the Bose distribution function. The tranformation coefficients uk = cosh θk and vk = sinh θk are determined by the condition (293) such that u2 k + v2 k = cosh 2θk = 1 p 1 −γ2 k (301) v2 k = 1 2 1 p 1 −γ2 k −1 ! , (302) such that the sulattice magnetization (299) becomes δMA N = 1 2 −1 N X k  nk + 1 2  1 p 1 −γ2 k (303) Remarks: 1. All the above “anomalies" in the AF case relative to the F case occur because we tried to expand about the wrong ground state, i.e. the classical Neél state, whereas we knew a priori that it was not an eigenstate of the quantum Heisenberg model. We were lucky: rather than breaking down entirely, the expansion signaled that including spin waves lowered the energy of the system; thus the ground state may be thought of as the classical Neél state and an admixture of spin waves. 2. The correction δMA is independent of S, and negative as it must be (see next point). However relative to the leading classical term S it becomes smaller and smaller as S increases, as expected. 3. The integral in (303) depends on the dimensionality of the system. It has a T-dependent part coming from nk and a T-independent part coming from the 1/2. At T = 0, where there are no spin waves excited thermally, nk = 0, and we find δMA N ≃          −0.078 D = 3 −0.196 D = 2 ∞ D = 1 (304) The divergence in D = 1 indicates the failure of spin-wave theory in one dimension. 4. The low temperature behavior of δM(T) must be calculated carefully due to the sin-gularities of the bose distribution function when ωk →0. If this divergence is cut off by introducing a scale k0 near k = 0 and k = (π/a, π/a), one finds that δMA diverges as 1/k0 in 1D, and as log k0 in 2D, whereas it is finite as k0 →0 in 3D. Thus on this 76 basis one does not expect long range spin order at any nonzero temperature in two dimensions (see discussion of Mermin-Wagner theorem below), nor even at T = 0 in one dimension. 3.4.3 Nonlinear spin wave theory 3.4.4 Frustrated models 3.5 1D & 2D Heisenberg magnets 3.5.1 Mermin-Wagner theorem 3.5.2 1D: Bethe solution 3.5.3 2D: Brief summary 3.6 Itinerant magnetism “Itinerant magnetism" is a catch-all phrase which refers to magnetic effects in metallic sys-tems (i.e., with conduction electrons). Most of the above discussion assumes that the spins which interact with each other are localized, and there are no mobile electrons. However we may study systems in which the electrons which magnetize, or nearly magnetize are mobile, and situations in which localized impurity spins interact with conduction electron spins in a host metal. The last problem turns out to be a very difficult many-body problem which stimulated Wilson to the develop of renormalization group ideas. 3.7 Stoner model for magnetism in metals The first question is, can we have ferromagnetism in metallic systems with only one relevant band of electrons. The answer is yes, although the magnetization/electron in the ferro-magnetic state is typically reduced drastically with respect to insulating ferromagnets. The simplest model which apparently describes this kind of transition (there is no exact solution in D > 1) is the Hubbard model we have already encountered. A great deal of attention has been focussed on the Hubbard model and its variants, particularly because it is the simplest model known to display a metal-insulator (Mott-Hubbard) transition qualitatively similar to what is observed in the high temperature superconductors. To review, the Hubbard model consists of a lattice of sites labelled by i, on which electrons of spin ↑or ↓may sit. The kinetic energy term in the Hamiltonian allows for electrons to hop between sites with matrix 77 element t, and the potential energy simply requires an energy cost U every time two opposing spins occupy the same site.75 Longer range interactions are neglected: H = −t X σ c† iσcjσ + 1 2U X σ niσni−σ, (305) where < ij > means nearest neighbors only. In its current form the kinetic energy, when Fourier transformed, corresponds, to a tight binding band in d dimensions of width 4dt, ϵk = −2t d X α=1 cos kαa, (306) where a is the lattice constant. The physics of the model is as follows. Imagine first that there is one electron per site, i.e. the band is half-filled. If U = 0 the system is clearly metallic, but if U →∞, double occupation of sites will be “frozen out". Since there are no holes, electrons cannot move, so the model must correspond to an insulating state; at some critical U a metal-insulator transition must take place. We are more interested in the case away from half-filling, where the Hubbard model is thought for large U and small doping (deviation of density from 1 particle/site) to have a ferromagnetic ground state.76 In particular, we would like to investigate the transition from a paragmagnetic to a ferromagnetic state as T is lowered. This instability must show up in some quantity we can calculate. In a ferromagnet the susceptibility χ diverges at the transition, i.e. the magnetization produced by the application of an infinitesimal external field is suddenly finite. In many-body language, the static, uniform spin susceptibility is the retarded spin density – spin density correlation function, 75What happened to the long-range part of the Coulomb interaction? Now that we know it is screened, we can hope to describe its effects by including only its matrix elements between electrons in wave functions localized on site i and site j, with |i −j| smaller than a screening length. The largest element is normally the i = j one, so it is frequently retained by itself. Note the Hamiltonian (305) includes only an on-site interaction for opposite spins. Of course like spins are forbidden to occupy the same site anyway by the Pauli principle. 76At 1/2-filling, one electron per site, a great deal is known about the Hubbard model, in particular that the system is metallic for small U (at least for non-nested lattices, otherwise a narrow-gap spin density wave instability is present), but that as we increase U past a critical value Uc ∼D a transition to an antiferromagnetic insulating state occurs (Brinkman-Rice transition). With one single hole present, as U →∞, the system is however ferromagnetic (Nagaoka state). 78 (for the discussion below I take gµB = 1) χ = χ(q = 0, ω = 0) = lim hz→0 ⟨Sz⟩ hz = Z d3r Z ∞ 0 dt⟨[Sz(r, t), Sz(0, 0)]⟩, (307) where in terms of electron number operators nσ = ψ† σψσ, the magnetization operators are Sz = (1/2)[n↑−n↓], i.e. they just measure the surplus of up over down spins at some point in space. k+q σ χ 0 F T T ~1/T Curie Pauli N 0 b) k U k' k'+q −σ −σ σ a) Figure 26: 1a) Hubbard interaction; 1b) Spin susceptibility vs. T for free fermions. Diagramatically, the Hubbard interaction Hint = U P i ni↑ni↓looks like figure 26a); note only electrons of opposite spins interact. The magnetic susceptibility is a correlation function similar to the charge susceptibility we have already calculated. At time t=0, we measure the magnetization Sz of the system, allow the particle and hole thus created to propagate to a later time, scatter in all possible ways, and remeasure Sz. The so-called “Stoner model" of ferromagnetism approximates the perturbation series by the RPA form we have already encountered in our discussion of screening,77 which gives χ = χ0/(1 −Uχ0).78 At sufficiently high T (χ0 varies as 1/T in the nondegenerate regime, Fig. 26c)) we will have Uχ0(T) < 1, but as T is lowered, Uχ0(T) increases. If U is large enough, such that UN0 > 1, there will 77In the static, homogeneous case it is equivalent to the self-consistent field (SCF) method of Weiss. 78 Hamiltonian after mean field procedure is H = X kσ ϵkc† kσckσ + U X kσ n−σc† kσckσ + H(nσ −n−σ) (308) which is quadratic & therefore exactly soluble. Also note units: H is really µBgH/2, but never mind. Hamiltonian is equivalent to a free fermion system with spectrum ˜ ϵkσ = ϵk + Un−σ + Hσ. The number of electrons with spin σ is therefore nσ = X k f(˜ ϵkσ) = X k f(ϵk + Un−σ + Hσ) (309) ≡ n0 σ + δnσ, (310) where n0 σ is the equilibrium number of fermions in a free gas with no U. We can expand δnσ for small field: 79 be a transition at Uχ0(Tc) = 1, where χ diverges. Note for Uχ0(T) > 1 (T < Tc), the susceptibility is negative, so the model is unphysical in this region. This problem arises in part because the ferromagnetic state has a spontaneously broken symmetry, ⟨Sz⟩̸= 0 even for hz →0. Nevertheless the approach to the transition from above and the location of the transition appear qualitatively reasonable. It is also interesting to note that for Uχ(0) = UN0 < 1, there will be no transition, but the magnetic susceptibility will be enhanced at low temperatures. So-called “nearly ferromagnetic" metals like Pd are qualitatively described by this theory. Comparing the RPA form χ = χ0(T) 1 −Uχ0(T) (315) to the free gas susceptibility in Figure 9b, we see that the system will look a bit like a free gas with enhanced density of states, and reduced degeneracy temperature T ∗ F.79 For Pd, the Fermi temperature calculated just by counting electrons works out to 1200K, but the susceptibility is found experimentally to be ∼10× larger than N0 calculated from band structure, and the susceptibility is already Curie-like around T ∼T ∗ F ≃300K. 3.7.1 Moment formation in itinerant systems We will be interested in asking what happens when we put a localized spin in a metal, but first we should ask how does that local moment form in the first place. If an arbitrary impurity is inserted into a metallic host, it is far from clear that any kind of localized moment δnσ = X k f(ϵk + Un−σ + Hσ) − X k f(ϵk + Hσ) ≃ X k d f dϵk d˜ ϵ dH H=0 H −(U →0) (311) = X k d f dϵk  U dn−σ dH + σ  H=0 H −(U →0) . → H→0 −χ0U dn−σ dH H=0 H where in the last step we set H = 0, in which case nσ = n−σ (paramagnetic state). I used χ0 = P k −d f/dϵk for Fermi gas. The magnetization is now m ≡χH = n↑−n↓= χ0H + δn↑−δn↓ (312) = χ0H + χ0UχH, (313) since χ = dm/dH. So total susceptibility has RPA form: χ = χ0 1 −Uχ0 . (314) 79Compare to the Fermi liquid form χ = m∗ m χ0 1 + F a 0 . (316) 80 will result: a donor electron could take its spin and wander off to join the conduction sea, for example. Fe impurities are known to give rise to a Curie term in the susceptibility when embedded in Cu, for example, but not in Al, suggesting that a moment simply does not form in the latter case. Anderson80 showed under what circumstances an impurity level in an interacting host might give rise to a moment. He considered a model with a band of electrons81 with energy ϵk, with an extra dispersionless impurity level E0. Suppose there are strong local Coulomb interactions on the impurity site, so that we need to add a Hubbard-type repulsion. And finally suppose the conduction (d) electrons can hop on and off the impurity with some matrix element V . The model then looks like H = X kσ ϵkc† kσckσ + E0 X σ n0σ + V X kσ (c† kσc0 + c† 0ckσ) + 1 2U X σ n0σn0−σ, (317) where n0σ = c† 0σc0σ is the number of electrons of spin σ on the impurity site 0. By the Fermi Golden rule the decay rate due to scattering from the impurity of a band state ϵ away from the Fermi level EF in the absence of the interaction U is of order ∆(ϵ) = πV 2 X k δ(ϵ −ϵk) ≃πV 2N0 (318) In the “Kondo" case shown in the figure, where E0 is well below the Fermi level, the scattering processes take place with electrons at the Fermi level ϵ = 0, so the bare width of the impurity state is also ∆≃πV 2N0. So far we still do not have a magnetic moment, since, in the absence of the interaction U, there would be an occupation of 2 antiparallel electrons. If one could effectively prohibit double occupancy, however, i.e. if U ≫∆, a single spin would remain in the localized with a net moment. Anderson obtained the basic physics (supression of double occupancy) by doing a Hartree-Fock decoupling of the interaction U term. Schrieffer and Wolff in fact showed that in the limit U →−∞, the virtual charge fluctuations on the impurity site (occasional double occupation) are eliminated, and the only degree of freedom left (In the so-called Kondo regime corresponding to Fig. 10a) is a localized spin interacting with the conduction electrons via an effective Hamiltonian HKondo = JS · σ, (319) where J is an antiferromagnetic exchange expressed in terms of the original Anderson model parameters as J = 2V 2 E0 , (320) 80PW Anderson, Phys. Rev. 124, 41 (1961) 81The interesting situation for moment formation is when the bandwidth of the "primary" cond. electron band overlapping the Fermi level is much larger than the bare hybridization width of the impurity state. The two most commonly considered situations are a band of s electrons with d-level impurity (transition metal series) and d-electron band with localized f-level (rare earths/actinides–heavy fermions). 81 S is the impurity spin-1/2, and σi = 1 2 X kk′αβ c† kα (τi)αβ ck′β, (321) with τi the Pauli matrices, is just the conduction electron spin density at the impurity site. E E N(E) E d-band E 0 Kondo Mixed valence Empty moment E 0 E 0 d-band d-band EF EF Figure 27: Three different regimes of large U Anderson model depending on position of bare level E0. In Kondo regime (E0 << EF ), large moments form at high T but are screened at low T. In mixed valent regime, occupancy of impurity level is fractional and moment formation is marginal. For E0 > EF , level is empty and no moment forms. 3.7.2 RKKY Interaction Kittel p. 360 et seq. 3.7.3 Kondo model The Hartree-Fock approach to the moment formation problem was able to account for the existence of local moments at defects in metallic hosts, in particular for large Curie terms in the susceptibility at high temperatures. What it did not predict, however, was that the low-temperature behavior of the model was very strange, and that in fact the moments present at high temperatures disappear at low temperatures, i.e. are screened completely by the conduction electrons, one of which effectively binds with the impurity moment to create a local singlet state which acts (at T = 0) like a nonmagnetic scatterer. This basic physical picture had been guessed at earlier by J. Kondo,82 who was trying to explain the existence of a resistance minimum in certain metallic alloys. Normally the resistance of 82J. Kondo, Prog. Theor. Phys. 32, 37 (64). 82 metals is monotonically decreasing as the temperature is lowered, and adding impurities gives rise to a constant offset (Matthiesen’s Rule) which does not change the monotonicity. For Fe in Cu, however, the impurity contribution δρimp increased as the temperature was lowered, and eventually saturated. Since Anderson had shown that Fe possessed a moment in this situation, Kondo analyzed the problem perturbatively in terms of a magnetic impurity coupling by exchange to a conduction band, i.e. adopted Eq. (319) as a model. Imagine that the system at t = −∞consists of a Fermi sea |0⟩and one additional electron in state |k, σ⟩at the Fermi level. The impurity spin has initial spin projection M, so we write the initial state as |i⟩= c† kσ|0; M⟩Now turn on an interaction H1 adiabatically and look at the scattering amplitude83 between |i⟩and |f⟩= c† k′σ′|0; M ′⟩ ⟨f|T|i⟩ = −2πi⟨f|H1 + H1 1 ϵk −H0 H1 + . . . |i⟩ (325) If H1 were just ordinary impurity (potential) scattering, we would have H1 = P kk′σσ′ c† kσVkk′ck′σ′, and there would be two distinct second-order processes k →k′ contributing to Eq. (325), as shown schematically at the top of Figure 28, of type a), ⟨0|ck′c† k′Vk′pcp 1 ϵk −H0 c† pVpkckc† k|0⟩= Vk′p 1 −f(ϵp) ϵk −ϵp Vpk (326) and type b), ⟨0|ck′c† pVpkck 1 ϵk −H0 c† k′Vk′pcpc† k|0⟩ = −Vpk f(ϵp) ϵk −(ϵk −ϵp + ϵk′)Vk′p = Vpk f(ϵp) ϵk′ −ϵp Vk′p (327) where I have assumed k is initially occupied, and k′ empty, with ϵk = ϵk′, whereas p can be either; the equalities then follow from the usual application of c†c and cc† to |0⟩. Careful checking of the order of the c’s in the two matrix elements will show you that the first process only takes place if the intermediate state p is unoccupied, and the second only if it is unoccupied. 83Reminder: when using Fermi’s Golden Rule (see e.g., Messiah, Quantum Mechanics p.807): dσ dΩ= 2π ℏv |T|2ρ(E) (322) we frequently are able to get away with replacing the full T-matrix by its perhaps more familiar 1st-order expansion dσ dΩ= 2π ℏv |H1|2ρ(E) (323) (Recall the T matrix is defined by ⟨ϕf|T|ϕi⟩= ⟨ϕf|T|ψ+ i ⟩, where the ϕ’s are plane waves and ψ+ is the scattering state with outgoing boundary condition.) In this case, however, we will not find the interesting log T divergence until we go to 2nd order! So we take matrix elements of T = H1 + H1 1 ϵk −H0 H1 + . . . (324) This is equivalent and, I hope, clearer than the transition amplitude I calculated in class. 83 Nnow when one sums the two processes, the Fermi function cancels. This means there is no significant T dependence to this order (ρimp due to impurities is a constant at low T), and thus the exclusion principle does not play an important role in ordinary potential scattering. Vpk Vk'p k p k' k p k' Vpk Vk'p k p k' kσ p k' σ σ' '' σ '' σ σ' Sµ ν S Sµ ν S potential scatt. spin scatt. a) b) b) a) Figure 28: 2nd-order scattering processes: a) direct and b) exchange scattering. Top: potential scattering; bottom: spin scattering. Two terms corresponding to Eq. 325 Now consider the analagous processes for spin-scattering. The perturbing Hamiltonian is Eq. 319. Let’s examine the amplitude for spin-flip transitions caused by H1, first of type a), ⟨0Ms′|H1|0Ms⟩ (328) = J2 4 SµSν⟨0Ms′|ck′σ′c† k′σ′τ ν σ′σ′′cpσ′′ 1 ϵk −H0 c† pσ′′τ µ σ′′σckσc† kσ|0Ms⟩ = J2 4 1 −f(ϵp) ϵk −ϵp ⟨M ′ S|SνSµτ ν σ′σ′′τ µ σ′′σ|MS⟩ and then of type b), J2 4 f(ϵp) ϵk −ϵp ⟨M ′ S|SνSµτ ν σ′′στ µ σ′σ′′|MS⟩. (329) Now note that τ µ σ′σ′′τ ν σ′′σ = (τ µτ ν)σ′σ, and use the identity τµτν = δνµ + iταϵαµν. The δµν pieces clearly will only give contributions proportional to S2, so they aren’t the important ones which will distinguish between a) and b) processes compared to the potential scattering case. The ϵανµ terms give results of differing sign, since a) gives ϵανµ and b) gives ϵαµν. Note the basic difference between the 2nd-order potential scattering and spin scattering is that the matrix elements in the spin scattering case, i.e. the Sµ, didn’t commute! When we add 84 a) and b) again the result is J2 ϵk −ϵp 1 4S(S + 1) −⟨M ′ Sσ′|S · σ|MS, σ⟩  +f(ϵp)⟨M ′ Sσ′|S · σ|MS, σ⟩} (330) so the spin-scattering amplitude does depend on T through f(ϵp). Summing over the inter-mediate states pσ′′ gives a factor X p f(ϵp) ϵk −ϵp ≃N0 Z dξp f(ξp) ξk −ξp = N0 Z dξp  −∂f ∂ξp  log |ξk −ξp|, (331) which is of order log T for states ξk at the Fermi surface! Thus the spin part of the interaction, to a first approximation, makes a contribution to the resistance of order J3 log T (ρ involves the square of the scattering amplitude, and the cross terms between the 1st and 2nd-order terms in perturbation theory give this result). Kondo pointed to this result and said, “Aha!", here is a contribution which gets big as T gets small. However this can’t be the final answer. The divergence signals that perturbation theory is breaking down, so this is one of those very singular problems where we have to find a way to sum all the processes. We have discovered that the Kondo problem, despite the fact that only a single impurity is involved, is a complicated many-body problem. Why? Because the spin induces correlations between the electrons in the Fermi sea. Example: suppose two electrons, both with spin up, try to spin flip scatter from a spin-down impurity. The first electron can exchange its spin with the impurity and leave it spin up. The second electron therefore cannot spin-flip scatter by spin conservation. Thus the electrons of the conduction band can’t be treated as independent objects. Summing an infinite subset of the processes depicted in Figure 28 or a variety of other techniques give a picture of a singlet bound state where the impurity spin binds one electron from the conduction electron sea with binding energy TK = De−1/(JN0), (332) where D is the conduction electron bandwidth. The renormalization group picture devel-oped by Wilson in the 70s and the exact Bethe ansatz solution of Wiegman/Tsvelick and Andrei/Lowenstein in 198? give a picture of a free spin at high temperatures, which hy-bridizes more and more strongly with the conduction electron sea as the temperature is lowered. Below the singlet formation temperature of TK, the moment is screened and the impurity acts like a potential scatterer with large phase shift, which approaches π/2 at T = 0. 85 4 Electron-phonon interaction 4.1 Hamiltonian The subtle interplay of electrons and phonons was explained in the 50’s by some of the earliest practitioners of quantum many-body theory, leading eventually to an understanding of the mechanism underlying superconductivity. Recall that the ions in a metal have two basic effects on the electronic states: 1) the static ionic lattice provides a periodic potential in which conduction electrons must move, leading to the evolution of plane wave states in the Fermi gas into Bloch waves in the crystal, and 2) the scattering of electrons by lattice vibrations, and vice versa. The first effect will be ignored here, as we are essentially interested in long-wavelength phenomena, where the differences between proper calculations using Bloch waves and simpler ones using plane waves are negligible. It suffices then to consider the phonons in a lattice interacting with a Fermi gas in which the most important effects of the long-range Coulomb interaction have been accounted for. Without the Coulomb interaction, the phonon frequencies are just those we would get from a classical model of balls of mass M (ionic mass) connected by springs. For a 3D solid with 1 atom per unit cell, there are 3N normal modes comprising 3 acoustic phonon branches ωλ k. When one includes the long-range Coulomb interaction but neglects the electron-phonon coupling, one finds that the longitudinal acoustic mode has been lifted to the ionic plasma frequency, ωion pl ≃(4πZ2e2n/M)1/2. The terms of the Goldstone theorem which insists on the existence of an acoustic mode for each spontaneously broken continuous symmetry are violated by the long-range nature of the Coulomb force, and the sloshing back and forth of the ion “fluid" at ωion pl occurs for the same reason and at the same frequency (up to the mass difference) that it does in the electron case. At this point we are seriously worried that we don’t understand how acoustic phonons ever exist in charged systems. If one now includes the electron- phonon coupling, however, the electronic medium screens the long-range Coulomb interaction, leading to a finite interaction length and the recovery of the Goldstone (acoustic) mode. Let’s give a brief overview of where we’re going. I first want to get to the point where we can write down the full Hamiltonian for the problem. We want to show that it makes sense to write the Hamiltonian describing the electron-phonon system as H = H0 el + H0 ph + Hcoul + Hint, (333) 86 where H0 el = X kσ ξkc† kσckσ (334) H0 ph = X kλ ωkλ(a† kλakλ + 1 2) (335) HCoul = 1 2 X kk′q σσ′ V (q)c† k′+qσ′c† kσck+qσck′σ′ (336) Hint = X kk′σλ gkk′c† kσck′σ(a† −qλ + aqλ) (337) where a† kλ creates a phonon with wave vector q ≡k′−k and polarization λ, and gkk′ ∝M −1/2 is the bare electron-phonon coupling. The unperturbed phonon Hamiltonian Hph is of course just the sum of 3N independent harmonic oscillators in 2nd quantized form, and the bare Coulomb matrix element in HCoul is V (q) = 4πe2/q2. The derivation of the electron-phonon Hamiltonian Hint and its quantization is relatively straightforward, and I will sketch it here. 4.1.1 Derivation of e−-ph coupling Assume the ion is located at position Ri, at a displacement ui from its equilibrium position R0 i . If the potential of the ion is assumed to be rigid, the interaction energy of the electronic charge density with the ions is simply84 Hint = X iσ Z d3rψ† σ(r)ψσ(r)V (r −Ri). (338) For small amplitude vibrations, we can expand in powers of ui: Hint = X iσ Z d3rψ† σ(r)ψσ(r)V (r −R0 i ) + X iσ Z d3rψ† σ(r)ψσ(r)ui · ∇RiV (r −Ri)|R0 i + . . . (339) Now expand the field operators ψσ in terms of Bloch waves: ψσ(r) = X k ckσϕk(r), (340) where ϕkσ(r + R0 i ) = eik·R0 i ϕk(r) (341) 84This is for a Bravais lattice. If there is a basis one has to be a bit careful about labelling the lattice sites with additional intracell indices, i.e. Riα, α = 1 . . . m, where m is number of atoms/cell. 87 so the quantity which appears in Eq. (339) may be recast by performing a shift by a Bravais lattice vector and using the periodicity of ∇R0 i V (r −R0 i ), Z d3rϕ∗ k′σ(r)ϕkσ(r)∇R0 i V (r −R0 i ) = Z d3rϕ∗ k′σ(r + R0 j)ϕkσ(r + R0 j)∇R0 i V (r −R0 i ) = ei(k−k′)·R0 j Z d3rϕ∗ k′σ(r)ϕkσ(r)∇R0 i V (r −R0 i ) | {z } (342) Wkk′ (343) Now let us 2nd-quantize the displacement u as we did when we were discussing the isolated phonon system,85 ui(t) = 1 √ NM X qλ Q(k, t)eλ(q)eiq·R0 i (345) with Qλ(q) = 1 p 2ωλ(q)  aλ(q) + a† λ(−q)  (346) so interaction Hamiltonian can be rewritten Hint = X kk′σ c† k′σckσ X j Wkk′ei(k−k′)·R0 j 1 √ NM × × X qλ Qλ(q)eλ(q)eiq·R0 j = X kk′σ X λ c† k′σckσ Wkk′ · eλ(q)  Qλ(q) r N M ≡ X kk′σλ gkk′λc† k′σckσ  aλ(q) + a† λ(−q)  (347) where now q due to momentum conservation (δ-function from summing over j) is to be interpreted as q = k −k′ + G (348) with G is a vector of reciprocal lattice (arose because q was defined to lie in 1st B-zone). The electron-phonon coupling constant is gkk′λ = Wkk′ · eλ(q)  s N 2Mωλ(q), (349) which contains the original ball-and-spring phonon frequencies ωλ. We have left out the Coulomb interactions, however, so at first sight it appears as though this expression should 85Before we dealt primarily with the 1D chain, so I suppressed the polarization indices. It is important to recall that in a 3 dimensional system there are 3N normal modes (3mN if there are m atoms per unit cell). For each value of k there are 3 acoustic (optical only if there are add’l atoms per unit cell requiring index α) modes denoted by different values of the branch index λ. The vectors eλ(k) are the polarization vectors of the modes and satisfy the orthogonality condition X α eλ α(k) · eλ′ α (k) = δλλ′ (344) 88 be modified by replacing ωλ by ωion pl if the mode is longitudinal and couples to the long-range Coulomb interaction. This does not include the ionic screening yet, however (see below). Regardless of the form of gkk′, the final result is that an electron in state k, σ can undergo a scattering process with amplitude gkk′, ending up in final state k′, σ by absorption (emission) of a phonon of momentum q. This form is useful, but calculating gkk′ from first principles is difficult because V is poorly known. 4.1.2 Jellium model We can get some dimensionally reasonable results in the so-called "jellium" model, where the ions are represented as a featureless, positively charged elastic continuum,86 we will simply replace the eigenfrequencies ωkλ of the neutral system by the constant ωion pl according to the arguments given above. Again we expand the crystal potential V (r −Rj) around the equi-librium sites R0 i . The dispacements u(R) in the jellium gives a charge density fluctuation87 −nZe∇· u. This interacts with the electron gas through the Coulomb interaction, leading to the interaction Hjellium int = Ze2 X σ Z d3rd3r′ ψ† σ(r)ψσ(r) 1 |r −r′|∇· u(r′), (350) and then quantizing the ionic displacements ui = Ri −R0 i as in Eq.(345), one finds Hjellium int = X kk′σ c† k′σckσ  aλ(q) + a† λ(−q)  4πiq · eλ(q) M 1/2 Ze2n q2 (351) Comparing with Eq. (347), we see that the effective e-ph coupling constant in an isotropic system is g(q) = 4πiZe2 q s N 2Mωion pl = 4πiZe2n1/2 qM 1 2 . (352) 4.1.3 Screening. The first point I would like to review is the renormalization of the electron-phonon coupling which leads to screening and the recovery of the acoustic spectrum. The main point is to realize that the singular behavior is due to the long-range Coulomb interaction, which will be screened. Any time a phonon of momentum q is excited, it creates charge density fluctuations because the ions are positively charged. These then interact with the electron gas. The bare potential created by the charge fluctuation is proportional to the electron-phonon coupling 86... justified by the large masses and correspondingly long timescales for ionic motion (Born-Oppenheimer) 87Recall from E& M the polarization charge density is ρP = −∇· P, where P is the polarization, and the polarization due to a density n of dipole moments p = Zeu is therefore nZeu. 89 constant, so screening the ionic charge amounts to replacing the bare coupling g with a screened coupling ¯ g(q, ωn) = g(q)/ϵ(q, ωn), (353) where q = k′−k and ϵ(q, ωn) = 1−V (q)χ(q, ωn) is the RPA dielectric constant. The frequency dependence of the dielectric constant will become important for us on scales of the Debye frequency ωD or smaller, and this is normally a very small scale by electronic standards! So for most purposes we keep a frequency-independent screened coupling ¯ g(q) ≃g(q)/ϵ(q, 0). We would like to see that this screening recovers acoustic modes as observed and expected. The bare interaction Hamiltonian may be written in the Jellium model as (see Eq. (347) Hint = X q g(q)n−qQq (354) with nq = P q c† k+qck. Consider now the entire Hamiltonian for the phonon coordinates, including the coupling to the electron gas (recall derivation of normal modes for linear chain in Section 1): Hph + Hint = 1 2 X q  1 M PqP−q + Mωion pl 2QqQ−q + 2gqQqn−q  (355) The Heisenberg equation of motion for the operator Qq becomes (check!) ¨ Qq + ωion pl 2Qq + g−qnq = 0 (356) We noted above that the ionic charge density fluctuation induced by an ionic displacement u was enion = −nZe∇· u; in Fourier space with Eq. (345) this reads nion q = −iZ r n M qQq (357) Also recall the definition of the dielectric constant ϵ = “external charge" total charge . (358) Now the total charge fluctuation is just electronic + ionic nq + nion q , so nq = nion q (1 −1/ϵ) = −iZ r n M qQq(1 −1/ϵ) (359) = iZ r n M qQq  k2 TF q2 + k2 TF  , (360) where in the last step we have used the Thomas-Fermi approximation for the dielectric function, ϵ = 1 + k2 TF/q2. ω2 phonon = ωion pl 2  1 − k2 TF q2 + k2 TF  . (361) 90 Equation of motion becomes 0 = ¨ Qq + ωion pl 2Qq + 4πiZe2n1/2 qM 1 2  | {z } iZ r n M qQq  k2 TF q2 + k2 TF  = ¨ Qq + ωion pl 2  1 −  k2 TF q2 + k2 TF  Qq. (362) So the phonon oscillation frequencies are ω2 phonon = ωion pl 2  q2 q2 + k2 TF  (363) Since k2 TF = 6πne2/EF, we do in fact recover acoustic phonons iωn = cq as q →0, with speed c = (m/3M)1/2vF. (364) So we have proven that sound propagates in a metal, which you already knew, but it is important to remember that screening makes it happen! 4.2 Polarons 4.2.1 Polaron in 1D. A polaron is formed when an electron has a bound state in the potential created by the ionic lattice distorted around itself due to its own charge. Consider a single electron in 1D, and represent the crystal lattice by “jellium", i.e. a positive continuum. The potential felt by the electron is V = −λdu/dx, where λ is the electron-phonon coupling constant and u(x) is the (smoothed) displacement of the ions from their equilibrium positions. The energy density associated with the deformation is (B/2)(du/dx)2, where B is the bulk modulus. 1. Write down the appropriate Schrödinger equation, assuming that the lattice distortion takes the form u = u0 tanh x/a0. Find the eigenenergies. (Hint: you may find the change of variables x →r = tanh x/a0 helpful.) Solution: Schrödinger equation is −1 2m ∂2ψ ∂x2 − V0 cosh2 x a0 ψ = Eψ, (365) with V0 = λu0/a0. Introduce new variable r = tanh x/a0. Eqn. becomes  ∂ ∂r(1 −r2) ∂ ∂r + 2ma2 0E 1 −r2 + 2ma2 0V0  ψ = 0. (366) Now define 2ma2 0V0 ≡ℓ(ℓ+ 1) and µ = p −2ma2 0E. Left with  ∂ ∂r(1 −r2) ∂ ∂r − µ2 1 −r2 + ℓ(ℓ+ 1)  ψ = 0 (367) 91 and we have Legendre’s equation. Eigenfunctions with appropriate falloff conditions at ∞are P µ ℓand eigenvalues are determined by condition µ −ℓ= n, with n integer n = 0, 1, 2 . . . . (⋆Note maximum n set by n = ℓ= (1/2)  −1 + p 1 + 8ma2 0V0  . See, e.g. Landau-Lifschitz, QM). Substituting back for E one obtains Eel = −1 8ma2 0  −(1 + 2n) + q 1 + 8ma2 0V0 2 , (368) with n = 0, 1, 2 . . . . The ground state energy is therefore Eel = −1 8ma2 0  −1 + q 1 + 8ma2 0V0 2 . (369) For attractive electron-phonon interactions (i.e. λ > 0 with current convention), ground state has negative energy, so electron can always gain energy by “digging a hole" in the lattice. Must also calculate lattice energy: Eph = 1 2 Z B ∂u ∂x 2 = 1 2 Bu2 0 a0 Z dy cosh4 y = 2 3 Bu2 0 a0 . (370) 2. Calculate the total energy of the coupled e−-ph system E = Eph + Ee. Solution: Introduce dimensionless variables s and t by u0 = 3λs/(2B2) and a0 = B2t/(12mλ2). Then the total energy is E = −( √ 1 + st −1)2 + s2/t (371) 3. By minimizing E with respect to u0 and a0, show that the system is always unstable to polaron formation if λ > 0. Calculate the size of the polaron a0. Solution: Energy minimized at s=1/3, t=24. 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 Untitled−1 1 t s Figure 29: Contours for total polaron energy. Transform back ⇒u0 = λ/B, a0 = B/(mλ) . 92 N.B. Physics: Size of polaron is larger for rigid lattices (large B) or weak electron-phonon interaction (small B). Note we have only taken a trial solution for the ionic distortion, not solved the whole problem! Even within this trial space, note our solution breaks down when u0/a0 = mλ3/B2 ≃1, because the harmonic approximation does. 4.3 Bloch resistivity 4.4 Effective e−−e−interaction For superconductivity it will be important to understand why one electron attracts each other effectively by polarizing the lattice. We now have at our disposal the effective screened electron-phonon interaction with coupling constant gkk′ and renormalized phonon frequencies ωλ(q). An interaction between electrons must involve the 1st electron polarizing the lattice, and the second interacting with this polarization, i.e. it must be a 2nd-order process in the coupling constant g. We can represent these processes in terms of Feynman-type pictures: The amplitude of the first process, according to 2nd-order perturbation theory, where the q -q k p k-q p+q k-q p+q k p g g g g Figure 30: Electron-electron interaction due to phonon exchange. Straight lines are electrons, wiggly are phonons, vertices are e-ph coupling constants g. 1st electron emits a phonon, is V 1 q = |gq|2 ϵk −ϵk−q −ωq (372) whereas for the process where it absorbs a phonon we get V 2 q = |gq|2 ϵp −ϵp+q −ωq . (373) Note as usual we have energy conservation at every scattering event depicted in these pro-cesses, i.e. ϵk + ϵp = ϵk−q + ϵp+q (374) 93 and I’ve assumed ωq = ω−q. According to quantum mechanics the two amplitudes contribute additively: Vq = V 1 q + V 2 q == |gq|2ωq (ϵk −ϵk−q)2 −ω2 q . (375) Let’s analyze this expression for the “effective electron-electron interaction" qualitatively. We will primarily be interested in ϵk−ϵk−q ≪ωq, since this situation corresponds to electrons with their momenta both quite close to the Fermi surface, well within a shell of width ωD, a typical phonon energy. • attractive • ind. of k ⇒isotropic • rapidly decreases when ϵk −ϵk−q ∼ωD. • energy space: interaction spread over ωD ⇒in time space it’s retarded • comparable when spread over time to Coulomb interactions: some metals supercon-ducting, some not. Check prefactor of deGennes calculation–should lead to V = 4πe2 q2 + k2 TF ω2 q ω2 q + k2 TF (376) 5 Superconductivity 5.1 Phenomenology Superconductivity was discovered in 1911 in the Leiden laboratory of Kamerlingh Onnes, who noticed that the resistivity of Hg metal vanished abruptly at about 4K. Although phenomenological models with predictive power were developed in the 30’s and 40’s, the mi-croscopic mechanism underlying superconductivity was not discovered until 1957 by Bardeen Cooper and Schrieffer. Superconductors have been studied intensively for their fundamen-tal interest and for the promise of technological applications which would be possible if a material which superconducts at room temperature were discovered. Until 1986, critical tem-peratures (Tc’s) at which resistance disappears were always less than about 23K. In 1986, Bednorz and Mueller published a paper, subsequently recognized with the 1987 Nobel prize, for the discovery of a new class of materials which currently include members with Tc’s of about 135K. 94 Figure 31: Properties of superconductors. Superconducting materials exhibit the following unusual behaviors: 1. Zero resistance. Below a material’s Tc, the DC electrical resistivity ρ is really zero, not just very small. This leads to the possibility of a related effect, 2. Persistent currents. If a current is set up in a superconductor with multiply connected topology, e.g. a torus, it will flow forever without any driving voltage. (In practice experiments have been performed in which persistent currents flow for several years without signs of degrading). 3. Perfect diamagnetism. A superconductor expels a weak magnetic field nearly com-pletely from its interior (screening currents flow to compensate the field within a surface layer of a few 100 or 1000 A, and the field at the sample surface drops to zero over this layer). 4. Energy gap. Most thermodynamic properties of a superconductor are found to vary as e−∆/(kBT), indicating the existence of a gap, or energy interval with no allowed eigenen-ergies, in the energy spectrum. Idea: when there is a gap, only an exponentially small number of particles have enough thermal energy to be promoted to the available unoccu-95 pied states above the gap. In addition, this gap is visible in electromagnetic absorption: send in a photon at low temperatures (strictly speaking, T = 0), and no absorption is possible until the photon energy reaches 2∆, i.e. until the energy required to break a pair is available. 5.2 Electron-phonon interaction Superconductivity is due to an effective attraction between conduction electrons. Since two electrons experience a repulsive Coulomb force, there must be an additional attractive force between two electrons when they are placed in a metallic environment. In classic supercon-ductors, this force is known to arise from the interaction with the ionic system. In previous discussion of a normal metal, the ions were replaced by a homogeneous positive background which enforces charge neutrality in the system. In reality, this medium is polarizable– the number of ions per unit volume can fluctuate in time. In particular, if we imagine a snapshot of a single electron entering a region of the metal, it will create a net positive charge density near itself by attracting the oppositely charged ions. Crucial here is that a typical electron close to the Fermi surface moves with velocity vF = ℏkF/m which is much larger than the velocity of the ions, vI = VFm/M. So by the time (τ ∼2π/ωD ∼10−13 sec) the ions have po-larized themselves, 1st electron is long gone (it’s moved a distance vFτ ∼108cm/s ∼1000 ◦ A, and 2nd electron can happen by to lower its energy with the concentration of positive charge before the ionic fluctuation relaxes away. This gives rise to an effective attraction between the two electrons as shown, which may be large enough to overcome the repulsive Coulomb interaction. Historically, this electron-phonon “pairing" mechanism was suggested by Frölich in 1950, and confirmed by the discovery of the “isotope effect", wherein Tc was found to vary as M −1/2 for materials which were identical chemically but which were made with different isotopes. Figure 32: Effective attraction of two electrons due to “phonon exchange" The simplest model for the total interaction between two electrons in momentum states k and k′, with q ≡k −k′, interacting both by direct Coulomb and electron-phonon forces, 96 is given by V (q, ω) = 4πe2 q2 + k2 s + 4πe2 q2 + k2 s ω2 q ω2 −ω2 q , (377) in the jellium model. Here first term is Coulomb interaction in the presence of a medium with dielectric constant ϵ = 1 + k2 s/q2, and ωq are the phonon frequencies. The screening length k−1 s is 1A or so in a good metal. Second term is interaction due to exchange of phonons, i.e. the mechanism pictured in the figure. Note it is frequency-dependent, reflecting the retarded nature of interaction (see figure), and in particular that the 2nd term is attractive for ω < ωq ∼ωD. Something is not quite right here, however; it looks indeed as though the two terms are of the same order as ω →0; indeed they cancel each other there, and V is seen to be always repulsive. This indicates that the jellium approximation is too simple. We should probably think about a more careful calculation in a real system as producing two equivalent terms, which vary in approximately the same way with kTF and ωq, but with prefactors which are arbitrary. In some materials, then, the second term might “win" at low frequencies, depending on details. The BCS interaction is sometimes referred to as a “residual" attraction, i.e. what is left when the long-range Coulomb interaction is accounted for. 5.3 Cooper problem A great deal was known about the phenomenology of superconductivity in the 1950’s, and it was already suspected that the electron phonon interaction was responsible, but the micro-scopic form of the wave function was unknown. A clue was provided by Leon Cooper, who showed that the noninteracting Fermi sea is unstable towards the addition of a single pair of electrons with attractive interactions. Cooper began by examining the wave function of this pair ψ(r1, r2), which can always be written as a sum over plane waves ψ(r1, r2) = X kq uk(q)eik·r1e−i(k+q)·r2ζ (378) where the uk(q) are expansion coefficients and ζ is the spin part of the wave function, either the singlet | ↑↓−↓↑> / √ 2 or one of the triplet, | ↑↑>, | ↓↓>, | ↑↓+ ↓↑> / √ 2. In fact since we will demand that ψ is the ground state of the two-electron system, we will assume the wave function is realized with zero center of mass momentum of the two electrons, uk(q) = ukδq,0. Here is a quick argument related to the electron-phonon origin of the attractive interaction.88 The electron-phonon interaction is strongest for those electrons with single-particle energies ξk within ωD of the Fermi level. In the scattering process depicted in Fig. 33, momentum is 88Thanks to Kevin McCarthy, who forced me to think about this further 97 k p -q k-q p+q k p g g ωD p+q k-q Figure 33: Electrons scattered by phonon exchange are confined to shell of thickness ωD about Fermi surface. explicitly conserved, i.e. the total momentum k + p = K (379) is the same in the incoming and outgoing parts of the diagram. Now look at Figure 34, and note that if K is not ∼0, the phase space for scattering (attraction) is dramatically reduced. So the system can always lower its energy by creating K = 0 pairs. Henceforth we will make this assumption, as Cooper did. -p K k ωD Figure 34: To get (attractive) scattering with finite cm momentum K, need both electron energies to be within ωD of Fermi level– very little phase space. Then ψ(r1, r2) becomes P k ukeik·(r1−r2). Note that if uk is even in k, the wave function has only terms ∝cos k ·(r1 −r2), whereas if it is odd, only the sin k ·(r1 −r2) will contribute. This is an important distinction, because only in the former case is there an amplitude for the two electrons to live "on top of each other" at the origin. Note further that in order to preserve the proper overall antisymmetry of the wave function, uk even (odd) in k implies the wave function must be spin singlet (triplet). Let us assume further that there is a general 98 two-body interaction between the two electrons (the rest of the Fermi sea is noninteracting in the model!) V (r1, r2), so that the Hamiltonian for the system is H = −∇2 1 2m −∇2 2 2m + V (r1, r2). (380) Inserting the assumed form of ψ into the Schrödinger equation Hψ = Eψ, and Fourier transforming both sides with respect to the relative coordinate, r = r1 −r2, we find (E −2ϵk)uk = X k>kF Vkk′uk′, (381) where ϵk = k2/2m and the Vkk′ = R d3rV (r)ei(k′−k)·r are the matrix elements of the two-body interaction. Recall k, k′ correspond to energies at the Fermi level ϵF in the absence of V . The question was posed by Cooper, is it possible to find an eigenvalue E < 2ϵF, i.e. a bound state of the two electrons? To simplify the problem, Cooper assumed a model form for Vkk′ in which Vkk′ =    −V ξk, ξk′ < ωc 0 otherwise (382) where as usual ξk ≡ϵk −ϵF. The BCS interaction Vkk′ is sometimes referred to as a “resid-ual" attractive interaction, i.e. the attractive, short-distance part left when the long-range Coulomb interaction has been subtracted out, as in (377). The bound state equation becomes uk = V P k′ ′uk′ 2ϵk −E , (383) where the prime on the summation in this context means sum only over k such that ϵF < ϵk < ϵF + ωc. Now uk may be eliminated from the equation by summing both sides P k ′, yielding 1 V = X k ′ 1 2ϵk −E (384) ≃ N0 Z ϵF +ωc ϵF dϵ 1 2ϵ −E = 1 2N0 log 2ϵF + 2ωc −E 2ϵF −E . (385) For a weak interaction N0V ≪1, we expect a solution (if at all) just below the Fermi level, so we treat 2ϵF −E as a small positive quantity, e.g. negligible compared to 2ωc. We then arrive at the pair binding energy ∆Cooper ≡2ϵF −E ≃2ωce−2/N0V . (386) There are several remarks to be made about this result. 1. Note (for your own information–Cooper didn’t know this at the time!) that the depen-dence of the bound state energy on both the interaction V and the cutoff frequency ωc 99 strongly resembles the famous BCS transition temperature dependence, with ωc identi-fied as the phonon frequency ωD, as given in equation (I.1). 2. the dependence on V is that of an essential singularity, i.e. a nonanalytic function of the parameter. Thus we may expect never to arrive at this result at any order in perturbation theory, an unexpected problem which hindered theoretical progress for a long time. 3. The solution found has isotropic or s-symmetry, since it doesn’t depend on the ˆ k on the Fermi surface. (How would an angular dependence arise? Look back over the calculation.) 4. Note the integrand (2ϵk −E)−1 = (2ξk +∆Cooper)−1 peaks at the Fermi level with energy spread ∆Cooper of states involved in the pairing. The weak-coupling (N0V ≪1) solution therefore provides a bit of a posteriori justification for its own existence, since the fact that ∆Cooper ≪ωc implies that the dependence of Vkk′ on energies out near the cutoff and beyond is in fact not terribly important, so the cutoff procedure used was ok. 5. The spread in momentum is therefore roughly ∆Cooper/vF, and the characteristic size of the pair (using Heisenberg’s uncertainty relation) about vF/Tc. This is about 100-1000A in metals, so since there is of order 1 electron/ unit cell in a metal, and if this toy calculation has anything to do with superconductivity, there are certainly many electron pairs overlapping each other in real space in a superconductor. 5.4 Pair condensate & BCS Wavefctn. Obviously one thing is missing from Cooper’s picture: if it is energetically favorable for two electrons in the presence of a noninteracting Fermi sea to pair, i.e. to form a bound state, why not have the other electrons pair too, and lower the energy of the system still further? This is an instability of the normal state, just like magnetism or charge density wave formation, where a ground state of completely different character (and symmetry) than the Fermi liquid is stabilized. The Cooper calculation is a T=0 problem, but we expect that as one lowers the temperature, it will become at some critical temperature Tc energetically favorable for all the electrons to pair. Although this picture is appealing, many things about it are unclear: does the pairing of many other electrons alter the attractive interaction which led to the pairing in the first place? Does the bound state energy per pair change? Do all of the electrons in the Fermi sea participate? And most importantly, how does the critical temperature actually depend on the parameters and can we calculate it? 100 5.5 BCS Model. A consistent theory of superconductivity may be constructed either using the full “effective interaction" or our approximation V (q, ω) to it. However almost all interesting questions can be answered by the even simpler model used by BCS. The essential point is to have an attractive interaction for electrons in a shell near the Fermi surface; retardation is sec-ondary. Therefore BCS proposed starting from a phenomenological Hamiltonian describing free electrons scattering via an effective instantaneous interaction à la Cooper: H = H0 −V X kk′q σσ′ ′c† kσc† −k+qσ′c−k′+qσ′ck′σ, (387) where the prime on the sum indicates that the energies of the states k and k′ must lie in the shell of thickness ωD. Note the interaction term is just the Fourier transform of a completely local 4-Fermi interaction ψ†(r)ψ†(r)ψ(r)ψ(r).89 Recall that in our discussion of the instability of the normal state, we suggested that an infinitesimal pair field could produce a finite amplitude for pairing. That amplitude was the expectation value ⟨c† kσc† −k−σ⟩. We ignore for the moment the problems with number conservation, and ask if we can simplify the Hamiltonian still further with a mean field approximation, again to be justified a posteriori. We proceed along the lines of generalized Hartree-Fock theory, and rewrite the interaction as c† kσc† −k+qσ′c−k′+qσ′ck′σ = [⟨c† kσc† −k+qσ′⟩+ δ(c†c†)] × ×[⟨c−k′+qσ′ck′σ⟩+ δ(cc)], (388) where, e.g. δ(cc) = c−k′+qσ′ck′σ −⟨c−k′+qσ′ck′σ⟩is the fluctuation of this operator about its expectation value. If a mean field description is to be valid, we should be able to neglect terms quadratic in the fluctuations when we expand Eq (20). If we furthermore make the assumption that pairing will take place in a uniform state (zero pair center of mass momen-tum), then we put ⟨c−k′+qσ′ck′σ⟩= ⟨c−k′σ′ck′σ⟩δq,0. The effective Hamiltonian then becomes (check!) H ≃H0 −(∆ X k c† k↑c† −k↓+ h.c.) + ∆⟨c† k↑c† −k↓⟩∗, (389) where ∆= V X k ′⟨c−k↓ck↑⟩, (390) and the primed sum indicates that the sum is taken only over the k-shell of width ωD over which V acts. What BCS (actually Bogoliubov, after BCS) did was then to treat the order 89Note this is not the most general form leading to superconductivity. Pairing in higher angular momentum channels requires a bilocal model Hamiltonian, as we shall see later. 101 parameter ∆as a (complex) number, and calculate expectation values in the approximate Hamiltonian (389), insisting that ∆be determined self-consistently via Eq. (390) at the same time.90 5.5.1 BCS wave function, gauge invariance, and number conservation. What BCS actually did in their original paper is to treat the Hamiltonian (387) variationally. Their ansatz for the ground state of (387) is a trial state with the pairs k ↑, −k ↓occupied with amplitude vk and unoccupied with amplitude uk, such that |uk|2 + |vk|2 = 1: |ψ >= Πk(uk + vkc† k↑c† −k↓)|0 > . (391) This is a variational wave function, so the energy is to be minimized over the space of uk, vk. Alternatively, one can diagonalize the Hartree-Fock (BCS) Hamiltonian directly, together with the self-consistency equation for the order parameter; the two methods turn out to be equivalent. I will follow the latter procedure, but first make a few remarks on the form of the wave function. First, note the explicit violation of particle number conservation: |ψ > is a superposition of states describing 0, 2, 4 , N-particle systems.91 In general a quantum mechanical system with fixed particle number N (like, e.g. a real superconductor!) manifests a global U(1) gauge symmetry, because H is invariant under c† kσ →eiθc† kσ. The state |ψ > is characterized by a set of coefficients {uk, vk}, which becomes {uk, e2iθvk} after the gauge transformation. The two states |ψ > and ψ(ϕ), where ϕ = 2θ, are inequivalent, mutually orthogonal quantum states, since they are not simply related by a multiplicative phase factor.92 Since H is independent of ϕ, however, all states |ψ(ϕ) > are continuously degenerate, i.e. the ground state has a U(1) gauge (phase) symmetry. Any state |ψ(ϕ) > is said to be a broken symmetry state, becaue it is not invariant under a U(1) transformation, i.e. the system has "chosen" a particular ϕ out of the degenerate range 0 < ϕ < 2π. Nevertheless the absolute value of the overall phase of the ground state is not an observable, but its variations δϕ(r, t) in space and time are. It is the rigidity of the phase, i.e. the energy cost of any of these fluctuations, which is responsible for superconductivity. Earlier I mentioned that it was possible to construct a number conserving theory. It is now instructive to see how: states of definite number are formed [Anderson 1958] by making 90If the pairing interaction is momentum dependent the self-consistency or “gap" equation reads ∆k = P′ k′ Vkk′⟨c−k′↓ck′↑⟩, which reduces to (390) if one sets Vkk′ = V . 91What happened to the odd numbers? In mesoscopic superconductors, there are actually differences in the properties of even and odd-number particle systems, but for bulk systems the distinction is irrelevant. 92In the normal state, |ψ > and ψ(ϕ) differ by a global multiplicative phase eiθ, which has no physical consequences, and the ground state is nondegenerate. 102 coherent superpositions of states of definite phase |ψ(N) >= Z 2π 0 dϕeiϕN/2|ψ(ϕ) > . (392) [The integration over ϕ gives zero unless there are in the expansion of the product contained in |ψ > precisely N/2 pair creation terms, each with factor exp iϕ.] Note while this state has maximal uncertainty in the value of the phase, the rigidity of the system to phase fluctuations is retained.93 It is now straightforward to see why BCS theory works. The BCS wave function |ψ > may be expressed as a sum |ψ >= P N aN|ψ(N) > [Convince yourself of this by calculating the aN explicitly!]. IF we can show that the distribution of coefficients aN is sharply peaked about its mean value < N >, then we will get essentially the same answers as working with a state of definite number N =< N >. Using the explicit form (391), it is easy to show ⟨N⟩= ⟨ψ| X kσ nkσ|ψ⟩= 2 X k |vk|2 ; ⟨(N −⟨N⟩)2⟩= X k u2 kv2 k. (393) Now the uk and vk will typically be numbers of order 1, so since the numbers of allowed k-states appearing in the k sums scale with the volume of the system, we have ⟨N⟩∼V , and ⟨(N −⟨N⟩)2⟩∼V . Therefore the width of the distribution of numbers in the BCS state is ⟨(N −⟨N⟩)2⟩1/2/⟨N⟩∼N −1/2. As N →1023 particles, this relative error implied by the number nonconservation in the BCS state becomes negligible. 5.5.2 Is the BCS order parameter general? Before leaving the subject of the phase in this section, it is worthwhile asking again why we decided to pair states with opposite momenta and spin, k ↑and −k ↓. The BCS argument had to do 1) with minimizing the energy of the entire system by giving the Cooper pairs zero center of mass momentum, and 2) insisting on a spin singlet state because the phonon mechanism leads to electron attraction when the electrons are at the same spatial position (because it is retarded in time!), and a spatially symmetric wavefunction with large amplitude at the origin demands an antisymmetric spin part. Can we relax these assumptions at all? The first requirement seems fairly general, but it should be recalled that one can couple to the pair center of mass with an external magnetic field, so that one will create spatially inhomogeneous (finite-q) states with current flow in the presence of a magnetic field. Even in zero external field, it has been proposed that systems with coexisting antiferromagnetic correlations could have pairing with finite antiferromagnetic nesting vector ⃗ Q [Baltensberger and Strässler 1963]. The requirement for singlet pairing can clearly be relaxed if there is 93The phase and number are in fact canonically conjugate variables, [N/2, ϕ] = i, where N = 2i∂/∂ϕ in the ϕ representation. 103 a pairing mechanism which disfavors close approach of the paired particles. This is the case in superfluid 3He, where the hard core repulsion of two 3He atoms supresses Tc for s-wave, singlet pairing and enhances Tc for p-wave, triplet pairing where the amplitude for two particles to be together at the origin is always zero. In general, pairing is possible for some pair mechanism if the single particle energies cor-responding to the states kσ and k′σ′ are degenerate, since in this case the pairing interaction is most attractive. In the BCS case, a guarantee of this degeneracy for k ↑and −k ↓in zero field is provided by Kramer’s theorem, which says these states must be degenerate because they are connected by time reversal symmetry. However, there are other symmetries: in a system with inversion symmetry, parity will provide another type of degeneracy, so k ↑, k ↓, −k ↑and −k ↓are all degenerate and may be paired with one another if allowed by the pair interaction. 5.6 Thermodynamics 5.6.1 Bogoliubov transformation We now return to (389) and discuss the solution by canonical transformation given by Bo-goliubov. After our drastic approximation, we are left with a quadratic Hamiltonian in the c’s, but with c†c† and cc terms in addition to c†c’s. We can diagonalize it easily, however, by introducing the quasiparticle operators γk0 and γk1 by ck↑ = u∗ kγk0 + vkγ† k1 c† −k↓ = −v∗ kγk0 + ukγ† k1. (394) You may check that this transformation is canonical (preserves fermion comm. rels.) if |uk|2 + |vk|2 = 1. Substituting into (389) and using the commutation relations we get HBCS = X k ξk[(|uk|2 −|vk|2)(γ† k0γk0 + γ† k1γk1) + 2|vk|2 +2u∗ kv∗ kγk1γk0 + 2ukvkγ† k1γ† k0] + X k [(∆kukv∗ k + ∆∗ ku∗ kvk)(γ† k0γk0 + γ† k1γk1 −1) +(∆kv∗2 k −∆∗ ku∗2 k )γk1γk0 + (∆∗ kv2 k −∆ku2 k)γ† k0γ† k1 +∆k⟨c† k↑c† −k↓⟩∗, (395) which does not seem to be enormous progress, to say the least. But the game is to eliminate the terms which are not of the form γ†γ, so to be left with a sum of independent number-type terms whose eigenvalues we can write down. The coefficients of the γ†γ† and γγ type terms 104 are seen to vanish if we choose 2ξkukvk + ∆∗ kv2 k −∆ku2 k = 0. (396) This condition and the normalization condition |uk|2 + |vk|2 = 1 are both satisfied by the solutions |uk|2 |vk|2 = 1 2  1 ± ξk Ek  , (397) where I defined the Bogolibov quasiparticle energy Ek = q ξ2 k + |∆k|2. (398) The BCS Hamiltonian has now been diagonalized: HBCS = X k Ek  γ† k0γk0 + γ† k1γk1  + X k  ξk −Ek + ∆k⟨c† k↑c† −k↓⟩∗ . (399) Note the second term is just a constant, which will be important for calculating the ground state energy accurately. The first term, however, just describes a set of free fermion excita-tions above the ground state, with spectrum Ek. 5.6.2 Density of states The Bogoliubov quasiparticle spectrum Ek is easily seen to have a minimum ∆k for a given direction k on the Fermi surface defined by ξk = 0; ∆k therefore, in addition to playing the role of order parameter for the superconducting transition, is also the energy gap in the 1-particle spectrum. To see this explicitly, we can simply do a change of variables in all energy integrals from the normal metal eigenenergies ξk to the quasiparticle energies Ek: N(E)dE = NN(ξ)dξ. (400) If we are interested in the standard case where the gap ∆is much smaller than the energy over which the normal state dos NN(ξ) varies near the Fermi level, we can make the replacement NN(ξ) ≃NN(0) ≡N0, (401) so using the form of Ek from (398) we find N(E) N0 =    E √ E2−∆2 E > ∆ 0 E < ∆ . (402) This function is sketched in Figure 35. 105 N(E) N 0 E Normal SC 1 0 1 ξ ξ| | ξ| | energy Ε ∆ Figure 35: a) Normalized density of states; b) Quasiparticle spectrum. 5.6.3 Critical temperature The critical temperature is defined as the temperature at which the order parameter ∆k vanishes. We can now calculate this with the aid of the diagonalized Hamiltonian. The self-consistency condition is ∆∗ k = V X k′ ′⟨c† k′↑c† −k′↓⟩∗ = V X k′ ′uk′v∗ k′⟨1 −γ† k0γk0 −γ† k1γk1⟩ = V X k ′ ∆∗ k 2Ek (1 −2f(Ek)) . (403) Since 1 −2f(E) = tanh[E/(2T)], the BCS gap equation reads ∆∗ k = V X k′ ′ ∆∗ k′ 2Ek′ tanh Ek′ 2T (404) This equation may now be solved, first for the critical temperature itself, i.e. the temperature at which ∆→0, and then for the normalized order parameter ∆/Tc for any temperature T. It is the ability to eliminate all normal state parameters from the equation in favor of Tc itself which makes the BCS theory powerful. For in practice the parameters ωD, N0, and particularly V are known quite poorly, and the fact that two of them occur in an exponential makes an accurate first principles calculation of Tc nearly impossible. You should always be suspicious of a theory which claims to be able to calculate Tc! On the other hand, Tc is easy to measure, so if it is the only energy scale in the theory, we have a tool with enormous predictive power. First note that at Tc, the gap equation becomes 1 N0V = Z ωD 0 dξk 1 ξk tanh ξk 2Tc . (405) This integral can be approximated carefully, but it is useful to get a sense of what is going on 106 by doing a crude treatment. Note that since Tc ≪ωD generally, most of the integrand weight occurs for ξ > T, so we can estimate the tanh factor by 1. The integral is log divergent, which is why the cutoff ωD is so important. We find 1 N0V0 ≃log ω Tc ⇒ Tc ≃ωDe−1/N0V (406) The more accurate analysis of the integral gives the BCS result Tc = 1.14ωDe−1/N0V (407) We can do the same calculation near Tc, expanding to leading order in the small quantity ∆(T)/T, to find ∆(T)/Tc ≃3.06(1 −T/Tc)1/2. At T = 0 we have 1 N0V = Z ωD 0 dξk 1 Ek = Z ωD ∆ dEN(E)/E (408) = Z ωD ∆ dE 1 √ E2 −∆2 ≃ln(2ωd/∆), (409) so that ∆(0) ≃2ωD exp −1/N0V , or ∆(0)/Tc ≃1.76. The full temperature dependence of ∆(T) is sketched in Figure 36). In the halcyon days of superconductivity theory, comparisons ∆ T Tc (T) 1.76 Tc Figure 36: BCS order parameter as fctn. of T. with the theory had to be compared with a careful table of ∆/Tc painstakingly calculated and compiled by Mühlschlegl. Nowadays the numerical solution requires a few seconds on a PC. It is frequently even easier to use a phenomenological approximate closed form of the gap, which is correct near T = 0 and = Tc: ∆(T) = δscTctanh{ π δsc r a δC CN (Tc T −1)}, (410) where δsc = ∆(0)/Tc = 1.76, a = 2/3, and δC/CN = 1.43 is the normalized specific heat jump.94 This is another of the “universal" ratios which the BCS theory predicted and which helped confirm the theory in classic superconductors. 94Note to evaluate the last quantity, we need only use the calculated temperature dependence of ∆near Tc, and insert into Eq. (47). 107 5.6.4 Specific heat. The gap in the density of states is reflected in all thermodynamic quantities as an activated behavior e−∆/T, at low T, due to the exponentially small number of Bogoliubov quasiparticles with thermal energy sufficient to be excited over the gap ∆at low temperatures T ≪∆. The electronic specific heat is particularly easy to calculate, since the entropy of the BCS superconductor is once again the entropy of a free gas of noninteracting quasiparticles, with modified spectrum Ek. The expression (II.6) then gives the entropy directly, and may be rewritten S = −kB Z ∞ 0 dEN(E){f(E)lnf(E) + [1 −f(E)]ln[1 −f(E)]}, (411) where f(E) is the Fermi function. The constant volume specific heat is just Cel,V = T[dS/dT]V , which after a little algebra may be written Cel,V = 2 T Z dEN(E)(−∂f ∂E )[E2 −1 2T d∆2 dT ]. (412) A sketch of the result of a full numerical evaluation is shown in Figure 31. Note the discon-tinuity at Tc and the very rapid falloff at low temperatures. It is instructive to actually calculate the entropy and specific heat both at low tempera-tures and near Tc. For T ≪∆, f(E) ≃e−E/T and the density of states factor N(E) in the integral cuts off the integration at the lower limit ∆, giving C ≃(N0∆5/2/T 3/2)e−∆/T.95 Note the first term in Eq. (47)is continuous through the transition ∆→0 (and reduces to the normal state specific heat (2π2/3)N0T above Tc), but the second one gives a discontinuity at Tc of (CN −CS)/CN = 1.43, where CS = C(T − c ) and CN = C(T + c ). To evaluate (412), we need the T dependence of the order parameter from a general solution of (404). 5.7 Electrodynamics 5.7.1 Linear response to vector potential The existence of an energy gap is not a sufficient condition for superconductivity (actually, it is not even a necessary one!). Insulators, for example, do not possess the phase rigidity which leads to perfect conductivity and perfect diamagnetism which are the defining characteristics 95To obtain this, try the following: • replace the derivative of Fermi function by exp-E/T • do integral by parts to remove singularity at Delta • expand around Delta E = Delta + delta E • change integration variables from E to delta E Somebody please check my answer! 108 of superconductivity. We can understand both properties by calculating the current response of the system to an applied magnetic or electric field.96 The kinetic energy in the presence of an applied vector potential A is just H0 = 1 2m X σ Z d3rψ† σ(r)[−i∇−(e c)A]2ψσ(r), (413) and the second quantized current density operator is given by j(r) = e 2m{ψ†(r)(−i∇−e cA)ψ(r) + [(i∇−e cA)ψ†(r)]ψ(r)} (414) = jpara −e2 mcψ†(r)ψ(r)A, (415) where jpara(r) = −ie 2m {ψ†(r)∇ψ(r) −(∇ψ†(r))ψ(r)}, (416) or in Fourier space, jpara(q) = e m X kσ kc† k−qσckσ (417) We would like to do a calculation of the linear current response j(q, ω) to the application of an external field A(q, ω) to the system a long time after the perturbation is turned on. Expanding the Hamiltonian to first order in A gives the interaction H′ = Z d3rjpara · A = e mc X kσ k · A(q)c† k−qσcqσ. (418) The expectation value < j > may now be calculated to linear order via the Kubo formula, yielding ⟨j⟩(q, ω) = K(q, ω)A(q, ω) (419) with K(q, ω) = −ne2 mc + ⟨[jpara, jpara]⟩(q, ω). (420) Note the first term in the current jdia(q, ω) ≡−ne2 mc A(q, ω) (421) is purely diagmagnetic, i.e. these currents tend to screen the external field (note sign). The second, paramagnetic term is formally the Fourier transform of the current-current correlation function (correlation function used in the sense of our discussion of the Kubo formula).97 Here are a few remarks on this expression: 96To see this, note that we can choose a gauge where E = −(1/c)∂A/∂t = −iωA/c for a periodic electric field. Then the Fourier component of the current is j(q, ω) = σ(q, ω)E(q, ω) = K(q, ω)A(q, ω), so σ(q, ω) = icK(q, ω)/ω. 97We will see that the first term gives the diamagnetic response of the system, and the second the temperature-dependent paramagnetic response. 109 • Note the simple product structure of (419) in momentum space implies a nonlocal rela-tionship in general between j and A., i.e. j(r) depends on the A(r′) at many points r′ around r.98 • Note also that the electric field in a gauge where the electrostatic potential is set to zero may be written E(q, ω) = −iωA(q, ω), so that the complex conductivity of the system defined by j = σE may be written σ(q, ω) = i ωK(q, ω) (423) • What happens in a normal metal? The paramagnetic second term cancels the diamag-netic response at ω = 0, leaving no real part of K (Im part of σ), i.e. the conductivity is purely dissipative and not inductive at ω, q = 0 in the normal metal. 5.7.2 Meissner Effect. There is a theorem of classical physics proved by Bohr99 which states that the lowest energy configuration of a system of charged particles in an external magnetic field carries zero current. The essential element in the proof of this theorem is the fact that the magnetic forces on the particles are always perpendicular to their velocities. In a quantum mechanical system, the three components of the velocity do not commute in the presence of the field, allowing for a finite current to be created in the ground state. Thus the existence of the Meissner effect in superconductors, wherein magnetic flux is expelled from the interior of a sample below its critical temperature, is a clear proof that superconductivity is a manifestation of quantum mechanics. The typical theorists’ geometry for calculating the penetration of an electromagnetic field into a superconductor is the half-space shown in Figure 37, and compared to schematics of practical experimental setups involving resonant coils and microwave cavities in Figs. 37 a)-c). In the gedanken experiment case, a DC field is applied parallel to the sample surface, and currents and fields are therefore functions only of the coordinate perpendicular to the surface, A = A(z), etc. Since we are interested in an external electromagnetic wave of very long wavelength compared to the sample size, and zero frequency, we need the limit ω = 0, q →0 of the response. We will assume that in this limit K(0, 0) →const, which we 98If we transformed back, we’d get the convolution j(r) = Z d3r′K(r, r′)A(r′) (422) 99See “The development of the quantum-mechanical electron theory of metals: 1928-33." L. Hoddeson and G. Baym, Rev. Mod. Phys., 59, 287 (1987). 110 λ H0 a) δ L λ(2 r) π b) I cavity c) Figure 37: a) Half-space geometry for penetration depth calculation; b) Resonant coil setup; c) Microwave cavity will call −(c/4π)λ−2 for reasons which will become clear! Equation (43) then has the form j = −c 4πλ−2A, (424) This is sometimes called London’s equation, which must be solved in conjunction with Maxwell’s equation ∇× B = −∇2A = 4π c j = −λ−2A, (425) which immediately gives A ∼e−z/λ, and B = B0e−z/λ. The currents evidently screen the fields for distances below the surface greater than about λ. This is precisely the Meiss-ner effect, which therefore follows only from our assumption that K(0, 0) = const. A BCS calculation will now tell us how the “penetration depth" λ depends on temperature. Evaluating the expressions in (420) in full generality is tedious and is usually done with standard many-body methods beyond the scope of this course. However for q = 0 the calculation is simple enough to do without resorting to Green’s functions. First note that the perturbing Hamiltonian H′ may be written in terms of the quasiparticle operators (394) as H′ = (426) −e mc X k k · A(q) h (ukuk+q + vkvk+q)(γ† k+q0γk0 −γ† k+q1γk1) +(vkuk+q −ukvk+q)(γ† k+q0γ† k1 −γk+q1γk0) i → q→0 −e mc X k k · A(0)(γ† k0γk0 −γ† k1γk1) (427) If you compare with the A = 0 Hamiltonian (399), we see that the new excitations of the 111 system are Ek0 →Ek −e mck · A(0) Ek1 →Ek + e mck · A(0) (428) We may similarly insert the quasiparticle operators (394) into the expression for the expec-tation value of the paramagnetic current operator(417): ⟨jpara(q = 0)⟩ = e m X k k⟨(γ† k0γk0 −γ† k1γk1)⟩ = e m X k k (f(Ek0) −f(Ek1)) . (429) We are interested in the linear response A →0, so that when we expand wrt A, the paramagnetic contribution becomes ⟨jpara(q = 0)⟩= 2e2 m2c X k [k · A(0)] k  −∂f ∂Ek  . (430) Combining now with the definition of the response function K and the diamagnetic current (421), and recalling P k →N0 R dξk(dΩ/4π), with N0 = 3n/(4ϵF)100 and R (dΩ/4π)kk = 1/3, we get for the static homogeneous response is therefore K(0, 0) = −ne2 mc {1 − Z dξk(−∂f ∂Ek )}1 (431) ≡ −ns(T)e2 mc 1 (432) where in the last step, I defined the superfluid density to be ns(T) ≡n−nn(T), with normal fluid density nn(T) ≡n Z dξk −∂f ∂Ek  . (433) Note at T = 0, −∂f/∂Ek →0, [Not a delta function, as in the normal state case–do you see why?], while at T = Tc the integral nn →1.101 Correspondingly, the superfluid density as defined varies between n at T = 0 and 0 at Tc. This is the BCS microscopic justification for the rather successful phenomenological two-fluid model of superconductivity: the normal fluid consists of the thermally excited Bogoliubov quasiparticle gas, and the superfluid is the condensate of Cooper pairs.102 Now let’s relate the BCS microscopic result for the static homogeneous response to the penetration depth appearing in the macroscopic electrodynamics calculation above. We find 100Here N0 is single-spin DOS! 101The dimensioness function nn(T/Tc)/n is sometimes called the Yoshida function, Y (T), and is plotted in Fig.38. 102The BCS theory and subsequent extensions also allow one to understand the limitations of the two-fluid picture: for example, when one probes the system at sufficiently high frequencies ω ∼∆, the normal fluid and superfluid fractions are no longer distinct. 112 Y(T) ~ exp-∆/ T 1 1 0 n /n n T/Tc a) T/Tc 1 1 0 1 0 λ T/Tc b) c) n /n s λ(T) Figure 38: a) Yoshida function; b) superfluid density ; c) penetration depth immediately λ(T) = ( mc2 4πns(T)e2) 1/2 . (434) At T = 0, the supercurrent screening excludes the field from all of the sample except a sheath of thickness λ(0). At small but finite temperatures, an exponentially small number of quasiparticles will be excited out of the condensate, depeleting the supercurrent and allowing the field to penetrate further. Both nn(T) and λ(T) −λ(0) may therefore be expected to vary as e−∆/T for T ≪Tc, as may be confirmed by explicit expansion of Eq. (433). [See homework.] Close to Tc, the penetration depth diverges as it must, since in the normal state the field penetrates the sample completely. 5.7.3 Dynamical conductivity. The calculation of the full, frequency dependent conductivity is beyond the scope of this course. If you would like to read an old-fashioned derivation, I refer you to Tinkham’s book. The main point to absorb here is that, as in a semiconductor with a gap, at T = 0 there is no process by which a photon can be absorbed in a superconductor until its energy exceeds 2∆, the binding energy of the pair. This “threshold" for optical absorption is one of the most direct measurements of the gaps of the old superconductors. If one is interested simply in the zero DC resistance state of superconductors, it is frustrating to find this is not discussed often in textbooks. In fact, the argument is somewhat oblique. One notes that the Ferrell-Tinkham-Glover sum rule R dωσ(0, ω) = πne2/(2m) requires that the integral under σ(q = 0, ω) be conserved when one passes through the superconducting transition. Thus the removal of spectal weight below 2∆(found in calculation of BCS conductivity, first by Mattis and Bardeen) implies that the lost spectral weight must be compensated by a delta-function in σ(ω) at ω = 0, i.e. infinite DC conductivity. 113 5.8 GL Free Energy While the BCS weak-coupling theory we looked at the last two weeks is very powerful, and provides at least a qualitatively correct description of most aspects of classic superconduc-tors,103 there is a complementary theory which a) is simpler and more physically transparent, although valid only near the transition; and b) provides exact results under certain circum-stances. This is the Ginzburg-Landau theory [ V.L. Ginzburg and L.D. Landau, Zh. Eksp. Teor. Fiz. 20, 1064 (1950)], which received remarkably little attention in the west until Gor’kov showed it was derivable from the BCS theory. [L.P. Gor’kov, Zh. Eksp. Teor Fiz. 36, 1918 (1959)]. The theory simply postulated the existence of a macrosopic quantum wave function ψ(r) which was equivalent to an order parameter, and proposed that on symmetry grounds alone, the free energy density of a superconductor should be expressible in terms of an expansion in this quantity: fs −fn V = a|ψ|2 + b|ψ|4 + 1 2m∗|(∇+ ie∗ c ⃗ A)ψ|2, (435) where the subscripts n and s refer to the normal and superconducting states, respectively. Let’s see why GL might have been led to make such a “guess". The superconducting-normal transition was empirically known to be second order in zero field, so it was natural to write down a theory analogous to the Landau theory of a ferromagnet, which is an expansion in powers of the magnetization, M. The choice of order parameter for the superconductor corresponding to M for the ferromagnet was not obvious, but a complex scalar field ψ was a natural choice because of the analogy with liquid He, where |ψ|2 is known to represent the superfluid density ns;104 a quantum mechanical density should be a complex wave function squared. The biggest leap of GL was to specify correctly how electromagnetic fields (which had no analog in superfluid He) would couple to the system. They exploited in this case the similarity of the formalism to ordinary quantum mechanics, and coupled the fields in the usual way to “charges" e∗associated with “particles" of mass m∗. Recall for a real charge in a magnetic field, the kinetic energy is: < Ψ|Hkin|Ψ > = −1 2m Z d3rΨ∗(∇+ ie c ⃗ A)2Ψ (436) = 1 2m Z d3r|(∇+ ie c ⃗ A)Ψ|2, (437) after an integration by parts in the second step. GL just replaced e, m with e∗, m∗to obtain the kinetic part of Eq. (435); they expected that e∗and m∗were the elementary electron charge and mass, respectively, but did not assume so. 103In fact one could make a case that the BCS theory is the most quantitatively accurate theory in all of condensed matter physics 104ψ in the He case has the microscopic interpretation as the Bose condensate amplitude. 114 T>T T<Tc c δ f |ψ| Figure 39: Mexican hat potential for superconductor. A system described by this free energy will undergo a second-order phase transition in zero field when a = 0: clearly when a is positive, the system can minimize δf by having ψ = 0 (no superconductivity), whereas if a is negative, δf has a minimum with ψ ̸= 0. The free energy (435) is a functional of the order parameter ψ, meaning the actual value of the order parameter realized in equilibrium satisfies δf/δψ = 0.105 Notice f is independent of the phase ϕ of the order parameter, ψ ≡|ψ|eiϕ, and so the ground state for a < 0 is equivalent to any state ψ related to it by multiplication by a pure phase. This is the U(1) gauge invariance of which we spoke earlier. This symmetry is broken when the system chooses one of the ground states (phases) upon condensation (Fig 1.). For a uniform system in zero field, the total free energy F = R d3rf is minimized when f is, so one finds for the order parameter at the minimum, |ψ|eq = [ −a 2b ]1/2, a < 0 (438) |ψ|eq = 0, a > 0. (439) When a changes sign, a minimum with a nonzero value becomes possible. For a second order transition as one lowers the temperature, we assume that a and b are smooth functions of T near Tc. Since we are only interested in the region near Tc, we take only the leading terms 105Thus you should not be perturbed by the fact that f apparently depends on ψ even for a > 0. The value of f in equilibrium will be fn = f[ψ = 0]. 115 in the Taylor series expansions in this region: a(T, H) = a0(T −Tc) and b = constant. Eqs. (438) and (439) take the form: |ψ(T)|eq = [ a0(Tc−T) 2b ]1/2, T < Tc (440) |ψ(T)|eq = 0, T > Tc. (441) Substituting back into Eqs.435, we find: fs(T) −fn(T) = −a2 0 4b(Tc −T)2, T < Tc (442) fs(T) −fn(T) = 0, T > Tc. (443) The idea now is to calculate various observables, and determine the GL coefficients for a given system. Once they are determined, the theory acquires predictive power due to its extreme simplicity. It should be noted that GL theory is applied to many systems, but it is in classic superconductors that it is most accurate, since the critical region, where deviations from mean field theory are large, is of order 10−4 or less. Near the transition it may be taken to be exact for all practical purposes. This is not the case for the HTSC, where the size of the critical region has been claimed to be as much as 10-20K in some samples. Supercurrents. Let’s now focus our attention on the term in the GL free energy which leads to supercurrents, the kinetic energy part: Fkin = Z d3r 1 2m∗|(∇+ ie∗ c ⃗ A)ψ|2 (444) = Z d3r 1 2m∗[(∇|ψ|)2 + (∇ϕ −e∗/cA)2|ψ|2]. (445) These expressions deserve several remarks. First, note that the free energy is gauge invariant, if we make the transformation ⃗ A →⃗ A+∇Λ, where Λ is any scalar function of position, while at the same time changing ψ →ψ exp(−ie∗Λ/c) . Second, note that in the last step above I have split the kinetic part of f into a term dependent on the gradients of the order parameter magnitude |ψ| and on the gradients of the phase ϕ. Let us use a little intuition to guess what these terms mean. The energy of the superconducting state below Tc is lower than that of the normal state by an amount called the condensation energy.106 From Eq. (435) in zero field this is of order |ψ|2 very close to the transition. To make spatial variations of the magnitude of ψ must cost a significant fraction of the condensation energy in the region of space in which it occurs.107 On the other hand, the zero-field free energy is actually invariant with respect to changes in ϕ, so fluctuations of ϕ alone actually cost no energy. 106We will see below from the Gorkov derivation of GL from BCS that it is of order N(0)∆2. 107We can make an analogy with a ferromagnet, where if we have a domain wall the magnetization must go to zero at the domain boundary, costing lots of surface energy. 116 With this in mind, let’s ask what will happen if we apply a weak magnetic field described by A to the system. Since it is a small perturbation, we don’t expect it to couple to |ψ| but rather to the phase ϕ. The kinetic energy density should then reduce to the second term in Eq. (445), and furthermore we expect that it should reduce to the intuitive two-fluid expression for the kinetic energy due to supercurrents, 1 2mnsv2 s. Recall from the superfluid He analogy, we expect |ψ|2 ≡n∗ s to be a kind of density of superconducting electrons, but that we aren’t certain of the charge or mass of the “particles". So let’s put fkin ≃ 1 2m∗|(∇+ ie∗ c ⃗ A)ψ|2. = Z d3r 1 2m∗(∇ϕ + e∗/cA)2|ψ|2 ≡1 2m∗n∗ sv2 s. (446) Comparing the forms, we find that the superfluid velocity must be ⃗ vs = 1 m∗(∇ϕ + e∗ c ⃗ A). (447) Thus the gradient of the phase is related to the superfluid velocity, but the vector potential also appears to keep the entire formalism gauge-invariant. Meissner effect. The Meissner effect now follows immediately from the two-fluid iden-tifications we have made. The supercurrent density will certainly be just ⃗ js = −e∗n∗ s ⃗ vs = −e∗n∗ s m∗(∇ϕ + e∗ c ⃗ A). (448) Taking the curl of this equation, the phase drops out, and we find the magnetic field: ∇× ⃗ js = −e∗2n∗ s m∗c ⃗ B. (449) Now recall the Maxwell equation ⃗ js = c 4π∇× ⃗ B, (450) which, when combined with (14), gives c 4π∇× ∇× ⃗ B = −c 4π∇2 ⃗ B = −e∗2ns m∗c ⃗ B, (451) or λ2∇2 ⃗ B = ⃗ B, (452) where λ = m∗c2 4πe∗2n∗ s 1/2 . (453) Notice now that if we use what we know about Cooper pairs, this expression reduces to the BCS/London penetration depth. We assume e∗is the charge of the pair, namely e∗= 2e, and similarly m∗= 2m, and |ψ|2 = n∗ s = ns/2 since n∗ s is the density of pairs. 117 Flux quantization. If we look at the flux quantization described in Part 1 of these notes, it is clear from our subsequent discussion of the Meissner effect, that the currents which lead to flux quantization will only flow in a small part of the cross section, a layer of thickness λ. This layer encloses the flux passing through the toroid. Draw a contour C in the interior of the toroid, as shown in Figure 40. Then vs = 0 everywhere on C. It follows js C λ Φ Figure 40: Quantization of flux in a toroid. that 0 = I C d⃗ ℓ· ⃗ vs = 1 m∗ I C d⃗ ℓ· (∇ϕ + e∗ c ⃗ A). (454) The last integral may be evaluated using I C d⃗ ℓ· ∇ϕ = 2π × integer, (455) which follows from the requirement that ψ be single-valued as in quantum mechanics. Having n ̸= 0 requires that one not be able to shrink the contour to a point, i.e. that the sample has a hole as in our superconducting ring. The line integral of the vector potential is e∗ c I C d⃗ ℓ· ⃗ A = e∗ c Z S d⃗ S · ∇× ⃗ A (456) = e∗ c Z S d⃗ S · ⃗ B (457) = e∗ c Φ. (458) Here S is a surface spanning the hole and Φ the flux through the hole. Combining these results, Φ = 2πℏc 2en = nhc 2e = nΦ0, (459) where n is a integer, Φ0 is the flux quantum, and I’ve reinserted the correct factor of ℏin the first step to make the units right. Flux quantization indeed follows from the fact that 118 the current is the result of a phase gradient.108 Derivation from Microscopic Theory. One of the reasons the GL theory did not enjoy much success at first was the fact that it is purely phenomenological, in the sense that the parameters a0, b, m∗are not given within any microscopic framework. The BCS theory is such a framework, and gives values for these coefficients which set the scale for all quantities calculable from the GL free energy. The GL theory is more general, however, since, e.g. for strong coupling superconductors the weak coupling values of the coefficients are simply replaced by different ones of the same order of magnitude, without changing the form of the GL free energy. In consequence, the dependence of observables on temperature, field, etc, will obey the same universal forms. The GL theory was derived from BCS by Gor’kov. The calculation is beyond the scope of this course, but can be found in many texts. 5.9 Type I and Type II superconductivity Now let’s look at the problem of the instability of the normal state to superconductivity in finite magnetic field H. At what magnetic field do we expect superconductivity to be destroyed, for a given T < Tc?109 Well, overall energy is conserved, so the total condensation energy of the system in zero field,fs−fn(T) of the system must be equal to the magnetic field energy R d3rH2/8π the system would have contained at the critical field Hc in the absence of the Meissner effect. For a completely homogeneous system I then have fs(T) −fn(T) = −H2 c /8π, (460) and from Eq. (8) this means that, near Tc, Hc = r 2πa2 0 b (Tc −T). (461) Whether this thermodynamic critical field Hc actually represents the applied field at which flux penetrates the sample depends on geometry. We assumed in the simplified treatment above that the field at the sample surface was the same as the applied field. Clearly for any realistic sample placed in a field, the lines of field will have to follow the contour of the sample if it excludes the field from its interior. This means the value of H at different points on the surface will be different: the homogeneity assumption we made will not quite hold. If we imagine ramping up the applied field from zero, there will inevitably come a point Happl = Happl,c where the field at certain points on the sample surface exceeds the critical 108It is important to note, however, that a phase gradient doesn’t guarantee that a current is flowing. For example, in the interior of the system depicted in Fig. 2, both ∇ϕ and ⃗ A are nonzero in the most convenient gauge, and cancel each other! 109Clearly it will destroy superconductivity since it breaks the degenerace of between the two componenets of a Cooper pair. 119 field, but at other points does not. For applied fields Happl,c < Happl < Hc, part of the sample will then be normal, with local field penetration, and other parts will still exclude field and be superconducting. This is the intermediate state of a type I superconductor. The structure of this state for a real sample is typically a complicated "striped" pattern of superconducting and normal phases. Even for small fields, edges and corners of samples typically go normal because the field lines bunch up there; these are called "demagnetizing effects", and must be accounted for in a quantitatively accurate measurement of, say, the penetration depth. It is important to note that these patterns are completely geometry dependent, and have no intrinsic length scale associated with them. In the 50’s, there were several materials known, however, in which the flux in sufficiently large fields penetrated the sample in a manner which did not appear to be geometry depen-dent. For example, samples of these so-called "type II" superconductors with nearly zero demagnetizing factors (long thin plates placed parallel to the field) also showed flux pene-tration in the superconducting state. The type-II materials exhibit a second-order transition at finite field and the flux B through the sample varies continuously in the superconducting state. Therefore the mixed state must have currents flowing, and yet the Meissner effect is not realized, so that the London equation somehow does not hold. The answer was provided by Abrikosov in 1957 [A.A.A., Sov. Phys. JETP 5, 1174 (1957).] in a paper which Landau apparently held up for several years because he did not believe it. Let us neglect the effects of geometry again, and go back to our theorist’s sample with zero demagnetizing factor. Can we relax any of the assumptions that led to the London equation (448)? Only one is potentially problematic, that n∗ s(r) = |ψ(r)|2 = constant independent of position. Let’s examine–as Abrikosov did–the energy cost of making spatial variations of the order parameter. The free energy in zero field is F = Z d3r[a|ψ|2 + 1 2m∗|∇ψ|2 + b|ψ|4], (462) or 1 −aF = Z d3r[−|ψ|2 + ξ2|∇ψ|2 + b −a|ψ|4], (463) where I’ve put ξ = [ 1 −2m∗a]1/2 = [ 1 −2m∗a0(Tc −T)]1/2. (464) Clearly the length ξ represents some kind of stiffness of the quantitiy |ψ|2, the superfluid density. [Check that it does indeed have dimensions of length!] If ξ, the so-called coherence length, is small, the energy cost of ns varying from place to place will be small. If the order parameter is somehow changed from its homogeneous equilibrium value at one point in space by an external force, ξ specifies the length scale over which it "heals". We can 120 then investigate the possibility that, as the kinetic energy of superfluid flow increases with increasing field, if ξ is small enough it might eventually become favorable to "bend" |ψ|2 instead. In typical type I materials, ξ(T = 0) is of order several hundreds or even thousands of Angstrom, but in heavy fermion superconductors, for example, coherence lengths are of order 50-100A. The smallest coherence lengths are attained in the HTSC, where ξab is of order 12-15A, whereas ξc is only 2-3A. The general problem of minimizing F when ψ depends on position is extremely difficult. However, we are mainly interested in the phase boundary where ψ is small, so life is a bit simpler. Let’s recall our quantum mechanics analogy once more so as to write F in the form: F = Z d3r[a|ψ|2 + b|ψ|4]+ < ψ| ˆ Hkin|ψ >, (465) where ˆ Hkin is the operator −1 2m∗(∇+ ie∗ c ⃗ A)2. (466) Now note 1) sufficiently close to the transition, we may always neglect the 4th-order term, which is much smaller; 2) to minimize F, it suffices to minimize < ˆ Hkin >, since the |ψ|2 term will simply fix the overall normalization. The variational principle of quantum mechanics states that the minimum value of < H > over all possible ψ is achieved when ψ is the ground state (for a given normalization of ψ). So we need only solve the eigenvalue problem ˆ Hkinψj = Ejψj (467) for the lowest eigenvalue, Ej, and corresponding eigenfunction ψj. For the given form of ˆ Hkin, this reduces to the classic quantum mechanics problem of a charged particle moving in an applied magnetic field. The applied field H is essentially the same as the microscopic field B since ψ is so small (at the phase boundary only!). I’ll remind you of the solution, due to to Landau, in order to fix notation. We choose a convenient gauge, A = −Hyˆ x, (468) in which Eq. 467 becomes 1 2m∗[(−i ∂ ∂x + y ℓ2 M )2 −∂2 ∂y2 −∂2 ∂z2]ψj = Ejψj, (469) where ℓM = (c/e∗H)1/2 is the magnetic length. Since the coordinates x and z don’t appear explicitly, we make the ansatz of a plane wave along those directions: ψ = η(y)eikxx+ikzz, (470) yielding 1 2m∗[(kx + y ℓ2 M )2 −∂2 ∂y2 + k2 z]η(y) = Eη(y). (471) 121 But this you will recognize as just the equation for a one-dimensional harmonic oscillator centered at the point y = −kxℓ2 M with an additional additive constant k2 z/2m∗in the energy. Recall the standard harmonic oscillator equation (−1 2m ∂2 ∂x2 + 1 2kx2)Ψ = EΨ, (472) with ground state energy E0 = ω0 2 = 1 2(k/m)1/2, (473) where k is the spring constant, and ground state wavefunction corresponding to the lowest order Hermite polynomial, Ψ0 ≈exp[−(mk/4)1/2x2]. (474) Let’s just take over these results, identifying ˆ Hkinψkx,kz = e∗H 2m∗cψkx,kz. (475) The ground state eigenfunction may then be chosen as ψkx,kz = ψ0(πℓ2 M L2 y )−1/4eikxx+ikzz exp[−(y + kxℓ2 M)2/2ℓ2 M)], (476) where Ly is the size of the sample in the y direction (LxLyLz = V = 1). The wave functions are normalized such that Z d3r|ψkx,kz|2 = ψ2 0 (477) (since I set the volume of the system to 1). The prefactors are chosen such that ψ2 0 represents the average superfluid denstity. One important aspect of the particle in a field problem seen from the above solution is the large degeneracy of the ground state: the energy is independent of the variable kx, for example, corresponding to many possible orbit centers. We have now found the wavefunction which minimizes < ˆ Hkin >. Substituting back into (465), we find using (475) F = [a0(T −Tc) + e∗H 2m∗c] Z d3r|ψ|2 + b Z d3r|ψ|4. (478) When the coefficient of the quadratic term changes sign, we have a transition. The field at which this occurs is called the upper critical field Hc2, Hc2(T) = 2m∗ca0 e∗ (Tc −T). (479) What is the criterion which distinguishes type-I and type II materials? Start in the normal state for T < Tc as shown in Figure 3, and reduce the field. Eventually one crosses either Hc or Hc2 first. Whichever is crossed first determines the nature of the instability in finite field, i.e. whether the sample expels all the field, or allows flux (vortex) penetration (see section C). 122 c1 H H H Meissner phase T H Normal state c2 c Figure 41: Phase boundaries for classic type II superconductor. In the figure, I have displayed the situation where Hc2 is higher, meaning it is encountered first. The criterion for the dividing line between type 1 and type II is simply |dHc dT | = |dHc2 dT | (480) at Tc, or, using the results (461) and (479), (m∗)2c2b π(e∗)2 = 1 2. (481) This criterion is a bit difficult to extract information from in its current form. Let’s define the GL parameter κ to be the ratio of the two fundamental lengths we have identified so far, the penetration depth and the coherence length: κ = λ ξ . (482) Recalling that λ2 = −m∗c2 4πe∗2n∗ s = −m∗c2b 2πe∗2a (483) and ξ2 = − 1 2m∗a. (484) The criterion (58) now becomes κ2 = m∗c2b/2πe∗2a 1/2m∗a = (m∗)2c2b πe∗2 = 1 2. (485) Therefore a material is type I (II) if κ is less than (greater than) 1 √ 2. In type-I superconduc-tors, the coherence length is large compared to the penetration depth, and the system is stiff with respect to changes in the superfluid density. This gives rise to the Meissner effect, where 123 ns is nearly constant over the screened part of the sample. Type-II systems can’t screen out the field close to Hc2 since their stiffness is too small. The screening is incomplete, and the system must decide what pattern of spatial variation and flux penetration is most favorable from an energetic point of view. The result is the vortex lattice first described by Abrikosov. 5.10 Vortex Lattice I commented above on the huge degeneracy of the wave functions (476). In particular, for fixed kz = 0, there are as many ground states as allowed values of the parameter kx. At Hc2 it didn’t matter, since we could use the eigenvalue alone to determine the phase boundary. Below Hc2 the fourth order term becomes important, and the minimization of f is no longer an eigenvalue problem. Let’s make the plausible assumption that if some spatially varying order parameter structure is going to form below Hc2, it will be periodic with period 2π/q, i.e. the system selects some wave vector q for energetic reasons. The x-dependence of the wave functions ψkx,kz=0 = ψ0(πℓ2 M L2 y )−1/4eikxx exp[−(y + kxℓ2 M)2/2ℓ2 M)]. (486) is given through plane wave phases, eikxx. If we choose kx = qnx, with nx =integer, all such functions will be invariant under x →x+2π/q. Not all nx’s are allowed, however: the center of the "orbit", kxℓ2 M should be inside the sample: −Ly/2 < kxℓ2 M = qℓ2 Mnx < Ly/2, (487) Thus nx is restricted such that −Ly 2qℓ2 = −nmax/2 < nx < nmax/2 = Ly 2qℓ2 (488) and the total number of degenerate functions is Ly/(qℓ2 M). Clearly we need to build up a periodic spatially varying structure out of wave functions of this type, with "centers" distributed somehow. What is the criterion which determines this structure? All the wave functions (486) minimize < ˆ Hkin >, and are normalized to R d3r|ψ|2 = |ψ0|2. They are all therefore degenerate at the level of the quadratic GL free energy, F = R d3r|ψ|2+ < ˆ Hkin >. The fourth order term must stabilize some linear combi-nation of them. We therefore write ψ(r) = X nx Cnxψnx(r), (489) with the normalization P nx |Cnx|2 = 1, which enforces R d3r|ψ(r)|2 = ψ2 0. Note this must minimize < ˆ Hkin >. Let’s therefore choose the Cnx and q to minimize the remaining terms in 124 F, R d3r[a|ψ|2+b|ψ|4]. Substituting and using the normalization condition and orthogonality of the different ψkz,kx, we find f = ˜ aψ2 0 + ˜ bψ4 0. (490) with ˜ a(H, T) = a0(T −Tc) + e∗H 2m∗c = e∗ 2m∗c(H −Hc2(T)), (491) ˜ b(H) = sb, (492) and s = (πℓ2 M L2 y )−1 nmax X nx1,nx2,nx3,nx4 C∗ nx1C∗ nx2Cnx3Cnx4 Z dz Z dx eiq(−nx1−nx2+nx3+nx4)x × Z dy e {− 1 2ℓ2 M [(y+qnx1ℓ2 M)2+(y+qnx2ℓ2 M)2+(y+qnx3ℓ2 M)2+(y+qnx4ℓ2 M)2]}. (493) The form of f[ψ0] is now the same as in zero field, so we immediately find that in equilibrium, ψ0|eq = (−˜ a 2˜ b )1/2. (494) and f = −˜ a2 4˜ b . (495) This expression depends on the variational parameters Cnx, q only through the quantity s appearing in ˜ b. Thus if we minimize s, we will minimize f (remember b > 0, so f < 0). The minimization of the complicated expression (493) with constraint P nx |Cnx|2 = 1 is difficult enough that A. Abrikosov made a mistake the first time he did it, and I won’t inflict the full solution on you. To get an idea what it might look like, however, let’s look at a very symmetric linear combination, one where all the Cnx’s are equal: Cn = n−1/2 max . (496) Then ψ(r) ∼ X n einqx exp[−(y + nqℓ2 M)2/2ℓ2 M], (497) which is periodic in x with period 2π/q, ψ(x + 2π/q, y) = ψ(x, y), (498) and periodic in y with period qℓ2 M, up to a phase factor ψ(x, y + qℓ2 M) = e−iqxψ(x, y). (499) 125 in a sufficiently large system. Note if q = √ 2π/ℓM, |ψ|2 forms a square lattice! The area of a unit cell is (2π/q) × (qℓ2 M) = 2πℓ2 M, and the flux through each one is therefore Φcell = 2πℓ2 MH = 2π c e∗H H = hc 2e = Φ0 (500) where I inserted a factor of ℏin the last step. We haven’t performed the minimization explicitly, but this is a characteristic of the solution, that each cell contains just one flux quantum. The picture is crudely depicted in Fig. 42a). Note by symmetry that the currents s=1.18 s=1.16 Figure 42: a) Square vortex lattice; b) triangular vortex lattice. must cancel on the boundaries of the cells. Since ⃗ js = −ens⃗ vs, integrating ∇ϕ + 2e ℏc ⃗ A = 0 around each square must give, as in our previous discussion of flux quantization in a toroid, Φcell = nΦ0, n = integer. (501) Somehow the vortex lattice consists of many such rings. The problem with this idea is that the only way H ∇ϕ · d⃗ ℓ= δϕ around the boundary can be nonzero and the usual argument about single-valuedness of the wave function carried through is if there is a “hole" in the wave function. If there is no hole, or region from which the wave function is excluded, the path can be shrunk to a point, but the value of the integral must remain the same since the integrand is the gradient of a scalar field. This is unphysical because it would imply a finite phase change along an infinitesimal path (and a divergence of the kinetic energy!) The only way out of the paradox is to have the system introduce its own “hole" in itself, i.e. have the amplitude of the order parameter density |ψ|2 go to zero at the center of each cell. Intuitively, the magnetic field will have an accompanying maximum here, since the screening tendency will be minimized. This reduction in order parameter amplitude, magnetic flux bundle, and winding of the phase once by 2π constitute a magnetic “vortex", which I’ll discuss in more detail next time. Assuming Cn = constant, which leads to the square lattice does give a relatively good (small) value for the dimensionless quantity s, which turns out to be 1.18. This was 126 Abrikosov’s claim for the absolute minimum of f[|ψ|2]. But his paper contained a (now famous) numerical error, and it turns out that the actual minimum s = 1.16 is attained for another set of the Cn’s, to wit Cn = n−1/2 max , n = even (502) Cn = in−1/2 max , n = odd. (503) This turns out to be a triangular lattice (Fig. 42b), for which the optimal value of q is found to be q = 31/4π1/2 ℓM , (504) Again the area of the unit cell is 2πℓ2 M, and there is one flux quantum per unit cell. 5.11 Properties of Single Vortex. Lower critical field Hc1 Given that the flux per unit cell is quantized, it is very easy to see that the lattice spacing d is actually of order the coherence length near Hc2. Using (479) and (464) we have Hc2 = c e∗ 1 ξ2 = Φ0 2πξ2. (505) On the other hand, as H is reduced, d must increase. To see this, note that the area of the triangular lattice unit cell is A = √ 3d2/2, and that there is one quantum of flux per cell, A = Φ0/H. Then the lattice constant may be expressed as d = 4π √ 3ξ(Hc2 H )1/2. (506) Since λ ≫ξ is the length scale on which supercurrents and magnetic fields vary, we expect the size of a magnetic vortex to be about λ. This means at Hc2 vortices are strongly overlapping, but as the field is lowered, the vortices separate, according to (500), and may eventually be shown to influence each other only weakly. To find the structure of an isolated vortex is a straightforward but tedious exercise in minimizing the GL free energy, and in fact can only be done numerically in full detail. But let’s exploit the fact that we are interested primarily in strongly type-II systems, and therefore go back to the London equation we solved to find the penetration depth in the half-space geometry for weak fields, allow ns to vary spatially, and look for vortex-like solutions. For example, equation (451) may be written −λ2∇× ∇× ⃗ B = ⃗ B. (507) Let’s integrate this equation over a surface perpendicular to the field ⃗ B = ⃗ B(x, y)ˆ z spanning one cell of the vortex lattice: −λ2 Z ∇× (∇× ⃗ B) · d⃗ S = Z ⃗ B · d⃗ S, (508) −λ24π c I ⃗ js · d⃗ ℓ = Φ0. (509) 127 But we have already argued that ⃗ js · d⃗ ℓshould be zero on the boundary of a cell, so the left side is zero and there is a contradiction. What is wrong? The equation came from assuming a two-fluid hydrodynamics for the superconductor, with a nonzero ns everywhere. We derived it, in fact, from BCS theory, but only for the case where ns was constant. Clearly there must be another term in the equation when a vortex-type solution is present, one which can only contribute over the region where the superfluid density is strongly varying in space, i.e. the coherence length-sized region in the middle of the vortex where the order parameter goes to zero (vortex “core"). Let’s simply add a term which enables us to get the right amount of flux to Eq. (507). In general we should probably assume something like λ2∇× ∇× ⃗ B + ⃗ B = Φ0g(⃗ r)ˆ z (510) where g(r) is some function which is only nonzero in the core. The flux will then come out right if we demand R d3rg(⃗ r) = 1. But let’s simplify things even further, by using the fact that ξ ≪λ: let’s treat the core as having negligible size, which means it is just a line singularity. We therefore put g(⃗ r) = δ(⃗ r). Then the modified London equation with line singularity acting as an inhomogeneous “source term" reads −λ2∇2 ⃗ B + ⃗ B = Φ0δ2(⃗ r)ˆ z (511) −λ21 ρ ∂ ∂ρ(ρ∂Bz ∂ρ ) + Bz = Φ0δ2(⃗ r), (512) where ρ is the radial cylindrical coordinate. Equation (91) has the form of a modified Bessel’s equation with solution: Bz = Φ0 2πλ2K0(ρ λ). (513) The other components of ⃗ B vanish. If you go to Abramowitz & Stegun you can look up the asymptotic limits: Bz = Φ0 2πλ2[log(λ ρ) + 0.116], ξ < ρ ≪λ (514) Bz = Φ0 2πλ2 s πλ 2ρ e−ρ/λ, ρ ≫λ. (515) Note the form (93) is actually the correct asymptotic solution to (91) all the way down to ρ = 0, but the fact that the solution diverges logarithmically merely reflects the fact that we didn’t minimize the free energy properly, and allow the order parameter to relax as it liked within the core. So the domain of validity of the solution is only down to roughly the core size, ρ ≃ξ, as stated. In Figure 43 I show schematically the structure of the magnetic and order parameter profiles in an isolated vortex. The solution may now be inserted into the free energy and the spatial integrals performed, with some interesting results: F = Lz Φ2 0 16π2λ2 log(κ). (516) 128 core ρ λ ξ 0 supercurrents |ψ| Bz Figure 43: Isolated vortex It is easy to get an intuitive feel for what this means, since if we assume the field is uniform and just ask what is the magnetic energy, we get roughly Fv = 1 8π × vortex volume × B2 (517) ≃ 1 8π × (πλ2Lz) × (Φ0/πλ2)2 (518) = Lz Φ2 0 8π2λ2, (519) the same result up to a slowly varying factor. Now the lower critical field Hc1 is determined by requiring the Gibbs free energies of the Meissner phase with no vortex be equal to the Gibbs free energy of the phase with a vortex.110 G differs from F through a term − R BH/4π. In the Meissner state G = F, so we may put F = F + ElineLz −1 4πHc1 Z Bd3r (520) = F + ElineLz −1 4πΦ0Lz, (521) where Eline is the free energy per unit length of the vortex itself. Therefore Hc1 = 4πEline Φ0 (522) is the upper critical field. But the line energy is given precisely by Eq. (95), Eline = Φ2 0 16π2λ2 log(κ), so Hc1(T) = Φ0 4πλ2 log(κ). (523) 110We haven’t talked about the Gibbs vs. Helmholtz free energy, but recall the Gibbs is the appropriate potential to use when the external field H is held fixed, which is the situation we always have, with a generator supplying work to maintain H. 129 5.12 Josephson Effect In 1962 Brian Josephson111, then a 22–year old graduate student, made a remarkable pre-diction that two superconductors separated by a thin insulating barrier should give rise to a spontaneous (zero voltage) DC current, Is = Ic sin ∆ϕ, where ∆ϕ is the difference in phase across the junction. And that if a finite (DC) voltage were applied, an AC current with frequency ω = 2eV/ℏwould flow. Ic is called the Josephson critical current. There is a myth that Brian Josephson did his calculation (1962) and won the Nobel prize (1973) as part of the solution to a homework problem of Phil Anderson’s. The truth is that Anderson was a lecturer on sabbatical at Cambridge in 1961-62, and he gave a series of lectures in which he mentioned the problem of tunneling between two superconductors, which Josephson then promptly solved. The idea was opposed at first by John Bardeen, who felt that pairing could not exist in the barrier region112. Thus much of the early debate centered on the nature of the tunneling process, whereas in fact today we know that the Josephson effect occurs in a variety of situations whenever two superconductors are separated by a “weak link", which can be an insulating region, normal metal, or short, narrow constriction.113 Let’s first consider the last example as the conceptually simplest. The Ginzburg–Landau equation appropriate for this situation may be written ξ2d2f dx2 + f −f 3 = 0 (524) where ξ = q ℏ 2m∗a(T) is the GL coherence length and f(x) ≡Ψ(x)/Ψ∞. Take L ≪ξ, so the deviations of Ψ coming from the bulk value Ψ1 of the first SC is small, and vice versa for the second SC. Changes of Ψ in the constriction occur over a length scale of L, so that the first term is of O((ξ/L)2) ≫f −f 3. So we must solve a Laplace equation for f, (d2f dx2 = 0) 111Phys. Lett. 1, 251 (1962) 112Physics Today, July 2001 113In Je. 2009 I received an email from Brian Josephson correcting this version of the history: Date: Wed, 10 Jun 2009 09:43:54 +0100 From: Brian Josephson bdj10@cam.ac.uk To: pjh@phys.ufl.edu Subject: the Josephson myth Dear Peter, While browsing I came across your mention of the ’myth’ that I discovered the effect because of a problem set by Anderson. Your correction is not completely correct either! It was Pippard, my supervisor, who drew my attention to Giaevar’s tunnelling expts. and his theory, which started me thinking (especially as to how one could get away without using coherence factors). Anderson on the other hand told me of the Cohen/Falicov/Phillps calculation involving a single superconductor when it came our in PRL, which gave me the idea of how to do the two-sc. case. Previously I had got the broken symmetry idea which was crucial from a number of papers including Anderson’s pseudospin model, and also expounded in his lecture course which I went to. Best regards, Brian J. 130 with B.C. f(0) = 1, f(L) = ei∆Φ. The solution will be f =  1 −x L  + x Lei∆ϕ. (525) The solution can be thought of as two terms, the first Ψ1, beginning to “leak" into the constriction from the left, the second Ψ2 leaking into the constriction from the right. The GL expression for the current will be j = e∗ℏ 2m∗i Ψ∗∇Ψ −Ψ∇Ψ∗ −e∗2 m∗cΨ∗ΨA | {z } zero = e∗ℏ 2m∗iΨ2 ∞  1 −x L + x Le−i∆ϕ  −1 L + 1 Lei∆ϕ  −c.c  = e∗ℏΨ2 ∞ m∗L sin ∆ϕ, (526) which means that the current will be I = Ic sin ∆ϕ, (527) Ic = e∗ℏΨ2 ∞ m∗L A, (528) where A is the cross–section. Given that we have two weakly coupled QM systems, it is “not unreasonable" (justified by microscopic theory) to write down coupled Schrödinger equations i∂Ψ1 ∂t = E1Ψ1 + αΨ2 (529) i∂Ψ2 ∂t = E2Ψ2 + αΨ1 (530) where H(i) 0 Ψi = EiΨi and E1 = E2 = E0 if the superconductors are identical. Take |Ψi|2 to be the density of pairs in SCi Ψi = √nieiϕi ⇒˙ Ψi = 1 2√ni ˙ nieiϕi + i√ni ˙ ϕieiϕi ⇒ ˙ n1 2√n1 + i√n1 ˙ ϕ1 = −iE1 √n1 −iα√n2ei(ϕ2−ϕ1) (531) ˙ n2 2√n2 + i√n2 ˙ ϕ2 = −iE2 √n2 −iα√n1ei(ϕ1−ϕ2). (532) If we take the real parts and use ˙ n1 = −˙ n2 we get ˙ n1 2√n1 = α√n2 sin(ϕ2 −ϕ1) − ˙ n1 2√n2 = α√n1 sin(ϕ1 −ϕ2) ⇒ ˙ n1 = 2α√n1n2 sin(ϕ2 −ϕ1), (533) Note I’ve put V=A=1. Then the current is just j = 2e ˙ n1. If we take the imaginary parts 131 1 2 Β x y z Figure 44: 132 we have √n1 ˙ ϕ1 = −E1 √n1 −α√n2 cos(ϕ2 −ϕ1) (534) √n2 ˙ ϕ2 = −E2 √n2 −α√n1 cos(ϕ1 −ϕ2), (535) and by subtracting and assuming n1 ≃n2 (let’s couple 2 identical superconductors at first) we get ˙ ϕ1 −˙ ϕ2 = E2 −E1 = 2e(V1 −V2) (536) where for the second equality we used the fact that the potential difference between the superconductors shifts the pair energies by −2eV . So we see that a finite voltage difference leads to a time changing phase difference ∆ϕ which means an AC current via Eq. (533). Magnetic fields. Now put a flux through the junction where the B field is along the −ˆ y direction and A = −Bxˆ z. The phase of the wave function Ψ must change by ϕ →ϕ −2e c Z dS · A (537) for the theory to be gauge invariant. Notice that ϕ is now space–dependent. So the Josephson equations will read j = 4eα√n1n2 | {z } jc sin  ∆ϕ −2e c Z 2 1 dS · A  , (538) 2e(V1 −V2) = ∂ ∂t  ϕ2 −ϕ1 −2e c Z 2 1 dS · A  , (539) and since Z 2 1 S · A = Z d 0 dz(−Bx) = −Bxd, (540) we will have J = Z L 0 dxj(x) = Z L 0 dxjc sin  ∆Φ −2e c Bxd  = Ljc 2πΦ/Φ0  cos ∆Φ −cos  ∆ϕ + 2πΦ Φ0  , (541) where Φ0 = 2πc 2e . What is the maximum current through the junction for all possible ∆ϕ? We have to calculate dJ d∆Φ = 0 which leads to the relation tan ∆Φ = cot πΦ/Φ0  (542) and with a bit of tedious trigonometry to Jc = Ljc sin πΦ/Φ0  πΦ/Φ0 . (543) This formula produces the Josephson–Fraunhofer interference pattern. DC SQUID (Super-conducting Quantum Interference Device) 133 Φ/Φ J 0 1 2 0 c Figure 45: 1 2 3 4 J J Φ Figure 46: 134 We ignore resistance and capacitance for now. The inside SC thickness is assumed much greater than λ and since vs = 0 we have ∇ϕ = 2A/Φ0. The flux will be Φ = I ds · A = Z 2 1 ds · A + Φ0 2 Z 3 2 ds · ∇ϕ + Z 4 3 ds · A + Φ0 2 Z 1 4 ds · ∇ϕ = Φ0 2 (ϕ3 −ϕ2) + Φ0 2 (ϕ1 −ϕ4) + Z 2 1 ds · A + Z 4 3 ds · A = Φ0 2 (ϕ1 −ϕ2) + Z 2 1 ds · A | {z } ≡−γ12 + Φ0 2 (ϕ3 −ϕ4) + Z 4 3 ds · A | {z } ≡−γ34 Φ = γ43 −γ12. (544) The Josephson current through the SQUID will be J = Jc sin γ12 + sin γ43  = Jc sin γ12 + sin(γ12 + Φ)  . (545) Which means that the current oscillates with the flux. And as a result of that the SQUID can be a sensitive measure of magnetic fields. In practice we include the capacitance and resistance of the device. 6 Topological States of Matter 6.1 Introduction The last 15 years have seen an explosion of interest in a variety of condensed matter models and materials that are studied under the framework of “topological matter”. In reality many of the concepts underlying this field go back much further, to William Shockley, who identified in the 1950s band inversions and edge states in semiconductors that we would now call topo-logical insulators, to the study of quantized vortices and vortex lattices in superconductors and superfluid He in the 60’s, to the discovery and explanation of the Berezinskii-Kosterlitz-Thouless transition in 2D films and Grisha Volovik’s ideas about the topological nature of superfluid 3He in the 1970s, to the discovery and theory of the integer quantum Hall effect in the early 1980s. It is fair to say, however, that it was a series of papers pointing out the similarities of these phenomena, and proposals for how nontrivial topological states could be realized in a variety of condensed matter systems, that created what was perceived as a coherent field unified by a set of ideas. The new concept, as articulated by Kane, Mele, Zhang, Fu and many others, was that a class of phase transitions involve ordered states where no symmetry is spontaneously broken. Instead, the observable distinction between 135 states is based on topology rather than symmetry. This description of the connection be-tween disparate topological systems was recognized by the Nobel prize in 2016 to Kosterlitz, Thouless, and Haldane114 6.2 Berry phase, connection, curvature and all that Invariant integrals characterizing the order in topological phases can often be expressed in terms of quantities identified by Michael Berry as essential to describe the global adiabatic evolution of quantum states. In some sense the adiabatic effects in systems in the ther-modynamic limit, limited to a particular topological sector because the energy barrier to other topological states is too great, can be thought of by analogy to electronic motion in molecules. The Born-Oppenheimer approach utilizes the high speeds of electrons compared to nuclei to suggest that the motion of the nuclei can be treated adiabatically. This works because the excitation energies of the electrons in the nuclear potential are very large and well separated, such that the slow motion of the nuclei never causes transitions between electronic states due to the exisitence of an energy gap. Consider a Hamiltonian H115, that depends on time through a set of parameters R1(t), R2(t)...RD(t). Note that D for now is just the number of independent parameters, and need not have anything to do with the dimension of space. For every ⃗ R, we have a set of orthonormal eigenvectors of H, H( ⃗ R)|n( ⃗ R)⟩= ϵn|n( ⃗ R)⟩. (546) Assume the simplest situation, that H has a spectrum that is always discrete and has no degeneracies. If we prepare the system initially in a particular t = 0 nth state |ψn(t = 0)⟩≡ |n(R(t = 0))⟩, the adiabatic theorem of ordinary quantum mechanics says that the system remains in the nth state at a later time, but can acquire a phase: |ψn(t)⟩= Cn(t)|n( ⃗ R(t))⟩, (547) where Cn is given by Cn(t) = eiγn(t) exp  −i Z t 0 dt′ϵn(t′)  . (548) The 2nd factor reduces to eiϵn(t) if H is time independent, and is thus just the ordinary evolution of an eigenfunction. The first factor is nontrivial and identical to the so-called 114Thouless shared 1/2 the prize, for contributions both to BKT transition and to edge states in quantum Hall effect. Haldane received 1/3 of the prize, for showing that differences in the excitation spectra of quantum spin chains of integer or half-integer spin were topological in nature. See 115Presentation here follows that of Girvin and Yang, Modern Condensed Matter Physics. 136 Berry phase if the path ⃗ R(t) taken in parameter space closes on itself. For now, plug our result into the Schrödinger equation i ∂ ∂t|ψn(t)⟩= H[ ⃗ R(t)]|ψn(t)⟩ (549) and take the inner product with ⟨ψn(t)| to get ∂ ∂tγn(t) = i⟨n( ⃗ R(t))| ∂ ∂t|n( ⃗ R(t))⟩, (550) so γn(t) = i Z t 0 dt′⟨n( ⃗ R(t′))| ∂ ∂t′|n( ⃗ R(t′))⟩ (551) = Z C ⃗ An( ⃗ R) · d ⃗ R, (552) where An( ⃗ R) = i⟨n( ⃗ R)| ∂ ∂⃗ R |n( ⃗ R)⟩ (Berry connection), (553) so we have replaced the time evolution by an integral over a path in parameter space C. In this sense γn depends on geometry (in Hilbert space) rather than actual dynamics. The geometric phase γn was ignored for many years for the same reason that the Aharonov-Bohm phase (a precursive example of the Berry phase phenomenon) was ignored: it is gauge dependent116, and it was assumed that all such phases accumulated in the wave function could be gauged away. Berry showed in 1984117 that if one traverses a closed loop in parameter space, γn was gauge independent and therefore observable (if non-zero), γn = I ⃗ An( ⃗ R) · d ⃗ R (Berry phase). (557) The gauge transformation properties of the Berry connection ⃗ An( ⃗ R) suggests an analogy with the vector potential of electromagnetism, and immediately implies there should exist a gauge independent 2nd rank tensor analogous to F µν. Define the “Berry curvature” as ωn µν( ⃗ R) = ∂RµAν n( ⃗ R) −∂RνAµ n( ⃗ R). (558) Stokes’ theorem then gives γn = Z ∂S ⃗ An( ⃗ R) · d ⃗ R = 1 2 Z S dRµ ∧dRν ωn µν( ⃗ R). (559) 116If we multiply by a ⃗ R-dependent phase, |n( ⃗ R)⟩ → eiϕ( ⃗ R)|n( ⃗ R)⟩ (554) so ⃗ An( ⃗ R) → ⃗ An( ⃗ R) − ∂ ∂⃗ R ϕ( ⃗ R) (555) and γn → γn + ϕ( ⃗ R(t = 0)) −ϕ( ⃗ R(t)), (556) so in general γn is gauge dependent. 117M.V. Berry, Proc. Roy. Soc. London A392, 45 (1984). 137 The integral on the left is taken around the boundary ∂S of a surface S, and the measure dRµ ∧dRν = −dRν ∧dRµ is the oriented surface area element. While all these concepts are familiar from E& M, it is important to remember that ⃗ R is not (necessarily) a position vector in a Cartesian space. IF the parameter space is 3D, we can express the curvature as a magnetic field, ⃗ bn ≡∇⃗ R × ⃗ An = i⟨∇⃗ Rn( ⃗ R)| × |∇⃗ Rn( ⃗ R)⟩ (Berry field or curvature (3D only)), (560) so γn = Z S ⃗ bn · d⃗ S (561) is in this case a kind of “Berry flux”. For example, in the case of the Aharonov-Bohm experiment with two electron beams passing on opposite sides of a solenoid, the parameter ⃗ R0 may be taken as a position vector taken on a path encircling the solenoid, in which case it may be shown that γ = I d ⃗ R0 · ⃗ A( ⃗ R0) = −2π Φ Φ0 , (562) so the Berry flux is directly related to the physical magnetic flux118. Before continuing, we note that although the Berry curvature is gauge invariant, it is not explicitly so since it contains A. There is another form of ωn µν that is often used, particularly in numerical calculations since the numerically generated eigenvector n( ⃗ R) is gauge dependent and may have an ⃗ R-dependent phase factor that varies discontinuously with ⃗ R. It is straightforward to show that ωn µν = i X n′̸=n ⟨n( ⃗ R)| ∂H ∂Rµ|n′( ⃗ R)⟩⟨n′( ⃗ R)| ∂H ∂Rν |n( ⃗ R)⟩ (ϵn( ⃗ R) −ϵn′( ⃗ R))2 (563) . 6.3 Berry curvature of Bloch Bands To describe the effects of topology on transport, we need to define Berry phase, etc. in a basis of Bloch states. Consider an electron moving in a magnetic field described by vector potential A and a periodic lattice potential V (r), H = 1 2m h p + e cA i2 + V (r) (564) Bloch eigenstates obey ψnk(r) = eik·runk(r) where by substituting ψnk into the Schrödinger equation, we see that unk(r) satisfies a “k-dependent Schrödinger equation” h(k)unk = ϵn(k)unk, (565) 118It’s easy to see that if the trajectories wind around the flux n times, a factor n appears on the right hand side. 138 and h(k) ≡ 1 2m h p + k + e cA i2 + V (r). (566) Here I make a few remarks on the magnetic Bloch bands. In the presence of a uniform magnetic field in a 2DEG, the states are the usual Landau levels. The standard treatment is in Landau gauge, which gives a “band index" corresponding to the Landau level, and a momentum ky (see our discussion of the vortex lattice in a superconductor), so it looks like band theory in a solid to a certain extent. But the wave vector ky is of course 1D and gauge-dependent. We can write ordinary Landau level wave functions for a translationally invariant 2DEG in Bloch form by using the fact that there is a huge degeneracy of each level, and taking a linear combination that has the Bloch structure. One might start by wondering if we can even label our eigenstates in the 2DEG with a momentum index, since although B is uniform A is not. The solution is found by examining how A transforms under lattice translations. Since we must have ∇× A(r) = ∇× A(r + a) = B (567) A(r) and A(r+a are related by a gauge transformation, A(r+a) = A(r)+∇fa(r), where the translated Hamiltonian H′ = eiϕa(r)He−iϕa(r), and ϕa(r) = efa/c. Even in the 2DEG we can define a magnetic lattice of unit cell area axay = 2πℓ2, where ℓ2 = c/(eB), as in the Abrikosov problem. Then one can construct Bloch like functions with momentum index k confined to a Brillouin zone associated with the magnetic unit cell119. There are some slight differences with ordinary Bloch states under translation. The magnetic translation operator that commutes with H has the phase factor ˜ Ta = eiϕaTa (568) so the magnetic Bloch functions satisfy generalized periodic boundary conditions ˜ Taunk = eiϕa(r)unk(r + a) = unk(r). (569) In the Landau gauge ϕa = −2πy/ay. The presence of a periodic lattice potential changes things a bit in that Landau subbands are introduced, i.e. the degeneracy of the Landau levels is broken. However the Bloch bands obey the same type of relation under translation, with general ϕa. For a more complete discussion, see e.g. Girvin and Wang, Ch. 12. Using our discussion of the Berry phase, we see that k can be thought of as a kind of parameter (restricted to the 1st Brillouin zone, since we carry along the band index n) of the Hamiltonian h(k). Therefore we can define the corresponding Berry curvature for each band. Recalling that in the case of a D = 3 parameter space we can formulate the theory in terms of a Berry flux, we have immediately for each band n ⃗ bn = i⟨∂kunk| × |∂kunk⟩. (570) It is important to examine the role of symmetry on the Berry curvature, particularly time reversal (T ) and space inversion or parity (P). Under (T ), v →−v, k →−k, and E →E. Under (P) we have v →−v, k →−k, and E →−E. Comparing with (570 ), or see later explicit (585) we have T : ⃗ bn(k) = −⃗ bn(−k) (571) P : ⃗ bn(k) = ⃗ bn(−k). (572) 119For full discussion see Girvin & Wang Ch. 12. 139 Thus, if both T and P are symmetries of the system, ⃗ bn(k) = 0. We now take up the subject of the semiclassical equations of motion of an electron wave packet in a solid. This is one of these lovely problems involving Berry phase where the solution was “well-known", to the extent that it was included in textbooks (cf.Chapter 13 of Ashcroft & Mermin) and used in countless research papers, but later turned out to be incorrect (in some special situations). The Berry flux (570) leads to a correction to the “standard" average velocity of Bloch states in the presence of a uniform electric field. The “standard" derivation for a wave packet characterized by r, k subject to slowly varying fields gives dr dt = ∂ϵk ∂k (573) dk dt = −eE −e c dr dt  × B(r). (574) In quantum mechanics the expectation value of the velocity operator in the absence of external fields is120 ⟨ψnk|v|ψnk⟩= ⟨unk|∂h ∂k|unk⟩= ∂ϵn(k) ∂k . (577) An external field modifies |ψnk⟩and will change ⟨v⟩. For example, if we add a uniform electric field to the Hamiltonian V = eE · r, it will change the wave functions by δ|ψnk⟩ = eE · X n′̸=n,k′ |ψn′k′⟩⟨ψn′k′|r|ψnk⟩ ϵnk −ϵn′k′ (578) = eE · X n′̸=n,k′ |ψn′k′⟩⟨ψn′k′|[r, H]|ψnk⟩ (ϵnk −ϵn′k′)2 (579) = ieE · X n′̸=n,k′ |ψn′k′⟩⟨ψn′k′|v|ψnk⟩ (ϵnk −ϵn′k′)2 , (580) where v = p/m = −i∇/m, and the |ψnk⟩are the eigenfunctions of the E = 0 Hamiltonian121. It is easy to check that the matrix elements ⟨ψn′k′|v|ψnk⟩are proportional to δkk′. This is because the integral over all space R d3r = P R R R d3r, and the integrand in each unit cell is identical except for a factor ei(k−k′)·R, which vanishes when summed over R unless k = k′. 120This follows from Hellman-Feynman theorem: if H[λ] is a Hamiltonian depending on a parameter λ, |n(λ)⟩is an eigenstate, and E(λ) is an eigenvalue, ∂E ∂λ = ⟨n(λ)| ∂H(λ) ∂λ |n(λ)⟩. (575) Note further that for our Hamiltonian, pψnk = eik·r(p + k)unk. so treating k as a parameter and taking A = 0, ⟨ψnk|v|ψnk⟩= ⟨ψnk| p m |ψnk⟩= ⟨unk| (p + k) m |unk⟩= ⟨unk| ∂h(k) ∂k |unk⟩= ∇kϵnk (576) 121Note that we are assuming a set of Bloch levels labelled by n, k and considering only the change in the wave function due to the electric field. There will be additional terms if we treat the magnetic field as a perturbation on the field-free levels. 140 So δ|unk⟩= ieE · X n̸=n′ |un′k⟩⟨un′k| ∂h ∂k|unk⟩ (ϵnk −ϵn′k)2 . (581) Therefore there are several contributions to the average velocity to 1st order in E, ⟨v⟩= ⟨unk|∂h ∂k|unk⟩+ (δ⟨unk|)∂h ∂k|unk⟩+ ⟨unk|∂h ∂kδ|unk⟩≡∂ϵn ∂k + va(n, k), (582) where the anomalous velocity is given by va(n, k) = ie X n′̸=n ⟨unk| ∂h ∂k|un′k⟩ E · ⟨un′k| ∂h ∂k|unk⟩ −c.c. (ϵnk −ϵn′k′)2 . (583) If the first vector in the numerator is a and the second term b·c, then since c.c. turns a ↔c this may be written va(n, k) = ieE × X n′̸=n ⟨unk| ∂h ∂k|un′k⟩× ⟨un′k| ∂h ∂k|unk⟩ (ϵnk −ϵn′k′)2 (584) = eE ×⃗ bn(k), (585) where in the last step we used (563) and specialized to 3D to define ⃗ bn(k). The addition of the anomalous velocity term makes the semiclassical equations symmet-ric122 (compare Eq. (574): dr dt = ∂ϵk ∂k + dk dt  ×⃗ bn(k) (586) dk dt = −eE −e c dr dt  × B(r). (587) 6.4 Integer Quantum Hall Effect We now specialize this rather general formalism to a 2D electron system in Hall geometry, i.e. in crossed electric and magnetic field perpendicular to the 2D system. This is the geometry where the Integer Quantum Hall Effect (IQHE) was discovered by von Klitizing in 1980 (Nobel prize 1985). I will describe the effect and some basic concepts of how it works before returning to the Berry formalism to show how it allows us to understand the IQHE and identify it as a fundamental paradigm of topological matter. This discussion is taken mostly from Chs. 12 and 13 in Girvin and Yang. In Fig. 47, we see a measurement of the IQHE in a MOSFET (Metal-Oxide-Semiconductor Field Effect Transistor), which includes some strange aspects. The Hall resistance exhibits plateaux at which the resistivity value is ρxy = h/ne2, n an integer (recall in 2D resistance 122I didn’t really calculate change in wave function perturbatively due to magnetic field, so you have to go back and complete this step as well. 141 Figure 47: IQHE: Hall resistivity vs. magnetic fields showing plateaux at ρxy = h/ne2. (n is labelled i in the figure). and resistivity are the same, RH = ρxy). The longitudinal resistivity ρxy goes to zero for wide regions of the field B, and then recovers to a finite value only over the transition region between plateaux. The data look extremely clean, and one could imagine that what one was observing was characteristic of a clean, idealized 2DEG123. However there is a problem with this picture. Ignore the background ionic lattice for the moment, and consider a perfectly clean 2DEG in the xy plane exposed to a perpendicular magnetic field. In the laboratory frame, v = 0, E = 0, and B = Bˆ z. If the system is quantum, the eigenvalues are the Landau levels (LL), E = nωc, with cyclotron frequency ωc = eB/mc. In a large system, each LL is massively degenerate with “filling" (degeneracy) ν = Ne/NΦ, with Ne the number of electrons in the system and NΦ = BL2/Φ0 the number of flux quanta ϕ0 = hc/e passing throug the system. Now in a frame moving with respect to the lab with constant velocity v electrons have (average) velocity v, E = −1 cv×B, and B = Bˆ z. We can define the current to be j = −env, where n is the density of electrons, in which case the induced electric field E = B necj × B, perpendicular to both the longitudinal current and the magnetic field. This is the Hall field. The linear response coefficients are defined jµ = σµνEν (588) Eµ = ρµνjµ (589) we get the response tensors ρ = B nec   0 1 −1 0   (590) σ = nec B  0 −1 1 0   (591) 123This was before higher mobility samples showed addiitonal plateaux at rational fractional indices, the fractional quantum Hall effect (FQHE). 142 In other words, ρxy ∝B, i.e. a linear behavior with no plateaux to speak of. The argu-ment depends only on Lorentz invariance, and should not depend on whether the system is quantum or classical. So what is missing? It must be that our assumption of a perfect, translationally invariant 2DEG is wrong. And of course it must be, because real samples always contain impurities at fixed positions that destroy this latter symmetry. When we think about this more deeply, it is even more puzzling. In two dimensions, Anderson localization at B = 0 is supposed to affect all bulk states, such that electrical conduction at T = 0 is not possible at all. Applying a magnetic field should localize states further. So how is it possible to have a longitudinal current and generate a Hall voltage at all? “It turns out" (see Girvin and Yang, chs. 12 and 13) that each Landau level is indeed broadened by disorder, as one would expect, but there is for each level “one" state that remains extended and can therefore carry a current. This turns out to be a quantum-mechanical edge state, the analog of a “skipping orbit" one might have in a semiclassical description, where the electron conducts half a circular orbit in the field, specularly reflects, and then continues skipping around the edge. Such states are confined to the edge of the sample, and as such are not included in the system considered to lead to localization of all states in 2D. In addition, “one can show" they are protected from backscattering, and therefore carry a dissipationless current. This is because they cannot mix with the bulk states, and other edge states are localized a distance the size of the sample away, so matrix elements for decay fall off exponentially with this distance. We will discuss this topological protection more below. For now, let us simply examine the consequences of this picture for the IQHE. As shown in Fig. 48, the extended states are at the center of the Landau level, as we would expect because they are immune to the effects of disorder. As the field is increased, the various LL’s pass through the Fermi energy one at a time. When the Fermi energy sits anywhere except for the center of a LL, it is in a region of localized states, so σxx = 0, and the states are occupied smoothly with no change in Hall voltage (plateau). When the Fermi energy sits in an extended state, a current can flow, so the longitudinal conductivity has a peak and the Hall voltage makes a transition. Thus the observation of the IQHE depends existentially on the existence of disorder. Let’s now return to the semiclassical equations of motion (587) including the anomalous velocity term. Imagine solving first the QM problem of the electron moving in a (not neces-sarily weak) uniform field perpendicular to a lattice. The Landau levels will be labelled by n, k with k = kx, ky. The “periodic” part of the Bloch wavefunctions will be unk(r). Now apply an electric field along x. We will examine the drift velocity of electrons in this electric field along y, which will be, according to (587), proportional to va ∝E ×⃗ bn, the fictitious 143 Figure 48: Broadened Landau Levels with localized bulk states everywhere except for extended edge states at center of level, and corresponding conductivities. Lorentz force. There is no need to consider an explicit magnetic field in the equations any more, since its effect is included in the new |nk⟩(T -breaking is reflected in nonzero ⃗ bn). Note va is along y since E||x and ⃗ bn(k) is along z since the vectors a, c discussed above are in the plane due to the 2 dimensionality of the system. We neglect the ∂ϵ ∂k component along y because when we sum over the BZ it will not yield a net current, and the ordinary E × B term since we take B = 0 since it is accounted for in the eigenstates unk already. Thus the drift velocity is just va, so we may write the current density from a given filled LL n as jn = −e A X k∈BZ va ≡σE (592) where A = a2 is the area of the magnetic unit cell, a = p c/eH is the magnetic length, and σ is the conductivity tensor. So the Hall conductivity is σxy = e2 A X k∈BZ,n bn(k) = e2 (2π)2 X n Z BZ dkxdky bn(k). (593) So quite generally the Hall conductance is proportional to the total Berry curvature of all bands. To understand at least the topological essence of the integer QHE, it remains only to show that 1 2π Z BZ dkxdky bn(k) = Cn, (594) where Cn is a topological integer called the Chern number, to find immediately the usual result that σxy = e2n/(2π) →(e2n)/h, where in the last step I put in ℏto get the correct dimensions. The Hall conductance is quantized in units of e2/h, as discovered by von Klitzing. 144 To prove this, start with the left hand side of (594) and use Stokes: 1 2π Z BZ ∇k × An(k) = 1 2π I An(k) · dk (595) = 1 2π I dk · Z d2r u∗ nk(r)∇kunk(r) (596) where the closed path is the edge of the Brillouin zone. Thus it becomes important to relate the wave function ψnk on both sides of the Brillouin zone, separated by a reciprocal lattice vector G. In general they are identical up to a phase: ψnk(r) = eiθψnk+G (597) In a normal periodic system Bloch says that θ = G · r, in which case θ has no k dependence; there will then be a cancellation of the integrand from opposite sides of the BZ due to the direction of the path, so we will get Cn = 0. But as we have seen above there is an additional phase factor that can occur in a magnetic system under translation. If this phase depends on a nontrivial way on k, we can get a different result. Inserting the phase factor into ψnk via ψnk(r) = e−iθn(k) ˜ ψnk(r), (598) with ˜ ψnk+G = ˜ ψnk, we find for the right hand side of (596) 1 2π I dk · ∇kθn(k) = Cn. (599) Note the left hand side is just the same mathematically as the circulation of the superfluid velocity in a vortex, and is quantized for the same reason. 6.5 Graphene Graphene is a single layer of graphite, identified as an interesting system early on by theorists, but considered unrealizable in practice until it was isolated using “scotch tape” in 2004 by Geim and Novoselov (Nobel Prize 2010). Graphene has a honeycomb lattice structure, which must be described as a Bravais lattice with a basis, hence has two primitive vectors which may be chosen, e.g. as ⃗ a1 = a 2(ˆ x + √ 3ˆ y), ⃗ a2 = a 2(−ˆ x + √ 3ˆ y), (600) where a is the lattice constant of the triangular Bravais lattice. The reciprocal space is spanned by ⃗ b1 = 1 a(ˆ x + ˆ y √ 3), ⃗ b2 = 1 a(−ˆ x + ˆ y √ 3), (601) 145 Figure 49: Graphene direct and reciprocal lattice structure so it is convenient to write any vector in 1st BZ as k = κ1⃗ b1 + κ2⃗ b2, π/a ≤κ1,2 ≤π/a. The tight-binding Hamiltonian is H = −t X ⟨iα,jβ⟩ (c† iαcjβ + h.c.) + X iα mαc† iαciα, (602) where i, j label different unit cells, while α, β = 1, 2 label the basis sites within each cell. For the moment I’ve suppressed the spin indices. mα is an on-site energy which can be different on the two sites within the cell. This might be the type of term you would include in the model Hamiltonian if you wanted to describe a plane of two different atoms on the A and B sublattice, e.g. boron nitride instead of graphene. Let’s consider in particular a site energy + on A and - on B, mα = (−1)αm. Now diagonalize H by transforming to the new basis c† iα = 1 √ N X k c† kαe−ik·Ri. (603) In this simplest tight-binding approximation, note ⟨iα, jβ⟩means hoppings only connect sublattices A(1) and B(2), H = −t X i  c† i1ci2 + c† i1ci−⃗ a1,2 + c† i1ci−⃗ a2,2 + h.c.  + X iα mαc† iαciα = −t X k h c† k1ck2 1 + e−ik·⃗ a1 + e−ik·⃗ a2 + c† k2ck1 1 + eik·⃗ a1 + eik·⃗ a2i + X kα mαc† kαckα. (604) Note that the sum k runs over the hexagonally shaped 1st Brillouin zone of the triangular lattice (Fig. 1). This can be represented in a simple way if we identify the sublattice degree of freedom 1,2 as a pseudospin υ, ↓, and rewrite 146 H = −t 2 X kαβ h σ+ αβc† kαckβ 1 + e−ik·⃗ a1 + e−ik·⃗ a2 + σ− αβc† kαckβ 1 + eik·⃗ a1 + eik·⃗ a2i + +m X kα σz ααc† kαckα, (605) where σ± = σx ± iσy. To compactify even further, introduce a vector d such that dz(k) = m d±(k) = −t 1 + e∓ik·⃗ a1 + e∓ik·⃗ a2 dx(k) = 1 2(d+(k) + d−(k)) = −t (1 + cos k · ⃗ a1 + cos k · ⃗ a2) dy(k) = 1 2i(d+(k) −d−(k)) = −t (sin k · ⃗ a1 + sin k · ⃗ a2) (606) such that the Hamiltonian takes the form H = X k,α,β d · ⃗ σαβc† kαckβ. (607) Now we can find the eigenvalues and eigenstates of H easily by using the properties of the Pauli matrices. Let H = P kαβ Hαβ(k)c† kαckβ with H(k) = d(k) · ⃗ σ and note that124 H2(k) = (d(k) · ⃗ σ)(d(k) · ⃗ σ) = didjσiσj = d(k) · d(k) + ⃗ σ · (d × d) = d(k) · d(k). (608) So eigenvalues are ϵ±(k) = ± p d(k) · d(k), (609) i.e. two bands symmetric around zero energy. If the “mass” m =0, all atoms are alike, dz = 0, and both dx(k) and dy(k) vanish at two distinct wave vectors in the Brillouin zone: k · ⃗ a1 = κ1 = 2π 3 , k · ⃗ a2 = κ2 = −2π 3 , (610) so that kx = (κ1 −κ2)/a and ky = (κ1 + κ2)/( √ 3a). Recalling k = κ1⃗ b1 + κ2⃗ b2 = κ1 1 a(ˆ x + ˆ y √ 3) + κ2 1 a(−ˆ x + ˆ y √ 3) (611) Now we can rewrite dx(k) = −t(1 + cos κ1 + cos κ2) = −t " 1 + 2 cos kxa 2 cos √ 3kya 2 # dy(k) = −t(sin κ1 + sin κ2) = −2t cos kxa 2 sin √ 3kya 2 (612) which vanish when kx = 4π 3a, ky = 0 (613) kx = −4π 3a , ky = 0 (614) 124Recall σi2 = 1, σiσj = iϵijkσk for i ̸= j, and (⃗ a · ⃗ σ)(⃗ b · ⃗ σ) = (⃗ a ·⃗ b)I + i(⃗ a ×⃗ b) · ⃗ σ 147 Figure 50: Graphene Dirac points. Of course these are just two of the 6 k related by 60◦rotations where the bands touch. Now let’s expand the Hamiltonian around the 2 Dirac points we’ve picked. Let kx = k0 x + δkx, ky = δky, and expand the cos and sin’s. Result is dx ± = ± √ 3ta 2 δkx (615) dy ± = √ 3ta 2 δky, (616) where ± just means the two points with kx = ± 4π 3a. Now let v = √ 3ta/2 (dimensions of velocity), and rewrite H+(k) = v(kxσx + kyσy) (617) H−(k) = v(−kxσx + kyσy), (618) and for fun (now; later it will be important), let’s add the mass term which distinguishes the two sublattices: H+(k) = v(kxσx + kyσy) + mσz H−(k) = v(−kxσx + kyσy) + mσz, (619) and again the eigenvalues can be found by constructing H2 ±, so that we have two bands ϵ(k) = ± √ v2k2 + m2, (620) i.e. we have the dispersion of a massive Dirac particle. When m = 0 the Dirac fermion has a linear spectrum; this corresponds to true graphene. 148 6.6 Quantum spin Hall effect and Kane-Mele model Reference: Kane and Mele, PRL 95, 226801 (2005). We will start by considering the Hamiltonian (619) as a model of something, and add some new physics to make it more interesting. One question is what types of terms can induce a gap in the graphene Hamiltonian. We have already seen that splitting the sublattice “masses” can do so, with a term mσz. There is another type of term that can do so, that was discovered by Haldane. Interesting effects, in particular a transition to a topological state, can arise from the competition between these two types of masses. Before we go further, let’s discuss the effects of parity (P : k →−k) and time reversal (T : k →−k, S →−S). Recall σi in (617) acts on the sublattice degree of freedom. Let’s call the two symmetry-distinct band touching wave vectors k± = ±(4π/(3a), 0), and introduce a new pseudospin variable (sometimes called “valley degeneracy”) ⃗ τ which acts on the k± degree of freedom. Then instead of writing H±(k) let’s just write H(k) with H(k) = v(kxτ zσx + kyσy) + mσz, (621) where I’ve included our sublattice mass modulation term. Let’s examine the effects of parity and time reversal on this H. First of all, parity takes sublattice 1 into 2 and vice versa, so that P : σz →−σz. (622) (Under P, σx →σx and σy →−σy, so d(k) · ⃗ σ (graphene) is invariant [check!]). But P also interchanges k± so P : τ z →−τ z. (623) So if we ever had a term like τ zσz it would be invariant under P, but violate time reversal (T ). To see this, note that since all momenta change sign under time reversal, T : τ z →−τ z. However time reversal does not affect the sublattice degree of freedom, T : σz →σz, so this term breaks T . A term mHτ zσz will therefore gap the Dirac points but H will break different symmetries than the H with sublattice mass m; this is the model invented by Haldane (Haldane Phys. Rev. Lett. 61, 2015 (1988)), as shown in Fig. 51, which consists of imaginary second-neighbor hoppings that flip sign upon reversal. Haldane showed that this model has a nonzero Chern number and therefore a nonzero quantum Hall effect without the need for any external field. Kane and Melé asked if one could generate a term which is invariant under both P and T . This is essentially two copies of the Haldane model, where the Haldane mass (hoppings) has one sign for spin up, and the other for spin down.This is a simplified version of the spin-orbit interaction in these materials. Let’s imagine adding a term to the nearest-neighbor 149 Figure 51: Sense of hoppings leading to positive hopping in Haldane Hamiltonian or spin-up Kane-Melé spin-orbit Hamiltonian. tight-binding Hamiltonian we have so far involving imaginary spin-dependent hoppings on the next-nearest neighbor bonds, as shown in Fig. 51. Remember that until now we have suppressed the true electron spin, but we will now bring it back in the form of an interaction σzτ zSz, (624) where Sz is the z-component of the electron spin. Since T : Sz →−Sz, and S doesn’t care about parity, the product of all three is invariant under T , P. Thus HSO = −it2 X ⟨⟨i,j⟩⟩αβ νij(Sz)αβc† iαcjβ + h.c., (625) where νij = −νji = ±1, depending on the orientation of the two nearest neighbor bonds d1 and d2 the electron traverses in going from site j to i. νji = +1(−1) if the electron makes a left (right) turn to get to the next nearest neighbor site (see arrows in Fig. 51). (Check that this tight-binding model can be put into a form H(k) = mHτ zSz for each Sz.) Now, including spin dependence, νij(Sz) can be written in a coordinate independent representation as d1 × d2 · S. Just as with the nearest-neighbor hopping for the tb graphene model, we can Fourier transform (625) and calculate corrections to d(k), and then linearize around the Dirac points. This gives a term in our compact notation (check!!!) HSO(k) = ∆SOσzτ zSz, (626) which is invariant under both T , P as discussed above. ∆SO turns out to be 3 √ 3t2.125 125Remember the origin of spin-orbit coupling in atoms: it’s a relativistic effect which can be understood crudely by boosting to a moving electron’s frame, and saying there is a magnetic field B due to the moving charged nucleus (or here, the ionic lattice), equal to B = (v × E)/c = (p × E)/(mc). B acts on the electron spin, thus coupling spin and momentum. 150 Remarks: • Notice that there is no term in the Hamiltonian at present which actually couples up and down real electron spin. Kane and Mele showed however that provided the gap does not close, the system is robust against the addition of terms Sx, Sy. • Since T and P are preserved, we expect no total Chern number and therefore no in-teger quantum Hall effect in this system. The spin-up and spin-down currents cancel. However there is a so-called “spin Hall effect”, as we will see below. 6.6.1 Sz = 1 Therefore let’s consider one spin sector at a time. For spins up, Sz = 1, we have H(k) = v(kxτ zσx + kyσy) + ∆SOσzτ z, (627) or for each “valley” (Dirac point) separately: H+(k) = v(kxσx + kyσy) + ∆SOσz (628) H−(k) = v(−kxσx + kyσy) −∆SOσz. (629) This is the same situation we analyzed for the sublattice mass problem, so we know ϵ(k) = ± q v2k2 + ∆2 SO. (630) Now let’s reintroduce the mass, which cared about sublattice but not about valley pseu-dospin. Therefore if we have both m and ∆SO we get H+(k) = v(kxσx + kyσy) + (m + ∆SO)σz (631) H−(k) = v(−kxσx + kyσy) + (m −∆SO)σz. (632) Now we can consider two extreme possibilities. First, imagine m >> ∆SO. As discussed above, this opens up a gap m at the Dirac points, so we have an insulator simply because we put a different potential on the 1 and 2 sublattices. This is called an “atomic” or “trivial” insulator. Now increase ∆SO relative to m. Nothing happens in the H+ block, but in the H−block the gap closes and reopens again when ∆SO > m. The ∆SO > m insulator is separated from the atomic insulator by a gap-closing phase transition. Therefore (see below) the distinction between the two is topological. Let’s investigate what this really means by looking at the response of the system to an applied field. We are really interested in the T invariant case without field, but it will help us to classify the states and then we will take the field strength to zero. To include the field we will replace k everywhere by −i∇+ e cA, and choose gauge A = xBˆ y for a field Bˆ z. This gives for valleys ±: 151 H+ = −iv ∂ ∂xσx +  −i ∂ ∂y + x ℓ2 B  σy + (m + ∆SO)σz (633) H−= iv ∂ ∂xσx +  −i ∂ ∂y + x ℓ2 B  σy + (m −∆SO)σz, (634) where ℓB = p c/eB is magnetic length. Again we will use the trick of squaring H± in order to find the eigenvalues. For example H2 + =  −iv ∂ ∂xσx + v  −i ∂ ∂y + x ℓ2 B  σy + (m + ∆SO)σz  ·  −iv ∂ ∂xσx + v  −i ∂ ∂y + x ℓ2 B  σy + (m + ∆SO)σz  = −v2 ∂2 ∂x2 + v2  i ∂ ∂y + x ℓ2 B 2 + (m + ∆SO)2 + cross −terms. and for the cross-terms we use the anticommutation of the Pauli matrices σiσj = σjσi for i ̸= j. Check then that these just reduce to −iv2iσz(1/ℓ2 B)[(∂/∂x), x] = (v2/ℓ2 B)σz, so H2 + = −v2 ∂2 ∂x2 + v2  i ∂ ∂y + x ℓ2 B 2 + (m + ∆SO)2 + ω2 Bσz, (635) where ωB = v/ℓB is the Dirac cyclotron frequency, and ℓB = p c/(eB) is the magnetic length. Note the first two terms have the form of H (not H2) for a regular 2DEG with “mass” 1/(2v2). The corresponding “cyclotron frequency” is ωeff c = eB mc ≡2v2 ℓ2 B = 2ω2 B. (636) Thus the eigenvalues of H2 + can be read off 2ω2 B(n + 1/2) + (m + ∆SO)2 + ω2 Bσz ; n = 0, 1, 2..., (637) and the spectrum of H+ itself is ϵn+ = ± q 2ω2 Bn + (m + ∆SO)2 ; n = 1, 2, .... (638) ϵ0+ = −(m + ∆SO). (639) Now one can go back and do the same thing for H−(still for real spin up!), and find H2 − = 2ω2 B(n + 1/2) + (m −∆SO)2 −ω2 Bσz (640) ϵn− = ± q 2ω2 Bn + (m −∆SO)2 ; n = 1, 2, .... (641) ϵ0− = m −∆SO. (642) 152 Eigenvalues −ϵ0+ and −ϵ0−do not occur although they are in principle square roots of eigenvalues of H2 ±. 126. Now when m > ∆SO, we have the same number of Landau levels above and below zero energy. The Hall conductivity with ϵF = 0 is therefore σxy = 0. However, once ∆SO > m, both ϵ0+ and ϵ0−are below ϵ = 0, so there is one extra filled Landau level in this case so that the Hall conductivity becomes σxy = e2/h.127 Now notice that the ordering of levels or the sign of their energies did not depend on the strength of the applied field B. Thus we can take B →0 and will be left with an insulator (quantum Hall insulator) which displays σxy = +e2/h for Sz = 1, i.e. spins ↑. 6.6.2 Sz = −1 Now follow exactly the same steps for Sz = −1. For completeness I’ll write it out explicitly, but basically only signs of ∆SO terms change: H(k) = v(kxτ zσx + kyσy) + mσz −∆SOσzτ z, (643) H+(k) = v(kxσx + kyσy) + (m −∆SO)σz (644) H−(k) = v(−kxσx + kyσy) + (m + ∆SO)σz. (645) so the Landau level structure is ϵn+ = ± q 2ω2 Bn + (m −∆SO)2 ; n = 1, 2, .... (646) ϵ0+ = −(m −∆SO) (647) ϵn− = ± q 2ω2 Bn + (m + ∆SO)2 ; n = 1, 2, .... (648) ϵ0− = m + ∆SO. (649) so when ∆SO > m, there is one more Landau level above ϵ = 0, so by the same argument the Hall conductivity should become σxy = −e2/h, again even for B →0. Thus the picture 126This is because the zero mode (eigenvalue) of the graphene Hamiltonian for a given valley is unique. For example for H+, the zero mode and is an eigenstate of σz with eigenvalue σz = −1 (A. H. Castro Neto, F. Guinea, N. M. R. Peres, K. S. Novoselov, and A. K. Geim Rev. Mod. Phys. 81, 109 (2009), pp. 126-127. I thank Ammar Jahin for discussing these aspects with me.) Thus adding the mass m + ∆SO shifts the zero mode down, hence ϵ0+ < 0. Now for the opposite valley τz = −1, the eigenvalues are shifted by (m −∆SO)σz, which is indeed up if m > ∆SO, since this zero mode is an eigenstate with σz = +1, but down if m < ∆SO. Only for m < ∆SO are there LL’s, that each contribute e2/2h to total σxy = e2/h for Haldane insulator, representing a spontaneous IQHE in zero applied field. This is the same as the Kane-Mele model for the Sz = 1 sector, but σxy is cancelled by −e2/h from Sz = −1 sector. 127In the integer quantum Hall effect we expect σxy = ne2/h for a filled Landau level. For graphene ϵF at the Dirac point, σxy = 0, due to two doubly degenerate levels corresponding to n = 0. 153 which emerges is that of an insulator which differs from the trivial one in that spins have finite, but opposite Hall conductivities even in zero field, due to the spin-orbit interaction. Thus spins up and down will accumulate on opposite sides of a Hall bar carrying a longitudinal electric current due to an applied electric field. This is not unique to topological insulators, of course, but is a property of semiconductors with spin-orbit coupling. Figure 52: Spin-carrying edge states in a topological insulator. From Qi and Zhang, Physics Today 63 (1), 33–38 (2010). 6.7 Edge states In fact we haven’t yet shown that the spin current is carried by edge states, as in Fig. 52. This can be done by a gauge argument analogous to that given by Laughlin for the integer QHE, or by explicit solution for a given geometry. The Schrödinger equation can be solved numerically in finite geometry by imposing open transverse boundary conditions. In this case Kane and Mele showed the solution (Fig. 5), which exhibits the bulk gapped Dirac like states and two characteristic states which cross at the Dirac point and carry the current; direct examination of the eigenfunctions shows that they are indeed edge states. We can show without involved numerics that such edge states exist if we put in the “edge” by a bit of sleight of hand. Consider for example the H−block for spin up Sz = 1. Assume the sample has an edge at y = 0, and the sample exists for y < 0, and y > 0 is vacuum. There will be some spatial variation along the y direction giving the edge state wave function, but we can assume translational invariance along x and take kx = 0. Then the Hamiltonian is H−(y) = −iv ∂ ∂yσy + (m −∆SO)σz ≡−iv ∂ ∂yσy + ˜ m(y)σz, (650) where I’m now considering a y-dependent potential given by ˜ m(y), which I will insist change sign at the edge, such that ˜ m is < 0 for y < 0, i.e. in the sample, ˜ m > 0 for y > 0. We’re looking for a zero-energy edge state wave function. Make the ansatz for the solution to the 154 Figure 53: Energy bands for a strip of graphene with SO coupling. The bands crossing the gap are spin filtered edge states. From Kane and Mele, PRL 2005. Schrödinger equation ψ(y) = iσyef(y)ϕ, (651) where ϕ is a 2-component spinor field. Plugging in, we get  v d f dy + ˜ m(y)σx  ϕ = 0, (652) which has the formal solution f(y) = −1 v Z y 0 dy′ ˜ m(y′) ; σxϕ = ϕ, (653) i.e. ϕ is an eigenstate of σx with e-value 1. Note that the effect of iσy = exp i π 2σy is to rotate by π around the y-axis. So the total solution is ψ(y) = exp − 1 v Z y 0 dy′ ˜ m(y′)  |σx = −1⟩. (654) One can be more explicit by assuming an “edge” like ˜ m(y) = ˜ m0 tanh(y/y0), in which case one finds ψ ∝exp −(y ˜ m0/v) log cosh(y/y0), which is a state of width v/ ˜ m0. Remarks: 1. The fact that the state is an eigenstate of σx apparently reflects the fact that it mixes the two sublattices by hopping along the boundary. 2. At finite kx, the same state has energy ϵ(kx) = −vkx, so that v(kx) = ∂ϵ(kx)/∂kx = −v. 3. For Sz = −1 we would take ˜ m →−˜ m in the large ∆SO limit to find an edge state with velocity in the opposite direction. 4. I chose arbitrarily one Dirac point H−. Could have chosen H+ as well. There should be two edge state solutions there as well, one for each spin. 155 6.8 Topological character of new insulating phase We’ve been dancing around the obvious question, what’s actually topological about topolog-ical insulators? This question is important because to qualify as a new state of matter they need to have something fundamentally different, and the claim is that such new insulating states do NOT (necessarily) break any symmetry, as do magnets, superconductors, crystals, and other phases we regard as distinct. The claim is that there is a topological invariant which characterizes such a phase, reflecting a finite energy gap towards deformation of the state into a new phase of trivial topological invariant, i.e. the state is robust against small perturbations. To see this explicitly, let’s again take Sz = 1 and considerH±(k) H+(k) = v(kxσx + kyσy) + (m + ∆SO)σz (655) H−(k) = v(−kxσx + kyσy) + (m −∆SO)σz, (656) recalling that we can write the Hamiltonian in terms of a d-vector, H±(k) = d±(k) · ⃗ σ, d±(k) = (±vkx, vky, m ± ∆SO), (657) and define ˆ d± = d±/|d±|. ˆ d(k) defines a mapping of 2D momentum space kx, ky to a unit sphere, for given v, m, ∆SO. This mapping can be assigned a topological index (Chern number) n = 1 4π X α Z d2k  ∂kx ˆ dα × ∂ky ˆ dα  . · ˆ dα, (658) Note this represents an integral of the Berry curvature, or the number of times that the mapping taking the 2D k-space128 into the unit sphere represented by ˆ d. n is the number of times the map wraps around the sphere as a function of k, and for smooth ˆ d it may be shown that n is always an integer. In Fig. 54 I sketch ˆ d± for the simple example given in Eq. 657. For m > ∆SO the dz component is a positive constant, i.e. dz > 0. Since ∂x ˆ dα × ∂y ˆ dα points along ˆ z, the Chern number is just the winding around a contour in the kx, ky plane weighted by sgn dz which is +1 for m > ∆SO so since the windings are ±1 for valleys + and −, the sum is n = 1 2 −1 2 = 0. For the case m < ∆SO, however, the ˆ dz component is reversed for ˆ d−, which means the winding numbers add, n = 1 2(+1) + (−1 2)(−1) = 1. This index therefore distinguishes a trivial insulator from a topological one. 128In principle the mapping is performed from the torus (1st Brillouin zone) to the unit sphere. However for the graphene case if the gap is small, the Berry curvature is concentrated in the neighborhood of the 2 Dirac points, so the edges of the zone don’t contribute. 156 Figure 54: d-vector for Haldane or K-M Hamiltonian. ˆ d± is plotted over a contour surrounding kx, ky = 0 and at kx, ky = 0. a) Case m > ∆SO showing different windings of d in plane for ± valleys. Winding numbers cancel for 2 valleys. b) Topologically nontrivial case ∆SO > m. dz has negative sign for −valley, leading to summing of two winding numbers to 1. 6.9 Kitaev model for 1D topological SC The revival of interest in topological superconductivity may be due in part to an influential paper written by Kitaev which introduced a model for a 1D p-wave superconducting chain, HKitaev = − X i t(c† ici+1) −µc† ici + ∆(cici+1 + h.c.), (659) where i labels the sites on a chain, t is the NN hopping, µ the chemical potential and ∆ the p wave pairing amplitude for the spinless fermions which move on the chain. Note that the Fourier transform of ∆cici+1 is i∆sin k ckc−k, so that the gap in momentum space is indeed of odd parity, which we will call p wave. An odd parity state of this type satisfies the Pauli principle with its orbital sign change under particle exchange without the need for spin degrees of freedom. Such a model is otherwise unphysical in many different ways, but has the virtue that one can exhibit very simply the existence of two distinct topological sectors. To see this, consider Fig. 56a): there are clearly two different situations with respect to the 157 Figure 55: Kitaev 1D p-wave superconducting chain chemical potential. It can lie inside the band, as in the standard BCS case; with the current convention for the Hamiltonian, this means |µ| < t. Alternatively, it can lie outside the band, |µ| > t; in this situation, BCS theory would say there is no superconductor. However, within the model, we can simply assert that the range in energy over which states are paired is very large, so that one can pair states away from the Fermi surface. It is interesting to examine the bulk theory first to see if there are differences between the two sectors. The ground state of the Hamiltonian is of course just the BCS wave function, which we write as Πk(1 + ϕ(k)c−kck)|0⟩, (660) where ϕ(k) ≡vk/uk is the form factor for the occupied pair, or Cooper wave function. In real space (1D), it is a function of the relative coordinate, ϕ(r) ∼    e−r/ζ |µ| > t strong pairing const. |µ| < t weak pairing We see that in the case labelled “strong pairing”, the pair size decays exponentially over a Figure 56: (a) Definition of strong and weak pairing sectors according to value of µ. (b) Mapping of ˆ d vector onto unit sphere under sweep of k from 0 to π. length scale ζ, ... (?), exactly as it does in the BCS theory when the interaction is made comparable or large compared to the Fermi energy, resulting in a very different theory cor-responding to the Bose-Einstein pairing of molecules. In 1D, the usual BCS case “weak 158 pairing” has an infinitely long range pair size. This distinction does not guarantee a topolog-ical difference between the two sectors by itself, but this can be shown as well. If we write the Hamiltonian (659) in terms of a Nambu representation129 HKitaev = 1 2 X k ⃗ c† kHk⃗ ck, Hk =  ϵk ∆k ∆∗ k −ϵk  , (661) with ⃗ ck = (c−k, c† k), ϵk = −t cos k −µ, ∆k = −i∆0 sin k, the matrix Hk may be written as a sum over Pauli matrices, Hk = h(k) · ⃗ τ. (662) It’s important that h(k) obey the symmetry hx,y(k) = −hx,y(−k) and hz(k) = hz(−k), which it does if the gap has odd parity. Now define a new unit vector ˆ h(k) ≡h(k)/|h(k)|. We can ask what the trajectory is on the unit sphere as k sweeps over the interval 0, π in the Brillouin zone. At both 0 and π, the gap vanishes and the direction of ˆ h is determined by the sign of the kinetic energy ϵk, which we’ll call s0 and sπ, respectively. We see that in the strong pairing sector we have either µ > t ⇒s0 = −1, sπ = −1 or µ < −t, such that s0 = 1, sπ = 1. Thus in the strong pairing case the mapping to the unit sphere is trivial: the vector ˆ h(k) follows a trajectory as k runs from 0 to π on the unit sphere that begins and ends at the same point. On the other hand if |µ| < t, s0 = −sπ, so the mapping goes from one pole to the other. These are topologically distinct trajectories distinguished by the Z2 invariant ν = s0sπ, which evidently can be 1 (trivial) or -1 (topological). We now follow Kitaev and show that the finite chain exhibits Majorana end-chain ex-citations. Majorana fermions130 can be thought of as half a Dirac Fermion. Each spin-less fermion in our original Hamiltonian (659) may be written as a sum of Majoranas, ci = (1/2)(βBi + iβAi), yielding HKitaev = −µ 2 N X i=1 (1 + iβBiβAi) −i 4 N−1 X i=1 [(∆0 + t)βBiβAi+1 + (∆0 −t)βAiβBi+1] . → −i t 2 N−1 X i=1 βBiβAi+1, (663) where in the last step we set µ = 0 and ∆= t, a special case where it can be easily shown that the solution supports Majorana zero modes. This can be seen by noting that H now 129In the topological superconductivity literature, the matrix representation of the Hamiltonian in particle-hole space is usually referred to as the Bogoliubov-de Gennes (BdG) representation, and the BdG symmetry, the antiunitary transformation which involves exchanging a particle for a hole and complex conjugating, ΞHBdGΞ−1 = −HBdG is one of the symmetries (along with time reversal and chiral symmetry) used to classify topological superconductors and insulators. 130Majorana fermions β are operators defined by Ettore Majorana as a possible description of the neutrino. Such objects are defined to be their own antiparticle, β = β†, and obey the commutation relations {βi, βj} = 2δij. 159 couples Majoranas A and B only at nearest neighbor sites. However there is no coupling for the B Majorana on site N, nor for the A Majorana on site 0. These therefore cost zero energy and are decoupled from the rest of the system. The rest of the system still has a gap for excitations, since it’s a superconductor. So we have system with quite usual excitations above a gap ∆= t, but in addition two independent zero modes located at the chain ends. We emphasize that the Majorana representation of the Hamiltonian was not necessary to obtain this result: we can diagonalize the finite system in the ordinary fermionic representation like any other mean field (BdG) Hamiltonian, and will find a zero-energy eigenstate whose wave function has weight only on the two end sites. The more general case with ∆̸= t and |µ| < t but µ ̸= 0 can also be solved and give the same qualitative picture, with the Majorana modes decaying exponentially into the bulk of the chain. 160
188277
https://www.britannica.com/science/phase-mechanics
SUBSCRIBE Ask the Chatbot Games & Quizzes History & Society Science & Tech Biographies Animals & Nature Geography & Travel Arts & Culture ProCon Money Videos phase mechanics Print Written by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Article History Related Topics: : wave : polyphase current : phase angle : phase velocity : multiphase current See all related content phase, in mechanics of vibrations, the fraction of a period (i.e., the time required to complete a full cycle) that a point completes after last passing through the reference, or zero, position. For example, the reference position for the hands of a clock is at the numeral 12, and the minute hand has a period of one hour. At a quarter past the hour the minute hand has a phase of one-quarter period, having passed through a phase angle of 90°, or π/2 radians. In this example the motion of the minute hand is a uniform circular motion, but the concept of phase also applies to simple harmonic motion such as that experienced by waves and vibrating bodies. If the position y of a point or particle changes according to a simple harmonic law, then it will change in time t according to the product of the amplitude, or maximum displacement, r, of the particle and a sine or cosine function composed of its angular speed, symbolized by the Greek letter omega (ω), the time t, and what is called the angle, symbolized by the Greek letter epsilon (ε): y = r sin (ωt + ε). The angle (ωt + ε) is called the phase angle at time t, which at zero time is equal to ε. Phase itself is a fractional value—the ratio of elapsed time t to the period T, or t/T—and is equal to the ratio of the phase angle to the angle of the complete cycle, 360°, or 2π radians. Thus, phase for uniform circular or harmonic motion has the value (ωt + ε)/2π. Applying this expression to the example of the moving minute hand cited above, ε is zero (zero phase angle at zero time), angular speed is 2π radians per hour, and time t is 1/4 hour, giving a phase of 1/4. When comparing the phases of two or more periodic motions, such as waves, the motions are said to be in phase when corresponding points reach maximum or minimum displacements simultaneously. If the crests of two waves pass the same point or line at the same time, then they are in phase for that position; however, if the crest of one and the trough of the other pass at the same time, the phase angles differ by 180°, or π radians, and the waves are said to be out of phase (by 180° in this case). More From Britannica electric generator: Phases The measurement of phase difference is of central importance in alternating-current technology. In the diagram, two curves represent the voltage (E) and the current (I) in an alternating-current (AC) circuit with pure inductance. The difference in phase angle between the voltage and the current is 90°, and the current is said to lag one-quarter cycle in phase. This lag may be seen from the diagram. In AC power transmission the terms multiphase and polyphase are applied to currents that are out of phase with one another. In a two-phase system there are two currents with a phase-angle difference of 90°; in a three-phase system the currents differ in phase angle by 120°. This article was most recently revised and updated by William L. Hosch. wave motion Introduction References & Edit History Quick Facts & Related Topics Images & Videos Quizzes Physics and Natural Law wave motion physics Print Also known as: wave propagation Written by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Article History Related Topics: : wave See all related content wave motion, propagation of disturbances—that is, deviations from a state of rest or equilibrium—from place to place in a regular and organized way. Most familiar are surface waves on water, but both sound and light travel as wavelike disturbances, and the motion of all subatomic particles exhibits wavelike properties. The study of waves therefore forms a topic of central importance in all physical science and engineering. The simplest types of wave motion are vibrations of elastic media, such as air, crystalline solids, or stretched strings. If, for example, the surface of a metal block is struck a sharp blow, the deformation of the surface material compresses the metal in the vicinity of the surface, and this transmits the disturbance to the layers beneath. The surface relaxes back to its initial configuration, and the compression propagates on into the body of the material at a speed determined by the stiffness of the material. This is an example of a compression wave. The steady transmission of a localized disturbance through an elastic medium is common to many forms of wave motion. In most systems of interest, two or more disturbances of small amplitude may be superimposed without modifying one another. Conversely, a complicated disturbance may be analyzed into several simple components. In radio transmission, for example, a high-frequency signal can be superimposed on a low-frequency carrier wave and then filtered out intact on reception. Britannica Quiz Physics and Natural Law In the simplest waves, the disturbance oscillates periodically with a fixed frequency and wavelength. These sinusoidal oscillations form the basis for the study of almost all forms of linear wave motion. In sound, for instance, a single sine wave produces a pure tone, and the distinctive timbre of different musical instruments playing the same note results from the admixture of sine waves of different frequencies. In electronics, the natural rhythmic oscillations of electric currents in tuned circuits are used to produce sinusoidal radio waves. Although the mathematical properties of all linear waves are common, the waves exhibit various physical manifestations. One important class—electromagnetic waves—represents oscillations of the electromagnetic field. These include infrared radiation, visible light, radio and television, microwave, ultraviolet, X-rays, and gamma rays. Electromagnetic waves are produced by moving electric charges and varying currents, and they can travel through a vacuum. Unlike sound waves, they are not, therefore, disturbances in any medium. Another difference between electromagnetic and sound waves is that the former are transverse, that is, the disturbance occurs in a direction perpendicular to that in which the wave is propagating. Sound waves are longitudinal: they vibrate along the path of their propagation. The propagation of a wave through a medium will depend on the properties of the medium. For example, waves of different frequencies may travel at different speeds, an effect known as dispersion. In the case of light, dispersion leads to the unscrambling of colours and is the mechanism whereby a prism of glass can produce a spectrum. In geophysics, the dispersive propagation of seismic waves can provide information about the constitution of Earth’s interior. Two important characteristics of all waves are the phenomena of diffraction and interference. When a wave disturbance is directed toward a small aperture in a screen or other obstacle, it emerges traveling in a range of directions. Thus, light rays, which normally follow straight paths, can bend upon passing through a small hole: this is the phenomenon known as diffraction. Access for the whole family! Bundle Britannica Premium and Kids for the ultimate resource destination. Subscribe Interference occurs when two waves are combined and the disturbances overlap. If the waves arrive at a point in phase, enhancement occurs and the disturbance is large. Where the waves are out of phase, their opposing motions cancel and the disturbance is small or nonexistent. The net effect is therefore a distinctive interference pattern of large and small disturbances. Mathematically less tractable is the study of nonlinear waves, which can be very important in many applications. These usually display a more complicated structure and behaviour; for example, water waves in a shallow channel can develop a humplike formation known as a soliton, which propagates as a coherent entity. Nonlinear waves are important in systems as diverse as nerve networks and the spiral arms of galaxies. The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by Erik Gregersen. Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style The Editors of Encyclopaedia Britannica. "phase". Encyclopedia Britannica, 14 Mar. 2025, Accessed 14 September 2025. Share Share to social media Facebook X External Websites Khan Academy - Phase diagrams UCLA Department of Mathematics - Phase Space (PDF) Indian Academy of Sciences - The fluid mechanics of phase change (PDF) Nature - Correlative image learning of chemo-mechanics in phase-transforming solids Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style The Editors of Encyclopaedia Britannica. "wave motion". Encyclopedia Britannica, 7 Jul. 2022, Accessed 14 September 2025. Share Share to social media Facebook X URL External Websites Hyperphysics - Wave Motion
188278
https://arxiv.org/abs/2207.01699
[2207.01699] Cycles of length 3 and 4 in edge-colored complete graphs with restrictions in the color transitions Skip to main content We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors.Donate >math> arXiv:2207.01699 Help | Advanced Search Search GO quick links Login Help Pages About Mathematics > Combinatorics arXiv:2207.01699 (math) [Submitted on 4 Jul 2022 (v1), last revised 11 May 2023 (this version, v2)] Title:Cycles of length 3 and 4 in edge-colored complete graphs with restrictions in the color transitions Authors:Hortensia Galeana-Sánchez, Felipe Hernández-Lorenzana, Rocío Sánchez-López View a PDF of the paper titled Cycles of length 3 and 4 in edge-colored complete graphs with restrictions in the color transitions, by Hortensia Galeana-S\'anchez and 2 other authors View PDF Abstract:Let Gbe an edge-colored graph, a walk in Gis said to be a properly colored walk iff each pair of consecutive edges have different colors, including the first and the last edges in case that the walk be closed. Let Hbe a graph possible with loops. We will say that a graph Gis an H-colored graph iff there exists a function c:E(G)\longrightarrow V(H). A path (v_1,\cdots,v_k)in Gis an H-path whenever (c(v_1v_2),\cdots,c(v_{k-1}v_k))is a walk in H, in particular, a cycle (v_1,\cdots,v_k,v_1)is an H-cycle iff (c(v_1 v_2),\cdots,c(v_{k-1}v_k),c(v_kv_1), c(v_1 v_2))is a walk in H. Hence, Hdecide which color transitions are allowed in a walk, in order to be an H-walk. Whenever His a complete graph without loops, an H-walk is a properly colored walk, so H-walk is a more general concept. In this paper, we work with H-colored complete graphs, with restrictions given by an auxiliary graph. The main theorems give conditions implying that every vertex in an H-colored complete graph, is contained in an H-cycle of length 3 and in an H-cycle of length 4. As a consequence of the main results, we obtain some well-known theorems in the theory of properly colored walks. Comments:19 pages, 3 figures Subjects:Combinatorics (math.CO) MSC classes:05C07, 05C15, 05C38 Cite as:arXiv:2207.01699 [math.CO] (or arXiv:2207.01699v2 [math.CO] for this version) Focus to learn more arXiv-issued DOI via DataCite Submission history From: Felipe Hernández-Lorenzana [view email] [v1] Mon, 4 Jul 2022 19:53:58 UTC (120 KB) [v2] Thu, 11 May 2023 06:36:37 UTC (122 KB) Full-text links: Access Paper: View a PDF of the paper titled Cycles of length 3 and 4 in edge-colored complete graphs with restrictions in the color transitions, by Hortensia Galeana-S\'anchez and 2 other authors View PDF TeX Source Other Formats view license Current browse context: math.CO <prev | next> new | recent | 2022-07 Change to browse by: math References & Citations NASA ADS Google Scholar Semantic Scholar export BibTeX citation Loading... BibTeX formatted citation × Data provided by: Bookmark Bibliographic Tools Bibliographic and Citation Tools [x] Bibliographic Explorer Toggle Bibliographic Explorer (What is the Explorer?) [x] Connected Papers Toggle Connected Papers (What is Connected Papers?) [x] Litmaps Toggle Litmaps (What is Litmaps?) [x] scite.ai Toggle scite Smart Citations (What are Smart Citations?) Code, Data, Media Code, Data and Media Associated with this Article [x] alphaXiv Toggle alphaXiv (What is alphaXiv?) [x] Links to Code Toggle CatalyzeX Code Finder for Papers (What is CatalyzeX?) [x] DagsHub Toggle DagsHub (What is DagsHub?) [x] GotitPub Toggle Gotit.pub (What is GotitPub?) [x] Huggingface Toggle Hugging Face (What is Huggingface?) [x] Links to Code Toggle Papers with Code (What is Papers with Code?) [x] ScienceCast Toggle ScienceCast (What is ScienceCast?) Demos Demos [x] Replicate Toggle Replicate (What is Replicate?) [x] Spaces Toggle Hugging Face Spaces (What is Spaces?) [x] Spaces Toggle TXYZ.AI (What is TXYZ.AI?) Related Papers Recommenders and Search Tools [x] Link to Influence Flower Influence Flower (What are Influence Flowers?) [x] Core recommender toggle CORE Recommender (What is CORE?) Author Venue Institution Topic About arXivLabs arXivLabs: experimental projects with community collaborators arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website. Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them. Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs. Which authors of this paper are endorsers? | Disable MathJax (What is MathJax?) About Help Contact Subscribe Copyright Privacy Policy Web Accessibility Assistance arXiv Operational Status Get status notifications via email or slack
188279
https://stackoverflow.com/questions/52066069/test-if-each-digit-of-a-number-is-prime-or-not
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Test if each digit of a number is prime or not Ask Question Asked 7 years, 1 month ago Modified7 years, 1 month ago Viewed 730 times This question shows research effort; it is useful and clear -1 Save this question. Show activity on this post. I want to test if each digit in a number entered by the user is prime or not. For example: For a number entered like 124: 1 is a prime number 2 is a prime number 4 is a composite number I have done as follows but clearly there is an error in this. ```python x=int(input("Enter the number you want to check\n")) primeflag=True lst=[] prime=[] com=[] while x>0: y=x%10 x=x//10 lst.append(y) l=(len(lst)) for i in (0,l-1): for j in range(2,lst[i]-1): if lst[i]%j==0: primeflag=False else: primeflag=True if primeflag==True: prime.append(lst[i]) else: com.append(lst[i]) print(prime,"are Prime Numbers") print(com,"are Composite Numbers") ``` python Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Aug 28, 2018 at 20:57 Thomas Weller 60.9k 23 23 gold badges 142 142 silver badges 262 262 bronze badges asked Aug 28, 2018 at 20:54 Prikshit SharmaPrikshit Sharma 3 2 2 bronze badges 9 1 We will not find the error for you. Please be specific.mad_ –mad_ 2018-08-28 20:55:56 +00:00 Commented Aug 28, 2018 at 20:55 1 An error typically has a message and a line number. It will be much easier for us to help you if you give us all information you have.Thomas Weller –Thomas Weller 2018-08-28 20:58:04 +00:00 Commented Aug 28, 2018 at 20:58 What makes you think that there is an error?Yunnosch –Yunnosch 2018-08-28 20:58:59 +00:00 Commented Aug 28, 2018 at 20:58 There are a total of 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Five of them are prime: 1, 2, 3, 5, 7. Just check for each digit: digit in '12357'zvone –zvone 2018-08-28 21:00:10 +00:00 Commented Aug 28, 2018 at 21:00 3 FYI: en.wikipedia.org/wiki/Prime_number#Primality_of_oneThomas Weller –Thomas Weller 2018-08-28 21:01:03 +00:00 Commented Aug 28, 2018 at 21:01 |Show 4 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. There are multiple problems with this loop: python for j in range(2,lst[i]-1): if lst[i]%j==0: primeflag=False else: primeflag=True The first problem is that you replace primeflag over and over, so ultimately this is just testing whether the last value that you checked was a divisor or not. What you want to check is whether any of the values was a divisor. So, you need to start off with True before the loop, and never set it back to True if you've ever set it to False: python primeflag=True for j in range(2,lst[i]-1): if lst[i]%j==0: primeflag=False While we're at it, once you find a factor, you can break, because you already know the number is composite: python primeflag=True for j in range(2,lst[i]-1): if lst[i]%j==0: primeflag=False break The second problem is that range(2,lst[i]-1) is all of the numbers up to but not includinglst[i]-1. Ranges are half-open in Python. So: python primeflag=True for j in range(2, lst[i]): if lst[i]%j==0: primeflag=False break This one, despite being wrong, doesn't actually break anything. Why? Well, you actually only need to test up to sqrt(lst[i])), and you're already getting 0, 1, and 2 wrong because of a different bug, and for every larger number, n-1 > sqrt(n), so it doesn't matter that you miss n-1. However, your outer loop has the same problem: python for i in (0,l-1): … and there, it is a problem—you never check the last digit. The third problem is that, for the digits 0, 1, and 2, range(2, lst[i]) is empty, so you're not going to loop at all. For your original code, that means you're just going to use the leftover value of primeflag from the last digit. With the fixes above, it means you're going to assume all three of those are prime, which is right for 1 (by your definition) and 2, but not for 0. The simplest fix is: python primeflag = lst[i] != 0 for j in range(2, lst[i]): if lst[i]%j==0: primeflag=False break But really, once we've implicitly put in hardcoded answers for 0, 1, and 2, why not (a) make it explicit, and (b) put in the answers for all 10 digits? Then you can just skip the whole loop, and you won't have all of these opportunities to get things wrong in the first place: python primeflag = lst[i] in {1, 2, 3, 5, 7} Finally, you can simplify the loop by just looping over the digits directly, instead of looping over a range up to the length of the list of digits—which again makes things simpler, and removes one of the opportunities for error (the one that caused you to miss the last digit): python for digit in lst: if digit in {1, 2, 3, 5, 7}: prime.append(digit) else: com.append(digit) One last thing: your code generates the digits in reverse order (so, e.g., input 124 gives you [4, 2, 1]), but your intended output seems to be the digits in forward order ("1 is a prime number 2 is a prime number 4 is a composite number"), so you may want to loop over lst[::-1]. Putting it all together, and cleaning things up a bit to fit PEP 8 style: ```python x = int(input("Enter the number you want to check\n")) lst = [] while x: y = x%10 x = x//10 lst.append(y) prime, com = [], [] for digit in lst[::-1]: if digit in {1, 2, 3, 5, 7}: prime.append(digit) else: com.append(digit) print(prime, "are Prime Numbers") print(com, "are Composite Numbers") ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Aug 28, 2018 at 21:20 answered Aug 28, 2018 at 21:15 abarnertabarnert 368k 54 54 gold badges 626 626 silver badges 691 691 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. Not at the best way to do it. Will remove after a while but might help to debug your issue. There were many errors (logical) one so I thought it might give you a push. There can be more I have not tested this myself and will remove after a while. Have a look I have modified your code only. python x=list(input("Enter the number you want to check\n")) primeflag=True lst=[] prime=[] com=[] lst=list(map(int,x)) l=(len(lst)) for i in range(0,l): primeflag=False print(lst[i]) if(lst[i]==2 ): prime.append(lst[i]) continue for j in range(2,lst[i]+1): if lst[i]%j==0: primeflag=False else: primeflag=True if primeflag: prime.append(lst[i]) else: com.append(lst[i]) print(prime,"are Prime Numbers") print(com,"are Composite Numbers") Wrap in a function or something and call it while iterating over the list. but frankly you can just define a dict for 0-9 numbers with status prime or not. It will be more efficient Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Aug 28, 2018 at 21:13 mad_mad_ 8,283 2 2 gold badges 31 31 silver badges 46 46 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! 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188280
https://arxiv.org/pdf/1210.5263
Published Time: Sat, 30 Nov 2024 01:13:09 GMT arXiv:1210.5263v1 [math.AG] 18 Oct 2012 Zero sets and factorization of polynomials of two variables Micki Balaich and Mihail Cocos ∗ November 27, 2024 Abstract The relationship between a polynomial’s zeros and factors is well known. If a ∈ R is a zero of f (x) ∈ R[x] then x − a is a factor of f (x). In this paper, we generalize this idea to R[x, y ]. We consider the zero sets of two variable polynomials in R[x, y ] and give criterion to when two polynomials with the same zero set in R[x, y ] have a common factor with the same zero set. When F is not a field, but the division algebra of Quaternions, we provide an example of two polynomials in F [x, y ] with the same zero set and no common factor. 1 Introduction Recall that a zero of a polynomial p(x) ∈ R[x] is a real number a such that p(a) = 0. It is a college algebra fact that if a real number a is a zero of a polynomial p(x) ∈ R[x] then x − a divides p(x). Many examples demonstrate this idea. For instance, if p(x) = x2 − 1 then p is divisible by x−1 and x+1 since the zeros of f are 1 and −1 and we can write the familiar p(x) = ( x − 1)( x + 1). This fundamental relationship guarantees that any two polynomials of one variable over the real numbers that share a zero set have a common factor with the same zero set. Although this relationship is generally attributed only to one variable polynomials, the same can be said for any two polynomials in n variables ∗ Mathematics Subject Classification: Primary 12D05 ; Secondary 12E05 1over the complex numbers, or any algebraically closed field. As motivation for our work, we will now establish this result using the cornerstone of algebraic geometry, the Hilbert Nullstellensatz Theorem. Theorem 1. Let F be an algebraically closed field. If p, g ∈ F [x1, x 2, . . . , x n] have the same zero set, then p and g have a common factor with the same zero set. Proof. Assume p and g share the same zero set and let I =< g > . Since F is algebraically closed, by the Nullstellensatz there exists a positive integer r such that pr = hg where h ∈ F [x1, x 2, . . . , x n]. Since F [x1, x 2, . . . , x n]is a UFD g is either irreducible or it can be written as a product of irreducible factors. If g is irreducible then because F [x1, x 2, . . . , x n] is a UFD it is also prime and must therefore divide p. If g is not irreducible then it can be written as the product of irreducible factors g1 · · · gk for some positive integer k. Now since each of these factors is irreducible they must also be prime, so they each divide p. Then the product of the distinct irreducible factors of g divides p.Thus p and g have a common factor. Now a repeated factor does not add any new zeros to the zero set of g so the common factor will have the same zero set as p and g. It follows that p and g have a common factor with the same zero set. The zero sets of one variable polynomials in R[x] also have a very nice property. The Fundamental Theorem of Algebra guarantees that not only are they are finite, but their maximum size is determined by the degree of the polynomial. Letting deg xp denote the degree of x of p(x, y ) ∈ R[x, y ]and deg yp denote the degree of y of p(x, y ), the following shows that, in a similar manner, the degree of the variables can, under certain circumstances, determine the type of zero set of a two variable polynomial in R[x, y ]. Theorem 2. Let p(x, y ) ∈ R[x, y ]. If deg xp or deg yp is odd, then p(x, y ) has an uncountabley infinite number of zeros. Proof. Suppose deg xp is odd. Write p(x, y ) in descending powers of x as p(x, y ) = qn(y)xn + qn−1(y)xn−1 + · · · + q0(y)x0 where each qi ∈ R[y]. Fix a real number y0 and consider the polynomial p(x, y 0). It is a polynomial of the variable x and has degree n which we assumed to be odd. Thus it must have at least one real root. This is true for each real number y0 so p(x, y )must have an uncountabley infinite number of zeros. If deg yp is odd, then write p(x, y ) in descending powers of y and fix an x0 ∈ R. Then use the same argument as before. 2Remark 3. This shows that if a polynomial in R[x, y ] is to have a finite, discrete zero set then deg xp and deg yp must both be even. If F is not algebraically closed the result of Theorem 1 breaks down. Take F = R, n = 2, and consider the polynomials p(x, y ) = x2 + y2 and g(x, y ) = x4 + y4 in R[x, y ]. The zero set of both polynomials is the single point (0 , 0), yet they do not have a common factor. The main result shows that with certain conditions on the zero set satisfied, the result of Theorem 1 can be extended to a large class of polynomials in R[x, y ]. In addition, when we replace R with the division algebra of the Quaternions there are in fact examples of two polynomials sharing an infinite zero set that do not share a factor, one of which we show in Section 3. 2 Main result In the setting of one variable polynomials it is natural use the division al-gorithm to establish divisibility. Given specific polynomials we can perform long division and look for the remainder to be zero. This idea can be extended to two variable polynomials. Write p and g in R[x, y ] by descending powers of x so as to view them as polynomials of the one variable x with coefficients in R[y] and then do long division just as if they were one variable polyno-mials, the result of which yields an equation similar to the division algorithm. For instance, if g(x, y ) = 5 x3 −2 and p(x, y ) = x−3y dividing g by p gives 5x2 + 15 yx + 45 y2 x − 3y) 5x3 − 2 − 5x3 + 15 yx 2 15 yx 2 − 15 yx 2 + 45 y2x 45 y2x − 2 − 45 y2x + 135 y3 (−2 + 135 y3)Thus g(x, y ) = p(x, y )(5 x2 + 15 yx + 45 y2) + 135 y3 − 2. 3If p(x, y ) = 2 x4 − 3x and g(x, y ) = yx 2 + yx then dividing p by g gives 2y−1x2 − 2y−1x + 2 y−1 yx 2 + yx ) 2x4 − 3x − 2x4 − 2x3 − 2x3 2x3 + 2 x2 2x2 − 3x − 2x2 − 2x − 5x So 2 x4 − 3x = ( yx 2 + yx )( 2x2 y − 2x y 2 y ) − 5x. Note that in the first example both the quotient and remainder are poly-nomials of two variables but in the latter the quotient is a polynomial with coefficients that are rational functions of the variable y. In general, if p and g are polynomials of two variables over the real numbers and n = deg xp then g = pq + r (4) where q and r are polynomials with coefficients in R(y) and deg xr ≤ n. If n= deg xp there are many examples of polynomials in R[x, y ] that have a zero set crossed by an infinite number of horizontal lines in at least n distinct places. Any polynomial that has its zero set on the curve y = x or y = xn naturally satisfies the conditions. Using this idea of long division we can show that these conditions ensure these polynomials will share the factor y − x or y − xn respectively with any polynomial that has the same zero set. Furthermore, Theorem 5 uses Equation 4 as a foundation to show that any two polynomials sharing a zero set with these conditions have a common factor with the same zero set. Theorem 5. Let p(x, y ), g (x, y ) ∈ R[x, y ] be two polynomials that have the same zero set and n= deg xp. If there are an infinite number of horizontal lines lines that intersect the zero set in at least n distinct places, then p and g have a common factor with the same zero set. 4Proof. Let n=deg xp and assume that there are an infinite number of parallel lines that intersect the zero set of p and g in at least n distinct places. Dividing g(x, y ) by p(x, y ) gives g(x, y ) = q(x, y )p(x, y ) + r(x, y ) (6) where r(x, y ) and q(x, y ) are polynomials of the variable x with coefficients in R(y) and deg xr ≤ n − 1. If we let h(y) be the common denominator of the coefficients in q and r and multiply both sides of (6) by h we then have hg =˜ qp + ˜ r where ˜ q = hq and ˜ r = hr are polynomials of two variables and since h ∈ R[y] deg x ˜r ≤ n − 1. This means that ˜ r(x, y ) = rn−1(y)xn−1 + rn−2(y)xn−2 + · · · + r0(y) where ri(y) ∈ R[y]. Now if ( x0, y 0) is a point in the zero set of p and g then g(x0, y 0) = p(x0, y 0) = 0. Thus 0 = ˜ q(x0, y 0) · 0 + ˜ r(x0, y 0). This implies that ˜ r(x0, y 0) = 0 . By assumption there are an infinite number of horizontal lines that inter-sect the zero set in at least n distinct places. Let y0 be the y coordinate of the points on one of these horizontal lines. It follows that there are at least n distinct x values such that ˜r(x, y 0) = 0 . But since deg x ˜r ≤ n − 1, it follows that rn−1(y0) = rn−2(y0) = rn−3(y0) = · · · = r0(y0) = 0 . Now, there are infinitely many values y0 for which the above relationship holds, and since all ri’s are polynomials of one variable it follows that they are identically equal to zero. Thus ˜ r(x, y ) = 0 , and therefore hg = ˜ qp. (7) This equation tells us that p divides hg. Since R[x, y ] is a UFD either p is irreducible or it can be written as the product of irreducible factors. We will now show that in either case p and g have a common factor with the same zero set. If p is irreducible it is also prime and must therefore divide either h or g.First suppose it divides h. This means that h = ˜hp for some ˜h ∈ R[x, y ]. Since 5h is a polynomial of the one variable y this implies that p is also a polynomial of the one variable y. But the zero set of a polynomial of one variable is a finite number of horizontal lines in R2, and it is therefore impossible for there to exist infinitely many horizontal lines that intersect this zero set nontrivially. It follows that p cannot divide h and so must divide g. If p is not irreducible then it can be written as the product of irreducible factors p1 · · · pk for some positive integer k and the zero set of p would be the union of the zero sets of these factors. Since each of these factors must also be prime, they each either divide g or divide h. Now if they all divide h then the zero set of p would be a finite number of horizontal lines, of which it is impossible for an infinite number of horizontal lines to cross. Thus one of these factors must divide g. If any of the other irreducible factors of p divide h then the zero set of each factor that is nontrivial is a horizontal line. The only way for the zero set of g to contain all the points on a horizontal line is for g to have a factor that is only in terms of the variable y. It follows that each of these distinct factors must also divide g. Thus p and g share a common factor with the same zero set. Please note that by writing p(x, y ) in descending powers of y one can similarly prove Theorem 8. Let p(x, y ), g (x, y ) ∈ R[x, y ] be two polynomials that have the same zero set and n= deg yp. If there are an infinite number of vertical lines that intersect the zero set in at least n distinct places, then p and g have a common factor with the same zero set. Recall that the degree of a term of a two variable polynomial is the sum of the exponents of the variables in the term and the degree is the maximum of the degrees of the terms which we will denote by deg [p]. Now based on these Theorems we give a slightly more general statement. Corollary 9. Let p(x, y ), g (x, y ) ∈ R[x, y ] be two polynomials that have the same zero set and n= deg [p]. If there are an infinite number of parallel lines that intersect the zero set in at least n distinct places, then p and g have a common factor with the same zero set. Proof. Let a b be the slope of the parallel(we may assume they are not vertical or horizontal) lines that intersect the zero set of the two polynomial lines in at least n distinct places. Consider the change of variables 6u = bx + ay and v = ax − by. This linear transformation is invertible and the two polynomials in the new variables u, v will share the same zero set. Let us denote these polynomials by P (u, v ) and G(u, v ). It is clear that deg uP (u, v ) = deg vP (u, v ) = n. The parallel lines that intersect the zero set of p(x, y ) and g(x, y ) are mapped into horizontal lines in the u, v plane. We are now under the conditions of Theorem 5. It follows that P (u, v ) divides G(u, v ) and applying the inverse linear transformation we conclude that p and g have a common factor with the same zero set. Note that no polynomial with a finite zero set will fall under the above theorems. This may in part explain why the polynomials p(x, y ) = x2 + y2 and g(x, y ) = x4 + y4 in R[x, y ] do not have a common factor. There are also polynomials with infinite zero sets that are left out. One example is any polynomial with a zero set on the curve x4 + y4 = 1 . In this case, n = 4 but one can show algebraically that this curve will be crossed at no more than two distinct places by any line. 3 On the non-commutative case Let us consider the case of polynomials over the division algebra of quater-nions H. If we look at p(x, y ) = xy − yx, then the zero set is the set of all commuting pairs of quaternions in H2. According to , the polynomial g(x, y ) = x2y + y2x − 2xyx, has the same exact zero set as p(x, y ). The following theorem shows that p and g do not have a common factor with the same zero set. Theorem 10. Let p(x, y ) = xy − yx and g(x, y ) = x2y + y2x − 2xyx. Then p and g do not have a common factor with the same zero set. 7Proof. We will first show that p is irreducible over H and then show that p does not divide g thus implying that p and g do not share a common factor. Suppose p is the product of two monomials. Then xy − yx = ( ax + by + c)( Ax + By + C) where a, b, c, A, B, C, ∈ H. Multiplying the right side out gives xy − yx = aAx 2 + aBxy + aCx + bAyx + bBy 2 + bCy + cAx + cBy + cC. Then aA = 0, aB = 1, and bA = −1. If aA = 0, then since H has no zero divisors either a = 0 or A = 0 . However this contradicts that fact that aB = 1 and bA = −1. It follows that p cannot be the product of two monomials and is therefore irreducible. Now suppose that p divides g. Then either g(x, y ) = p(x, y )h(x, y ) or g(x, y ) = h(x, y )f (x, y ) for some h(x, y ) ∈ H[x.y ]. Since h(x, y ) cannot be a constant suppose that it is linear, that h(x, y ) = ax + by + c. If g = ph then x2y +y2x−2xyx = ( xy −yx )( ax +by +c). Multiplying the right side out gives x2y + y2x − 2xyx = axyx + bxy 2 + cxy − ayx 2 − byxy − cyx. Since there is no x2y term it follows that g 6 = ph. If g = hp then a similar argument shows that there cannot be a y2x term. It follows that p does not divide g and that they have no common factor with the same zero set. References M.Cocos: On the commutativity of an element and its derivative in a fi-nite dimensional algebra, International Journal of Algebra, Vol.5,2011, no. 29-32 Department of Mathematics, Davinci Academy, Ogden, UT 84401, USA, e-mail: micki.balaich@davinciacademy.org Department of Mathematics, Weber State University, Ogden, UT 84408, USA, e-mail: mihailcocos@weber.edu 8
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https://www.chegg.com/homework-help/questions-and-answers/definition-1-multiplicative-order--multiplicative-order-element-modulo-m-written-ordma-non-q57313392
Solved Definition 1 (Multiplicative Order). The | Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Advanced Math Advanced Math questions and answers Definition 1 (Multiplicative Order). The multiplicative order of an element modulo m, written ordma, is a nonnegative integer defined as follows: ordmax at 31 (mod m). In other words, qordma = 1 (mod m) and ordm a > 0; ak = 1 (mod m) ordma | r. Definition 2 (Primitive Roots). An element g is a primitive root modulo m iff ordm 9 4(m). (d) Prove: almordma Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: Definition 1 (Multiplicative Order). The multiplicative order of an element modulo m, written ordma, is a nonnegative integer defined as follows: ordmax at 31 (mod m). In other words, qordma = 1 (mod m) and ordm a > 0; ak = 1 (mod m) ordma | r. Definition 2 (Primitive Roots). An element g is a primitive root modulo m iff ordm 9 4(m). (d) Prove: almordma Show transcribed image text Here’s the best way to solve it.Solution Share Share Share done loading Copy link View the full answer Previous questionNext question Transcribed image text: Definition 1 (Multiplicative Order). The multiplicative order of an element modulo m, written ordma, is a nonnegative integer defined as follows: ordmax at 31 (mod m). In other words, qordma = 1 (mod m) and ordm a > 0; ak = 1 (mod m) ordma | r. Definition 2 (Primitive Roots). An element g is a primitive root modulo m iff ordm 9 4(m). (d) Prove: almordma (m). (e) Prove: ordma" = (ordm a) Kr. (f) Prove: ordm ab | [ordma, ordm b]. (g) Show with an explicit counterexample that the above cannot be strengthened to an equation. (h) Find the order of every element modulo m for each modulus with 0 <m<9. Display your final results in table form; show all work on separate pages. You may use previous results in your work. (i) Find the order of every element modulo 15. Display your final results in table form; show all work on separate pages. You may use previous results in your work. (j) Find all primitive roots for the above moduli, where they exist. (k) Prove that if there exists a primitive root modulo m, there are exactly (4(m)) such roots. Not the question you’re looking for? Post any question and get expert help quickly. 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https://www.researchgate.net/publication/245369398_Contact_Mechanics_Approach_to_Determine_Effective_Radius_in_Bolted_Joints
Published Time: 2005-01-01 (PDF) Contact Mechanics Approach to Determine Effective Radius in Bolted Joints Article PDF Available Contact Mechanics Approach to Determine Effective Radius in Bolted Joints January 2005 Journal of Tribology 127(1) DOI:10.1115/1.1829717 Authors: Qian Zou Oakland University T. S. Sun T. S. Sun This person is not on ResearchGate, or hasn't claimed this research yet. Sayed Nassar Oakland University Gary Barber Oakland University Show all 6 authors Hide Download full-text PDFRead full-text Download full-text PDF Read full-text Download citation Copy link Link copied Read full-textDownload citation Copy link Link copied Citations (33)References (12)Figures (7) Abstract and Figures The effective radius is a very important factor in determining frictional torque under bolt head/nut is bolted joints. Since the effective radius is hard to be precisely determined in practice, the mean radius, which is the mean value of the inner and outer radii of the contact area under bolt head/nut, is used. In this paper, contact mechanics analysis is used to determine the real contact pressure distribution between bolt head/nut and the joint surface. Based on the pressure distribution, the effective radius can be precisely calculated. Effect of the surface roughness, underhead load distribution, underhead load value, and the ratio of the maximum to minimum contact radii are investigated. The results are compared with the result from theoretical formulas and the value of mean radius. These data will be very useful in enhancing the reliability and safety of bolted joints. Torque components … Flow chart for numerical procedure … Sample surface profile … Effect of surface roughness … +2 Underhead load distributions … Figures - uploaded by Dong Zhu Author content All figure content in this area was uploaded by Dong Zhu Content may be subject to copyright. Discover the world's research 25+ million members 160+ million publication pages 2.3+ billion citations Join for free Public Full-text 1 Content uploaded by Dong Zhu Author content All content in this area was uploaded by Dong Zhu on Feb 17, 2016 Content may be subject to copyright. Q. Zou T. S. Sun S. Nassar G. C. Barber H. El-Khiamy Fastening and Joining Research Institute, Department of Mechanical Engineering, Oakland University, Rochester, MI 48309 D. Zhu Eaton Corporation, Southfield, MI 48037 Contact Mechanics Approach to Determine Effective Radius in Bolted Joints The effective radius is a very important factor in determining frictional tor que under the bolt head and nut in bolted joints.Since the effective radius is har d to be precisely determined in practice,the mean radius,which is the mean value of the inner and outer radii of the contact ar ea under the bolt head and nut,is used.In this paper,contact mechanics analysis is used to determine the real contact pr essure distribution between the bolt head and nut and the joint surface.Based on the pr essure distribution,the effective radius can be precisely calculated.The effect of the surface r oughness,underhead load distribution,underhead load value,and the ratio of the maximum to minimum contact radii are investigated.The results ar e compar ed with the r esult fr om theoretical formulas and the value of mean radius.These data will be very useful in enhancing the reliability and safety of bolted joints.关 DOI:10.11 15/1.1829717 兴 Keywor ds:Bolted Joint;Effective Bearing Radius;T or que-T ension Relationship;Con- tact Mechanics Introduction Bolted joints are widely used in mechanical systems. T o ensure the safety,reliability,and durability of the bolted joints,a proper clamping force is required.However,in most engineering appli- cations,it is very difficult to measure the clamping force directly. The torque applied during the initial tightening of the joint is usually used as a controlling factor. Therefore, it is very important to precisely determine the relation between the applied torque and the clamping force,i.e.,torque-tension relationship.The torque required to tighten a bolt consists of three components: 共 i 兲 T p ,the pitch torque component that creates fastener tension and clamping force,共 ii 兲 T t ,the torque component that overcomes the friction between male and female threads,and 共 iii 兲 T b ,the bearing fric- tion torque component that overcomes friction between the turn- ing bolt head or nut and the clamped surfaces.It can be described by the following equation: T⫽T p ⫹T t ⫹T b (1) where T p ⫽ p t 2 ␲ F,T t ⫽ ␮ t r t cos ␤ F,T b ⫽ ␮ b r b F In the equation,T is the input tightening torque applied to the fastener head or nut,F is the fastener tension,p t is the thread pitch, ␮ t is the coefficient of friction between male and female threads, ␮ b is the coefficient of friction between the bearing sur- faces under the turning fastener head or nut,r t is an effective contact radius between threads, r b is an effective bearing radius of the bearing contact area under the turning head or nut,and ␤ is half of the thread profile angle,which is 30 deg for standard UN and ISO threads 关 1 兴. In most applications,a very high percentage of the input torque is consumed in overcoming the combined effect of the two fric- tional torque components.Only a much smaller percentage of the input torque power is the useful part that creates the fastener elon- gation and tension and,simultaneously,produces the clamping force in the joint.The torque-tension relationship in threaded fas- tener applications is highly sensitive to friction-torque variations. T ypically about 90%of the work on a nut or the head of bolt is converted to heat,owing to the frictional resistance between the face of the nut and the surface of the joint and between male and female threads.About 50%is lost under the head and nut and about 40%within the threads.Only 10%of the input work ends up as bolt preload or as clamping force between joint members, which creates the fastener elongation and tension.The torque- tension relationship in threaded fastener applications is highly sensitive to friction-torque variations. A small variation in friction will have a significant impact on achieving the clamp load from torque.From Fig.1 we can see that a 20%decrease in underhead friction will result in 100%increase in clamping force. T o determine the bearing friction-torque component T b ,the effective bearing radius r b of the bearing contact area under the turning head or nut must be precisely determined.Because the effective radius is hard to precisely determine in practice,design- ers usually use the mean radius,which is the mean value of the inner and outer radii of the contact area under the bolt head and nut,to replace it.The mean contact radius is given by r m ⫽ 共 ␥ ⫹1 兲 D 4 (2) where D is nominal size of the fastener and ␥ is the ratio of the outside-to-inside radii of the contact area.However,this assump- tion for effective radius is very crude. A number of studies has been conducted to investigate the ef- fect of the friction components on the torque-tension relation. Ex- perimental studies were carried out by Bray and Levi 关 2 兴,Sakai 关 3 兴,Jiang 关 4 兴,etc.Different test devices were set up to measure the clamping force,the torque needed to overcome the thread friction,and the bearing friction,separately.The mechanical be- haviors of bolted joints during tightening and under repeated tightening-loosening action were also investigated 关 4–6 兴.Theo- retical analysis was done by Nassar 关 7,8 兴,who derived an equa- tion to calculate the effective underhead bearing friction radius r b and the thread friction radius r t . In this paper,contact mechanics analysis is used to determine the real contact pressure distribution between the bolt head and nut and the joint surface.Based on the pressure distribution,the effective radius can be precisely calculated.The effect of the clamping load value and distribution,surface roughness,and the ratio of the maximum to minimum contact radii are investigated. Manuscript received February 25,2004;revision received July 19,2004.Review conducted by:M.Lovell. 30 Õ V ol.127,JANUARY 2005 Copyright©2005 by ASME T ransactions of the ASME The results are compared with the results from theoretical equa- tions and the value of mean radius. These data will be very useful in enhancing the reliability and safety of bolted joints. Contact Mechanics Model The contact mechanics model was developed by Hu,Barber, and Zhu 关 9 兴.When two rough surfaces are in contact,the vertical surface displacement u(x,y)due to a distributed normal pressure p( ␰ , ␩ )and tangential force q( ␰ , ␩ )can be obtained by the fol- lowing equation: u 共 x,y 兲 ⫽ 2 ␲ E 冕冕 S p 共 ␰ , ␩ 兲 冑 共 x⫺ ␰ 兲 2 ⫹ 共 y⫺ ␩ 兲 2 d ␰ d ␩ ⫺ 1 ␲ G 冕冕 S q 共 ␰ , ␩ 兲 共 ␰ ⫺x 兲 共 x⫺ ␰ 兲 2 ⫹ 共 y⫺ ␩ 兲 2 d ␰ d ␩ (3) where E is the equivalent Y oung’s modulus and G is the equiva- lent shear modulus 1 E ⫽ 1 2 冉 1⫺ ␯ 1 2 E 1 ⫹ 1⫺ ␯ 2 2 E 2 冊 (4) 1 G ⫽ 1 2 冋 共 1⫹ ␯ 1 兲 共 1⫺2 ␯ 1 兲 E 1 ⫺ 共 1⫹ ␯ 2 兲 共 1⫺2 ␯ 2 兲 E 2 册 (5) Here,we assume that the friction force is proportional to the nor- mal force,and the coefficient of friction is a constant.Therefore, the force-displacement relation can be expressed by u 共 x,y 兲 ⫽ 2 ␲ E 冕冕 S p 共 ␰ , ␩ 兲 ⫻ 冉 1 冑 共 ␰ ⫺x 兲 2 ⫹ 共 ␩ ⫺y 兲 2 ⫺ E 2 G ␮ 共 ␰ ⫺x 兲 共 ␰ ⫺x 兲 2 ⫹ 共 ␩ ⫺y 兲 2 冊 ⫻d ␰ d ␩ (6) The calculation of displacement from a known pressure distri- bution is carried out through a fast Fourier transformation 共 FFT 兲 based approach 关 9 兴.And the pressure distribution is solved by a conjugate gradient method 共 CGM 兲关 10 兴.Using the model,the real contact pressure distribution between the bolt head and nut and the joint surface can be obtained.Based on the pressure distribu- tion,the effective underhead bearing friction radius r b can be calculated by r b ⫽ 兰 r i r o pr 2 dr 兰 r i r o prdr ⫽ 兺 j p j r j 2 兺 j p j r j (7) where p j is the contact pressure for a given point and r j is the distance from the point to the center of the contact area.The flowchart of the numerical procedure is shown in Fig.2. Based on the pressure distribution,the real area of contact can also be calculated,which is the sum of the area with positive pressure acting on it. The real contact ratio is the ratio between the real area of contact and the nominal area of contact. Results and Discussions Here,contact mechanics analysis is used to determine the ef- fective bearing radius precisely.The effects of the surface rough- ness,underhead load distribution,underhead load value,and the ratio of the maximum to minimum contact radii are investigated. Results are compared to the value of mean radius and also to the results for the theoretical equation. Effect of Surface Roughness.Three surface roughnesses and their combinations are used.For surface 1,R a ⫽9.288 ␮ m.For surface 2,R a ⫽0.9288 ␮ m.The surface 3 is assumed to be an ideal smooth surface.Therefore,R a of surface 3 is equal to zero. The roughness was measured by a WYKO optical profiling sys- tem, which can be used to measure three-dimensional surface pro- files precisely. The sample surface profile of surface 2,before and after contact,is shown in Figs.3 共 a 兲 and 3 共 b 兲.In the calculation, Fig.1 T orque components Fig.2 Flow chart for numerical procedure Journal of T ribology JANUARY 2005,V ol.127 Õ 31 we divide the entire contact area into many small pieces.Each piece was treated,approximately,as a rectangle because the area is very small,typically 0.5 mm⫻0.5 mm. A contact mechanics model is used to determine the effective bearing radius.Five cases are analyzed,i.e.,surface 1 contacting to surface 1 共 rough surface to rough surface 兲,surface 2 contacting to surface 2 共 moderately rough surface to moderately rough sur- face 兲,surface 1 contacting to surface 3 共 rough surface to smooth surface 兲,surface 2 contacting to surface 3 共 moderately rough sur- face to smooth surface 兲,and surface 3 contacting to surface 3 共 smooth surface to smooth surface 兲. The bolt used here is an M12 bolt with inner radius of 6 mm and ␥ ratio of 1.5.The bolt and joint material is steel.Normal load applied on the bolt is 98,960 N. The contact mechanics simulation results of ef fective bearing radius under different surface roughness are obtained.The results are compared to the value of mean radius,and the error between them is calculated.On the one hand,a positive percent error in- dicates an underestimation of the bearing friction torque compo- nent when the mean radius r m is used instead of r b .This will result in a lower torque-tension ratio for the fastener and a lower clamping force in a bolted joint.On the other hand,a negative error indicates an overestimation of the bearing friction that ulti- mately leads to a higher torque-tension ratio in the fastener and a higher clamping force in the joint. The real contact ratio,which is the ratio between the real area of contact and the nominal contact area,can also be obtained from the contact mechanic analysis. All of the results are listed in T able 1 and shown in Fig.4.From these results we can see that the surface roughness has a certain effect on the effective bearing radius.When the surface roughness is small,the real contact ratio is almost equal to 100%.The simula- tion results are quite close to the mean radius.Under this condi- tion,the mean radius gives a very good prediction of the effective bearing radius.However,when the surface roughness is high,the real contact ratio dropped significantly.There is some difference between the simulation results and the mean radius.The error changes from a negative value to a positive value,and its absolute value is almost doubled.Therefore,for rough underhead and joint surfaces,using the mean radius will cause an error on the friction torque. Fig.3 Sample surface profile Fig.4 Effect of surface roughness T able 1 Effect of surface roughness 32 Õ V ol.127,JANUARY 2005 T ransactions of the ASME Effect of Underhead Load Distribution.Four scenarios of underhead load distribution are discussed 共 i.e.,uniform underhead load,linearly decreasing load,exponentially decreasing load,and sinusoidal underhead load 兲 and shown in Fig.5.A uniform pres- sure distribution simulates the use of a strong steel bolt and nut to clamp onto a much weaker clamping surface,such as plastic sur- faces or soft washers.Other pressure distributions considered would simulate contact stress concentration at the edge of the fastener hole or represent the zero pressure conditions at mini- mum or maximum radii of the contact area. A contact mechanics model is used to determine the effective bearing radius.The contact mechanics simulation results of the effective bearing radius under dif ferent underhead load distribu- tions are obtained.The bolt used here is an M12 bolt with inner radius of 6 mm and ␥ ratio of 1.5.The bolt and joint material is steel.The normal load applied on the bolt is 98,960 N.Both surfaces are surface 2. The results are compared to the value of mean radius and the calculation results from Nassar’s theoretical equation 关 7 兴共 listed below 兲,where r b⫺t means the theoretical calculation result of r b , Fig.5 Underhead load distributions Journal of T ribology JANUARY 2005,V ol.127 Õ 33 Uniform underhead load:r b⫺t ⫽ 0.656 共 ␥ 2.9 ⫺1 兲 r i 共 ␥ 1.9 ⫺1 兲 (8) Linearly decreasing load: r b⫺t ⫽ 0.95 r i 共 ␥ 2 ⫺1 兲 ␥ 1.9 ⫺1 冉 6 ␥ 3.9 ⫺23.4 ␥ ⫹17.4 11.31 ␥ 3 ⫺33.93 ␥ ⫹22.62 冊 (9) Exponentially decreasing load: r b⫺t ⫽ 0.95 共 ␥ 2 ⫺1 兲 r i 0.1 ␥ 1.9 ⫺1 冋 兰 r i ␥ r i r 1.9 e ⫺a 共 r⫺r i 兲 dr 兰 r i ␥ r i re ⫺a 共 r⫺r i 兲 dr 册 (10) where a⫽ ln 共 p max /p min 兲 r i 共 ␥ ⫺1 兲 Sinusoidal underhead load: r b⫺t ⫽ 0.95 共 ␥ 2 ⫺1 兲 r i 0.1 ␥ 1.9 ⫺1 冋 兰 r i ␥ r i r 1.9 sin 冋 ␲ 共 r⫺r i 兲 r i 共 ␥ ⫺1 兲 册 dr 兰 r i ␥ r i r sin 冋 ␲ 共 r⫺r i 兲 r i 共 ␥ ⫺1 兲 册 dr 册 (11) The error between the simulation result and the mean radius is calculated. The real contact ratio is also obtained from the contact mechanics analysis.All of the results are listed in T able 2 and shown in Fig.6.From these results we can see that the underhead load distribution has an obvious effect on the effective bearing radius.Under uniform and sinusoidal underhead load distribution, the simulation results,mean radius,and theoretical calculation results are very close.But when the load distribution is linearly and exponentially decreasing,both the simulation result and the- oretical calculation result are much smaller than the mean radius. Therefore,under these load distributions,if the mean radius is used to calculate the friction torque,then an error will occur.A negative error indicates an overestimation of the bearing friction, which ultimately leads to a higher torque-tension ratio in the fas- tener and to a higher clamping force in the joint. Effect of Underhead Load V alue.Under the uniform load distribution,the effect of different underhead load values is inves- tigated.The loads are 70,685,98,960,and 127,234 N,respec- tively.T wo contacting pairs,surface 1 to surface 1 and surface 2 to surface 2, are used. The bolt is an M12 with a hexane head. The results are shown in T able 3 and Fig.7. From these results we can see that when the surface roughness is small,the load value does not have much effect on the effective bearing radius.From the real contact ratio data we can see that increasing the load value only increases the real contact area slightly because the real contact area is almost equal to the nomi- nal contact area.When the surface roughness is large,the load value has greater effect on the effective bearing radius.W ith an increase in the load,the effective bearing radius decreases.This is because the real contact area between two rough surfaces is only a small portion of the nominal contact area.When the load in- creases, the real contact area will increase as well. This will affect the friction force between the two surfaces.Here again,we can see that the surface roughness played an important role in deter- mining the effective bearing radius. Effect of Maximum to Minimum Contact Radii Ratio ␥ . For most fasteners,the value of the maximum to minimum con- tact radii ratio ␥ is within the range of 1.5⬃2.5.For example,for standard hexane head bolts, ␥ ⫽1.5,and for the flange head bolts, ␥ ⫽2.5.Here the ef fective bearing radius under a different ␥ ratio Fig.6 Effect of underhead load distribution T able 2 Effect of underhead load distribution 34 Õ V ol.127,JANUARY 2005 T ransactions of the ASME is calculated.The load distribution is assumed to be uniform with a value of 98,960 N.The two surfaces are assumed to be smooth. The results are compared to the mean radius value and the theoretical calculation results. All of the results are listed in T able 4 and shown in Fig.8.From these results we can see that for standard hexane head bolts,the simulation result is close to the mean radius value.W ith an increase of the ␥ ratio,the error be- tween them also increases.For flange head bolts,a larger error is observed.Therefore,the mean radius cannot be used for flange head bolts. Conclusions In this paper,contact mechanics analysis is used to determine effective radius.The effect of surface roughness,underhead load distribution,underhead load value,and the ratio of the maximum to minimum contact radii are investigated.The results are com- pared to those from theoretical equations and the mean radius value.The following conclusions can be drawn: 1.When the surface roughness is small,the mean radius gives a very good prediction of the effective bearing radius.How- ever,when the surface roughness is high,there is some dif- ference between the simulation results and the mean radius. The error changes from a negative value to positive value, and its absolute value is almost doubled. 2.Under uniform and sinusoidal underhead load distributions, the simulation results,the mean radius,and theoretical cal- culation results are very close.But when the load distribu- tion is linearly and exponentially decreasing,both simula- tion results and theoretical calculation results are smaller than the mean radius. The results show that the modeling of the fastener underhead load has some effect on the bearing friction torque component and on the overall fastener torque-tension relationship.This has direct impact on the level of the clamping force achieved in the joint when the fastener is initially tightened to a given torque level. 3.For smooth surfaces,the load value does not have much effect on the effective bearing radius. But for rough surfaces, the load value has a greater effect on the effective bearing radius.With the increase of the load,the effective bearing radius decreases. Fig.7 Effect of underhead load value T able 3 Effect of underhead load value Journal of T ribology JANUARY 2005,V ol.127 Õ 35 4.With the increase of the ␥ ratio,the error between the simu- lation result and the mean radius also increases. From these results we can see that contact mechanics analysis provided a more accurate calculation of the fastener underhead bearing friction-torque component.The use of the mean radius of the contact area should be replaced by using the contact mechan- ics model developed in this paper to determine the actual bearing friction radius. This will enhance the quality, safety, and reliability of many mechanical and structural components. Nomenclature D⫽nominal diameter of the fastener E⫽equivalent modulus of elasticity F⫽fastener tension G⫽equivalent shear modulus H(m,n)⫽transfer function in frequency domain U(m,n)⫽Fourier transform of u(x,y) R a ⫽average roughness T⫽total torque applied on the head or nut of the fas- tener T b ⫽bearing friction torque T p ⫽pitch torque T t ⫽thread friction torque p j ⫽contact pressure for a given point p t ⫽thread pitch p( ␰ , ␩ )⫽contact pressure distribution q( ␰ , ␩ )⫽tangential force r b ⫽bearing friction radius 共 effective radius of the con- tact area under the turning head or nut of the fas- tener 兲 r i ⫽inner radius of contact area r j ⫽distance from the given point to the center of the contact area r m ⫽mean of the contact area under the turning fas- tener head 共 or nut 兲 r o ⫽outer radius of contact area r t ⫽thread friction radius 共 effective thread contact ra- dius 兲 u(x,y)⫽vertical surface displacement or deformation ␤ ⫽half of the thread profile angle ␦ ⫽surface approach ␥ ⫽ratio of the maximum to minimum contact radii ␮ ⫽coefficient of friction ␮ t ⫽coefficient of friction between male and female threads ␮ b ⫽coefficient of friction between the bearing surfaces under the turning fastener head or nut ␯ ⫽Poisson ratio References 关 1 兴 Motosh,N.,1976,‘‘Development of Design Charts for Bolts Preloaded up to the Plastic Range,’’ ASME J.Eng.Ind.,98 共 3 兲,pp.849–851. 关 2 兴 Bray, A.,and Levi,R.,1966,‘‘Factors Affecting Tightening Characteristics of Bolts,’’Exp.Mech.,6,pp.409–419. 关 3 兴 Sakai, T.,1978,‘‘The Friction Coefficient of Fasteners,’’ Bull. JSME, 21 共 152 兲, pp.333–340. 关 4 兴 Jiang,Y.,Chang,J.,and Lee,C.,2000,‘‘An Experimental Study of T orque- T ension Relationship for Bolted Joints,’’ Int. J.Mater. Prod.T echnol.,16 共 4/5 兲, pp.417–429. 关 5 兴 Hagiwara,M.,and Ohashi,N.,1994,‘‘A New Tightening T echnique for Threaded Fasteners,’’ T rans. ASME,Ser.C:J.Heat Transfer,11 6,pp.64–69. 关 6 兴 Fukuoka,T.,and T akaki,T.,1998,‘‘Mechanical Behaviors of Bolted Joint During T ightening Using T orque Control,’’JSME Int.J.,Ser.A,41, pp.185–191. 关 7 兴 Nassar,S.,Barber,G.C.,and Zuo,D.,2004,‘‘Bearing Friction T orque in Bolted Joints,’’ 59th STLE Annual Meeting,May 17–20,T oronto,Canada. 关 8 兴 Nassar,S.,Matin,P.,and Barber,G.C.,2004,‘‘Thread Friction T orque in Bolted Joints,’’ASME Pressure V essels and Piping Conference,LaJolla,Cali- fornia. 关 9 兴 Hu, Y.,Barber, G.C.,and Zhu,D.,1999,‘‘Numerical Analysis for the Elastic Contact of Real Rough Surfaces,’’ Tribol.Trans.,42 共 3 兲,pp.443–452. 关 10 兴 Johnson,K.L.,1985,Contact Mechanics,Cambridge University Press,Cam- bridge,UK. T able 4 Effect of maximum to minimum contact radii ratio Fig.8 Effect of contact radii ratio 36 Õ V ol.127,JANUARY 2005 T ransactions of the ASME Citations (33) References (12) ... In the study of friction radius, Zou et al. proposed an integration method of friction radius based on two forms of contact pressure distribution-exponential and sinusoidal functions-and found that the method had higher accuracy when the surface roughness was small by experimental comparison. Nassar et al. proposed a friction radius integration method based on two forms of uniform and linear contact pressure distribution, considering the effects of tightening speed, three-dimensional thread geometry, and dynamic-static friction factors, by means of a tightening test system that can measure the decomposition torque; they concluded that the tightening speed was an important factor affecting the uncertainty of the thread friction radius, as shown in Figure 4. ... ... In the study of friction radius, Zou et al. proposed an integration method of friction radius based on two forms of contact pressure distribution-exponential and sinusoidal functions-and found that the method had higher accuracy when the surface roughness was small by experimental comparison. Nassar et al. proposed a friction radius integration method based on two forms of uniform and linear contact pressure distribution, considering the effects of tightening speed, three-dimensional thread geometry, and dynamic-static friction factors, by means of a tightening test system that can measure the decomposition torque; they concluded that the tightening speed was an important factor affecting the uncertainty of the thread friction radius, as shown in Figure 4. Furthermore, orthogonal tests were conducted on steel and aluminum connectors to study the variation in the tightening torque-preload transformation relationship under different scales of surface roughness, different contact surface materials, and different thread parameters. ... A Review of Aeroengines’ Bolt Preload Formation Mechanism and Control Technology Article Full-text available Mar 2023 Zhaoyu Li Xiaoqiang Li Yujie Han Gang Zhao The bolt connection structure is widely used in the connection of aeroengine parts, and its connection quality is very important, as it can directly affect the geometric and dynamic performance of the aeroengine. Taking the bolt preload as the entry point, the research status of bolted aeroengine connections in recent years is summarized based on the bolt preload formation mechanism and control technology in this review. For the research of the preload formation mechanism, single and multiple bolts of aeroengine bolt arrays are analyzed. According to the characteristics of aeroengine bolt connections, the research of aeroengine bolt preload control technology is divided into two parts: bolt tightening technology, and bolt tightening equipment. Finally, based on the summarization of the research, the future research prospects are evaluated and discussed. View Show abstract ... Hence, Zou et al. performed contact mechanics analysis to determine the actual contact pressure distribution between the bolt head/nut and joint surface for various contact surface roughness. Based on the pressure distribution, the effective friction radius was calculated precisely. ... Preload Control Method of Threaded Fasteners: A Review Article Full-text available Sep 2024 Xing Yan Zhifeng Liu Mingpo Zheng Wentao Chen Threaded fasteners are one of the most commonly used connection methods for mechanical structures. Its primary function is to generate appropriate clamping forces and fasten the connected parts. An inappropriate preload can cause loosening, fatigue fracture, and other problems. This will affect the safety and reliability of mechanical equipment. The precise control of the preload has become a critical issue in mechanical assembly processes. Over the past few decades, various tightening measures and methods have been proposed to address this issue. However, many problems continue to exist with practical applications that have not been reviewed comprehensively and systematically. First, various control methods were summarized systematically, and their advantages and disadvantages in engineering applications were analyzed. Torque control is the most widely used tightening method owing to its simple operation and low cost. Therefore, the research on the torque control method was summarized systematically from three aspects: the torque-preload correlation formula, effective friction radius, and friction characteristics during tightening. In addition, the special circumstances that may increase preload uncertainty were discussed. Finally, based on a summary of the current research status, the prospects for future research were discussed. This study would aid researchers in extensively understanding the problems in preload control. View Show abstract ... 2.1) to the contact pressure simulation, the interfacial pressure distribution data can be extracted. Considering that the interfacial pressure gradually decreases along the distance from the bolt , the pressure distribution is fitted to a fourth-order polynomial concerning the distance from the bolt hole, d, as represented in Eq. (31). ... Interfacial micromechanics study on contact modeling for bolted joints Article Full-text available Apr 2023 ACTA MECH Yu Chang Jianguo Ding H. Fan Bolted joints represent the discontinuity of the assembled structures, so their contact characteristics contribute significantly to the overall static and dynamic performances of the mechanical system. For the multi-scale geometrical properties of contacting surfaces, an interfacial micromechanics modeling method is proposed to predict contact characteristics using the fractal theory. Meanwhile, the interaction effect caused by the successive tightening of multiple bolts is incorporated into the contact analysis, which characterizes the residual preload of bolts to improve the contact load model. Three contact models for the bolted joint are examined by combining the interfacial micromechanics model with the transfer matrix method for multi-body systems, the finite element method, and the virtual material method. A comparison with the experimental data of a dumbbell-shaped bolted structure is conducted to validate the contact models and estimate their practicality and accuracy. The models show their advantages and drawbacks, which depend on the complexity of the bolted structure, the requirements of computational efficiency, and the research focus. View Show abstract ... In the assembly of high-end aerospace precision products, the dispersion of the preload is very likely to cause the problem of out-of-tolerance of product accuracy. Many scholars have studied the influencing factors of preload dispersion through theoretical and finite element analysis and experiments . The research of Wettstein shows that up to 30-40 parameters can influence the torque-preload relationship in threaded fasteners, such as material , surface finish , etc. Grabon et al. studied the effect of the thread manufacturing process on the dispersion of preload through experiments. ... Research on Preload and Stress Distribution of Small-Size Precision Threaded Connection Considering Geometric Deviation Article Full-text available Dec 2022 Yang Yang Bo Yuan Qinchao Sun Binbin Zhao The small-size threaded connection is one of the main connection methods for complex and precision electromechanical products such as aerospace equipment. The geometric deviation is unavoidable in the thread manufacturing process, which will lead to the problem of excessive dispersion of the connection preload and uneven stress distribution, resulting in a low product qualification rate and a short stability period. In this paper, the influence of thread geometric deviation on preload and stress distribution is studied by taking the small-size M2 thread commonly used in high-end aerospace precision products. Firstly, the thread engagement model with geometric parameters is established, and the influence of different types of geometric deviations on the preload is analyzed. Secondly, the mechanism of non-uniform stress distribution on the connector in the process of thread engagement is analyzed. Finally, the accuracy of the simulation model and the analysis results are verified by the small-size thread-tightening experiment. The results show that the pitch diameter deviation, profile angle deviation, and pitch deviation of precision thread affect the preload and stress distribution, among which the profile angle deviation has a significant influence on the preload and stress distribution. View Show abstract Optimization Methodology Development of FEA and Design Analysis on High Voltage Bolted Joints for Electric Vehicles Conference Paper Apr 2025 Joon Ha Lee Zhijun Wu Marco Gerini-Romagnoli Sayed Nassar div class="section abstract"> To promote the electric performance and safety of development for EV mobility, optimization methodology and design guide of high voltage bolted joint should be newly developed. This paper describes the development process of multi-physics (electrical, mechanical, thermal) FEA methodology, various experimental tests and establishment of optimization methodology of busbar bolted joint design in terms of bolt preload validation and joint temperature rise. The various key factors on high voltage joint tightening are quantitatively studied by utilizing this optimized methodology. </div View Show abstract A novel theoretical analysis on mechanical properties of bolted joints under torsional loads considering service torque Article Feb 2025 ENG STRUCT Mingpo Zheng Ying Li Jinfei Guo Magd Abdel Wahab View Analysis of Torque Variation in Bolt Fastening on Coated Steel Surfaces Article Nov 2024 TRIBOL INT Sukkyung Kang Seounghee Yun Hyena Hwang Sanha Kim As the number of parts ever increases, the push for automation in assembly processes is intensifying in advanced manufacturing industries. The drive to improve productivity through manufacturing automation is relentless, and minimizing torque variation in automated bolt fastening is crucial for achieving successful assembly. Therefore, a precise understanding of the mechanisms of interfacial friction between bolts and coated steel plates, along with a torque analysis based on the interactions of various involved parameters, is essential. This study examines the interactions between surface topography and mechanical properties of bolts and coated steel plates. We explain three primary friction modes (plowing, sliding, and penetrating) within the defined contact areas and propose a prediction model based on contact mechanics to estimate the frictional torque under the bolt head on coated surfaces. Additionally, we quantitatively evaluate how the proportions of these friction modes change under various axial loads and coating conditions, and how these variations affect torque values. We also assess how variations in key parameters influence changes in friction modes, providing insights into the critical factors that should be monitored for effective control of fastening torque. This study presents methods for analyzing and calculating friction torque, offering guidelines for monitoring bearing surfaces to address bolt topography deviations and solving key industrial challenges in automated assembly. View Show abstract Cıvatalı Bağlantı Elemanlarının Tekrarlı Kullanımı Durumunda Sürtünme Değişikliklerini Entegre Eden Cıvatalı Bağlantı Simülasyon Modelinin Geliştirilmesi Article Sep 2024 Barış Tanrıkulu Ramazan Karakuzu Cıvatalı bağlantı elemanları günümüzde birçok mekanik uygulamada sıklıkla kullanılmaktadır. Özellikle kritik uygulamalarda, kilitleme yükü değeri, montajlanan parçaların yaşam ömrü için önem arz etmektedir. Cıvatalı bağlantı elemanları tekrar kullanılabilir elemanlar olmasına rağmen sürtünme katsayılarındaki değişim, her sıkım esnasında kilitleme yükünün giderek düşmesine sebebiyet vermektedir. Bu durum, özellikle kilitleme yükünün önem arz ettiği uygulamalarda, beklenmedik sonuçların meydana gelebileceğini ortaya koymaktadır. Çalışma kapsamında, bağlantı elemanlarının tekrar kullanımı durumundaki sürtünme katsayısı değişimi deneysel olarak belirlenmiş ve bu veriler ışığında yüzey kayma mesafesi bazlı sürtünme değişim modeli geliştirilmiştir. Sayısal simülasyon çalışmalarında tekrar kullanım etkisinin göz ardı edilmesinden dolayı, bu etkinin tahmin edilebilmesine olanak sağlayan modellerin geliştirilmesine ihtiyaç duyulmaktadır. Elde edilen kayma mesafesi bazlı sürtünme modeli, cıvatanın 5. sıkım sonucunda göstermiş olduğu davranışı, sayısal simülasyon yazılımı ile tahmin edebilmiştir. Model sonucunda elde edilen tork çıktıları, deneysel veri sonucunda elde edilen veriler ile kıyaslanarak hata payları çıkarılmıştır. Kayma mesafesi bazlı sürtünme modelinin, sürtünme katsayısının sabit kaldığı varsayımına dayanarak gerçekleştirilen simulasyonlardan elde edilen çıktı torkundaki hata payını, %19,5’ den %4,48’e düşürdüğü gözlemlenmiştir. View Show abstract Effect of different underhead shot-peening and lubrication conditions on high-strength screws undergoing multiple tightenings Article Aug 2023 TRIBOL INT Dario Croccolo Massimiliano De Agostinis Stefano Fini Giorgio Olmi View Degradation and failure analysis of bolted joints under transverse loading Article Nov 2022 TRIBOL INT Mingpo Zheng Zhifeng Liu Magd Abdel Wahab Chao Li There are a huge number of bolted joints used in engineering applications. However, unreasonable design and use may lead to degradation and failure of the connected structure. Herein, finite element analysis method is used to investigate the influence of various factors on the mechanical properties of bolted joints. To simulate the actual tightening and assembly process, different analysis steps were considered, given as, tightening step, initial lose step and the loading step. In the loading step, transverse load is applied to obtain the force-displacement curve. Structural parameters, i.e. clamping force, effective clamp length and assembly clearance, and friction coefficients of all the contact surface are considered in the finite element analysis. The results showed that the performance of the bolted joint under transverse loading exhibits different stages, which is the same as our previous experimental results. Through the analysis of the force-displacement curves, the characterization parameters for the mechanical properties of the bolted joint can be extracted, such as curve form, tangential stiffness, slip distance, transverse load threshold, etc. The characterization and evaluation of the mechanical properties integrates the process from design to service, which helps to provide reference for the reliability design and performance improvement of bolted joints. View Show abstract Show more An experimental study of the torque-tension relationship for bolted joints Article Full-text available Jan 2001 Yanyao Jiang Jianjun Chang Chu-Hwa Lee An experimental approach to determining the torque-tension relationship for a typically used bolted joint was developed. In addition to the bulk torque-tension relationship, the experiments measured the thread friction and the bearing friction between the nut surface and the clamped surface. The influence of the use of washers made of different materials under repeated tightening was studied. A statistics analysis was conducted to allow for a reliability assessment of the bolted joints in terms of the clamping force control. View Show abstract Bearing Friction Torque in Bolted Joints Article Full-text available Jan 2005 Sayed Nassar Gary Barber Dajun Zuo Formulas are developed for calculation of the effective radius of the bearing friction forces on the rotating contact surface in threaded fasteners. These formulas provide a more accurate estimation of the underhead bearing friction torque component in threaded fastener applications. This enhances the reliability, safety, and quality of bolted assemblies, especially in critical applications. It is well known that the torque-tension correlation in threaded fasteners, and the resulting joint clamping force, is highly sensitive to friction torque components: under the turning head and between threads. This analysis focuses on the bearing friction torque component under the turning head of a threaded fastener. Furthermore, it analyzes the error contained in the current practice when an approximate value, equal to the mean contact surface radius, is used instead of the actual bearing radius. New formulas for the bearing friction radius are developed for a mathematical model of a bolted joint using four different scenarios of the contact pressure distribution under the rotating fastener head or nut. The effect of the radially varying sliding speed over the rotating contact surface is analyzed and compared with a constant-friction-coefficient scenario. Numerical results and error analysis are presented in terms of a single nondimensional variable, namely, the radii ratio between the outside and the inside bearing area. View Show abstract Development of Design Charts for Bolts Preloaded up to the Plastic Range Article Aug 1976 J Eng Ind Trans ASME Nabil Motosh A mathematical formulation of the tightening process in bolts is presented, and from this design charts were developed which enable the determination of the necessary tightening torques for the maximum usable preloads. An apparatus which measures the friction coefficients between the threads and under the nut is described. For each bolt and friction coefficient under the nut it was found that there was a critical value of thread friction coefficient which gave a maximum tightening torque. (V.B.D.) View Show abstract The Friction Coefficient of Fasteners Article Feb 1978 Tomotsugu Sakai (1) The influence of the tightening velocity on the friction coefficient of fastener μ was investigated. (2) The influence on μ of the lubrication state, the surface treatment and the material combination of bolts and nuts was examined from the viewpoints of the central tendency and the dispersion. (3) The screw-thread friction coefficient μS and the bearing-surface friction coefficient μW were measured separately. Both the mean value and the standard deviation of μ were calculated statistically with the data of μS and μW, and a comparison between the calculated and the experimental results was made. (4) Some coating materials to reduce the standard deviation of μ, which are called "μ-stabilizers", were tested, and it was found that they are useful for decreasing the variation among subgroups. View Show abstract Mechanical Behaviors of Bolted Joint during Tightening Using Torque Control Article May 1997 Toshimichi FUKUOKA Tomohiro Takaki In the tightening process of a bolted joint using torque control, the effect of friction on axial force and torque is an issue of great interest. In this study, the mechanical behaviors of a bolted joint during tightening, such as variations of axial force and torque, are investigated both experimentally and numerically. Hence, friction coefficients on pressure flank of screw thread and the nut bearing surface are estimated by measuring the total torque, axial force and torque applied to the bolt, and compared with variations of axial force and torque. The validity of the finite element approach proposed here for evaluating changes in torque and axial force is ascertained by comparing the numerical results to the experimental ones. It is shown that the size of grip length has some influence on the reduction of torque and axial force. View Show abstract Mechanical behaviors of bolted joint during tightening using torque control Article Apr 1998 Toshimichi FUKUOKA Tomohiro Takaki In the tightening process of a bolted joint using torque control, the effects of friction on axial tension of bolt and torque is an issue of great interest. In this study, the mechanical behaviors of a bolted joint during tightening, such as variations of axial tension and torque, are investigated both experimentally and numerically. Hence, friction coefficients on pressure flank of screw thread and the nut loaded surface are estimated by measuring the total torque applied to nut, axial tension and thread friction torque, and compared with variations of axial tension and torque. The validity of the finite element approach proposed here for evaluating changes in torque and axial tension is ascertained by comparing the numerical results to the experimental ones. It is also shown that the size of grip length influences, to some extent, the reduction rates of torque and axial tension. View Show abstract Thread Friction Torque in Bolted Joints Article Nov 2005 Sayed Nassar Payam H. Matin Gary Barber In this paper, formulas are developed for the calculation of the effective thread friction radius in fasteners, in order to determine the thread friction torque component. Results provided by these formulas are compared with those given by current practice that uses the average value of the minor and major thread radii. It is well known that the torque-tension relationship in threaded fastener applications is highly sensitive to the friction torque components: between threads, and under the turning fastener head or nut. Even moderate variations or inaccuracies in determining the friction torque components would significantly impact the fastener tension and the joint clamp load. This makes critical the accuracy in the estimation of the friction torque components, as it directly affects the reliability, safety, and the quality of bolted assemblies, especially in critical applications. This analysis focuses on the thread friction torque component. The new formulas for the thread friction radius are developed for a mathematical model of a bolted joint using three assumed scenarios of the contact pressure between male and female threads. Because of the fact that the variation in the sliding speed of various points on thread surface is insignificant, a uniform thread friction coefficient is used in the analysis. However, a contact area weighted average value is used for the thread friction coefficient. Numerical results and error analysis are presented in terms of a single non-dimensional variable; namely, the ratio between the major and minor thread radii. View Show abstract A New Tightening Technique for Threaded Fasteners Article May 1994 M. Hagiwara N. Ohashi A new technique for tightening threaded fasteners was proposed aiming to assure the quality and the reliability in the tightening operations. The technique is based on the torque difference and the relative rotation angle in tightening and loosening process, and no preliminary experiment is required. Repeated tightening (and loosening) experiments were performed for different types of lubricant to verify the applicability of this technique. Experimental results show that the preload provided to the joint can be estimated with high accuracy and precision by this technique, and the possibility to boost the reliability of tightening control is confirmed. View Show abstract Numerical Analysis for the Elastic Contact of Real Rough Surfaces Article Jul 1999 Yuanzhong Hu Gary Barber Dong Zhu The elastic contact of rough surfaces and the subsurface stresses caused by the contact have been analyzed by means of a numerical model based on fast Fourier transforms (FFT) and minimization of complementary energy. The elastic contact has been modeled mathematically as a linear complementarity problem and solved by a robust algorithm, Conjugate Gradient Method, while the force-displacement relation is determined through a FFT approach. After solving for the pressure distribution, the subsurface stress field is obtained by calculating the stresses due to the application of a point force, and then integrating over the contact region. In comparison with the matrix based method published in recent years, the numerical approach presented in this study is more efficient, more stable and requires less memory. It has great potential in application to problems with general contact geometry and three-dimensional surface roughness. The results show that high frequency roughness could lead to very sharp impulses and significant oscillations in pressure distribution. The amplitude and location of the maximum shear stress in the subsurface region are constantly changing when the contacting rough surfaces are in relative motion. Presented at the 53rd Annual Meeting in Detroit, Michigan May 17–21, 1998 View Show abstract Factors affecting the tightening characteristics of bolts Article Jan 1966 A. Bray Raffaello Levi A study on the tightening characteristics of bolt-nut-washer assemblies was undertaken after a preliminary literature survey showed different opinions to exist on the effect of several commonly encountered factors on the torque-tension relationship.The test program included tightening bolts up to the yield point under a planned series of combination of factors, and recording nut-tightening torque, bolt torque and tension.The main factors tested were:(a) Plating of the mating surfaces of bolt, nut and washer (b) Lubrication of the interfaces. Among the subsidiary factors considered were the hardness and yield stress of the materials and the class of fit of the threads.The tests were planned and analyzed with statistical techniques in order to obtain estimates of the effects of each factor and the more important interactions (singled out of the total number in a series of pilot tests) as well as of the errors to be considered for the tests of significance. The analysis was carried out with mathematical and graphical methods giving accurate results in an easily understandable form from a reasonably small number of experimental data. The isolation of the main effects and interactions was achieved with no loss of general applicability of the results, taking advantage of the peculiar properties of the factorial experimental design adopted.A comparison between the torque wrench and the turn-of-nut method of bolt-tightening is made in terms of closeness of bolt tension control.A simple bolt-testing machine was built for the tests. A motor-driven wrench with a strain-gage bridge measuring the input torque tightened the nut at low speed while a two-component strain-gage transducer measured bolt torque and tension. Provision was also made for measuring the angle of rotation of the nut and the torsion of the bolt's shank. View Show abstract Show more Powered By 00:00/01:09 10 Why a Commitment to Open Science Makes a Great Job Candidate Share Next Stay Recommended publications Discover more Conference Paper Study of Two-Stage Tightening of Threaded Fasteners Using Various Tool Speed Combinations January 2007 Sayed Nassar Saravanan Ganeshmurthy This paper investigates the effect of various tool speed combinations on the torque-tension relationship in a two-stage torque control process for the tightening of threaded fasteners. The tightening speed has a considerable affect on the thread and bearing friction coefficients. Hence, the amount of clamp load generated by a specified level of the tightening torque is significantly influenced by ... [Show full abstract] the tool speed combinations in many high volume applications. In two-stage processes, the fastener is first tightened to an intermediate torque level beyond which the tool speed is changed but the tightening is continued until the full target torque is reached. The effect of tightening speed combinations (for stages one and two) on the torque-tension relationship is investigated. Data on the coefficients of thread and bearing friction as well as the Nut Factor K is investigated. The findings of this study are helpful in predicting the clamp load generated in bolted joints, which enhance its reliability and safety. Additionally, the study provides helpful data that may be utilized for developing reliable assembly strategies in mass production applications such as the automotive industry. Read more Article An Experimental Study of Bearing and Thread Friction in Fasteners April 2005 · Journal of Tribology Sayed Nassar Gary Barber T. S. Sun [...] H. El-Khiamy An experimental procedure is proposed for studying the underhead and thread friction in fasteners. The effective bearing friction radius, the underhead friction coefficient, and the thread friction coefficient are experimentally determined for fasteners with standard hexagonal heads and for flanged head fasteners. Hence, greater accuracy has been achieved in determining the value of the torque ... [Show full abstract] components that are consumed in overcoming friction in threaded fasteners. This would lead to a more reliable torque-tension correlation and would enhance the safety and quality of bolted assemblies. A Design of Experiment (DOE) procedure is presented in order to investigate the effect of fastener material class, the thread pitch, and the fastener size on thread friction coefficient. For the underhead bearing friction, an experimental model is presented in order to determine the effect of the contact area radii ratio on the bearing friction radius. Read more Article Surface roughness effect on the torque-tension relationship in threaded fasteners March 2007 · Proceedings of the Institution of Mechanical Engineers Part J Journal of Engineering Tribology Sayed Nassar T. S. Sun An experimental study is presented in order to investigate the effect of surface roughness on the torque-tension relationship in threaded fasteners. Three levels of surface roughness are considered for the fastener underhead and the joint surface; namely, low, medium, and high levels of surface roughness. The study is conducted for two joint materials, two fastener classes, and for coarse and ... [Show full abstract] fine threads. In this study, the torque-tension data are expressed in terms of the value of the nut factor as well as its scatter. The effect of the number of tightening on surface roughness and on the torque-tension relationship is investigated as well. The surface roughness is measured before tightening, and after each loosening using a WYKO optical profiling system. An M12 fastener is used in this study. Both fine and coarse threads and fastener material Classes 8.8 and 10.9 for M12 fasteners are used in this study. The torque-tension data is analysed for both steel and aluminium joints. The safety and reliability of bolted assemblies are mainly determined by the level and the stability of the clamp load provided by the initial tightening of the threaded fastener. The value of initial clamp load, which is achieved by a specific level of tightening torque, is highly sensitive to the friction torque components. This study provides an insight into the reliability of the existing engineering practices for estimating the clamp load level from the tightening torque. Hence, the findings of the study would help enhance the reliability and the safety of bolted assemblies, especially in critical applications. Read more Article Full-text available Effect of Lubrication on Friction and Torque-Tension Relationship in Threaded Fasteners April 2007 · Tribology Transactions Qian Zou T. S. Sun Sayed Nassar [...] A. K. Gumul In this research, three types of lubricants are investigated for their effects on the friction and torque-tension relationships in threaded fastener applications. Investigated lubricants included oils, greases, and solid film lubricants that were applied to the threads of the fastener and the nut, and the turning bolt underhead surface as well as the joint surface in contact with the bolt head. ... [Show full abstract] The oils tested were mineral oils with different viscosities, i.e. 21cst, 115cst, 1000cst respectively at 40°C. Three types of grease were tested; namely, regular grade, pure nickel special, and heavy metal free grease. For the solid lubricants, an olefinic type, an olefinic base with a polytetrafluoroethylene additive, and an olefinic base with an inorganic additive (Molybdenum disulfide), were applied. Class 8.8 M12 fasteners with both fine and coarse threads were used. The experiments were carried out using a Laboratory Fastener Testing System which measures the input torque, the fastener tension, the thread friction torque, and the angular rotation of the bolt head. The system computes the thread and bearing friction coefficients from the measured data. In each test, the fastener was repeatedly tightened and loosened five times. The tests were also conducted under different tightening speeds, varying from 1 rpm to 100 rpm. The nut factor, the coefficients of thread and the underhead friction were obtained from the experiments. The effects of the number of tightening and loosening cycles, the tightening speed, and the type of lubricant on friction and nut factor were investigated. It was found that lubrication had a significant effect on the friction and the torque-tension relationship in threaded fasteners. Presented at the STLE/ASME International Joint Tribology Conference in San Antonio, Texas, October 23-25, 2006Review led by Thierry Blanchet View full-text Article The Effect of Coating and Tightening Speed on the Torque-Tension Relationship in Threaded Fasteners April 2006 · SAE Technical Papers Sayed Nassar Qian Zou Tianshu Sun This study investigates the effect of fastener coating, tightening speed, and repeated tightening on the torque-tension relationship, and on the coefficients of thread and underhead friction. Selected coatings provide three distinctly different levels of friction between threads and under the rotating fastener head; these coatings are assigned a low, medium, or high friction designation. Due to ... [Show full abstract] the fact that the torque-tension relationship is highly sensitive to normal variations in friction, the findings of this study would improve the reliability and safety of bolted assemblies, especially in critical applications. For the three selected coatings, the experimental procedure determines the coefficients of thread and underhead bearing friction as well as the overall fasteners torque-tension relationship at two tightening speeds for three fastener sizes with fine and coarse threads. Additionally, the effect of repeated tightening is examined. Read more Last Updated: 07 Feb 2025 Discover the world's research Join ResearchGate to find the people and research you need to help your work. Join for free ResearchGate iOS App Get it from the App Store now. Install Keep up with your stats and more Access scientific knowledge from anywhere or Discover by subject area Recruit researchers Join for free Login Email Tip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password? - [x] Keep me logged in Log in or Continue with Google Welcome back! Please log in. Email · HintTip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password? - [x] Keep me logged in Log in or Continue with Google No account? Sign up Company About us News Careers Support Help Center Business solutions Advertising Recruiting © 2008-2025 ResearchGate GmbH. All rights reserved. Terms Privacy Copyright Imprint Consent preferences
188283
https://www.tutorocean.com/questions-answers/newtons-second-law
Search For Tutors Find tutors who specialize in coursesRequest a Tutor Let our team find you the perfect tutorHow It Works Find out how we can help you get great gradesQuestions & Answers Ask questions and get answers from expertsFor ParentsHow It Works Find tutors who can help your child succeedSearch for K-12 Tutors Find tutors who teach elementary and high school educationSearch for University Tutors Let our team find your child the perfect tutor How It Works Get paid teaching students from homeHow Much Can I Earn Your set your rate and your hours Higher Education For universities and collegesK-12 For schools and districtsEnterprise For employee benefits Learn arrow_drop_down Teach arrow_drop_down Business arrow_drop_down EN arrow_drop_down Back to feed Newton'S Second Law What Is Newton'S Second Law Of Motion? Physics Alexander 3 years ago 0 votes Vote Comment View answers (5) Aryanna S. Compassionate & Experienced - Math/Physics Tutor Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force applied and inversely proportional to its mass. It can be expressed mathematically as F = ma. Newton's Second Law Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. Mathematically, this can be expressed as: F = ma Where F is the net force applied to the object, m is the mass of the object, and a is the acceleration of the object. This law is often referred to as the "law of inertia" because it states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This means that an object will not change its velocity unless a force is applied to it. The Second Law of Motion is one of the most important laws of physics and is used to explain a wide variety of phenomena, from the motion of planets to the motion of everyday objects. It is also used to calculate the forces required to move objects, such as the force required to lift a weight or the force required to accelerate a car. 3 years ago 0 votes Vote Comment Muhammad Nouman A. Physics teacher with two years of experience in teaching primary, secondary, high school, GCSE/IGCSE, O-Level and A-Level Students. Newton's second law of motion, also known as the law of acceleration, states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. This law can be stated mathematically as F = ma, where F is the net force applied to an object, m is the mass of the object, and a is the acceleration of the object. In other words, the greater the force applied to an object, the greater its acceleration will be, and the more massive an object is, the less it will accelerate under the same amount of force. This law is fundamental to understanding how forces affect the motion of objects, and it is often used to calculate the motion of objects in various physical scenarios. For example, if a force of 10 N is applied to a 2 kg object, then its acceleration can be calculated as follows: a = F/m a = 10 N / 2 kg a = 5 m/s^2 This means that the object will accelerate at a rate of 5 meters per second squared under the force of 10 N. 2 years ago 0 votes Vote Comment Reefat R. Redefine your learning experience Newton's second law states that the net resultant force is proportional to the acceleration of the body. Mathematically, can be written as, F = ma, where m is the proportionality constant also known as the mass of the body. 2 years ago 0 votes Vote Comment Usman U. Expert Physics and Mathematics Tutor - Physics M.Phil Student Newton’s second law of motion: is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned. 2 years ago 0 votes Vote Comment Mohamed C. Experienced Math Tutor (All levels) Newton's second law of motion, also known as the law of acceleration, states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be represented as: F = ma Where: F represents the net force acting on the object, m represents the mass of the object, and a represents the acceleration produced. This equation indicates that the force exerted on an object is equal to the mass of the object multiplied by the acceleration it experiences. Furthermore, it implies that the object will accelerate in the direction of the net force applied to it. In simpler terms, Newton's second law can be summarized as follows: The acceleration of an object depends on the strength of the force applied to it and the object's mass. If a larger force is applied to an object, it will experience a greater acceleration, while a smaller force will result in a smaller acceleration. Similarly, if the mass of an object is increased, the same force will produce a smaller acceleration, whereas a smaller mass will lead to a greater acceleration for the same force. 2 years ago 0 votes Vote Comment Get Instant Answers with AI Tutor Try For Free Popular Topics Math English Calculus French Chemistry Physics Computer Science ESL Statistics Python Trending Tutors Math Tutors English Tutors Calculus Tutors French Tutors Chemistry Tutors Physics Tutors Computer Science Tutors ESL Tutors Statistics Tutors Python Tutors
188284
https://medlineplus.gov/download/genetics/condition/lesch-nyhan-syndrome.pdf
Reprinted from MedlinePlus Genetics ( 1 Lesch-Nyhan syndrome Description Lesch-Nyhan syndrome is a condition that occurs almost exclusively in males. It is characterized by neurological and behavioral abnormalities and the overproduction of uric acid. Uric acid is a waste product of normal chemical processes and is found in blood and urine. Excess uric acid can be released from the blood and build up under the skin and cause gouty arthritis (arthritis caused by an accumulation of uric acid in the joints). Uric acid accumulation can also cause kidney and bladder stones. The nervous system and behavioral disturbances experienced by people with Lesch-Nyhan syndrome include abnormal involuntary muscle movements, such as tensing of various muscles (dystonia), jerking movements (chorea), and flailing of the limbs ( ballismus). People with Lesch-Nyhan syndrome usually cannot walk, require assistance sitting, and generally use a wheelchair. Self-injury (including biting and head banging) is the most common and distinctive behavioral problem in individuals with Lesch-Nyhan syndrome. Frequency The prevalence of Lesch-Nyhan syndrome is approximately 1 in 380,000 individuals. This condition occurs with a similar frequency in all populations. Causes Mutations in the HPRT1 gene cause Lesch-Nyhan syndrome. The HPRT1 gene provides instructions for making an enzyme called hypoxanthine phosphoribosyltransferase 1. This enzyme is responsible for recycling purines, a type of building block of DNA and its chemical cousin RNA. Recycling purines ensures that cells have a plentiful supply of building blocks for the production of DNA and RNA. HPRT1 gene mutations that cause Lesch-Nyhan syndrome result in a severe shortage ( deficiency) or complete absence of hypoxanthine phosphoribosyltransferase 1. When this enzyme is lacking, purines are broken down but not recycled, producing abnormally high levels of uric acid. For unknown reasons, a deficiency of hypoxanthine phosphoribosyltransferase 1 is associated with low levels of a chemical messenger in the brain called dopamine. Dopamine transmits messages that help the brain control physical movement and emotional behavior, and its shortage may play a role in the Reprinted from MedlinePlus Genetics ( 2 movement problems and other features of this disorder. However, it is unclear how a shortage of hypoxanthine phosphoribosyltransferase 1 causes the neurological and behavioral problems characteristic of Lesch-Nyhan syndrome. Some people with HPRT1 gene mutations produce some functional enzyme. These individuals are said to have Lesch-Nyhan variant. The signs and symptoms of Lesch-Nyhan variant are often milder than those of Lesch-Nyhan syndrome and do not include self-injury. Learn more about the gene associated with Lesch-Nyhan syndrome • HPRT1 Inheritance This condition is inherited in an X-linked recessive pattern. The gene associated with this condition is located on the X chromosome, which is one of the two sex chromosomes. In males (who have only one X chromosome), one altered copy of the gene in each cell is sufficient to cause the condition. In females (who have two X chromosomes), a mutation would have to occur in both copies of the gene to cause the disorder. Because it is unlikely that females will have two altered copies of this gene, males are affected by X-linked recessive disorders much more frequently than females. A characteristic of X-linked inheritance is that fathers cannot pass X-linked traits to their sons. Other Names for This Condition • Choreoathetosis self-mutilation syndrome • Complete HPRT deficiency • Complete hypoxanthine-guanine phosphoribosyltransferase deficiency • Deficiency of guanine phosphoribosyltransferase • Deficiency of hypoxanthine phosphoribosyltransferase • HGPRT deficiency • Hypoxanthine guanine phosphoribosyltransferase deficiency • Hypoxanthine phosphoribosyltransferase deficiency • Juvenile gout, choreoathetosis, mental retardation syndrome • Juvenile hyperuricemia syndrome • Lesch-Nyhan disease • LND • LNS • Primary hyperuricemia syndrome • Total HPRT deficiency • Total hypoxanthine-guanine phosphoribosyl transferase deficiency Reprinted from MedlinePlus Genetics ( 3 • X-linked hyperuricemia • X-linked primary hyperuricemia • X-linked uric aciduria enzyme defect Additional Information & Resources Genetic Testing Information • Genetic Testing Registry: Lesch-Nyhan syndrome ( conditions/C0023374/) Genetic and Rare Diseases Information Center • Lesch-Nyhan syndrome ( Patient Support and Advocacy Resources • National Organization for Rare Disorders (NORD) ( Clinical Trials • ClinicalTrials.gov ( syndrom e%22) Catalog of Genes and Diseases from OMIM • LESCH-NYHAN SYNDROME; LNS ( • HYPERURICEMIA, HPRT-RELATED; HRH ( Scientific Articles on PubMed • PubMed ( MAJR%5D%29+AND+%28Lesch-Nyhan+syndrome%5BTIAB%5D%29+AND+englis h%5Bla%5D+AND+human%5Bmh%5D+AND+%22last+1800+days%22%5Bdp%5D ) References • Ceballos-Picot I, Mockel L, Potier MC, Dauphinot L, Shirley TL, Torero-Ibad R, Fuchs J, Jinnah HA. Hypoxanthine-guanine phosphoribosyl transferase regulatesearly developmental programming of dopamine neurons: implications for Reprinted from MedlinePlus Genetics ( 4 Lesch-Nyhandisease pathogenesis. Hum Mol Genet. 2009 Jul 1;18(13):2317-27. doi: 10.1093/hmg/ddp164. Epub 2009 Apr 2. Citation on PubMed ( m.nih.gov/19342420) or Free article on PubMed Central ( ov/pmc/articles/PMC2694685/) • Deutsch SI, Long KD, Rosse RB, Mastropaolo J, Eller J. Hypothesized deficiencyof guanine-based purines may contribute to abnormalities of neurodevelopment, neuromodulation, and neurotransmission in Lesch-Nyhan syndrome. ClinNeuropharmacol. 2005 Jan-Feb;28(1):28-37. doi:10.1097/01.wnf.0000152043. 36198.25. Citation on PubMed ( • Ernst M, Zametkin AJ, Matochik JA, Pascualvaca D, Jons PH, Hardy K, HankersonJG, Doudet DJ, Cohen RM. Presynaptic dopaminergic deficits in Lesch-Nyhandisease. N Engl J Med. 1996 Jun 13;334(24):1568-72. doi:10.1056/ NEJM199606133342403. Citation on PubMed ( 8337) • Jinnah HA, De Gregorio L, Harris JC, Nyhan WL, O'Neill JP. The spectrum ofinherited mutations causing HPRT deficiency: 75 new cases and a review of 196previously reported cases. Mutat Res. 2000 Oct;463(3):309-26. doi:10.1016/ s1383-5742(00)00052-1. Citation on PubMed ( 8746) • Jinnah HA, Visser JE, Harris JC, Verdu A, Larovere L, Ceballos-Picot I,Gonzalez-Alegre P, Neychev V, Torres RJ, Dulac O, Desguerre I, Schretlen DJ,Robey KL, Barabas G, Bloem BR, Nyhan W, De Kremer R, Eddey GE, Puig JG, Reich SG; Lesch-Nyhan Disease International Study Group. Delineation of the motor disorderof Lesch-Nyhan disease. Brain. 2006 May;129(Pt 5):1201-17. doi:10.1093/brain/ awl056. Epub 2006 Mar 20. Citation on PubMed ( 549399) or Free article on PubMed Central ( s/PMC3508431/) • Nyhan WL. Dopamine function in Lesch-Nyhan disease. Environ Health Perspect. 2000 Jun;108 Suppl 3(Suppl 3):409-11. doi: 10.1289/ehp.00108s3409. Citation on PubMed ( or Free article on PubMed Central ( • Nyhan WL. The recognition of Lesch-Nyhan syndrome as an inborn error of purinemetabolism. J Inherit Metab Dis. 1997 Jun;20(2):171-8. doi:10.1023/a: 1005348504512. Citation on PubMed ( • Puig JG, Torres RJ, Mateos FA, Ramos TH, Arcas JM, Buno AS, O'Neill P. Thespectrum of hypoxanthine-guanine phosphoribosyltransferase (HPRT) deficiency. Clinical experience based on 22 patients from 18 Spanish families. Medicine( Baltimore). 2001 Mar;80(2):102-12. doi: 10.1097/00005792-200103000-00003. Citation on PubMed ( • Saito Y, Takashima S. Neurotransmitter changes in the pathophysiology ofLesch-Nyhan syndrome. Brain Dev. 2000 Sep;22 Suppl 1:S122-31. doi:10.1016/s0387-7604(00)00143-1. Citation on PubMed ( • Visser JE, Bar PR, Jinnah HA. Lesch-Nyhan disease and the basal ganglia. BrainRes Brain Res Rev. 2000 Apr;32(2-3):449-75. doi: 10.1016/s0165-0173(99) 00094-6. Citation on PubMed ( Reprinted from MedlinePlus Genetics ( 5 Last updated February 1, 2013
188285
https://kids.wordsmyth.net/we/?rid=56711&as_level=2
| | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- | online dictionary-thesaurus | | | | | | --- --- | | | | | | --- | Log In | | | Register | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- | home | | | subscription | | | feedback | | | about us | | | blog | | | widget | | | FAQ | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | --- --- --- | | | | | | | | | Dictionaries | | | Comprehensive | | Children's | | WILD (Illustrated) | | Word Exploration | | | Word Explorer | | Browse | | Search Filter | | Reverse Search | | A-Z Word Parts | | Puzzle Solvers | | | Anagram Solver | | Crossword Solver | | Teacher Tools | | | Classes | | Students | | Lessons | | Assignments | | Reports | | Vocabulary Center | | | Activities | | Wordlist Maker | | Writing Tool (Beta) | | Legacy activities | | My Wordsmyth | | | Lookup History | | My Wordlists | | Legacy activities | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- | Word Explorer Children's Dictionary Suite | | | --- | | | Elementary dictionary | | | Intermediate dictionary | Help | | | --- | | Help | Help | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | Intermediate Dictionary | | | See entry in Elementary Dictionary | | | More results | | | Show multi-word results | | Browse in wordlist | | See entries that contain "fire truck" | | | Display options | | | --- | | | Show syllables | | | Show Lookup History | | | Double-Click Lookup | | | Show Word Parts | | | Show Spanish support | | | Show Chinese support | | | Pronunciation | | | --- | | | Wordsmyth | | | Phonics | | | IPA | | | | | | | Lookup History | | fire truck | | | | | --- | | fire truck | | pronunciation: faIr truhk | | | --- | | part of speech: | noun | | definition: | a large truck that firefighters ride in to go to a fire and that carries the things firefighters need to put out fires; fire engine. Several firefighters got out of the fire truck and worked to put out the fire. | | | | | | | | | | | | | | | | | | | | | | | | | | | Subscribe for ad-free Wordsmyth and more Learn more | | | | | | | | | | | | | | | | | | --- --- --- --- --- | Home | Send Feedback | Having a problem? | Suggest a Word | Privacy Policy | Privacy Manager | | | | | --- | | | | | ©2025 Wordsmyth | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- | online dictionary-thesaurus | | | | | | --- --- | | | | | | --- | Log In | | | Register | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- | home | | | subscription | | | feedback | | | about us | | | blog | | | widget | | | FAQ | | | | | | | | --- --- | | | | | | --- | Log In | | | Register | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- | home | | | subscription | | | feedback | | | about us | | | blog | | | widget | | | FAQ | | | | | | --- | Log In | | | Register | | | | | | | | | | | | | | | --- --- --- --- --- --- | home | | | subscription | | | feedback | | | about us | | | blog | | | widget | | | FAQ | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | --- --- --- | | | | | | | | | Dictionaries | | | Comprehensive | | Children's | | WILD (Illustrated) | | Word Exploration | | | Word Explorer | | Browse | | Search Filter | | Reverse Search | | A-Z Word Parts | | Puzzle Solvers | | | Anagram Solver | | Crossword Solver | | Teacher Tools | | | Classes | | Students | | Lessons | | Assignments | | Reports | | Vocabulary Center | | | Activities | | Wordlist Maker | | Writing Tool (Beta) | | Legacy activities | | My Wordsmyth | | | Lookup History | | My Wordlists | | Legacy activities | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- | Word Explorer Children's Dictionary Suite | | | --- | | | Elementary dictionary | | | Intermediate dictionary | Help | | | --- | | Help | Help | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | Intermediate Dictionary | | | See entry in Elementary Dictionary | | | More results | | | Show multi-word results | | Browse in wordlist | | See entries that contain "fire truck" | | | Display options | | | --- | | | Show syllables | | | Show Lookup History | | | Double-Click Lookup | | | Show Word Parts | | | Show Spanish support | | | Show Chinese support | | | Pronunciation | | | --- | | | Wordsmyth | | | Phonics | | | IPA | | | | | | | Lookup History | | fire truck | | | | | --- | | fire truck | | pronunciation: faIr truhk | | | --- | | part of speech: | noun | | definition: | a large truck that firefighters ride in to go to a fire and that carries the things firefighters need to put out fires; fire engine. Several firefighters got out of the fire truck and worked to put out the fire. | | | | | | | | | | | | | | | | | | | | | | | | | | | Subscribe for ad-free Wordsmyth and more Learn more | | | | | | | | | | | --- --- --- | | | | | | | | | Dictionaries | | | Comprehensive | | Children's | | WILD (Illustrated) | | Word Exploration | | | Word Explorer | | Browse | | Search Filter | | Reverse Search | | A-Z Word Parts | | Puzzle Solvers | | | Anagram Solver | | Crossword Solver | | Teacher Tools | | | Classes | | Students | | Lessons | | Assignments | | Reports | | Vocabulary Center | | | Activities | | Wordlist Maker | | Writing Tool (Beta) | | Legacy activities | | My Wordsmyth | | | Lookup History | | My Wordlists | | Legacy activities | | | | | | | | | | | | Comprehensive | | Children's | | WILD (Illustrated) | | | | Word Explorer | | Browse | | Search Filter | | Reverse Search | | A-Z Word Parts | | | | Anagram Solver | | Crossword Solver | | | | Classes | | Students | | Lessons | | Assignments | | Reports | | | | Activities | | Wordlist Maker | | Writing Tool (Beta) | | Legacy activities | | | | Lookup History | | My Wordlists | | Legacy activities | | | | | | | | | | | --- --- --- --- | Word Explorer Children's Dictionary Suite | | | --- | | | Elementary dictionary | | | Intermediate dictionary | Help | | | --- | | Help | Help | | | | | | | | | | --- | | | Elementary dictionary | | | Intermediate dictionary | | | | --- | | Help | Help | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | Intermediate Dictionary | | | See entry in Elementary Dictionary | | | More results | | | Show multi-word results | | Browse in wordlist | | See entries that contain "fire truck" | | | Display options | | | --- | | | Show syllables | | | Show Lookup History | | | Double-Click Lookup | | | Show Word Parts | | | Show Spanish support | | | Show Chinese support | | | Pronunciation | | | --- | | | Wordsmyth | | | Phonics | | | IPA | | | | | | | Lookup History | | fire truck | | | | | See entry in Elementary Dictionary | | | | Show multi-word results | | Browse in wordlist | | See entries that contain "fire truck" | | | | --- | | | Show syllables | | | Show Lookup History | | | Double-Click Lookup | | | Show Word Parts | | | Show Spanish support | | | Show Chinese support | | | Pronunciation | | | --- | | | Wordsmyth | | | Phonics | | | IPA | | | | | --- | | | Wordsmyth | | | Phonics | | | IPA | | | | Lookup History | | fire truck | | | | --- | | fire truck | | fire truck | | | --- | | part of speech: | noun | | definition: | a large truck that firefighters ride in to go to a fire and that carries the things firefighters need to put out fires; fire engine. 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188286
https://www1.cmc.edu/pages/faculty/lenny/papers/Optimization_lecture_notes.pdf
Topics in Discrete Optimization Lenny Fukshansky Contents Chapter 1. Preface i 1.1. What is discrete optimization? i 1.2. Asymptotic notation and computational complexity ii Chapter 2. Knapsack and Frobenius problems 1 2.1. Complexity of knapsack problems 1 2.2. Approximating knapsack problem by relaxation 6 2.3. LP-polytope 9 2.4. Integer knapsack and counting integer lattice points in polytopes 12 2.5. Frobenius problem 18 2.6. Problems 25 Chapter 3. Geometry of Numbers 27 3.1. Lattices 27 3.2. Theorems of Blichfeldt and Minkowski 35 3.3. Successive minima 38 3.4. Inhomogeneous minimum 43 3.5. Problems 46 Chapter 4. Lattice Problems, Connections and Applications 48 4.1. Sphere packing, covering and kissing number problems 48 4.2. Lattice packings in dimension 2 54 4.3. Algorithmic problems on lattices 59 4.4. CVP is NP-hard 62 4.5. Geometry of the Frobenius problem 63 4.6. Lattice point counting 66 4.7. Applications of lattices in coding theory and cryptography 69 4.8. Euclidean frames 71 4.9. Problems 76 Bibliography 77 v CHAPTER 1 Preface 1.1. What is discrete optimization? Let f(x) be a multi-variable function defined on some domain D. An optimiza-tion problem defined by f on D can then be formulated as: Maximize / minimize f(x) on D. An optimization problem like this is called discrete if the domain D is a discrete set inside of some topological space, i.e. if every point of D is an isolated point (i.e., no open neighborhood of a point in D contains any other points of D). The condition of a point belonging to the domain D can often be formulated as a collection of certain constraints or inequalities. In these notes, we will discuss several central discrete optimization problems, coming from different (but related) areas of discrete mathematics. Our goal will be to describe the problems with the necessary context and background they come from while focusing on their interpretation in the scope of discrete geometry. It is this geometric framework that naturally connects all the problems we will discuss and brings them together. Here are the specific problems we will be interested in: • The Knapsack Problems: given a collection of objects with assigned weight and cost, maximize the objective function (i.e. total cost) while keeping the weight under the specified threshold (subject to possibly some additional constraints). In addition to its intrinsic mathematical signifi-cance, this problem often comes up in resource allocation. • The Frobenius Problem: given a collection of relatively prime positive integers, find the largest positive integer that cannot be represented as their nonnegative integer linear combination. This problem appears in many areas of mathematics and is related to the knapsack problems. • The Main Problem of Coding Theory: maximize the error-correcting capability of a linear code while keeping its codeword length bounded. This problem is central in the study of accurate data transmission over potentially noisy channels. • Optimization Problems on Lattices: optimize packing density, cov-ering thickness and kissing number of a Euclidean lattice in n dimensions. This is the main problem of lattice theory, a branch of mathematics at the intersection of number theory and discrete geometry. In addition to its theoretical value, it has numerous applications, for instance in digital and wireless communications. • Coherence Minimization on Euclidean Frames: find frames (overde-termined spanning sets) in Euclidean vector spaces of large cardinality and small coherence. Such frames allow for sufficiently fast data transmission with efficient erasure-recovery capabilities. i ii 1. PREFACE 1.2. Asymptotic notation and computational complexity The main measure of “hardness” of different problems that we will discuss is given by their computational complexity. Here, we briefly and somewhat informally recall some basic notions of computational complexity. To start with, we need some asymptotic notation. Given two functions f, g : R →R, we write f(x) = O(g(x)) if there exists a real constant C so that f(x) ≤Cg(x) as x →∞. This is called big-O notation. We use big-O notation to assess the running time of an algorithm. The model computer used for algorithmic analysis is a Turing machine, as introduced by Alan Turing in 1936. Roughly speaking, this is an abstract computational device, a good practical model of which is a modern computer. It consists of an infinite tape subdivided into cells which passes through a head. The head can do the following four elementary operations: write a symbol into one cell, read a symbol from one cell, fast forward one cell, rewind one cell. These correspond to elementary operations on a computer, which uses symbols from a binary alphabet 0, 1. The number of such elementary operations required for a given algorithm is referred to as its running time. Running time is usually measured as a function of the size of the input, that is the number of cells of the infinite tape required to store the input. If we express this size as an integer n and the running time as a function f(n), then an algorithm is said to run in polynomial time if f(n) = O(nk) for some constant exponent k independent of n. We refer to the class of problems that can be solved in polynomial time as the P class. This is our first example of a computational complexity class. For some problems we may not know whether it is possible to solve them in polynomial time, but given a potential answer we can verify whether it is correct or not in polynomial time. Such problems are said to lie in the NP computational complexity class, where NP stands for non-deterministic polynomial. One of the most important open problems in contemporary mathematics (and arguably the most important problem in theoretical computer science) asks whether P = NP? In other words, if an answer to a problem can be verified in polynomial time, can this problem be solved by a polynomial-time algorithm? Most frequently this question is asked about decision problem, that is problems the answer to which is YES or NO. This problem, commonly known as P vs NP, was originally posed in 1971 independently by Stephen Cook and by Leonid Levin. It is believed by most experts that P ̸= NP, meaning that there exist problems answer to which can be verified in polynomial time, but which cannot be solved in polynomial time. For the purposes of thinking about the P vs NP problem, it is quite helpful to introduce the following additional notions. A problem is called NP-hard if it is “at least as hard as any problem in the NP class”, meaning that for each problem in the NP class there exists a polynomial-time algorithm using which our problem can be reduced to it. A problem is called NP-complete if it is NP-hard and is know to lie in the NP class. Now suppose that we wanted to prove that P = NP. One way to do this would be to find an NP-complete problem which we can show is in the P class. Since it is NP, and is at least as hard as any NP problem, this would mean that all NP problems are in the P class, and hence the equality would be proved. Although this equality seems unlikely to be true, this argument still presents serious motivation to study NP-complete problems. CHAPTER 2 Knapsack and Frobenius problems 2.1. Complexity of knapsack problems The problems we will discuss here fit into the general linear programming or linear optimization paradigm. A linear program (LP) is a problem that can be stated in the following form: Given vectors c, b and a matrix A, find a vector x ≥0 that maximizes the objective function c⊤x subject to the constraint Ax ≤b. A linear program is called an integer linear program (ILP) or simply an integer program (IP) if the solution vector x is required to have integer coordinates. Suppose we have a knapsack that can hold weight no more than W. We want to pack it with objects of types 1 through n where an object of type i has weight wi and price pi. Our objective is to maximize the value of the knapsack. If we write xi for the number of objects of type i that we take, we have the following optimization problem: Maximize the objective function n X i=1 pixi under the constraint n X i=1 wixi ≤W. This is the basic prototype of a knapsack problem. Putting on additional constraints distinguishes different types of knapsack problems, for instance: • Binary knapsack problem (BKP): the variables xi can take values 0, 1 only • Bounded knapsack problem (BndKP): for each i, xi is an integer with 0 ≤xi ≤bi for some upper bounds bi. • Unbounded knapsack problem (UbndKP): for each i, xi is an inte-ger. • Subset-sum problem (SSP): for each i, wi = pi and xi = 0, 1. All of these problems are NP-hard. To show this, first observe that SSP is the special case of BKP with pi = wi for each i; also, BKP is the special case of BndKP with bi = 1 for each i. On the other hand, UbndKP is clearly at least as hard as BndKP. Hence, NP-hardness of all of these problems follows from NP-completeness of SSP. We will show that SSP is NP-complete, more specifically we will deal with the the decision version of SSP: 1 2 2. KNAPSACK AND FROBENIUS PROBLEMS Given a set of weights S = {w1, . . . , wn} and a target value t, is there a subset S′ ⊆S such that P wi∈S′ wi = t? We first need some notation from Boolean logic. A Boolean formula is an expression built from Boolean variables (taking values TRUE = 1 or FALSE = 0) and operators AND (∧), OR (∨), NOT (¬) and parentheses separating different clauses of the formula. A Boolean formula is said to be satisfiable if it can be made TRUE by an appropriate assignment of variables. Example 2.1.1. The formula x ∧¬y is satisfiable: setting x = TRUE, y = FALSE makes this formula TRUE. On the other hand, the formula x ∧¬x is not satisfiable. The (unrestricted) Boolean satisfiability problem SAT is the problem of determining if a given Boolean formula is satisfiable or not. It can be formally stated as the following decision problem: INPUT: Boolean formula OUTPUT: YES (satisfiable) or NO (not satisfiable) Theorem 2.1.1 (S. Cook (1971), L. Levin (1973)). SAT is NP-complete. This was the first provable instance of an NP-complete problem – the notion did not properly exist before the work of Cook and Levin, who independently established this result. The fact that SAT is NP is not difficult to see: given any assignment of the variables, it can be verified in polynomial time whether they make the given formula TRUE or not. To show that it is NP-complete, one needs to prove that any NP problem can be reduced to an instance of SAT by a polynomial time algorithm. We do not prove this result here, however we will mention a (restricted) variation of the SAT problem, called 3-SAT which is also known to be NP-complete: 3-SAT is the Boolean satisfiability problem, where every clause consists of no more than 3 literals (a literal is either a variable x, or negation of a variable ¬x). In fact, every instance of a SAT formula can be transformed into a 3-SAT formula as follows. First notice that every formula can be rewritten in a way that the clauses are joined by ∧operator: for example, a formula like (x1 ∧y1) ∨(x2 ∧y2) can be transformed into (x1 ∨x2) ∧(y1 ∨x2) ∧(x1 ∨y2) ∧(y1 ∨y2). Such a form is called a generalized conjunctive normal form for a given Boolean formula. Suppose now that there is a clause in an unrestricted SAT formula that looks like ℓ1 ∨· · · ∨ℓn, where ℓ1, . . . , ℓn are literals. Introducing new variables x1, . . . , xn−2 we can rewrite this formula as (ℓ1 ∨ℓ2 ∨x1) ∧(¬x1 ∨ℓ3 ∨x2) ∧· · · ∧(¬xn−3 ∨ℓn−2 ∨xn−2) ∧(¬xn−2 ∨ℓn−1 ∨ℓn). 2.1. COMPLEXITY OF KNAPSACK PROBLEMS 3 This new formula is satisfiable if and only if the original is, and its length is at most 3 times longer than the original, which means that the reduction from SAT to 3-SAT implies only polynomial growth in the length of the formula. Hence SAT and 3-SAT have the same order of computational complexity, i.e. they are both NP-complete by the Cook-Levin theorem. We are now ready to show (somewhat informally) that the subset-sum problem is NP-complete by constructing a polynomial time reduction algorithm from 3-SAT to it. Theorem 2.1.2. The decision version of SSP is NP-complete. Sketch of proof. Recall that the running time is measured as a function of the input size. It is easy to see that our problem is NP. Indeed, let S = {w1, . . . , wm} be the set of weights and T the target sum. Given a specific subset S′ ⊆S, it is simply a summation problem to verify whether S′ sums to T – this summation algorithm runs in polynomial (in fact, linear) time. Now we show that the decision version of SSP is NP-hard by constructing a polynomial-time reduction from 3-SAT to it. Let us write n for the number of variables and k for the number of clauses in our Boolean formula. We demonstrate this reduction on an example. Consider the Boolean formula (x1 ∨x2 ∨x3) ∧(¬x1 ∨¬x2 ∨x3) ∧(x1 ∨¬x2 ∨x3) ∧(x1 ∨x2 ∨¬x3). There are four clauses which we label as c1, c2, c3, c4, so in this example n = 3, k = 4. This formula is satisfied if and only if each of the clauses is TRUE. Let us introduce two variables vi1, vi2 for each of the Boolean variables xi and two variables wi1, wi2 for each of the clauses ci along with n −1 auxiliary variables si1, . . . , si(n−1) for each ci. With this notation, let us build a table consisting of four blocks as follows: x1 x2 x3 c1 c2 c3 c4 v11 1 0 0 1 0 1 1 v12 1 0 0 0 1 0 0 v21 0 1 0 1 0 0 1 v22 0 1 0 0 1 1 0 v31 0 0 1 1 1 1 0 v32 0 0 1 0 0 0 1 s11 0 0 0 1 0 0 0 s12 0 0 0 1 0 0 0 s21 0 0 0 0 1 0 0 s22 0 0 0 0 1 0 0 s31 0 0 0 0 0 1 0 s32 0 0 0 0 0 1 0 s41 0 0 0 0 0 0 1 s42 0 0 0 0 0 0 1 T 1 1 1 3 3 3 3 The values in the cells of the table are assigned as follows. • The variable vi1 stands for the TRUE value of xi and vi2 stands for the FALSE value of xi. Hence the cell corresponding to vi1, xj or vi2, xj gets a 1 if i = j and 0 if i ̸= j. 4 2. KNAPSACK AND FROBENIUS PROBLEMS • The cell corresponding to vi1, cj gets a 1 if setting xi = TRUE makes cj TRUE, and 0 otherwise. The cell corresponding to vi2, cj gets a 1 if setting xi = FALSE makes cj TRUE, and 0 otherwise. • The cell corresponding to sil, xj gets a 0 for all i, l, j. • The cell corresponding to sil, cj gets a 1 if i = j and 0 otherwise for each 1 ≤l ≤n −1. • The cells in the row labeled T (target sum) corresponding to a variable xi get a 1 and those corresponding to a clause ci get an n = number of variables. Now let S be the (multi-) set of numbers as written in rows except for the last one and the last row be the target sum T, so in our example S = {1001011, 1000100, 101001, 100110, 11110, 10001, 1000, 1000, 100, 100, 10, 10, 1, 1} and T = 1113333. We claim that the Boolean formula represented by this table is satisfiable if and only if there exists a subset S′ of S which sums up to T. Indeed, notice that each row in the upper part of the table corresponds to a TRUE or FALSE value of the variable xi (we would always pick precisely one of the rows vi1 and vi2, since xi must be assigned TRUE or FALSE but not both at the same time, and hence the corresponding digit of T would always be 1). Then we pick the rows in the bottom part of the table to compensate for those positions that are < n: the formula is not satisfiable if and only if there is a column corresponding to some ci whose entries add up to a number < n (this happens precisely when there is no choice of the variables making the clause ci TRUE). In our example, the Boolean formula is satisfiable, since for instance the rows corresponding to v11, v21, v31 and s21, s22, s31, s41 add up to T: 1001011 + 101001 + 11110 + 100 + 100 + 10 + 1 = 1113333. Indeed, this choice of the subset S′ corresponds to the assignment x1 = TRUE, x2 = TRUE, x3 = TRUE, which makes each of the clauses TRUE, hence making the formula TRUE. Notice that each element of S has at most n + k digits in it and there are at most 2n + (n −1)k elements in S. This ensures that this reduction procedure runs in polynomial time in the size of the input, which is itself a function of n and k. This completes the proof. □ Many of the problems mentioned in this section (e.g. ILP, 3-SAT and SSP) are among the original Karp’s 21 NP-complete problems. In 1972, Richard Karp published a paper [Kar72] in which he showed the NP-completeness of 21 different natural combinatorial and graph theoretic problems. His main tool was Cook-Levin Theorem: he constructed polynomial time reductions from SAT to several problems and then used the Cook-Levin Theorem to establish NP-completeness of these problems. He then used these few original problems to show NP-completeness of the rest of his list of 21. To close this section, let us also mention a simple but curious reformulation of the SAT problem in terms of polynomial vanishing. For a Boolean formula in conjunctive normal form (CNF), we construct a CNF-polynomial in the variables x1, . . . , xn corresponding to this formula as follows: 2.1. COMPLEXITY OF KNAPSACK PROBLEMS 5 (1) A literal xi becomes a linear factor xi and a literal ¬xi becomes a linear factor 1 −xi. (2) Each clause becomes a product of linear factors corresponding to its liter-als, so disjunction becomes multiplication. We refer to such products as literal-monomials. (3) Conjunction becomes addition. Let us consider the following example: (2.1) (x1 ∨x2 ∨x3) ∧(¬x1 ∨¬x2 ∨x3) ∧(x1 ∨¬x2 ∨x3) ∧(x1 ∨x2 ∨¬x3), then the formula (2.1) correspond to the CNF-polynomial (2.2) f(x1, x2, x3) = x1x2x3 + (1 −x1)(1 −x2)x3 + x1(1 −x2)x3 + x1x2(1 −x3). Assign the values 0 = TRUE and 1 = FALSE. A CNF formula is satisfiable if and only if there exists a TRUE / FALSE assignment of the Boolean variables such that each clause is TRUE. This happens if and only if there exists a 0 / 1 assignment of the variables in the corresponding CNF-polynomial which makes every monomial 0. This property is equivalent to the CNF-polynomial vanishing at some vertex of the unit cube [0, 1]n in Rn. For example, the polynomial in (2.2) vanishes at (0, 1, 1); this corresponds to the assignment of the Boolean variables x1 = TRUE, x2 = FALSE, x3 = FALSE, which indeed satisfies the formula (2.1). We can use this construction to prove the following observation. Proposition 2.1.3. The problem of determining whether a given multilinear poly-nomial of degree k ≥3 in n ≥2 variables vanishes at a vertex of the unit cube [0, 1]n in Rn is NP-complete. Proof. It is clear that this problem is NP: the procedure of evaluating a poly-nomial at a given point has polynomial complexity. The construction above shows a polynomial-time reduction from k-SAT, the Boolean k-satisfiability problem, to our problem for the corresponding CNF-polynomial. Since k-SAT is NP-hard for every k ≥3, so must be our problem. □ 6 2. KNAPSACK AND FROBENIUS PROBLEMS 2.2. Approximating knapsack problem by relaxation The Binary knapsack problem (BKP) can be formulated as follows: maximize n X i=1 pixi subject to n X i=1 wixi ≤W, xi ∈{0, 1} ∀1 ≤i ≤n. Let us write p = (p1, . . . , pn) for the profit vector and w = (w1, . . . , wn) for the weight vector. This is arguably the most important of the knapsack problems. Indeed, SSP is a special case of BKP and BndKP can be reduced to BKP (in a larger number of variables) as we now show. BndKP can be formulated as follows: maximize n X i=1 pixi subject to n X i=1 wixi ≤W, xi ∈Z, 0 ≤xi ≤bi ∀1 ≤i ≤n. For each 1 ≤i ≤n, let Bi := {bi1, . . . , bili} be a minimal (with respect to size) partition of bi so that every integer between 0 and bi is representable as a sum of some subcollection of Bi: such a partition always exists, since in the worst case scenario we can always take Bi = {1, . . . , 1}, but in general it will smaller. For example, if bi = 10 we can take Bi = {1, 2, 3, 4}. Then we can introduce new variables yik, 1 ≤i ≤n, 1 ≤k ≤li, and rewrite the BndKP above as the following instance of BKP: maximize n X i=1 pi(bi1yi1 + · · · + biliyili) subject to n X i=1 wi(bi1yi1 + · · · + biliyili) ≤W, yik ∈{0, 1} ∀1 ≤i ≤n, 1 ≤k ≤li. These two problems are equivalent since there is bijection between all values of xi between 0 and bi and all values (written without repetition) of the sum Pli k=1 bikyik as the variables yik assume values in the set {0, 1}. From Section 2.1 we know that there is no known polynomial time algorithm to solve BKP. However, we can look for an approximate solution to BKP via certain relaxations. First such relaxation is the continuous knapsack problem (CKP): maximize n X i=1 pixi subject to n X i=1 wixi ≤W, 0 ≤xi ≤1 ∀1 ≤i ≤n. 2.2. APPROXIMATING KNAPSACK PROBLEM BY RELAXATION 7 In other words, we no longer require xi to take only integer values. Let us assume that the items are ordered so that (2.3) p1 w1 ≥p2 w2 ≥· · · ≥pn wn . The following additional assumptions can be made for CKP: (1) Every weight wi ≤W. If this is not the case for some weight wi, this weight (and its corresponding price pi) can be eliminated, hence reducing the number of variables. (2) Pn i=1 wi > W. If this is not the case, then taking xi = 1 for each i will maximize the objective function. (3) The inequalities in (2.3) are all strict. Suppose not, then there exist some indices 1 < j < k < n such that c := pj wj = · · · = pk wk , which means that Pk i=j pixi = c Pk i=j wixi. Then set t = Pk i=j wi and define a new variable y = 1 t Pk i=j wixi; observe that 0 < y < 1. In this case we can restate our instance of CKP as follows: maximize j−1 X i=1 pixi + cty + n X i=k+1 pixi subject to j−1 X i=1 wixi + ty + n X i=k+1 wixi ≤W, 0 ≤xi ≤1 ∀i, 0 ≤y ≤1. To solve CKP, we define the critical index s = min ( j : j X i=1 wi > W ) . Then 1 < s ≤n and we have the following result. Theorem 2.2.1. [Dantzig, 1957] The optimal solution x∗to CKP is given by setting (2.4) x∗ i =  1 if 1 ≤i ≤s −1 0 if s + 1 ≤i ≤n, and x∗ s = 1 ws  W −Ps−1 j=1 wj  . Proof. Our proof follows [MT90]. First observe that a vector x = (x1, . . . , xn) maximizing the objective function must satisfy the condition (2.5) n X i=1 wixi = W, since otherwise some coordinates of x can be increased still under the weight re-striction, which will increase the value of the objective function. Arguing towards a contradiction, suppose the optimal solution x is not of the form x∗, say xi < 1 for some i < s. Then there must exist some index j ≥s such that xj > x∗ j. Now, for a sufficiently small ε > 0, replace xi by xi + ε and xj by xj −εwi/wj, hence 8 2. KNAPSACK AND FROBENIUS PROBLEMS still preserving condition (2.5). However, this change will increase the objective function by ε  pi −pjwi wj  , which is positive, since pi/wi > pj/wj. This contradicts the optimality of the solution x. The assumption xj > 0 for some j > s is handled analogously, also leading to a contradiction. Hence we must have the condition (2.4) satisfied for the optimal solution, and the formula for x∗ s follows from maximality. This completes the proof. □ The maximal value of the objective function in CKP is then easy to compute: it is s−1 X i=1 pi + ps ws  W − s−1 X j=1 wj  . This immediately implies an upper bound on the maximal value of the objective function for the associated instance of BKP: U := s−1 X i=1 pi +  ps ws  W − s−1 X j=1 wj    . There are other known relaxations of BKP (such as the Lagrangian relaxation, stemming from an application of the method of Lagrange multipliers) leading to other upper bounds on the objective function. There is also a lot of literature on the algorithmic complexity of computing these bounds. Some of the known algorithms rely on the greedy approach (making the locally optimal choice at each stage of the algorithm) as well as the branch-and-bound method (recursively splitting the search space into smaller pieces and optimizing on each of those). We do not get into this material here, but mention a book by Martello and Toth [MT90] as a comprehensive source of algorithmic information about knapsack problems. 2.3. LP-POLYTOPE 9 2.3. LP-polytope Our goal in this section is to present a geometric interpretation of the knapsack problems. First we need some geometric notation. Recall that a compact (i.e. closed and bounded) subset X ⊂Rn is called convex if for any pair x, y ∈X, tx + (1 −t)y ∈X for any 0 ≤t ≤1. An important special class of convex sets is convex hulls: the convex hull of a set X ⊂Rn is Co(X) = ( X x∈X txx : tx ≥0 ∀x ∈X, X x∈X tx = 1 ) . This is the smallest convex set (with respect to inclusion) containing X, so X is convex if and only if X = Co(X). A convex polytope is the convex hull of a finite collection of points. There is also a related notion of a convex polyhedron. A halfspace in Rn is a set H = ( x ∈Rn : n X i=1 aixi ≤b ) for some a1, . . . , an, b ∈R, and the set H = ( x ∈Rn : n X i=1 aixi = b ) is called a bounding hyperplane of H. A convex polyhedron is a compact intersection of a finite collection of halfspaces. Hence P is a convex polyhedron if and only if (2.6) P = {x ∈Rn : Ax ≤b} for an m × n real matrix A and a vector b ∈Rm such that this set is bounded. Theorem 2.3.1 (Minkowski-Weyl). A set P ⊂Rn is a convex polytope if and only if it is a convex polyhedron. While we do not prove this theorem here, we point out one of its important conse-quence. A point v in a convex set X ⊂Rn is called a vertex if there exists some c ∈Rn such that for all c⊤v < c⊤x for all x ∈X. Then every convex polyhedron is the convex hull of its vertices, of which there are only finitely many. More generally, we can define a k-dimensional face of an n-dimensional polytope P, 1 ≤k < n, to be a k-dimensional subset F ⊂P such that for some c ∈Rn, c⊤v = c⊤u ∀v, u ∈F and c⊤v < c⊤x ∀v ∈F, x ∈P. Here, by dimension of a subset F we mean dimR (spanR F) −1, so vertices are 0-dimensional faces of P and every face of P contains at least one vertex. The polytope P can then be represented as the disjoint union of its interior P o and its boundary ∂P, where ∂P is the union of all of the faces of P and P o = P \ ∂P. Given a linear program maximize p(x) = n X i=1 pixi subject to Ax ≤b 10 2. KNAPSACK AND FROBENIUS PROBLEMS for an m × n matrix A and a vector b ∈Rm we can define the corresponding LP-polytope as in (2.6). Then the linear program can be reformulated as maximize p(x) = n X i=1 pixi on P. In general, the polytope P can be unbounded, but we will focus specifically on the situations when it is compact – we refer to such linear programs as bounded LPs. Notice that our knapsack problems BKP, BndKP and SSP are all bounded LPs. With this notation, we can prove an important theorem. Theorem 2.3.2. The objective function p(x) is maximized at a vertex of P. In other words, there exists a vertex v ∈P such that p(v) ≥p(x) for all x ∈P. Proof. Suppose that v ∈P is a point such that p(v) ≥p(x) for all x ∈P, define b to be this maximal value, i.e. b = p(v). Define the hyperplane Hp(b) = {x ∈Rn : p(x) = b} and two halfspaces H1 = {x ∈Rn : p(x) ≤b} , H2 = {x ∈Rn : p(x) ≥b} . By our assumption, we must have P ⊂H1 and hence Hp(b) cannot intersect the interior of P, i.e. F = P ∩Hp(b) ̸= ∅must be some (union of) face(s) of P. Then p(x) is constant on F and F contains a vertex u of P, so p(u) = p(v) is a maximal value of p(x) on P. □ Therefore Theorem 2.3.2 implies that to solve a given linear program we need to find all the vertices of the corresponding LP-polytope and identify an optimal one among them. This is done by George Dantzig’s simplex algorithm. The main idea of the simplex algorithm is to start at a vertex of the LP-polytope and move along an edge (1-dimensional face) to a neighboring vertex corresponding to a larger value of the objective function. The algorithm terminates when no such vertex exists. While we will not get into the details of the algebraic implementation of this algorithm, we will demonstrate a geometric example with an instance of CKP. Example 2.3.1. Consider the following instance of CKP: maximize 3x1 + 5x2 + 2x3 subject to 2x1 + 5x2 + 7x3 ≤11, x1, x2, x3 ∈[0, 1]. Define the corresponding LP-polytope to be P =  x ∈R3 ≥0 : Ax ≤b , where A =     2 5 7 1 0 0 0 1 0 0 0 1    , b =     12 1 1 1    . We want to maximize the objective function p(x) = 3x1 + 5x2 + 2x3 on P. Notice that P is the intersection of the unit cube C = [0, 1]3 with the halfspace H = {x ∈R3 : 2x1 + 5x2 + 7x3 ≤12}, and hence vertices of P lie on the edges of the unit cube C. These vertices are not hard to find in this case – they are the origin 0, the standard basis vectors e1, e2, 2.3. LP-POLYTOPE 11 e3, their pairwise sums e1 +e2, e1 +e3, e2 +e3 (this last one lying in the bounding hyperplane of H), as well as the two more points in the bounding hyperplane of H: v1 = (1, 1, 5/7), v2 = (1, 3/5, 1). We can now describe the geometric idea of the simplex algorithm in this example. (1) Start at the vertex 0 and pick the direction towards any vertex connected to it by an edge of P, since p(0) = 0. Say, we pick e1. (2) Move to e1, where p(e1) = 3. Pick a neighboring vertex with a larger value of p, say e1 + e2. (3) Move to e1 +e2, where p(e1 +e2) = 8 > p(e1). Pick a neighboring vertex with a larger value of p, which is v1. (4) Move to v1, where p(v1) = 66/7 > p(e1 + e2). No neighboring vertex of v1 gives a larger value of p, thus we stop. (5) Return the maximum value of p on P, which is p(v1) = 66/7. Let us compare this result to the result yielded by Dantzig’s Theorem 2.2.1 (the assumptions for this theorem are satisfied here). Ordering the items in our instance of CKP so that (2.3) is satisfied, we have: p1 w1 = 3 2 > p2 w2 = 5 5 > p3 w3 = 2 7. Then the critical index is s = 3. Hence the theorem guarantees that the optimal solution is  1, 1, 1 7(12 −(2 + 5))  = (1, 1, 5/7), as expected. This example demonstrates that in case the case our linear program is an in-stance of CKP the simplex algorithm essentially reduces to Theorem 2.2.1, however it applies far more generally than just CKP which is a big advantage. This being said, it still only applies to instance of LP, not ILP, and hence it does not directly help us with the knapsack problems. On the other hand, a knapsack problem can be formulated as maximization problem for an objective function on the set of in-teger lattice points (i.e. points of Zn) inside of the specific compact LP-polytope defined by the corresponding constraints. Such a set is finite, so if we could find all these integer lattice points, we could simply evaluate our objective function at all of them and pick the largest value. While not necessarily efficient, this would lead to a solution. The problem is that integer lattice points in polytope are difficult not only to find, but even to count. We will discuss such a counting problem in more details in the next section. 12 2. KNAPSACK AND FROBENIUS PROBLEMS 2.4. Integer knapsack and counting integer lattice points in polytopes Let us start by defining a certain variation of the knapsack problems that is somewhat different from the previous versions we were discussing. This is a decision problem known as the integer knapsack problem (IKP): Given a set of weights S = {w1, . . . , wn} and the target value t, do there exist x1, . . . , xn ∈Z≥0 such that Pn i=1 wixi = t? Notice that this is a generalization of the decision version of SSP, where the variables x1, . . . , xn were only allowed to take on values 0 or 1. Thus this problem is NP-hard. We can reformulate this problem geometrically by introducing the knapsack polytope P(S, t) := ( x ∈Rn ≥0 : n X i=1 wixi = t ) . The problem then is to determine whether P(S, t) ∩Zn = ∅. In other words, we can ask whether the counting function |P(S, t) ∩Zn| > 0, which naturally leads to the question of counting integer lattice points in polytopes. This is the main focus of this section. We discuss the following general question: given a compact convex polytope P ⊂Rn, what is the number of integer lattice points in it? In other words, we want to find the quantity |P ∩Zn|. Let us start with the two-dimensional situation, where we can prove a beautiful formula for even a somewhat more general situation. Let P be a simple polygon (with no holes or self-intersections) in R2, not necessarily convex, with integer vertices. Let us write A(P) for the area of P, I(P) for the number of integer lattice points in the interior of P and B(P) for the number of integer lattice points on the boundary of P. The following famous theorem was proved by Georg Alexander Pick in 1899. Theorem 2.4.1 (Pick’s Theorem). A(P) = I(P) + 1 2B(P) −1. Sketch of Proof. Let x ∈P ∩Z2 and define αP (x) to be the visibility angle of P from x, i.e. it is the angle of the cone C(x) ∩P where C(x) is a unit circle centered at x. Notice that αP (x) =  2π if x is an interior point of P, π if x is on the boundary of P but not a vertex, and αP (x) is the corresponding interior angle of P if x is a vertex. Define the weight enumerator W(P) = X x∈P ∩Z2 αP (x) 2π . We will now sketch a proof of the formula A(P) = W(P). First observe that W(P) is additive, i.e. if P = P1 ∪P2 where P1 and P2 are polytopes sharing a piece of boundary then (2.7) W(P) = W(P1) + W(P2). To see this, notice that interior points of P1, P2 remain interior points of P, bound-ary points on non-overlapping parts of the boundary remain boundary points, 2.4. INTEGER KNAPSACK AND COUNTING INTEGER LATTICE POINTS IN POLYTOPES 13 whereas a boundary point on the common part of the boundary of P1 and P2 is either a vertex or becomes an interior point. If such a point x is a vertex, then αP (x) = αP1(x) + αP2(x); if x was on the joint boundary and became an interior point then it was counted with αP1(x) = π in W(P1), αP2(x) = π in W(P2) and will now be counted with αP (x) = 2π in W(P). The verification of the formula A(P) = W(P) for rectangles and triangles is done in Problem 2.3. Now notice that any polygon P can be split into a union of triangles with non-overlapping interiors but possibly joint boundaries. This obser-vation together with Problem 2.3 and (2.7) implies A(P) = W(P) for all polygons. Now let n = number of vertices of P, m = B(P) −n = number of boundary integer lattice points that are not vertices, and k = I(P) = number of internal integer lattice points. Let {x1, . . . , xn}, {y1, . . . , ym}, {z1, . . . , zk} be these sets of points, respectively. Recall that the sum of internal angles of P is (n −2)π – this formula holds in general whether P is convex or not. Then W(P) = 1 2π n X i=1 αP (xi) + m X i=1 αP (yi) + k X i=1 αP (zi) ! = n −2 2 + m 2 + k = k + n + m 2 −1 = I(P) + 1 2B(P) −1. This completes the proof. □ Next we discuss the problem of counting integer lattice points in convex poly-topes in dimensions ≥3. Specifically, we address the following question: how can we count the number of integer lattice points in homogeneous expansions of poly-topes? An area of mathematics that aims to answer this question is called Ehrhart theory. Let P ⊆Rn be a convex polytope such that Vol(P) > 0, and vertices of P are points of Zn: such P is called a lattice polytope. Write GP (t) = |tP ∩Zn| . We want to understand the behavior of GP (t) for all t ∈Z>0; specifically, we will prove a famous theorem of Ehrhart, which states that GP (t) is a polynomial in t. Our presentation closely follows [Ewa96]. First we consider a special case of polytopes, namely simplices. Lemma 2.4.2. Let a1, . . . , an ∈Zn be linearly independent, and define the simplex S = Co(0, a1, . . . , an) = ( n X i=1 tiai : ti ≥0 ∀1 ≤i ≤n, n X i=1 ti ≤1 ) . Then there exist β1, . . . , βn ∈Z≥0 such that for every t ∈Z>0, we have G(tS) = |tS ∩Zn| = n + t n  + n X i=1 n + t −i n  βi. Proof. Let A be the half-open parallelotope spanned by the vectors a1, . . . , an, i.e. A = ( n X i=1 tiai : 0 ≤ti < 1 ∀1 ≤i ≤n ) . 14 2. KNAPSACK AND FROBENIUS PROBLEMS For every y ∈tS ∩Zn there exists a unique representation of y of the form (2.8) y = x + n X i=1 αiai, where x ∈A ∩Zn and α1, . . . , αn ∈Z≥0. For each 0 ≤j ≤t, let Hj be the hyperplane which passes through the points ja1, . . . , jan. We will determine the number of points of Zn in Hj ∩tS, and the number of points of Zn ∩tS in the strips of space bounded by Hj−1 and Hj for each 1 ≤j ≤t; notice that H0 = {0}. First, let x = 0 in (2.8). Then y as in (2.8) lies in Hj if and only if (2.9) n X i=1 αi = j, 0 ≤αi ≤j ∀1 ≤i ≤n. We will prove now that there are precisely n+j−1 n−1  possibilities for α1, . . . , αn sat-isfying (2.9) for each j. We argue by induction on n. If n = 1, then there is only 1 = j 0  possibility. Suppose the claim is true for n−1. Then there are n+(j−αn)−2 n−2  possibilities for α1, . . . , αn−1 such that n−1 X i=1 αi = j −αn for each value of 0 ≤αn ≤j. Then the number of possibilities for α1, . . . , αn satisfying (2.9) is (2.10) j X αn=0 n + (j −αn) −2 n −2  = j X i=0 n + i −2 n −2  . Then our claim follows by combining (2.10) with the result of Problem 2.6: j X i=0 n + i −2 n −2  = n + j −1 n −1  . Now to find the number of points y as in (2.8) with x = 0 on St j=0 Hj, we sum over j, using the result of Excercise 2.6 once again: t X j=0 n + j −1 n −1  = n + t n  . If x in (2.8) lies properly between H0 and H1, then the number of possible y as given by (2.8) that lie in St j=0 Hj reduces to n+t−1 n  . Similarly, the number of possibilities for y as in (2.8) with x lying properly between Hi−1 and Hi or on Hi is n+t−i n  for each 1 ≤i ≤n. Therefore, if βi is the number of points x ∈A ∩Zn which lie properly between Hi−1 and Hi or on Hi, then the number of corresponding points y as in (2.8) is n + t −i n  βi. Finally, in the case t < n, we let βi = 0 for each t + 1 ≤i ≤n. The statement of the lemma follows. □ 2.4. INTEGER KNAPSACK AND COUNTING INTEGER LATTICE POINTS IN POLYTOPES 15 Let a1, . . . , an ∈Zn be linearly independent, and let S = Co(0, a1, . . . , an) be the simplex as in Lemma 2.4.2. Define the pseudo-simplex associated with S S0 = S \ (Co(0, a1, . . . , an−1) ∪. . . ∪Co(0, a2, . . . , an)) . Lemma 2.4.3. G(tS0) is a polynomial in t ∈Z≥0. Proof. We argue by induction on dimension of S0. If dim(S0) = 0, there is nothing to prove, so assume the lemma is true for pseudo-simplices of dimension < n. Let F (1), . . . , F (s) be proper faces of S which contain 0 and satisfy 0 < dim(F (i)) < n, ∀1 ≤i ≤s. Then S \ S0 = {0} ∪F (1) 0 ∪. . . ∪F (s) 0 is a disjoint union. By induction hypothesis, G(t(S \ S0)) = 1 + G(tF (1) 0 ) + · · · + G(tF (s) 0 ) is a polynomial in t. Hence, by Lemma 2.4.2, G(tS0) = G(tS) −G(t(S \ S0)) = G(tS) −1 −G(tF (1) 0 ) −· · · −G(tF (s) 0 ) is a polynomial in t. □ We are now ready to prove a theorem of Eugene Ehrhart’s from the 1960s. Theorem 2.4.4 (Ehrhart). Let P be a lattice polytope in Rn. Then GP (t) is a polynomial in t ∈Z≥0. Proof. We can assume 0 to be a vertex of P, since such translation would not change the number of integer lattice points. Notice that each (n −1)-dimensional face of P which does not contain 0 can be given a decomposition as a simplicial complex whose 0-cells are the vertices of this face. We can then join each simplex, obtained in this manner, to 0 resulting in a decomposition of P into a simplicial complex whose 0-cells are precisely the vertices of P. Then P can be represented as a disjoint union P = {0} ∪S(1) 0 ∪. . . ∪S(r) 0 , where S(1) 0 , . . . , S(r) 0 are precisely the cells of this simplicial complex which contain 0, but are not equal to {0}. The theorem follows by Lemma 2.4.3. □ GP (t) as in Theorem 2.4.4 is called Ehrhart polynomial of P. An excellent ref-erence on Ehrhart polynomials, their many fascinating properties, and connections to other important mathematical objects is [BR06]. For a general lattice polytope P very little is known about the coefficients of its Ehrhart polynomial GP (t). Let GP (t) = n X i=0 ci(P)ti, then it is known that the leading coefficient cn(P) is equal to Vol(P), and cn−1(P) is (n −1)-dimensional volume of the boundary ∂P, which is normalized by the 16 2. KNAPSACK AND FROBENIUS PROBLEMS determinants of the sublattices induced by the corresponding faces of P. Also, c0(P) is the combinatorial Euler characteristic χ(P): χ(P) = n X i=0 (−1)i(number of i −dimensional faces of P). The rest of the coefficients of GP (t) are in general unknown, however there are known relations and identities that they satisfy; see [BR06] for further details. Let us present the first simple example of Ehrhart polynomial. Consider the n-dimensional cube of sidelength 2 centered at the origin: (2.11) Cn = {x ∈Rn : |x| ≤1}, then for each t ∈Z>0 |tCn ∩Zn| = (2t + 1)n = n X i=0 2i n k  ti is the corresponding Ehrhart polynomial. We will give two more explicit examples of Ehrhart polynomial. The first one is for an open simplex, which is precisely the interior of the simplex S of Lemma 2.4.2 with ai = ei for each 1 ≤i ≤n; the following observation along with the proof is due to S. I. Sobolev. Proposition 2.4.5. Define an open simplex S◦= ( x ∈Rn : xi > 0 ∀1 ≤i ≤n, n X i=1 xi < 1 ) . Then GS◦(t) = 0 if t ≤n, and for every t ∈Z>n, (2.12) GS◦(t) = t −1 n  . Proof. Let t > n, and notice that the simplex tS◦can be mapped by an affine transformation to the simplex tS◦ 1 = {x ∈Rn : 0 < x1 < · · · < xk < t} . This transformation is volume-preserving and maps Zn to itself. Integral points of tS◦ 1 correspond to increasing sequences of integers 0 < y1 < · · · < yn < t. The number of such sequences is precisely t−1 n  , which is the number of all possible n-element subsets of the set {1, ..., t −1}. □ Notice that (2.12) can be thought of as a geometric interpretation of binomial coefficients. The next example is closely related to the one in Proposition 2.4.5: it has been established in [BCKV00]. Proposition 2.4.6. Let Sn = ( x ∈Rn : n X i=1 |xi| ≤1 ) . Then for every t ∈Z>0 (2.13) GSn(t) = min{t,n} X i=0 2i n i t i  . 2.4. INTEGER KNAPSACK AND COUNTING INTEGER LATTICE POINTS IN POLYTOPES 17 Proof. Notice that for each 0 ≤i ≤min{t, n} the number of points in tSn∩Zn with precisely i nonzero coordinates is 2i n i t i  . Indeed, the number of choices of which coordinates are nonzero is n i  ; for each such choice there are 2i choices of ± signs, and t i  choices of absolute values. Summing over all 0 ≤i ≤min{t, n} completes the proof. □ Remark 2.4.1. A remarkable property of the polynomial in Proposition 2.4.6 is that the right hand side (2.13) is symmetric in t and n. This means that |tSn ∩Zn| = |nSt ∩Zt|. 18 2. KNAPSACK AND FROBENIUS PROBLEMS 2.5. Frobenius problem In this section we introduce a problem closely related to the IKP. For n ≥2, consider integers (2.14) 1 < a1 < · · · < an such that gcd(a1, . . . , an) = 1, and define S(a1, . . . , an) := ( n X i=1 aixi : x1, . . . , xn ∈Z≥0 ) . This is an example of a numerical semigroup, i.e. a subset of Z≥0 containing 0 which is closed under addition. The set N \ S(a1, . . . , an) is called the set of gaps of S(a1, . . . , an). Theorem 2.5.1. The set of gaps of the numerical semigroup S(a1, . . . , an) under the condition (2.14) is finite. Proof. Since the n-tuple a1, . . . , an is relatively prime, there exist integers m1, . . . , mn such that a1m1 + · · · + anmn = 1. This sum has some positive and some negative terms, hence it can be written A −B = 1, where A and B are both nonnegative integer linear combinations of a1, . . . , an. Therefore A, B ∈S(a1, . . . , an). Now, let z be any positive integer, then Euclid’s division lemma implies that z = qa1 + r, q, r ∈Z≥0, 0 ≤r < a1. On the other hand, r = r × 1 = r(A −B). Notice that qa1 ∈S(a1, . . . , an), and hence z+(a1−1)B = qa1+r(A−B)+(a1−1)B = qa1+rA+(a1−r−1)B ∈S(a1, . . . , an), since a1 −r −1 ≥0. This implies that every integer ≥(a1 −1)B is in S(a1, . . . , an), and hence the number of gaps at most (a1 −1)B −1. □ Remark 2.5.1. The presentation of the above argument closely followed [Ram05]. The Frobenius number g(a1, . . . , an) is defined to be the largest gap of S(a1, . . . , an), i.e. the largest integer t that cannot be expressed in the form t = Pn i=1 aixi for some nonnegative integers x1, . . . , xn. With this notation, we can formulate the Frobenius Problem (FP): Given an n-tuple a1, . . . , an satisfying (2.14) find g(a1, . . . , an). Notice that FP can be stated in terms of the knapsack polytopes as follows: writing a = (a1, . . . , an), find the smallest positive integer g so that P(a, t) ∩Zn ̸= ∅for every t > g. Alternatively, we can write it as follows: Find min{g ∈Z>0 : |P(a, t) ∩Zn| > 0 ∀b > t}. Notice also that n X i=1 aixi = t ⇐ ⇒ n X i=1 ai xi t  = 1, i.e. x ∈P(a, t) if and only if 1 t x ∈P(a, 1), meaning that P(a, t) = tP(a, 1) is a homogeneous expansion of a polytope that we discussed in the previous section. This, however, does not necessarily apply directly to the integer lattice points, since 2.5. FROBENIUS PROBLEM 19 for x ∈Zn the rescaled points 1 t x may no longer be in Zn. Due to this knapsack connection, it is not surprising that FP is known to be NP-hard (specifically, there a polynomial-time algorithm that reduces IKP to FP). We begin our discussion of FP with the simplest case n = 2. Let us start with a simple binary linear Diophantine equation of the form (2.15) ax + by = c, in which a, b, c are nonzero integers. There are always rational solutions to (2.15). For which values of a, b, c does it have solutions in integers x, y? The greatest common divisor provides a criterion for the existence of solutions. Lemma 2.5.2. Let a, b, c be nonzero integers. Then (2.15) has a solution in integers x, y if and only if gcd(a, b)|c. Proof. (⇒) Suppose that ax + by = c for some x, y ∈Z. Since gcd(a, b) divides a and b, it divides ax + by = c. (⇐) If gcd(a, b)|c, write c = d gcd(a, b), in which d ∈Z. By Euclid’s Division Lemma, there exist x′, y′ ∈Z such that ax′+by′ = gcd(a, b). Thus, a(dx′)+b(dy′) = d(ax′ + by′) = d gcd(a, b) = c and hence (2.15) has integer solutions x = dx′ and y = dy′. □ In fact, we can classify all integer solutions to (2.15). Theorem 2.5.3. Let a, b, c be nonzero integers, and let d = gcd(a, b). Assume d|c. Then the equation ax + by = c has infinitely many integer solutions. In fact, if x0, y0 is one such solution pair, then all solutions are given by (2.16) xt = x0 −t b d, yt = y0 + ta d as t ranges over all the integers. Proof. First let t ∈Z and xt, yt be as in (2.16). Then axt + byt = a  x0 −t b d  + b  y0 + ta d  = (ax0 + by0) + t ab d −ab d  = c, hence our pair x, y is a solution to (2.15) for any t ∈Z. We now show that any solution is of this form. Indeed, suppose x, y is a solution pair, then ax0 + by0 = c = ax + by, and so a(x0 −x) = b(y −y0). Let us divide both sides of the above equation by d and write a′ = a/d, b′ = b/d, then gcd(a′, b′) = 1 and a′(x0 −x) = b′(y −y0). Then Euclid’s Lemma implies that a′|y −y0 and b′|x0 −x, say a′ = y−y0 t and b′ = x0−x s for some integers t and s. Then we have (y −y0)(x0 −x) t = (x0 −x)(y −y0) s , and so s = t. Therefore we obtain y = y0 + a′t, x = x0 −b′t, which is precisely what we wanted. □ 20 2. KNAPSACK AND FROBENIUS PROBLEMS Corollary 2.5.4. If gcd(a, b) = 1, then for any c the equation ax + by = c has infinitely many solutions. Furthermore, if x0, y0 is one such solution pair, then all solutions are of the form xt = x0 −tb, yt = y0 + ta for t ∈Z. Example 2.5.1. Let a = 4, b = 6, c = 9. Since gcd(a, b) = 2 ∤9, the equations 4x+6y = 9 has no integer solutions. On the other hand, if c = 10, then gcd(a, b)|c, and so the equation 4x+6y = 10 has infinitely many integer solutions. Since x = 1, y = 1 is one such solution, all solutions are of the form xt = 1 −3t, yt = 1 + 2t as t ranges over all the integers. These observations also have a simple geometric interpretation. Notice that the set of integer solution pairs to (2.15)  (x, y) ∈Z2 : ax + by = c is the set of all integer lattice points on the line given by the equation (2.15) in the Euclidean plane. For instance, the set of all such points in the case a = 4, b = 6, c = 10 of Example 2.5.1 is {(1 −6t, 1 + 4t) : t ∈Z}. Assume now that c > 0 and gcd(a, b) divides c, so the line ax + by = c contains infinitely many integer lattice points, but does it necessarily contain any such points with nonnegative coordinates? Upon a quick inspection, we can see for instance that the line (2.17) 3x + 5y = c contains integer lattice points for any c, but no such points with x, y ≥0 when c = 1, 2, 4. For which values of c is our line guaranteed to have nonnegative integer lattice points? Here is an initial observation, which follows from Theorem 2.5.3 via a geometric argument. 2.5. FROBENIUS PROBLEM 21 Corollary 2.5.5. Let a, b, c be positive integers with d := gcd(a, b) dividing c. If c ≥ab/d, then the equation (2.15) has integer solution pairs x, y ≥0. Proof. Let t, s ∈Z and consider the solution pairs (xt, yt) and (xs, ys), as in (2.16), where (x0, y0) is some fixed solution pair. Notice that the Euclidean distance between the points (xt, yt) and (xs, ys) is p (xt −xs)2 + (yt −ys)2 = r b2 d2 (t −s)2 + a2 d2 (t −s)2 = |t −s| √ a2 + b2 d , which is minimized when |t −s| = 1. Let ℓa,b(c) be the line ax + by = c in the Euclidean plane, then the minimal distance between two integer lattice points on ℓa,b(c) is √ a2+b2 d , which is assumed for any neighboring pair of integer lattice points (xt, yt) and (xt+1, yt+1). Notice that the intersection of the line ℓa,b(c) with the positive quadrant {(x, y) ∈R2 : x, y ≥0} is a line segment with endpoints (c/a, 0) and (0, c/b), so the length of this line segment is r c2 a2 + c2 b2 = c √ a2 + b2 ab . If the length of this line segment is no less than the distance between the neighboring integer lattice points, then the line segment must contain at least one integer lattice point. This means that when c √ a2 + b2 ab ≥ √ a2 + b2 d , the equation (2.15) has integer solution pairs x, y ≥0. This happens when c ≥ ab/d. □ Going back to the example of equation (2.17) and applying Corollary 2.5.5, we are guaranteed that there are nonnegative solutions at least for all c ≥15. Checking by hand, we quickly see that in fact there are nonnegative solutions already for all c ≥8, suggesting that the bound of Corollary 2.5.5 may not be very good. Indeed, we can obtain more precise results. Let a, b be relatively prime positive integers, and suppose that we have un-limited supply of coins of denominations a and b. What is the maximal amount of change which we cannot give with such coins? This is precisely the Frobenius number g(a, b) and we know from Corollary 2.5.5 that g(a, b) < ab. But is there an exact formula? This problem, although possibly in different terms was mentioned in the lectures of a famous German mathematician Ferdinand Georg Frobenius in the late 1800s, although Frobenius himself never published anything in these regards. Nonetheless, this problem became known as the (binary) Frobenius coin exchange problem with the maximal impossible amount of change g(a, b) being the Frobenius number of a and b. Interestingly, closely related problems also ap-pear in recreational mathematical literature under different names, such as postage stamp problem or the chicken McNugget problem. The origins of the latter name are curious: apparently, in the 1980s chicken McNuggets were sold by McDonalds in the UK in boxes of 3, 6 and 20 pieces, prompting a mathematician Henri Picciotto to ask what is the maximal number of nuggets that cannot be purchased (and then 22 2. KNAPSACK AND FROBENIUS PROBLEMS answering his own question – it is 43). Let us now derive a formula for the binary Frobenius number. Theorem 2.5.6. Let gcd(a, b) = 1, then g(a, b) = (a −1)(b −1) −1. In other words, this is the largest number that cannot be represented as ax+by with x, y nonnegative integers. Proof. Since a and b are relatively prime, for every c ∈Z there exist x, y ∈Z such that c = ax + by. We will say that c is representable in terms of a and b if there exist such x, y ≥0. Notice in fact that we can assume without loss of generality that 0 ≤x < b: if x ≥b, then x = nb + x′ for some n, x′ ∈Z with 0 ≤x′ < b, and so c = a(nb + x′) + by = ax′ + b(an + y), meaning that we can replace x with x′ by replacing y with an + y, if necessary. Now, if 0 ≤x < b, then for every c there is a unique pair (x, y) such that c = ax + by, and so c is representable if and only if y ≥0. Notice then that the largest non-representable c corresponds to the largest choice of x (namely, x = b−1) and the largest negative choice of y (namely, y = −1). This means that the largest non-representable integer is g(a, b) = a(b −1) + b(−1) = ab −a −b = (a −1)(b −1) −1. □ Theorem 2.5.6 therefore guarantees that for every c > ab −a −b the line ax + by = c contains a nonnegative integer lattice point, however for c < ab −a −b such a point may or may not exist. Revisiting for instance our example (2.17), we see that while g(3, 5) = 7, the equation 3x+5y = c has nonnegative integer solutions for c = 3, 5, 6, but does not for c = 1, 2, 4, 7, i.e. these are gaps of S(3, 5). Given a and b, we can ask how many gaps are there? This natural question was asked as a challenge problem in a journal called Educational Times by James Joseph Sylvester in 1884. Specifically, Sylvester, who has already obtained and published the answer himself in 1882, asked for a proof that this number is equal to 1 2(a −1)(b −1); in other words, out of (a −1)(b −1) −1 integers between 1 and the Frobenius number g(a, b) about half are non-representable. A clever solution was produced by W. J. Curran Sharp. We prove this result here. Theorem 2.5.7. The number of gaps with respect to a relatively prime pair of positive integers a and b is 1 2(a −1)(b −1). Proof. Let 0 ≤c ≤g(a, b), and define c′ = g(a, b) −c = ab −a −b −c. By our argument in the proof of Theorem 2.5.6, there must exist the unique integers x, y with 0 ≤x < b such that c = ax + by, then c′ = ab −a −b −c = ab −a −b −ax −by = ax′ + by′, 2.5. FROBENIUS PROBLEM 23 where x′ = b −x −1 and y′ = −y −1. Since 0 ≤x′ < b, we see that y′ must also be unique. Suppose that c is representable by a and b (including c = 0), then y ≥0, and y′ < 0, hence c′ is not representable. On the other hand, assume that c is not representable, then y < 0, and so y′ ≥0, meaning that c′ is representable. It is clear that c and c′ are in a bijection with each other, and c = c′ if and only if c = 1 2(ab −a −b), but this cannot be an integer, since a and b cannot both be even. Hence precisely a half of g(a, b) + 1 integers between 0 and g(a, b) are representable and the rest are gaps, meaning that there are 1 2(g(a, b) + 1) = 1 2(a −1)(b −1) gaps. □ The Frobenius number has also been defined more generally. Let n ≥2 be an integer and let (2.18) 1 < a1 < · · · < an be relatively prime integers. We say that a positive integer t is representable by the n-tuple a := (a1, . . . , an) if (2.19) t = a1x1 + · · · + anxn for some nonnegative integers x1, . . . , xn, and we call each such solution x := (x1, . . . , xn) of (2.19) a representation for t in terms of a. Let s ≥0 be an in-teger, then the s-Frobenius number of this n-tuple, gs(a), as defined by Beck and Robins in [BR04], is the largest positive integer that has at most s distinct repre-sentations in terms of a. In the binary case (n = 2), Beck and Robins proved the following natural generalization of Theorem 2.5.6. Theorem 2.5.8. Let gcd(a, b) = 1 and s ≥0, then gs(a, b) = (s + 1)ab −(a + b). In the case s = 0, the formula of Theorem 2.5.6 is recovered. This is a generalization of the classical Frobenius number g0(a), i.e., the largest positive integer that has no such representations. The Frobenius number has been studied extensively by a variety of authors, starting as early as late 19th century; see [Ram05] for a detailed account and bibliography. Generalizing Theorem 2.5.1, the condition (2.20) gcd(a1, . . . , an) = 1 implies that gs(a) exists for every s, but the NP-hardness of FP (and the fact that P vs NP is an open problem) in particular implies that no general closed form formulas for the Frobenius numbers is known, sparking interest in upper and lower bounds. Frobenius numbers and their various generalizations tend to play an important role in several areas of mathematics, including theory of numerical semi-groups, commutative algebra, algebraic geometry, number theory, combinatorics, operations research, and theoretical computer science, to name a few. The liter-ature on this subject is vast with a large number of relevant references available 24 2. KNAPSACK AND FROBENIUS PROBLEMS in the bibliography to the book [Ram05]. We will talk more about the Frobenius number and its beautiful geometric connections in Section 4.5. 2.6. PROBLEMS 25 2.6. Problems Problem 2.1. Consider the Boolean formula B = (x ∧y) ∨(¬x ∧¬y) ∨(¬x ∧z) ∨(y ∧¬z). Part a. Rewrite B in generalized conjunctive normal form. Part b. Rewrite the formula you obtained in part a in the form with clauses con-sisting of three literals each. Part c. Construct the table to reduce your formula from part b to an instance of SSP, as in the proof of Theorem 2.1.2. Part d. Decide if the instance of SSP you obtained is solvable or not. If so, what values of the Boolean variables make the formula from part b satisfiable? How about the original formula B? Problem 2.2. Consider the following instance of BKP: maximize 4x1 + 5x2 + 7x3 subject to 6x1 + 3x2 + 5x3 ≤12, x1, x2, x3 ∈{0, 1}. Part a. Use Theorem 2.2.1 to solve the CKP relaxation of this problem. Part b. Define the corresponding LP-polytope for the CKP problem in part a and find its vertices. Part c. Use the geometric description of the simplex method as in Example 2.3.1 to find the solution. Make sure it is consistent with part a. Part d. List all the integer lattice points in the LP-polytope from part b and solve BKP by evaluating the objective function at each one of them and comparing. Was the solution to CKP that you found a good approximation to the BKP solution? Problem 2.3. Prove the formula A(P) = W(P) as in the proof of Pick’s Theorem (Theorem 2.4.1) for rectangles and triangles. Hint: First prove it for a rectangle, then for a right triangle – splitting rectangle into two of them and applying additivity, and then for an arbitrary triangle by embedding it into a rectangle and applying additivity. Problem 2.4. Suppose P is a polygon with integer vertices and h simple polygonal holes, each also with integer vertices. Can you generalize Pick’s theorem to P? Problem 2.5. Let n be a positive integer. The Farey series Fn of order n is the set of all reduced nonnegative rationals in the interval [0, 1] with denominators no bigger than n written in increasing order, e.g. F5 = 0 1, 1 5, 1 4, 1 3, 2 5, 1 2, 3 5, 2 3, 3 4, 4 5, 1 1  . Let a b , c d ∈Fn, where n = max{b, d}. Let P be the parallelogram with vertices (0, 0), (a, b), (c, d), (a + c, b + d). Use Pick’s theorem to prove that a b , c d are neighbors in Fn if and only if the area of P is equal to 1. 26 2. KNAPSACK AND FROBENIUS PROBLEMS Problem 2.6. Prove that j X i=0 n + i −2 n −2  = n + j −1 n −1  . Problem 2.7. Compute Ehrhart polynomials of a rectangle, a right triangle, and a right trapezoid. Problem 2.8. Let a, b be positive relatively prime integers. Express the Frobenius number g(a, b) in terms of areas of parallelograms with a side of length √ a2 + b2. Problem 2.9. Let a1, . . . , an be positive relatively prime integers. Part a. Express the s-Frobenius number gs(a1, . . . , an) in terms of the restricted partition function pa1,...,an(t) = # ( (x1, . . . , xn) ∈Zn ≥0 : X i=1 aixi = t ) . Part b. Prove the recursive formula pa1,...,an(t) = X m≥0 pa1,...,an−1(t −man). Part c. Interpret the recursive formula from part b in terms of the s-Frobenius numbers. Problem 2.10. Let a1, . . . , an be positive relatively prime integers. For each s ≥0, define Ss(a1, . . . , an) to be the set of all integers that have more than s representa-tions in the form Pn i=1 aixi with n-tuple of nonnegative integers x1, . . . , xn ∈Z≥0. Prove that · · · ⊆Ss(a1, . . . , an) ⊆Ss−1(a1, . . . , an) ⊆· · · ⊆S0(a1, . . . , an) = S(a1, . . . , an) is a sequence of numerical semigroups. Problem 2.11. Let a, b and c, d be two pairs of positive relatively prime integers. Suppose g(a, b) ≥g(c, d). Does this mean that g1(a, b) ≥g1(c, d)? Prove or give a counterexample. CHAPTER 3 Geometry of Numbers 3.1. Lattices We start with an algebraic definition of lattices. Let a1, . . . , ar be a collection of linearly independent vectors in Rn. Definition 3.1.1. A lattice Λ of rank r, 1 ≤r ≤n, spanned by a1, . . . , ar in Rn is the set of all possible linear combinations of the vectors a1, . . . , ar with integer coefficients. In other words, Λ = spanZ {a1, . . . , ar} := ( r X i=1 niai : ni ∈Z for all 1 ≤i ≤r ) . The set a1, . . . , ar is called a basis for Λ. There are usually infinitely many different bases for a given lattice. Notice that in general a lattice in Rn can have any rank 1 ≤r ≤n. We will often however talk specifically about lattices of rank n, that is of full rank. The most obvious example of a lattice is the set of all points with integer coordinates in Rn: Zn = {x = (x1, . . . , xn) : xi ∈Z for all 1 ≤i ≤n}. Notice that the set of standard basis vectors e1, . . . , en, where ei = (0, . . . , 0, 1, 0, . . . , 0), with 1 in i-th position is a basis for Zn. Another basis is the set of all vectors ei + ei+1, 1 ≤i ≤n −1. If Λ is a lattice of rank r in Rn with a basis a1, . . . , ar and y ∈Λ, then there exist m1, . . . , mr ∈Z such that y = r X i=1 miai = Am, where m =    m1 . . . mr   ∈Zr, and A is an n×r basis matrix for Λ of the form A = (a1 . . . ar), which has rank r. In other words, a lattice Λ of rank r in Rn can always be described as Λ = AZr, where A is its m×r basis matrix with real entries of rank r. As we remarked above, bases are not unique; as we will see later, each lattice has bases with particularly nice properties. 27 28 3. GEOMETRY OF NUMBERS An important property of lattices is discreteness. To explain what we mean more notation is needed. First notice that Euclidean space Rn is clearly not com-pact, since it is not bounded. It is however locally compact: this means that for every point x ∈Rn there exists an open set containing x whose closure is compact, for instance take an open unit ball centered at x. More generally, every subspace V of Rn is also locally compact. A subset Γ of V is called discrete if for each x ∈Γ there exists an open set S ⊆V such that S ∩Γ = {x}. For instance Zn is a discrete subset of Rn: for each point x ∈Zn the open ball of radius 1/2 centered at x contains no other points of Zn. We say that a discrete subset Γ is co-compact in V if there exists a compact 0-symmetric subset U of V such that the union of translations of U by the points of Γ covers the entire space V , i.e. if V = [ {U + x : x ∈Γ}. Here U + x = {u + x : u ∈U}. Recall that a subset G is a subgroup of the additive abelian group Rn if it satisfies the following conditions: (1) Identity: 0 ∈G, (2) Closure: For every x, y ∈G, x + y ∈G, (3) Inverses: For every x ∈G, −x ∈G. By Problems 3.3 and 3.4 a lattice Λ of rank r in Rn is a discrete co-compact subgroup of V = spanR Λ. In fact, the converse is also true. Theorem 3.1.1. Let V be an r-dimensional subspace of Rn, and let Γ be a discrete co-compact subgroup of V . Then Γ is a lattice of rank r in Rn. Proof. In other words, we want to prove that Γ has a basis, i.e. that there exists a collection of linearly independent vectors a1, . . . , ar in Γ such that Γ = spanZ{a1, . . . , ar}. We start by inductively constructing a collection of vectors a1, . . . , ar, and then show that it has the required properties. Let a1 ̸= 0 be a point in Γ such that the line segment connecting 0 and a1 contains no other points of Γ. Now assume a1, . . . , ai−1, 2 ≤i ≤r, have been selected; we want to select ai. Let Hi−1 = spanR{a1, . . . , ai−1}, and pick any c ∈Γ \ Hi−1: such c exists, since Γ ̸⊆Hi−1 (otherwise Γ would not be co-compact in V ). Let Pi be the closed parallelotope spanned by the vectors a1, . . . , ai−1, c. Notice that since Γ is discrete in V , Γ∩Pi is a finite set. Moreover, since c ∈Pi, Γ ∩Pi ̸⊆Hi−1. Then select ai such that d(ai, Hi−1) = min y∈(Pi∩Γ)\Hi−1{d(y, Hi−1)}, where for any point y ∈Rn, d(y, Hi−1) = inf x∈Hi−1{d(y, x)}. Let a1, . . . , ar be the collection of points chosen in this manner. Then we have a1 ̸= 0, ai / ∈spanZ{a1, . . . , ai−1} ∀2 ≤i ≤r, which means that a1, . . . , ar are linearly independent. Clearly, spanZ{a1, . . . , ar} ⊆Γ. 3.1. LATTICES 29 We will now show that Γ ⊆spanZ{a1, . . . , ar}. First of all notice that a1, . . . , ar is certainly a basis for V , and so if x ∈Γ ⊆V , then there exist c1, . . . , cr ∈R such that x = r X i=1 ciai. Notice that x′ = r X i=1 [ci]ai ∈spanZ{a1, . . . , ar} ⊆Γ, where [ ] stands for the integer part function (i.e. [ci] is the largest integer which is no larger than ci). Since Γ is a group, we must have z = x −x′ = r X i=1 (ci −[ci])ai ∈Γ. Then notice that d(z, Hr−1) = (cr −[cr]) d(ar, Hr−1) < d(ar, Hr−1), but by construction we must have either z ∈Hr−1, or d(ar, Hr−1) ≤d(z, Hr−1), since z lies in the parallelotope spanned by a1, . . . , ar, and hence in Pr as in our construction above. Therefore cr = [cr]. We proceed in the same manner to conclude that ci = [ci] for each 1 ≤i ≤r, and hence x ∈spanZ{a1, . . . , ar}. Since this is true for every x ∈Γ, we are done. □ From now on, until further notice, our lattices will be of full rank in Rn, that is of rank n. In other words, a lattice Λ ⊂Rn will be of the form Λ = AZn, where A is a non-singular n × n basis matrix for Λ. Theorem 3.1.2. Let Λ be a lattice of rank n in Rn, and let A be a basis matrix for Λ. Then B is another basis matrix for Λ if and only if there exists an n × n integral matrix U with determinant ±1 such that B = AU. Proof. First suppose that B is a basis matrix. Notice that, since A is a basis matrix, for every 1 ≤i ≤n the i-th column vector bi of B can be expressed as bi = n X j=1 uijaj, where a1, . . . , an are column vectors of A, and uij’s are integers for all 1 ≤j ≤n. This means that B = AU, where U = (uij)1≤i,j≤n is an n × n matrix with integer entries. On the other hand, since B is also a basis matrix, we also have for every 1 ≤i ≤n ai = n X j=1 wijbj, 30 3. GEOMETRY OF NUMBERS where wij’s are also integers for all 1 ≤j ≤N. Hence A = BW, where W = (wij)1≤i,j≤n is also an n × n matrix with integer entries. Then B = AU = BWU, which means that WU = In, the n × n identity matrix. Therefore det(WU) = det(W) det(U) = det(In) = 1, but det(U), det(W) ∈Z since U and W are integral matrices. This means that det(U) = det(W) = ±1. Next assume that B = UA for some integral n×n matrix U with det(U) = ±1. This means that det(B) = ± det(A) ̸= 0, hence column vectors of B are linearly independent. Also, U is invertible over Z, meaning that U −1 = (wij)1≤i,j≤n is also an integral matrix, hence A = U −1B. This means that column vectors of A are in the span of the column vectors of B, and so Λ ⊆spanZ{b1, . . . , bn}. On the other hand, bi ∈Λ for each 1 ≤i ≤n. Thus B is a basis matrix for Λ. □ Corollary 3.1.3. If A and B are two basis matrices for the same lattice Λ, then | det(A)| = | det(B)|. Definition 3.1.2. The common determinant value of Corollary 3.1.3 is called the determinant of the lattice Λ, and is denoted by det(Λ). We now talk about sublattices of a lattice. Let us start with a definition. Definition 3.1.3. If Λ and Ωare both lattices in Rn, and Ω⊆Λ, then we say that Ωis a sublattice of Λ. There are a few basic properties of sublattices of a lattice which we outline here – their proofs are left to exercises. (1) A subset Ωof the lattice Λ is a sublattice if and only if it is a subgroup of the abelian group Λ. (2) For a sublattice Ωof Λ two cosets x + Ωand y + Ωare equal if and only if x −y ∈Ω. In particular, x + Ω= Ωif and only if x ∈Ω. (3) If Λ is a lattice and µ a real number, then the set µΛ := {µx : x ∈Λ} is also a lattice. Further, if µ is an integer then µΛ is a sublattice of Λ. From here on, unless stated otherwise, when we say Ω⊆Λ is a sublattice, we always assume that it has the same full rank in Rn as Λ. Lemma 3.1.4. Let Ωbe a subattice of Λ. There exists a positive integer D such that DΛ ⊆Ω. Proof. Recall that Λ and Ωare both lattices of rank n in Rn. Let a1, . . . , an be a basis for Ωand b1, . . . , bn be a basis for Λ. Then spanR{a1, . . . , an} = spanR{b1, . . . , bn} = Rn. 3.1. LATTICES 31 Since Ω⊆Λ, there exist integers u11, . . . , unn such that      a1 = u11b1 + · · · + u1nbn . . . . . . . . . an = un1b1 + · · · + unnbn. Solving this linear system for b1, . . . , bn in terms of a1, . . . , an, we easily see that there must exist rational numbers p11 q11 , . . . , pnn qnn such that      b1 = p11 q11 a1 + · · · + p1n q1n an . . . . . . . . . bn = pn1 qn1 a1 + · · · + pnn qnn an. Let D = q11 ×· · ·×qnn, then D/qij ∈Z for each 1 ≤i, j, ≤n, and so all the vectors        Db1 = Dp11 q11 a1 + · · · + Dp1n q1n an . . . . . . . . . Dbn = Dpn1 qn1 a1 + · · · + Dpnn qnn an are in Ω. Therefore spanZ{Db1, . . . , Dbn} ⊆Ω. On the other hand, spanZ{Db1, . . . , Dbn} = D spanZ{b1, . . . , bn} = DΛ, which completes the proof. □ We can now prove that a lattice always has a basis with “nice” properties with respect to any given basis of a given sublattice, and vice versa. Theorem 3.1.5. Let Λ be a lattice, and Ωa sublattice of Λ. For each basis b1, . . . , bn of Λ, there exists a basis a1, . . . , an of Ωof the form        a1 = v11b1 a2 = v21b1 + v22b2 . . . . . . . . . . . . . . . . . . . . . . . . an = vn1b1 + · · · + vnnbn, where all vij ∈Z and vii ̸= 0 for all 1 ≤i ≤n. Conversely, for every basis a1, . . . , an of Ωthere exists a basis b1, . . . , bn of Λ such that the relations as above hold. Proof. Let b1, . . . , bn be a basis for Λ. We will first prove the existence of a basis a1, . . . , an for Ωas claimed by the theorem. By Lemma 3.1.4, there exist integer multiples of b1, . . . , bn in Ω, hence it is possible to choose a collection of vectors a1, . . . , an ∈Ωof the form ai = i X j=1 vijbj, for each 1 ≤i ≤n with vii ̸= 0. Clearly, by construction, such a collection of vectors will be linearly independent. In fact, let us pick each ai so that |vii| is as small as possible, but not 0. We will now show that a1, . . . , an is a basis for Ω. Clearly, spanZ{a1, . . . , an} ⊆Ω. 32 3. GEOMETRY OF NUMBERS We want to prove the inclusion in the other direction, i.e. that (3.1) Ω⊆spanZ{a1, . . . , an}. Suppose (3.1) is not true, then there exists c ∈Ωwhich is not in spanZ{a1, . . . , an}. Since c ∈Λ, we can write c = k X j=1 tjbj, for some integers 1 ≤k ≤n and t1, . . . , tk. In fact, let us select a c like this with minimal possible k. Since vkk ̸= 0, we can choose an integer s such that (3.2) |tk −svkk| < |vkk|. Then we clearly have c −sak ∈Ω\ spanZ{a1, . . . , an}. Therefore we must have tk−svkk ̸= 0 by minimality of k. But then (3.2) contradicts the minimality of |vkk|: we could take c−sak instead of ak, since it satisfies all the conditions that ak was chosen to satisfy, and then |vkk| is replaced by the smaller nonzero number |tk −svkk|. This proves that c like this cannot exist, and so (3.1) is true, hence finishing one direction of the theorem. Now suppose that we are given a basis a1, . . . , an for Ω. We want to prove that there exists a basis b1, . . . , bn for Λ such that relations in the statement of the theorem hold. This is a direct consequence of the argument in the proof of Theorem 3.1.1. Indeed, at i-th step of the basis construction in the proof of Theorem 3.1.1, we can choose i-th vector, call it bi, so that it lies in the span of the previous i −1 vectors and the vector ai. Since b1, . . . , bn constructed this way are linearly independent (in fact, they form a basis for Λ by the construction), we obtain that ai ∈spanZ{b1, . . . , bi} \ spanZ{b1, . . . , bi−1}, for each 1 ≤i ≤n. This proves the second half of our theorem. □ In fact, it is possible to select the coefficients vij in Theorem 3.1.5 so that the matrix (vij)1≤i,j≤n is upper (or lower) triangular with non-negative entries, and the largest entry of each row (or column) is on the diagonal: we leave the proof of this to Problem 3.9. Remark 3.1.1. Let the notation be as in Theorem 3.1.5. Notice that if A is any basis matrix for Ωand B is any basis for Λ, then there exists an integral matrix V such that A = BV . Then Theorem 3.1.5 implies that for a given B there exists an A such that V is lower triangular, and for for a given A exists a B such that V is lower triangular. Since two different basis matrices of the same lattice are always related by multiplication by an integral matrix with determinant equal to ±1, Theorem 3.1.5 can be thought of as the construction of Hermite normal form for an integral matrix. Problem 3.9 places additional restrictions that make Hermite normal form unique. Here is an important implication of Theorem 3.1.5. 3.1. LATTICES 33 Theorem 3.1.6. Let Ω⊆Λ be a sublattice. Then det(Ω) det(Λ) is an integer; moreover, the number of cosets of Ωin Λ, i.e. the index of Ωas a subgroup of Λ is [Λ : Ω] = det(Ω) det(Λ). Proof. Let b1, . . . , bn be a basis for Λ, and a1, . . . , an be a basis for Ω, so that these two bases satisfy the conditions of Theorem 3.1.5, and write A and B for the corresponding basis matrices. Then notice that B = AV, where V = (vij)1≤i,j≤n is an n × n triangular matix with entries as described in Theorem 3.1.5; in particular det(V ) = Qn i=1 |vii|. Hence det(Ω) = | det(A)| = | det(B)|| det(V )| = det(Λ) n Y i=1 |vii|, which proves the first part of the theorem. Moreover, notice that each vector c ∈Λ is contained in the same coset of Ωin Λ as precisely one of the vectors q1b1 + · · · + qnbn, 0 ≤qi < vii ∀1 ≤i ≤n, in other words there are precisely Qn i=1 |vii| cosets of Ωin Λ. This completes the proof. □ There is yet another, more analytic interpretation of the determinant of a lattice. Definition 3.1.4. A fundamental domain of a lattice Λ of full rank in Rn is a convex set F ⊆Rn containing 0, so that Rn = [ x∈Λ (F + x), and for every x ̸= y ∈Λ, (F + x) ∩(F + y) = ∅. In other words, a fundamental domain of a lattice Λ ⊂Rn is a full set of coset representatives of Λ in Rn (see Problem 3.10). Although each lattice has infinitely many different fundamental domains, they all have the same volume, which is equal to the determinant of the lattice. This fact can be easily proved for a special class of fundamental domains (see Problem 3.11). Definition 3.1.5. Let Λ be a lattice, and a1, . . . , an be a basis for Λ. Then the set F = ( n X i=1 tiai : 0 ≤ti < 1, ∀1 ≤i ≤n ) , is called a fundamental parallelotope of Λ with respect to the basis a1, . . . , an. It is easy to see that this is an example of a fundamental domain for a lattice. Fundamental parallelotopes form the most important class of fundamental domains, which we will work with most often. Notice that they are not closed sets; we will often write F for the closure of a fundamental parallelotope, and call them closed fundamental domains. Another important convex set associated to a lattice is its Voronoi cell, which is the closure of a fundamental domain; by a certain abuse of notation we will often refer to it also as a fundamental domain. 34 3. GEOMETRY OF NUMBERS Definition 3.1.6. The Voronoi cell of a lattice Λ is the set V(Λ) = {x ∈Rn : ∥x∥≤∥x −y∥∀y ∈Λ}. It is easy to see that V(Λ) is (the closure of) a fundamental domain for Λ: two translates of a Voronoi cell by points of the lattice intersect only in the boundary. The advantage of the Voronoi cell is that it is the most “round” fundamental domain for a lattice; we will see that it comes up very naturally in the context of sphere packing and covering problems. Notice that everything we discussed so far also has analogues for lattices of not necessarily full rank. We mention this here briefly without proofs. Let Λ be a lattice in Rn of rank 1 ≤r ≤n, and let a1, . . . , ar be a basis for it. Write A = (a1 . . . ar) for the corresponding n × r basis matrix of Λ, then A has rank r since its column vectors are linearly independent. For any r × r integral matrix U with determinant ±1, AU is another basis matrix for Λ; moreover, if B is any other basis matrix for Λ, there exists such a U so that B = AU. For each basis matrix A of Λ, we define the corresponding Gram matrix to be M = A⊤A, so it is a square r ×r nonsingular matrix. Notice that if A and B are two basis matrices so that B = UA for some U as above, then det(B⊤B) = det((AU)⊤(AU)) = det(U ⊤(A⊤A)U) = det(U)2 det(A⊤A) = det(A⊤A). This observation calls for the following general definition of the determinant of a lattice. Notice that this definition coincides with the previously given one in case r = n. Definition 3.1.7. Let Λ be a lattice of rank 1 ≤r ≤n in Rn, and let A be an n × r basis matrix for Λ. The determinant of Λ is defined to be det(Λ) = q det(A⊤A), that is the determinant of the corresponding Gram matrix. By the discussion above, this is well defined, i.e. does not depend on the choice of the basis. With this notation, all results and definitions of this section can be restated for a lattice Λ of not necessarily full rank. For instance, in order to define fundamental domains we can view Λ as a lattice inside of the vector space spanR(Λ). The rest works essentially verbatim, keeping in mind that if Ω⊆Λ is a sublattice, then index [Λ : Ω] is only defined if rk(Ω) = rk(Λ). 3.2. THEOREMS OF BLICHFELDT AND MINKOWSKI 35 3.2. Theorems of Blichfeldt and Minkowski In this section we will discuss some of the famous theorems related to the following very classical problem in the geometry of numbers: given a set M and a lattice Λ in Rn, how can we tell if M contains any points of Λ? Theorem 3.2.1 (Blichfeldt, 1914). Let M be a compact convex set in Rn. Suppose that Vol(M) ≥1. Then there exist x, y ∈M such that 0 ̸= x −y ∈Zn. Proof. First suppose that Vol(M) > 1. Let P = {x ∈Rn : 0 ≤xi < 1 ∀1 ≤i ≤n}, and let S = {u ∈Zn : M ∩(P + u) ̸= ∅}. Since M is bounded, S is a finite set, say S = {u1, . . . , ur0}. Write Mr = M ∩(P + ur) for each 1 ≤r ≤r0. Also, for each 1 ≤r ≤r0, define M ′ r = Mr −ur, so that M ′ 1, . . . , M ′ r0 ⊆P. On the other hand, Sr0 r=1 Mr = M, and Mr ∩Ms = ∅for all 1 ≤r ̸= s ≤r0, since Mr ⊆P + ur, Ms ⊆P + us, and (P + ur) ∩(P + us) = ∅. This means that 1 < Vol(M) = r0 X r=1 Vol(Mr). However, Vol(M ′ r) = Vol(Mr) for each 1 ≤r ≤r0, r0 X r=1 Vol(M ′ r) > 1, but Sr0 r=1 M ′ r ⊆P, and so Vol r0 [ r=1 M ′ r ! ≤Vol(P) = 1. Hence the sets M ′ 1, . . . , M ′ r0 are not mutually disjoined, meaning that there exist indices 1 ≤r ̸= s ≤r0 such that there exists x ∈M ′ r ∩M ′ s. Then we have x + ur, x + us ∈M, and (x + ur) −(x + us) = ur −us ∈Zn. Now suppose M is closed, bounded, and Vol(M) = 1. Let {sr}∞ r=1 be a sequence of numbers all greater than 1, such that lim r→∞sr = 1. By the argument above we know that for each r there exist xr ̸= yr ∈srM such that xr −yr ∈Zn. Then there are subsequences {xrk} and {yrk} converging to points x, y ∈M, respectively. Since for each rk, xrk −yrk is a nonzero lattice point, it must be true that x ̸= y, and x −y ∈Zn. This completes the proof. □ As a corollary of Theorem 3.2.1 we can prove the following version of Minkowski Convex Body Theorem. 36 3. GEOMETRY OF NUMBERS Theorem 3.2.2 (Minkowski). Let M ⊂Rn be a compact convex 0-symmetric set with Vol(M) ≥2n. Then there exists 0 ̸= x ∈M ∩Zn. Proof. Notice that the set 1 2M = 1 2x : x ∈M  =      1/2 0 . . . 0 0 1/2 . . . 0 . . . . . . ... . . . 0 0 . . . 1/2     M is also convex, 0-symmetric, and by Problem 3.12 its volume is det      1/2 0 . . . 0 0 1/2 . . . 0 . . . . . . ... . . . 0 0 . . . 1/2     Vol(M) = 2−n Vol(M) ≥1. Thererfore, by Theorem 3.2.1, there exist 1 2x ̸= 1 2y ∈1 2M such that 1 2x −1 2y ∈Zn. But, by symmetry, since y ∈M, −y ∈M, and by convexity, since x, −y ∈M, 1 2x −1 2y = 1 2x + 1 2(−y) ∈M. This completes the proof. □ Remark 3.2.1. This result is sharp: for any ε > 0, the cube C =  x ∈Rn : max 1≤i≤n |xi| ≤1 −ε 2  is a convex 0-symmetric set of volume (2 −ε)n, which contains no nonzero integer lattice points. Problem 3.13 extends Blichfeldt and Minkowski theorems to arbitrary lattices as follows: • If Λ ⊂Rn is a lattice of full rank and M ⊂Rn is a compact convex set with Vol(M) ≥det Λ, then there exist x, y ∈M such that 0 ̸= x−y ∈Λ. • If Λ ⊂Rn is a lattice of full rank and M ⊂Rn is a compact convex 0-symmetric set with Vol(M) ≥2n det Λ, then there exists 0 ̸= x ∈M ∩Λ. As a first application of these results, we now prove Minkowski’s Linear Forms Theorem. Theorem 3.2.3. Let B = (bij)1≤i,j≤n ∈GLn(R), and for each 1 ≤i ≤n define a linear form with coefficients bi1, . . . , bin by Li(X) = n X j=1 bijXj. Let c1, . . . , cn ∈R>0 be such that c1 . . . cn = | det(B)|. Then there exists 0 ̸= x ∈Zn such that |Li(x)| ≤ci, 3.2. THEOREMS OF BLICHFELDT AND MINKOWSKI 37 for each 1 ≤i ≤n. Proof. Let us write b1, . . . , bn for the row vectors of B, then Li(x) = bix, for each x ∈Rn. Consider parallelepiped P = {x ∈Rn : |Li(x)| ≤ci ∀1 ≤i ≤n} = B−1R, where R = {x ∈Rn : |xi| ≤ci ∀1 ≤i ≤n} is the rectangular box with sides of length 2c1, . . . , 2cn centered at the origin in Rn. Then by Problem 3.12, Vol(P) = | det(B)|−1 Vol(R) = | det(B)|−12nc1 . . . cn = 2n, and so by Theorem 3.2.2 there exists 0 ̸= x ∈P ∩Zn. □ 38 3. GEOMETRY OF NUMBERS 3.3. Successive minima Let us start with a certain restatement of Minkowski’s Convex Body theorem. Corollary 3.3.1. Let M ⊂Rn be a compact convex 0-symmetric and Λ ⊂Rn a lattice of full rank. Define the first successive minimum of M with respect to Λ to be λ1 = inf {λ ∈R>0 : λM ∩Λ contains a nonzero point } . Then 0 < λ1 ≤2  det Λ Vol(M) 1/n . Proof. The fact that λ1 has to be positive readily follows from Λ being a discrete set. Hence we only have to prove the upper bound. By Theorem 3.2.2 for a general lattice Λ (Problem 3.13), if Vol(λM) ≥2n det(Λ), then λM contains a nonzero point of Λ. On the other hand, by Problem 3.12, Vol(λM) = λn Vol(M). Hence as long as λn Vol(M) ≥2n det(Λ), the expanded set λM is guaranteed to contain a nonzero point of Λ. The conclusion of the corollary follows. □ The above corollary thus provides an estimate as to how much should the set M be expanded to contain a nonzero point of the lattice Λ: this is the meaning of λ1, it is precisely this expansion factor. A natural next question to ask is how much should we expand M to contain 2 linearly independent points of Λ, 3 linearly independent points of Λ, etc. To answer this question is the main objective of this section. We start with a definition. Definition 3.3.1. Let M be a convex, 0-symmetric set M ⊂Rn of non-zero volume and Λ ⊆Rn a lattice of full rank. For each 1 ≤i ≤n define the i-th succesive minimum of M with respect to Λ, λi, to be the infimum of all positive real numbers λ such that the set λM contains at least i linearly independent points of Λ. In other words, λi = inf {λ ∈R>0 : dim (spanR{λM ∩Λ})} ≥i. Since Λ is discrete in Rn, the infimum in this definition is always achieved, i.e. it is actually a minimum. Remark 3.3.1. Notice that the n linearly independent vectors u1, . . . , un corre-sponding to successive minima λ1, . . . , λn, respectively, do not necessarily form a basis. It was already known to Minkowski that they do in dimensions n = 1, . . . , 4, but when n = 5 there is a well known counterexample. Let Λ =       1 0 0 0 1 2 0 1 0 0 1 2 0 0 1 0 1 2 0 0 0 1 1 2 0 0 0 0 1 2       Z5, 3.3. SUCCESSIVE MINIMA 39 and let M = B5, the closed unit ball centered at 0 in Rn. Then the successive minima of B5 with respect to Λ is λ1 = · · · = λ5 = 1, since e1, . . . , e5 ∈B5 ∩Λ, and x = 1 2, 1 2, 1 2, 1 2, 1 2 ⊤ / ∈B5. On the other hand, x cannot be expressed as a linear combination of e1, . . . , e5 with integer coefficients, hence spanZ{e1, . . . , e5} ⊂Λ. An immediate observation is that 0 < λ1 ≤λ2 ≤· · · ≤λn and Corollary 3.3.1 gives an upper bound on λ1. Can we produce bounds on all the successive minima in terms of Vol(M) and det(Λ)? This question is answered by Minkowski’s Successive Minima Theorem. Theorem 3.3.2. With notation as above, 2n det(Λ) n! Vol(M) ≤λ1 . . . λn ≤2n det(Λ) Vol(M) . Proof. We present the proof in case Λ = Zn, leaving generalization of the given argument to arbitrary lattices as an excercise. We start with a proof of the lower bound following [GL87], which is considerably easier than the upper bound. Let u1, . . . , un be the n linearly independent vectors corresponding to the respective successive minima λ1, . . . , λn, and let U = (u1 . . . un) =    u11 . . . un1 . . . ... . . . u1n . . . unn   . Then U = UZn is a full rank sublattice of Zn with index | det(U)|. Notice that the 2n points ±u1 λ1 , . . . , ±un λn lie in M, hence M contains the convex hull P of these points, which is a generalized octahedron. Any polyhedron in Rn can be decomposed as a union of simplices that pairwise intersect only in the boundary. A standard simplex in Rn is the convex hull of n points, so that no 3 of them are co-linear, no 4 of them are co-planar, etc., no k of them lie in a (k −1)-dimensional subspace of Rn, and so that their convex hull does not contain any integer lattice points in its interior. The volume of a standard simplex in Rn is 1/n! (Problem 3.14). Our generalized octahedron P can be decomposed into 2n simplices, which are obtained from the standard simplex by multiplication by the matrix    u11 λ1 . . . un1 λn . . . ... . . . u1n λ1 . . . unn λn   , 40 3. GEOMETRY OF NUMBERS therefore its volume is (3.3) Vol(P) = 2n n! det    u11 λ1 . . . un1 λn . . . ... . . . u1n λ1 . . . unn λn    = 2n| det(U)| n! λ1 . . . λN ≥ 2n n! λ1 . . . λn , since det(U) is an integer. Since P ⊆M, Vol(M) ≥Vol(P). Combining this last observation with (3.3) yields the lower bound of the theorem. Next we prove the upper bound. The argument we present is due to M. Henk [Hen02], and is at least partially based on Minkowski’s original geometric ideas. For each 1 ≤i ≤n, let Ei = spanR{e1, . . . , ei}, the i-th coordinate subspace of Rn, and define Mi = λi 2 M. As in the proof of the lower bound, we take u1, . . . , un to be the n linearly inde-pendent vectors corresponding to the respective successive minima λ1, . . . , λn. In fact, notice that there exists a matrix A ∈GLn(Z) such that A spanR{u1, . . . , ui} ⊆Ei, for each 1 ≤i ≤n, i.e. we can rotate each spanR{u1, . . . , ui} so that it is contained in Ei. Moreover, volume of AM is the same as volume of M, since det(A) = 1 (i.e. rotation does not change volumes), and Aui ∈λ′ iAM ∩Ei, ∀1 ≤i ≤n, where λ′ 1, . . . λ′ n is the successive minima of AM with respect to Zn. Hence we can assume without loss of generality that spanR{u1, . . . , ui} ⊆Ei, for each 1 ≤i ≤n. For an integer q ∈Z>0, define the integral cube of sidelength 2q centered at 0 in Rn Cn q = {z ∈Zn : |z| ≤q}, and for each 1 ≤i ≤n define the section of Cn q by Ei Ci q = Cn q ∩Ei. Notice that Cn q is contained in real cube of volume (2q)n, and so the volume of all translates of M by the points of Cn q can be bounded (3.4) Vol(Cn q + Mn) ≤(2q + γ)n, where γ is a constant that depends on M only. Also notice that if x ̸= y ∈Zn, then int(x + M1) ∩int(y + M1) = ∅, where int stands for interior of a set: suppose not, then there exists z ∈int(x + M1) ∩int(y + M1), 3.3. SUCCESSIVE MINIMA 41 and so (z −x) −(z −y) = y −x ∈int(M1) −int(M1) = {z1 −z2 : z1, z2 ∈M1} = int(λ1M), (3.5) which would contradict minimality of λ1. Therefore (3.6) Vol(Cn q + M1) = (2q + 1)n Vol(M1) = (2q + 1)n λ1 2 n Vol(M). To finish the proof, we need the following lemma. Lemma 3.3.3. For each 1 ≤i ≤n −1, (3.7) Vol(Cn q + Mi+1) ≥ λi+1 λi n−i Vol(Cn q + Mi). Proof. If λi+1 = λi the statement is obvious, so assume λi+1 > λi. Let x, y ∈Zn be such that (xi+1, . . . , xn) ̸= (yi+1, . . . , yn). Then (3.8) (x + int(Mi+1)) ∩(y + int(Mi+1)) = ∅. Indeed, suppose (3.8) is not true, i.e. there exists z ∈(x + int(Mi+1)) ∩(y + int(Mi+1)). Then, as in (3.5) above, x −y ∈int(λi+1M). But we also have u1, . . . , ui ∈int(λi+1M), since λi+1 > λi, and so λiM ⊆int(λi+1M). Moreover, u1, . . . , ui ∈Ei, meaning that ujk = 0 ∀1 ≤j ≤i, i + 1 ≤k ≤n. On the other hand, at least one of xk −yk, i + 1 ≤k ≤n, is not equal to 0. Hence x −y, u1, . . . , ui are linearly independent, but this means that int(λi+1M) contains i+1 linearly independent points, contradicting minimality of λi+1. This proves (3.8). Notice that (3.8) implies Vol(Cn q + Mi+1) = (2q + 1)n−i Vol(Ci q + Mi+1), and Vol(Cn q + Mi) = (2q + 1)n−i Vol(Ci q + Mi), since Mi ⊆Mi+1. Hence, in order to prove the lemma it is sufficient to prove that (3.9) Vol(Ci q + Mi+1) ≥ λi+1 λi n−i Vol(Ci q + Mi). Define two linear maps f1, f2 : Rn →Rn, given by f1(x) = λi+1 λi x1, . . . , λi+1 λi xi, xi+1, . . . , xn  , f2(x) =  x1, . . . , xi, λi+1 λi xi+1, . . . , λi+1 λi xn  , and notice that f2(f1(Mi)) = Mi+1, f2(Ci q) = Ci q. Therefore f2(Ci q + f1(Mi)) = Ci q + Mi+1. 42 3. GEOMETRY OF NUMBERS This implies that Vol(Ci q + Mi+1) = λi+1 λi n−i Vol(Ci q + f1(Mi)), and so to establish (3.9) it is sufficient to show that (3.10) Vol(Ci q + f1(Mi)) ≥Vol(Ci q + Mi). Let E⊥ i = spanR{ei+1, . . . , en}, i.e. E⊥ i is the orthogonal complement of Ei, and so has dimension n −i. Notice that for every x ∈E⊥ i there exists t(x) ∈Ei such that Mi ∩(x + Ei) ⊆(f1(Mi) ∩(x + Ei)) + t(x), in other words, although it is not necessarily true that Mi ⊆f1(Mi), each section of Mi by a translate of Ei is contained in a translate of some such section of f1(Mi). Therefore (Ci q + Mi) ∩(x + Ei) ⊆(Ci q + f1(Mi)) ∩(x + Ei)) + t(x), and hence Vol(Ci q + Mi) = Z x∈E⊥ i Voli((Ci q + Mi) ∩(x + Ei)) dx ≤ Z x∈E⊥ i Voli((Ci q + f1(Mi)) ∩(x + Ei)) dx = Vol(Ci q + f1(Mi)), where Voli stands for the i-dimensional volume. This completes the proof of (3.10), and hence of the lemma. □ Now, combining (3.4), (3.6), and (3.7), we obtain: (2q + γ)n ≥ Vol(Cn q + Mn) ≥  λn λn−1  Vol(Cn q + Mn−1) ≥. . . ≥  λn λn−1  λn−1 λn−2 2 . . . λ2 λ1 n−1 Vol(Cn q + M1) = λn . . . λ1 Vol(M) 2n (2q + 1)n, hence λ1 . . . λn ≤ 2n Vol(M) 2q + γ 2q + 1 n → 2n Vol(M), as q →∞, since q ∈Z>0 is arbitrary. This completes the proof. □ We can talk about successive minima of any convex 0-symmetric set in Rn with respect to the lattice Λ. Perhaps the most frequently encountered such set is the closed unit ball Bn in Rn centered at 0. We define the successive minima of Λ to be the successive minima of Bn with respect to Λ. Notice that successive minima are invariants of the lattice. 3.4. INHOMOGENEOUS MINIMUM 43 3.4. Inhomogeneous minimum Here we exhibit one important application of Minkowski’s successive minima theorem. As before, let Λ ⊆Rn be a lattice of full rank, and let M ⊆Rn be a convex 0-symmetric set of nonzero volume. Throughout this section, we let λ1 ≤· · · ≤λn to be the successive minima of M with respect to Λ. We define the inhomogeneous minimum of M with respect to Λ to be µ = inf{λ ∈R>0 : λM + Λ = Rn}. The main objective of this section is to obtain some basic bounds on µ. We start with the following result of Jarnik [Jar41]. Lemma 3.4.1. µ ≤1 2 n X i=1 λi. Proof. Let us define a function F(x) = inf{a ∈R>0 : x ∈aM}, for every x ∈Rn. This function is a norm (Problem 3.15). Then M = {x ∈Rn : F(x) ≤1} can be thought of as the unit ball with respect to this norm. We will say that F is the norm of M. Let z ∈Rn be an arbitrary point. We want to prove that there exists a point v ∈Λ such that F(z −v) ≤1 2 n X i=1 λi. This would imply that z ∈ 1 2 Pn i=1 λi  M + v, and hence settle the lemma, since z is arbitrary. Let u1, . . . , un be the linearly independent vectors corresponding to successive minima λ1, . . . , λn, respectively. Then F(ui) = λi, ∀1 ≤i ≤n. Since u1, . . . , un form a basis for Rn, there exist a1, . . . , an ∈R such that z = n X i=1 aiui. We can also choose integer v1, . . . , vn such that |ai −vi| ≤1 2, ∀1 ≤i ≤n, and define v = Pn i=1 viui, hence v ∈Λ. Now notice that F(z −v) = F n X i=1 (ai −vi)ui ! ≤ n X i=1 |ai −vi|F(ui) ≤1 2 n X i=1 λi, since F is a norm. This completes the proof. □ 44 3. GEOMETRY OF NUMBERS Using Lemma 3.4.1 along with Minkowski’s successive minima theorem, we can obtain some bounds on µ in terms of the determinant of Λ and volume of M. A nice bound can be easily obtained in an important special case. Corollary 3.4.2. If λ1 ≥1, then µ ≤2n−1n det(Λ) Vol(M) . Proof. Since 1 ≤λ1 ≤· · · ≤λn, Theorem 3.3.2 implies λn ≤λ1 . . . λn ≤2n det(Λ) Vol(M) , and by Lemma 3.4.1, µ ≤1 2 n X i=1 λi ≤n 2 λn. The result follows by combining these two inequalities. □ A general bound depending also on λ1 was obtained by Scherk [Sch50], once again using Minkowski’s successive minima theorem (Theorem 3.3.2) and Jarnik’s inequality (Lemma 3.4.1) He observed that if λ1 is fixed and λ2, . . . , λn are subject to the conditions λ1 ≤· · · ≤λn, λ1 . . . λn ≤2n det(Λ) Vol(M) , then the maximum of the sum λ1 + · · · + λn is attained when λ1 = λ2 = · · · = λn−1, λn = 2n det(Λ) λn−1 1 Vol(M). Hence we obtain Scherk’s inequality for µ. Corollary 3.4.3. µ ≤n −1 2 λ1 + 2n−1 det(Λ) λn−1 1 Vol(M). One can also obtain lower bounds for µ. First notice that for every σ > µ, then the bodies σM + x cover Rn as x ranges through Λ. This means that µM must contain a fundamental domain F of Λ, and so Vol(µM) = µn Vol(M) ≥Vol(F) = det(Λ), hence (3.11) µ ≥  det(Λ) Vol(M) 1/n . In fact, by Theorem 3.3.2,  det(Λ) Vol(M) 1/n ≥(λ1 . . . λn)1/n 2 ≥λ1 2 , 3.4. INHOMOGENEOUS MINIMUM 45 and combining this with (3.11), we obtain (3.12) µ ≥λ1 2 . Jarnik obtained a considerably better lower bound for µ in [Jar41]. Lemma 3.4.4. µ ≥λn 2 . Proof. Let u1, . . . , un be the linearly independent points of Λ corresponding to the successive minima λ1, . . . , λn of M with respect to Λ. Let F be the norm of M, then F(ui) = λi, ∀1 ≤i ≤n. We will first prove that for every x ∈Λ, (3.13) F  x −1 2un  ≥1 2λn. Suppose not, then there exists some x ∈Λ such that F x −1 2un  < 1 2λn. Since F is a norm, we have F(x) ≤F  x −1 2un  + F 1 2un  < 1 2λn + 1 2λn = λn, and similarly F(un −x) ≤F 1 2un −x  + F 1 2un  < λn. Therefore, by definition of λn, x, un −x ∈spanR{u1, . . . , un−1}, and so un = x + (un −x) ∈spanR{u1, . . . , un−1}, which is a contradiction. Hence we proved (3.13) for all x ∈Λ. Further, by Problem 3.16, µ = max z∈Rn min x∈Λ F(x −z). Then lemma follows by combining this observation with (3.13). □ We define the inhomogeneous minimum of Λ to be the inhomogeneous minimum of the closed unit ball Bn with respect to Λ, since it will occur quite often. This is another invariant of the lattice. 46 3. GEOMETRY OF NUMBERS 3.5. Problems Problem 3.1. Let a1, . . . , ar ∈Rn be linearly independent points. Prove that r ≤n. Problem 3.2. Prove that if Λ is a lattice of rank r in Rn, 1 ≤r ≤n, then spanR Λ is a subspace of Rn of dimension r (by spanR Λ we mean the set of all finite real linear combinations of vectors from Λ). Problem 3.3. Let Λ be a lattice of rank r in Rn. By Problem 3.2, V = spanR Λ is an r-dimensional subspace of Rn. Prove that Λ is a discrete co-compact subset of V . Problem 3.4. Let Λ be a lattice of rank r in Rn, and let V = spanR Λ be an r-dimensional subspace of Rn, as in Problem 3.3 above. Prove that Λ and V are both additive groups, and Λ is a subgroup of V . Problem 3.5. Let Λ be a lattice and Ωa subset of Λ. Prove that Ωis a sublattice of Λ if and only if it is a subgroup of the abelian group Λ. Problem 3.6. Let Λ be a lattice and Ωa sublattice of Λ of the same rank. Prove that two cosets x + Ωand y + Ωof Ωin Λ are equal if and only if x −y ∈Ω. Conclude that a coset x + Ωis equal to Ωif and only if x ∈Ω. Problem 3.7. Let Λ be a lattice and Ω⊆Λ a sublattice. Suppose that the quotient group Λ/Ωis finite. Prove that rank of Ωis the same as rank of Λ. Problem 3.8. Given a lattice Λ and a real number µ, define µΛ = {µx : x ∈Λ}. Prove that µΛ is a lattice. Prove that if µ is an integer, then µΛ is a sublattice of Λ. Problem 3.9. Prove that it is possible to select the coefficients vij in Theorem 3.1.5 so that the matrix (vij)1≤i,j≤n is upper (or lower) triangular with non-negative entries, and the largest entry of each row (or column) is on the diagonal. Problem 3.10. Prove that for every point x ∈Rn there exists uniquely a point y ∈F such that x −y ∈Λ, i.e. x lies in the coset y + Λ of Λ in Rn. This means that F is a full set of coset representatives of Λ in Rn. Problem 3.11. Prove that volume of a fundamental parallelotope is equal to the determinant of the lattice. 3.5. PROBLEMS 47 Problem 3.12. Let S be a compact convex set in Rn, A ∈GLn(R), and define T = AS = {Ax : x ∈S}. Prove that Vol(T) = | det(A)| Vol(S). Hint: If we treat multiplication by A as coordinate transformation, prove that its Jacobian is equal to det(A). Now use it in the integral for the volume of T to relate it to the volume of S. Problem 3.13. Prove versions of Theorems 3.2.1 - 3.2.2 where Zn is replaced by an arbitrary lattice Λ ⊆Rn or rank n and the lower bounds on volume of M are multiplied by det(Λ). Hint: Let Λ = AZn for some A ∈GLn(R). Then a point x ∈A−1M ∩Zn if and only if Ax ∈M ∩Λ. Now use Problem 3.12 to relate the volume of A−1M to the volume of M. Problem 3.14. Prove that a standard simplex in Rn has volume 1/n!. Problem 3.15. Let M ⊂Rn be a compact convex 0-symmetric set. Define a function F : Rn →R, given by F(x) = inf{a ∈R>0 : x ∈aM}, for each x ∈Rn. Prove that this is a norm, i.e. it satisfies the three conditions: (1) F(x) = 0 if and only if x = 0, (2) F(ax) = |a|F(x) for every a ∈R and x ∈Rn, (3) F(x + y) ≤F(x) + F(y) for all x, y ∈Rn. Problem 3.16. Let F be a norm like in Problem 3.15. Prove that the inhomo-geneous minimum of the corresponding set M with respect to the full-rank lat-tice Λ ⊂Rn satisfies µ = max z∈Rn min x∈Λ F(x −z). CHAPTER 4 Lattice Problems, Connections and Applications 4.1. Sphere packing, covering and kissing number problems Lattices play an important role in discrete optimization from classical problems to the modern day applications, such as theoretical computer science, digital com-munications, coding theory and cryptography, to name a few. We start with an overview of three old and celebrated problems that are closely related to the tech-niques in the geometry of numbers that we have so far developed, namely sphere packing, sphere covering and kissing number problems. An excellent comprehen-sive, although slightly outdated, reference on this subject is the well-known book by Conway and Sloane [CS99]. Let n ≥2. Throughout this section by a sphere in Rn we will always mean a closed ball whose boundary is this sphere. We will say that a collection of spheres {Bi} of radius r is packed in Rn if int(Bi) ∩int(Bj) = ∅, ∀i ̸= j, and there exist indices i ̸= j such that int(B′ i) ∩int(B′ j) ̸= ∅, whenever B′ i and B′ j are spheres of radius larger than r such that Bi ⊂B′ i, Bj ⊂B′ j. The sphere packing problem in dimension n is to find how densely identical spheres can be packed in Rn. Loosely speaking, the density of a packing is the proportion of the space occupied by the spheres. It is easy to see that the problem really reduces to finding the strategy of positioning centers of the spheres in a way that maximizes density. One possibility is to position sphere centers at the points of some lattice Λ of full rank in Rn; such packings are called lattice packings. Alhtough clearly most packings are not lattices, it is not unreasonable to expect that best results may come from lattice packings; we will mostly be concerned with them. Definition 4.1.1. Let Λ ⊆Rn be a lattice of full rank. The density of correspond-ing sphere packing is defined to be ∆= ∆(Λ) := proportion of the space occupied by spheres = volume of one sphere volume of a fundamental domain of Λ = rnωn det(Λ), 48 4.1. SPHERE PACKING, COVERING AND KISSING NUMBER PROBLEMS 49 where r is the packing radius, i.e. radius of each sphere in this lattice packing, and ωn is the volume of a unit ball in Rn, given by (4.1) ωn = ( πk k! if n = 2k for some k ∈Z 22k+1k!πk (2k+1)! if n = 2k + 1 for some k ∈Z. Hence the volume of a ball of radius r in Rn is ωnrn. It is easy to see that the packing radius r is precisely the radius of the largest ball inscribed into the Voronoi cell V of Λ, i.e. the inradius of V. Clearly ∆≤1. The first observation we can make is that the packing radius r must depend on the lattice. In fact, it is easy to see that r is precisely one half of the length of the shortest non-zero vector in Λ, in other words r = λ1 2 , where λ1 is the first successive minimum of Λ. Therefore ∆= λn 1ωn 2n det(Λ). It is not known whether the packings of largest density in each dimension are necessarily lattice packings, however we do have the following celebrated result of Minkowski (1905) generalized by Hlawka in (1944), which is usually known as Minkowski-Hlawka theorem. Theorem 4.1.1. In each dimension n there exist lattice packings with density (4.2) ∆≥ζ(n) 2n−1 , where ζ(s) = P∞ k=1 1 ks is the Riemann zeta-function. All known proofs of Theorem 4.1.1 are nonconstructive, so it is not generally known how to construct lattice packings with density as good as (4.2); in particular, in dimensions above 1000 the lattices whose existence is guaranteed by Theorem 4.1.1 are denser than all the presently known ones. We refer to [GL87] and [Cas59] for many further details on this famous theorem. Here we present a very brief outline of its proof, following [Cas53]. The first observation is that this theorem readily follows from the following result. Theorem 4.1.2. Let M be a convex bounded 0-symmetric set in Rn with volume < 2ζ(n). Then there exists a lattice Λ in Rn of determinant 1 such that M contains no points of Λ except for 0. Now, to prove Theorem 4.1.2, we can argue as follows. Let χM be the characteristic function of the set M, i.e. χM(x) =  1 if x ∈M 0 if x ̸∈M for every x ∈Rn. For parameters T, ξ1, . . . , ξn−1 to be specified, let us define a lattice Λ = ΛT (ξ1, . . . , xn−1) := n T(a1 + ξ1b), . . . , T(an−1 + ξn−1b), T −(n−1)b  : a1, . . . , an−1, b ∈Z o , 50 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS in other words (4.3) Λ =        T 0 . . . 0 ξ1 0 T . . . 0 ξ2 . . . . . . ... . . . . . . 0 0 . . . T ξn−1 0 0 . . . 0 T −(n−1)        Zn. Hence determinant of this lattice is 1 independent of the values of the parameters. Points of Λ with b = 0 are of the form (Ta1, . . . , Tan−1, 0), and so taking T to be sufficiently large we can ensure that none of them are in M, since M is bounded. Thus assume that T is large enough so that the only points of Λ in M have b ̸= 0. Notice that M contains a nonzero point of Λ if and only if it contains a primitive point of Λ, where we say that x ∈Λ is primitive if it is not a scalar multiple of another point in Λ. The number of symmetric pairs of primitive points of Λ in M is given by the counting function ηT (ξ1, . . . , ξn−1) = X b>0 X a1,...,an−1 gcd(a1,...,an−1,b)=1 χM  T(a1 + ξ1b), . . . , T(an−1 + ξn−1b), T −(n−1)b  . The argument of [Cas53] then proceeds to integrate this expression over all 0 ≤ ξi ≤1, 1 ≤i ≤n −1, obtaining an expression in terms of the volume of M. Taking a limit as T →∞, it is then concluded that since this volume is < 2ζ(n), the average of the counting function ηT (ξ1, . . . , ξn−1) is less than 1. Hence there must exist some lattice of the form (4.3) which contains no nonzero points in M. In general, it is not known whether lattice packings are the best sphere packings in each dimension. In fact, the only dimensions in which optimal packings are currently known are n = 2, 3, 8, 24. In case n = 2, Gauss has proved that the best possible lattice packing is given by the hexagonal lattice (4.4) Λh := 1 1 2 0 √ 3 2  Z2, and in 1940 L. Fejes T´ oth proved that this indeed is the optimal packing (a previous proof by Axel Thue. Its density is π √ 3 6 ≈0.9068996821. In case n = 3, it was conjectured by Kepler that the optimal packing is given by the face-centered cubic lattice   −1 −1 0 1 −1 0 0 1 −1  Z3. The density of this packing is ≈0.74048. Once again, it has been shown by Gauss in 1831 that this is the densest lattice packing, however until recently it was still not proved that this is the optimal packing. The famous Kepler’s conjecture has been settled by Thomas Hales in 1998. Theoretical part of this proof is published only in 2005 [Hal05], and the lengthy computational part was published in a series of papers in the journal of Discrete and Computational Geometry (vol. 36, no. 1 (2006)). 4.1. SPHERE PACKING, COVERING AND KISSING NUMBER PROBLEMS 51 Dimensions n = 8 and n = 24 were settled in 2016, a week apart from each other. Maryna Viazovska [Via17], building on previous work of Cohn and Elkies [CE03], discovered a “magic” function that implied optimality of the ex-ceptional root lattice E8 for packing density in R8. Working jointly with Cohn, Kumar, Miller and Radchenko [CKM+17], she then immediately extended her method to dimension 24, where the optimal packing density is given by the famous Leech lattice. Detailed constructions of these remarkable lattices can be found in Conway and Sloane’s book [CS99]. This outlines the currently known results for optimal sphere packing configurations in general. On the other hand, best lattice packings are known in dimensions n ≤8, as well as n = 24. There are dimensions in which the best known packings are not lattice packings, for instance n = 11. Next we give a very brief introduction to sphere covering. The problem of sphere covering is to cover Rn with spheres such that these spheres have the least possible overlap, i.e. the covering has smallest possible thickness. Once again, we will be most interested in lattice coverings, that is in coverings for which the centers of spheres are positioned at the points of some lattice. Definition 4.1.2. Let Λ ⊆Rn be a lattice of full rank. The thickness Θ of corresponding sphere covering is defined to be Θ(Λ) = average number of spheres containing a point of the space = volume of one sphere volume of a fundamental domain of Λ = Rnωn det(Λ), where ωn is the volume of a unit ball in Rn, given by (4.1), and R is the covering radius, i.e. radius of each sphere in this lattice covering. It is easy to see that R is precisely the radius of the smallest ball circumscribed around the Voronoi cell V of Λ, i.e. the circumradius of V. Clearly Θ ≥1. Notice that the covering radius R is precisely µ, the inhomogeneous minimum of the lattice Λ. Hence combining Lemmas 3.4.1 and 3.4.4 we obtain the following bounds on the covering radius in terms of successive minima of Λ: λn 2 ≤µ = R ≤1 2 n X i=1 λi ≤nλn 2 . The optimal sphere covering is only known in dimension n = 2, in which case it is given by the same hexagonal lattice (4.4), and is equal to ≈1.209199. Best possible lattice coverings are currently known only in dimensions n ≤5, and it is not known in general whether optimal coverings in each dimension are necessarily given by lattices. Once again, there are dimensions in which the best known coverings are not lattice coverings. In summary, notice that both, packing and covering properties of a lattice Λ are very much dependent on its Voronoi cell V. Moreover, to simultaneously optimize packing and covering properties of Λ we want to ensure that the inradius r of V is largest possible and circumradius R is smallest possible. This means that we want 52 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS to take lattices with the “roundest” possible Voronoi cell. This property can be expressed in terms of the successive minima of Λ: we want λ1 = · · · = λn. Lattices with these property are called well-rounded lattices, abbreviated WR; an-other term ESM lattices (equal successive minima) is also sometimes used. Notice that if Λ is WR, then by Lemma 3.4.4 we have r = λ1 2 = λn 2 ≤R, although it is clearly impossible for equality to hold in this inequality. Sphere packing and covering results have numerous engineering applications, among which there are applications to coding theory, telecommunications, and image processing. WR lattices play an especially important role in these fields of study. Another closely related classical question is known as the kissing number prob-lem: given a sphere in Rn how many other non-overlapping spheres of the same radius can touch it? In other words, if we take the ball centered at the origin in a sphere packing, how many other balls are adjacent to it? Unlike the packing and covering problems, the answer here is easy to obtain in dimension 2: it is 6, and we leave it as an exercise for the reader (Problem 4.2). Although the term “kissing number” is contemporary (with an allusion to billiards, where the balls are said to kiss when they bounce), the 3-dimensional version of this problem was the subject of a famous dispute between Isaac Newton and David Gregory in 1694. It was known at that time how to place 12 unit balls around a central unit ball, however the gaps between the neighboring balls in this arrangement were large enough for Gregory to conjecture that perhaps a 13-th ball can some how be fit in. Newton thought that it was not possible. The problem was finally solved by Sch¨ utte and van der Waerden in 1953 [SvdW53] (see also [Lee56] by J. Leech, 1956), con-firming that the kissing number in R3 is equal to 12. The only other dimensions where the maximal kissing number is known are n = 4, 8, 24. More specifically, if we write τ(n) for the maximal possible kissing number in dimension n, then it is known that τ(2) = 6, τ(3) = 12, τ(4) = 24, τ(8) = 240, τ(24) = 196560. In many other dimensions there are good upper and lower bounds available, and the general bounds of the form 20.2075...n(1+o(1)) ≤τ(n) ≤20.401n(1+o(1)) are due to Wyner, Kabatianski and Levenshtein; see [CS99] for detailed references and many further details. A more specialized question is concerned with the maximal possible kissing number of lattices in a given dimension, i.e. we consider just the lattice packings instead of general sphere packing configurations. Here the optimal results are known in all dimensions n ≤8 and dimension 24: al of the optimal lattices here are also known to be optimal for lattice packing. Further, in all dimensions where the overall maximal kissing numbers are known, they are achieved by lattices. Let Λ ⊂Rn be a lattice, then its minimal norm |Λ| is simply its first successive minimum, i.e. |Λ| = min {∥x∥: x ∈Λ \ {0}} . 4.1. SPHERE PACKING, COVERING AND KISSING NUMBER PROBLEMS 53 The set of minimal vectors of Λ is then defined as S(Λ) = {x ∈Λ : ∥x∥= |Λ|} . These minimal vectors are the centers of spheres of radius |Λ|/2 in the sphere packing associated to Λ which touch the ball centered at the origin. Hence the number of these vectors, |S(Λ)| is precisely the kissing number of Λ. One immediate observation then is that to maximize the kissing number, same as to maximize the packing density, we want to focus our attention on WR lattices: they will have at least 2n minimal vectors. A matrix U ∈GLn(R) is called orthogonal if U −1 = U ⊤, and the subset of all such matrices in GLn(R) is On(R) = {U ∈GLn(R) : U −1 = U ⊤}. This is a subgroup of GLn(R) (Problem 4.5). Discrete optimization problems on the space of lattices in a given dimension, as those discussed above, are usually considered up to the equivalence relation of similarity: two lattices L and M of full rank in Rn are called similar, denoted L ∼M, if there exists α ∈R and an orthogonal matrix U ∈On(R) such that L = αUM. This is an equivalence relation on the space of all full-rank lattices in Rn (Problem 4.3), and we refer to the equivalence classes under this relation as similarity classes. If lattices L and M are similar, then they have the same packing density, covering thickness, and kissing number (Problem 4.4). We use the perspective of similarity classes in the next section when considering lattice packing density in the plane. 54 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS Figure 1. Hexagonal lattice with Voronoi cell translates and as-sociated circle packing 4.2. Lattice packings in dimension 2 Our goal here is to prove that the best lattice packing in R2 is achieved by the hexagonal lattice Λh as defined in (4.4) above (see Figure 1). Specifically, we will prove the following theorem. Theorem 4.2.1. Let L be a lattice of rank 2 in R2. Then ∆(L) ≤∆(Λh) = π 2 √ 3 = 0.906899 . . . , and the equality holds if any only if L ∼Λh. This result was first obtain by Lagrange in 1773, however we provide a more con-temporary proof here following [Fuk11]. Our strategy is to show that the problem of finding the lattice with the highest packing density in the plane can be restricted to the well-rounded lattices without any loss of generality, where the problem be-comes very simple. We start by proving that vectors corresponding to successive minima in a lattice in R2 form a basis. Lemma 4.2.2. Let Λ be a lattice in R2 with successive minima λ1 ≤λ2 and let x1, x2 be the vectors in Λ corresponding to λ1, λ2, respectively. Then x1, x2 form a basis for Λ. Proof. Let y1 ∈Λ be a shortest vector extendable to a basis in Λ, and let y2 ∈Λ be a shortest vector such that y1, y2 is a basis of Λ. By picking ±y1, ±y2 if necessary we can ensure that the angle between these vectors is no greater than π/2. Then 0 < ∥y1∥≤∥y2∥, and for any vector z ∈Λ with ∥z∥< ∥y2∥the pair y1, z is not a basis for Λ. Since x1, x2 ∈Λ, there must exist integers a1, a2, b1, b2 such that (4.5) (x1 x2) = (y1 y2) a1 b1 a2 b2  . Let θx be the angle between x1, x2, and θy be the angle between y1, y2, then π/3 ≤θx ≤π/2 by Problem 4.7. Moreover, π/3 ≤θy ≤π/2: indeed, suppose 4.2. LATTICE PACKINGS IN DIMENSION 2 55 θy < π/3, then by Problem 4.6, ∥y1 −y2∥< ∥y2∥, however y1, y1 −y2 is a basis for Λ since y1, y2 is; this contradicts the choice of y2. Define D = det a1 b1 a2 b2  , then D is a positive integer, and taking determinants of both sides of (4.5), we obtain (4.6) ∥x1∥∥x2∥sin θx = D∥y1∥∥y2∥sin θy. Notice that by definition of successive minima, ∥x1∥∥x2∥≤∥y1∥∥y2∥, and hence (4.6) implies that D = ∥x1∥∥x2∥ ∥y1∥∥y2∥ sin θx sin θy ≤ 2 √ 3 < 2, meaning that D = 1. Combining this observation with (4.5), we see that (x1 x2) a1 b1 a2 b2 −1 = (y1 y2) , where the matrix a1 b1 a2 b2 −1 has integer entries. Therefore x1, x2 is also a basis for Λ, completing the proof. □ As we know from Remark 3.3.1 in Section 3.3, the statement of Lemma 4.2.2 does not generally hold for d ≥5. We will call a basis for a lattice as in Lemma 4.2.2 a minimal basis. The goal of the next three lemmas is to show that the lattice packing density function ∆attains its maximum in R2 on the set of well-rounded lattices. Lemma 4.2.3. Let Λ and Ωbe lattices of full rank in R2 with successive minima λ1(Λ), λ2(Λ) and λ1(Ω), λ2(Ω) respectively. Let x1, x2 and y1, y2 be vectors in Λ and Ω, respectively, corresponding to successive minima. Suppose that x1 = y1, and angles between the vectors x1, x2 and y1, y2 are equal, call this common value θ. Suppose also that λ1(Λ) = λ2(Λ). Then ∆(Λ) ≥∆(Ω). Proof. By Lemma 4.2.2, x1, x2 and y1, y2 are minimal bases for Λ and Ω, respectively. Notice that λ1(Λ) = λ2(Λ) = ∥x1∥= ∥x2∥ = ∥y1∥= λ1(Ω) ≤∥y2∥= λ2(Ω). Then ∆(Λ) = πλ1(Λ)2 4 det(Λ) = λ1(Λ)2π 4∥x1∥∥x2∥sin θ = π 4 sin θ ≥ λ1(Ω)2π 4∥y1∥∥y2∥sin θ = λ1(Ω)2π 4 det(Ω) = ∆(Ω). (4.7) □ 56 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS The following lemma is a converse to Problem 4.7. Lemma 4.2.4. Let Λ ⊂R2 be a lattice of full rank, and let x1, x2 be a basis for Λ such that ∥x1∥= ∥x2∥, and the angle θ between these vectors lies in the interval [π/3, π/2]. Then x1, x2 is a minimal basis for Λ. In particular, this implies that Λ is WR. Proof. Let z ∈Λ, then z = ax1 + bx2 for some a, b ∈Z. Then ∥z∥2 = a2∥x1∥2 + b2∥x2∥2 + 2abx⊤ 1 x2 = (a2 + b2 + 2ab cos θ)∥x1∥2. If ab ≥0, then clearly ∥z∥2 ≥∥x1∥2. Now suppose ab < 0, then again ∥z∥2 ≥(a2 + b2 −|ab|)∥x1∥2 ≥∥x1∥2, since cos θ ≤1/2. Therefore x1, x2 are shortest nonzero vectors in Λ, hence they correspond to successive minima, and so form a minimal basis. Thus Λ is WR, and this completes the proof. □ Lemma 4.2.5. Let Λ be a lattice in R2 with successive minima λ1, λ2 and corre-sponding basis vectors x1, x2, respectively. Then the lattice ΛWR =  x1 λ1 λ2 x2  Z2 is WR with successive minima equal to λ1. Proof. By Problem 4.7, the angle θ between x1 and x2 is in the interval [π/3, π/2], and clearly this is the same as the angle between the vectors x1 and λ1 λ2 x2. Then by Lemma 4.2.4, ΛWR is WR with successive minima equal to λ1. □ Now combining Lemma 4.2.3 with Lemma 4.2.5 implies that (4.8) ∆(ΛWR) ≥∆(Λ) for any lattice Λ ⊂R2, and (4.7) readily implies that the equality in (4.8) occurs if and only if Λ = ΛWR, which happens if and only if Λ is well-rounded. Therefore the maximum packing density among lattices in R2 must occur at a WR lattice, and so for the rest of this section we talk about WR lattices only. Next observation is that for any WR lattice Λ in R2, (4.7) implies: sin θ = π 4∆(Λ), meaning that sin θ is an invariant of Λ, and does not depend on the specific choice of the minimal basis. Since by our conventional choice of the minimal basis and Problem 4.7, this angle θ is in the interval [π/3, π/2], it is also an invariant of the lattice, and we call it the angle of Λ, denoted by θ(Λ). Lemma 4.2.6. Let Λ be a WR lattice in R2. A lattice Ω⊂R2 is similar to Λ if and only if Ωis also WR and θ(Λ) = θ(Ω). Proof. First suppose that Λ and Ωare similar. Let x1, x2 be the minimal basis for Λ. There exist a real constant α and a real orthogonal 2 × 2 matrix U such that Ω= αUΛ. Let y1, y2 be a basis for Ωsuch that (y1 y2) = αU(x1 x2). 4.2. LATTICE PACKINGS IN DIMENSION 2 57 Then ∥y1∥= ∥y2∥, and the angle between y1 and y2 is θ(Λ) ∈[π/3, π/2]. By Lemma 4.2.4 it follows that y1, y2 is a minimal basis for Ω, and so Ωis WR and θ(Ω) = θ(Λ). Next assume that Ωis WR and θ(Ω) = θ(Λ). Let λ(Λ) and λ(Ω) be the respective values of successive minima of Λ and Ω. Let x1, x2 and y1, y2 be the minimal bases for Λ and Ω, respectively. Define z1 = λ(Λ) λ(Ω)y1, z2 = λ(Λ) λ(Ω)y2. Then x1, x2 and z1, z2 are pairs of points on the circle of radius λ(Λ) centered at the origin in R2 with equal angles between them. Therefore, there exists a 2 × 2 real orthogonal matrix U such that (y1 y2) = λ(Λ) λ(Ω)(z1 z2) = λ(Λ) λ(Ω)U(x1 x2), and so Λ and Ωare similar lattices. This completes the proof. □ We are now ready to prove the main result of this section. Proof of Theorem 4.2.1. The density inequality (4.8) says that the largest lattice packing density in R2 is achieved by some WR lattice Λ, and (4.7) implies that (4.9) ∆(Λ) = π 4 sin θ(Λ), meaning that a smaller sin θ(Λ) corresponds to a larger ∆(Λ). Problem 4.7 implies that θ(Λ) ≥π/3, meaning that sin θ(Λ) ≥ √ 3/2. Notice that if Λ is the hexagonal lattice Λh = 1 1 2 0 √ 3 2  Z2, then sin θ(Λ) = √ 3/2, meaning that the angle between the basis vectors (1, 0) and (1/2, √ 3/2) is θ = π/3, and so by Lemma 4.2.4 this is a minimal basis and θ(Λ) = π/3. Hence the largest lattice packing density in R2 is achieved by the hexagonal lattice. This value now follows from (4.9). Now suppose that for some lattice Λ, ∆(Λ) = ∆(Λh), then by (4.8) and a short argument after it Λ must be WR, and so ∆(Λ) = π 4 sin θ(Λ) = ∆(Λh) = π 4 sin π/3. Then θ(Λ) = π/3, and so Λ is similar to Λh by Lemma 4.2.6. This completes the proof. □ While we have only settled the question of best lattice packing in dimension two, we saw that well-roundedness is an essential property for a lattice to be a good contender for optimal packing density. There are, however, infinitely many WR lattices in the plane, even up to similarity, and only one of them worked well. One can then ask what properties must a lattice have to maximize packing density? 58 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS A full-rank lattice Λ in Rn with minimal vectors x1, . . . , xm is called eutactic if there exist positive real numbers c1, . . . , cm such that (4.10) ∥v∥2 = m X i=1 ci(v⊤xi)2 for every vector v ∈spanR Λ. If c1 = · · · = cn, Λ is called strongly eutactic. A lattice is called perfect if the set of symmetric matrices {xix⊤ i : 1 ≤i ≤m} spans the real vector space of n × n symmetric matrices. These properties are preserved on similarity classes (Problem 4.8), and up to similarity there are only finitely many perfect or eutactic lattices in every dimension. For instance, up to similarity, the hexagonal lattice is the only one in the plane that is both, perfect and eutactic (Problem 4.9). Suppose that Λ = AZn is a lattice with basis matrix A, then, as we know, B is another basis matrix for Λ if and only if B = AU for some U ∈GLn(Z). In this way, the space of full-rank lattices in Rn can be identified with the set of orbits of GLn(R) under the action by GLn(Z) by right multiplication. The packing density ∆is a continuous function on this space, and hence we can talk about its local extremum points. A famous theorem of Georgy Voronoi (1908) states that a lattice is a local maximum of the packing density function in its dimension if and only if it is perfect and eutactic. Hence, combining Problem 4.9 with Voronoi’s theorem gives another proof of unique optimality of the hexagonal lattice for lattice packing in the plane. Further, Voronoi’s theorem suggests a way of looking for the maximizer of the lattice packing density in every dimension: identify the finite set of perfect and eutactic lattices, compute their packing density and choose the largest. Unfortunately, this approach is not very practical, since already in dimension 9 the number of perfect lattices is over 9 million. 4.3. ALGORITHMIC PROBLEMS ON LATTICES 59 4.3. Algorithmic problems on lattices There is a class of algorithmic problems studied in computational number the-ory, discrete geometry and theoretical computer science, which are commonly re-ferred to as the lattice problems. One of their distinguishing features is that they are provably known to be very hard to solve in the sense of computational complexity of algorithms involved. As usual, we write Λ ⊂Rn for a lattice of full rank and 0 < λ1 ≤· · · ≤λn for its successive minima. A lattice can be given in the form its basis matrix, i.e. a matrix A ∈GLn(R) such that Λ = AZn. There are several questions that can be asked about this setup. We formulate them in algorithmic form. Shortest Vector Problem (SVP). Input: A matrix A ∈GLn(R). Output: A vector x1 ∈Λ = AZn such that ∥x1∥= λ1. Shortest Independent Vector Problem (SIVP). Input: A matrix A ∈GLn(R). Output: Linearly independent vectors x1, . . . , xn ∈Λ = AZn such that ∥xi∥= λ1 ∀1 ≤i ≤n. Closest Vector Problem (CVP). Input: A matrix A ∈GLn(R) and a vector y ∈Rn. Output: A vector x ∈Λ = AZn such that ∥x −y∥≤∥z −y∥∀z ∈Λ. Shortest Basis Problem (SBP). Input: A matrix A ∈GLn(R). Output: A basis b1, . . . , bn for Λ = Zn such that ∥bi∥= min{∥x∥: x ∈Λ is such that b1, . . . , bi−1, x is extendable to a basis} for all 1 ≤i ≤n. Notice that SVP is a special case of CVP where the input vector y is taken to be 0: indeed, a vector corresponding to the first successive minimum is precisely a vector that is closer to the origin than any other point of Λ. On the other hand, SIVP and SBP are different problems: as we know, lattices in dimensions 5 higher may not have a basis of vectors corresponding to successive minima. All of these algorithmic problems are all known to be NP-hard. In fact, even the problem of determining the first successive minimum of the lattice is already NP-hard. We can also ask for γ-approximate versions of these problems for some approximation factor γ. In other words, for the same input we want to return an answer that is bigger than the optimal by a factor of no more than γ. For instance, the γ-SVP would ask for a vector x ∈Λ such that ∥x∥≤γλ1. It is an open problem to decide whether the γ-approximate versions of these prob-lems are in the P class for any values of γ polynomial in the dimension n. 60 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS On the other hand, γ-approximate versions of these problems for γ exponential in n are known to be polynomial. The most famous such approximation algo-rithm is LLL, which was discovered by A. Lenstra, H. Lenstra and L. Lovasz in 1982 [LLL82]. LLL is a polynomial time reduction algorithm that, given a lattice Λ, produces a basis b1, . . . , bn for Λ such that min 1≤i≤n ∥bi∥≤2 n−1 2 λ1, and (4.11) n Y i=1 ∥bi∥≤2 n(n−1) 4 det(Λ). We can compare this to the upper bound given by Minkowski’s Successive Minima Theorem (Theorem 3.3.2): (4.12) n Y i=1 λi ≤2n ωn det(Λ). For instance, when n = 2k the bound (4.11) gives n Y i=1 ∥bi∥≤2 k(2k−1) 2 det(Λ), while (4.12) gives n Y i=1 λi ≤4kk! πk det(Λ). Let us briefly describe the main idea behind LLL. The first observation is that an orthogonal basis, if one exists in a lattice, is always the shortest one. Indeed, suppose u1, . . . , un is such a basis, then for any a1, . . . , an ∈Z, n X i=1 aiui 2 = n X i=1 a2 i ∥ui∥2, which implies that the shortest basis vectors can only be obtained by taking one of the coefficients ai = ±1 and the rest 0. Of course, most lattices do not have orthogonal bases, in which case finding a short basis is much harder. Still, the basic principle of constructing a short basis is based on looking for vectors that would be “close to orthogonal”. We observed in Section 4.2 (in particular, see Problems 4.6, 4.7, Lemma 4.2.4) that the angle between a pair of shortest vectors must be between [π/3, 2π/3], i.e. these vectors are “near-orthogonal”: in fact, these vectors have to be as close to orthogonal as possible within the lattice. This is the underlying idea behind the classical Lagrange-Gauss Algorithm for finding a shortest basis for a lattice in R2. Specifically, an ordered basis b1, b2 for a planar lattice Λ consists of vectors corresponding to successive minima λ1, λ2 of Λ, respectively, if and only if µ := b⊤ 1 b2 ∥b1∥2 ≤1 2. On the other hand, if |µ| > 1/2, then replacing b2 with b2 −⌊µ⌉b1, 4.3. ALGORITHMIC PROBLEMS ON LATTICES 61 where ⌊µ⌉stands for the nearest integer to µ, produces a shorter second basis vector. We leave the proof of this as an exercise (Problem 4.10). Hence we can formulate the Gauss-Lagrange Algorithm: Input: b1, b2 ∈R2 such that ∥b1∥≤∥b2∥ Compute µ: µ = b⊤ 1 b2 ∥b1∥2 Check µ: if |µ| ≤1/2, output b1, b2; else set b2 ←b2 −⌊µ⌉b1 and repeat the algorithm (swapping b1, b2, if necessary, to ensure ∥b1∥≤∥b2∥) This algorithm terminates in a finite number of steps (Problem 4.11). Let us demonstrate this algorithm on an example. Suppose Λ = spanZ{b1, b2}, where b1 = 1 5  , b2 = 1 0  . We notice that ∥b1∥> ∥b2∥, so we swap the vectors: b1 ↔b2. We then compute µ = b⊤ 1 b2 ∥b1∥2 = 1 > 1/2. The nearest integer to µ is 1, so we set b2 ←b2 −b1 = 0 5  . We still have ∥b1∥< ∥b2∥, so no need to swap the vectors. With the new basis b1, b2 we again compute µ, which is now equal to 0 < 1/2. Hence we found a shortest basis for Λ: 1 0  , 0 5  . LLL is based on a generalization of this idea. We can start with a basis b1, . . . , bn for a lattice Λ in Rn and use the Gram-Schmidt orthogonalization proce-dure to compute a corresponding orthogonal (but not normalized) basis b′ 1, . . . , b′ n for Rn. For any pair of indices i, j with 1 ≤j < i ≤i, let us compute the Gram-Schmidt coefficient µij = b⊤ i b′ j ∥b′ j∥2 . If this coefficient is > 1/2 in absolute value, we swap bi ←bi −⌊µ⌉bj: this ensures the length reduction, but one other condition is also needed. Formally speaking, a resulting basis b1, . . . , bn is called LLL reduced if the following two conditions are satisfied: (1) For all 1 ≤j < i ≤n, |µij| ≤1/2 (2) For some parameter δ ∈[1/4, 1), for all 1 ≤k ≤n, δ∥b′ k−1∥2 ≤∥b′ k∥+ µ2 k,(k−1)∥b′ k−1∥2. Traditionally, δ is taken to be 3/4. While we will not go into further details about the LLL, some good more detailed references on this subject include the original paper [LLL82], as well as more recent books [Coh00], [Bor02], and [HPS08]. 62 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS 4.4. CVP is NP-hard In this section we discuss the complexity of the decision version of the CVP for sublattices of the integer lattice Zn. Specifically, here is the problem we are considering: Given an n × m integer basis matrix B, m ≤n, a target vector t ∈Zn and a (usually rational) number r > 0, does there exist a vector x ∈BZm such that ∥x −t∥≤r? We will now explicitly show that this problem is NP-hard. Theorem 4.4.1. The decision version of CVP is NP-complete. Proof. First notice that, given a vector x ∈BZm, checking whether ∥x−t∥≤ r is a polynomial problem: it comes down to computing the difference vector, evaluating its Euclidean norm, and comparing it to r. Hence our problem is NP. To show that it is NP-hard, we will construct a polynomial-time reduction algorithm from SSP (the subset sum problem) to decision CVP. Since we know that SSP is NP-hard (Theorem 2.1.2), the result will follow. Let (4.13) a = (a1, . . . , an), s be an instance of SSP, i.e. a is the n-tuple of weights and s is the target sum. Define the (n + 1) × n basis matrix B for a lattice BZn ⊂Zn+1 by B =  a 2In  =        a1 a2 . . . an 2 0 . . . 0 0 2 . . . 0 . . . . . . ... . . . 0 0 . . . 2        , and let t = (s, 1, . . . , 1)⊤∈Zn+1 be the target vector. The we can consider the instance of decision CVP with B, t and r = √n, i.e. (4.14) does there exist a vector x ∈BZn so that ∥x −t∥≤√n? Assume (4.13) is a YES instance of SSP, i.e. Pn i=1 aixi = s for some choice of coefficients x1, . . . , xn ∈{0, 1}, then Bx −t =      Pn i=1 aixi −s 2x1 −1 . . . 2xn −1     , and so ∥Bx −t∥2 = Pn i=1 |2xi −1|2 = n, since 2xi −1 = ±1 for every i, and thus the answer to (4.14) is also YES. Conversely, a YES instance of (4.14) with B and t as above gives a vector y = Bx ∈BZn such that ∥Bx −t∥2 = n X i=1 aixi −s 2 + n X i=1 |2xi −1|2 ≤n, which can only be true if Pn i=1 aixi −s = 0, since again 2xi −1 = ±1 for every i. Thus we obtain a YES instance of (4.13). Hence we have a reduction from SSP to decision CVP, which is polynomial-time by construction. □ 4.5. GEOMETRY OF THE FROBENIUS PROBLEM 63 4.5. Geometry of the Frobenius problem In this section we will apply the newly-acquired knowledge of lattices and their geometric invariants to the Frobenius problem. A geometric approach to the clas-sical Frobenius problem has been pioneered in the influential paper of R. Kan-nan [Kan92], leading to a polynomial-time algorithm to find the Frobenius number for each fixed n. Bounds on the classical Frobenius number stemming from fur-ther geometry of numbers applications have been obtained in [FR07] and [AG07]. These ideas have also been extended to the more general s-Frobenius problem in [FS11] and [AFH12]. A higher-dimensional analogue of the Frobenius problem has also been considered in the recent years by several authors, notably in [AH10], [AHL13], and [ALL16]. A generalization of this problem to certain number fields has been studied in [FS20]. Let us briefly describe Kannan’s approach to the Frobenius problem. Let La = ( x ∈Zn−1 : n−1 X i=1 aixi ≡0 (mod an) ) , then La is a sublattice of Zn−1 of full rank. Define also a simplex Sa = ( x ∈Rn−1 ≥0 : n−1 X i=1 aixi ≤1 ) . With this notation, Kannan proved the following remarkable identity. Theorem 4.5.1. (4.15) g0(a) = µ(Sa, La) − n X i=1 ai. where µ(Sa, La) is the inhomogeneous minimum (also known as the covering radius) of Sa with respect to L, namely (4.16) µ(Sa, La) = inf  t ∈R>0 : tSa + La = Rn−1 . Proof. Kannan’s argument consists of an upper and a lower bound on the inhomogeneous minimum. First we show that (4.17) µ(Sa, La) ≤g0(a) + n X i=1 ai. Assume that y ∈Zn−1 is such that Pn−1 i=1 aiyi ≡m (mod an). Let tm be the smallest positive integer congruent to m modulo an that is representable as a nonnegative integer linear combination of a1, . . . , an−1. Then there exist coefficients x1, . . . , xn ∈ Z≥0 such that tm = a1x1 + · · · + an−1xn−1 + anxn = m + anxn. Let x′ = (x1, . . . , xn−1) for this choice of the coefficients, and observe that y −x′ ∈ La. Further, x′ ∈mSa ⊆tmSa and so y = (y −x′) + x′ ∈La + tmSa. Since the choice of y ∈Zn−1 was arbitrary, we conclude that Zn−1 ⊆La+tmSa. Additionally, tm ≤g0(a) + an, and thus Zn−1 ⊆La + (g0(a) + an)Sa. 64 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS Also notice that for any point z ∈Rn−1, the integer part [z] = ([z1], . . . , [zn−1]) ∈ Zn−1 and the point z′ = z −[z] has each coordinate ≤1, and so satisfies the inequality n−1 X i=1 aiz′ i ≤ n−1 X i=1 ai. This means that z = [z] + z′ ∈Zn−1 + n−1 X i=1 ai ! Sa. Thus Rn−1 ⊆Zn−1 + n−1 X i=1 ai ! Sa ⊆La + g0(a) + n X i=1 ai ! Sa, which implies (4.17). Next we establish that (4.18) µ(Sa, La) ≥g0(a) + n X i=1 ai. For this, we first need an auxiliary lemma. Lemma 4.5.2. g0(a) = max1≤m≤an−1 tm −an. Proof. If a positive integer T is congruent to 0 modulo an, then T is just a multiple of an. Otherwise, T ≡m (mod an) for some 1 ≤m ≤an −1 and thus it is representable as a nonnegative linear combination of a1, . . . , an if and only if it is ≥tm. □ Back to the proof of (4.18), let us consider the set (g0(a)+an)Sa +La. We will show that g0(a)+an is the smallest positive real value of t so that tSa+La contains Zn−1. Suppose not, then there exists some t′ < g0(a)+an so that Zn−1 ⊂t′Sa+La. Pick any 1 ≤m ≤an −1 and take y ∈Zn−1 be such that n−1 X i=1 aiyi ≡m (mod an). Since y ∈t′Sba + x for some x ∈La, we must have y −x ∈t′Sa. However, n−1 X i=1 ai(yi −xi) ≡m (mod an) and yi −zi ≥0 ∀1 ≤i ≤n implies that tm ≤t′. This is true for any choice of m, Lemma 4.5.2 implies g0(a) = max 1≤m≤an−1 tm −an ≤t′ −an < g0(a) by our assumption on t′. This is a contradiction, hence g0(a) + an = min  t > 0 : Zn−1 ⊂tSa + La . Therefore there must exist y ∈Zn−1 such that for any x ∈La with yi −xi ≥0 for all i, we have (4.19) n−1 X i=1 ai(yi −xi) ≥g0(a) + an. 4.5. GEOMETRY OF THE FROBENIUS PROBLEM 65 Let ε ∈(0, 1) and define the point p ∈Rn−1 by pi = yi+(1−ε) for every i. Suppose x ∈La is such that xi ≥pi for every i. Since all xi’s are integers, we must have xi ≤yi for every i, and so n−1 X i=1 ai(pi −xi) = (1 −ε) n−1 X i=1 ai + n−1 X i=1 ai(yi −xi) ≥(1 −ε) n−1 X i=1 ai + (g0(a) + an), by (4.19). Now, µ(Sa, Lba) is ≥than the left hand side of this inequality, which holds for any ε ∈(0, 1). Thus we must have µ(Sa, La) ≥ n−1 X i=1 ai + (g0(a) + an) = g0(a) + n X i=1 ai. This completes the proof. □ On the other hand, Kannan showed that in every fixed dimension n there is a polynomial-time algorithm to find the covering radius, given Sa and La (which is to say, given a). This result, along with his identity (4.15) implies a polynomial-time algorithm for the Frobenius number in fixed dimension. Kannan’s Theorem 4.5.1 has been extended to the s-Frobenius numbers in [AFH12]. For integer s ≥1, define (4.20) µs(Sa, La) = min{t > 0 : ∀x ∈Rn ∃b1, . . . , bs ∈La s.t. x ∈bi + tSa} be the smallest positive number t such that any x ∈Rn is covered by at least s lattice translates of tSa: this µs(Sa, La) is called the s-covering radius of Sa with respect to La. If s = 1, this is precisely the classical covering radius as in (4.16). With this notation, the following theorem is established in [AFH12]. Theorem 4.5.3. gs(a) = µs+1(Sa, La) − n X i=1 ai. Such geometric ideas have also been used by different authors to give expected val-ues of Frobenius numbers with respect to the uniform probability distribution on ensembles of vectors in Zn defined with respect to different norms; see [Arn99], [Arn06], [AH09], [AHH11], [BS07], [Li15], [Mar10], [Str12], [SSU09], [Ust10], and [AFH12] for results on average behavior of Frobenius numbers. 66 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS 4.6. Lattice point counting All of the lattice point counting results in the previous chapters were specifically for integer lattice points in polytopes, which is a rather special class of convex bodies in Rn and only one lattice. What can be said for more general convex bodies and lattices? Let M ⊆Rn be closed, bounded, and Jordan measurable with Vol(M) > 0, and let Λ ⊆Rn be a lattice of full rank. Suppose we homogeneously expand M by a positive real parameter t, i.e. for each positive real value of t we will consider the set tM. How many points of Λ are there in tM as t grows? To partially answer this question, we will be interested in the asymptotic behavior of the function GM,Λ(t) = |tM ∩Λ| as t →∞. In general, this is a very difficult question. We will need to make some additional assumptions on M in order to study GM,Λ(t). Definition 4.6.1. Let S be a subset of some Eucildean space. A map ϕ : S →Rn is called a Lipschitz map if there exists C ∈R>0 such that for all x, y ∈S ∥ϕ(x) −ϕ(y)∥2 ≤C∥x −y∥2. We say that C is the corresponding Lipschitz constant. Let Cn be the cube as in (2.11). We say that a set S ⊆Rn is Lipschitz parametrizable if there exists a finite number of Lipschitz maps ϕj : Cn →S, such that S = S j ϕj(Cn). Definition 4.6.2. Let f(t) and g(t) be two functions defined on R. We will say that f(t) = O(g(t)) as t →∞ if there exists a positive real number B and a real number t0 such that for all t ≥t0, |f(t)| ≤B|g(t)|. We usually use the O-notation to emphasize the fact that f(t) behaves similar to g(t) when t is large. This is quite useful if g(t) is a simpler function than f(t); in this case, such a statement helps us to understand the asymptotic behavior of f(t), namely its behavior as t →∞. Let ∂M be the boundary of M, and assume that ∂M is (n−1)-Lipschitz parametriz-able. Notice that for t ∈R>0, ∂(tM) = t∂M. The following result is Theorem 2 on p. 128 of [Lan94]. Theorem 4.6.1. Let t ∈R>0, then GM,Λ(t) = Vol(M) det(Λ) tn + O(tn−1), where the constant in O-notation depends on Λ, n, and Lipschitz constants. 4.6. LATTICE POINT COUNTING 67 Proof. Let x1, . . . , xn be a basis for Λ, and let F be the corresponding fun-damental parallelotope, i.e. F = ( n X i=1 tixi : 0 ≤ti < 1, ∀1 ≤i ≤n ) . For each point x ∈Λ we will write Fx for the translate of F by x: Fx = F + x. Notice that if x ∈tM ∩Λ, then Fx ∩tM ̸= ∅. Moreover, either Fx ⊆int(tM), or Fx ∩∂(tM) ̸= ∅. Let m(t) = |{x ∈Λ : Fx ⊆int(tM)}| , b(t) = |{x ∈Λ : Fx ∩∂(tM) ̸= ∅}| . Then clearly m(t) ≤GM,Λ(t) ≤m(t) + b(t). Moreover, since Vol(F) = det(Λ) m(t) det(Λ) ≤Vol(tM) = tn Vol(M) ≤(m(t) + b(t)) det(Λ), hence m(t) ≤Vol(M) det(Λ) tn ≤m(t) + b(t). Therefore to conclude the proof we only need to estimate b(t). Let ϕ : Cn−1 →∂M be one of the Lipschitz paramterizing maps for a piece of the boundary of M, and let C be the maximum of all Lipschitz constants corresponding to these maps. Then tϕ parametrizes a corresponding piece of ∂(tM) = t∂M. Cut up each side of Cn−1 into segments of length 1/[2t], then we can represent Cn−1 as a union of [t]n−1 small cubes with sidelength 1/[2t] each, call them C1, . . . , C[t]n−1. For each such Ci, we have ∥ϕ(x) −ϕ(y)∥2 ≤C∥x −y∥2 ≤C√n −1 [2t] , for each x, y ∈Ci, i.e. the image of each such Ci under ϕ has diameter at most C√n−1 [2t] . Hence image of each such Ci under the map tϕ has diameter at most C √ n −1 t [2t] ≤2 C √ n −1. Clearly therefore the number of x ∈Λ such that the corresponding translate Fx has nonempty intersection with tϕ(Ci), for each 1 ≤i ≤[t]n−1, is bounded by some constant C′ that depends only on Λ, C, and n. Hence b(t) ≤C′[t]n−1. This completes the proof. □ 68 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS Theorem 4.6.1 provides an asymptotic formula for GM,Λ(t), demonstrating an important general principle, namely that as t →∞, GM,Λ(t) grows like Vol(M) det(Λ) tn, which is what one would expect. However, it does not give any explicit information about the constant in the error term O(tn−1). Can this constant be somehow bounded, i.e. what can be said about the quantity GM,Λ(t) −Vol(M) det(Λ) tn ? A large amount of work has been done in this direction (see for instance pp. 140 - 147 of [GL87] for an overview of results and bibliography). This subject essen-tially originated in a paper of Davenport [Dav51], who used a principle of Lipschitz [Lip65]; also see [Thu93] for a nice overview of Davenport’s result and its gener-alizations and [Wid12] for further recent results. We present here without proof a result of P. G. Spain [Spa95], which is a refinement of Davenport’s bound, and can be thought of as a continuation of Theorem 4.6.1. Theorem 4.6.2. Let the notation be as in Theorem 4.6.1, and let C be the maximal Lipschitz constant corresponding to parametrization of ∂M. Then for each t ∈R>0, GM,Λ(t) −Vol(M) det(Λ) tn ≤2n(Ct + 1)n−1. Finally, for very explicit inequalities in the case of counting lattice points in rect-angular boxes see [Fuk06a], [Fuk06b] and [FH13]. 4.7. APPLICATIONS OF LATTICES IN CODING THEORY AND CRYPTOGRAPHY 69 4.7. Applications of lattices in coding theory and cryptography Here we very briefly mention two applications of lattices in digital communi-cations. First of this is to coding theory. The theory of error correcting codes assumes transmission of a signal from transmitter to receiver over a potentially noisy channel. There is a possibility of two types of errors in the channel: (1) Erasure: a character in the signal codeword was erased in transmission. (2) Alteration: a character in the signal codeword was alterated in transmis-sion. We will briefly talk about erasures in the next section. Here, we will say a few words about how lattices can be used to deal with alterations. The main idea of constructing good error correcting codes is to ensure large distance between the codewords (here distance can be defined in different ways, most commonly the Hamming distance, which we do not discuss here). Imagine, for instance, that we use points of a full-rank lattice L ⊂Rn as our codewords. Specifically, let r be a sufficiently large integer and let Lr = {x ∈L : ∥x∥≤r} . We can use Lr as our code-space for transmission of information, i.e. signals are converted to points of Lr for transmission. If a codeword x ∈Lr is transmitted, an error due to alteration in a noisy channel may result in the introduction of a sufficiently small error vector ε so that the received codeword is x + ε. The correction mechanism then needs to strip-offthe error and return x. Write λ1 for the first successive minimum of L. Assuming that ∥ε∥< λ1/2, we see that x is the solution of CVP on L with the input point x + ε, i.e. the error correction comes down to solving an instance of CVP. While CVP is hard in general, it can be made much easier provided we know a shortest basis for our lattice L: in that case, it can be solved by Babai’s nearest hyperplane algorithm (see [MG02]) for details). Another use of lattices in coding theory comes from design of transmitter net-works. Given a lattice Λ ∈Rn, we can regard its nonzero points as transmitters which interfere with the transmitter at the origin, and then a standard measure of the total interference of Λ is given by EΛ(2), where (4.21) EΛ(s) = X x∈Λ{0} 1 ∥x∥2s is the Epstein zeta-function of Λ, and the signal-to-noise ratio of Λ is defined by (4.22) SNR(Λ) = 10 log10 1 9EΛ(2), as in [BSW97]. Suppose that we have a network of transmitters positioned at the points of a planar lattice Λ. The plane is tiled with translates of the Voronoi cell of Λ, which are the cells serviced by the corresponding transmitters at their centers. The packing density of Λ is precisely the proportion of the plane covered by the transmitter network. WR lattices allow for transmitters of the same power. To maximize SNR(Λ) on the set of all planar WR lattices of a fixed determinant ∆is 70 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS the same as to minimize EΛ(2) on this set. This optimization problem is discussed in [BSW97] and [FHL+12]. Another application of lattices comes from cryptography, resulting in a sub-area knows as lattice cryptography. The book [MG02] is an excellent introduction to this exciting and active area of research. Here we only mention a basic connec-tion. Cryptography assumes transmission of information over an unsecured channel, which allows for for intruders to intercept the message. The goal is to encode a message in a way that allows the intended receiver to easily decode it, but makes decoding very hard for intruders. Asymmetric cryptography then recognizes that the transmitter does not need to be able to decode the message, only the receiver needs to be able to do this. Encoding is done with use of a public key, i.e. a piece of publicly available information, while the decoding requires a private key known only to the receiver. The security of such a scheme is based on the assumption that decoding without the private key is a computationally hard problem. We describe one example of a lattice crypto-system based on CVP: this is the GGH encryption scheme, named after its creators O. Goldreich, S. Goldwasser and S. Halevi. Let L be a lattice in Rn, and define: • Private key: a shortest basis B for L and a matrix U ∈GLn(Z), • Public key: a basis B′ = BU for L. Let m ∈Zn be message text and let ε be a small error vector. To encrypt m, take m′ = B′m + ε. If we know the private key B and U, we can compute B−1 and U −1 and decrypt as follows: B−1m′ = B−1BUm + B−1ε = Um + B−1ε then use Babai’s nearest hyperplane algorithm to solve the corresponding instance of CVP retrieving Um and compute m multiplying by U −1. On the other hand, the intruders possessing only the public key would attempt to do the same and receive: (B′)−1m′ = m + (B′)−1ε = m + (B′)−1ε, where ∥(B′)−1ε∥can be sufficiently big to produce an incorrect CVP solution, i.e. the resulting lattice vector would be different from the message text m. Unfortu-nately, this algorithm does have some security issues as demonstrated by P. Nguyen in 1999. This being said, it still serves as a good illustration of the lattice encryp-tion idea. There is also a good number of other more secure encryption schemes based on lattices and high complexity of lattice problems. 4.8. EUCLIDEAN FRAMES 71 4.8. Euclidean frames An (n, k)-frame in a Euclidean space Rk is a set {x1, . . . , xn}, n ≥k, that satisfies the following property: there exist positive real constants γ1, γ2 such that for every vector v ∈Rk, γ1∥v∥2 ≤ n X i=1 ⟨v, xi⟩2 ≤γ2∥v∥2, where ⟨, ⟩denotes the usual Euclidean inner product. It is not difficult to see (Problem 4.12) that a frame is a spanning set for Rk. In this section we will discuss some particular types of frames, their properties and their connections to lattices. In particular, we will be interested in uniform frames, meaning that ∥x1∥= · · · = ∥xn∥ (if this common value is 1, we call such a frame unit). Further, we will say that a uniform (n, k)-frame {x1, . . . , xn} is tight if γ1 = γ2, i.e. if there exists a positive real constant c such that for every vector v ∈Rk, (4.23) ∥v∥2 = c n X i=1 ⟨v, xi⟩2 . Compare (4.23) to (4.10) and observe the following fact: a full-rank lattice Λ ⊂Rk with the set of minimal vectors S(Λ) = {x1, . . . , xn} is strongly eutactic if and only if S(Λ) forms a uniform tight (n, k)-frame in Rk. Problem 4.12 implies that such a lattice must also be WR (a fact that we mentioned before, as it is true for eutactic and for perfect lattices too). We will be especially interested in a more specialized class of tight frames. A uniform tight (n, k)-frame {x1, . . . , xn} is called equiangular (abbreviated ETF = equiangular tight frame) if there exists a real constant α such that α = | ⟨xi, xj⟩| ∥xi∥∥xj∥∀i ̸= j, in other words if absolute value of the cosine of the angle between any distinct pair of frame vectors xi, xj is the same. By a certain abuse of notation, we will refer to α as the angle of this ETF. ETF’s are a generalization of an orthonormal basis: indeed, an orthonormal basis for Rk is a unit (k, k)-ETF of angle 0. The first non-trivial example of an ETF is the Mercedes-Benz frame of three equiangular unit vectors in R2: 1 0  ,  1/2 √ 3/2  , −1/2 √ 3/2  . The angle of this (3, 2)-ETF is 1/2, and it is equal to S′(Λh) where Λh is the hexagonal lattice and S′(Λh) is the subset of the set of minimal vectors S(Λh) obtained by choosing one vector from each ± pair. More generally, we can ask how large can an ETF in Rk be? To this end, there is the following bound. Theorem 4.8.1 (Gerzon). If X = {x1, . . . , xn} is an ETF in Rk, then n ≤k(k + 1) 2 . 72 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS Proof. Normalizing, if necessary, we can assume that X is a unit frame of angle α, then α = | ⟨xi, xj⟩| ∀i ̸= j, | ⟨xi, xi⟩| = 1 ∀1 ≤i ≤n. For each 1 ≤i ≤n, consider the k × k symmetric matrix xix⊤ i as a vector in Rk2, then we can compute the inner products xix⊤ i , xjx⊤ j = ⟨xi, xj⟩2 =  α2 if i ̸= j, 1 if i = j. Consider a symmetric k × k matrix A = n X i=1 aixix⊤ i , then the squared norm of A is ⟨A, A⟩ = n X i=1 n X j=1 aiaj xix⊤ i , xjx⊤ j = n X i=1 n X j=1 ⟨xi, xj⟩2 aiaj = n X i=1 a2 i + 2α2 n X i=1 n X j=i+1 aiaj, which is a positive definite quadratic form in the variables a1, . . . , an. Hence A = 0 if and only if ⟨A, A⟩= 0, which happens if and only if a1 = · · · = an = 0. Thus the set X∗=  xix⊤ i : 1 ≤i ≤n is linearly independent, meaning that its cardinality n cannot be larger than the dimension of the space of all real symmetric k × k matrices, which is k(k+1) 2 . □ Notice that Gerzon’s bound is sharp, as demonstrated by the Mercedes-Benz example: 2(2+1) 2 = 3. ETF’s achieving Gerzon’s bound are called maximal. One can ask in which dimensions do maximal ETFs occur? In fact, only four examples are known: (3, 2), (6, 3), (28, 7) and (276, 23). Out of these examples, besides the hexagonal lattice, the (28, 7)-ETF is S′(Λ) of a certain perfect strongly eutactic lattice in R7 and the (276, 23)-ETF appears among the set of minimal vectors of the famous 24-dimensional Leech lattice. There are many other examples of ETFs (although not maximal) appearing as sets of minimal vectors of strongly eutactic lattices, however these lattices are not perfect: perfection would require |S′(Λ)| ≥k(k+1) 2 , which implies maximal ETF; see [BFG+16], [BF17] for details. Frames have applications in information transmission, for instance in recov-ering erasures (as defined in Section 4.7 above) in signal transmission. If X = {x1, . . . , xn} ⊂Rk is an (n, k)-frame, then an arbitrary vector v ∈Rk can be written in these frame coordinates as v(X) = (v1(X), . . . , vn(X)), where vi(X) = ⟨v, xi⟩for each 1 ≤i ≤n. This generalizes the notion of coordinates of a vector with respect to a basis. We can now transmit the message encoded by v by instead transmitting its vector of coordinates v(X). The advantage of using an overdetermined frame is that the vector v can be reconstructed with a certain degree of accuracy even if some of the coordinates were lost in transmission, i.e. if erasures 4.8. EUCLIDEAN FRAMES 73 occurred (see [HP04] for more information on this). Hence, ideally we want the cardinality n of the frame to be large as compared to the dimension k: this gives more coordinates, an so a better potential chance at accurate reconstruction of a signal in the presence of erasures. On the other hand, the accuracy of reconstruction is increased if the frame vectors are not “aligned” with each other, i.e. if the angles between them are large. To this end, let us define the frame coherence. Given an arbitrary (n, k)-frame X = {x1, . . . , xn} ⊂Rk, we define its coherence to be C(X) = max 1≤i̸=j≤n | ⟨xi, xj⟩| ∥xi∥|xj∥, i.e. maximal absolute value of cosines of the angles between pairs of frame vectors. This is a measure of how aligned with each other frame vectors are. For instance, coherence of an orthogonal basis is 0 and coherence of an ETF is its angle. Thinking of frame vectors as frequencies encoding a signal, coherence represents the measure of interference between different frequencies used in transmission: the lower this interference is the better it is for accurate transmission. Notice, however, that the larger is the cardinality of a frame the more aligned its vectors would have to be, and so the higher would be its coherence. Hence we have the following optimization problem. Problem 4.1. Construct (n, k)-frame X ⊂Rk with large cardinality n as compared to the dimension k and low coherence C(X). The first question to ask then is how small can coherence be? To this end, we have the following inequality. Theorem 4.8.2. (Welch) Given an (n, k)-frame X = {x1, . . . , xn} ⊂Rk, C(X) ≥ s n −k k(n −1). The equality is achieved if and only if X is an ETF. Proof. Rescaling the vectors, if necessary, we can assume that X is a unit frame: rescaling does not change coherence. Let us write A = (x1 . . . xn) for the k×n matrix whose columns are the frame vectors. Define the corresponding n × n Gram matrix to be G = A⊤A, then the ij-entry of G is equal to ⟨xi, xj⟩. In particular, all the diagonal entries of G are equal to 1, and so its trace Tr(G) = n. On the other hand, trace of G is equal to the sum of its eigenvalues λ1, . . . , λn. Notice that G is a positive semi-definite matrix of rank k, so we can assume that λ1, . . . , λk > 0, λk+1 = · · · = λn = 0. Then (4.24) n2 = Tr(G)2 = k X i=1 λi !2 ≤k k X i=1 λ2 i = k n X i=1 λ2 i . 74 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS by Cauchy-Schwartz inequality. Recall that the Frobenius norm of this Gram matrix is simply its Euclidean norms viewed as a vector in Rn2, i.e. ∥G∥2 = n X i=1 n X j=1 | ⟨xi, xj⟩|2. The symmetric matrix G is diagonalizable by some orthogonal matrix U, i.e. UGU ⊤=    λ1 . . . 0 . . . ... . . . 0 . . . λn   , and the Frobenius norm is invariant under such diagonalization, i.e. (4.25) ∥G∥2 = n X i=1 n X j=1 | ⟨xi, xj⟩|2 = n X i=1 λ2 i . Combining (4.24) and (4.25), we obtain (4.26) n X i=1 n X j=1 | ⟨xi, xj⟩|2 ≥n2 k . On the other hand, notice that | ⟨xi, xj⟩| ≤C(X) (with equality if an only if X is an ETF) for i ̸= j and | ⟨xi, xi⟩| = 1, hence n2 k ≤n + X i̸=j | ⟨xi, xj⟩|2 ≤n + (n2 −n)C(X)2, with the last inequality being equality if and only if X is an ETF, therefore C(X)2 ≥ 1 n2 −n n2 k −n  = n −k k(n −1), which establishes the Welch bound. Further, if X is an ETF all nonzero eigenvalues of G are the same, and so n = Tr(G) = kλ1, meaning that n X i=1 λ2 i = kλ2 1 = k n2 k2  = n2 k , in other words there is an equality in (4.26). This establishes equality in the Welch bound if and only if X is an ETF. □ We then have an immediate consequence of Welch’s bound for ETFs. Corollary 4.8.3. If X is an (n, k) ETF of angle α, then α = s n −k k(n −1). In fact, there are some additional remarkable properties an (n, k) ETF of angle α must possess. • (Neumann) If n > 2k, then 1/α is an odd integer. 4.8. EUCLIDEAN FRAMES 75 • (Sustik, Tropp, Dhillon, Heath) Let 1 < k < n −1. Suppose n ̸= 2k, then 1/α is an odd integer and the quantity r (n −k)(n −1) k is also an odd integer. If n = 2k, then k must be an odd integer and n−1 the sum of two squares. These and related properties are used to eliminate the pairs (n, k) for which ETFs cannot exist. One of the main goals here is to find more maximal ETFs beyond the four mentioned above. 76 4. LATTICE PROBLEMS, CONNECTIONS AND APPLICATIONS 4.9. Problems Problem 4.2. Prove that the optimal kissing number in R2 is equal to 6. Problem 4.3. Prove that similarity is an equivalence relation on the set of all lattices of full rank in Rn. Problem 4.4. Assume two full-rank lattices L and M in Rn are similar. Prove that they have the same packing density, covering thickness and kissing number. Problem 4.5. Prove that the set of all real orthogonal n × n matrices On(R) is a subgroup of GLn(R). Problem 4.6. Let x1 and x2 be nonzero vectors in R2 so that the angle θ between them satisfies 0 < θ < π 3 . Prove that ∥x1 −x2∥< max{∥x1∥, ∥x2∥}. Problem 4.7. Let Λ ⊂R2 be a lattice of full rank with successive minima λ1 ≤λ2, and let x1, x2 be the vectors in Λ corresponding to λ1, λ2, respectively. Let θ ∈ [0, π/2] be the angle between x1 and x2. Prove that π/3 ≤θ ≤π/2. Problem 4.8. Let L and M be two similar lattices. Prove that if L is eutactic (respectively, strongly eutactic, perfect), then so is M. Problem 4.9. Prove that the hexagonal lattice Λh is both, perfect and eutactic. 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Biomolecules 2023, 13(2), 197; Submission received: 29 November 2022 / Revised: 11 January 2023 / Accepted: 15 January 2023 / Published: 18 January 2023 (This article belongs to the Special Issue GABA Receptors in Pharmacology and Neurobiology) Download keyboard_arrow_down Download PDF Download PDF with Cover Download XML Download Epub Download Supplementary Material Browse Figures n = 8; HARM 15 mg/kg, n = 8; MP 3.2 mg/kg, n = 8; MP 10 mg/kg, n = 8; MP 3.2 mg/kg + HARM 15 mg/kg, n = 10; MP 10 mg/kg + HARM 15 mg/kg, n = 10. Statistics: ANOVA for repeated measures with regard to AP2 (HARM effect: F[1,47] = 27.111, p = 0.001; MP effect: F[2,47] = 1.459, p = 0.243; time effect: F[1,47] = 54.740, p = 0.001), Tremor index (HARM effect: F[1,47] = 217.767, p = 0.001; MP effect: F[2,47] = 3.576, p = 0.036; time effect: F[1,47] = 2.004, p = 0.163), AP1 (HARM effect: F[1,47] = 83.421, p = 0.001; MP effect: F[2,47] = 2.381, p = 0.104; time effect: F[1,47] = 275.344, p = 0.001) and Distance (HARM effect: F[1,47] = 3.537, p = 0.067; MP effect: F[2,47] = 5.414, p = 0.007; time effect: F[1,47] = 279.195, p = 0.001). LSD post-hoc test: p ≤ 0.05, p ≤ 0.01, p ≤ 0.001, a p = 0.077, b p = 0.060 vs. SOLV; c p = 0.090 vs. HARM 15 mg. " href=" n = 7; PIM 1 mg/kg, n = 8; MP 3.2 mg/kg, n = 8; MP 10 mg/kg, n = 8, PIM 1 mg/kg + MP 3.2 mg/kg, n = 6; PIM 1 mg/kg + MP 10 mg/kg, n = 7. Statistics: Factorial ANOVA (PIM effect: F[1,38] = 41.736, p = 0.001; MP effect: F[2,38] = 3.919, p = 0.028) + LSD post-hoc test ( p ≤ 0.01, p ≤ 0.001). " href=" n = 11; HARM 15 mg/kg, n = 15; ZOL 0.34 mg/kg, n = 5; ZOL 0.67 mg/kg, n = 6; ZOL 1.01 mg/kg, n = 6; ZOL 0.34 mg/kg + HARM 15 mg/kg, n = 8; ZOL 0.67 mg/kg + HARM 15 mg/kg, n = 6; ZOL 1.01 mg/kg + HARM 15 mg/kg, n = 8. Statistics: Factorial ANOVA with regard to AP2 (HARM effect: F[1,57 = 21,391, p = 0.001; ZOL effect: F[3,57] = 0.546, p = 0.653), Tremor index (HARM effect: F[1,57] = 103.001, p = 0.001; ZOL effect: F[3,57] = 3.142, p = 0.032), AP1 (HARM effect: F[1,57] = 49.307, p = 0.001; ZOL effect: F[3,57] = 19.014, p = 0.001), and Distance (HARM effect: F[1,57] = 5.421, p = 0.023; ZOL effect: F[3,57] = 4.481, p = 0.007). LSD post-hoc test: p ≤ 0.01, p ≤ 0.001. " href=" n = 11; PIM 1 mg/kg, n = 10; TBZ 2 mg/kg, n = 11; ZOL 0.67 mg/kg, n = 10; PIM 1 mg/kg +ZOL 0.67 mg/kg, n = 10; TBZ 2 mg/kg + ZOL 0.67 mg/kg, n = 10. Statistics: Factorial ANOVA for PIM (PIM effect: F[1,37] = 26.019, p = 0.001; ZOL effect: F[1,37] = 0.446, p = 0.508) and TBZ (TBZ effect: F[1,38] = 51.114, p = 0.001; ZOL effect: F[1,38] = 0.410, p = 0.526) + LSD post-hoc test ( p ≤ 0.01, p ≤ 0.001). " href=" Review Reports Versions Notes Abstract Treatment of tremors, such as in essential tremor (ET) and Parkinson’s disease (PD) is mostly ineffective. Exact tremor pathomechanisms are unknown and relevant animal models are missing. GABA-A receptor is a target for tremorolytic medications, but current non-selective drugs produce side effects and have safety liabilities. The aim of this study was a search for GABA-A subunit-specific tremorolytics using different tremor-generating mechanisms. Two selective positive allosteric modulators (PAMs) were tested. Zolpidem, targeting GABA-A α1, was not effective in models of harmaline-induced ET, pimozide- or tetrabenazine-induced tremulous jaw movements (TJMs), while the novel GABA-A α2/3 selective MP-III-024 significantly reduced both the harmaline-induced ET tremor and pimozide-induced TJMs. While zolpidem decreased the locomotor activity of the rats, MP-III-024 produced small increases. These results provide important new clues into tremor suppression mechanisms initiated by the enhancement of GABA-driven inhibition in pathways controlled by α2/3 but not α1 containing GABA-A receptors. Tremor suppression by MP-III-024 provides a compelling reason to consider selective PAMs targeting α2/3-containing GABA-A receptors as novel therapeutic drug targets for ET and PD-associated tremor. The possibility of the improved tolerability and safety of this mechanism over non-selective GABA potentiation provides an additional rationale to further pursue the selective α2/3 hypothesis. Keywords: resting tremor; action tremor; GABAA alpha 1; GABAA alpha 2/3; Parkinson’s disease; essential tremor; MP-III-024; zolpidem; positive allosteric modulator; tetrabenazine 1. Introduction Tremors are uncontrollable, rhythmic, and oscillating movements produced by alternating or synchronous contractions of antagonistic muscles in different parts of the body. Tremor is a common symptom of many neurological diseases, such as Parkinson’s disease (PD), or can constitute an independent disease, such as essential tremor (ET). Tremor significantly hinders daily functioning and has a direct negative impact on the patient’s quality of life. Besides some progress in the treatment of those diseases, tremor remains their key, but still untreated, symptom. Currently, there are no specific and effective enough anti-tremor pharmacotherapies and we still do not understand the exact mechanism underlying tremor pathology. Therefore, there is an immediate need to focus attention on exploring new molecular pathways and searching for improved tremorolytics drug targets. 1.1. Essential Tremor ET is a chronic progressive disease that increases in prevalence with age, in which postural or kinetic tremor is present with a frequency range of 4–12 Hz, affecting various parts of the body, mainly the hands, head, tongue, legs, and less frequently other parts such as the face, torso, or voice [1,2]. Pathophysiological mechanisms underlying ET are complex and poorly understood. Results of post-mortem studies are inconsistent, some of them demonstrating Lewy Body pathology , cerebellar Purkinje cell (PCs) degeneration [3,4,5,6] or even normal brain morphology without PCs loss . Results of clinical and post-mortem studies and research in animal models indicate that brain structures such as the inferior olive nuclei, cerebellum, thalamus, and cerebral cortex forming the cerebello–thalamo–cortical network are involved, and that hyperactivity of the glutamatergic neurons of the olivocerebellar circuit seems to be crucial for evoking tremor [2,8,9,10,11]. There are several different hypotheses regarding the mechanisms of ET generation, the most popular of which are: (1) the neurodegenerative hypothesis; (2) the central oscillatory network hypothesis; and (3) the GABAergic hypothesis [2,8,11]. A lack of understanding of ET pathophysiology makes it difficult to develop effective pharmacotherapy, and there is hardly any drug specifically developed for treating tremors. Potentiating GABA-A receptors in the CNS or inhibiting beta-adrenergic systems most consistently improves tremor in the clinic. Current symptomatic treatment of ET with propranolol (beta-antagonist) and primidone (an antiepileptic drug), is ineffective in 30–50% patients and induces a large number of side effects [2,12,13,14]. Other drugs likely to be effective in ET include topiramate or benzodiazepines (clonazepam and alprazolam) and analogues of GABA (pregabalin and gabapentin). Interestingly, ethanol also binds to GABA-A receptors and decreases ET. However, the specific anti-tremor mechanisms of actions of these drugs has not been defined. In general, it seems that substances that act via GABA-A and increase the duration of receptor opening but not the drugs that increase GABA availability are associated with tremor reduction . Nevertheless, the drugs used are not tremor-specific agents but have a broad range of action, are not effective enough and show multiple harmful side effects. Another major problem in the development of anti-tremor therapies is the lack of reliable models. One of them is GABA-A α1 global or cerebellar PCs selective knock-out mice [15,16], while the most common is the harmaline (a β-carboline derivative) model of ET [2,17,18,19]. Many substances that were shown to reduce human tremors are also effective in the harmaline model (ethanol, benzodiazepines, primidone, and other GABA-A receptor potentiators) [14,17], thus validating the model’s predictive utility for human translation, though it still has drawbacks. 1.2. PD Tremor Besides akinesia, bradykinesia, and muscle rigidity, tremor at rest is one of the main symptoms of idiopathic PD, yet most resistant to currently-available therapy. Tremor is also observed in neuroleptic-induced parkinsonism [20,21]. Bradykinesia and rigidity result from the degeneration of dopaminergic nigrostriatal neurons and disturbances in the basal ganglia-thalamo-cortico-basal ganglia neuronal circuitry [22,23,24]. Neuropathological mechanisms underlying tremors are still not fully understood but seem to be distinct from other PD motor symptoms [20,21,25]. Importantly, there is no correlation between tremor magnitude and the degree of DA deficiency in the striatum and a ‘gold standard’ therapy with L-DOPA and dopamine agonists is very poor in treating tremors . Other PD therapies include deep brain stimulation (DBS) of subcortical nuclei such as the ventral intermediate nucleus of the thalamus (VIM) and subthalamic nucleus (STN), which is limited to drug-resistant, advanced forms of tremor . Besides DA, other neurotransmitters including GABA and glutamate in the basal ganglia have also been suggested to be involved in the mechanism of parkinsonian tremor induction [27,28,29]. Results of neuroimaging human studies have suggested that in addition to the basal ganglia and cerebellum, which are the most important structures for the generation of parkinsonian tremor, the ventral thalamus, which receives glutamatergic projections from the cerebellum and GABAergic projections from the basal ganglia, is a relay station for neurotransmission to cortical regions [30,31,32,33]. The key structures in those neuronal networks seemingly responsible for tremor express GABA-A receptors [34,35,36]. Importantly, a phase 2, open-label clinical study (NCT03000569) with 14 PD patients explored Zuranolone (SAGE-217), an investigational oral neuroactive steroid GABA-A PAM as an oral adjunct to the treatment of Parkinsonian tremor and showed a reduction of tremor symptoms after 7 days of treatment. The MDS-UPDRS Part II/III tremor score significantly improved by 40% from baseline and persisted for 7 days after drug discontinuation . This report strongly supports the role of GABA-A in tremor management, although Zuranolone is not a selective GABA-A PAM as it potentiates both synaptic and extrasynaptic GABA-A receptors, enhancing phasic and tonic inhibitory currents. Interestingly, there are reports on the putative PD tremor-reducing effect of zolpidem, a PAM of GABA-A receptors which displays a high affinity to the α1 subunit. Daniele et al. described several case studies in which zolpidem administered to patients with PD as a sleep inducer showed a visible improvement in their motor symptoms (such as akinesia, rigidity, but also resting tremor). The authors hypothesized that such effects were mediated by the high density of zolpidem binding sites in the two main output structures of the basal ganglia that are abnormally overactive in PD (globus pallidus internal, GPi, and substantia nigra reticulata, SNr), through GABA-A α1 receptors. Currently, there are no ideal models of Parkinsonian resting tremor recapitulating the full dynamics and complexity of its pathophysiological mechanisms. The only available way of studying it in rodents is the analysis of Tremulous Jaw Movements (TJMs), defined as “rapid vertical deflections of the lower jaw that resemble chewing but are not directed at any particular stimulus” . TJMs are triggered by similar conditions that lead to parkinsonism in humans e.g., by striatal DA depletion (for example by tetrabenazine and reserpine) and acute or subchronic administration of typical antipsychotics (haloperidol and pimozide) [27,39], and are suppressed by known antiparkinsonian drugs, including L-DOPA and DA agonists, as well as by DBS of the STN [39,40,41,42,43]. Multiple studies have demonstrated that TJMs depend on the ventrolateral striatum, but the involvement of both “indirect” striatopallidal and “direct” striatonigral GABAergic downstream pathways has also been suggested [35,39,44,45,46,47]. Although ET and PD tremors have different characteristics, it is possible that they also share some similar mechanisms. There is current evidence that ET may precede the onset of PD in a subset of patients. The risk of ET is significantly increased in relatives of PD patients, suggesting a shared hereditary predisposition. As in PD, the dopaminergic deficit and Lewy bodies as well as smell deficits are observed in some ET patients [48,49,50]. Therefore, a direct comparison of ET and PD tremors, as in the present study, could provide new insight into their mechanistic commonalities. 1.3. GABA-A α2/3 Subunit Specific Receptors as Novel Drug Targets against Tremor As summarized above, convergent input from anatomical, pharmacological, and neurochemical studies have strongly implicated the enhancement of GABAergic neurotransmission through GABA-A receptors as an important anti-tremor mechanism both in ET and PD [35,38,46,51,52,53,54,55]. Although enhanced inhibition by GABA-A receptor PAMs that non-selectively interact with α1, α2, α3, and α5 subunits produces anti-tremorogenic effects, such PAMs are also associated with sedation, ataxia, tolerance development, memory impairment, and abuse liability . Amrutkar et al. showed that, in addition to diazepam, NS16085—a GABA-A α2/3 subtype selective modulator with reduced modulation at α1 and negligible efficacy at α5 receptors [57,58]—could also suppress tremor in a model using harmaline as a tremor inducer. This finding is potentially a highly important clue in defining a new drug target for the discovery of improved medications for treating tremors. There are several selective compounds in development that selectively target α2/3-containing GABA-A receptors [58,59]. Both preclinical and, in some cases, clinical data support the proposition of improvement in the side-effect profile of these compounds relative to non-selective GABA-A receptor PAMs [56,59]. One such α2/3-selective compound, KRM-II-81, an imidazodiazepine, already showed antiseizure, antinociceptive, and anxiolytic activity in rodent models without sedation, tolerance, or abuse liability. Therefore, the aim of the present study is to further test the hypothesis that the potentiation of α2/3-selective GABA-A receptors will have a tremorolytic effect. In the present work, we used the α2/3-selective PAM, MP-III-024, a compound without signficicant effect on α1- and α5-containing GABA-A receptors [59,60]. As for comparison and in relation to the described-above PD case studies, zolpidem, which selectively potentiates α1-containing GABA-A receptors , was used. To ensure that there is broad generality to the anti-tremor effects of this mechanism, and due to a lack of ideal models, it is necessary to employ multiple tremor inducers. The previous single study targeting GABA-A α2/3-selective mechanisms used only harmaline as an inducer. In the present study, we employed three different animal models of tremor. ET was modelled by harmaline injection while dopamine-related TJMs were induced by pimozide or tetrabenazine. The results of this study show that zolpidem in subhypnotic doses was not effective in reducing either ET or TJM, but MP-III-024 targeting GABA-A α2/3 receptor subunits has the therapeutic potential both in the harmaline and TJMs animal models. Moreover, MP-III-024 was not sedating. We conclude that α2/3-containing GABA-A receptors comprise a new molecular target for the discovery of an improved treatment for the suppression of tremors in PD and ET patients. 2. Materials and Methods 2.1. Animals Adult male Wistar Han rats (Charles River, Germany), weighing 250–340 g prior to the experiment, were kept in an animal room at 12 h dark/light cycle (the light on from 7 AM to 7 PM) with a temperature of 22 ± 2 °C and relative humidity of 55 ± 5%, with 4–5 rats per cage with free access to food and water. All experiments were carried out during the light period and behavioral tests were performed during the light phase between 8 AM and 4 PM. The experiments were carried out in compliance with the Act on Experiments on Animals of 21 January 2005 amended on 15 January 2015 (published in Journal of Laws no. 23/2015 item 266, Poland), and according to the Directive of the European Parliament and of the Council of Europe 2010/63/EU of 23 September 2010 on the protection of animals used for scientific purposes. They also received the approval of the Local Ethics Committee at the Maj Institute of Pharmacology, Polish Academy of Sciences, Cracow (permission no: 271/2018, 89/2020, 90/2021). All efforts were designed to minimize both the suffering and the number of animals used. 2.2. Compounds Harmaline hydrochloride dihydrate (Sigma-Aldrich, Merck & Co. Inc., Rahway, NJ, USA) was dissolved in sterile redistilled water and administered acutely at a dose of 15 mg/kg ip. Pimozide (Sigma-Aldrich, Merck & Co. Inc., Rahway, NJ, USA), antagonist of dopamine receptors, mainly D2, but also D3 i D4, was dissolved in warm 0.3% tartaric acid and administered subchronically (7 injections, once a day) at a dose of 1 mg/kg. Tetrabenazine (TBZ), a reversible type 2 vesicular monoamine transporter (VMAT) inhibitor, was dissolved in 20% DMSO in 0.9% NaCl (1N HCl was added to get the drug completely into solution, according to [61,62]; final pH = 3.5) and given acutely at a dose of 2 mg/kg. This treatment procedure was based on previous experiments that employed the pimozide-induced TJMs model [47,63], see also Supplementary Data Figures S1–S3. Zolpidem hemitartare (synthesized and provided by Dr. M. Marcinkowska, Jagiellonian University, Medical College, Kraków, Poland), according to the previously reported procedure , a positive allosteric modulator of GABA-A receptors containing α1 subunit was dissolved in sterile redistilled water and administered acutely at doses of 0.34, 0.67, 1.01 mg/kg. MP-III-024 (methyl 8-ethynyl-6-(pyridin-2-yl)-4H-benzo[f]imidazo[1,5-a][1,4]diazepine-3-carboxylate, synthesized in the laboratory of J.M. Cook, University of Wisconsin-Milwaukee, (Milwaukee, WI, USA), a positive allosteric modulator selective for α2/3 subunits of GABA-A receptor, was suspended in 0.5% methyl cellulose in 0.9% NaCl and given at doses of 3.2 and 10 mg/kg in a volume of 2 mL/kg. All compounds were injected in a volume of 2 mL/kg ip. Control animals received appropriate vehicles: water, 0.3% tartaric acid, 20% DMSO in 0.9% NaCl (pH = 3.5) or 0.5% methyl cellulose in 0.9% NaCl instead of harmaline/zolpidem, pimozide, TBZ and MP-III-024, respectively. The times of administration of the above compounds are shown in Figure 1. The doses and pre-treatment times were based on the previous literature [60,65,66,67,68] and pharmacokinetic data provided by substance authors. MP-III-024 was administered 30 min prior to harmaline, 3.5 h after the last pimozide injection. Zolpidem was administered subsequently with harmaline, 3 h 45 min after the last pimozide injection or 1 h 45 min after TBZ. 2.3. Experimental Procedures 2.3.1. Harmaline-Induced Tremor and Locomotor Activity Analysis The measurement of tremor and locomotor activity was performed automatically using Force Plate Actimeters (FPA; BASi, West Lafayette, IN, USA) according to [54,69,70], and started immediately after harmaline administration. FPA consists of a measuring cage placed in a sound-attenuating chamber. Four force transducers placed under the corners of the measuring cage‘s floor allow for recording the animal position on a Cartesian plane, tracking its movement across the floor and measuring the force exerted on the plate. Data were collected during time units of 10.24 s (“frames”) with the sampling frequency of 100 points/s. Tremor was analyzed using Fast Fourier Transform on each frame of the experiment. The resulting power spectra were subjected to logarithmic transformation (log10) and averaged over one (experiments with zolpidem) or two (study with MP-III-024) consecutive 180-frame series creating two intervals (0–30.72 and 30.73–61.44 min; to simplify, further times are designated as 0–30 and 30–60 min) to give the following parameters: AP1—averaged power in frequency band I (0–8 Hz), AP2—averaged power in frequency band II (9–15 Hz), Tremor Index (TI)–the difference in power between AP2 and AP1. The total distance travelled during one or two intervals, measured in millimeters, was used as a measure of locomotor activity. 2.3.2. Tremulous Jaw Movements (TJMs) Analysis The measurement of TJMs started 4 h after the last dose of pimozide or 2 h after TBZ administration. All rats used in these experiments responded to TBZ and subchronic pimozide with TJMs. Observations of rats were made in a clear Plexiglass cylinder (30 cm diameter, 40 cm high) placed in front of an angled mirror. TJMs were defined as rapid vertical deflections of the lower jaw that resembled chewing but were not directed at any particular stimulus . Each individual deflection of the jaw was recorded for 10 min using a mechanical hand counter by a trained observer, who was blind to the experimental condition of the rat being observed. The observation was also video recorded for later verification with live counting. 2.4. Statistical Analysis Results are presented as means ± standard errors of mean (SEM). The statistical analysis of the results was performed using STATISTICA v. 13 software (StatSoft Inc., Tulsa, OK, USA); p ≤ 0.05 was considered statistically significant and 0.1 ≥ p ≥ 0.05 were considered trends. For the results obtained using FPA (harmaline-induced tremor and locomotor activity), factorial ANOVA (zolpidem experiments) or ANOVA for repeated measures (experiments with MP-III-024; two time points: 0–30 min and 30–60 min) was used. TJMs were analyzed by factorial ANOVA. For individual comparisons between groups, the LSD post hoc test was used. The number of animals in each group is given in the figure descriptions. 3. Results 3.1. MP-III-024 Inhibits Harmaline-Induced Tremor and Increases the Locomotor Activity In order to establish whether the GABA-A α2/3 subunit specific PAM MP-III-024 has the potential to decrease tremors related to ET, we tested its effect on harmaline-induced tremor and locomotor activity. Harmaline (15 mg/kg) caused a generalized tremor in rats, which was observed as a significant increase in the AP2 parameter (power in the range of 8–15 Hz) (Figure 2B) and the tremor index (Figure 2C) compared to the control group and both these changes persisted for the entire 60 min of tremor measurement. Harmaline additionally altered the locomotor activity of rats, measured by the total distance travelled, decreasing it in the first 30 min, and then increasing walking between 30–60 min after administration as compared to the control rats (Figure 2E). Harmaline itself lowered AP1 (power within 0–8 Hz frequency band) within the first 30 min of measurement (Figure 2D). This experiment was performed up to 60 min after harmaline injection to observe the length of effect (Figure 1A). MP-III-024, a positive allosteric modulator of α2/3 subunit containing GABA-A receptors, given alone had no effect on AP2 and distance parameters, but showed a trend to decrease tremor index (both doses) as compared to solvent. MP-III-024 itself also increased animal mobility. The lower dose (3.2 mg/kg) extended the total distance travelled by the animals during the first 30 min of measurement and raised the AP1 parameter at the same time. The increase in AP1 was also noted for the dose of 10 mg/kg in comparison to solvent control (Figure 2D). MP-III-024 at both doses used (3.2 and 10 mg/kg), significantly attenuated the tremor induced by harmaline, which is illustrated by the power spectrum and AP2 graphs (Figure 2B). The anti-tremor effect of MP-III-024 at the higher dose was weaker, as it affected AP2 but not the tremor index parameter and lasted for the entire 60 min of measurement. On the other hand, the effect of MP-III-024 at the lower dose was stronger and more visible, as it decreased both the AP2 and the tremor index but lasted for a shorter duration (0–30 min). Additionally, following harmaline, MP-III-024 in both doses increased the locomotor activity (distance parameter) in comparison to animals treated with harmaline alone during 0–30 min (after the dose of 3.2 mg/kg) or 0–60 min of measurement (10 mg/kg) (Figure 2E). Generally, the observed effects of MP-III-024 on harmaline tremor were not dose-dependent in the tested range. 3.2. MP-III-024 Inhibits the Pimozide-Induced TJMs To determine if a GABA-A α2/3 subunit specific PAM is effective in reducing tremor in other types of tremor, pimozide was administered at a dose of 1 mg/kg once a day for 7 days. Pimozide induced TJMs in all animals. There were no differences between the effect of MP-III-024 given alone and the solvent (SOLV) control group, but MP-III-024 at both doses used (3.2 and 10 mg/kg) significantly inhibited pimozide-induced TJMs (Figure 3). 3.3. Zolpidem Has No Effect on Either Harmaline-Induced Tremor or TJMs after Pimozide or TBZ To verify if GABA-A α1 subunit specific drug zolpidem could affect tremor related to ET, it was tested in a harmaline model. Harmaline (15 mg/kg), similar to our previous experiments [54,69,70], induced whole-body tremor. It appeared a few minutes after administration and was manifested by an increase in tremor power within the frequency band of 8–15 Hz (AP2) (Figure 4B), with the peak for frequencies around 10 Hz (Figure 4A), and in increasing the tremor index parameter (AP2-AP1, Figure 4C) compared to the control group for the entire 30 min of measurement. At the same time, harmaline decreased the AP1 parameter representing animal movements in the lower range of frequency (0–8 Hz) but without affecting the locomotor activity of the animals (distance) (Figure 4D,E). Zolpidem, a positive allosteric modulator of GABA-A α1 receptors, was tested in three subhypnotic doses [65,66]. Given alone, zolpidem had no major effect on tremor or activity parameters as compared to control solvent (SOLV) group, but the profile of changes was dose-dependent. In particular, at the highest dose (1.01 mg/kg), zolpidem itself increased the tremor index, while in the middle and the highest doses (0.67 and 1.01 mg/kg) it also lowered the AP1 and distance parameters compared to controls (Figure 4C,D). Zolpidem at the lowest (0.34 mg/kg) and the highest dose (1.01 mg/kg), administered together with harmaline, seemed to increase the power between 8–15 Hz above the harmaline-induced peak when looking at the power spectrum graph (Figure 4A), but no such effect was observed when analyzing the values of individual tremor parameters (AP2 or tremor index) (Figure 4B,C). Coadministration of zolpidem (at any dose) with harmaline had no effect on tremor parameters, AP2 and tremor index, or on locomotor activity compared to animals treated with harmaline alone. Pimozide administered at a dose of 1 mg/kg once a day for 7 days, as well as TBZ given as a single dose of 2 mg/kg, induced TJMs in all animals subjected to those treatments. Control experiment showing Pimozide or TBZ influence on locomotor activity and catalepsy is shown in Supplementary Figures S1–S3. Zolpidem administered alone in the middle dose (0.67 mg/kg) had no effect, and coadministration with pimozide or TBZ did not affect TJMs in either model (Figure 5A,B). 4. Discussion GABA-A receptors are ionotropic, rapidly-responding ligand-gated ion channels formed as heteropentamers. They can be assembled from 19 different subunits forming Cl channels, thus presenting enormous heterogeneity of receptor subtypes. The α1-containing GABA-A receptors are the most widely distributed GABA-A receptors in the brain (48% of the global gene expression for GABA-A receptors), expressed predominantly in the cerebral cortex and cerebellum, while α2 and α3 subunits are mostly present in the thalamus and molecular layers of the cerebellum (see also below). Depending on the subunit composition and arrangement, as well as distribution in the various brain regions, these receptors exhibit different pharmacology and functional output of tested substances. Two GABA binding sites are formed by an α and β subunits and one benzodiazepine binding site formed by an α and γ subunits. Whereas the GABA neurotransmitter opens the Cl channel, ligands acting via the benzodiazepine binding site cannot directly open it, but only allosterically enhance (positive allosteric modulators-PAMs) or reduce (negative allosteric modulators) GABA-induced currents [56,57]. 4.1. The Tremorolytic Effect of GABA-A α2/3 Subtype Selective Positive Allosteric Modulator MP-III-024 In this study we tested novel GABA-A α2/3 selective receptor PAM, MP-III-024 and showed that it significantly reduced the harmaline-induced tremor which is in line with the previous reports using other GABA-A α2/3 targeting but less selective substances of different pharmacological profiles . So far, multiple studies showed tremorolytic effect of non-selective GABA-A α1/2/3/5 receptor PAMs both in ET patients and experimental rodent studies. Anti-tremor effects of benzodiazepines, ethanol, alprazolam, diazepam, muscimol, and used as first-line therapy for ET–primidone, strongly support a role for the GABAergic system in ET. However, their clinical use is limited due to side effects such as sedation, ataxia, tolerance development, and memory impairment [2,51,56]. With such a non-selective approach, sedation, ataxia, and dependence are related to the activity at the α1 subunit, muscle relaxation to α2, anxiolysis and analgesia and anticonvulsant effects to α2/3, and memory impairment depends on the α5 subunit . Therefore, Amrutkar et al. were the first to hypothesize that subtype selective GABA-A receptor modulators acting selectively via the α2 and α3 subunits may have an improved side effect profile while retaining the tremorolytic beneficial effects in ET. They showed experimentally in the animal model of ET, similar to that induced by harmaline, a tremorolytic effect of NS16085 , which is a GABA-A α2/3 subtype selective modulator with low level of negative modulation at α1 and negligible efficacy at α5 receptors [57,58]. By comparison of NS16085 with NS11394-GABA-A α2/3/5 receptor PAM, they showed that it was the α2/3, but not the α5, subunit which played a role in the anti-tremor efficacy . Our study, using another GABA-A α2/3 selective compound, MP-III-024, supports their observations. Another study on ET patients using GABA-A α2/3/5 subtype-selective PAM, TPA023, have shown its effectiveness “superior to placebo, but not statistically significant” in reducing kinetic tremors in ET patients . Therefore, GABA-A α2/3/5 subtype-selective PAMs have tremor-reducing potential both in human and in an animal ET model. The absence of α5 potentiation by MP-III-024 strongly suggests that potentiation of α2/3-containing GABA-A receptors is sufficient. Since α5 has potential burdens on cognition, sedation, and tolerance , further study of the potential therapeutic value of α2/3-selective PAMs are warranted. Our studies went a step further, and we tested GABA-A α2/3-specific PAM without α1 or α5 efficacy, MP-III-024, and confirmed the previous observation in harmaline-induced ET-like tremor. That α2/3, but not α1, subunit-containing GABA-A receptors are sufficient for anti-tremor efficacy is based on the MP-III-024 pharmacological profile proven previously in vitro in functional GABA receptor assays and the fact that zolpidem, which is α1-selective, was not tremorolytic and NS16085’s effectiveness was shown even with its small α1 antagonism. Additionally, other selective α2/3-receptor PAMs have shown activity in tremor patients (discussed above). Moreover, we show for the first time that MP-III-024 was also effective not only in the ET model but also against pimozide-induced TJMs. MP-III-024 is a positive allosteric modulator with preference for GABA-A α2 and α3 receptors relative to α1 and α5, as well as an improved metabolic profile as compared to another subtype-selective PAMs. Although MP-III-024 is an ester, it is more metabolically stable than its analog, HZ-166 [60,67]. In addition, other analogs such as KRM-II-81 that substitute a bioisostere for the ester function are known to provide markedly-improved drug-like properties . MP-III-024 has a good metabolic stability, with 75% and 76% remaining after one hour of mouse or human liver microsome assay. The half-life of MP-III-024 is 141.5 min with an intrinsic clearance of 0.491 µL/min/mg and a metabolic rate of 9.815 nmol/min/mg . As an ionotropic receptor ligand, its effects are observed very fast after administration. Our study timeline fits in the maximum drug bioavailability window. MP-III-024 produced significant decreases in tremor but these effects were not always dose-dependent. This finding contrasts with the dose-dependent changes in nociception observed with this compound . The data show that the lowest dose of MP-III-024 tested (3.2 mg/kg) produced almost complete reversal of harmaline-induced tremor in the AP2 parameter (Figure 2B). Doses lower than 3.2 mg/kg should be tested in future studies to fully realize the dose range of activity of this compound. Nonetheless, the data suggest an important possibility that tremor is more potently controlled by α2/3-containing GABA-A receptors than nociception. Fischer et al. showed that MP-III-024 dose- and time-dependently reversed mechanical hyperalgesia (10 and 32 mg/kg) and did not affect locomotor activity or operant behavior. It was reported by Rahman et al. that MP-III-024 can reduce the amount of opioids needed to control pain when dosed together with morphine, therefore showing synergistic pain reduction. As predicted, substances of this type are non-sedative, do not impair motor functions and do not develop tolerance or addiction. An example of this was described using another compound of similar pharmacology, KRM-II-81, which was disclosed as a non-sedating anxiolytic-like agent. Preclinical data demonstrated its efficacy in animal models of anxiety, depression, acute and chronic pain, epilepsy, and traumatic brain injury, along with reduced sedation, motor-impairment, tolerance development, and abuse liability [56,73]. Our results indicate that GABA-A α2/3 could become a starting point for studies to understand tremor generation and extinction mechanisms and indicate a new biological target for therapy. 4.2. Different Tremor Models, Same Effect—Why? Interestingly, for comparison, we also tested MP-III-024 in pimozide-induced model of TJMs and it also generated a significant tremorolytic effect. Since MP-III-024 was effective in tremor reduction in harmaline and pimozide-induced models, GABA-A α2/3-related mechanisms could be the common point of tremor regulation. Despite that both models are induced by differently acting substances, they potentially share common effector. Thus, MP-III-024 effect is probably located downstream of striatum and at least partially dependent on GABA-A α2/3. Tremors in ET are mostly postural or kinetic with a frequency range of 4–12 Hz, while resting tremor in PD occurs in a frequency range of 3–7 Hz. Pimozide acts as an antagonist of dopamine D2, D3, and D4 receptors; the 5-HT7 receptor and SNr and GPi appear to be a major basal ganglia output region through which the thalamo-cerebellar circuit and tremor are regulated . Interaction between the basal ganglia and cerebellum was also indicated as the most important for the generation of parkinsonian tremor. This concept has been summarized as the “dimmer-switch” theory of PD tremor, which proposes that tremor is initiated in basal ganglia structures, but the cerebellum modulates its amplitude and rhythmicity [25,31,33,34]. Recent studies indicated that other brain structures, such as the ventral thalamus, which receives glutamatergic projections from the cerebellum and GABAergic projections from the basal ganglia, appear to be a region of convergence of neuronal impulses, where they are processed and relayed to the cerebral cortex. These basal ganglia-thalamo-cortical and cerebello-thalamo-cortical circuits were assumed to be completely independent, but recent data suggested some points of their contact, such as the motor cortex [30,32]. Therefore, the pathophysiology of parkinsonian tremor also needs further research. The suggested mechanisms of harmaline-induced tremors are that it enhances synaptic activity of the climbing fibers originating in the inferior olive nuclei. Their pathological activation causes excessive glutamate release in the cerebellum [74,75] and affects GABAergic Purkinje cells. This leads to disinhibition of deep cerebellar neurons consequently increasing glutamate release in the thalamus . Rhythmic activity of this structure, results in generation of action tremor, the clinical feature of ET [2,11]. The exact mechanisms underlying tremor generation by harmaline, or in ET are still unclear. Based on the results of this study, we can only roughly hypothesize but it seems that both in pimozide- and harmaline-induced tremors and by approximation probably in ET and PD, the common tremor regulating pathways include the cerebellum, thalamus and motor cortex. 4.3. Lack of Tremorolytic Effect of Zolpidem in Experimental Studies Zolpidem is the most selective known GABA-A α1 PAM used in the clinic as a hypnotic drug for the short-term treatment of sleeping problems. Previous clinical observations by Daniele et al. in a pilot study with ten PD patients showed remarkable improvement of motor symptoms, surprisingly including tremor, even after a single dose of zolpidem. A case report by Farver & Khan described the beneficial effects of zolpidem in patient suffering from antipsychotic-induced parkinsonian-like tremors. In addition, Růzicka et al. and Hall et al. reported improved patients’ UPDRS scores after zolpidem. The hypothesis behind those observations was that in PD overactivity of the GABAergic neurons in the GP can lead to overinhibition of the thalamus and the cerebral cortex. With a selective inhibition of GABA-A α1 in the main structures involved in movement disorder zolpidem might counteract it. In contrast, a recent study by Diamond et al. indicated no change in GABA-A α1 expression in the GP of PD patients as compared to healthy humans, counteracting the classical hypothesis of pathophysiology of movement disorder evoked by dopamine loss. Besides clinical case reports and promising hypothesis and few optimistic reviews, there are limited studies attempting to validate the GABA-A α1 role in PD tremor besides the motor dysfunction. Even reports by Daniele et al. and zolpidem use in the clinic were focused on motor impairment while tremor was indicated only as one of the symptoms. Recent metanalysis by Laifenfeld et al. suggested repurposing zolpidem for use in slowing PD progression but their analysis was focused on dementia, falls and psychosis, not tremor. There are two registered clinical trials on zolpidem in PD (clinicaltrials.gov). One was withdrawn and second assessed motor function including tremor (NCT03621046), although the data are still not released. The same group published interesting results on zolpidem positively affecting bradykinesia and movement initiation by the measurement of beta oscillations in 17 PD patients but again, not tremor. Despite promising preliminary effects in PD patients, and an attractive working hypothesis based on GABA-A α1 expression in the key brain structures related to PD and tremor and strong selectivity of the tested drug, in our experimental study zolpidem did not affect tremor in either of the tested doses and none of the different animal tremor models. The studies of Diamond et al. are in line with our suggestion that α1 subunit does not play a role in CNS-mediated PD tremor. One of the reasons for the lack of zolpidem effectivity could be due to an overly widespread expression of GABA-A α1 in the brain to evoke specific effects on the tremor, despite its localization in key regions of interest. GABA-A receptors and associated α1 subunit are expressed in the SNr, GPi, GPe, STN, thalamus, motor cortex and many other structures, thus possibly counteracting each other function. The second, more reliable reason could be the role of advanced dopaminergic neuron degeneration and resulting adaptation/dysfunction of multiple downstream regulatory loops, as well as the involvement of other neuronal pathways affected in severe stages of late PD as it was described in the clinical case reports. The complexity and heterogeneity of patient PD tremor is not fully reflected by the available simplified rat models. In our test models using pimozide and tetrabenazine for induction of TJM, the nigrostriatal system was only functionally and temporarily blocked. Therefore, although it is possible to evoke dopamine-related tremor in rodents even without permanent nigrostriatal lesion our results in rat models indicate that tremor induction by pimozide or tetrabenazine is not identical with human tremor and is based on mechanisms far downstream from striatum and probably not GABA-A α1 selectively–related process. Importantly, besides just a few studies on non-human, old age primates, the available rodent models of PD-related nigrostriatal degeneration and/or proteinopathy evoking loss of dopamine (6-OHDA, MPTP/MPP+, α-synuclein, rotenone, paraquat, genetic models, etc.), even those most severe, do not exhibit tremors. Therefore, the direct role of dopamine in the pathophysiology of PD-tremors is speculative, especially given that L-DOPA treatment only partially alleviates tremor in patients. Although tremor is considered to be the most specific marker of PD, its severity has not been shown to parallel disease progression or the degree of nigrostriatal neuron loss. However, recent study showed that rest tremor correlates with reduced contralateral striatal dopamine transporter binding in PD patients . Therefore, research is inconclusive in this aspect. PD is also a very heterogenous disease itself, recently being divided into multiple more specific subtypes. Thus, the tremor-dominant PD subtype has a more benign course and it is hypothesized that tremor itself could be the sign of compensatory mechanisms counteracting akinesia/bradykinesia. All these prompts to hypothesize that additional anatomical regions, downstream affected networks, and neurotransmitters contribute to parkinsonian tremor. In the exploratory small clinical study by Daniele et al. , the best effects of zolpidem were observed in the patients with the most advanced stages of PD. Therefore, long-term use of antiparkinsonian pharmacotherapy and system adaptation in late-phase PD patients should be taken under consideration . Our results indicate that human late PD-tremor and tetrabenazine- or pimozide-induced dopamine-related rat tremor may not have overlapping pathomechanisms with respect to GABA-A α1 receptor subunit. The question remains which physioanatomical or functional aspects of the human, but not rodent, brain is responsible for tremor generation. This discrepancy could be a starting point for a search for tremor-specific mechanisms in PD. Therefore, if the tested models represent at least partially-relevant mechanisms underlying tremor generation, our results are not in line with repurposing zolpidem as an anti-tremor medication. The mechanistic role of the effects zolpidem on akinesia, bradykinesia, stiffness, and other PD symptoms should be studied otherwise. 4.4. Lack of Tremorolytic Effect of Zolpidem in ET Model Our results show also that zolpidem has no influence also on harmaline-induced ET-like tremor in rats in neither of the three tested doses. In the report by Assini & Abercrombie , the first dose affecting motor function tested on rotarod in naive mice was 1 mg/kg of zolpidem. A dose of 0.5 mg/kg did not evoke any significant impairment. In our study, although we tested a similarly low, sub-hypnotic doses of zolpidem, it decreased locomotor activity measured in FPA as distance travelled and general rat mobility measured as AP1 parameter in the two higher doses (0.67 and 1.01 mg/kg) 30 min after treatment. The effect was dose-dependent. This is in line with known observations that GABA-A α1 subunit is responsible for sedation. Therefore, those results point out that if harmaline induced tremor is relevant to ET physiology then GABA-A α1 is not specifically involved in ET pathology. 4.5. GABA-A α1 vs. α2/3 Subunit Localisation In this study, tremor was decreased by GABA-A α2/3 but not GABA-A α1 selective PAMs. The pharmacological output of tested substances depends on receptor subunit composition and distribution in the specific brain regions. Therefore, based on our results, one could hypothesize that structures with dominant α2/3 but not α1 expression could be the regions of interest for further detailed studies of tremor. One of the possible explanations is the dominating expression of α2 and lack of α1 GABA-A subunit in the inferior olivary complex, a structure which enhanced synaptic activity which is directly responsible for tremor generation, at least in ET and harmaline-induced models. Activation of inhibitory GABAergic transmission in this structure is probably responsible for decreasing tremor. Other potential targets lacking GABA-A α1, but with high expression of α2, are striatum and cerebellar Purkinje cell layers with no α1 or α2 subunits but numerous α3 protein signals. Strong immunoreactivity for α2- and α 3-containing GABA-A receptors has been detected in the parvalbumin positive neurons in the thalamus and molecular layers of the cerebellum [85,86,87,88,89]. Hypothetically, structures expressing α2/3 but not α1, GABA-A receptor subunits could be the target for future tremor therapies. 4.6. Effect on Locomotor Activity Side effects characteristics of benzodiazepines are daytime drowsiness and impaired motor coordination. In fact, GABA-A α2/3, but not α1, specific PAMs have an improved profile in this regard. Previous studies already showed that MP-III-024 did not affect locomotor activity (3.2, 10, 32 mg/kg) during 60 min of open field tests or operant behavior (3.2, 10, 32, 100 mg/kg) during 160 min in the nose-poke operant chamber test. In our study MP-III-024 alone tested in a lower dose (3.2 mg/kg) significantly increased distance travelled (walking), and all animal movements (not only locomotion but also body movements in place) quantified as AP1 parameters in both doses (3.2 and 10 mg/kg). This increased animal mobility was observed during the first 30 min of the test in FPA, which corresponds to the first 60 min after MP-III-024 injection. As compared to the results of Fischer et al. , their animals at the corresponding doses did not show significant increases in the two types of movement behavior analyses, although some tendencies might be visible. The FPA analysis used in our study is a much more sensitive method of quantification than open field. When MP-III-024 was administered before harmaline it also increased locomotor activity as compared to harmaline alone group. Interestingly, harmaline itself has bidirectional effects on locomotor activity in rats due to tremor generation, when analyzed directly after i.p. injection . In the first 30 min of analysis, it decreases distance travelled then increases it in the following 30 min as compared to controls. It seems that harmaline treatment and induction of tremor in rats postpones their exploratory activity and it was also observed in the combined treatment group. In conclusion, MP-III-024 not only does not impair motor activity in rats but temporarily increases it in time-points mostly corresponding with exploratory activity. Further cognitive- and motivation-driven tests would bring more information on the role of GABA-A α2/3 specific mechanisms. Nevertheless, additional mechanisms facilitating movement could be of the highest interest in potential PD pharmacotherapy using selective GABA-A receptor α2/3 subtype but not α1 specific PAMs. 5. Conclusions Traditional GABA-A PAM drugs have safety and tolerability concerns that include sedation, motor-impairment, respiratory depression, tolerance, and dependence. The search for improved therapeutic agents focuses on more selective ligands that potentiate GABA-A receptors . There is a huge void in pharmacotherapy of tremor, therefore, it is necessary to target it as the main study subject. Here we have shown for the first time that GABA-A α2/3, but not GABA-A α1, is a promising target for tremorolytic therapies both in ET and PD tremor disorders. Our results indicate that zolpidem, a positive allosteric modulator selective to GABA-A α1 subunits did not affect tremor induced by harmaline, tetrabenazine, or pimozide, while MP-III-024—a positive allosteric modulator selective to GABA-A receptor α2/3 subunits—decreased tremor parameters in both ET and PD related types of tremors. The effectiveness of MP-III-024 suggests at least partially overlapping mechanism of tremor pathophysiology in those two neuropathologies. The results of this work offer an important hint in elucidating tremor suppression mechanisms. The data suggest that GABA-A α2/3 should be considered a novel drug-target for PD and ET tremor therapy, with additional potential towards movement facilitation in PD. Supplementary Materials The following supplementary data can be downloaded at: Figure S1. Tremolous Jaw Movements and Catalepsy induced in Wistar rat by 7 doses of Pimozide. Figure S2. Tremolous Jaw Movements and Catalepsy induced in Wistar rat by 1 dose of Tetrabenazine. Figure S3. Walking distance affected by Pimozide or Tetrabenazine, measured in Force Plate Actimeters (BASi, West Lafayette, IN, USA) Author Contributions Conceptualization: K.Z.K., B.K., J.W., M.M., J.M.W. and J.M.C.; methodology and experimental part—data analysis: B.K., K.Z.K., M.P., D.B., J.K. and D.S.; statistical analysis: B.K.; resources, M.M., J.M.W. and J.M.C.; data curation: B.K. and K.Z.K.; writing—original draft preparation: K.Z.K., B.K. and J.W.; writing—review: K.Z.K., B.K., J.W., M.M. and J.M.W.; visualization: B.K. All authors have read and agreed to the published version of the manuscript. Funding This study was supported by the statutory funds of Department of Neuropsychopharmacology, Maj Institute of Pharmacology Polish Academy of Sciences, Kraków, Poland and the statutory funds of Dept. of Pharmaceutical Chemistry, Jagiellonian University Medical College, Kraków, Poland: N42/DBS/000176. We thank the following granting agencies for support: DA-043204 and NS-076517 and the National Science Foundation, Division of Chemistry [Grant CHE-1625735]. We also acknowledge the UW-Milwaukee Shimadzu Laboratory for Advanced and Applied Analytical Chemistry and support from the Milwaukee Institute of Drug Discovery and the University of Wisconsin-Milwaukee Research Foundation. Institutional Review Board Statement The experiments were carried out in compliance with the Act on Experiments on Animals of 21 January 2005 amended on 15 January 2015 (published in Journal of Laws no. 23/2015 item 266, Poland), and according to the Directive of the European Parliament and of the Council of Europe 2010/63/EU of 23 September 2010 on the protection of animals used for scientific purposes. They also received an approval of the Local Ethics Committee at the Maj Institute of Pharmacology, Polish Academy of Sciences, Cracow (permission no: 271/2018, 89/2020, 90/2021). Informed Consent Statement Not applicable. Data Availability Statement Research data are available upon request to the first or corresponding author at mroz@if-pan.krakow.pl or kuter@if-pan.krakow.pl. Acknowledgments James Cook is named as inventors on patents describing KRM-II-81 and analogues, certain rights to which have been licensed to RespireRx Pharmaceuticals Inc., James Cook, and Jeffrey Witkin are members of the research advisory group for RespireRx Pharmaceuticals Inc. Conflicts of Interest The authors declare that there are no conflict of interest. References Elble, R.J. What Is Essential Tremor? Curr. Neurol. Neurosci. Rep. 2013, 13, 353. [Google Scholar] [CrossRef] [PubMed] Kosmowska, B.; Wardas, J. The Pathophysiology and Treatment of Essential Tremor: The Role of Adenosine and Dopamine Receptors in Animal Models. Biomolecules 2021, 11, 1813. [Google Scholar] [CrossRef] [PubMed] Rajput, A.H.; Adler, C.H.; Shill, H.A.; Rajput, A. 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Timelines of drug administration and harmaline-induced tremor measurement (A,B) or observation and counting of TJMs induced by pimozide (C,D) or TBZ (E). FPA measurement of harmaline-induced tremor lasted 30 min or 60 min for experiments with zolpidem and MP-III-024, respectively. Abbreviations: FPA—Force Plate Actimeter; TBZ—tetrabenazine; TJMs—Tremulous Jaw Movements. Figure 1. Timelines of drug administration and harmaline-induced tremor measurement (A,B) or observation and counting of TJMs induced by pimozide (C,D) or TBZ (E). FPA measurement of harmaline-induced tremor lasted 30 min or 60 min for experiments with zolpidem and MP-III-024, respectively. Abbreviations: FPA—Force Plate Actimeter; TBZ—tetrabenazine; TJMs—Tremulous Jaw Movements. Figure 2. The effect of MP-III-024 (3.2, 10 mg/kg) on the harmaline-induced tremor: (A) power spectrum, (B) AP2, (C) tremor index, (D) AP1, and (E) locomotor activity (distance) of rats. The power spectrum within a range of 0–20 Hz, averaged over the two intervals (0–30 and 30–60 min) for all animals in each group is shown. Tremor parameter data are shown as the means ± SEM. AP1—power in the 0–8 Hz band, AP2—power in the 8–15 Hz band. The number of animals: SOLV, n = 8; HARM 15 mg/kg, n = 8; MP 3.2 mg/kg, n = 8; MP 10 mg/kg, n = 8; MP 3.2 mg/kg + HARM 15 mg/kg, n = 10; MP 10 mg/kg + HARM 15 mg/kg, n = 10. Statistics: ANOVA for repeated measures with regard to AP2 (HARM effect: F[1,47] = 27.111, p = 0.001; MP effect: F[2,47] = 1.459, p = 0.243; time effect: F[1,47] = 54.740, p = 0.001), Tremor index (HARM effect: F[1,47] = 217.767, p = 0.001; MP effect: F[2,47] = 3.576, p = 0.036; time effect: F[1,47] = 2.004, p = 0.163), AP1 (HARM effect: F[1,47] = 83.421, p = 0.001; MP effect: F[2,47] = 2.381, p = 0.104; time effect: F[1,47] = 275.344, p = 0.001) and Distance (HARM effect: F[1,47] = 3.537, p = 0.067; MP effect: F[2,47] = 5.414, p = 0.007; time effect: F[1,47] = 279.195, p = 0.001). LSD post-hoc test: p ≤ 0.05, p ≤ 0.01, p ≤ 0.001, a p = 0.077, b p = 0.060 vs. SOLV; c p = 0.090 vs. HARM 15 mg. Figure 2. The effect of MP-III-024 (3.2, 10 mg/kg) on the harmaline-induced tremor: (A) power spectrum, (B) AP2, (C) tremor index, (D) AP1, and (E) locomotor activity (distance) of rats. The power spectrum within a range of 0–20 Hz, averaged over the two intervals (0–30 and 30–60 min) for all animals in each group is shown. Tremor parameter data are shown as the means ± SEM. AP1—power in the 0–8 Hz band, AP2—power in the 8–15 Hz band. The number of animals: SOLV, n = 8; HARM 15 mg/kg, n = 8; MP 3.2 mg/kg, n = 8; MP 10 mg/kg, n = 8; MP 3.2 mg/kg + HARM 15 mg/kg, n = 10; MP 10 mg/kg + HARM 15 mg/kg, n = 10. Statistics: ANOVA for repeated measures with regard to AP2 (HARM effect: F[1,47] = 27.111, p = 0.001; MP effect: F[2,47] = 1.459, p = 0.243; time effect: F[1,47] = 54.740, p = 0.001), Tremor index (HARM effect: F[1,47] = 217.767, p = 0.001; MP effect: F[2,47] = 3.576, p = 0.036; time effect: F[1,47] = 2.004, p = 0.163), AP1 (HARM effect: F[1,47] = 83.421, p = 0.001; MP effect: F[2,47] = 2.381, p = 0.104; time effect: F[1,47] = 275.344, p = 0.001) and Distance (HARM effect: F[1,47] = 3.537, p = 0.067; MP effect: F[2,47] = 5.414, p = 0.007; time effect: F[1,47] = 279.195, p = 0.001). LSD post-hoc test: p ≤ 0.05, p ≤ 0.01, p ≤ 0.001, a p = 0.077, b p = 0.060 vs. SOLV; c p = 0.090 vs. HARM 15 mg. Figure 3. Reversal of the pimozide-induced tremulous jaw movements (TJMs) by MP-III-024 (3.2, 10 mg/kg). The data are shown as the mean (± SEM) number of tremulous jaw movements (TJMs) per 10 min observation period. The number of animals: SOLV, n = 7; PIM 1 mg/kg, n = 8; MP 3.2 mg/kg, n = 8; MP 10 mg/kg, n = 8, PIM 1 mg/kg + MP 3.2 mg/kg, n = 6; PIM 1 mg/kg + MP 10 mg/kg, n = 7. Statistics: Factorial ANOVA (PIM effect: F[1,38] = 41.736, p = 0.001; MP effect: F[2,38] = 3.919, p = 0.028) + LSD post-hoc test ( p ≤ 0.01, p ≤ 0.001). Figure 3. Reversal of the pimozide-induced tremulous jaw movements (TJMs) by MP-III-024 (3.2, 10 mg/kg). The data are shown as the mean (± SEM) number of tremulous jaw movements (TJMs) per 10 min observation period. The number of animals: SOLV, n = 7; PIM 1 mg/kg, n = 8; MP 3.2 mg/kg, n = 8; MP 10 mg/kg, n = 8, PIM 1 mg/kg + MP 3.2 mg/kg, n = 6; PIM 1 mg/kg + MP 10 mg/kg, n = 7. Statistics: Factorial ANOVA (PIM effect: F[1,38] = 41.736, p = 0.001; MP effect: F[2,38] = 3.919, p = 0.028) + LSD post-hoc test ( p ≤ 0.01, p ≤ 0.001). Figure 4. The effect of zolpidem (0.34, 0.67, 1.01 mg/kg) on harmaline-induced tremor: (A) power spectrum, (B) AP2, (C) tremor index, (D) AP1, and (E) locomotor activity (distance) of rats. The power spectrum within a range of 0–20 Hz, averaged over the whole measurement period (0–30 min) for all animals in each group is shown. Tremor parameters data are shown as the means ± SEM. AP1—power in the 0–8 Hz band, AP2—power in the 8–15 Hz band. The number of animals: SOLV, n = 11; HARM 15 mg/kg, n = 15; ZOL 0.34 mg/kg, n = 5; ZOL 0.67 mg/kg, n = 6; ZOL 1.01 mg/kg, n = 6; ZOL 0.34 mg/kg + HARM 15 mg/kg, n = 8; ZOL 0.67 mg/kg + HARM 15 mg/kg, n = 6; ZOL 1.01 mg/kg + HARM 15 mg/kg, n = 8. Statistics: Factorial ANOVA with regard to AP2 (HARM effect: F[1,57 = 21,391, p = 0.001; ZOL effect: F[3,57] = 0.546, p = 0.653), Tremor index (HARM effect: F[1,57] = 103.001, p = 0.001; ZOL effect: F[3,57] = 3.142, p = 0.032), AP1 (HARM effect: F[1,57] = 49.307, p = 0.001; ZOL effect: F[3,57] = 19.014, p = 0.001), and Distance (HARM effect: F[1,57] = 5.421, p = 0.023; ZOL effect: F[3,57] = 4.481, p = 0.007). LSD post-hoc test: p ≤ 0.01, p ≤ 0.001. Figure 4. The effect of zolpidem (0.34, 0.67, 1.01 mg/kg) on harmaline-induced tremor: (A) power spectrum, (B) AP2, (C) tremor index, (D) AP1, and (E) locomotor activity (distance) of rats. The power spectrum within a range of 0–20 Hz, averaged over the whole measurement period (0–30 min) for all animals in each group is shown. Tremor parameters data are shown as the means ± SEM. AP1—power in the 0–8 Hz band, AP2—power in the 8–15 Hz band. The number of animals: SOLV, n = 11; HARM 15 mg/kg, n = 15; ZOL 0.34 mg/kg, n = 5; ZOL 0.67 mg/kg, n = 6; ZOL 1.01 mg/kg, n = 6; ZOL 0.34 mg/kg + HARM 15 mg/kg, n = 8; ZOL 0.67 mg/kg + HARM 15 mg/kg, n = 6; ZOL 1.01 mg/kg + HARM 15 mg/kg, n = 8. Statistics: Factorial ANOVA with regard to AP2 (HARM effect: F[1,57 = 21,391, p = 0.001; ZOL effect: F[3,57] = 0.546, p = 0.653), Tremor index (HARM effect: F[1,57] = 103.001, p = 0.001; ZOL effect: F[3,57] = 3.142, p = 0.032), AP1 (HARM effect: F[1,57] = 49.307, p = 0.001; ZOL effect: F[3,57] = 19.014, p = 0.001), and Distance (HARM effect: F[1,57] = 5.421, p = 0.023; ZOL effect: F[3,57] = 4.481, p = 0.007). LSD post-hoc test: p ≤ 0.01, p ≤ 0.001. Figure 5. Zolpidem (0.67 mg/kg) had no effect on pimozide-induced (A) or TBZ-induced (B) tremulous jaw movements (TJMs). The data are shown as the mean (±SEM) number of tremulous jaw movements (TJMs) per 10 min observation period. The number of animals: SOLV, n = 11; PIM 1 mg/kg, n = 10; TBZ 2 mg/kg, n = 11; ZOL 0.67 mg/kg, n = 10; PIM 1 mg/kg +ZOL 0.67 mg/kg, n = 10; TBZ 2 mg/kg + ZOL 0.67 mg/kg, n = 10. Statistics: Factorial ANOVA for PIM (PIM effect: F[1,37] = 26.019, p = 0.001; ZOL effect: F[1,37] = 0.446, p = 0.508) and TBZ (TBZ effect: F[1,38] = 51.114, p = 0.001; ZOL effect: F[1,38] = 0.410, p = 0.526) + LSD post-hoc test ( p ≤ 0.01, p ≤ 0.001). Figure 5. Zolpidem (0.67 mg/kg) had no effect on pimozide-induced (A) or TBZ-induced (B) tremulous jaw movements (TJMs). The data are shown as the mean (±SEM) number of tremulous jaw movements (TJMs) per 10 min observation period. The number of animals: SOLV, n = 11; PIM 1 mg/kg, n = 10; TBZ 2 mg/kg, n = 11; ZOL 0.67 mg/kg, n = 10; PIM 1 mg/kg +ZOL 0.67 mg/kg, n = 10; TBZ 2 mg/kg + ZOL 0.67 mg/kg, n = 10. Statistics: Factorial ANOVA for PIM (PIM effect: F[1,37] = 26.019, p = 0.001; ZOL effect: F[1,37] = 0.446, p = 0.508) and TBZ (TBZ effect: F[1,38] = 51.114, p = 0.001; ZOL effect: F[1,38] = 0.410, p = 0.526) + LSD post-hoc test ( p ≤ 0.01, p ≤ 0.001). | | | Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. | © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( Share and Cite MDPI and ACS Style Kosmowska, B.; Paleczna, M.; Biała, D.; Kadłuczka, J.; Wardas, J.; Witkin, J.M.; Cook, J.M.; Sharmin, D.; Marcinkowska, M.; Kuter, K.Z. GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats. Biomolecules 2023, 13, 197. AMA Style Kosmowska B, Paleczna M, Biała D, Kadłuczka J, Wardas J, Witkin JM, Cook JM, Sharmin D, Marcinkowska M, Kuter KZ. GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats. Biomolecules. 2023; 13(2):197. Chicago/Turabian Style Kosmowska, Barbara, Martyna Paleczna, Dominika Biała, Justyna Kadłuczka, Jadwiga Wardas, Jeffrey M. Witkin, James M. Cook, Dishary Sharmin, Monika Marcinkowska, and Katarzyna Z. Kuter. 2023. "GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats" Biomolecules 13, no. 2: 197. APA Style Kosmowska, B., Paleczna, M., Biała, D., Kadłuczka, J., Wardas, J., Witkin, J. M., Cook, J. M., Sharmin, D., Marcinkowska, M., & Kuter, K. Z. (2023). GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats. Biomolecules, 13(2), 197. Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here. Article Metrics No Article Access Statistics For more information on the journal statistics, click here. Multiple requests from the same IP address are counted as one view. Supplementary Material Supplementary File 1: ZIP-Document (ZIP, 242 KB) clear Zoom | Orient | As Lines | As Sticks | As Cartoon | As Surface | Previous Scene | Next Scene Export citation file: BibTeX) MDPI and ACS Style Kosmowska, B.; Paleczna, M.; Biała, D.; Kadłuczka, J.; Wardas, J.; Witkin, J.M.; Cook, J.M.; Sharmin, D.; Marcinkowska, M.; Kuter, K.Z. GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats. Biomolecules 2023, 13, 197. AMA Style Kosmowska B, Paleczna M, Biała D, Kadłuczka J, Wardas J, Witkin JM, Cook JM, Sharmin D, Marcinkowska M, Kuter KZ. GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats. Biomolecules. 2023; 13(2):197. Chicago/Turabian Style Kosmowska, Barbara, Martyna Paleczna, Dominika Biała, Justyna Kadłuczka, Jadwiga Wardas, Jeffrey M. Witkin, James M. Cook, Dishary Sharmin, Monika Marcinkowska, and Katarzyna Z. Kuter. 2023. "GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats" Biomolecules 13, no. 2: 197. APA Style Kosmowska, B., Paleczna, M., Biała, D., Kadłuczka, J., Wardas, J., Witkin, J. M., Cook, J. M., Sharmin, D., Marcinkowska, M., & Kuter, K. Z. (2023). GABA-A Alpha 2/3 but Not Alpha 1 Receptor Subunit Ligand Inhibits Harmaline and Pimozide-Induced Tremor in Rats. Biomolecules, 13(2), 197. 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188288
https://www.cuemath.com/algebra/standard-notation/
Standard Notation Standard Notation Julia asked her kids, "Do you know how old the Earth is?" They thought and said that the Earth was older than her, them, and their grandparents. Julia told them that the Earth was believed to have been formed over 4.5 billion years ago. She asked them if they could try counting this number on their fingers. When they replied that they couldn't, she told them that a number as huge as this can be expressed mathematically in its standard notation only. In this mini-lesson, we will explore the world of scientific and standard notations of the numbers by finding the answers to the questions like what is a standard notation and how to convert scientific notation to standard notation. Let us explore the chapter Standard Notation in detail. Lesson Plan | | | --- | | 1. | What Is Standard Notation? | | 2. | Important Notes on Standard Notation | | 3. | Solved Examples on Standard Notation | | 4. | Challenging Questions on Standard Notation | | 5. | Interactive Questions on Standard Notation | What Is Standard Notation? A standard notation is a form of writing a given number, an equation, or an expression in a form that follows certain rules. For example, 4.5 billion years is written as 4,500,000,000 years As you can see here, writing a large number like 4.5 billion in its number form is not just ambiguous but also time-consuming and there are chances that we may write a few zeros less or more while writing in the number form. So, to represent very large or very small numbers concisely, we use the standard notation. In Britain, standard notation is also known as scientific notation where a large number is written in the form of power of 10. Depending upon which mathematical concept we are dealing with, the standard notation will vary. The standard form has different meanings depending on which country you are in. In the United States and countries using US conventions, the standard form is the usual way of writing numbers in decimal notation. Standard form = (3890) Expanded form = (3000 + 800 + 90) Written form = Three thousand eight hundred and ninety Let us now look at how to convert standard notations to scientific notation and vice-versa. How To Convert Scientific Notation To Standard Notation? Why in the above section 800 is written as (8\times10^2) in scientific notation? (800) = (8\times100) (100=10\times10\ or 10^2) Hence, 800 can be written in both the ways. Power to the 10 shows the exact place of decimal to be moved. To find the place of decimal you should follow the below points: a) If the power on 10 is greater than or equal to 10. The decimal point will move to the left side. This case implies only if the power of 10 is positive. For example, (10^{2}) b) If the power on 10 is lesser than 1. The decimal point will move to the right side. This case implies only if the power of 10 is negative. For example, (10^{-2}) Let us look at the sample problem, (6.5\times10^{-3}) can be written as (6.5\times0.001) or (0.0065) Here one decimal before (5) represents one zero. If exponent in standard notation or scientific notations are in minus (10^{-3}) it is represented as (\dfrac{1}{1000}). Important Notes Degree means the highest power of the variable present in the equation. Calculators often use the letter "E" for example, (1E+5) which represents (1\times10^5). This is known as scientific e notation. In scientific notation, the exponential part helps in fixing the movement of the decimal point. Solved Examples Here are a few standard notation examples for you to understand how to use exponents in standard notation and how to convert standard notations into scientific notations in detail. | | | Example 1 | The mass of Neptune is (83,000,000,000,000,000,000,000) (lb). Help Jamie to write this in scientific notation. Solution Here, standard notation (83,000,000,000,000,000,000,000) is written as (83) followed by 21 zeroes. To write the number in scientific notation, Zoe needs to change (83) to (8.3). Therefore, Jamie has to think of how many decimal places to be moved to the left side, i.e., 22 places. So, the scientific notation for the number is (8.3 \times 10^{22}) (lb). | | | (\therefore) (8.3 \times 10^{22}) (lb). | | | | Example 2 | Can you help Zoe to write (7.56\times10^{11}) in the standard notation? Solution Here, (7.56) is (756). Now, Zoe has to move the decimal point 11 places to the right and add zeroes accordingly. The standard notation for (7.56\times10^{11}) is (756,000,000,000). | | | (\therefore) (756,000,000,000) | | | | Example 3 | The speed of a sound in the air is (1.53645\times10^{5}) miles per hour. Help Tim to write this in standard notation. Solution The power of (10) is (5), so move the decimal point (5) places to the right: (1.53645) (→) (15.3645) (→) (153.645) (→) (1536.45) (→) (15364.5) (→) (153645) So the speed of sound in the air is (153645\ mph). | | | (\therefore) (153645) miles per hour. | Challenging Questions Write the following in scientific notationa) 8E+7b) 7.12E4 Interactive Questions on Standard Notation Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result. Let's Summarize The mini-lesson targeted the fascinating concept of standard notation. The math journey around standard notation starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath. About Cuemath At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in. Frequently Asked Questions (FAQs) 1. What is the difference between standard and scientific notation? Standard and scientific notation are the ways to represent numbers mathematically. We write numbers in standard and scientific notations using the rules for respective mathematical concepts. For example, (7.56\times10^{11}) is (756,000,000,000) Here, (7.56\times10^{11}) is a scientific notation. (756,000,000,000) is standard notation. When a number is represented with a power of 10 it is termed to be scientific notation. For big numbers with more than 10 zeroes, we prefer to write the number in scientific notation as it is easy to read large numbers and also helpful in quick calculations. 2. What is function notation? An easy method of describing a function in a simple manner such as f(x) which can be read as “f” of “x” 3. What is the expanded form of a whole number? An expanded form is a method to write a number with expansion to understand the logic behind one's, ten's, hundred's places. For example, whole number = (3890), expanded form = (3000 + 800 + 90) More Important Topics NumbersAlgebraGeometryMeasurementMoneyDataTrigonometryCalculus More Important Topics NumbersAlgebraGeometryMeasurementMoneyDataTrigonometryCalculus FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Events MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement MATH TEST Math Kangaroo AMC 8 MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 ABOUT US Our Mission Our Journey Our Team MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math Terms and ConditionsPrivacy Policy
188289
https://physics.nist.gov/cgi-bin/cuu/Value?eqmmn
CODATA Value: neutron molar mass Constants Topics: Values Energy Equivalents Searchable Bibliography Background Constants Bibliography Constants, Units & Uncertainty home page Click equation to show only symbol neutron molar mass Numerical value1.008 664 917 12 x 10-3 kg mol-1 Standard uncertainty0.000 000 000 51 x 10-3 kg mol-1 Relative standard uncertainty5.1 x 10-10 Concise form1.008 664 917 12(51)x 10-3 kg mol-1 Click here for correlation coefficient of this constant with other constants Source: 2022 CODATArecommended valuesDefinition of uncertaintyCorrelation coefficient with any other constant
188290
https://cse.sc.edu/~javidian/Notes_Presentations/BackFrontDoor.pdf
The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example An Overview of The Back-Door and Front-Door Criteria A Presentation Based On Sections 3.3 and 3.4 of Pearl’s Causality Mohammad Ali Javidian1 Marco Valtorta1 1Department of Computer Science University of South Carolina June, 2018 Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 1. (Back-Door) Outline 1 The Back-Door Criterion Definition 1. (Back-Door) Theorem 1. (Back-Door Adjustment) Proof of Theorem 1. 2 The Front-Door Criterion Definition 2. (Front-Door) Theorem 2. (Front-Door Adjustment) 3 do Calculus 4 Symbolic Derivation 5 Example Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 1. (Back-Door) Back-Door Criterion Definition A set of variables Z satisfies the back-door criterion relative to an ordered pair of variables (Xi, Xj) in a DAG G if: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 1. (Back-Door) Back-Door Criterion Definition A set of variables Z satisfies the back-door criterion relative to an ordered pair of variables (Xi, Xj) in a DAG G if: (i) no node in Z is a descendant of Xi; and Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 1. (Back-Door) Back-Door Criterion Definition A set of variables Z satisfies the back-door criterion relative to an ordered pair of variables (Xi, Xj) in a DAG G if: (i) no node in Z is a descendant of Xi; and (ii) Z blocks every path between Xi and Xj that contains an arrow into Xi. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 1. (Back-Door) Back-Door Criterion Definition A set of variables Z satisfies the back-door criterion relative to an ordered pair of variables (Xi, Xj) in a DAG G if: (i) no node in Z is a descendant of Xi; and (ii) Z blocks every path between Xi and Xj that contains an arrow into Xi. Figure: S1 = {X3, X4} and S2 = {X4, X5} would qualify under the back-door criterion, but S3 = {X4} would not because X4 does not d-separate Xi from Xj along the path (Xi , X3, X1, X4, X2, X5, Xj ). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Theorem 1. (Back-Door Adjustment) Outline 1 The Back-Door Criterion Definition 1. (Back-Door) Theorem 1. (Back-Door Adjustment) Proof of Theorem 1. 2 The Front-Door Criterion Definition 2. (Front-Door) Theorem 2. (Front-Door Adjustment) 3 do Calculus 4 Symbolic Derivation 5 Example Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Theorem 1. (Back-Door Adjustment) Back-Door Criterion Back-Door Adjustment Theorem If a set of variables Z satisfies the back-door criterion relative to (X, Y ), then the causal effect of X on Y is identifiable and is given by the formula Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Theorem 1. (Back-Door Adjustment) Back-Door Criterion Back-Door Adjustment Theorem If a set of variables Z satisfies the back-door criterion relative to (X, Y ), then the causal effect of X on Y is identifiable and is given by the formula P(y|ˆ x) = X z P(y|x, z)P(z). (1) Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. Outline 1 The Back-Door Criterion Definition 1. (Back-Door) Theorem 1. (Back-Door Adjustment) Proof of Theorem 1. 2 The Front-Door Criterion Definition 2. (Front-Door) Theorem 2. (Front-Door Adjustment) 3 do Calculus 4 Symbolic Derivation 5 Example Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. The proof originally offered in Pearl [2, 1993] is based on the observation that, when Z blocks all back-door paths from X to Y , setting (X = x) or conditioning on X = x has the same effect on Y . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. The proof originally offered in Pearl [2, 1993] is based on the observation that, when Z blocks all back-door paths from X to Y , setting (X = x) or conditioning on X = x has the same effect on Y . The effect of an atomic intervention do(Xi = x′ i ) is encoded by adding to G a link Fi →Xi (see the following Figure), where Fi is a new variable taking values in {do(x′ i ), idle}, x′ i ranges over the domain of Xi, and idle represents no intervention. Thus, the new parent set of Xi in the augmented network is PA′ i = PAi ∪{Fi}, and it is related to Xi by the conditional probability: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. The proof originally offered in Pearl [2, 1993] is based on the observation that, when Z blocks all back-door paths from X to Y , setting (X = x) or conditioning on X = x has the same effect on Y . The effect of an atomic intervention do(Xi = x′ i ) is encoded by adding to G a link Fi →Xi (see the following Figure), where Fi is a new variable taking values in {do(x′ i ), idle}, x′ i ranges over the domain of Xi, and idle represents no intervention. Thus, the new parent set of Xi in the augmented network is PA′ i = PAi ∪{Fi}, and it is related to Xi by the conditional probability: Figure: Representing external intervention Fi by an augmented network G′. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. P(xi|pa′ i) =    P(xi|pai) if Fi = idle 0 if Fi = do(x′ i ) and xi ̸= x′ i 1 if Fi = do(x′ i ) and xi = x′ i (2) Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. P(xi|pa′ i) =    P(xi|pai) if Fi = idle 0 if Fi = do(x′ i ) and xi ̸= x′ i 1 if Fi = do(x′ i ) and xi = x′ i (2) The effect of the intervention do(x′ i ) is to transform the original probability function P(x1, . . . , xn) into a new probability function P(x1, . . . , xn|ˆ x′ i ), given by Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. P(xi|pa′ i) =    P(xi|pai) if Fi = idle 0 if Fi = do(x′ i ) and xi ̸= x′ i 1 if Fi = do(x′ i ) and xi = x′ i (2) The effect of the intervention do(x′ i ) is to transform the original probability function P(x1, . . . , xn) into a new probability function P(x1, . . . , xn|ˆ x′ i ), given by P(x1, . . . , xn|ˆ x′ i ) = P′(x1, . . . , xn|Fi = do(x′ i )) (3) Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. P(xi|pa′ i) =    P(xi|pai) if Fi = idle 0 if Fi = do(x′ i ) and xi ̸= x′ i 1 if Fi = do(x′ i ) and xi = x′ i (2) The effect of the intervention do(x′ i ) is to transform the original probability function P(x1, . . . , xn) into a new probability function P(x1, . . . , xn|ˆ x′ i ), given by P(x1, . . . , xn|ˆ x′ i ) = P′(x1, . . . , xn|Fi = do(x′ i )) (3) By writing P(y|ˆ x) in terms of the augmented probability function P′ in accordance with eq. (3) and conditioning on Z we obtain: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. P(xi|pa′ i) =    P(xi|pai) if Fi = idle 0 if Fi = do(x′ i ) and xi ̸= x′ i 1 if Fi = do(x′ i ) and xi = x′ i (2) The effect of the intervention do(x′ i ) is to transform the original probability function P(x1, . . . , xn) into a new probability function P(x1, . . . , xn|ˆ x′ i ), given by P(x1, . . . , xn|ˆ x′ i ) = P′(x1, . . . , xn|Fi = do(x′ i )) (3) By writing P(y|ˆ x) in terms of the augmented probability function P′ in accordance with eq. (3) and conditioning on Z we obtain: P(y|ˆ x) = P′(y|Fx) = X z P′(y|z, Fx)P′(z|Fx) = X z P′(y|z, x, Fx)P′(z|Fx). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. According to the condition (i) of the definition of the back-door criterion, no node in Z is a descendant of x. So, all paths between Fx and each node z ∈Z have (at least) a collider i.e., Fx ⊥ ⊥z (see the following figure). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. According to the condition (i) of the definition of the back-door criterion, no node in Z is a descendant of x. So, all paths between Fx and each node z ∈Z have (at least) a collider i.e., Fx ⊥ ⊥z (see the following figure). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. According to the condition (i) of the definition of the back-door criterion, no node in Z is a descendant of x. So, all paths between Fx and each node z ∈Z have (at least) a collider i.e., Fx ⊥ ⊥z (see the following figure). Therfore, P′(z|Fx) = P′(z) = P(z). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. If all back-door paths from X to Y are blocked, then all paths from FX to Y must go through the children of X, and those would be blocked if we condition on X. The implication is that Y is independent of FX given X, Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. If all back-door paths from X to Y are blocked, then all paths from FX to Y must go through the children of X, and those would be blocked if we condition on X. The implication is that Y is independent of FX given X, P(y|x, FX = do(x)) = P(y|x) = P(y|x, FX = idle) (4) Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. If all back-door paths from X to Y are blocked, then all paths from FX to Y must go through the children of X, and those would be blocked if we condition on X. The implication is that Y is independent of FX given X, P(y|x, FX = do(x)) = P(y|x) = P(y|x, FX = idle) (4) According to the back-door condition (ii), Z blocks every path between X and Y that contains an arrow into X. This observation together with eq. (4) implies that Y ⊥ ⊥Fx|(X, Z) i.e., Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Proof of Theorem 1. If all back-door paths from X to Y are blocked, then all paths from FX to Y must go through the children of X, and those would be blocked if we condition on X. The implication is that Y is independent of FX given X, P(y|x, FX = do(x)) = P(y|x) = P(y|x, FX = idle) (4) According to the back-door condition (ii), Z blocks every path between X and Y that contains an arrow into X. This observation together with eq. (4) implies that Y ⊥ ⊥Fx|(X, Z) i.e., P′(y|z, x, Fx) = P′(y|z, x) = P(y|z, x). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 2. (Front-Door) Outline 1 The Back-Door Criterion Definition 1. (Back-Door) Theorem 1. (Back-Door Adjustment) Proof of Theorem 1. 2 The Front-Door Criterion Definition 2. (Front-Door) Theorem 2. (Front-Door Adjustment) 3 do Calculus 4 Symbolic Derivation 5 Example Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 2. (Front-Door) Front-Door Criterion Definition A set of variables Z satisfies the front-door criterion relative to an ordered pair of variables (X, Y ) in a DAG G if: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 2. (Front-Door) Front-Door Criterion Definition A set of variables Z satisfies the front-door criterion relative to an ordered pair of variables (X, Y ) in a DAG G if: (i) Z intercepts all directed paths from X to Y ; Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 2. (Front-Door) Front-Door Criterion Definition A set of variables Z satisfies the front-door criterion relative to an ordered pair of variables (X, Y ) in a DAG G if: (i) Z intercepts all directed paths from X to Y ; (ii) there is no unblocked back-door path from X to Z; and Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 2. (Front-Door) Front-Door Criterion Definition A set of variables Z satisfies the front-door criterion relative to an ordered pair of variables (X, Y ) in a DAG G if: (i) Z intercepts all directed paths from X to Y ; (ii) there is no unblocked back-door path from X to Z; and (iii) all back-door paths from Z to Y are blocked by X. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Definition 2. (Front-Door) Front-Door Criterion Definition A set of variables Z satisfies the front-door criterion relative to an ordered pair of variables (X, Y ) in a DAG G if: (i) Z intercepts all directed paths from X to Y ; (ii) there is no unblocked back-door path from X to Z; and (iii) all back-door paths from Z to Y are blocked by X. Figure: A diagram representing the front-door criterion. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Theorem 2. (Front-Door Adjustment) Outline 1 The Back-Door Criterion Definition 1. (Back-Door) Theorem 1. (Back-Door Adjustment) Proof of Theorem 1. 2 The Front-Door Criterion Definition 2. (Front-Door) Theorem 2. (Front-Door Adjustment) 3 do Calculus 4 Symbolic Derivation 5 Example Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Theorem 2. (Front-Door Adjustment) Front-Door Criterion Front-Door Adjustment Theorem If a set of variables Z satisfies the front-door criterion relative to (X, Y ) and if P(x, z) > 0, then the causal effect of X on Y is identifiable and is given by the formula Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Theorem 2. (Front-Door Adjustment) Front-Door Criterion Front-Door Adjustment Theorem If a set of variables Z satisfies the front-door criterion relative to (X, Y ) and if P(x, z) > 0, then the causal effect of X on Y is identifiable and is given by the formula P(y|ˆ x) = X z P(z|x) X x′ P(y|x′, z)P(x′). (5) Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Rules of do Calculus Preliminary Notation Figure: Subgraphs of G used in the derivation of causal effects. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Inference Rules Rules of do Calculus Rule 1 (Insertion/deletion of observations): P(y|ˆ x, z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Inference Rules Rules of do Calculus Rule 1 (Insertion/deletion of observations): P(y|ˆ x, z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX . Rule 2 (Action/observation exchange): P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Inference Rules Rules of do Calculus Rule 1 (Insertion/deletion of observations): P(y|ˆ x, z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX . Rule 2 (Action/observation exchange): P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . Rule 3 (Insertion/deletion of actions): P(y|ˆ x, ˆ z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX,Z(W ). where Z(W ) is the set of Z-nodes that are not ancestors of any W -node in GX. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 1: Compute P(z|ˆ x) X ⊥ ⊥Z in GX because the path from X to Z is blocked by the converging arrows at Y . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 1: Compute P(z|ˆ x) X ⊥ ⊥Z in GX because the path from X to Z is blocked by the converging arrows at Y . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 1: Compute P(z|ˆ x) X ⊥ ⊥Z in GX because the path from X to Z is blocked by the converging arrows at Y . G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 1: Compute P(z|ˆ x) X ⊥ ⊥Z in GX because the path from X to Z is blocked by the converging arrows at Y . G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . In Rule 2, set y = z, x = ø, z = x, w = ø: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 1: Compute P(z|ˆ x) X ⊥ ⊥Z in GX because the path from X to Z is blocked by the converging arrows at Y . G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . In Rule 2, set y = z, x = ø, z = x, w = ø: P(z|ˆ x) = P(z|x) because (Z ⊥ ⊥X)GX . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2: Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2: Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z). X ⊥ ⊥Z in GZ because the path from X to Z is blocked by the converging arrows at Y . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2: Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z). X ⊥ ⊥Z in GZ because the path from X to Z is blocked by the converging arrows at Y . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2: Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z). X ⊥ ⊥Z in GZ because the path from X to Z is blocked by the converging arrows at Y . G satisfies the applicability condition for Rule 3: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX,Z(W ). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2: Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z). X ⊥ ⊥Z in GZ because the path from X to Z is blocked by the converging arrows at Y . G satisfies the applicability condition for Rule 3: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX,Z(W ). In Rule 3, set y = x, x = ø, z = z, w = ø: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2: Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z). X ⊥ ⊥Z in GZ because the path from X to Z is blocked by the converging arrows at Y . G satisfies the applicability condition for Rule 3: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX,Z(W ). In Rule 3, set y = x, x = ø, z = z, w = ø: P(x|ˆ z) = P(x) because (Z ⊥ ⊥X)GZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2 (continued): Compute P(y|ˆ z) (Z ⊥ ⊥Y |X)GZ because the path from Z to Y is blocked by X. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2 (continued): Compute P(y|ˆ z) (Z ⊥ ⊥Y |X)GZ because the path from Z to Y is blocked by X. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2 (continued): Compute P(y|ˆ z) (Z ⊥ ⊥Y |X)GZ because the path from Z to Y is blocked by X. G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2 (continued): Compute P(y|ˆ z) (Z ⊥ ⊥Y |X)GZ because the path from Z to Y is blocked by X. G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . In Rule 2, set y = y, x = ø, z = z, w = x: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2 (continued): Compute P(y|ˆ z) (Z ⊥ ⊥Y |X)GZ because the path from Z to Y is blocked by X. G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . In Rule 2, set y = y, x = ø, z = z, w = x: P(y|x, ˆ z) = P(y|x, z) because (Z ⊥ ⊥Y |X)GZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2 (continued): Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z) = P x P(y|x, z)P(x). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 2 (continued): Compute P(y|ˆ z) P(y|ˆ z) = P x P(y|x, ˆ z)P(x|ˆ z) = P x P(y|x, z)P(x). This formula is a special case of the back-door formula in Theorem 1. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3: Compute P(y|ˆ x) P(y|ˆ x) = P z P(y|z, ˆ x)P(z|ˆ x). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3: Compute P(y|ˆ x) P(y|ˆ x) = P z P(y|z, ˆ x)P(z|ˆ x). (Y ⊥ ⊥Z|X)GXZ because there is no outgoing edge from Z and no incoming edge to X in GXZ. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3: Compute P(y|ˆ x) P(y|ˆ x) = P z P(y|z, ˆ x)P(z|ˆ x). (Y ⊥ ⊥Z|X)GXZ because there is no outgoing edge from Z and no incoming edge to X in GXZ. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3: Compute P(y|ˆ x) P(y|ˆ x) = P z P(y|z, ˆ x)P(z|ˆ x). (Y ⊥ ⊥Z|X)GXZ because there is no outgoing edge from Z and no incoming edge to X in GXZ. G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3: Compute P(y|ˆ x) P(y|ˆ x) = P z P(y|z, ˆ x)P(z|ˆ x). (Y ⊥ ⊥Z|X)GXZ because there is no outgoing edge from Z and no incoming edge to X in GXZ. G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . In Rule 2, set y = y, x = x, z = z, w = ø : Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3: Compute P(y|ˆ x) P(y|ˆ x) = P z P(y|z, ˆ x)P(z|ˆ x). (Y ⊥ ⊥Z|X)GXZ because there is no outgoing edge from Z and no incoming edge to X in GXZ. G satisfies the applicability condition for Rule 2: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, z, w) if (Y ⊥ ⊥Z|X, W )GXZ . In Rule 2, set y = y, x = x, z = z, w = ø : P(y|z, ˆ x) = P(y|ˆ z, ˆ x) because (Y ⊥ ⊥Z|X)GXZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3 (continued): Compute P(y|ˆ x) (Y ⊥ ⊥X|Z)GXZ because there is no incoming edge to X and no outgoing edge from X in GXZ. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3 (continued): Compute P(y|ˆ x) (Y ⊥ ⊥X|Z)GXZ because there is no incoming edge to X and no outgoing edge from X in GXZ. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3 (continued): Compute P(y|ˆ x) (Y ⊥ ⊥X|Z)GXZ because there is no incoming edge to X and no outgoing edge from X in GXZ. G satisfies the applicability condition for Rule 3: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX,Z(W ). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3 (continued): Compute P(y|ˆ x) (Y ⊥ ⊥X|Z)GXZ because there is no incoming edge to X and no outgoing edge from X in GXZ. G satisfies the applicability condition for Rule 3: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX,Z(W ). In Rule 3, set y = y, x = z, z = x, w = ø: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3 (continued): Compute P(y|ˆ x) (Y ⊥ ⊥X|Z)GXZ because there is no incoming edge to X and no outgoing edge from X in GXZ. G satisfies the applicability condition for Rule 3: P(y|ˆ x, ˆ z, w) = P(y|ˆ x, w) if (Y ⊥ ⊥Z|X, W )GX,Z(W ). In Rule 3, set y = y, x = z, z = x, w = ø: P(y|ˆ z, ˆ x) = P(y|ˆ z) because (Y ⊥ ⊥Z|X)GXZ . Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Symbolic Derivation of Causal Effects: An Example Step 3 (continued): Compute P(y|ˆ x) P(y|ˆ x) = P z P(y|z, ˆ x)P(z|ˆ x) = P z P(z|x) P x′ P(y|x′, z)P(x′). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Example: Smoking and the Genotype Theory The tobacco industry has managed to forestall antismoking legislation by arguing that the observed correlation between smoking and lung cancer could be explained by some sort of carcinogenic genotype (U) that involves inborn craving for nicotine. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Example: Smoking and the Genotype Theory The tobacco industry has managed to forestall antismoking legislation by arguing that the observed correlation between smoking and lung cancer could be explained by some sort of carcinogenic genotype (U) that involves inborn craving for nicotine. Figure: A diagram representing the story of smoking and the genotype (X=S=Smoking, Z=T=Tar, Y=C=Cancer, and U=G=Genotype (unobserved)). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Example: Smoking and the Genotype Theory continued Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Example: Smoking and the Genotype Theory continued These results seem to prove that smoking is a major contributor to lung cancer. However, the tobacco industry might argue that the table tells a different story that smoking actually decreases ones risk of lung cancer. Their argument goes as follows. If you decide to smoke, then your chances of building up tar deposits are 95%, compared to 5% if you decide not to smoke. In order to evaluate the effect of tar deposits, we look separately at two groups, smokers and nonsmokers. The table shows that tar deposits have a protective effect in both groups: in smokers, tar deposits lower cancer rates from 90% to 85%; in nonsmokers, they lower cancer rates from 10% to 5%. Thus, regardless of whether I have a natural craving for nicotine, I should be seeking the protective effect of tar deposits in my lungs, and smoking offers a very effective means of acquiring those deposits. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Example: Smoking and the Genotype Theory continued To settle the dispute between the two interpretations, we now apply the front-door formula (eq. (5)) to the data in the Table of the previous slide: Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Example: Smoking and the Genotype Theory continued To settle the dispute between the two interpretations, we now apply the front-door formula (eq. (5)) to the data in the Table of the previous slide: P(y|ˆ x) = X z P(z|x) X x′ P(y|x′, z)P(x′). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Example: Smoking and the Genotype Theory continued To settle the dispute between the two interpretations, we now apply the front-door formula (eq. (5)) to the data in the Table of the previous slide: P(y|ˆ x) = X z P(z|x) X x′ P(y|x′, z)P(x′). Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria The Back-Door Criterion The Front-Door Criterion do Calculus Symbolic Derivation Example Reference For Further Reading J. Pearl. Causality. Models, reasoning, and inference. Cambridge University Press, 2009. J. Pearl. Comment: Graphical Models, Causality and Intervention. Statistical Science, 8(3):266–269, 1993. Mohammad Ali Javidian, Marco Valtorta University of South Carolina An Overview of The Back-Door and Front-Door Criteria
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https://arxiv.org/pdf/2104.05962
arXiv:2104.05962v2 [math.CO] 4 Jul 2021 SHELAH’S PARTITION FUNCTIONS AND THE HALES-JEWETT NUMBERS MOHAMMAD GOLSHANI AND MOSTAFA MIRABI Abstract. In this paper we study several partition relations, defined by Saharon She-lah, and relate them to the Hales-Jewett numbers. In particular we give an upper bound for the Hales-Jewett numbers using the primitive recursive function f8,∗ which belongs to the class E5 of the Grzegorczyk hierarchy and grows slower than the function f13 .This improves the recent result of the first author and Shelah. Introduction The Hales-Jewett theorem is the bread and butter of Ramsey theory. In the words of , “the Hales-Jewett theorem strips van der Waerden’s theorem of its unessential elements and reveals the heart of Ramsey theory. It provides a focal point from which many results can be derived and acts as a cornerstone for much of the more advanced work”. Van der Waerden’s theorem states that a finite coloring of N contains arbitrarily long monochromatic arithmetic progressions. In fact, in 1927, he proved a seemingly stronger statement (which is actually equivalent to the previous statement, by compact-ness) as follows. Theorem 1.1 (Van der Waerden) . For all positive integers n and r there exists the least positive integer W( r, n ) such that any r-coloring of {1, . . . , W( r, n )} contains a monochromatic n-term arithmetic progression. In 1963, Hales and Jewett made the crucial observation that van der Waerden’s famous “double-induction proof” can be adapted to higher-dimensions. 2010 Mathematics Subject Classification. 05A17, 03D20. Key words and phrases. Hales-Jewett theorem, partition theorems, Gallai-Witt numbers, Van der Waerden numbers. The first author’s research has been supported by a grant from IPM (No. 1400030417). The authors would like to thank Saharon Shelah for his help, in particular the main result of Section 3 is motivated by some of his hints. The authors also would like to thank Alex Kruckman for his very careful reading of the paper and for the helpful comments and corrections. 12M. GOLSHANI AND M. MIRABI Theorem 1.2 (Hales and Jewett) . For all positive integers n, r there exists the least positive integer HJ( r, n ) such that any r-coloring of nHJ( r,n ) contains a monochromatic line. Roughly speaking, the k-dimensional Hales-Jewett theorem says that for a given finite set Λ of alphabets and a finite set C of colors, every coloring of the combinatorial space UM, Λ = {η | η : M → Λ} has, under some appropriate assumptions, a monochromatic k-dimensional subspace 1. How large is HJ( r, n )? This is a famous open problem. The best known upper bound on HJ( r, n ) was proved by Shelah in 1988. It is a primitive recursive 2 function which is far better than the original Hales-Jewett threshold. The original van der Waerden–Hales–Jewett proof led to the notorious Ackermann function. However, Shelah’s improved bound is still far too large for “tractable combinatorics”. To get an idea about the order of magnitude of Shelah’s bound, look at the following example: HJ(2 , 4) ≤ 222...2 where the height of the tower is 24. In , Matet streamlines Shelah’s argument, in particular replacing a brute force argument with a clever use of Ramsey’s theorem, to give a proof that takes less than a page. We refer to for a compact version of Shelah’s proof. In the other direction, finding lower bounds for Hales-Jewett numbers is important. In their original paper, Hales and Jewett also proved a linear lower bound to the Hales-Jewett numbers by an explicit construction. In , Shelah introduced a long list of primitive recursive functions of the form f−−(−, −) and f−,∗− (−, −) and compared them with the Hales-Jewett numbers in order to approach a tight upper (lower) bound. The partition function f13 was established by Shelah and the first author in . By using this function they give an upper bound for the Hales-Jewett numbers, and as an application they acquire an alternative proof of the fact that the Hales-Jewett numbers belong to the class E5 of the Grzegorczyk hierarchy. Also a list of similar partition functions is provided in to give a primitive recursive upper bound for the generalized version of the Hales-Jewett numbers. In this paper we study some partition relations defined in and and relate them to the Hales-Jewett numbers. In particular, we give an upper bound for the Hales-Jewett numbers using the primitive recursive function f8,∗, which belongs to the class E5 of the Grzegorczyk hierarchy and grows slower than f13 . 1See Section 2 for undefined notions 2In computer science, a primitive recursive function can be described by only do-loops. SHELAH’S PARTITION FUNCTIONS AND THE HALES-JEWETT NUMBERS 3 The relationships between these partition functions can be summarized in the following diagrams, where by −→ we mean ≤, and dashed arrows ( 99K ) indicate the new results in this paper. Also HJ and WC denote the Hales-Jewett number and the Gallai-Witt number respectively, and the partition functions f8, f8,∗, f9, f9,∗ and f13 are defined in and respectively. HJ C(m, Λ) f13 Λ (m, C) f8,∗ Λ (m, C) f9,∗ Λ (m, C) m · HJ C(1 , |Λ|m) f8Λ(m, C) f9Λ(m, C) and f8,∗ Λ ( |Λ|2 · WC(|Λ|, 1) , C ) f9,∗ Λ ( |Λ|2 · WC(|Λ|, 1) , C ) f13 Λ ( |Λ|2 · WC(|Λ|, 1) , C ) HJ C(1 , Λ) f13 Λ ( |Λ| · WC(|Λ|, 1) , C ) The paper is organized as follows. In Section 2 we express some preliminaries which will be used for the rest of the paper and also some elementary results. Then in Section 3 we prove a main theorem that gives an upper bound for the Hales-Jewett numbers in terms of the function f8,∗ ∈ E 5.2. Some preliminaries and basic results This section is devoted to some preliminaries including notation and conventions and also basic defintions and results. We refer the reader to , , and for some related results. 2.1. Notation and conventions. Throughout this paper we use the following notation and conventions. (1) i, j, k, l, m, n denote natural numbers. (2) For a natural number n, [n] denotes the set {1, . . . , n }.(3) Given a natural number n, we identify it with n = {0, . . . , n − 1}.(4) Λ denotes a finite non-empty set of alphabets, whose elements are usually denoted by α, β, . . . .(5) C denotes a finite non-empty set; the set of colors. 4 M. GOLSHANI AND M. MIRABI (6) M, N, . . . denote finite non-empty linear orders. (7) We use lowercase bold letters d, e, . . . to denote coloring functions. (8) If (M, ≤M ) is a linear order and r > 0, then [M]r denotes the set of all r-element sets {u0, . . . , u r−1} of M such that u0 <M · · · <M ur−1. Also, [M]<l = ⋃ 0<r<l [M]r .(9) UM, Λ = {η | η : M → Λ is a function }.2.2. Primitive recursive functions. By a number theoretic function we mean a func-tion from Nk to N. Let g, h and f be number theoretic functions of arities k, k + 2 and k + 1 respectively. Recall that f is said to be defined by primitive recursion from g and h whever for every x ∈ Nk and every n ∈ N we have:  f (0 , x ) = g(x),f (n + 1 , x ) = h(f (n, x ), n, x ). There is a simpler kind of primitive recursion which is appropriate to define unary func-tions, namely  f (0) = m, f (n + 1) = h(f (n), n ). where m ∈ N and h : N2 → N. Definition 2.1. The class of primitive recursive functions is the smallest set of number theoretic functions that contains the constant zero function, the successor function and the projection functions, and is closed under composition and primitive recursion. The simplest function which is not primitive recursive was constructed by Ackermann in . 2.3. The Grzegorczyk hierarchy. For each n ∈ N, we define the function En as follows E0(x, y ) = x + y, E1(x) = x2 + 2 ,En+2 (0) = 2 , and En+2 (x + 1) = En+1 (En+2 (x)) . Observe that each En is primitive recursive. Definition 2.2. For each n ∈ N we define the nth Grzegorczyk’s class En to be the smallest set of number theoretic functions that contains the functions Ek for k < n , the constant zero function, the successor function and the projection functions, and is closed SHELAH’S PARTITION FUNCTIONS AND THE HALES-JEWETT NUMBERS 5 under composition and limited primitive recursion (that is, if g, h, j ∈ E n and f is defined by primitive recursion from g and h, has the same arity as j and is pointwise bounded by j, then f belongs to En as well). Fact 2.1. The following properties hold. (1) A number theoretic function f is primitive recursive if and only if f ∈ E n for some n ∈ N. (2) For every n ∈ N we have En ⊂ E n+1 . (3) If g, h ∈ E n for some n ∈ N and f is defined by primitive recursion from g and h, then f ∈ E n+1 . (4) For every n ∈ N and every f ∈ E n of arity k there exists F ∈ E n of arity k which dominates f pointwise and satisfies F (x1, . . . , x k) ≤ F (y1, . . . , y k) for every x1, . . . , x k, y 1, . . . , y k ∈ N with xi ≤ yi for all i ∈ [k]. See [, Appendix A] and for more details on the Grzegorczyk hierarchy. 2.4. Van der Waerden numbers . Van der Waerden’s theorem states that for any positive integers r and m there exists a positive integer N such that if the integers {1, 2, . . . , N } are colored, each with one of r different colors, then there are at least m integers in arithmetic progression all of the same color. The smallest such N is the van der Waerden number W( r, m ).Here we first define the multidimensional van der Waerden numbers (also called Gallai-Witt numbers) which is denoted by WC(h, m ). Definition 2.3. The Gallai-Witt number WC(h, m ) is defined to be the minimal n such that if d : Uh,n → C is a C-coloring of Uh,n , then we can find d > 0 and a sequence 〈me | e < h 〉 of natural numbers such that for each e < h, m e + d · m < n and d is constant on {〈 me + d · ie | e < h 〉 | i0 ≤ m, · · · , i h−1 ≤ m}. The existence of the Gallai-Witt numbers is guaranteed by Gallai-Witt theorem, proved independently by Gallai (Grünwald), as quoted in 3, and Witt . Remark 2.4. Given natural numbers m, r , the van der Waerden number W( r, m ) is de-fined to be the least n such that for any coloring d : n → r, there exists a d-monochromatic arithmetic progression of length m. Then W( r, m + 1) = W C(1 , m ), where |C| = r. 3There is no publication by Gallai that contains it; however, Rado in proved it and credits Gallai. 6M. GOLSHANI AND M. MIRABI By a celebrated result of Gowers we have W( r, m ) ≤ 22r22m+9 , in particular, W( r, m ) ∈ E 3. 2.5. The Hales-Jewett Theorem. The notions of line and m-dimensional subspace play an important role in the Hales-Jewett theorem. Definition 2.5. A set L ⊆ U M, Λ is a line of UM, Λ if there exist a non-empty subset M0 ⊆ M and ρ : M \ M0 → Λ such that for every η ∈ U M, Λ, η ∈ L whenever (a) η↾M \ M0 = ρ. (b) η↾M0 is constant. Definition 2.6. A set S ⊆ U M, Λ is an m-dimensional subspace of UM, Λ if there are pairwise disjoint non-empty subsets Mℓ, ℓ < m, of M and a map ρ : M \ ⋃ ℓ<m Mℓ → Λ such that for every ν ∈ U M, Λ, ν ∈ S if and only if (1) ν↾M \ ⋃ ℓ<m Mℓ = ρ. (2) for every ℓ < m, ν ↾Mℓ is constant. It is evident that a line of UM, Λ is just a 1-dimensional subspace of UM, Λ. Notation 2.7. Given an m-dimensional subspace S of UM, Λ as in Definition 2.6, we denote it by S = S(〈 Ml : l < m 〉, ρ ). Definition 2.8. The Hales-Jewett number, HJ C(m, Λ) , is defined to be the minimal k such that for any linear order M of size k and any C-coloring d : UM, Λ → C, there exists a d-monochromatic m-dimensional subspace of UM, Λ. By Hales-Jewett , HJ C(m, Λ) is finite. The original proof by Hales and Jewett was based on double induction and did not give any primitive recursive bounds for HJ C(m, Λ) .In , Shelah showed that the function HJ C(m, Λ) is primitive recursive, and indeed HJ C(m, Λ) ∈ E 5. Let us now borrow a long list of definitions from , which are essentially some variants of the Hales-Jewett numbers. Definition 2.9. ( ) (1) f9Λ(m, C), where |Λ| divides m, is defined to be the minimal k divisible by |Λ| such that for any linear order M of size k and any coloring d : UM, Λ → C, there are pairwise disjoint subsets Ml, l < m, of M and a map ρ : M \ ⋃ l<m Ml → Λ such that: SHELAH’S PARTITION FUNCTIONS AND THE HALES-JEWETT NUMBERS 7 (a) for all l0 < l 1 < m, |Ml0 | = |Ml1 | > 0, (b) d is constant on T , where T consists of those ν ∈ S (〈 Ml : l < m 〉, ρ ) such that for every α ∈ Λ, ∣∣ {l < m : ν↾Ml ≡ α} ∣∣ = m |Λ| . (2) f8Λ(m, C) is defined similarly without clause (a). We now define a modified version of the above partition functions. Definition 2.10. ( ) (1) f9,∗ Λ (m, C), where |Λ| divides m, is defined as in f9Λ(m, C), where clause 2.9(b) is replaced by: (b∗): for ν1, ν 2 ∈ S (〈 Ml : l < m 〉, ρ ), d(ν1) = d(ν2) provided that for all α ∈ Λ ∣∣{l < m : ν1↾Ml ≡ α}∣ ∣ = ∣∣{l < m : ν2↾Ml ≡ α}∣ ∣. (2) f8,∗ Λ (m, C) is defined similarly, using f8Λ(m, C) by replacing clause 2.9(b) by ( b∗). The next lemma is evident from the above definitions. Lemma 2.11. (1) f8Λ(m, C) ≤ f9Λ(m, C). (2) f8,∗ Λ (m, C) ≤ f9,∗ Λ (m, C). (3) Let ℓ ∈ { 8, 9}. Then fℓ Λ (m, C) ≤ fℓ, ∗ Λ (m, C). (4) f8,∗ Λ (m, C) ≤ HJ C(m, Λ) . In order to compare these partition functions with the Hales-Jewett numbers, we define HJ = C (m, Λ) which is above all of these functions and still not far from HJ C(m, Λ) . Definition 2.12. The partition function HJ = C (m, Λ) is defined to be the minimal k such that for any linear order M of size k and any C-coloring d : UM, Λ → C, there exists a m-dimensional subspace S = S(〈 Ml : l < m 〉, ρ ) such that (1) d is constant on S, and (2) for every l, l ′ < m , |Ml| = |Ml′ |. Note that HJ = is essentially HJ with the additional condition that Ml’s have the same size. Lemma 2.13. HJ = C (m, Λ) ≤ m · HJ C(1 , |Λ|m).Proof. Let n = HJ C(1 , |Λ|m), and set M = n × m. Let d : UM, Λ → C be a C-coloring. Define F : Un, Λm → U M, Λ as follows: let η : n → Λm, then F (η) : M → Λ is defined by F (η)( i, ℓ ) = η(i)( ℓ).8 M. GOLSHANI AND M. MIRABI Let e : Un, Λm → C be defined as e(η) = d(F (η)) . UM, Λ C Un, Λm d Fe By our assumption, there are N ⊆ n and ρ : n \ N → Λm which witness HJ C(1 , |Λ|m) with respect to e. Thus e is constant on {η : n → Λm : n↾N is constant and n ⊇ ρ}.For ℓ < m set Mℓ = N × { ℓ} ⊆ M. Define also ̺ : M \ ⋃ ℓ<m Mℓ → Λ by ̺ (i, ℓ ) = ρ(i)( ℓ). Note that for all ℓ, ℓ ′ < m we have |Mℓ| = |M′ ℓ | > 0. Now suppose that v1, v 2 ∈ S (〈Mℓ : ℓ < m 〉,̺ ). Define η1, η 2 : n → Λm such that for all i < n and ℓ < m : η1(i)( ℓ) = v1(i, ℓ ),η2(i)( ℓ) = v2(i, ℓ ). Now it follows that • η1, η 2 ⊇ ρ (as v1, v 2 ⊇̺ ) • η1↾N is constant; to see this suppose i1, i 2 ∈ N ⊆ n. Then for all ℓ < m η1(i1)( ℓ) = v1(i1, ℓ )= v1(i2, ℓ )= η1(i2)( ℓ). Thus η1(i1) = η1(i2). • similarly η2↾N is constant. Thus e(η1) = e(η2) which implies that d(v1) = d(F (η1)) = e(η1) = e(η2) = d(F (η2)) = d(v2). The lemma follows. Remark 2.14. Note that since clearly HJ C(m, Λ) ≤ HJ = C (m, Λ) and f9,∗ Λ (m, C) ≤ HJ = C (m, Λ) ,Lemma 2.13 shows that (1) f9,∗ Λ (m, C) ≤ m · HJ C(1 , |Λ|m). (2) HJ C(m, Λ) ≤ m · HJ C(1 , Λm) (see [2, Page 26] ). SHELAH’S PARTITION FUNCTIONS AND THE HALES-JEWETT NUMBERS 9 Let M be a finite linear order and let Sym (M) be the set of all permutations of M.This induces an equivalence relation EM on UM, Λ defined by ρ1EM ρ2 ⇐⇒ ∃ π ∈ Sym (M), ρ 2 = ρ1 ◦ π. Remark 2.15. Suppose η1, η 2 ∈ U N, Λ. Then it is easily seen that η1EN η2 ⇐⇒ for all α ∈ Λ, |η−11 {α}| = |η−12 {α}| . We borrow the following definitions from . Definition 2.16. f13 Λ (m, C) is defined to be the minimal k such that for any linear order M of size k and any C-coloring d : UM, Λ → C, there are N ⊆ M of size m and a function ρ : M \ N → Λ such that for all η1, η 2 ∈ U N, Λ,η1EN η2 =⇒ d(η1 ∪ ρ) = d(η2 ∪ ρ). The next theorem gives the relation between f13 Λ (m, C) and HJ C(n, Λ) . Theorem 2.17 (see ) . The following statements hold. (1) f13 Λ (m, C) ∈ E 5. (2) HJ C(m, Λ) ≤ f13 Λ ( |Λ| · WC (|Λ|, m ), C ) . The next lemma compares f9,∗ Λ (m, C) and f13 Λ (m, C). Lemma 2.18. f9,∗ Λ (m, C) ≤ f13 Λ (m, C). In particular f9,∗ Λ (m, C) ∈ E 5.Proof. Let k = f13 Λ (m, C). Suppose M is a linear order of size k and let d : UM, Λ → C be a C-coloring. Let N ⊆ M of size m and ρ : M \ N → Λ witness k = f13 Λ (m, C). For each l < m let Ml = {al}, where N = 〈al : l < m 〉 is the increasing enumeration of N. Now suppose ν1, ν 2 ∈ S (〈Ml : l < m 〉, ρ ) and suppose they satisfy the requirement 2.10 (c∗).It then immediately follows that ν1EN ν2 and hence d(ν1 ∪ ρ) = d(ν2 ∪ ρ). In the above proof, the sets Ml, l < m all have size 1. By a similar argument, we can prove the following lemma. Lemma 2.19. Let n > 0 and f9,∗ Λ,n (m, C) ≤ f13 Λ (m · n, C), where f9,∗ Λ,n (m, C) is defined as in f9,∗ Λ (m, C) with the additional requirement that for all ℓ < m, |Mℓ| = n. Bounding Hales-Jewett numbers with f8,∗ This section contains a main theorem which compares Hales-Jewett numbers with f8,∗. Main Theorem 1. HJ C(1 , Λ) ≤ f8,∗ Λ ( |Λ|2 · WC(|Λ|, 1) , C ) .10 M. GOLSHANI AND M. MIRABI Proof. First we prove the following two lemmas. Lemma 3.1. HJ C(1 , Λ) ≤ f8,∗ Λ (m∗, C), where m∗ satisfies the following: (⊕): assume Ω = {¯ℓ = 〈ℓα : α ∈ Λ〉 ∈ U Λ,m ∗ : ∑ α∈Λ ℓα = m∗} and d : Ω → C is a C-coloring. Then there are ℓ∗ > 0 and 〈ℓ∗ α : α ∈ Λ〉 such that letting, ¯ℓα(β) =  ℓ∗ β β 6 = αℓ∗ α ℓ∗ β = α then ¯ℓα ∈ Ω for α ∈ Λ and d↾{¯ℓα : α ∈ Λ} is constant. Proof. Let m∗∗ = f8,∗ Λ (m∗, C) and assume d : Um∗∗ ,Λ → C is a C-coloring. Claim. There is a C-coloring e : Um∗ ,Λ → C such that (1) if η1, η 2 ∈ U m∗ ,Λ and for all α ∈ Λ, |η−11 {α}| = |η−12 {α}| , then e(η1) = e(η2), and (2) if there is an e-monochromatic line in Um∗,Λ, then there is a d-monochromatic line in Um∗∗ ,Λ.Proof. We abuse notation to identify m∗ and m∗∗ with linear orders of size m∗ and m∗∗ .By definition of f8,∗ Λ (m∗, C), there are pairwise disjoint subsets 〈Mℓ : ℓ < m ∗〉 of m∗∗ and a map ρ : m∗∗ \ ⋃ ℓ<m ∗ Mℓ → Λ such that for every ν1, ν 2 ∈ S (〈 Mℓ : ℓ < m ∗〉, ρ ), d(ν1) = d(ν2) provided that for each α ∈ Λ ∣∣{ℓ < m ∗ : ν1↾Mℓ ≡ α}∣ ∣ = ∣∣{ℓ < m ∗ : ν2↾Mℓ ≡ α}∣ ∣. Define F : Um∗ ,Λ → U m∗∗ ,Λ with η 7 → F (η) : m∗∗ → Λ where F (η)( t) =  η(ℓ) t ∈ Mℓ, (ℓ < m ∗) ρ(t) t ∈ m∗∗ \ ⋃ Mℓ Now, define a C-coloring e : Um∗ ,Λ → C with e(η) = d(F (η)) . We show that it satisfies the conditions of the claim. First, let η1, η 2 ∈ U m∗ ,Λ and for all α ∈ Λ, |η−11 {α}| = |η−12 {α}| .Then for all α ∈ Λ, ∣∣{ℓ < m ∗ : F (η1)↾Mℓ ≡ α}∣ ∣ = ∣∣{ℓ < m ∗ : F (η2)↾Mℓ ≡ α}∣ ∣ It follows that e(η1) = d(F (η1)) = d(F (η2)) = e(η2).Second, let N ⊆ m∗ and τ : m∗ \ N → Λ be a e-monochromatic line in Um∗ ,Λ.Let N′ = ⋃ ℓ∈N Mℓ and τ ′ : m∗∗ \ N′ → Λ with τ ′↾Mℓ = τ (ℓ) for ℓ ∈ m∗ \ N, and τ ′↾m∗∗ \ ⋃ ℓ<m ∗ Mℓ = ρ. We claim that (N′, τ ′) is a d-monochromatic line in Um∗∗ ,Λ. Let ξ1, ξ 2 ∈ U m∗∗ ,Λ such that ξ1↾N′ ≡ α and ξ2↾N′ ≡ β. We show that d(ξ1 ∪ τ ′) = d(ξ2 ∪ τ ′).It suffices to find η1, η 2 : N → Λ such that ξi ∪ τ ′ = F (ηi ∪ τ ) for i = 1 , 2. Let η1 ≡ αSHELAH’S PARTITION FUNCTIONS AND THE HALES-JEWETT NUMBERS 11 and η2 ≡ β. Then d(ξ1 ∪ τ ′) = d(F (η1 ∪ τ )) = e(η1 ∪ τ ) = e(η2 ∪ τ ) = d(F (η2 ∪ τ )) = d(ξ2 ∪ τ ′). Now, we show that there is an e-monochromatic line in Um∗,Λ. Define a C-coloring c : Ω → C with η 7 → e(G(η)) , where for every η = 〈ℓα : α ∈ Λ, ∑ α∈Λ ℓα = m∗〉 ∈ Ω, G(η) : m∗ → Λ is a map such that if {Pα : α ∈ Λ} is a partition of m∗ with |Pα| = ℓα, then G(η)↾Pα ≡ α. Notice that c does not depend on the partition and so c is well-defined. By using (⊕) for coloring c, there are ℓ∗ and 〈ℓ∗ α : α ∈ Λ〉 such that if ¯ℓα(β) =  ℓ∗ β β 6 = αℓ∗ α ℓ∗ β = α then for each α ∈ Λ, ¯ℓα ∈ Ω and c↾{¯ℓα : α ∈ Λ} is constant. Now, let {P ′ α : α ∈ Λ}∪{ P ′} partition m∗ such that |P ′| = ℓ∗ and for every α ∈ Λ, |P ′ α | = ℓ∗ α . Set N′′ = P ′ and let τ ′′ : m∗ \N′′ → Λ such that τ ′′ ↾P ′ α ≡ α. Then (N′′ , τ ′′ ) is a witness of an e-monochromatic line in Um∗,Λ, and the result follows. Lemma 3.2. Let m∗ = |Λ|2 · WC(|Λ|, 1) . Then m∗ satisfies (⊕) from Lemma 3.1. Proof. Set h = |Λ|, n = W C(h, 1) and k = h2 · n. Let d : Uh,k → C4 be a C-coloring. Define F : Uh,n → U h,k as follows. Fix η ∈ U h,n , then for each e < h , F (η)( e) = h · η(e) + h · n − ∑ i<h η(i).F is easily seen to be well-defined and that ∑ e<h F (η)( e) = h ∑ e<h η(e) + h2 · n − ∑ e<h ∑ i<h η(i) = k. Define the coloring e : Uh,n → C by e(η) = d(F (η)). By definition of WC , we can find ~m = 〈me : e < h 〉 and d > 0 such that e is constant on 〈ϑe = ~m + d · ie : e < h 〉, where ~m + d · ie = 〈m0, · · · , m e−1, m e + d, m e+1 , · · · , m h−1 〉. Let ℓ = h · n − (∑ e<h me ) − d. This means that d is constant on 〈ϑ∗ e : e < h 〉, where ϑ∗ e = 〈h · m0 + ℓ, · · · , h · me−1 + ℓ, h · me + d + ℓ, h · me+1 + ℓ, · · · , h · mh−1 + ℓ〉 Now let ℓ∗ e = h · me + ℓ and ℓ∗ = d. It is easily seen they are as required by m∗. To complete the proof of the main theorem, combine Lemma 3.1 and Lemma 3.2. 4Note that only members of Ωmatter. In fact we apply dto elements of the form F(η), which are always in Ω. 12 M. GOLSHANI AND M. MIRABI Remark 3.3. Notice that Main Theorem 1 provides an alternative and apparently a better upper bound for HJ C(1 , Λ) in comparison with Theorem 2.17(2). For primitive recursive functions f and g, set f ≤E g ⇐⇒ min {n : f ∈ E n} ≤ min {n : g ∈ E n}, and f =E g is defined similarly. Our results show that f8 =E f8,∗ =E f9 =E f9,∗ =E HJ ≤E f13 . Now we can ask the following natural question. Question 3.4. Is f13 ≤E HJ ? References Wilhelm Ackermann. Zum Hilbertschen aufbau der reellen zahlen. Math. Ann. , 99:118–133, 1928. Pandelis Dodos and Vassilis Kanellopoulos. Ramsey theory for product spaces , volume 212 of Mathematical Surveys and Monographs . American Mathematical Society, Providence, RI, 2016. Mohammad Golshani and Saharon Shelah. Iterated Ramsey bounds for the Hales-Jewett numbers. Submit-ted , 2020. Timothy Gowers. A new proof of szemerédi’s theorem. Geom. Funct. Anal. GAFA , 11:465–588, 2001. Ronald L. Graham, Bruce L. Rothschild, and Joel H. Spencer. Ramsey Theory . John Wiley and Sons, New York, 2 edition, 1990. Alfred W. Hales and Robert I. Jewett. Regularity and positional games. Trans. Amer. Math. Soc. , 106:222– 229, 1963. Pierre Matet. Shelah’s proof of the Hales–Jewett theorem revisited. European Journal of Combinatorics ,28:1742–1745, 2007. Alon Nilli. Mathematics of Ramsey Theory , volume 5 of Algorithms and Combinatorics , chapter Shelah’s proof of the Hales–Jewett theorem, pages 150–151. Springer, Berlin, Heidelberg, 1990. Richard Rado. Note on combinatorial analysis. Proc. London Math. Soc. , 48:122–160, 1943. Harvey E. Rose. Subrecursion: functions and hierarchies . Oxford Logic Guides. The Clarendon Press, Oxford University Press, New York, 1984. Saharon Shelah. Primitive recursive bounds for van der Waerden numbers. J. Amer. Math. Soc. , 1:683–697, 1988. Saharon Shelah. On what I do not understand (and have something to say), model theory. Math. Japon. ,51:329–377, 2000. Saharon Shelah. On what I do not understand (and have something to say) part I. Saharon Shelah’s an-niversary issue. Fund. Math. , 166:1–82, 2000. Saharon Shelah. A partition theorem. Sci. Math. Jpn. , 56:413–438., 2002. Bartel L. van der Waerden. Beweis einer Baudetschen Vermutung. Nieuw Arch. Wisk. , 15:212–216, 1927. Ernst Witt. Ein kombinatorisches Satz der Elementargeometrie. Mathematische Nachrichten , 6:261–262, 1951. SHELAH’S PARTITION FUNCTIONS AND THE HALES-JEWETT NUMBERS 13 Mohammad Golshani, School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: 19395–5746, Tehran, Iran. Email address : golshani.m@gmail.com URL : Department of Mathematics and Computer Science, Wesleyan University, Middletown, CT 06459, USA. Email address : mmirabi@wesleyan.edu URL :
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This collection of printable worksheets on inequality is sure to remove all hurdles and help students of 6th grade and above gain integral practice in solving inequalities. Inequalities generally deal with situations that have multiple solutions represented as intervals. Manipulate and solve the inequality, paying attention to the properties of inequalities and the rules for solving them. Assist students to ascend the ladder of inequalities with ease as they progress from solving one-step to multi-step inequalities, followed by solving compound, absolute and quadratic inequalities. These free worksheets are specially designed for students of grade 6 through high school. Translating Inequality Phrases Worksheets Gain practice on the parts of algebraic inequalities with verbal phrases or sentences in these pdf worksheets. Study the verbal descriptions, identify the operations, coefficients, variable, and translate them as inequalities. 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Chapter 1: Introduction and Overview - Chapter 1: Introduction and Overview Powering the Future Fusion & Plasmas Search for: Preface Executive Summary Chapter 1: Introduction and Overview Fusion and plasmas: Powering the future Technology and Science Drivers Programs and facilities to execute the strategic plan Process and Prioritization Criteria Chapter 2: Recommendations and Budget Scenarios Overarching Recommendations Program and Project Specific Recommendations Budget Scenarios Scenario Table Appendix A: Recommendations Appendix B: Public-Private Partnerships Appendix C: DEI, Workforce, and Outreach Appendix D: Charge Letter Appendix E: Membership Appendix F: Process and Meetings Appendix G: Acronyms ↓Download Report (PDF: 8.5MB)↓ Download Report Summary (PDF: 233 KB) Powering the Future Fusion & Plasmas Preface Executive Summary Chapter 1: Introduction and Overview Fusion and plasmas: Powering the future Technology and Science Drivers Programs and facilities to execute the strategic plan Process and Prioritization Criteria Chapter 2: Recommendations and Budget Scenarios Overarching Recommendations Program and Project Specific Recommendations Budget Scenarios Scenario Table Appendix A: Recommendations Appendix B: Public-Private Partnerships Appendix C: DEI, Workforce, and Outreach Appendix D: Charge Letter Appendix E: Membership Appendix F: Process and Meetings Appendix G: Acronyms Search for: Home Page Chapter 1 Introduction and Overview Fusion and plasmas: Powering the future The US is at a critical moment in the effort to develop fusion as a carbon-neutral, sustainable source of energy. The past decade has seen significant progress in the physics and engineering necessary to confine high-temperature plasmas for fusion. Important technological breakthroughs include high-temperature superconductors that enable the advances in magnet technology required to achieve that confinement. We are on the verge of entering an era of burning plasmas, with the international ITER experiment set to begin operation this decade. At the same time, privately financed fusion research and development (R&D) has experienced rapid growth that has spurred an emerging fusion energy industry. For US fusion research, these developments have created a unique and ambitious path toward a low-capital-cost fusion pilot plant (FPP) that will form the basis for economically attractive fusion electricity. Fusion energy and plasmas are inextricably linked. A fusion reactor requires a confined, controlled, burning plasma at its core. For that reason, fusion research has historically been an important driver for the development of plasma physics as a fundamental field. The link between the two fields is strong but does not fully define either one. Fusion energy requires R&D into materials resistant to neutron irradiation, into technologies for breeding fusion fuel, and into enabling technologies like magnets. The field of plasma science and engineering is intellectually diverse, is highly interdisciplinary, and has myriad applications beyond fusion energy. Plasma science and engineering has advanced significantly over the past decade, and future opportunities abound. Extreme states of matter have been produced and studied using the world’s most intense lasers developed from Nobel Prize-winning research in chirped pulse amplification. Understanding of the most energetic events in the universe requires deep knowledge of plasma physics. Such research is essential to interpreting electromagnetic signatures from events like black hole mergers in this era of multimessenger astronomy. Plasmas enable technologies essential to our everyday lives, including plasmaprocessing of semiconductor devices, which is key to the trillion-dollar information technology industry. There is potential to expand these applications with significant societal benefit; for example, plasma-enhanced chemistry could help address energy security and climate change by providing ways to make products from carbon-free electricity, purifying water and developing new medical treatments. This report details opportunities to accelerate the development of practical fusion energy and to advance the frontiers of plasma science and engineering. Importantly, it outlines a strategy for the Department of Energy (DOE) Fusion Energy Sciences (FES) to act on these opportunities. Embracing opportunities to form partnerships that accelerate progress in R&D is an important theme of this report. Partnership opportunities exist within the federal government, internationally, and with industry. DOE FES is the primary federal sponsor for fusion research, but other agencies have made important investments, including DOE Advanced Research Project Agency-Energy (ARPA-E), DOE Advanced Scientific Computing Research (ASCR), and the DOE National Nuclear Security Administration (NNSA). Because the field is interdisciplinary in nature and offers a multitude of applications, many federal agencies invest in broader plasma science and engineering, including the National Science Foundation (NSF), NASA, DOE High Energy Physics (DOE HEP), the Office of Naval Research (ONR), and the Air Force Office of Science Research (AFOSR). Better coordination among agencies involved in various aspects of fusion and plasma research could result in more efficient use of federal resources and enable more rapid progress in advancing plasma science and engineering and in developing fusion energy. Fusion energy and plasma science research are global endeavors. Many nations recognize the promise of fusion energy and have made significant investments in R&D. International collaboration has been critically important to progress. This has been particularly true in the quest to address a top priority for the global fusion research community: Experimental access to a burning plasma, in which energy released by fusion reactions is the dominant heating mechanism. The international community, with the US as a key partner, is collaborating to construct the ITER experiment in France to achieve this goal. At the time of this writing, the ITER project is more than 70% complete toward first plasma. The Burning Plasma Report from the National Academies of Sciences, Engineering, and Medicine (NASEM) highlighted the importance of the ITER project to the US fusion program and stated that it provides the most compelling path to accessing a burning plasma at reactor scale. However, significant R&D is required in addition to ITER to produce electricity from fusion. Additional investment supporting that R&D is needed to advance the science and technology of a fusion pilot plant in a timely manner. While other international parties are considering a reactor scaled directly from ITER, the NASEM report recognized that this approach is too large and expensive to be economically competitive in the US market when compared with other carbon-neutral energy technologies. Consequently, the NASEM report instead set forth a unique US vision for fusion energy using scientific and technological innovations to target the development of a low-capital-cost FPP. That emphasis on developing innovative, world-leading solutions makes the near-term investments in R&D even more critical as other nations continue to invest in new fusion facilities that advance their own approaches to fusion energy development. Research in fundamental plasma science is also vibrant and growing internationally, with activity spanning scales ranging from the subatomic to the cosmic, from low-temperature atmospheric plasmas to the most extreme conditions in the universe. Over the past few decades, shrewd investments by DOE in worldclass facilities have placed the US at the forefront of pioneering plasma research. However, such scientific leadership requires agility and continuous nurturing. In some instances, the US is losing its leadership position. For example, the 2018 NASEM report Opportunities in Intense Ultrafast Lasers: Reaching for the Brightest Light, highlights how the US has already lost leadership in the highintensity lasers that are essential for high-energy-density science. Although the chirped-pulse laser amplification technology that led to petawatt lasers was developed in the US, the vast majority of high-intensity laser systems are now being built in Europe and Asia. This long-range plan describes a path to regain a leadership position in fundamental plasma science and its applications in the US. Fusion energy and plasma science research advances fundamental science, but also translates to direct commercial application. The ultimate goal of fusion energy research is the development of commercial fusion power. The fusion energy industry is already taking root, but realizing the ultimate goal of producing power will require additional support to help it become firmly established. The past decade has seen about $2 billion invested worldwide in fusion energy development in the private sector. Internationally, the United Kingdom and China have already established multi-hundred-million-dollar partnership programs to attract private fusion energy companies. Therefore, it is imperative that the US strengthen partnerships in the private sector to accelerate the development of fusion power in the US and maintain a leadership position in the emerging fusion energy industry. For decades, plasma technologies have played a ubiquitous role in manufacturing, crucial for the fabrication of microelectronic circuits, lighting, optics, advanced materials, materials processing, and much more. The future looks even more promising. Recent research suggests that plasmas will influence much of the future economy; they will play a decisive role in technologies that convert electricity from carbon-free sources to the products that drive society, and in future medical treatments, aerospace, particle accelerators, advanced X-ray sources, and agriculture. Countries that can solve the science questions that make these technologies possible and can facilitate technology transfer from academic research to commercial applications will position themselves to lead tomorrow’s economy. This long-range plan outlines ways in which the US can take the lead in both commercial fusion energy and other plasma-based technologies. This report marks the first time a strategic planning process for DOE FES has been undertaken that addresses both fusion energy and plasma science and that has had a significant community-led portion of the process. The strategic planning process involved two stages: a community-driven stage followed by a stage led by the Fusion Energy Sciences Advisory Committee (FESAC), using input from the community process. The year-long Community Planning Process (CPP), was organized by the American Physical Society’s Division of Plasma Physics. The process was invaluable and resulted in the consensus CPP report that not only enumerates scientific and technological opportunities, but also provides guidance for prioritization. The CPP report formed the basis for this strategic plan and remains an essential companion to this report for those looking for more technical detail on specific initiatives. The technical bases for the considerations in both reports were made based on white papers submitted to CPP and the expert groups that evaluated them. This report presents a strategic plan based on the resulting new program elements and facilities. Technology and Science Drivers As acknowledged by the recent burning plasma and plasma decadal reports by NASEM and by the CPP report, fusion science and technology has reached a level of maturity that calls for FES to broaden its focus from the plasma core of a fusion reactor toward a comprehensive energy mission. At the same time, these reports show that plasma science and technology outside the fusion energy mission deepens our understanding of the universe and lays the foundation for creating transformative technologies ranging from microelectronics and medicine to particle accelerators and new materials such as advanced alloys, ceramics, and materials for magnets. The energy mission is driven by the urgent desire to address climate change and energy security on a time scale that requires activities to resolve the critical challenges of fusion energy in the next two decades. This mission-driven program is founded on the steady progress in plasma science, ITER construction, predictive integrated-modeling capabilities, and a burgeoning investment in private fusion enterprises. However, the least developed domain in the mission portfolio is in fusion materials and technology (FM&T). Fulfilling the energy mission demands a shift in the balance of research toward FM&T, which connects the three science drivers: Sustain a Burning Plasma, Engineer for Extreme Conditions, and Harness Fusion Energy. The program’s renewed attention to economic viability distinguishes it from other ITER partners. It leverages US innovation, leadership, and technology advances to address the key gaps in fusion plasma science, nuclear science, materials science, and the enabling technology that will be required to construct an FPP, anticipated to be the key remaining step to enable commercial fusion energy. Critical gaps in FM&T will have to be closed for any choice of plasma core in an FPP, and without immediate investment those gaps could become pace-limiting. Such a program will create US leadership in a broad range of disciplines through innovation and rigorous scientific inquiry. A critical need in the quest for fusion energy production is the ability to sustain a burning plasma by controlling and predicting its dynamics. Burning plasmas, in which the heating is primarily due to the energy released from fusion reactions, pose challenges to stability and control that are not fully addressable in current experiments and for which significant uncertainty exists. Addressing those challenges requires establishing scenarios for maintaining high performance in a burning regime and preventing damage associated with transient events through the development of tools to predict, avoid, and mitigate such events. The tokamak approach for the plasma core is the most technically advanced and mature confinement concept. A tokamak FPP will require completing critical research on existing domestic facilities, and significant participation in the ITER research program. ITER participation will increase knowledge in burning plasma physics and in materials science and technology. New collaborations with industry potentially offer pathways to accelerate access to burning plasma conditions. Complementing these priority areas is research into non-tokamak confinement approaches, including stellarators, inertial fusion energy, and other alternate confinement approaches. Investment in the alternate approaches is important both as a risk-mitigation strategy for the tokamak approach and to support innovations that could accelerate progress toward an FPP and commercial fusion energy. An FPP will produce heat, particle, and neutron fluxes that significantly exceed those in present confinement facilities, and new approaches and materials need to be developed and engineered for the anticipated extreme reactor conditions. Those intense conditions affect all regions of the reactor in distinct ways, including the plasma-facing components (PFCs); structural, functional, magnet, and diagnostic materials; and ex-vessel components. In an FPP, high fluxes of 14 MeV neutrons produce damaging and poorly understood effects in materials. A scientific understanding of how the properties of materials evolve and degrade due to fusion neutron exposure is needed to safely predict the behavior of materials in fusion reactors. Even those components not directly exposed to high fluxes from the plasma still experience a complex multifactor environment that includes high temperatures, tritium migration and trapping, material interfaces, and high stresses. Innovative approaches and new developments will lead to integrated solutions to those harsh conditions. Interlinked with a burning plasma and materials are the key systems required to harness fusion power, breed fuel, and ensure the safe operation of a reactor. Before an FPP is constructed, materials and components must be qualified and a system design must ensure the compatibility of all components. Just as the plasma and materials in a fusion reactor will need to advance beyond today’s capabilities, the balance of plant equipment, remote handling, tritium breeding, and safety systems will also require significant advances. The research encompassed by these three technology and science drivers is essential to lowering the risks to an acceptable level for an FPP and will allow the US to pursue a swift, innovative, and economically attractive path to fusion energy production. The societal benefit of establishing a new carbon-neutral power source and developing the industry that supports it cannot be understated. Such a power source would be one of the most transformational technologies in the field of plasma science. On the road to achieving this scientific grand challenge are myriad additional spinoff technologies and fundamental investigations that can reveal new knowledge about the universe. The field of plasma science and technology is a rich and diverse landscape, from the search for accurate theoretical descriptions of the complex emergent behavior of the plasma state to the production of matter at extreme conditions that exceed even those at the core of giant planets or stars. Low-temperature plasma science can also play a critical role in the development of new technologies. Expanding the fundamental understanding of plasmas and their interactions with their surroundings across wide ranges of temperature and density underpins not only fusion physics but the practical application of plasmas for manufacturing, medicine, and agriculture. The plasma science and technology component of the FES mission is impelled by three main drivers: Strengthen the Foundations, Understand the Plasma Universe, and Create Transformative Technologies. Together these drivers tackle the plasma questions of highest scientific impact and urgency, and they foster innovation by spurring exploration as dynamic as the processes in plasmas themselves. The programs, initiatives, and facilities identified here represent an opportunity to increase US leadership by strengthening investment in research areas of high potential, while moving forward with new capabilities and facilities and tapping the collective wisdom of the scientific community through a series of networks, collaborations, and partnerships. Strengthening the foundations of plasma science deepens our fundamental understanding of nature. Exciting new experimental capabilities are unlocking unprecedented plasma regimes, while new theories and computational methods provide the insight to decipher them. Extremely intense lasers are making compact particle accelerators possible and may soon reach nonlinear quantum electrodynamic (QED) regimes in which pair plasmas will be created directly from light. Pulsed-power facilities compress matter to such high density that the behavior of the resulting warm dense plasma is fundamentally different from known states of condensed matter or plasmas. Because the plasma has high electrical conductivity, magnetic fields can be compressed to approach strengths only found in astrophysical objects such as white dwarfs. Coupling these drivers with X-ray free electron lasers allows exquisite measurements of these novel states of plasma. At the same time, tabletop-scale experiments create and trap exotic states of antimatter and strongly correlated plasmas, which can be so sensitively diagnosed that they can be used to test fundamental symmetries of nature. Strengthening the foundations of plasma science will require facilities and computational hardware at a range of scales, theoretical research that charts next steps, and a hierarchy of computational techniques that connect the microscopic to the macroscopic. Understanding the plasma universe is essential to learning about the origins and the evolution of the universe. Nearly every aspect of the cosmos is influenced by plasma, from lightning and aurora in Earth’s atmosphere to stellar winds that fill the space between planets and stars; from accretion disks surrounding supermassive black holes at the centers of the galaxies to the particle jets launched from the most distant and ancient quasars. All these systems are strongly affected by plasma behaviors that are not yet understood, including magnetic reconnection, turbulence, and particle energization. Viewing astrophysics through the lens of plasma physics is crucial, given recent advances in multimessenger astronomy and spacecraft missions. As spacecraft such as the Parker Solar Probe and Solar Orbiter “touch the Sun,” knowledge of plasma mechanisms will play a key role in interpreting this frontier of space exploration. In addition to theoretical and computational studies, exploration of the plasma universe can be conducted through experiments on Earth. The breadth of conditions observed in the plasma universe requires a wide-ranging laboratory approach, from high-energy-density laser experiments to magnetized plasma facilities at multiple scales. Plasma science and technology lays the foundation for creating transformative technologies unique in implementation and application. The realization of an FPP opens the door to ubiquitous carbon-free electricity. Plasma-based technologies promise unique pathways to bring that electricity to the products that power society. That power could revolutionize the way chemicals are manufactured. Such technologies promise the realization of novel materials that cannot be manufactured by conventional means, such as functionalized nanoparticles for drug delivery and new materials relevant to quantum information systems. The next generation of rockets, powered by fusion, may enable human exploration of the solar system and beyond with faster transit times. The next generation of ultrafast, compact electronic devices, such as cell phones and computers, will rely on plasma science to fuel advances in semiconductor manufacturing. Novel, precise therapies for cancer and for antibiotic-resistant bacterial infections are now within reach, buoyed by advances in atmospheric-pressure plasmas and plasma-based ultracompact accelerators. Technology and Science Drivers Fusion Technology & Science Drivers ------------------------------------- The Fusion Science and Technology area should focus on establishing the scientific and technical basis for a fusion pilot plant by the 2040s. → Learn More Plasma Technology & Science Drivers ------------------------------------- The Plasma Science and Technology area should focus on new opportunities to advance fundamental understanding and, translate these advances into applications that benefit society. → Learn More Programs and facilities to execute the strategic plan Aligning the program with the six technology and science drivers will require redirection of programs and development of new facilities. Collaborations with international and privately funded research programs are important components of the strategy, and participation in ITER is considered essential for obtaining access to a high-gain burning plasma. Rigorous scientific inquiry is cultivated by leveraging current leadership, partnerships, and priority research areas that advance general plasma science and high-energy-density physics while emphasizing the potential of plasma-based technology for translational research. Success in all of these areas will require robust support for foundational cross-cutting research in theory, modeling, and computation; diagnostic development; and transformative enabling technologies. The multidisciplinary workforce needed for fusion energy and plasma science requires that the community commit to the creation and maintenance of a healthy climate of diversity, equity, and inclusion, which will benefit the community as a whole and the mission of FES. Research Program Areas New or expanded research program areas are urgently needed to fulfill the mission of developing our fundamental understanding of plasmas and to move toward a fusion energy source—with FPP readiness by the 2040s. These research program elements are described here at a high level, targeting the specific technology and science drivers identified above, and are not in priority order (prioritization is provided in the budget scenarios in Chapter 2). FPP System Design and Integration: A central overarching priority is to form a domestic multi-institutional, collaborative FPP mission, design, and study program. This effort will provide the resources and coordination to integrate critical research advances made across the FES portfolio into attractive FPP concepts. It will need to merge advances in the understanding of burning plasma physics with the capabilities of new fusion materials and technologies. Attention also needs to be paid to licensing and safety-related issues (e.g., tritium and activation product transport and stored energy sources including the plasma, magnets, and cryogens). An integrated plant design requires consideration for balance of plant equipment and remote handling capabilities and should address the reliability, availability, maintainability and inspectability (RAMI) of the plant. Participation by private and public stakeholders is essential to ensure economic attractiveness. Innovations made outside the public program are appropriately considered in developing these concepts. An essential component underpinning this effort is a strong theory and computation program, including the advancement of multiscale, multiphysics theory and modeling capabilities necessary to predict the complex interactions between numerous plasma, material, and engineering processes that will occur within an FPP. A vital part of the program is the continued development of validated models at a range of complexities and experimental fidelities, along with the predictive integrated modeling capabilities that utilize them. Creating such models will require continued close partnership between FES and ASCR to fully leverage US investments in highperformance computing, including coming exascale machines. Moreover, accelerated progress and increased readiness of multiple systems are needed to safely design and operate a fusion reactor; those components include advances in diagnostics, instrumentation, data handling, and automated real-time decision making. This design effort should give significant attention to activities contrasting tokamak-based concepts with concept studies for different plasma cores like stellarators, alternates, or inertial confinement fusion energy (IFE). It should also include activities agnostic to the plasma core. Additionally, designs for concept exploration or for devices aiming to extend the performance of successfully tested innovative concepts should be pursued to provide an information basis for the design, decision, and pursuance of new facilities. Fusion Materials and Technology: Critical developments are needed in fusion materials, magnets, and heating and current drive actuators. Technology advances are needed to handle the extreme conditions expected in future fusion reactors and to harness fusion energy and breed fuel. In addition to advancing key research on existing facilities such as linear plasma devices and in-pile fission irradiation, resource enhancement must allow timely resolution of critical FPP design questions. Because of the significant time scales involved in facility development and subsequent research, immediate action is needed. Increased investment in theory and simulation supporting the research on these facilities is also needed. Focus is given to the development of plasma-facing materials and components, structural and functional materials, and fusion blanket and fuel cycle elements needed for an FPP. Diagnostic advances for fusion materials studies are needed to understand the interaction of materials with the fusion environment. Magnets are an integral feature of magnetic fusion configurations, and it is desirable to develop magnets with higher fields, operating temperatures, and reliability, which are constructed with streamlined manufacturing processes and reduced production costs. All of these factors improve the performance and/or lower the costs of an FPP. Private industry has made significant progress in developing the relevant magnet technology, including high-temperature superconducting magnets, and the federal program should complement and, when possible, collaborate with those activities. Launching structures for radio-frequency plasma heating and current drive actuators must be made of new materials in order to withstand the neutron and plasma environment, have integrated steady-state cooling, and have more acceptable long-pulse reliability. Efficiency improvements in the source, the transmission, and the plasma coupling must be developed to enhance FPP competitiveness. The development of materials and technology appropriate for the nuclear conditions of a fusion reactor is a critical need in the international effort to develop fusion energy. The US is poised for leadership in this area through targeted investments in unique facilities. Collaboration and partnering with the DOE Office of Nuclear Energy in the areas of materials development, generation of qualification-level data, and improved technologies for materials and component irradiation should be cultivated. Fusion Plasma Core: The tokamak is the most technically advanced approach for use as a fusion reactor power core. The ITER international experiment is the largest single investment by DOE FES, and a US ITER research team needs to be formed to leverage it. That team will make essential contributions to achieving the high gain mission for ITER, exploit unique access to a burning plasma at the reactor scale, and enable US scientists to close the nuclear science and engineering gaps in order to build an FPP. Access to burning plasmas could also be possible in the US-based privately funded SPARC tokamak as early as mid-decade. SPARC will be parallel and complementary to international fusion efforts, including ITER, and to other ongoing private-sector fusion endeavors. The existing DIII-D and NSTX-U national tokamak facilities are key to preparation for the study of burning plasmas in ITER and in other planned and future private devices. Additional research on these facilities, in combination with private and international collaborations, continuing support of existing university tokamak programs, and utilization of US expertise in theory and simulation, is needed to find solutions to remaining technical gaps. These gaps include disruption prediction, avoidance, and mitigation; plasma-facing component integration; and FPPrelevant scenario development. Advances in technology and in our understanding of plasma physics have opened paths to lower capital cost tokamak FPPs, but have also brought scientific and technical challenges that must be overcome. These challenges motivate the construction of a new world-leading domestic tokamak, which would be uniquely situated to develop integrated solutions in a useful time frame. In order to mitigate risks associated with the tokamak approach, alternative pathways to fusion are also pursued, which could lead to more economic fusion power in the longer term by capitalizing on US expertise. Quasi-symmetric stellarators are considered, as are alternate plasma core solutions beyond the tokamak and stellarator. These alternate pathways are supported at three levels, from basic validation of the physics, through development of self-consistent solutions, to demonstration of integrated solutions. A reestablished IFE program takes advantage of US leadership in high-energydensity physics and progress that the NNSA has made toward high yield in inertial confinement fusion. General Plasma Science Program (GPS): GPS research explores the fundamental behaviors of plasmas. This includes foundational theoretical descriptions of plasma dynamics, numerical methods to model multiscale behavior, and experiments that test whether our understanding of plasmas is accurate. Such foundational research serves as the basis for all areas of plasma science and technology, ranging from the laboratory to astrophysics. Although motivated primarily by the desire to understand nature, many of the physics processes studied have direct relevance to fusion and other technological applications. The GPS program funds research at a range of scales, including operations and construction of the Basic Plasma Science Facility at UCLA, the Wisconsin Plasma Physics Laboratory, the Magnetized Dusty Plasma Experiment at Auburn, and the Facility for Laboratory Reconnection Experiments at Princeton Plasma Physics Laboratory. A major component of the GPS research program is the long-standing NSF–DOE Partnership in Plasma Science and Engineering. High-Energy-Density Laboratory Plasmas (HEDLP): HEDLP research explores and applies novel regimes resulting from the extraordinary ability to concentrate power—in many cases more power than the world’s total electric generating capacity in an area smaller than the end of a human hair—for a brief fraction of a second. That ability creates new states of matter that include condensed matter, warm and hot dense matter, and plasmas relevant to astrophysical phenomena, stellar properties and processes, and fusion reactors. Self-organized, far-from-equilibrium plasmas are probed and controlled, enabling unique applications such as new accelerators and materials. This program has a successful history of partnering with DOE NNSA, NSF, and DOE HEP to fund research on several midscale laser, pulsed-power, and X-ray free electron (XFEL) facilities. Plasma-Based Technology Program: Technologies in the plasma science and technology (PST) portfolio include low-temperature plasmas and plasma-based accelerators. These technologies benefit the public by enabling cell phones, computers, advanced drinking water purification, and security and medical methods. They underlie key industries such as semiconductor manufacturing and materials processing, which directly fuel the economy through innovation and maintaining core competence and leadership in those industries. A plasmabased technology program that consolidates and focuses critical efforts will facilitate technology transfer and realize the promise of this area. Networks: Collaborative networks of researchers and facilities can provide enormous value as a coordinating organization and mechanism for leveraging resources and capabilities. LaserNetUS is a successful model that brings together 10 unique midscale laser facilities and opens up opportunities to a large number of new users. In a similar vein, the establishment of a MagNet centered around basic magnetized plasma and laboratory space/astrophysics, a ZNet for pulsed-power science and technology, and an LTP-Net for low-temperature plasmas could similarly support growth and enable collaborative research in their respective areas. These networks can encourage cross-fertilization as researchers work on multiple facilities and will facilitate the training of students. Coordination and access to computational/theoretical models, diagnostics, and other resources in support of experiments can also be established. These network structures also position the US to be more competitive, because investments, technology development, and future planning can be implemented more strategically by engaging the full community. Facilities New mid- to large-scale facilities are urgently needed to meet the goal of FPP readiness by the early 2040s and to realize the goals of plasma science and technology. The elements of the following list are grouped by topical area and are not in priority order Fusion Prototypic Neutron Source (FPNS): The science of material exposure to fusion neutron fluxes is a key gap in the international fusion program. No facility exists that can generate the necessary fluence, energy spectrum, and helium production level in the lattice of candidate materials. FPNS concepts that utilize existing facilities like accelerators or commercial units, combined into a costeffective system, can be a fast track forward. FPNS provides leadership opportunities based on existing expertise in nuclear materials in the US program by enabling the fundamental explorations of fusion nuclear material science, which needs to be combined with a reinvigorated neutron theory and computation program. Moreover, accelerated access to fusion neutron exposure is an area of extreme interest to the fusion industry and has significant opportunities for near-term public–private partnerships. Material Plasma Exposure eXperiment (MPEX): MPEX is under construction and will provide a unique capability to study plasma-material interactions under conditions that are prototypical for a reactor divertor regime as far as the near-wall plasma-material interface is concerned. The ability to expose irradiated materials to these plasma conditions and conduct rapid turnaround in-situ and ex-situ material characterization are the most important project elements that need to be met as key program deliverables toward an FPP. High-Heat-Flux (HHF) testing facilities: Testing capabilities to explore properties of materials and plasma-facing components, both solid and liquid, under high heat fluxes address a key gap toward FPP material definitions. Experimental capabilities to conduct fundamental testing on coupon levels (centimeter scale) are a necessary testbed for model validation of material properties. The couponlevel testing is a prerequisite for component-level testing (tens of centimeters to meters scale) to qualify components for an FPP. Accordingly, testing facilities for both levels of high-heat-flux materials research are required. EXhaust and Confinement Integration Tokamak Experiment (EXCITE): High-magneticfield approaches to a tokamak-based FPP raise specific scientific and engineering challenges. High-divertor-power exhaust solutions need to be integrated with sustainment of high-power-density plasma cores, which are needed for generation of significant fusion power. Both the NASEM Burning Plasma Report and the CPP report identify the need to address these challenges in an integrated fashion, rather than at separate facilities. This requirement motivates the need for construction of a new domestic tokamak, previously referred to as NTUF (New Tokamak User Facility) in the CPP report. Blanket Component Test Facility (BCTF): The CPP report outlines an R&D program on blanket materials and transport phenomena that culminates in the design and fabrication of blanket-section prototypes, which undergo staged testing in a Blanket Component Test Facility (BCTF) and Volumetric Neutron Source (VNS). The CPP report describes a BCTF that integrates all non-nuclear features of a fusion blanket and its ancillary systems (prototypic, at-scale complex structures and coolants) under prototypic conditions of temperature, pressure, magnetic field, and mechanical stress, with surrogate surface and volumetric heating and injected hydrogen or deuterium in place of tritium. Concepts successfully vetted in the BCTF, and fission and/or fusion irradiations, could potentially proceed to full nuclear testing and tritium production in the VNS. Further definition and development of these facilities and research plans should be undertaken by the program and the community. Midscale Stellarator: A proof of concept experiment is needed to demonstrate improved steady-state plasma confinement in combination with a novel nonresonant divertor. Development of this research line provides risk mitigation for the mainline tokamak approach and could lead to a commercially more attractive fusion system. This stellarator facility would therefore be a discovery-oriented facility that could stimulate a great deal of innovation. Volumetric Neutron Source (VNS): Recognizing the critical need for integral-effect irradiation testing of components or subcomponents, such as blanket modules, the CPP report recommended pursuit of a VNS for this purpose without specifying particular metrics or a confinement concept that would provide fusion neutrons. Multiple VNS concepts have been proposed and a concept assessment study should evaluate any plasma physics developments required to realize each concept, determine the relevance of these configurations to tokamak/FPP components, and assess them against quantitative metrics (e.g., on neutron flux or fluence) to be achieved in advance of FPP operation. This initial assessment activity should identify either a suitable concept for further development, construction, and operation, or identify an alternate approach (e.g., fission reactor irradiation or early phase testing in FPP) that best meets this mission need. MEC-Upgrade: An upgrade to the Matter in Extreme Conditions (MEC) end-station at the Linac Coherent Light Source (LCLS) would enable the co-location of a PW-laser operating at 1–10 Hz repetition rate and a multi-kJ long pulse laser with our only domestic XFEL. This would enable us to tackle physical and chemical changes at fundamental time scales and explore new regimes of dense material physics, astrophysics, planetary physics, and short-pulse laser-plasma interactions. The MEC-U proposal has achieved Critical Decision 0 and is currently in preparation for CD-1, also having received line-item status in the FY 2020 Congressional budget. Solar Wind Facility: How the solar wind is accelerated, heated, and driven turbulent is among the most persistent and important open questions in plasma science. It is an opportune moment to develop, in concert with advanced space missions, a next-generation experimental facility to isolate, control, and diagnose plasma phenomena responsible for the complex solar wind behavior, at relevant scales. This facility would leverage and coordinate existing laboratory space/astrophysics research groups, as the experimental conditions needed to pursue solar-wind-related questions can also benefit research in broader astrophysical contexts. Such a venture would be a prime opportunity to coordinate among interested funding agencies, primarily NSF and NASA, but also ONR and AFOSR. Multi-Petawatt Laser Facility: Tens-of-petawatt laser systems can produce light pressures in the exapascal regime, copious amounts of radiation, and extremely bright beams of energetic particles, including electrons, ions, neutrons, or antimatter. The novel capabilities enabled by multi-PW lasers open new frontiers in R&D such as particle acceleration and advanced light sources, high-field physics and nonlinear quantum electrodynamics (QED), and laser-driven nuclear physics. As identified in the BLI report, there is a need for the US to develop ultrahigh-intensity technology and build an open-access laser user facility with multiple beamlines at 10–100 PW peak powers. High-Repetition-Rate Laser Facility: New high-repetition-rate (10 Hz to kHz) laser systems coming online represent a fundamentally new system architecture for high energy density (HED). The greater than 1000 times increase in shot rate over today’s systems, coupled to emerging technologies such as machine learning and additive manufacturing, will result in an enormous acceleration in the rate of knowledge acquisition. Such high-rep-rate high-energy lasers further open the door to unprecedented temporal and spatial resolution of HEDP phenomena, including GeV-class electron beams and precision HED pumps and probes. Recent community reports from NASEM and BLI have clearly outlined the urgent science case and FES mission-relevant needs for a short-pulse, highpeak-power, high-average-power laser system. This may be an area for partnering with DOE HEP, which may take the lead on this facility. Midscale Z pinch: Extremely strong magnetic fields over macroscopic volumes are only accessible via pulsed-power facilities, which open up the physics of plasmas in a way that other plasma drivers cannot. Current US facilities are either very large and complex (the 26 MA Sandia Z-Machine with < 1 shot/day) or too small (~1 MA or less) to address the breadth of science expressed by the community. There is clear interest in establishing a pulsed-power facility at an intermediate size (up to 10 MA) accessible to the academic community, with a higher shot rate than Z, yet still capable of fielding fusion-relevant and HED experiments. Further, such a facility could explore driver technologies and pulsed-power science for next-generation larger-scale pulsed-power devices such as a 60 MA “Z-Next.” This facility would provide an opportunity for FES to partner with another agency, such as NNSA or NSF, which might take the lead. Process and Prioritization Criteria The following criteria express the principles used to prioritize projects and programs discussed in this report. Consensus criteria and guidance for prioritization within program areas were developed by the research community during the CPP process. That guidance is incorporated in the criteria below, which were used for prioritization of the entire portfolio. In applying the criteria and following the charge language, we assume that the ITER construction project will be successful, and we thus focus on the non-ITER portion of the budget. Alignment: Align projects and programs with the technology and science drivers to achieve the fusion mission, specifically the path to an FPP, and to advance fundamental plasma science and enable societally beneficial plasma applications. Balance technological development with scientific discovery, recognizing the importance of both as the sources of innovations that benefit the entire program. Urgency: Prioritize the most expeditious path to fusion energy and other plasma technologies that provide compelling solutions to urgent issues such as sustainable, carbon-free power production, advanced medical therapies, and more efficient industrial processes. Innovation: Embrace innovative research, new developments in technology, and interdisciplinary connections to address key challenges. Reduce the time and cost to develop usable fusion energy and other plasma applications. Impact: Implement a logical sequence of programs that increases scientific and technological progress relative to investment, reduces the risks associated with the FPP mission and the technology and science objectives, and takes into account time constraints and impacts on the overall program. Leadership: Establish and maintain US leadership, including world-leading facilities, science, and industries that attract international participation. Recognize federal, industry, and international efforts in fusion and plasma development and form partnerships whenever possible. Stewardship: As experimental capabilities are developed and program transitions occur, ensure the continued productivity of an essential workforce to maintain scientific and technological progress. Engage all stakeholders in executing the program, including national laboratories, industry, and universities. ↓Download Report (PDF: 8.5MB)↓ Download Report Summary (PDF: 233 KB) Additional Resources →DOE FESAC Long Range Planning Activity →DOE Changes/Reports →APS-DPP Community Planning Process A long-range plan to deliver fusion energy and to advance plasma science Site Credits
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https://www.youtube.com/watch?v=b5b2a9rKxGE
Learn how to label the discontinuity of a piecewise function by graphing Brian McLogan 1590000 subscribers Description 23874 views Posted: 5 May 2019 👉 Learn how to graph piecewise functions. A piecewise function is a function which have more than one sub-functions for different sub-intervals(sub-domains) of the function's domain. To graph a piecewise function, we graph the different sub-functions for the different sub-intervals of the function's domain. 👏SUBSCRIBE to my channel here: ❤️Support my channel by becoming a member: 🙋‍♂️Have questions? Ask here: 🎉Follow the Community: Organized Videos: ✅ Piecewise Functions ✅ Graph a Piecewise Function | Hard ✅ Graph a Piecewise Function | Easy ✅ Evaluate a Piecewise Function ✅ Find the Value that Makes the Functions Continuous ✅ Write the Piecewise Function From a Graph ✅ Write the Piecewise Functions From a Table 🗂️ Organized playlists by classes here: 🌐 My Website - 🎯Survive Math Class Checklist: Ten Steps to a Better Year: Connect with me: ⚡️Facebook - ⚡️Instagram - ⚡️Twitter - ⚡️Linkedin - 👨‍🏫 Current Courses on Udemy: 👨‍👩‍👧‍👧 About Me: I make short, to-the-point online math tutorials. I struggled with math growing up and have been able to use those experiences to help students improve in math through practical applications and tips. Find more here: functions #brianmclogan 9 comments Transcript: where this would be expected that you can do this without being distracted by talking other class students in class so the the main idea guys is you have two lines in this equation again we're trying to identify if it's continuous or not and then describe the discontinuity um so we basically have two lines here let's just go and graph 2x minus 1. so forget about the restriction for a second let's just graph y equals 2x minus 1. all right so i go down that's my y intercept up 2 over 1. and i can basically go in the positive direction and i can go in the negative direction so that is the graph now i'm going to go ahead and apply the constraint and the constraint says i only want you to graph this equation for x values that are less than 0. so that means 0 is not defined so therefore i'm going to put a nice little open circle there and then it's only for x values that are less than 0. well here's the x axis here's the negative x values here's the positive x values right so our exercises are less than 0 are only going to be to the left that means i'm going to erase the rest of the graph that's in the positive for a positive x values anybody have any questions or clarifying questions there again this is not a time that you should be looking at your calculator it's just a time for you to be looking up here i know the stimulation of electronics is very difficult for many of you to overcome but if you need me to give you a different seat that you can go ahead and pay attention to then i'll be more than happy to do that for you guys the next one is let's go and graph negative x plus three this one has a different y intercept which is positive three and then we have a negative slope so down into the right one or up into the left one either way you guys are going to graph that looks something like that we're not trying to be exact guys but what we want to recognize here is this is for only x values that are greater than or equal to zero so therefore it's going to be only for positive values all right um and then yeah that's supposed to cross at three so i'll just be able but anyway you can look at this graph and say is it continuous can i draw the graph without lifting my pen or marker off of the board and the obvious answer is no it is not continuous so therefore we say well then what type of discontinuity is it and you can say that is a jump discontinuity at x is equal to zero right and one thing with you guys remember if there is going to be a discontinuity if you have two functions that are continuous that is a continuous line that's a continuous line right they are restricted but both of these functions are continuous correct so if there is going to be a discontinuity where is that discontinuity going to occur at the restriction change right it's the only thing that make would make sense now not always are we going to have continuous functions for each of the functions but when we do that's where it will occur okay all right
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https://brilliant.org/wiki/partition-of-an-integer/
Partition of an Integer | Brilliant Math & Science Wiki HomeCourses Sign upLog in The best way to learn math and computer science. Log in with GoogleLog in with FacebookLog in with email Join using GoogleJoin using email Reset password New user? Sign up Existing user? Log in Partition of an Integer Sign up with FacebookorSign up manually Already have an account? Log in here. Patrick Corn, Voon Yee Vee, Elek Lamoureux, and 3 others Chew-Seong Cheong Jimin Khim Eli Ross contributed A partition of a positive integer n n n is an expression of n n n as the sum of one or more positive integers (or parts). The order of the integers in the sum "does not matter": that is, two expressions that contain the same integers in a different order are considered to be the same partition. The number of different partitions of n n n is denoted p(n) p(n) p(n). This function is called the partition function. The partitions of 5 5 5 are as follows: 5 4+1 3+2 3+1+1 2+2+1 2+1+1+1 1+1+1+1+1.\begin{aligned} &5 \ &4+1 \ &3+2 \ &3+1+1 \ &2+2+1 \ &2+1+1+1 \ &1+1+1+1+1. \end{aligned}​5 4+1 3+2 3+1+1 2+2+1 2+1+1+1 1+1+1+1+1.​ So p(5)=7 p(5)=7 p(5)=7. Contents Recursions and Closed-form Formulas Pictorial Representations Conjugate Partitions Generating Functions Recursions and Closed-form Formulas Combinatorial functions such as p(n) p(n) p(n) often lend themselves to recursions that make them easier to compute. For instance, consider the number of decompositions of n n n as the sum of positive integers in which order does matter (sometimes called compositions). Let c(n) c(n) c(n) be the number of compositions of n n n. For instance, c(3)=4 c(3) = 4 c(3)=4 because 3=1+1+1=1+2=2+1=3 3 = 1+1+1 = 1+2 = 2+1 = 3 3=1+1+1=1+2=2+1=3. Then there is a recursion for c(n) c(n) c(n): if the first term of a composition of n n n is k k k, the remaining terms are a composition of n−k n-k n−k. So the number of such compositions is c(n−k) c(n-k) c(n−k). This leads to the recursive formula c(n)=c(n−1)+c(n−2)+⋯+c(1)+1. c(n) = c(n-1) + c(n-2) + \cdots + c(1) + 1. c(n)=c(n−1)+c(n−2)+⋯+c(1)+1. Solving this recursion is not hard: note that c(n+1)=c(n)+c(n−1)+⋯+c(1)+1=c(n)+c(n)=2 c(n) c(n+1) = c(n)+c(n-1)+\cdots+c(1)+1 = c(n)+c(n) = 2c(n) c(n+1)=c(n)+c(n−1)+⋯+c(1)+1=c(n)+c(n)=2 c(n), and c(1)=1 c(1)=1 c(1)=1, so c(n)=2 n−1 c(n) = 2^{n-1} c(n)=2 n−1. (Exercise for the reader: there are direct, non-recursive proofs of this formula as well.) However, proceeding in the same way for the partition function is not as straightforward, because of the possibility of over-counting partitions. For instance, a partition of 4 4 4 that starts with 1 1 1 is 1+2+1 1+2+1 1+2+1, which corresponds to the partition 2+1 2+1 2+1 of 3 3 3, and a partition of 4 4 4 that starts with 2 2 2 is 2+1+1 2+1+1 2+1+1, which corresponds to the partition 1+1 1+1 1+1 of 2 2 2; but these are the same partition, so any recursive formula must avoid counting them twice. In fact, there is no closed-form formula for p(n) p(n) p(n) in any meaningful sense. Many down-to-earth questions about the partition function are quite difficult and mysterious: for instance, it is not even known whether there are infinitely many n n n such that p(n) p(n) p(n) is divisible by 3 3 3. Reveal the answer p(17)=297 p(18)=385 p(19)=490 p(20)=627 p(21)=792 \begin{aligned} p(17) &= 297 \ p(18) &= 385 \ p(19) &= 490 \ p(20) &= 627 \ p(21) &= 792 \end{aligned} p(17)p(18)p(19)p(20)p(21)​=297=385=490=627=792​ Let p(n) p(n) p(n) be the number of partitions of an integer n n n. The values of p(17),p(18),p(19),p(20),p(21)p(17), p(18),p(19),p(20), p(21) p(17),p(18),p(19),p(20),p(21) are as shown above. How many partitions of 20 are there that do not contain any parts equal to 1? The correct answer is: 137 Pictorial Representations Partitions are represented pictorially in Ferrers diagrams: each part of the permutation is represented by a row of dots, where the number of dots equals the part. The rows, like the parts, are laid out in descending order of size: ∙∙∙∙∙∙∙∙\begin{aligned} &\bullet\, \bullet\, \bullet \ &\bullet\, \bullet \ &\bullet\, \bullet \ &\bullet \end{aligned}​∙∙∙∙∙∙∙∙​ This diagram represents the partition 8=3+2+2+1 8 = 3+2+2+1 8=3+2+2+1. Young diagrams are similar, but use squares instead of dots. Here is the Young diagram for 10=5+4+1 10 = 5+4+1 10=5+4+1: Conjugate Partitions One concept that is most easily understood from diagrams of partitions is the notion of conjugacy. Let π \pi π be a partition of n n n, and let a k a_k a k​ be the number of parts of π \pi π that are ≥k \ge k ≥k. Suppose the largest part is d d d. Then the conjugate of π \pi π is the partition n=a 1+a 2+⋯+a d.n = a_1 + a_2 + \cdots + a_d.n=a 1​+a 2​+⋯+a d​. There is something to prove in the definition, namely that the sum of the a i a_i a i​ actually equals n n n. To see this, note that a part of size b b b is counted once in a 1,a 2,…,a b a_1, a_2, \ldots, a_b a 1​,a 2​,…,a b​. So, a part of size b b b contributes b b b to the sum, and therefore the sum is the same as the sum of the parts of π \pi π. This is not the easiest way to think about conjugate partitions. They are most simply described in terms of Ferrers diagrams: the Ferrers diagram of the conjugate partition is obtained from the Ferrers diagram of the original partition via reflection along the top-left to bottom-right diagonal. The operation is similar to a matrix transpose; the columns of the diagram change to rows and the rows change to columns. The partition 8=3+2+2+1 8=3+2+2+1 8=3+2+2+1 has the following Ferrers diagram: ∙∙∙∙∙∙∙∙\begin{aligned} & \bullet\, \bullet\, \bullet \ & \bullet\, \bullet \ & \bullet\, \bullet \ & \bullet \end{aligned}​∙∙∙∙∙∙∙∙​ Its conjugate has the following Ferrers diagram: ∙∙∙∙∙∙∙∙\begin{aligned} & \bullet\, \bullet\, \bullet\, \bullet \ & \bullet\, \bullet\, \bullet \ & \bullet \end{aligned}​∙∙∙∙∙∙∙∙​ So, it is 8=4+3+1 8 = 4+3+1 8=4+3+1. Note that the description via Ferrers diagrams shows without any work that conjugating twice returns the original partition, which was not necessarily immediately clear from the definition of conjugation. Conjugate partitions are used in many bijective proofs of results about partitions; here is one basic example. Show that the number p(n,k) p(n,k) p(n,k) of partitions of a positive integer n n n into exactly k k k parts equals the number of partitions of n n n whose largest part equals k k k. For instance, 7=5+1+1=4+2+1=3+2+2=3+3+1 7 = 5+1+1 = 4+2+1 = 3+2+2 = 3+3+1 7=5+1+1=4+2+1=3+2+2=3+3+1 and 7=3+3+1=3+2+2=3+2+1+1=3+1+1+1+1 7 = 3+3+1 = 3+2+2 = 3+2+1+1 = 3+1+1+1+1 7=3+3+1=3+2+2=3+2+1+1=3+1+1+1+1, so p(7,3)=4 p(7,3) = 4 p(7,3)=4. Let S(n,k) S(n,k)S(n,k) be the set of partitions of n n n into k k k parts and let T(n,k) T(n,k) T(n,k) be the set of partitions of n n n whose largest part equals k k k. Then there is a bijection f ⁣:S(n,k)→T(n,k) f\colon S(n,k) \to T(n,k) f:S(n,k)→T(n,k) defined by f(π)= f(\pi) = f(π)= the conjugate of π \pi π. This is because the Ferrers diagram of a partition in S(n,k) S(n,k) S(n,k) has a first column of length k k k, so its conjugate has top row of length k k k. And f f f is a bijection because it has an inverse map, namely conjugation. So the two sets have the same size. □_\square □​ For example, for n=7,k=3 n=7,k=3 n=7,k=3, the function f f f is f(5+1+1)=3+1+1+1+1 f(4+2+1)=3+2+1+1 f(3+2+2)=3+3+1 f(3+3+1)=3+2+2. \begin{aligned} f(5+1+1) &= 3+1+1+1+1 \ f(4+2+1) &= 3+2+1+1 \ f(3+2+2) &= 3+3+1 \ f(3+3+1) &= 3+2+2. \end{aligned} f(5+1+1)f(4+2+1)f(3+2+2)f(3+3+1)​=3+1+1+1+1=3+2+1+1=3+3+1=3+2+2.​ Other examples are given in the wiki on bijective proofs. Generating Functions Main Article: Generating Functions Many theorems about partitions that have complicated combinatorial proofs are easier and more accessible via generating functions. It is often the case that formal power series of the form ∑a n x n \sum a_n x^n ∑a n​x n, where the a n a_n a n​ count certain types of partitions, can be expressed as infinite products in useful ways. Express ∑n=0∞p(n)x n\displaystyle \sum_{n=0}^\infty p(n) x^n n=0∑∞​p(n)x n as an infinite product of power series. Consider the product (1+x+x 2+x 3+⋯)(1+x 2+x 4+x 6+⋯)(1+x 3+x 6+x 9+⋯)(⋯). \big(1+x+x^2+x^3+\cdots\big)\big(1+x^2+x^4+x^6+\cdots\big)\big(1+x^3+x^6+x^9+\cdots\big)(\cdots). (1+x+x 2+x 3+⋯)(1+x 2+x 4+x 6+⋯)(1+x 3+x 6+x 9+⋯)(⋯). The coefficient of x n x^n x n in this expansion is p(n) p(n) p(n): suppose a partition of n n n has a 1 a_1 a 1​ 1s, a 2 a_2 a 2​ 2s, and so on. Then n=a 1+2 a 2+⋯+n a n n = a_1+2a_2+\cdots+na_n n=a 1​+2 a 2​+⋯+n a n​, and this corresponds to the term in the expansion of the infinite product given by x a 1 x^{a_1} x a 1​ from the first sum, times x 2 a 2 x^{2a_2} x 2 a 2​ from the second sum, and so on. Rewriting this using geometric series gives ∑n=0∞p(n)x n=∏k=1∞1 1−x k.□ \sum_{n=0}^\infty p(n) x^n = \prod_{k=1}^{\infty} \frac1{1-x^k}.\ _\square n=0∑∞​p(n)x n=k=1∏∞​1−x k 1​.□​ Euler's Pentagonal Number Theorem The product ∏k=1∞(1−x k)=1−x−x 2+x 5+x 7−x 12−⋯ \prod_{k=1}^{\infty} \big(1-x^k\big) = 1 - x - x^2 + x^5+x^7- x^{12}-\cdots k=1∏∞​(1−x k)=1−x−x 2+x 5+x 7−x 12−⋯ equals ∑m=−∞∞(−1)m x m(3 m−1)2. \sum_{m=-\infty}^{\infty} (-1)^m x^{\frac{m(3m-1)}2}. m=−∞∑∞​(−1)m x 2 m(3 m−1)​. Note: Numbers of the form m(3 m−1)2 \frac{m(3m-1)}2 2 m(3 m−1)​ are called pentagonal numbers. The theorem can be proved using partitions as well: the coefficient of x k x^k x k in the product counts the number of partitions of k k k with an even number of parts minus the partitions of k k k with an odd number of parts. It can be shown (using Ferrers diagrams) that there is a bijection between these two sets of partitions unless k k k is a pentagonal number, in which case there is a bijection between the sets with one specific partition removed. Yes No Undecidable in ZFC Reveal the answer Let the partition function P(n)P(n)P(n) enumerate the ways n n n can be expressed as a distinct sum of positive integers, e.g. P(4)=5 P(4) = 5 P(4)=5 since 4=3+1=2+2=2+1+1=1+1+1+1 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1 4=3+1=2+2=2+1+1=1+1+1+1 are the only ways to represent 4 4 4. ∏p prime[∑n=0∞P(n)p−n]\prod_{p \ \text{prime}} \left[ \sum_{n=0}^{\infty} P(n)p^{-n} \right]p prime∏​[n=0∑∞​P(n)p−n] Does the above product converge? The correct answer is: No Euler's pentagonal number theorem implies the following beautiful recurrence relation for p(n) p(n) p(n): p(n)=p(n−1)+p(n−2)−p(n−5)−p(n−7)+p(n−12)⋯, p(n) = p(n-1)+p(n-2)-p(n-5)-p(n-7)+p(n-12)\cdots, p(n)=p(n−1)+p(n−2)−p(n−5)−p(n−7)+p(n−12)⋯, where the sign of p(n−k) p(n-k) p(n−k) is the opposite of the sign of x k x^k x k in the sum expansion in Euler's pentagonal number theorem. Cite as: Partition of an Integer. Brilliant.org. Retrieved 07:27, September 28, 2025, from Join Brilliant The best way to learn math and computer science.Sign up Sign up to read all wikis and quizzes in math, science, and engineering topics. Log in with GoogleLog in with FacebookLog in with email Join using GoogleJoin using email Reset password New user? Sign up Existing user? Log in
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https://www.ldh.la.gov/assets/medicaid/MCPP/4_2_25/1214_LHCC_LA.CP.MP.163_Total_Parenteral_Nutrition_and_Intradialytic_Parenteral_Nutrition.pdf
Page 1 of 9 Clinical Policy: Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Reference Number: LA.CP.MP.163 Coding Implications Date of Last Revision: 03/25 Revision Log See Important Reminder at the end of this policy for important regulatory and legal information. Description Parenteral nutrition (PN) is the intravenous administration of an artificially prepared solution of nutrients that bypasses the gastrointestinal tract and meets the nutritional requirements of a patient. PN is necessary when enteral nutrition is incapable of meeting the needs of the patient’s gastrointestinal tract. This policy describes the medical necessity requirements for two types of PN, (A) total parenteral nutrition (TPN), in which all of the necessary macronutrients and micronutrients are supplied to the patient, and (B) intradialytic parenteral nutrition (IDPN), in which nutrition is supplied to end-stage renal disease (ESRD) patients undergoing dialysis as an alternative to regularly scheduled TPN. Policy/Criteria I. It is the policy of Louisiana Healthcare Connections that the following are medically necessary for members/enrollees when meeting all of the associatedfollowing indications: A. Total Parenteral Nutrition (TPN), when all the following criteria are met: 1. Documentation of nutritional insufficiency, in the absence of TPN, as shown by any of the following: a. Weight For adults, involuntary weight loss >of 10% of ideal body weight over three months or less, or > 20% of usual body weight; within six months or 5% within one month; a. Total protein < 6 g/dL in the past four weeks; b. Serum albumin < 3.4 g/dL in the past four weeks; b. ii. For children, weight for length, weight for height, or sex < 10th percentile or inadequate weight gain or a significant decrease in usual growth percentile; c. iii. For neonates, extremely low birth weight < 1000 g; 2. Evidence of structural or functional bowel disease that makes oral or tube feedings inappropriate, or a condition in which the gastrointestinal tract is non-functioning for a period of time, including, but not necessarily limited to, any of the following: a. Crohn’s disease; b. Short bowel syndrome; c. Single or multiple fistulae (enterocolic, enterovesical, or enterocutaneous); c. Central nervous system (CNS) disorder resulting in swallowing difficulties and high risk of aspiration; d. Obstructing stricture; e. Motility disorder; f. Newborn anomalies of the gastrointestinal tract which prevent or contraindicate oral feedings such as tracheoesophageal fistula, gastroschisis, omphalocele, or massive intestinal atresia; g. Infants and young children who fail to thrive due to cardiac or respiratory disease, short bowel syndrome, malabsorption, or chronic idiopathic diarrhea; h. ProlongedParalytic ileus in children or prolonged paralytic ileus following a major surgical procedure or multiple injuries; i. Radiation enteritis; j. Liver failure in children approved for liver transplants, who fail to grow while CLINICAL POLICY Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Page 2 of 9 receiving enteral nutritional support; k. Liver failure in adults who have hepatic encephalopathy and cannot tolerate a protein source consisting of standard amino acids or enteral nutritional support (TPN used for the administration of a liver-specific amino acid mixture); l. Acute necrotizing pancreatitis in adults with an inadequate oral intake for longer than a week, where enteral feedings exacerbate abdominal pain, ascites, or fistulous output.; m. Necrotizing enterocolitis;17 n. Small bowel ischemia.17 Initial approval duration for TPN is for three months. Continued approval duration is six months, given that the member/enrollee has no evidence of unacceptable complications from treatment, and documentation supports positive response to therapy. B. Intradialytic Parenteral Nutrition (IDPN), when all the following criteria are met: 1. Meets TPN criteria in section A; 2. Member/enrollee has stage 5 chronic kidney disease; 3. Member/enrollee is undergoing hemodialysis; 4. IDPN is offered as an alternative to regularly scheduled TPN. Initial approval duration for IDPN is for three months. Continued approval duration is six months, given that the member/enrollee has no evidence of unacceptable complications from treatment and documentation supports positive response to therapy. II. It is the policy of Louisiana Healthcare Connections that the following indications are not proven safe and effective: A. TPN: 1. Children who were previously well nourished or mildly malnourished, who are undergoing oncologic treatment associated with a low nutrition risk (e.g., less advanced disease, less intense cancer treatments, advanced disease in remission during maintenance treatment); 2. Members/enrollees with advanced cancer whose malignancy is documented as unresponsive to chemotherapy or radiation therapy; 3. Members/enrollees for whom liver transplantation is not feasible and whose prognosis will not change in spite of TPN therapy; B. IDPN, when any of the following criteria are met: 1. IDPN treatments offered in addition to regularly scheduled infusions of TPN; 2. IDPN treatments in members/enrollees who are suffering from acute kidney injury and who do not have ESRD. Background Total Parenteral Nutrition (TPN) TPN is the delivery of macronutrients (i.e., proteins, fats, and carbohydrates) and micronutrients (i.e., vitamins, minerals, and trace elements) intravenously. TPN is indicated in situations for which the gastrointestinal tract is incapable of digesting nutrients through enteral (oral or feeding tube) nutrition. Short-term TPN is delivered peripherally through a subclavian, internal jugular, or a femoral central venous catheter, while long-term TPN requires a tunneled central venous catheter, such as a Hickman or Groshong catheter, or an implanted infusion port.1 CLINICAL POLICY Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Page 3 of 9 Some advantages of TPN include the ease of administration, the ability to correct fluid and electrolyte imbalances, and the ability to manage nutrition in the setting of mucositis. However, some disadvantages of TPN include catheter-associated infections, fluid overload, hyperglycemia, catheter-associated thrombosis, hepatic thrombosis, hepatic dysfunction, blood electrolyte abnormalities, and enterocyte atrophy.2,14 Long-term complications associated with longer parenteral therapy and home parenteral therapy could include hepatobiliary and bone disease.14 American Gastroenterological Association Long-term parenteral nutrition is indicated for patients with prolonged gastrointestinal tract failure that prevents the absorption of adequate nutrients to sustain life.6 Intradialytic Parenteral Nutrition (IDPN) Malnutrition presents an ongoing concern with patients receiving chronic hemodialysis or peritoneal dialysis affecting between 20 to 70% of patients. There is a positive association between length of time on dialysis and increasing decline in nutritional parameters. The administration of IDPN through the patient’s dialysis access is advantageous since this approach eliminates the need for additional venous catheter placement.10 IDPN is delivered during dialysis for patients who continue to lose weight or have very low serum albumin levels (< 3.4 g/dL) despite oral supplements and for those with severe gastroparesis who may be unable to tolerate oral supplements.6 However, IDPN only provides 70% of the nutrients to the patient because of loss into the dialysate.3 A Hayes evaluation of peer-reviewed literature demonstrated findings of low-quality evidence that IDPN is relatively safe and is associated with improvements in baseline laboratory measures (serum albumin, serum prealbumin, creatinine), body mass index/body weight, and mortality rates compared with conventional therapies. Findings also reflect individual study limitations, heterogeneity among the studies in IDPN formulation, and remaining questions regarding patient selection criteria for IDPN and long-term benefits.6 Several societies have published position guidelines supporting the use of IDPN in specific situations. American Society for Parenteral and Enteral Nutrition IDPN should be reserved for patients that are incapable of meeting their nutritional needs orally and who are not candidates for enteral nutrition or TPN because of gastrointestinal intolerance, venous access problems, or other reasons.4 European Society for Clinical Nutrition and Metabolism IDPN is indicated in undernourished patients undergoing hemodialysis with poor compliance to oral nutritional supplements and not requiring TPN.5,6 National Kidney Foundation’s Kidney Disease Outcomes Quality Initiative (KDOQI) and the Academy of Nutrition and Dietetics KDOQI recommended in a 2020 clinical guideline update that a trial of IDPN should be administered to adults on maintenance hemodialysis with stage 5 chronic kidney disease (CKD).6 CLINICAL POLICY Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Page 4 of 9 Coding Implications This clinical policy references Current Procedural Terminology (CPT®). CPT® is a registered trademark of the American Medical Association. All CPT codes and descriptions are copyrighted 20232024, American Medical Association. All rights reserved. CPT codes and CPT descriptions are from the current manuals and those included herein are not intended to be all-inclusive and are included for informational purposes only. Codes referenced in this clinical policy are for informational purposes only and may not support medical necessity. Inclusion or exclusion of any codes does not guarantee coverage. Providers should reference the most up-to-date sources of professional coding guidance prior to the submission of claims for reimbursement of covered services. NOTE: Coverage is subject to each requested code’s inclusion on the corresponding LDH fee schedule. Non-covered codes are denoted () and are reviewed for Medical Necessity for members under 21 years of age on a per case basis. CPT® Codes Description N/A HCPCS Codes Description B4164 through B5200 Parenteral nutrition solutions and supplies B9004 Parenteral nutrition infusion pump, portable B9006 Parenteral nutrition infusion pump, stationary S9364 Home infusion therapy, total parenteral nutrition (TPN); administrative services, professional pharmacy services, care coordination, and all necessary supplies and equipment including standard TPN formula (lipids, specialty amino acid formulas, drugs other than in standard formula and nursing visits coded separately), per diem (do not use with home infusion codes S9365 through S9368 using daily volume scales) S9365 Home infusion therapy, total parenteral nutrition (TPN); one liter per day, administrative services, professional pharmacy services, care coordination, and all necessary supplies and equipment including standard TPN formula (lipids, specialty amino acid formulas, drugs other than in standard formula and nursing visits coded separately), per diem S9366 Home infusion therapy, total parenteral nutrition (TPN); more than one liter but no more than two liters per day, administrative services, professional pharmacy services, care coordination, and all necessary supplies and equipment including standard TPN formula (lipids, specialty amino acid formulas, drugs other than in standard formula and nursing visits coded separately), per diem CLINICAL POLICY Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Page 5 of 9 S9367 Home infusion therapy, total parenteral nutrition (TPN); more than two liters but no more than three liters per day, administrative services, professional pharmacy services, care coordination, and all necessary supplies and equipment including standard TPN formula (lipids, specialty amino acid formulas, drugs other than in standard formula and nursing visits coded separately), per diem S9368 Home infusion therapy, total parenteral nutrition (TPN); more than three liters per day, administrative services, professional pharmacy services, care coordination, and all necessary supplies and equipment including standard TPN formula (lipids, specialty amino acid formulas, drugs other than in standard formula and nursing visits coded separately), per diem CLINICAL POLICY Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Page 6 of 9 Reviews, Revisions, and Approvals Revision Date Approval Date Converted corporate to local policy. 08/15/20 Added indications for radiation enteritis, liver failure in children, liver failure in adults, and acute necrotizing pancreatitis in adults, in I.A.2.j – I.A.2.m., along with relevant ICD-10 codes (i.e., K52.0, K72.00-K72.91, K85.01, K85.02, K85.11, K85.12, K85.31, K85.32, K85.81, K85.82, K85.91, K85.92 and Z76.82. In I.B.2, changed “end-stage renal disease” to “stage 5 chronic kidney disease.” Replaced member with member/enrollee in all instances. Replaced “experimental/investigational” with “not proven safe and effective” in section II. Annual review. References reviewed and updated to AMA format. Changed “review date” in the header to “date of last revision” and “date” in the revision log header to “revision date." Added “and may not support medical necessity” to coding implications. Background updated with no impact to criteria. Specialist reviewed. 7/22 9/26/22 Annual review. Minor rewording in Criteria section with no impact on criteria. Clarifying language added to Criteria I.A.1.a. Background updated with no impact on criteria. Minor rewording to HCPCS codes with no clinical significance. ICD-10 codes removed. References reviewed and updated. 5/23 7/21/23 Annual review completed. Minor rewording in Criteria section with no clinical significance. Background updated with no impact to criteria. References reviewed and updated. External specialist reviewed. 3/24 5/22/24 Annual review. Updated criteria I.A.1.a. regarding low body weight to include details by age group and expanded to I.A.1.a. through c. Removed previous criteria I.A.1.b. and c. regarding total protein and serum albumin. Removed previous criteria I.A.2.d. CNS disorders from list of conditions that make oral or tube feedings inappropriate. Updated criteria I.A.2.h. to include children with paralytic ileus and added I.A.2.m and n. to list conditions that make oral or tube feedings inappropriate. References reviewed and updated. 3/25 CLINICAL POLICY Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Page 7 of 9 References 1. Seres D. Nutrition support in critically ill patients: Parenteral nutrition. UpToDate. www.uptodate.com. Updated March 08, 2021.November 19, 2024. Accessed January 3December 20, 2024. 2. Rubin H, Mehta J, Fong JL, Greenberg D, GrusChak S, Trifilio S. Revisiting Infectious Complications Following Total Parenteral Nutrition Use During Hematopoietic Stem Cell Transplantation. J Adv Pract Oncol. 2020;11(7):675 to 682. doi:10.6004/jadpro.2020.11.7.2 3. Beddhu S, Cho ME, BansaiBansal S. Pathogenesis and treatment of malnutrition in patients on maintenance dialysis. UpToDate. www.uptodate.com. Updated January 30, 2023. Accessed January 3December 23, 2024. 4. American Society for Parenteral and Enteral Nutrition (A.S.P.E.N.) Board of Directors. Clinical Guidelines for the Use of Parenteral and Enteral Nutrition in Adult and Pediatric Patients, 2009. JPEN J Parenter Enteral Nutr. 2009;33(3):255 to 259. doi:10.1177/0148607109333115 5. Cano N, Fiaccadori E, Tesinsky P, et al. ESPEN Guidelines on Enteral Nutrition: Adult renal failure. Clin Nutr. 2006;25(2):295 to 310. doi:10.1016/j.clnu.2006.01.023 6. Health Technology Assessment. Intradialytic parenteral nutrition (IDPN) for end-stage renal disease in adults. Hayes. www.hayesinc.com. Published December 29, 2020 (annual review December 606, 2023). Accessed January 3December 23, 2024. 7. Worthington P, Balint J, Bechtold M, et al. When Is Parenteral Nutrition Appropriate?. JPEN J Parenter Enteral Nutr. 2017;41(3):324 to 377. doi:10.1177/0148607117695251 8. McClave SA, DiBaise JK, Mullin GE, Martindale RG. ACG Clinical Guideline: Nutrition therapy in the Adult Hospitalized Patient. Am J Gastroenterol. 2016;111(3):315 to 335. doi:10.1038/ajg.2016.28 9. Khan FA, Selvaggi, G. Overview of intestinal and multivisceral transplantation. UpToDate. www.uptodate.com. Updated August 17, 2022.July 03, 2024. Accessed January 3December 23, 2024. 10. Ikizler TA, Burrowes JD, Byham-Gray LD, et al. KDOQI Clinical Practice Guideline for Nutrition in CKD: 2020 Update [published correction appears in Am J Kidney Dis. 2021 Feb;77(2):308]. Am J Kidney Dis. 2020;76(3 Suppl 1):S1 to S107. doi:10.1053/j.ajkd.2020.05.006 11. Mihatsch W, Fewtrell M, Goulet O, et al. ESPGHAN/ESPEN/ESPR/CSPEN guidelines on pediatric parenteral nutrition: Calcium, phosphorus and magnesium. Clin Nutr. 2018;37(6 Pt B):2360 to 2365. doi:10.1016/j.clnu.2018.06.950 12. Wang N, Cui L, Liu Z, et al. Optimizing parenteral nutrition to achieve an adequate weight gain according to the current guidelines in preterm infants with birth weight less than 1500 g: a prospective observational study. BMC Pediatr. 2021;21(1):303. Published 2021 Jul 7. doi:10.1186/s12887-021-02782-1 13. Brown RO, Compher C. American Society for Parenteral and Enteral Nutrition Board of Directors. A.S.P.E.N. Clinical guidelines: nutrition support in adult acute and chronic renal failure. JPEN J Parenter Enteral Nutr. 2010;34(4):366 to 377. doi:10.1177/0148607110374577 14. Berlana D. Parenteral Nutrition Overview. Nutrients. 2022;14(21):4480. Published 2022 Oct 25. doi:10.3390/nu14214480 15. Compher C, Bingham AL, McCall M, et al. Guidelines for the provision of nutrition support therapy in the adult critically ill patient: The American Society for Parenteral and Enteral Nutrition [published correction appears in JPEN J Parenter Enteral Nutr. 2022 Aug;46(6):1458-1459]. JPEN J Parenter Enteral Nutr. 2022;46(1):12 to 41. doi:10.1002/jpen.2267 16. National Institutes of Health (NIH), National Library of Medicine. Albumin blood (serum) test. CLINICAL POLICY Total Parenteral Nutrition and Intradialytic Parenteral Nutrition Page 8 of 9 Updated February 28, 2023. Accessed December 23, 2024. 17. Baker RD, Baker SS, Bojczuk G. Parenteral nutrition in infants and children. UpToDate. www.uptodate.com. Updated May 31, 2024. Accessed December 23, 2024. 18. American Society for Parenteral and Enteral Nutrition. Definitions. ons/. Accessed December 23, 2024. 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Introduction to Exponents and Surds | Cambridge IGCSE Maths The article title is Introduction to Exponents and Surds | Cambridge IGCSE Mathematics. [Please watch the video attached at the end of this blog for a visual explanation of exponents and surds, as well as how to use the calculator when solving such questions at the exam] Square root this, square root that, cube root, radicals, and surds. All these terms might seem a bit odd to some people, but they should certainly not be odd concepts for a student sitting for Cambridge IGCSE Mathematics exams. This blog article is focused on exponents and surds, a lesson that can seem confusing but is pretty easy once you get the hang of it! Exponents/ Powers To begin with, we will look at exponents, which can also be called powers or indices (singularly index). This shows us the number of times a number is multiplied by itself. Ex: 2 2 2 2 2 If we look at the example given above, we can see that 2 is repeatedly multiplied by itself five times. This is difficult to write and is also time-consuming. Therefore instead of that, what you can simply do is write it as 25 (read as 2 to the power of 5). Once you see this expression, you will know that 2 is being multiplied by itself 5 times. Or when you have to write an expression like this, you can see that the 2 is multiplied twice, so you write that as 22, 3 is multiplied twice so you write it as 32, and then there is the 5 at the end. Or you have ⅓ being multiplied 4 times, which you can write as (⅓)4 (one-third to the power of four). Rules regarding operations with powers There are four rules regarding operations with powers. Important: In order for these rules to apply, the bases must be equal. When multiplying, the indices with equal bases, the powers are added. Ex: 25 23 Since both of them are based on two, the powers can be added (5 + 3), and this will add up to 2 to the power 8. 2. When dividing the indices with equal bases, the powers are subtracted. Ex: 25 ÷ 23 Here, since both bases are 2, you can subtract the powers (5-3), which then gives you 2 to the power 2. Remember: When multiplying, you have to add the powers, when dividing you have to subtract the powers,but this is only if the bases are equal. 3. Power of a power is obtained by multiplying the powers Ex: (23)2 In order to find the answer to this example, you simply multiply the two powers together.(3 2 = 6), therefore the answer is 2 to the power 6. 4. Negative power or an index can be converted into positive by taking the reciprocal. Ex: 10-3 In order to make this a positive fraction, we need to make 10-3 the denominator. This will result in (1⁄103). Similarly, when making a positive power a denominator, it will become a negative power. Ex: 104 becomes (1⁄10-4 ) Refer to the video for calculator tips! Roots Roots are basically the inverses of powers. The root of a number is a number that when multiplied by itself gives the original number. They are usually denoted by the symbol √. Ex: PIC ∛, ∜, 5√ The above example will be read as 3 roots, 4 roots, and 5 roots. If you have the square root of a number, you can convert that as a power of half. Ex: √4 The square root of four can also be written as 4½. This can be simplified further. 4 = 2 2, which, when written as a power of a number, is 22. Using the multiplication of powers rule, (2 ½ ) will give us 21 which is simply 2. Therefore, the square root of 4 is 2. Ex: ∛27 The cubic root of 27 can also be denoted as 27⅓. This can be simplified further. 27 = 33, and using the power multiplication, (3 ⅓ ), this can be simplified into 31, which is simply 3. Hence, the cubic root of 27 is 3. If it is a square root, it is to the power of ½ (half), if it is a cube root, it is to the power of ⅓ (one-third). Refer to the video for calculator tips here too! Surds A surd is a radical which will give us an irrational number as the answer (numbers that cannot be written as a fraction with both numerator and denominator being integers). A sure, therefore, is any number in which when you find the square root, instead of getting an integer value as an answer, all you get is a decimal answer. Ex: Therefore, the square root of a multiplied by the square root of b will be the same as finding the square root of a multiplied by b (ab). This is the first rule you must keep in mind. Simplifying surds Ex: √80 Since finding the square root of 80 straightaway would be a bit tricky, we can try simplifying it.If we consider 80 in multiples, 80 is the same as 16 5. This means, we can break √80 into √(165), which further simplifies into √16 √5. AThe square root of 16 is 4 and we can then leave the answer as 4 √5. Sometimes we also get to simplify surds that look like the example below. Ex: √27 ÷ √3 To simplify √27 ÷ √3, we can also write this as √(27/3). Once we divide 27 by 3, we get the answer to be 9. Hence, the square root of 9 is what we must find, resulting in 3 being the final answer. Rationalising Surds In rationalizing surds, if we have a square root in the denominator, we try to get rid of it. To do this, we multiply the top value and the bottom value by the same square root value. Ex: In the example below, in order to get rid of √b, we must multiply both the denominator and numerator by √b. This will result in a√b being the numerator, and √b2 becoming the denominator. The square root of b2 is going to be (2 ½) , therefore the denominator simplifies itself to be just b. A numerical example can be seen below. Adding and subtracting surds Adding and subtracting surds is similar to adding and subtracting terms in algebraic expressions. If you have two radicands (the number under the root symbol) that are the same, you can add them together. If you are required to subtract radicands, then provided the radicand values are the same, then they can be subtracted. However, if the radicands are different, then the surds cannot be added or subtracted. Helpful points for practicing exponents and surds for your upcoming Cambridge IGCSE Mathematics exams. Mathematics is a subject that relies heavily on how well you prepare in advance for your exams. To make things easier, here are a few pointers that may help you get ready for your upcoming Cambridge IGCSE Mathematics exams. Pay close attention to the powers and the roots. It is easy to get confused but this will result in you losing marks if you write the wrong answer. Take your time to attempt these questions, especially if they are long ones. Practice sample questions – Practise, practice, practice. Questions on surds and exponents come in all shapes and sizes. They can sometimes be quite tricky, so do attempt all the questions you can get your hands on before exams. Our question banks have questions of different difficulty levels that you can practice before your actual exams. Practice past paper questions to familiarise yourself with the format of the questions and paper. Ask for help – You are never too late to ask for assistance if you’re struggling. Reach out to us at Tutopiya to find the right tutor to help you pass your exams with flying colors! Watch the video below for a visual explanation of surds and exponents and attempt the quiz to challenge yourself! For Schools Book a FREE Trial Courses Close Courses Open Courses 1-on-1 Tuition Personalized classes tailored to each student's learning pace. IGCSE TuitionsComprehensive preparation for all IGCSE subjects. IB TuitionExpert-led support for IB curriculum success. 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A surd is a radical which results in an irrational number. [Please watch the video attached at the end of this blog for a visual explanation of exponents and surds, as well as how to use the calculator when solving such questions at the exam] Square root this, square root that, cube root, radicals, and surds. All these terms might seem a bit odd to some people, but they should certainly not be odd concepts for a student sitting for Cambridge IGCSE Mathematics exams. This blog article is focused on exponents and surds, a lesson that can seem confusing but is pretty easy once you get the hang of it! Exponents/ Powers To begin with, we will look at exponents, which can also be called powers or indices (singularly index). This shows us the number of times a number is multiplied by itself. Ex: 2 2 2 2 2 If we look at the example given above, we can see that 2 is repeatedly multiplied by itself five times. This is difficult to write and is also time-consuming. Therefore instead of that, what you can simply do is write it as 2 5 (read as 2 to the power of 5). Once you see this expression, you will know that 2 is being multiplied by itself 5 times. Or when you have to write an expression like this, you can see that the 2 is multiplied twice, so you write that as 2 2, 3 is multiplied twice so you write it as 3 2, and then there is the 5 at the end. Or you have ⅓ being multiplied 4 times, which you can write as (⅓)4 (one-third to the power of four). Rules regarding operations with powers There are four rules regarding operations with powers. Important: In order for these rules to apply, the bases must be equal. When multiplying, the indices with equal bases, the powers are added. Ex: 2 5 2 3 Since both of them are based on two, the powers can be added (5 + 3), and this will add up to 2 to the power 8. When dividing the indices with equal bases, the powers are subtracted. Ex: 2 5 ÷ 2 3 Here, since both bases are 2, you can subtract the powers (5-3), which then gives you 2 to the power 2. Remember: When multiplying, you have to add the powers, when dividing you have to subtract the powers,but this is only if the bases are equal. Power of a power is obtained by multiplying the powers Ex: (2 3)2 In order to find the answer to this example, you simply multiply the two powers together.(3 2 = 6), therefore the answer is 2 to the power 6. Negative power or an index can be converted into positive by taking the reciprocal. Ex: 10-3 In order to make this a positive fraction, we need to make 10-3 the denominator. This will result in (1⁄10 3). Similarly, when making a positive power a denominator, it will become a negative power. Ex: 10 4 becomes (1⁄10-4 ) Refer to the video for calculator tips! Roots Roots are basically the inverses of powers. The root of a number is a number that when multiplied by itself gives the original number. They are usually denoted by the symbol √. Ex: PIC ∛, ∜,5√ The above example will be read as 3 roots, 4 roots, and 5 roots. If you have the square root of a number, you can convert that as a power of half. Ex: √4 The square root of four can also be written as 4 ½. This can be simplified further. 4 = 2 2, which, when written as a power of a number, is 2 2. Using the multiplication of powers rule, (2 ½ ) will give us 2 1 which is simply 2. Therefore, the square root of 4 is 2. Ex: ∛27 The cubic root of 27 can also be denoted as 27 ⅓. This can be simplified further. 27 = 3 3, and using the power multiplication, (3 ⅓ ), this can be simplified into 3 1, which is simply 3. Hence, the cubic root of 27 is 3. If it is a square root, it is to the power of ½ (half), if it is a cube root, it is to the power of ⅓ (one-third). Refer to the video for calculator tips here too! Surds A surd is a radical which will give us an irrational number as the answer (numbers that cannot be written as a fraction with both numerator and denominator being integers). A sure, therefore, is any number in which when you find the square root, instead of getting an integer value as an answer, all you get is a decimal answer. Ex: Therefore, the square root of a multiplied by the square root of b will be the same as finding the square root of a multiplied by b (ab). This is the first rule you must keep in mind. Simplifying surds Ex: √80 Since finding the square root of 80 straightaway would be a bit tricky, we can try simplifying it.If we consider 80 in multiples, 80 is the same as 16 5. This means, we can break √80 into √(165), which further simplifies into √16√5.AThe square root of 16 is 4 and we can then leave the answer as4√5. Sometimes we also get to simplify surds that look like the example below. Ex: √27÷√3 To simplify √27÷√3, we can also write this as √(27/3).Once we divide 27 by 3, we get the answer to be 9. Hence, the square root of 9 is what we must find, resulting in 3 being the final answer. Rationalising Surds In rationalizing surds, if we have a square root in the denominator, we try to get rid of it. To do this, we multiply the top value and the bottom value by the same square root value. Ex: In the example below, in order to get rid of √b, we must multiply both the denominator and numerator by √b. This will result ina√bbeing the numerator, and √b2 becoming the denominator. The square root of b2 is going to be (2 ½), therefore the denominator simplifies itself to be just b. A numerical example can be seen below. Adding and subtracting surds Adding and subtracting surds is similar to adding and subtracting terms in algebraic expressions. If you have two radicands (the number under the root symbol) that are the same, you can add them together. If you are required to subtract radicands, then provided the radicand values are the same, then they can be subtracted. However, if the radicands are different, then the surds cannot be added or subtracted. Helpful points for practicing exponents and surds for your upcoming Cambridge IGCSE Mathematics exams. Mathematics is a subject that relies heavily on how well you prepare in advance for your exams. To make things easier, here are a few pointers that may help you get ready for your upcoming Cambridge IGCSE Mathematics exams. Pay close attention to the powers and the roots. It is easy to get confused but this will result in you losing marks if you write the wrong answer. Take your time to attempt these questions, especially if they are long ones. Practice sample questions – Practise, practice, practice. Questions on surds and exponents come in all shapes and sizes. They can sometimes be quite tricky, so do attempt all the questions you can get your hands on before exams. 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https://stackoverflow.com/questions/40345469/converting-to-and-from-a-number-system-that-doesnt-have-a-zero-digit
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Converting to and from a number system that doesn't have a zero digit Ask Question Asked 8 years, 11 months ago Modified8 years, 11 months ago Viewed 695 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Consider Microsoft Excel's column-numbering system. Columns are "numbered" A, B, C, ... , Y, Z, AA, AB, AC, ... where A is 1. The column system is similar to the base-10 numbering system that we're familiar with in that when any digit has its maximum value and is incremented, its value is set to the lowest possible digit value and the digit to its left is incremented, or a new digit is added at the minimum value. The difference is that there isn't a digit that represents zero in the letter numbering system. So if the "digit alphabet" contained ABC or 123, we could count like this: (base 3 with zeros added for comparison) base 3 no 0 base 3 with 0 base 10 with 0 ----------- ------------- -------------- - - 0 0 A 1 1 1 B 2 2 2 C 3 10 3 AA 11 11 4 AB 12 12 5 AC 13 20 6 BA 21 21 7 BB 22 22 8 BC 23 100 9 CA 31 101 10 CB 32 102 11 CC 33 110 12 AAA 111 111 13 Converting from the zeroless system to our base 10 system is fairly simple; it's still a matter of multiplying the power of that space by the value in that space and adding it to the total. So in the case of AAA with the alphabet ABC, it's equivalent to (13^2) + (13^1) + (13^0) = 9 + 3 + 1 = 13. I'm having trouble converting inversely, though. With a zero-based system, you can use a greedy algorithm moving from largest to smallest digit and grabbing whatever fits. This will not work for a zeroless system, however. For example, converting the base-10 number 10 to the base-3 zeroless system: Though 9 (the third digit slot: 3^2) would fit into 10, this would leave no possible configuration of the final two digits since their minimum values are 13^1 = 3 and 13^0 = 1 respectively. Realistically, my digit alphabet will contain A-Z, so I'm looking for a quick, generalized conversion method that can do this without trial and error or counting up from zero. Edit The accepted answer by n.m. is primarily a string-manipulation-based solution. For a purely mathematical solution see kennytm's links: What is the algorithm to convert an Excel Column Letter into its Number? How to convert a column number (eg. 127) into an excel column (eg. AA) algorithm radix greedy base-conversion number-systems Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited May 23, 2017 at 10:27 CommunityBot 1 1 1 silver badge asked Oct 31, 2016 at 15:35 Taylor LopezTaylor Lopez 873 1 1 gold badge 10 10 silver badges 29 29 bronze badges 8 I'm having trouble following all of your explanations, but isn't this just equivalent to a base-3 system, plus 1? To go from base-10-with-0 to base-3-without-0, would it not work to add 1, convert first to base-3-with-zero, and subtract 1? Or something like that. I've got a nasty head cold right now and my thoughts are not so clear. xP B. Eckles –B. Eckles 2016-10-31 15:44:01 +00:00 Commented Oct 31, 2016 at 15:44 Maybe one thing I'm missing-- are you saying that you're excluding numbers with zero in them from being representable?B. Eckles –B. Eckles 2016-10-31 15:45:49 +00:00 Commented Oct 31, 2016 at 15:45 @B.Eckles No, The system simply has no zero digit. so instead of 0, 1, 2, it has 1, 2, 3 in each digit position.Taylor Lopez –Taylor Lopez 2016-10-31 15:46:45 +00:00 Commented Oct 31, 2016 at 15:46 2 Excel to number: stackoverflow.com/questions/667802/…; Number to Excel: stackoverflow.com/questions/181596/…kennytm –kennytm 2016-10-31 15:47:25 +00:00 Commented Oct 31, 2016 at 15:47 2 @TaylorLopez: That's the second link in the comment...kennytm –kennytm 2016-10-31 15:51:40 +00:00 Commented Oct 31, 2016 at 15:51 |Show 3 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Convert to base-3-with-zeroes first (digits 0AB), and from there, convert to base-3-without-zeroes (ABC), using these string substitutions: A0 => 0C B0 => AC C0 => BC Each substitution either removes a zero, or pushes one to the left. In the end, discard leading zeroes. It is also possible, as an optimisation, to process longer strings of zeros at once: A000...000 = 0BBB...BBC B000...000 = ABBB...BBC C000...000 = BBBB...BBC Generalizable to any base. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Oct 31, 2016 at 16:13 answered Oct 31, 2016 at 16:02 n. m. could be an AIn. m. could be an AI 122k 14 14 gold badges 140 140 silver badges 266 266 bronze badges 3 Comments Add a comment Taylor Lopez Taylor LopezOver a year ago I really like this solution because it's clever, and it's strictly based on the string patters. It's probably still a little less efficient than the Excel link above since this would require you to make several passes, and gets more complicated the larger your alphabet is (if it's a true string replacement), but I like it nonetheless. 2016-10-31T16:23:42.303Z+00:00 0 Reply Copy link n. m. could be an AI n. m. could be an AIOver a year ago No, one pass is enough. It's basically "replace 0 wirh C and subtract 1 from the digit on the left". If that digit is 0 again, you reolace it with B and subtract 1 from the next one on the left, etc. So you make one right to left pass. 2016-10-31T16:27:55.897Z+00:00 0 Reply Copy link Taylor Lopez Taylor LopezOver a year ago Ah I see. I misunderstood. 2016-10-31T16:28:25.227Z+00:00 0 Reply Copy link Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 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https://hal.science/hal-03592377/document
Revisiting the Cobalt(II) hydration from Molecular Dynamics and X-ray Absorption Spectroscopy Daniel Z. Caralampio a and Benjamin Reeves b and Maria R. Beccia b and Jos´ e M. Mart´ ınez a and Rafael R. Pappalardo a and Christophe den Auwer b and Enrique S´ anchez Marcos a a Department of Physical Chemistry, University of Seville, 41012 Seville, Spain b Universit´ e Cˆ ote d’Azur, Institut de Chimie de Nice, CNRS, UMR 7272, 06108 Nice, France ARTICLE HISTORY Compiled July 18, 2019 ABSTRACT Solution chemistry of Co(II) is receiving a renewal attention due to the high interest for knowing the speciation in seawater of its 60 Co radioactive isotope which appeared in the Japan sea as a consequence of the Fukushima-Daichii nuclear power plant accident. Experimental EX-AFS and XANES spectra of a dilute Co(II) aqueous solutions have been recorded and structural data derived from their analysis. Based on QM calculations, an ab-initio intermolecular poten-tial has been generated for the Co(II)–H 2O interaction using the hydrated ion model that uses a polarizable and flexible solvent description through the MCDHO2 model. Classical molecular dynamics simulations of Co(II) in water have been performed and X-ray Absorption spectra have been simulated and compared with the experimental ones. Energetic, structural, dynam-ical and spectroscopical properties of the cobalt cation in solution have been computed and compared with previous experimental and theoretical data. These comparisons have assessed the good performance of the developed intermolecular potential. KEYWORDS Intermolecular potential and Polarizable and Flexible model and MCDHO2 potential and MD simulation and power spectrum 1. Introduction The Fukushima-Daichii nuclear central accident, which happened in 2011, has stimulated the interest in different ion-containing aqueous solutions due to the speciation of some radioactive isotopes in the seawater surrounding the nuclear plant shores.[1, 2] The main challenge is to establish the physicochemical properties of these radioactive cations in aqueous media containing relatively high concentrations of marine salts. The latter are able to alter the behavior of radioactive material present in low concentration both, by electrostatic interactions, as supporting electrolytes, and by chemical bonding, as anion complexation.[3, 4] One of the involved radioactive cations is the 60 Co that appears in its divalent oxidation state as a sub-product of the 127 Cs decay. The aim of this work is to establish a double experimental-theoretical reference of Co(II) in dilute aqueous solutions to be used in further studies of Co(II) solvated in the more complex seawater medium. To undertake this type of involved studies, it is convenient to establish reference data of simpler systems both at the XAS instrumental level and at the computer simulation level. This allows monitoring of the changes of physicochemical properties of the Co(II) from a dilute solution without other ions than its salt counterion to a seawater solution. Table 1 collects a wide set of experimental and theoretical properties of Co(II) aqueous solutions which appeared in the literature. There is a general agreement about the Co(II) hexa-coordination at highly dilute aqueous solutions, because its octahedral aqua ion forms a stable high-spin [Co(H 2O) 6]2+ (d 7) complex. An infrared study together with a combined electronic spectroscopy and multi-reference quantum mechanical study support the six-fold CONTACT E. S´ anchez. Author. Email: sanchez@us.es coordination as the most representative aqua ion structure in solution.[6, 7] X-ray diffraction studies[8, 9] proposed a first-shell Co-O I distance range of 2.00-2.10 ˚ A and Levochkin et al. a second hydration shell centered at 4.05 ˚ A. EXAFS fittings found hexacoordination for the Co(II) aqua ion and a Co-O I range of 2.06-2.09 ˚ A with Debye-Waller factors in the range 0.0038–0.0062 ˚ A2.[10–12] D’Angelo et al., from the fitting of the XANES spectrum, proposed a hexahydrated ion with the Co-O I distance equals to 2.06 ˚ A. This methodology was proposed originally by Benfatto and Della Longa. Several groups have performed computer simulations of Co(II)-containing aqueous solutions using classical and quantum mechanics Molecular Dynamics (MD) simulations.[12, 15, 16] There is a general agreement on the six-fold coordination of the first shell with a range of Co-O I distance of 2.07-2.27˚ A. For the second-shell hydration number, the range is 12-23 with Co-O II distances between 4.1 and 4.6 ˚ A. 2. Methodology 2.1. Co 2+ -H 2O interaction potential building The exchangeable hydrated ion model[17, 18], a refinement of our original hydrated ion approach, allows the release of first-shell water molecules from the hydrated ions, i.e. water exchange is allowed. A flexible and polarizable Co 2+ -H 2O interaction potential has been built. Following the idea performed in a previous Scandium hydration study, we have developed the cobalt-water intermolecular potential considering Co(II) hexahydrate structures, [Co(H 2O) 6]2+ , derived from the quantum-mechanical optimized structure. Distorted structures where Co-H 2O distances and angles are changed according to the stretching and bending normal modes of the minimum energy octahedral structure were used to build the potential. As Co(II) transition metal cation is an open-shell system, the use of its hexahydrate instead of the bare cation to extract the quantum-mechanical interaction energies is particularly appropriate. The octahedral coordination complex formed by its aqua ion guarantees that its electronic state (quartet multiplicity) is preserved along the geometrical deformations considered for the sampling, something that it would not have been preserved if bare Co 2+ -H 2O interactions were considered. The QM calculations were carried out at the B3LYP level, using the ECP10MDF pseudopotential and the associated basis set of Stuttgart group for Co[21, 22] and the aug-cc-pVTZ basis set for O and H. To prospect the cobalt(II)-water potential energy surface, together with the [Co(H 2O) 6]2+ minimum energy structure, the extraction of one water molecule from the Co 2+ aqua ion structure, a bending and the symmetric and assymmetric stretchings of the aqua ion were included in the intermolecular potential generation. Figure 1displays some of the representative structure deformations used for the fitting. All calculation were performed with Gaussian 09. Figure 1. Representative set of deformations chosen from Co 2+ hexahydrate used to build the intermolecular potential. Arrows indicates the type of distortion (normal mode) considered in the hydrates. From left to right, asymmetric stretching mode, bending mode and shortening and lengthening of only one Co–OH 2distance The metal cation is defined as polarizable and the water molecules are defined by the MCDHO2 potential, a polarizable and flexible model, derived from the original mobile charge density harmonic oscillator model MCDHO. Details about the potential employed 2as well as the set of fitted coefficients for the new Co 2–H 2O potential are collected in the Suplementary Material (SM). The obtained potential gives an accurate description of both energetic and structural properties of the mimimum energy hexahydrate: Eint = -342.7 kcal/mol (QM) vs. -342.5 kcal/mol (Pot) and a Co-O I distance of 2.110 ˚ A (QM) vs. 2.113 ˚ A(Pot). The whole set of structures considered in the fitting were 50 and the mean error of the fit was 1.5 kcal/mol. 2.2. Molecular dynamics simulations MD simulations in the canonical ensemble (NVT) at 300 K employing the Noose-Hoover thermostat with a relaxation time of 0.5 ps were carried out. A new modified version of the DL POLY program (Classic version) that incorporates the force field was employed. The polarizable character was described by means of the adiabatic shell model. The simulation box contained 1 Co 2+ cation and 1000 water molecules, with a box length of 31.098 ˚ A in order to fit the experimental water density, 0.997 g/cm 3. Long-range interactions were treated by means of the Ewald sum. A trajectory with a 100 ps of equilibration time and 1 ns of production time was employed for the analysis. 2.3. Experimental XAS spectra Co K-edge XAS fluorescence spectra of a CoCl 2 0.005M in HCl 0.05M aqueous solution were recorded in the MARS beamline at the SOLEIL Synchrotron Radiation source (Saclay, France). In this concentration range the stability of the aqua ion is high enough to prevent the formation of chloro-aquo complexes. This is confirmed by the vis-UV electronic spectrum of the sample, as well as by the EXAFS and XANES spectra which would had been highly modified if chloride ligands had replaced water molecules in the Co 2+ first coordination sphere. The change of EXAFS and XANES spectra of Cr 3+ aqua ion when chloride ligands replace water molecules in its first coordination shell has been shown in a previous study. The storage ring operated at an energy of 2.5 GeV with a top electron current of 200 mA. The optics of the beamline essentially consist of a water-cooled double-crystal monochromator (FMB Oxford), which is used to select the incident energy of the X-ray beam and for horizontal focalization, and two large water-cooled reflecting mirrors (IRELEC/SESO) that are used for high-energy rejection (harmonic part) and vertical collimation and focalization. In this case, the monochromator was set with the Si(111) crystals. Energy calibration of the monochromator was done using Ni foil as reference at maximum first derivative equal to 8.333 keV. Because of the low concentration of the samples, EXAFS measurements were performed in fluorescence mode using a 13-element high purity germanium detector (ORTEC). The sample was kept in a specifically designed cell for diluted liquids (Ets CANAPLE) with kapton window. Three scans were carried out and the signal averaged. EXAFS extraction was performed in k2χ(k) with Athena code using a spline function. Data adjustment was performed with Artemis code (Demeter version 0.9.25) in R space after Fourier transform between 2.4 and 12.5 ˚ A-1. Adjustment was performed with one shell of Co-O using the FEFF6 code that is included in the Demeter version. k2-weigthed EXAFS spectrum and its Fourier Transform together with their fitting are plotted in Figure 2. XANES spectrum is plotted in Figure 7, it was recorded with an energy step of 0.5 eV in the edge. 2.4. Simulated XAS spectra Simulated Co K-edge EXAFS and XANES spectra have been computed using the FEFF 9.6 code . They have been built by averaging the individual spectra of 500 evenly distributed snapshots. The Hedin-Lundqvist exchange-correlation potential was used to compute the electron density distribution within the SCF approach. The cutoff radius used to select the extension of the hydration water shell around the Co absorber atom was large enough to include well beyond the second hydration shell ( Rcut = 8 ˚ A, see RDF in Figure 3). The ∆ E0 applied for the simulated EXAFS spectrum was chosen such as it matches the first EXAFS oscillation at 3.5 ˚ A−1. Previous examples of the methodology applied to other highly-charged metal cations in water can be found elsewhere [17, 18, 31, 32]. A sphere of 6 ˚ Aaround the cation was applied in the full multiple scattering formalism (FMS) for the XANES calculation.The same cutoff radius for selecting the extension of the hydration structure around the Co absorber atom than for EXAFS, i.e. 8 ˚ A, was used. The SCF procedure provides a very good estimation of the edge position, thus the simulated spectrum has been shifted only -2.0 eV to overimpose the white line with the experimental one. FEFF input files for the EXAFS and XANES calculations are given in the SM. 3Figure 2. (a) Co K-edge k2-weighted EXAFS spectrum (dotted line) and fit (blue line); (b) Modulus of Fourier Transform EXAFS spectrum (black line) and its fit (blue line). ( E0 = 7727 eV; inner correction potential, ∆ E0 = −2.9 eV; amplitude reduction factor, S20 = 0 .9; χ2 red = 4 .74; quality factor = 1.6%.) 3. Results and discussion Figure 2 shows the experimental Co K-edge EXAFS spectrum (top Figure, dotted line) and its fit (top Figure, blue line), as well as the Fourier Transform of them (bottom Figure). The spectrum is a rather simple function with a good signal/noise ratio up to 13.5 ˚ A−1. The fit of the spectrum is obtained with the use of only a first-shell formed by 6 water molecules with a Co-O distance of 2.08 ± 0.01 ˚ A and a Debye-Waller value (DW) of 0.0071 ± 0.0004 ˚ A2.The Co-O and Co-H RDFs for Co 2+ in water obtained by the MD simulation are shown in Figure 3. The Co-O RDF (black line) has two well-defined peaks with maxima at 2.09 ˚ Aand 4.24 ˚ A. The value for the Co-O I distance is similar to the value provided by the EXAFS fit, and within the range of values proposed in the literature. The Co-O II distance is shorter than the theoretical values previously proposed,[12, 15, 16] in the range 4.1-4.6 ˚ A, but larger than the experimentally reported from X-ray diffraction measurements, 4.05 ˚ A. The depletion zone between the first and second peak indicates there are no water exchanges between these hydration shells during the simulation time. The running integration numbers are 6 and ∼ 15 for the first and second shell, respectively. Average tilt angle of the first-shell water molecules is 35 ◦, what means that there is a significant trend to ion-dipole orientation, mediated by temperature effects and water interactions with second-shell water molecules. M-H RDF shows two maxima, one at 2.73 ˚ A and the other at 4.87 ˚ A, the integration of these peaks being ∼ 13.5 and ∼ 41.5. First-shell DW computed as the average of the mean-square displacements of R(Co-O I), [σ2 I =< (RI − RI)2 >], gives a value of 0.0082 ˚ A2, slightly higher than the value obtained from the EXAFS fit, and those experimental values previously reported, 0.0042-0.0062 ˚ A2.[10–12] Another interesting structural parameter is the eccentricity index, ε =< | ~RM n+ − ~RCM | >,. This index accounts for the assymmetry of the hydrated ion by calculating the average distance between the mass center of the first hydration shell and the central metal cation. For the Co 2+ hexahydrate, ε = 0 .11 ˚ A, a small value compared to what has been reported for other more labile ions, such as Rb + and Cs +, that reflects the robustness of this hydrated ion. The polarization of the closer water molecules by the cation presence is high in the first shell increasing their dipole moment by 0.7 D, whereas the increase in the second hydration shell is small, only 0.1 D (see Table 1). An additional structural information to understand the hydration structure around Co 2+ is the hydrogen bond (HB) network formed by the successive hydration shells. Total average of HBs per first-shell water molecule is 2.15, being formed exclusively between first and second shell water molecules. 89% of these HBs are donated by the first-shell molecules to the second-shell molecules and the rest are HBs accepted by the first shell molecules. No HBs 4Figure 3. Radial distribution function Cobalt-Oxygen and Cobalt-Hydrogen. among first-shell molecules are found. This can be understood on the basis of the average tilt angle, ∼ 34 ◦, that shows a marked ion-dipole orientation of the first shell water molecules, although fluctuations allow the occasional HB acceptor behavior of the first-shell water molecules(11%), and consequently more than two HBs per water molecule can be found. Experimental mean residence time (MRT) values for first-shell water molecules are of ∼1ms [34, 35]. These times are much longer than the simulation time, 1 ns, accordingly no water exchange was observed between first and second shell along the simulation. At this point it is worth noting that although the intermolecular potential built is based on the hydrated ion model, first-shell water molecules are treated as independent particles. This means that the water exchange is not prevented by any type of constraint but rather by the correct behavior of the employed potentials. The use of periodic boundary conditions causes an interparticle coupling with their neighboring images that reduces their motion and then their difusion coefficient. We have applied the Yeh and Hummer expression to correct D0 for a cubic cell. The corrected value obtained for Co 2+ is (0.6 ± 0.1) 10 −5 cm 2 s−1, this value is similar to what was determined experimentally, 0.732 ×10 −5 cm 2 s−1. This result on the Co 2+ dynamical behavior emphasizes the idea of the Co 2+ aqua ion as the species responsible of the ion mobility, since MRTs guarantees that the ion and its first hydration shell define the dynamic entity describing the correct diffusional behavior in solution. From the energetic point of view, the hydration enthalpy obtained in this work, -479 kcal/mol, agrees fairly well with the experimental data, -487 kcal/mol. Water reorientational times for the first and second order correlation functions associated to the motion of the dipole moment vector μ, the hydrogen-hydrogen vector HH, the normal vector to the molecular plane ⊥ and the oxygen-hydrogen axis OH have been computed. Reorientational times ( τ1 and τ2) have been calculated from the correlation functions defined in equations 1 and 2 C1,i (t) = 〈~ui(0) ~ui(t)〉 (1) C2,i (t) = 12 〈 3( ~ui(0) ~ui(t)) 2 − 1 〉 (2) ~ui is the unit vector representing any of the reference directions previously mentioned. The re-orientational correlation functions were calculated from a 1 ns simulation with a 0.01 ps time interval between structures. Re-orientational times of the first shell water molecules are shown in Table 2. Values for the MCDHO2 water molecule are also included. The relative re-orientational times τ1,n,f irst −shell /τ 1,n,bulk of this work are close to the relative values obtained by Rode and col. in their QM/MM simulation. The mobility restriction suffered 5Table 1. Properties of Co 2+ aqueous solution. Standard deviation in parenthesis. Property this work Literature EXAFS MD RM-O I (˚ A) 2.08 2.09 2.0, 2.10, 2.092 2.097, 2.07-2.14, 2.08[10, 38] 2.06, 2.17-2.27 CN I 6.0 6.0 6.0[9, 10, 12, 13, 16], 6.0-5.3, 6.7 6.5 DW (˚ A2) 0.0071 0.0082 0.012-0.015, 0.0038-0.0062, 0.0065 tilt angle I (◦) 34(4) 18 ε (˚ A) 0.11 RM-O II (˚ A) 4.24 4.05, 4.40, 4.10-4.28 4.6 CN II 15.0 8, 12.7, 12.48-13.18 15.9-22.7 tilt angle II (◦) 64(4) 37 ∆μI (D) 0.7(0.3) ∆μII (D) 0.1(0.3) MRT( t∗ = 0) (ps) ∞ ∞, ∼ 1×10 6 MRT( t∗ = 2) (ps) ∞ ∆Hhyd (kcal/mol) –479(18) –487 Do (10 −5 cm 2/s) 0.6(0.1) 0.732 by first-shell water molecules with respect to a bulk water molecule is clearly observed in both first- and second-order re-orientational times whose values for first-shell molecules are roughly three times longer than for those of bulk. However, in the case of τi,μ , i.e. C2 molecular axis, the ratio becomes higher than ten times. This shows how first-shell water molecules hardly modify the ion-dipole orientation. Table 2. First, τ1,n , and second order, τ2,n , re-orientational times of first shell water molecules with respect to the main rotational water axis ( μ, dipole moment; HH direction, ⊥, normal to molecular plane; OH direction τ (ps) Water Molecule τ1,μ τ2,μ τ1,HH τ2,HH τ1,⊥ τ2,⊥ τ1,OH τ2,OH 1st. shell 74 25 16 10 12 4 23 8Bulk 5 2 6 3 4 2 6 3 The power spectra of the Co 2+ aqua ion have been computed by means of the Fourier Transform of the velocity autocorrelation functions (FT-VAC) of cobalt and first-shell oxygen atoms derived from MD simulation. Figure 4 shows the power spectra of the aqua ion isolated, i.e. in gas phase, (top figure) and in aqueous solution (bottom figure). The hexahydrate power spectrum in gas phase presents two peaks at low frequency, 105 and 160 cm −1, associated to intermolecular bending modes, whereas at higher frequencies two bands, which can be associated to the Co-O stretching normal mode region, appear at ∼300 and 420 cm −1. Analysis of dynamical variables associated to internal coordinates combination indicates that the peak at ∼ 300 cm −1 corresponds to the E2g mode, and the symmetric breathing Ag mode corresponds to the band centered at 420 cm −1. The power spectrum in solution shows how the inclusion of second-shell interactions with the aqua ion widens the sharp gas-phase peaks. These solvent effects induce slight red-shifts of the bands leading to a symmetric stretching normal modes at ∼ 390 cm −1, value close to the experimental value derived from Raman measurements by Kano, ∼ 380 cm −1. Figure 5 shows the Co K-edge k2-weigthed EXAFS spectrum corresponding to the experimental measurement of the very dilute CoCl 2 aqueous solution and the simulated EXAFS spectrum derived from evenly-spaced snapshots taken from the Co 2+ MD simulation (red line). This simulated spectrum was calculated including all atoms, i.e. cobalt, oxygen and 6Figure 4. Power spectra of the cation (red line) and oxygen atoms of the Co 2+ aqua ion (black line) in gas phase (top) and in solution (bottom). 7hydrogen, in the SCF calculation of backscattering potentials and in the scattering phenomena, using a Rcutoff = 8 ˚ A for the snapshots. This criterium guarantees that all contributions of the second hydration shell to the scattering were included in the EXAFS. It is observed a general good agreement in the frequency and the intensity of the signal up to 13.5 ˚ A−1. Bearing in mind that the developed potential is based on quantum-mechanical calculations, the ensemble of structures contributing are completely independent of empirical information. This agreement gives confidence in the structural information derived from the intermolecular potential developed. Figure 5. k2-weighted Co K-edge EXAFS spectra of Co 2+ in aqueous solution: experimental spectrum (black dots) and simulated spectra including the first hydration shell (blue line) or the first and second hydration shells (red line). The good experimental-theoretical agreement allows us to revisit in this well-defined sample the role of the second shell water molecules and the hydrogen atoms acting as backscatterers. There are previous studies of ions in solution where the simulated EXAFS spectrum considered the inclusion of hydrogen atoms in the calculation, whereas in others it was considered that its inclusion exagerated the EXAFS signal intensity[40–42]. It has also been discussed about the capabilities of the EXAFS technique to characterize the second hydration shell.[12, 31, 41, 43, 44]. To study the role of the second hydration shell, we have computed the simulated spectrum including only the first hydration shell atoms in the calculation, a Rcutoff = 3 ˚ A was applied to the snapshots, then only the [Co(H 2O) 6]2+ structure was accounted for computing the spectrum (see blue line in Figure 5 ). There is a slight difference, with respect to the spectrum simulated with two shells, in intensity at low k values (2.5-4 ˚ A−1) but at higher k the spectra overlap. This subtle difference indicates that for a divalent aqua ion such as that of Co 2+ it is not possible to extract information about the second hydration shell, differing from what has been concluded in the case of stable and light trivalent transition metal cations.[31, 43, 45] To check hydrogen effects on the EXAFS signal we have computed the spectrum excluding the scattering paths involving hydrogen atoms (green line in Figure 6). The comparison with the whole simulated spectrum (red line in Figure 6) shows that both spectra almost overimpose . This indicates that the new formalism of FEFF9.6 code[30, 46] has improved the description of hydrogen backscattering contribution. Analysis of XANES spectrum becomes an additional and complementary information to that derived from EXAFS, particularly when it is possible to combine it with microscopic data obtained from an external source to the XAS technique, i.e. computer simulations based on statistical methods. Evenmore, a minimization procedure has been proposed to find a structure that reproduces the experimental XANES spectrum. Figure 7 shows the comparison between the Co K-edge experimental XANES recorded in this study and the simulated spectra derived from the snapshot set taken from the 1 ns MD simulation. The FEFF9.6 code was used for the computation of the XANES spectrum of each individual snapshot, considering the two already used Rcutoff values for the EXAFS computations (8 ˚ A and 3 ˚ A). It is worth pointing out that we are not trying to simulate the pre-edge 1 s → 3d forbidden transition. Because of the SCF nature of FEFF computations, the simulated spectra have been only shifted by ∼ -5 eV in order to overimpose the absolute position of the computed Co K-edge. It is remarkable the reproduction of the experimental XANES spectrum by the simulated ones in several items: the intensity of the white line and 8Figure 6. k2-weighted Co K -edge simulated EXAFS spectra of Co 2+ in aqueous solution: including (green line) or excluding (red line) hydrogen atoms as backscaterers in the computation. the second resonance, the coherent positions of their respective peaks and the presence of a small hump in between these two maxima. According to Natoli’s rule, where the energy gap between the white line and its second resonance is related to the absorber-backscaterer distance, we must conclude that the Co-O I distance provided by the MD simulation is strongly supported by independent experiments. Figure 7. K-edge XANES spectra of a Co 2+ aqueous solution, experimental spectrum (black dots) and simulated spectrum with the first hydration shell (blue line) or the first and second hydration shells (red line). A second interesting XANES feature is the small hump at ∼ 7737 eV. Figure 7 only shows slight differences between the spectrum considering one hydration shell (blue line) and that using two hydration shells (red line). The hump after the white line seems to be better reproduced by the one-shell spectrum, conversely the region above 7760 eV is better reproduced by the two-shell spectrum. In previous studies on some trivalent cations we had assigned this post-edge feature to the presence of a robust second hydration shell[31, 32, 48], as well as D’Angelo et al. assigned it in the case of Ni 2+ . One can observe that the 9inclusion of the second hydration shell mainly changes the hump region rising this feature, although a change of slope is already present when the computation only includes the first hydration shell. This suggests that the possibility of extracting second hydration shell structural information from XANES is a difficult matter. This information could be only extracted if a well-tempered experimental-theoretical synergy was applied. 4. Concluding Remarks Co 2+ -containing dilute aqueous solution has been studied from both experimental and theoretical techniques. On one side, Co K-edge EXAFS and XANES spectra have been recorded and analyzed by standard fitting procedure as well as by combining them with theoretical results derived from classical MD simulations. A new intermolecular potential to describe Co 2+ –H 2O interaction based on the exchangeable hydrated ion model has been developed. The potential was built taking as reference quantum-mechanical interaction energies. The potential performance has been tested against experimental and theoretical results, showing a fair description of energetic, structural, dynamical and spectroscopical properties. In addition to the well known hexacoordination of its first-shell, a Co-O I distance of 2.08 ±0.01 or 2.09 ±0.01˚ A is proposed depending on EXAFS fitting or MD simulation, respectively. From MD simulations, Co-O RDF shows a well-defined second peak corresponding to a second hydration shell composed by ∼ 15 water molecules centered at 4.24 ˚A. The fair agreement between the experimental EXAFS and XANES spectra with those simulated from the statistical average of a large number of MD snapshots leads to get strong confidence in the previous structural data. The use of cuttoff radii in the snapshots used for the simulated XAS spectra allows us to test the impact of the second hydration shell on EXAFS and XANES features. It is shown that for this divalent transition metal cation aqueous solution, in both the EXAFS and XANES cases this effect is almost negligeable, suggesting that the experimental detection of the second hydration shell could hardly be achieved. 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