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Our Rugby Mouth Guards are the best mouthpieces in the industry. Our custom mouth guards for rugby help keep you performing at the top of your game. Each rugby mouth guard is hand crafted for better breathability, maximum protection, and performance. All our rugby Mouth guards are made with 3 layers of heavy duty protection that all our pro mouth guards are made of and is between 3.5-5mm of thickness.
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{
"redpajama_set_name": "RedPajamaC4"
}
| 4,900
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Q: java object hierarchy, and passing objects to functions I have created a TreeMap like so:
TreeMap<Integer, ArrayList<MyClass>> wrap = new TreeMap<Integer, ArrayList<MyClass>>();
I have created a constructor like so:
public foo (TreeMap<Integer, Collection<Spot> > objects) {
this.field = objects;
}
However, eclipse gives me a red squigly when I use the constructor, with my wrap variable as the single parameter:
The constructor foo(TreeMap<Integer,ArrayList<Spot>>) is undefined
An ArrayList is a type of Collection...yes? So why is this not working?
A: Generics don't work as you think they do in this case.
What you need is something similar to:
public foo (TreeMap<Integer, ? extends Collection<Spot> > objects) {
this.field = objects;
}
The ? is called a wild card. It will allow you to pass in a Collection, or anything that extends/implements Collection.
The line ? extends Collection<Spot> reads like:
Something that extends a Collection.
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,671
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{"url":"https:\/\/www.nature.com\/articles\/s41467-021-26355-z?error=cookies_not_supported&code=22c3c688-acd0-4a1d-a0cd-b125784fd08d","text":"## Introduction\n\nStabilizing mean global temperatures requires a global transition to energy systems with near-zero (or net-negative) carbon dioxide equivalent emissions1,2,3. In cost-optimized scenarios that accomplish this transition, solar and wind resources often supply a large share (e.g., >60%) of electricity4,5,6,7,8,9,10. Designing and operating a highly reliable electricity system that is dependent on such large shares of wind and solar generation can be a challenge, however, due to the variable and uncertain nature of solar and wind resources11,12. The efficacy of meeting electricity demands with generation from solar and wind resources depends on factors such as location and weather; the area over which generating assets are distributed; the mix and magnitude of solar and wind generation capacities; the availability of energy storage; and firm generation capacity11,12,13,14,15,16. Meanwhile, reliability standards in industrialized countries are typically very high (e.g., targeting <2\u20133\u2009h of unplanned outages per year, or ~99.97%17). Resource adequacy planning standards for \u201c1-in-10\u201d are also high: in North America (BAL-502-RF-03)18, generating resources must be adequate to provide no more than 1 day of unmet electricity demand\u2014or in some cases 1 loss of load event\u2014in 10 years (i.e., 99.97% or 99.99%, respectively)19.\n\nHere, we present a systematic analysis of the ability of specified amounts of solar and wind generation to meet electricity demands in 42 major countries across a range of assumptions associated with transmission, energy storage, and generation amounts. In particular, we assess spatial and temporal gaps between electricity demand and the availability of solar and wind resources, which represent gaps that must be filled by other non-emitting generation technologies or operating strategies in reliable electricity systems based on zero-carbon sources. The complementarity of renewable energy sources for this study is defined as a hybridization of solar-wind resources over a given area (here, countries), which we estimate by the Kendall correlation coefficient of these resources across 39-years of resource data20. Our goal is to identify the opportunities, complementarity, and challenges of variable renewable resources in greater detail than can be done by integrated assessment models that have multi-year time steps. Our results do not account for realistic power system specifications. Rather, we examine fundamental geophysical constraints on wind- and solar-dominated power systems independent of cost estimates. Note that we do not mean to suggest that the temporal variability of such resources would ever make it physically impossible to meet a given electricity demand (with enough capacity the solar and wind resources would be able to meet demand), but rather the extent to which such variability may determine the economic or socio-political feasibility of reliable systems. Our results will thus continue to be informative even as technological and socio-political feasibility evolves.\n\nDetails of our analytical approach are in the \u201cMethods\u201d section. In summary, we use 39 years (1980\u20132018) of gridded (0.5\u00b0\u2009\u00d7\u20090.625\u00b0) and hourly reanalysis data21,22 and actual\/projected hourly electricity demand from a single recent year to evaluate the adequacy of solar and wind resources to meet electricity demand in each of 42 major countries (data sources and countries are listed in Supplementary Data\u00a01). First, hourly, area-weighted capacity factors for both solar and wind resources are calculated over each country (or region), assuming perfect transmission within the country or region. Then we exogenously specify (1) the mix of solar and wind generation, (2) the overall level of annual generation from these sources, and (3) the capacity of energy storage, and analyze the ability of the specified technologies to meet hourly demand. We analyze systems ranging from 100% solar (no wind) to 100% wind (no solar), in which total annual generation ranges from equal to annual demand (\u201c1x generation\u201d) to up to three times annual demand (\u201c3x generation\u201d), and in which available energy storage ranges from none (\u201c0\u2009h\u201d) to 12\u2009h of mean demand (\u201c12\u2009h\u201d). In addition, we simulate the impacts of different demands (i.e., demand load profiles) and technologies (i.e., single-axis and dual-axis solar tracking systems) on electricity system reliabilities as sensitivity tests. The number of countries, years of reanalysis data, and different system configurations we analyze require computation and analysis of ~300,000 year-long simulations.\n\n## Results\n\n### Resources and demand variability\n\nFigure\u00a01 shows the seasonal and daily variability of solar and wind resources and electricity demand in the six countries with the greatest electricity demand on every continents except Antarctica\u00a0(results from six other major countries and continent-level aggregated regions are shown in Supplementary Figs.\u00a01 and 2, respectively). Solar and wind consistently peak in summer and winter, respectively,\u00a0in countries of the Northern Hemisphere (seasons are reversed in countries of the Southern Hemisphere; Fig.\u00a01a\u2013f). The seasonal cycles of solar and wind thus suggest potential complementarity in many countries (e.g., China, Fig.\u00a01a; and Germany, Fig.\u00a01b). However, during the 39-year period, interannual variability of wind is consistently much greater than that of solar in most countries (Fig.\u00a01a\u2013f), though the magnitude of these resources\u2019 variability differs substantially between two particular countries. For example, Germany\u2019s small area (0.36\u2009million\u2009km2) and high latitude (centroid 51.2\u2009\u00b0N) result in large interannual variations in both solar (measured by the robust coefficient of variation23; RCoV\u2009=\u200958.8%) and wind resources (RCoV\u2009=\u200947.2%, Fig.\u00a01b), whereas solar resource variability is very low (RCoV\u2009=\u20096.6%) in the larger and tropical country of Brazil (8.52\u2009million\u2009km2 and centroid 14.2\u2009\u00b0S; Fig.\u00a01e). Wind resources are also more variable than solar resources on the time scale of days to weeks in each country, which acts to limit and undermine the resources\u2019 seasonal complementarity. Electricity demand profiles for each country are determined by factors such as economic conditions, prevailing weather conditions and consumer usage patterns24. Therefore, electricity demand for two countries can have unique seasonal shapes and a range of variabilities even if they have similar wind and solar resources. For example, seasonal variability of demand in France (RCoV\u2009=\u200914.4%; Supplementary Fig.\u00a01e) is greater than that in Germany (RCoV\u2009=\u20097.4%; Fig.\u00a01b), despite the countries\u2019 similar wind and solar resource profiles.\n\nDaily cycles of solar and wind resources in each country are also somewhat complementary. Wind power usually peaks at night and rarely falls to zero when resources are aggregated over an entire country. This daily cycle is not substantially different during the summer and winter months (comparing Fig.\u00a01g\u2013l with Fig.\u00a01m\u2013r). Thirty-four (of the 42) countries have higher average wind power availability during the nighttime than during the daytime. Solar power peaks in the middle of the day and drops off sharply to zero at dusk. The amplitude and duration of the daily cycles for solar power availability is consistently different during the summer and winter months across countries (Fig.\u00a01g\u2013l versus Fig.\u00a01m\u2013r). The daily cycle of solar resources is a barrier to realizing reliable solar-dominated electricity systems without energy storage and\/or complementary wind generation to meet demand during the hours when the solar resource is not available. In addition, given our assumption of single-axis solar tracking, available solar power tends to be flat for several hours around its daytime peak during the daily cycles (Fig.\u00a01g\u2013r), though in some countries (e.g., Germany, South Africa, Australia) there is a consistent dip near noon, perhaps related to our adjustments of the direct radiation (details in Supplementary Note\u00a01). Kendall\u2019s correlation coefficients of solar and wind resources in the 42 main countries range from \u22120.91 to \u22120.83 (see Supplementary Data\u00a02), another indication of good complementarity (where \u22121 is the best possible complementarity)20.\n\n### The most reliable generation systems\n\nThe colors in Fig.\u00a02 show the reliability of electricity systems (i.e., the average percentage of electricity demand that is met each year from 1980 to 2018) based only on solar and wind resources for 18 major countries (4 from each of Asia, Europe, Africa, and the Americas, and 2 from Oceania; horizontal axes of each panel), according to: the mix of solar and wind generation (vertical axes), the level of annual generation relative to annual demand (1x in left panels and 1.5x in right panels), and the capacity of energy storage relative to mean electricity demand (0, 3, and 12\u2009h in the first, second, and third rows of panels, respectively). Results for 24 other countries are shown in Supplementary Fig.\u00a03 and Supplementary Data\u00a03. Figure\u00a02a shows that without any excess annual generation or energy storage (assuming perfect national transmission), the most reliable mixes (white circles) of solar and wind generation could potentially meet 72\u201391% (average 83%) of electricity demand in these countries. Under these generation and storage assumptions, the most reliable solar-wind generation mixes range from 65 to 85% wind power (73% on average), with countries with substantial desert (like Algeria, Egypt, South Africa) favoring slightly more solar and less wind (65\u201370% wind) and with higher-latitude countries like Russia and Canada favoring more wind (85% wind; Fig.\u00a02a).\n\nAdding 3\u2009h of energy storage, but still without excess annual generation, increases the reliability so that the most reliable mixes (white circles) meet 78\u201393% (average 87%) of electricity demand. The share of solar generation in these most reliable mixes increases to 15\u201350% (36% on average; Fig.\u00a02b). However, the share of solar generation increases less, or even decreases, in higher-latitude countries like Russia, Canada, and Germany (Fig.\u00a02b). These trends continue as more storage is added, so that with 12\u2009h of energy storage and no excess annual generation, 83\u201394% (average 90%) of electricity demand is met with mixes of 10\u201370% solar power (49% on average; Fig.\u00a02c).\n\nIf generating capacities are instead increased so that annual generation exceeds annual demand in each country by 50% (i.e., 1.5x generation), but without energy storage, the most reliable mixes meet 83\u201399% (average 94%) of electricity demand. The 1.5x generation most reliable mixes are substantially more reliable than in the 1x generation systems but include more wind power: 70\u201390% wind power (78% on average; Fig.\u00a02d). These \u201coverbuilt\u201d systems are more reliable in all of these 18 countries than the systems with 12\u2009h of energy storage but no excess generation (Fig.\u00a02c). Adding energy storage to systems whose generation is 1.5x annual demand again increases both the system reliability (89\u2013100%, average 98%) and the share of solar generation (most reliable mixes have 10\u201360% solar power, 36% on average; Fig.\u00a02e, f).\n\n### The unmet demand\n\nThe scatter plots in Fig.\u00a03 show the relationships among reliability, energy storage, excess annual generation, and countries\u2019 land area for the most reliable solar-wind mixes of all 42 countries analyzed (see relationships with a log y-axis in Supplementary Fig.\u00a04). The linear fits in each panel show that solar-wind systems are generally less reliable in countries with smaller land areas (e.g., Fig.\u00a03a). Specifically, our results across countries indicate that the reliability of solar-wind systems that lack energy storage increases by 7.2% for every factor of 10 increase in land area; this relationship further suggests the improvement in system reliability that might be expected by expanding transmission systems within large countries. However, excess annual generation tends to alleviate the disadvantage of small country area more than energy storage (this can be seen by comparing the slopes of the linear fits in panels of Fig.\u00a03c and d). In addition, within each country, to compare the gains in reliability from excess annual generation and energy storage, a nonlinear function was fit to the reliability given the land area, the level of annual generation, and the capacity of energy storage (see Supplementary\u00a0Information). Our results indicate that a 10% increase in excess annual generation is equivalent to 3.9\u2009h of storage (Supplementary Note\u00a02).\n\nFigure\u00a03 also points to the nature of systems\u2019 unreliability: the color of bubbles indicates the average number of events in which there would be unmet demand in each of at least 24 contiguous hours (i.e., \u201clong-duration gaps\u201d). In systems that meet >95% of a countrie\u2019s demand, dozens of such long-duration gaps often remain each year (yellow and green circles). In some countries, excess annual generation reduces the number of such long-duration gaps more than adding 12\u2009h of energy storage (e.g., compare Sweden, Australia, Canada, and Russia in Fig.\u00a03c and d).\n\nFigure\u00a04 further characterizes the magnitude and duration of unmet demand in 16 major countries (removing two African countries from the 18 countries shown in Fig.\u00a01 for figure symmetry; in descending order of their land area), with curves showing gaps of different system configurations sorted by their magnitude and according to the number of hours each year that such a gap occurred (power supply gap represents the fraction of unmet demand to the total demand in that hour averaging over 1980\u20132018; see relationships with a log y-axis in Supplementary Fig.\u00a05). For example, the pale purple curves show that systems with no excess annual generation and 12\u2009h of energy storage consistently have gaps in which >50% of demand is unmet for >1000\u2009h per year (Fig.\u00a04). Pale green curves show that systems with 50% excess annual generation and 12\u2009h of energy storage may have much smaller and shorter gaps in some countries (e.g., <10% of demand unmet in fewer than 100\u2009h per year in Russia, China, and Australia), but the gaps may still be >20% of demand for tens of hours or more in countries with relatively large land areas (e.g., Canada, Brazil, India, and Mexico) and >60% of demand for several hundred hours per year in countries with smaller areas (e.g., France, Japan, Germany, New Zealand, the U.K., and South Korea; Fig.\u00a04). Indeed, in smaller countries, substantial gaps (>30% of demand for >20\u2009h per year; pale orange curves in Fig.\u00a04)\u00a0remain in systems even with 12\u2009h of energy storage and annual generation that is 3x annual demand.\n\n### Benefits from sharing resources of multiple nations\n\nWe also evaluate the reliability benefits of regional electricity interconnections whereby the solar and wind resources of multiple nations are pooled and shared, again assuming perfect transmission within these regions. The maps in Fig.\u00a05 present the effects of such spatial aggregation, showing the highest reliability of solar-wind generation with no excess annual generation or energy storage at the national level (Supplementary Data\u00a07; Fig.\u00a05a), as well as when a system is aggregated into 19 separate, contiguous multinational regions (Fig.\u00a05b; categorization in Supplementary Data\u00a04) and 6 continents (Supplementary Data 7;\u00a0Fig.\u00a05c). Each step produces substantial improvements in reliability, with >89.8% of hourly demand met everywhere when resources are aggregated at the continental level (Fig.\u00a05c). Figure\u00a05c also indicates the additional reliability gains in these systems that would be achieved as a consequence of specific intercontinental connections. Supplementary Fig.\u00a06 shows that the supply gaps in continental-scale solar-wind systems might be entirely eliminated in Africa, Asia, and South America, and limited to <2% of demand and 49, 26, and 13\u2009h in Europe, Oceania, and North America, respectively, given excess annual generation of 50% and 12\u2009h of storage. Substantial supply gaps remain for continental-scale systems when excess annual generation and energy storage are not available (Supplementary Fig.\u00a07).\n\n## Discussion\n\nOur results suggest that, neglecting transmission constraints, with systems sized to meet time-integrated annual electricity demand, major countries\u2019 solar and wind resources could meet at least 72% of instantaneous electricity demand without excess annual generation or energy storage. For instance, in the contiguous U.S., a solar and wind power system could provide ~85% of total electricity demand, which is consistent with the prior studies and reports12,25. Solar and wind resources can achieve greater levels of reliability by adding energy storage, increasing deployed capacities (i.e., generating electricity in excess of annual demand), or pooling resources of contiguous, multinational regions26. However, the marginal improvements in reliability related to these options differ considerably across countries and regions, according to their land area, location, and geophysical resources (Supplementary Figs.\u00a08 and 9).\n\nIn small, high-latitude countries, the highest reliability systems are usually wind-heavy (e.g., as high as 95% wind power), with particularly large reliability gains achievable by regional aggregation. In contrast, the most reliable systems in temperate\/tropical countries tend to include more solar. Meanwhile, the most reliable systems are not always the same systems that would minimize the frequency of long-duration (\u226524\u2009h) power supply gaps (Supplementary Fig.\u00a09). In general, more solar in the wind-solar mix reduces the frequency of long-duration gaps. Although reasonably high levels of reliability can be reached by solar-wind resources alone, the defining challenge of such systems are the longer-duration gaps, often associated with extreme weather episodes. For instance, historical solar and wind resources data in Germany reveal that there were nearly 2 weeks in which dispatchable generation had to cover practically all of the demand because of a period with very low solar and wind power availability (called \u201cdark doldrums\u201d)27. Although with vast enough wind and solar capacity it might still be possible to meet demand in all hours, the required capacity increases exponentially after a point that depends on the renewable resources of that country, and it is this geophysically-dependent point that will largely determine the cost-effectiveness of highly-reliable, renewables-based electricity systems. Although dispatchable fossil fuel generators with 100% effective carbon capture storage (CCS) could provide system reliability without emissions2, such underutilized and capital-intensive backup electricity would require higher investments and variable costs. In contrast, combustion turbines or combined cycle plants burning carbon-neutral biogas, syngas, or hydrogen might have comparatively low capital costs, but would require additional and large capital investments to produce such fuels (e.g., biodigestion, direct air capture, Fischer-Tropsch, and\/or electrolysis). Sector-coupling or right-sizing of these net-zero emissions fuel-production facilities could nonetheless make infrequent operation of generators feasible28. More firm generation would mean less solar and wind capacity in a given system, which might or might not be cost-effective depending on technology costs. But many jurisdictions and advocates are interested in \u201cmaxing out\u201d solar and wind. Our results are especially relevant in that context, highlighting the implications of country-level differences in the variability of solar and wind resources, including how much storage and firm generation might be required to ensure resource adequacy. Although our\u00a0methods are simple and transparent, our goals and findings are remarkably consistent with much more complex approaches. For example, the recently published Net-Zero America report includes a cost-optimized \u201call-renewables\u201d scenario which decarbonizes U.S. electricity without nuclear or CCS: by 2050, ~81.6% of primary energy in the E\u2009+\u2009RE\u2009+\u2009scenario is from solar and wind29.\n\nOur analysis has important limitations and uncertainties. To improve the generality of our results, our analysis focuses exclusively on geophysical constraints and does not consider economic feasibility. As noted throughout, our reliability estimates are a best case given the assumption that electricity can be transmitted losslessly throughout a region of interest. Also, we use area-weighted averages for solar and wind generation potential without regard to existing protections or uses. This use of area-weighted averages affects our estimates in two important ways. First, our estimates may include areas where currently\u00a0generation cannot be sited. Second, our derivation of solar and wind capacity factors implies uniform distribution of wind and solar generation technology (i.e., a horizontal single-axis tracking system applied in this work), which does not allow us to select locations with particularly high capacity factors or to strategically select a set of locations whose generating potential is mutually negatively correlated. This second point has the effect of making our estimates for the efficacy of solar and wind resources to meet electricity demand more conservative by using the horizontal single-axis tracking system compared to the dual-axis solar tracking systems. For this case, dual-axis solar tracking systems are added to test the impacts on the system reliabilities (see Supplementary Note\u00a03), we find that different solar tracking systems have very small impacts on the electricity system reliabilities and the reliability change ratios are within \u00b15% under the 1x generation system and less sensitive under 3x generation system (Supplementary Fig.\u00a010). However, either method to calculate capacity factors of national and regional area-weighted averages may also reduce the resource variability and thereby increase estimates of reliability. Third, hourly variations of solar and wind capacity factors in the reanalysis data MERRA-2 we used may be biased. A new analysis based on a different and independent reanalysis product, ERA530,31, has been added and compared to\u00a0the original\u00a0results (see Supplementary Note 3 and Supplementary Figs.\u00a011-12). Our estimates of the system reliabilities by using ERA5 data in the 42 major countries are in good agreement with results of MERRA-2: under 1x generation and the most reliable mixes without storage, reliability under the different loads varies on average from \u22129.4 to 1.3% (see Supplementary Fig.\u00a09a). The differences are similar in systems with excess generation (Figs. S11b-c). We also compared the magnitude and duration of unmet demand in 16 major countries like Fig.\u00a04 (see Supplementary Fig.\u00a012). The data products of MERRA-2 and ERA5 both can essentially capture the number of hours each year that such a gap occurred. By contrast, the MERRA-2 data has a better performance of meeting hourly demand in larger countries (i.e., Russia and Canada) but a similar performance in small countries (i.e., United Kingdom). The somewhat different patterns of resource variability in the two datasets do not alter our main conclusions.\n\nOur estimates show that the marginal reliability benefit of increased capacity of storage or increased overbuild of wind\/solar declines steadily. Under a given capacity of energy storage (e.g., 3\u2009h), our results of 1x, 1.5x, and 3x generation show that the first 10% excess generation increase is larger than the next 10% excess generation increase (i.e., the marginal benefit for system reliability decreases as excess generation increases). As might be expected, the diminishing marginal benefits between excess generation and increased storage apply in both directions. Our fitting model performs well across different nations, overbuild levels, and storage levels. The differences in reliability between the estimates and the model predicted values are between \u22125.5 and 5.8% and ~80% of the differences are within \u00b12%, with no systematic bias related to region or the magnitude of overbuild or storage. Nonetheless, our model and conclusions are limited by our experimental design and the discrete levels of excess generation (1x, 1.5x, and 3x) and storage (0, 3, and 12\u2009h) we evaluated.\n\nWe compare the reliability improvements obtainable by energy storage, excess capacity, and regional aggregation but not the relative costs of the different options. For example, the energy storage capacities we consider are in some cases quite large: energy storage equal to 12\u2009h of mean electricity demand in the contiguous U.S., Germany, and Japan represents 5.6, 0.7, and 1.4 TWh, respectively (Supplementary Data\u00a05). These combined storage capacities represent ~35 times the capacity of Li-ion batteries produced globally to date32, and more than 200 times the pumped hydro storage capacities that now exist in those countries33. The feasibility of 12 or more hours of energy storage may depend on continued innovation and learning related to the associated materials and technologies34,35,36,37. Similarly, the feasibility of pooling solar and wind resources over national or multinational regions may depend on both technological advances that reduce the costs, losses, and risks of power transmission38,39,40 as well as shifts in the socio-political support for such infrastructure41,42. In addition, setting up purely solar and wind supplied electricity systems requires a large number of solar panels and wind turbines to be installed, and we do not incorporate the impacts or interactions (e.g., wakes) from these hypothetical build-outs. Last, in this work, only 1-year of demand data is employed to assess the geophysical constraints of 39-year solar and wind resources. On one hand, we understand that the hourly patterns of countries electricity demand will of course change over time with changes in population, economic activities, power generation structure, and technology. For example, stronger positive correlation between solar\/wind availability and demand may be observed as renewable energy gradually dominates the power system. However, our analysis compares resources and demand in different years and at the country-level, which should preclude any bias related to specific subnational weather events. On the other hand, electricity demand profile may also dramatically change with future high electrification. We therefore perform additional analysis using the demand pattern from the future high electrification scenario (i.e., combining the high electrification scenario and rapid technology advancement)43 and use the results to discuss the sensitivity of our results to such different load profiles (see Supplementary Fig.\u00a013), and the results of this test for the U.S. show that reliability is not especially sensitive to the high electrification demand profile: under 1x generation without storage, reliability under the different renewable mixes varies on average from \u22121 to 2.5%. The differences are even smaller in solar-heavy systems with excess generation. In addition, we test the sensitivity of our results to such changes in demand by simulating the reliability of U.S. resources in meeting current loads from each other region (see Supplementary Note\u00a04). These tests show that reliability is not especially sensitive to demand profiles: under 1x generation and the most reliable mixes without storage, reliability under the different loads varies on average from \u22129 to 2% (see Supplementary Fig.\u00a014a). The differences become even smaller in systems with excess generation (Supplementary Fig.\u00a014).\n\nDespite these simplifying assumptions, our results offer insights from those provided by multi-year time step integrated assessment modes (IAMs) or hourly, cost-optimized energy system models. Specifically, hourly resolution over several decades allows us to evaluate the adequacy of regional solar and wind resources independent of costs. For example, cost-optimizing models which either require renewables sources to meet a very high share of demand or else assume extremely cheap renewable costs generally find substantial increases in system costs related to, e.g., energy storage. Our geophysically-focused results help to explain such results irrespective of cost assumptions. Indeed, we compare the estimates of reliability and capacities in this study with several techno-economic studies that have used independent approaches to model regional solar- and wind-dominated electricity systems in detail29,44,45. In each case, focusing on the U.S., these studies find that the share of non-emitting (or carbon neutral) electricity contributed by solar and wind in cost-optimized systems is typically ~80%, with the residual demand for non-emitting generation met by firmer renewables such as biomass, hydroelectricity, and geothermal29,44,45.\n\nVariable solar and wind energy are projected by many to be the dominant sources of electricity in net-zero emissions energy systems of the future. With solar and wind capacities sized such that total annual generation meets total annual demand, seasonal and daily complementarities of these resources make them capable of meeting three-quarters of hourly electricity demand in larger countries. Increasing the share of demand that can be met by solar and wind generation will require either \u201coverbuilding\u201d (i.e., excess annual generation), the introduction of large-scale energy storage, and\/or aggregating resources across multinational regions (Supplementary Data\u00a06). We highlight the geophysical considerations related to these options, but economics and geopolitics will also strongly influence which strategies are ultimately adopted and are therefore important topics for further research. Our analysis for the 39-year record of solar and wind resources is in part to obtain a statistically significant analysis of interannual variability and rare events (such as prolonged storms). Establishing estimates for interannual variability and the frequency of rare events that impact solar and wind generation potential is important when considering the lifetime of the capital asset stock in an electricity grid and requires the use of many years of data. Our normalized analysis of the reliability for purely solar and wind supplied electricity system would apply as well to a system with other slowly time-varying generation (e.g., coal, hydro, geothermal, or nuclear) because the variability of solar and wind generation and related long-duration gaps in electricity supply will have to be managed either by ramping backup technologies up and down or by curtailing excess solar and wind generation. Our results reveal national and regional differences in solar and wind resources that may be useful to decision makers and researchers prioritizing their investments in pursuit of reliable and cost-effective electricity systems based predominantly on solar and wind energy.\n\n## Methods\n\n### Hourly solar and wind capacity factors\n\nThe capacity factor describes the actual energy output as compared to the systems\u2019 rated energy output (power capacity multiplied by 1\u2009h)12. To calculate the wind and solar capacity factors for this study, we first obtained the hourly climatology data from the Modern-Era Retrospective analysis for Research and Application, Version-2 (MERRA-2) reanalysis product, which spans 39 years (1980\u20132018) and has a horizontal resolution of 0.5\u00b0 by latitude [\u221290\u201390\u00b0] and 0.625\u00b0 by longitude [\u2212180\u2013179.375\u00b0] with 361\u2009\u00d7\u2009576 grid cells worldwide21,22. Here we used the surface incoming shortwave flux [W\u2009m\u22122] (variable name: SWGDN), top-of-atmosphere incoming shortwave flux [W\u2009m\u22122] (variable name: SWTDN), and surface air temperature [K] (variable name: T) for deriving solar capacity factors; and wind speed at 100\u2009m [m\u2009s\u22121], estimated based on wind speed at 10\u2009m and 50\u2009m (variable names: U10M, V10M, U50M, and V50M) and a power-law relationship, to derive wind capacity factors. Wind and solar capacity factors were calculated for each grid cell and each hour. Each raw data point (an hourly energy density (solar) or wind speed (wind) value at a specific location and time) was then converted into the corresponding capacity factor based on the following procedures.\n\nFor solar capacity factor, we first calculated the solar zenith angle and the solar incidence angle for each grid based on the latitude\/longitude location and local time46,47, and then estimated the in-panel solar radiation48. Here we separated the direct and diffuse solar radiation components based on an empirical piecewise model49 that takes into account both ratios of surface to top-of-atmosphere solar radiation (i.e., the clearness index) and the local zenith angle. We assumed a horizontal single-axis tracking system (north-south direction) with a tilt of the solar panel to be 0\u00b0 and a maximum tuning angle of 45\u00b0. Solar power output from a given panel is calculated using the performance model described by Huld et al.50 and Pfenninger and Staffell51, which considers both the surrounding temperature and the effect of irradiance. It is noted that we assumed the single-axis trackers for calculating solar capacity factors, which may be unsuitable enough for the small countries such as Japan, South Korea, and United Kingdom. These small countries do not have enough uncommitted land area for that and are going very likely to have to favor no tracking with rooftop photovoltaic system. Therefore, we further assessed the impacts of different solar tracking systems (i.e., single-axis and dual-axis for both a horizontal and a vertical axis) on the electricity system reliability. The detailed comparisons are shown in the SI (Supplementary Note\u00a03 and Supplementary Fig.\u00a010).\n\nFor wind capacity factor, by assuming a wind turbine hub height of 100\u2009m, the raw wind speed data is first interpolated to 100\u2009m by employing a power law, based on wind speed at 10 and 50\u2009m. The 100-m-height wind speed is estimated by employing the following Eqs. (1) and (2):\n\n$$\\alpha =\\frac{{{{{{\\rm{log }}}}}}({U}_{50,i})-{{{{{\\rm{log }}}}}}({U}_{10,i})}{{{{{{\\rm{log }}}}}}\\left(50\\right)-{{{{{\\rm{log }}}}}}(10)}$$\n(1)\n$${U}_{100,i}={U}_{10,i}* \\left(\\frac{100}{10}\\right)^{\\alpha }$$\n(2)\n\nwhere $$i,\\alpha$$ represent grid and alpha exponent for wind profile, and $${U}_{10},{U}_{50},$$ and $${U}_{100}$$ represent wind speed at 10, 50, and 100\u2009m.\n\nThe wind capacity factor calculation employed a piecewise function consisting of four parts: (i) below a cut-in speed ($${u}_{{ci}}$$) of 3\u2009m\u2009s\u22121 the capacity factor is zero, (ii) between the cut-in speed of 3\u2009m\u2009s\u22121 and rated speed ($${u}_{r}$$) of 12\u2009m\u2009s\u22121 the capacity factor is $${{u}_{{ci}}}^{3}\/{{u}_{r}}^{3}$$, (iii) between the rated speed of 12\u2009m\u2009s\u22121 and the cut-out speed ($${u}_{{co}}$$) of 25\u2009m\u2009s\u22121 the capacity factor is 1.0, and (iv) above the cut-out speed of 25\u2009m\u2009s\u22121 the capacity factor is zero12,52,53. The process yielded the solar and wind capacity factors for each grid cell and each hour.\n\nAn area-weighed mean hourly energy generation profile was created for the solar and wind resources individually for each region of interest. In this work, hourly solar and wind capacity factors for 168 countries\/regions were produced. Capacity factors derived from reanalysis data were known to differ from real-world systems12,54, and thus these calculated capacity factors from the reanalysis dataset were rescaled. That is, the reanalysis data were used herein only for reflecting the temporal and spatial characteristics of the resource. For consistency, we normalized capacity factor values using the 25th percentile calculated capacity factor data for a region of interest due to data availability of real-world wind and solar capacity factors from public datasets or reports for all the countries and regions of interest. Our estimates represent real-world wind and solar capacity factors that are in good agreement with available observational data55. We then obtained the time-series hourly normalized wind and solar capacity factor dataset at the country\/region level.\n\n### Country-level hourly electricity demand data\n\nIn this work, country-level hourly electricity demand data were estimated in various ways, such as from government and electricity market websites, public power systems datasets, and previous studies (Supplementary Data\u00a01). As shown in Supplementary Data\u00a01, we compiled 168 countries and regions\u2019 demand data, including real-word hourly demand data of 62 countries and regions, and projected hourly demand data of the rest due to data availability56. Toktarova et al. developed a multiple linear regression model to project electricity demand in hourly resolution for all countries globally by incorporating 57 real load data profiles of diverse countries to analyze the cyclical pattern of the data. In addition, given the different self-consistent continuous gapless time series of hourly electricity demand among different countries and regions, a single latest year of hourly electricity demand data was used in our following simulations to investigate the impact of diversity of solar and wind resources across years on power system reliability for countries and regions with available real-world electricity data. For the rest of the countries and regions, we chose the hourly electricity demand data of the most recent year of future (i.e., the year of 2020) from the projection model56. We herein obtained the country-level demand dataset by joining the 1-year hourly demand data together 39 times to form a 39-year record consistent with the resource data. In addition, for regional demand data, we combined all the demand data of available countries within the according region at hourly scale to represent the temporal characteristics.\n\n### Simulation design\n\nA set of forward simulations were performed to track the ability of wind, solar installed capacity, and energy storage, if present, to meet demand in every hour. In this study, we used a Macro Energy Model (MEM), which is developed for optimizing electricity system (or electricity and fuels) without considering any spatial variation, policy, capacity markets57. Without considering any power system cost, generation technology, and transmission loss, we modeled the idealized hourly power supply process through dispatching wind and solar energy, as well as charging or discharging of storage, if present. Here we specified the wind and solar installed capacity, and storage capacity under the various capacity mixes of solar and wind fractions (i.e., every 5% change of solar fraction from 0% solar and 100% wind to 100% solar and 0% wind) and different levels of excess annual generation (i.e., 1x, 1.5x, and 3x generation) and energy storage (i.e., maximun\u00a03 and 12\u2009h of charging time) assumptions. The installed capacities for solar, wind, and storage for individual countries\/regions are estimated using the Eqs. (3)\u2013(5).\n\n$${{{{Capacity}}}}_{{{{solar}}},y}={{{{{\\rm{SF}}}}}}\\times {{{{{\\rm{OB}}}}}}\\times {{{{{{{{\\rm{Pwr}}}}}}\\_{{{{{\\rm{avg}}}}}}}}}_{y}\\times {{{{{{\\rm{Hrs}}}}}}}_{y}\/\\mathop{\\sum}\\limits_{y}{{{{{{\\rm{CF}}}}}}}_{{{{solar}}}}\\qquad$$\n(3)\n$${{{{Capacity}}}}_{{{{wind}}},y}=(1-{{{{{\\rm{SF}}}}}})\\times {{{{{\\rm{OB}}}}}}\\times {{{{{{{\\rm{Pwr}}}}}}}_{{{{{{\\rm{avg}}}}}}}}_{y}\\times {{{{{{\\rm{Hrs}}}}}}}_{y}\/\\mathop{\\sum}\\limits_{y}{{{{{{\\rm{CF}}}}}}}_{{{{wind}}}}$$\n(4)\n$${{{{Capacity}}}}_{{{{{{{\\mathrm{storage}}}}}}},y}={{{{{{{{\\rm{Pwr}}}}}}\\_{{{{{\\rm{avg}}}}}}}}}_{y}\\times {{{{{{{{\\mathrm{Bat}}}}}}}}}_{s}\\qquad\\qquad\\qquad\\qquad\\qquad\\quad$$\n(5)\n\nwhere $$y$$ and $$s$$ represent the year and size, respectively. $${{{{Capacity}}}}_{{{{solar}}}}$$, $${{{{Capacity}}}}_{{{{wind}}}}$$, and $${{{{Capacity}}}}_{{{{{{{\\mathrm{storage}}}}}}}}$$ represent the solar, wind, and storage capacities, respectively. $${{{{{\\rm{SF}}}}}}$$ represents the fraction of energy generated from solar (from 0 to 100% at intervals of 5%); $${{{{{\\rm{OB}}}}}}$$ represents the overbuilding of capacity, equaling 1, 1.5, or 3; $${{{{{\\rm{Pwr}}}}}}\\_{{{{{\\rm{avg}}}}}}$$ represents the mean power demand; $${{{{{\\rm{Hrs}}}}}}$$ represents the total hours in the year; $${{{{{\\rm{CF}}}}}}_{{solar}}$$ and $${{{{{{\\rm{CF}}}}}}}_{{{{wind}}}}$$ represent normalized capacity factors of solar and wind, respectively. And $${{{{{\\rm{Bat}}}}}}$$ represents battery storage, equaling 0 (i.e., no storage), 3, or 12.\n\nWhen storage was assumed to be available, we assumed the initial status of storage was the same as the final status for each year, which means the charging and discharging process is balanced. We also assumed a storage charging round-trip efficiency and storage decay rate of, respectively, 90% and 1.14\u2009\u00d7\u200910\u22126 per hour (i.e., 1% of stored electricity lost per month)2, reflecting the high-end performance of current batteries58,59. Dispatchable energy used to charge a battery (called the maximum hourly storage charging) was no more than the storage power rating, equaling storage capacity divided by storage charging time.\n\nGiven the restriction of computing resources, we chose ten major countries by comprehensively considering the electricity demand and growth domestic product (GDP) from each continent except Oceania (i.e., Asia, Africa, Europe, and the\u00a0America), within which only two main countries were selected (i.e., Australia and New Zealand). For each main country, 21 sets of the solar and wind mix from 0% solar and 100% wind to 100% solar and 0% wind with 5% change under 3 groups of overbuilt (1x, 1.5x, and 3x generation) and 3 groups of storage (no storage, 3\u2009h, and 12\u2009h of storage) were simulated, totaling 7938 simulations for all the main countries. To investigate the ability to supply power at multinational regions, continental, and intercontinental scales, we further applied the same simulation design for the main countries to multinational regions, continents, and multi-continental regions (Supplementary Data\u00a04). In addition, except the abovementioned main countries, 103,194 one-year simulations consisting of 21 sets of the solar and wind mix with no excess generation or energy storage, were added for each of the remaining 126 countries worldwide.\n\n### Hourly electricity supply process\n\nFor only solar-wind electricity systems without storage, in a given hour, the MEM model estimates the ability of power to be produced by assessing whether dispatchable solar and wind energy is no less than electricity demand. Excess solar and wind energy can be curtailed due to no available storage. 100% reliability results if the solar and wind power supply system can meet all the electricity demand in every hour of the simulation.\n\nWhen storage is assumed to be available in a given hour, if the solar and wind energy could meet the electricity demand, storage would be charged with excess solar and wind generation, if available, until the storage is full under the constraint of the maximum hourly storage charging, after which solar and wind energy can be curtailed. In contrast, if wind and solar energy cannot meet electricity demand, storage would be discharged to fill the power supply gap until storage is emptied or the power supply gap is filled.\n\nHere, we define reliability assuming electricity systems use only wind\/solar\/storage resources to meet current demand for electricity services. If one allows for other backup electricity (e.g., using natural gas with or without CCS), then issues of reliability with excess annual generation and\/or storage are largely moot.","date":"2023-02-05 05:57:31","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 2, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5508375763893127, \"perplexity\": 2198.905438426839}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-06\/segments\/1674764500215.91\/warc\/CC-MAIN-20230205032040-20230205062040-00628.warc.gz\"}"}
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Q: Can I pause itertools on python, and resume later? I need to create a list of strings with all the possible combinations of all letters uppercase and lowercase, with non repeating characters, of lenght 14, this is massive and I know it will take a lot of time and space.
My code right now is this:
import itertools
filename = open("strings.txt", "w")
for com in itertools.permutations('abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ', 14):
filename.write("\n"+"1_"+"".join(com)+"\n"+"0_"+"".join(com))
print ("".join(com))
pretty basic, it does the job and I have not found a faster way as of yet (tried a java algorithm I found that seemed faster but python was faster)
Since this will take a long time, from time to time I need to turn off my computer, so I need to be able to save somewhere where I left and continue, else I will start from the beginning each time it crashes/turn off my pc / anything happen.
Is there any way to do that?
A: You can pickle that iterator object. Its internal state will be stored in the pickle file. When you resume it should start from where it left off.
Something like this:
import itertools
import os
import pickle
import time
# if the iterator was saved, load it
if os.path.exists('saved_iter.pkl'):
with open('saved_iter.pkl', 'rb') as f:
iterator = pickle.load(f)
# otherwise recreate it
else:
iterator = itertools.permutations('abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ', 14)
try:
for com in iterator:
# process the object from the iterator
print(com)
time.sleep(1.0)
except KeyboardInterrupt:
# if the script is about to exit, save the iterator state
with open('saved_iter.pkl', 'wb') as f:
pickle.dump(iterator, f)
Which results in:
>python so_test.py
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'o')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'p')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'q')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'r')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 's')
>python so_test.py
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 't')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'u')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'v')
('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'w')
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,978
|
{"url":"https:\/\/www.semanticscholar.org\/paper\/On-shrinking-targets-and-self-returning-points-Kirsebom-Kunde\/4c43cad30e1ac96489e4dbeac9aacd0a618a3e5e","text":"Corpus ID: 211817963\n\n# On shrinking targets and self-returning points\n\n@article{Kirsebom2020OnST,\ntitle={On shrinking targets and self-returning points},\nauthor={M. Kirsebom and P. Kunde and T. Persson},\njournal={arXiv: Dynamical Systems},\nyear={2020}\n}\n\u2022 Published 2020\n\u2022 Mathematics\n\u2022 arXiv: Dynamical Systems\n\u2022 We consider the set $\\mathcal{R}_\\mathrm{io}$ of points returning infinitely many times to a sequence of shrinking targets around themselves. Under additional assumptions we improve Boshernitzan's pioneering result on the speed of recurrence. In the case of the doubling map as well as some linear maps on the $d$ dimensional torus, we even obtain a dichotomy condition for $\\mathcal{R}_\\mathrm{io}$ to have measure zero or one. Moreover, we study the set of points eventually always returning and\u2026\u00a0CONTINUE READING\n1 Citations\n\n#### Figures from this paper\n\nDynamical Borel-Cantelli lemma for recurrence theory\n\u2022 Mathematics\n\u2022 2020\n\u2022 Highly Influenced\n\u2022 PDF\n\n#### References\n\nSHOWING 1-10 OF 16 REFERENCES","date":"2021-02-28 19:53:21","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8405610918998718, \"perplexity\": 3070.083319205381}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-10\/segments\/1614178361723.15\/warc\/CC-MAIN-20210228175250-20210228205250-00251.warc.gz\"}"}
| null | null |
Hi, I'm looking for a partner who's mostly willing to work on a project for the love of the game. I have a steady job, and can easily pay for most of the things needed, but someone who's ready to help a little financially would be of great help to get the project moving. We would of course share the donations within the staff, if we ever decide to implement any, which would only be statless hats for the only purpose of looking good! I already have a whole server setuped on Digital Ocean and I also own a Web Host at StableHost. We really need someone dedicated and who's comfortable doing a little bit of everything: Basic scripting Basic scripts modifications Basic SRC files modifications We also need someone who's ready and motivated to learn how to use platforms such as: Ubuntu linux droplet, on which the servers runs. You only need to learn some basic commands which I can teach you in 5 minutes. Github, to modify files right from the server's repository instead of directly on the SFTP, where the server files are hosted. Bitvise, which is used to access the sftp for the server files, the SSH window to access ubuntu on which you use the commands to start, stop the server and many other things. Also, the web host to modify any files related to the website. I'm very patient and open minded, I will gladly guide you through it all if you don't know how to use any this. If you're interested or you know someone who might, you can contact me on Discord: Sirique#9034 & If you don't use Discord just PM here or send me a good old email at [email protected] Thank you for your time and hopefully you'll be interested to join me on this great adventure that is the making of an awesome Ragnarok Online Private server!
Here at EggRO we're looking for many staff positions to be filled, but for now, let me tell you a little more about our project entitled "EggRO". First of all, we are well informed that RO is not as huge as it used to be. There's clearly predominant pre-renewal lr servers right now, but our goal is to make an impact by doing things differently with the good old memories we all hold dear of our first steps as RO players. "Just like when my mama used to cook my eggs on a cold december morning..." That's a good memory right?! This is exactly where we want to hit as hard as we can... Nostalgia of course... Not my lovely mama! We have a lot of ideas and we want to share them with our whole staff. There's no games of hide and seek in our team, everyone will be well aware of everything we plan to do. Here's some information you might find useful about what we've done so far: 1. I used Ubuntu 16.04.3 x64 on DigitalOcean to setup the server. 2. We have a full functional RO folder that match our server's setup. 3. We also have a professional website and thor patcher's design. As for the staff positions we're looking to fill, they're are the following: -Event GMs -Experienced scripter -Someone with developing experiences to support me -Support GMs -Someone to make/manage our wiki and facebook page. (It can wait) **Right now we mostly want to have a small group to build this server from the ground up and then expand the staff when the server is up and running at his full potential. Of course if you want to be an Event GM or Support GM, feel free to apply, we'll keep you informed of everything our developing team is working on. We are very dedicated and we want to make this project come to fruition. We try to be as active as much as we can, because we know from past experiences that when you to start a project like that and that your partners suddenly transform into mute ghosts, it can be very frustrating. For more information feel free to contact me on skype: My email is [email protected] If we decide we're on the same page I'll grant you access to our staff Discord's page.
Hi everyone, First, I need to know what client are stable for Classic RO which only have Knight, Priest and etc job only and lvl 99/50. Second, I need to know is that any problem if I'm using latest Hercules server file? Or I need to do my own modification for making Classic RO server? Third, If anyone have classic server file, could someone share it here? Thank You.
Hello. I'd like to request for a classic quest npc please. maybe around 5 npc's if possible or any. Thanks in advance!
I am looking for a classic Hercules package that all NPCs, maps, monsters, etc. are in classic time. Not sure which episode (guess around 8.2) it was but there was no high class and third class (only upto 2-2). Where can I find those packages? Thank you!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 3,006
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{"url":"http:\/\/stats.stackexchange.com\/tags?page=4&tab=popular","text":"# Tags\n\nA tag is a keyword or label that categorizes your question with other, similar questions. Using the right tags makes it easier for others to find and answer your question.\n\n Type to find tags:\n matrix\u00d7\u00a0180 a rectangular array of numbers, symbols, or expressions arranged in rows and columns. The individual items in a matrix are called its elements or entries. terminology\u00d7\u00a0179 Indicates questions asking about the use and meaning of specific technical words\/concepts in statistics. sample\u00d7\u00a0178 a subset of a population. Statistics, in general, is concerned with using samples to make inference about the parameters governing a larger (possibly infinite) population. books\u00d7\u00a0175 a generic tag for requests of any kind of resources: books, textbooks, manuals, papers, presentations, video lectures, scripts, etc. multicollinearity\u00d7\u00a0168 Multicollinearity means predictor variables are correlated with each other, making it harder to determine the role each of the correlated variables is playing. Mathematically, it means the standard er\u2026 random-generation\u00d7\u00a0168 The act of generating a sequence of numbers or symbols randomly, or (more often) pseudo-randomly; i.e., with lack of any predictability or pattern. power-analysis\u00d7\u00a0168 An inquiry into the quality of a statistical test by calculating the power - the probability of rejecting the null hypothesis given that it is false - under certain circumstances. Power analysis is of\u2026 model\u00d7\u00a0166 a formalization of relationships between variables in the form of mathematical equations. The model is statistical as the variables are not deterministically but stochastically \u2026 k-means\u00d7\u00a0163 a family of cluster analysis methods in which you specify the number of clusters you expect. This is as opposed to hierarchical cluster analysis methods. meta-analysis\u00d7\u00a0162 Meta-analysis refers to methods focused on contrasting and combining results from different studies, in the hope of identifying patterns among study results, sources of disagreement among those result\u2026 assumptions\u00d7\u00a0160 Refers to the conditions under which a statistics procedure yields valid estimates and\/or inference. E.g., many statistical techniques require the assumption that the data are randomly sampled in some\u2026 normality\u00d7\u00a0160 Refers to the normal distribution, the Gaussian continuous probability distribution. reliability\u00d7\u00a0158 said to have a high reliability if it produces similar results under consistent conditions. DO NOT confuse reliability with validity (see tag wiki). aic\u00d7\u00a0157 AIC stands for the Akaike Information Criterion, which is one technique used to select the best model from a class of models using a penalized likelihood. A smaller AIC implies a better model. binary-data\u00d7\u00a0156 In broader sense - synonym of \"dichotomous data\": any data that can take on only one of two values. In narrower sense - dichotomous data coded as 1 or 0; furthermore, sometimes \"1\" is supposed to mean\u2026 sem\u00d7\u00a0154 a multivariate technique popular in social sciences. It is based on formulating a set of linear relations between variables, some of which may be latent, and estimating\u2026 dimensionality-reduction\u00d7\u00a0153 Dimensionality reduction refers to techniques for reducing many variables into a smaller number while keeping as much information as possible. One prominent method is [tag pca] excel\u00d7\u00a0151 a commercial spreadsheet program created by Microsoft. summary-statistics\u00d7\u00a0151 A brief numerical description of a set of data. ancova\u00d7\u00a0150 Analysis of Covariance. effect-size\u00d7\u00a0146 \"a measure of the strength of a phenomenon or a sample-based estimate of that quantity\" [Wikipedia]. fitting\u00d7\u00a0145 power\u00d7\u00a0145 Is a property of a hypothesis testing method: the probability of rejecting the null hypothesis given that it is false, i.e. the probability of not making a type II error. The power of a test depends o\u2026 r-squared\u00d7\u00a0143 In linear regression, the coefficient of determination, usually symbolized by $R^2$, is the proportion of the total response variance explained by the regression model. count-data\u00d7\u00a0143 non-negative integers representing whole amounts. When such data are the dependent variable in a regression, Poisson or negative binomial regression may be appropriate methods. One comm\u2026 basic-concepts\u00d7\u00a0143 scales\u00d7\u00a0142 bias\u00d7\u00a0141 Bias, in a statistical framework, means that an estimate of a parameter has an expected value that is not equal to the actual parameter value. There is often a tradeoff between bias and variance - low\u2026 computational-statistics\u00d7\u00a0141 Refers to the interface of statistics and computing; the use of algorithms and software for statistical purposes. logit\u00d7\u00a0141 A name given to the log-odds function, which maps probabilities to the real line. curve-fitting\u00d7\u00a0140 prior\u00d7\u00a0139 In Bayesian statistics a prior distribution formalizes information or knowledge (often subjective), available before a sample is seen, in the form of a probability distribution. A distribution with la\u2026 roc\u00d7\u00a0136 Receiver Operating Characteristic, also known as ROC curve. robust\u00d7\u00a0134 Robustness in general refers to a statistic's insensitivity to deviations from its underlying assumptions (Huber and Ronchetti, 2009). large-data\u00d7\u00a0133 'Large data' refers to situations where the number of observations is so large that it necessitates changes in the way the the data analyst thinks about or conducts the analysis. group-differences\u00d7\u00a0133 Group differences broadly refer to statistics which quantify the differences between two or more subpopulations.","date":"2014-04-18 03:03:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.674981415271759, \"perplexity\": 796.864691532371}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-15\/segments\/1397609532480.36\/warc\/CC-MAIN-20140416005212-00382-ip-10-147-4-33.ec2.internal.warc.gz\"}"}
| null | null |
{"url":"http:\/\/mathoverflow.net\/questions\/132265\/a-reference-for-kummer-theory-with-proofs","text":"a reference for Kummer theory, with proofs ?\n\nWhat is a standard reference for Kummer theory of semi-Abelian varieties ? I need a complete exposition with detailed proofs. Also in prime characteristic, although I am not sure what the statement there exactly is.\n\nBelow I give an example of what I mean by Kummer theory in zero characteristic.\n\nLet $G = A\\times L$ be a product of an Abelian variety by a torus L so that after a finite extension of k it satisfies Poincare's complete reducibility theorem (as a variety over k).\n\nLet $l$ be a prime. Let $G_l=\\{ x \\in > G(\\bar k) : \\exists n l^nx=0\\}$ be the $l$-torsion of G. For a point $P\\in > G(\\bar k)$, let $G_P$ be the smallest algebraic subgroup of G containing P, i.e. Zariski closure of subgroup ${\\Bbb Z}P$ of $G$, and let $G_P$ be its connected component through the origin,\n\n$$\\zeta_l(P) : Gal(\\bar > k\/k(G_l,P))\\longrightarrow T_l(A\\times > L)$$ $$\\sigma \\mapsto > \\sigma(P_l)-P_l$$ where $P_l$ is a compatible sequence of division points, $P_1 = P$. By Kummer theory I mean the statement that ([Bertrand, Theorem 2]) the image of of the map $\\prod \\zeta_l$ is of finite index in $NT(G_P^o)$ for a large $N$.\n\nAbove is taken from Bertrand, D.: Galois representations and transcendental numbers. In: New advances in transcendence theory (Durham, 1986), Cambridge University Press, Cambridge (1988); but proofs there are a little too sketchy for me. 13. Ribet, K.: Kummer theory on extensions of varieties by tori. Duke Math. J. 46(4), (1979) has a slightly weaker statement.\n\n-","date":"2015-09-03 19:14:18","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8976345062255859, \"perplexity\": 270.71599611316674}, \"config\": {\"markdown_headings\": false, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2015-35\/segments\/1440645323734.78\/warc\/CC-MAIN-20150827031523-00289-ip-10-171-96-226.ec2.internal.warc.gz\"}"}
| null | null |
{"url":"https:\/\/brilliant.org\/problems\/area-of-cubic\/","text":"# Area of Cubic\n\nLevel pending\n\nLet $$S$$ be the area of the region bounded by the curve $$y=x(x-3)(x+2)$$ and the $$x$$-axis. What is $$S?$$\n\n\u00d7","date":"2018-03-23 15:01:23","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8263980746269226, \"perplexity\": 244.18819749574632}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-13\/segments\/1521257648313.85\/warc\/CC-MAIN-20180323141827-20180323161827-00676.warc.gz\"}"}
| null | null |
Wonoploso is een bestuurslaag in het regentschap Mojokerto van de provincie Oost-Java, Indonesië. Wonoploso telt 2774 inwoners (volkstelling 2010).
Plaats in Oost-Java
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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| 2,515
|
Chris Gutierrez (born Christopher Juno Balbin on 9 May 1992) is a Filipino actor. He is a member of ABS-CBN's Star Magic Batch 13. He is a grandson of two Philippine showbiz greats, Gloria Romero and Juancho Gutierrez.
Biography
Family background
Christopher Gutierrez is a grandson of veteran Filipino actress and former movie queen Gloria Romero. His grandfather is actor Juancho Gutierrez, another Philippine showbiz veteran actor. He is the only son of Maritess, a professional chef, the only daughter of Gloria and Juancho.
Chris was born in Delaware, United States on May 9, 1992. His parents divorced when he was young, and then moved to the Philippines with his mother, leaving behind his dad in Seattle. They now live together with his grandmother Gloria Romero.
Career
Until the offer from Star Magic came, Chris never considered entering showbiz. He was only 13 when he was contacted by ABS-CBN's talent management arm. In 2006, he was launched as a member of Star Magic Batch 13 along with 23 other aspiring actors.
The same year, he made guest appearances in ABS-CBN shows: Homeboy, Wowowee and ASAP. He was later cast in ABS-CBN's show, Star Magic Presents: Abt Ur Luv. He plays role of Rickson a love interest to Giselle, played by Zia Marquez. This is the start of the Chris-Zia team up. He and Zia later reprised their respective roles in the follow-up season, Star Magic Presents: Abt Ur Luv Ur Lyf 2.
When Abt Ur Luv ended in 2008. He then starred in the new season of Star Magic Presents, Star Magic Presents: Astigs. The show only went on for two mini-seasons, Astigs in Haay...School Life and Astigs in Luvin' Lyf. In Astigs in Luvin' Lyf, Chris was again paired with Marquez.
Filmography
Notes
References
External links
Chris Gutierrez Official Forum Site
1992 births
Living people
21st-century Filipino male actors
Filipino male child actors
Filipino male television actors
Filipino television personalities
Filipino people of American descent
Star Magic
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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| 303
|
Q: RestAssured_Class Not Found_com/github/fge/jsonschema/main/JsonSchemaFactory Please help me with validating schema of the response body. I am facing a runtime error.:
Exception in thread "main" java.lang.NoClassDefFoundError:
com/github/fge/jsonschema/main/JsonSchemaFactory
import io.restassured.RestAssured;
import io.restassured.http.ContentType;
import io.restassured.response.Response;
import io.restassured.response.ValidatableResponse;
import static io.restassured.module.jsv.JsonSchemaValidator.matchesJsonSchemaInClasspath;
import static io.restassured.RestAssured.*;
public class Sample_JsonSchemaValidation {
public static void main(String[] args) {
// TODO Auto-generated method stub
//ClassLoader loader = Sample_JsonSchemaValidation.class.getClassLoader();
//System.out.println(loader.getResource("Sample_JsonSchemaValidation.class"));
RestAssured.baseURI = "https://reqres.in/";
given().contentType(ContentType.JSON).queryParam("id", 2).
when().get("api/users/"). then().body(matchesJsonSchemaInClasspath("JsonSchemavalidator.json"));
}
}
Please help resolve the error.
A: Do you have the below in your POM ?
<dependency>
<groupId>io.rest-assured</groupId>
<artifactId>json-schema-validator</artifactId>
<version>4.3.0</version>
</dependency>
The compile dependencies include 2 artifacts, If you do not have them then include those as well in your POM
<dependency>
<groupId>com.github.java-json-tools</groupId>
<artifactId>json-schema-validator</artifactId>
<version>2.2.10</version>
</dependency>
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,684
|
Bassano del Grappa ( or Bassan, ) is a city and comune, in the Vicenza province, in the region of Veneto, in northern Italy. It bounds the communes of Cassola, Marostica, Solagna, Pove del Grappa, Romano d'Ezzelino, Campolongo sul Brenta, Conco, Rosà, Cartigliano and Nove. Some neighbourhoods of these communes have become in practice a part of the urban area of Bassano, so that the population of the whole conurbation totals around 70,000 people.
The 16th century painter Jacopo Bassano was born, worked, and died in Bassano, and took the town name as his own surname.
History
Prehistoric and Roman periods
The city was founded in the 2nd century BC by a Roman called Bassianus, whence the name, as an agricultural estate. However, an ancient bronze sword (called "spada di Riccardo"), found in 2009
and dating back to the 7th century BC, possibly between the 18th and 15th century BC, suggests that the area of Bassano was already inhabited not just in the pre-Roman period, but possibly even in the pre-Venetic period, as confirmed by some artifacts found in a necropolis located in the neighbourhood of San Giorgio di Angarano.
From the Middle Ages to Venice
The first news of the existence of the medieval city dates from 998, while the castle is mentioned first in 1150. In 1175 Bassano was conquered by Vicenza, but the city maintained a semi-autonomous status as a free comune in the 13th century also, when it was home to the family of the Ezzelini, who first unified the various territories of Veneto.
In 1278, according to Giovanni da Nono, Matteo of the Cortusi family of Padua was elected podestà. In 1281, the city came under Paduan control. In 1368 Bassano was acquired by the Visconti of Milan and was given the status of "separate land" (terra separata).
In 1404, Bassano became a part of the Stato da Tera 'Mainland State' of the Venetian Republic, which granted the Bassanese district the status of autonomous podesteria, "free and separate from whatever city and from the jurisdiction of whatever city" (sit ipsa terra exempta et separata a quacumque civitate et iurisdictione cuiuscumque civitatis) and subordinate only to Venice. The autonomous district included Bassano properly and the villas of Cartigliano, Cismon and Primolano, Rossano, San Nazario, Pove, Solagna plus Cassola (on lands previously belonging to Pove and Solagna) and Tezze and Rosà (on lands previously part of Bassano). In addition to this, Valstagna and Campese (then belonging to Vicenza and the Seven Communes) and Romano and Mussolente (then belonging to Treviso) had strong commercial and political ties with the district as they were located very close to Bassano and its port on the river Brenta connected with Venice. In 1760 Doge Francesco Loredan granted Bassano the title of City, subsequently retained under the Austrian and the Italian States. The Serenissima did not alter the town's magistratures, limiting itself to impose a Captain chosen by the Venetian Senate. The city became home to a flourishing industry producing wool, silk, iron and copper, and mainly for ceramics; in the 18th became especially famous in all Europe for the presence of the printer company.
From the fall of Venice to modern times
During the French Revolutionary Wars the city was the site of the Battle of Bassano. In 1815 it was included in the Kingdom of Lombardy–Venetia, and became part of the unified Kingdom of Italy in 1866. Napoleon Bonaparte remained in Bassano del Grappa for many months.
The original name of the town was Bassano Veneto. After the terrible battles on Mount Grappa in World War I, where thousands of soldiers died, a decision was made to change the name of the town. In 1928, the name was changed to Bassano del Grappa, meaning Bassano of Mount Grappa, as a memorial to the soldiers killed.
Ernest Hemingway during his days as an ambulance driver in the war spent many days in Bassano and eventually settled there as part of A Farewell to Arms. Also other American writers spent some days in Bassano during World War I such as Scott Fitzgerald and Dos Passos.
During World War I Bassano was in the front area, and all industrial activities were halted.
In the last days of World War II, Bassano del Grappa was bombed by USAF B-24s and B-17s.
The symbol of the town is the covered Ponte Vecchio, which was designed by the architect Andrea Palladio in 1569. The wooden pontoon bridge was destroyed many times, the last time during World War II. The Alpine soldiers, or Alpini have always revered the wooden bridge and Bassano del Grappa. After the destruction of the bridge, they took up a private collection and had the bridge completely rebuilt. Often soldiers flock to the bridge to remember and sing songs from their days as alpine soldiers. The grappa shop of Nardini Distillery is located on the bridge, known as Ponte degli Alpini.
Bassano del Grappa is also the long residence town of Renzo Rosso, the founder and President of Diesel. Since Diesel began to expand in the mid-1980s, the company has become an important source of business for the city and its surrounding region. As thanks for the support that Rosso has received locally, he has invested personally in the city's professional soccer team, Bassano Virtus 55 S.T.
Geography
Bassano is located at above sea level and has an area of . Its highest point is at , whereas the lowest point is at . The city lies at the foothills of the Venetian Prealps, where river Brenta comes out the southern end of Canal di Brenta (also called Valbrenta 'Brenta valley') and flows in the lowlands at the borders of Vicenza, Treviso and Padua provinces.
Main sights
The cathedral (Duomo), built around the year 1000 but renovated in 1417. It has works by Leandro da Bassano, Ottavio Marinali and others
The Castello Superiore (Upper Castle)
The church of St. John the Baptist, built in the 14th century and restored in the 18th century.
San Francesco: with a Crucifix by Guariento (14th century) and remains of contemporary frescoes. Next to the church is the Town Museum, with ancient archaeological remains, works by Antonio Canova and the Tiepolos, and drawings by Gian Lorenzo Bernini, Spagnoletto, Albrecht Dürer and Rembrandt
The wooden covered Bridge, or Ponte degli Alpini, on the Brenta River, designed in 1569 by the architect Andrea Palladio to replace a pre-existing construction existing at least from 1209. The bridge was destroyed in 1748, and was rebuilt three years later. The Nardini tavern on the bridge is unchanged since 1779.
Palazzo Michieli-Bonato, with a façade frescoed by Jacopo da Bassano.
The Palazzo del Municipio (Town Hall), erected from 1404. It has a noteworthy loggia and a fresco attributed to Jacopo da Bassano.
The Monte di Pietà, a Renaissance edifice with 15th-century coats of arms.
The Palazzo Sturm, home to the Ceramics Museum
The Torre Civica (Civic Tower, 14th Century) 43 metres, in Piazza Garibaldi.
In the neighbourhood are the Villa Rezzonico, designed by Baldassarre Longhena, Art Nouveau's Villa Agnesina, designed by Francesco Bonfanti in 1923, and the 17th century Villa Bianchi-Michiel, with a garden decorated by statues.
Administrative subdivisions
The municipal statute (art.6, par.2) of Bassano, recognizes only Rubbio as frazione and defines Campese, Marchesane, San Michele, Sant'Eusebio and Valrovina as contrade. The other existing neighbourhoods of Bassano are not mentioned in the statute. However, in practice, all the administrative subdivisions have the same prerogatives and are named quartieri.
Frazioni
Rubbio is a frazione and quartiere located at an altitude of on the Asiago plateau. This hamlet is contiguous with another hamlet, also named Rubbio, which is part of the commune of Conco. Thus, in practice, the two hamlets form one village (named Rubbio), even though they belong to two different communes from the administrative point of view.
Contrade
Officially, the contrade (in ven. contrae) are Campese, Marchesane, San Michele, Sant'Eusebio and Valrovina. From an administrative point of view these are also quartieri. However, in practice, some of these neighbourhoods themselves contain smaller inhabited areas (as streets, groups of houses) also called contrade: there are thus contrade within contrade. Besides, some places known as contrade exist also within other neighbourhoods which are officially simply defined as quartieri, but not contrade.
Quartieri
All the administrative subdivision (quartieri) of Bassano are: Centro Storico, Margnan-Conca d'oro, San Marco, San Vito, Ca'Baroncello, Quartiere Firenze, Nuovo Ospedale, San Lazzaro, San Fortunato, Borgo Zucco, Marchesane, Rondò Brenta, Angarano, Quartiere XXV Aprile, Sant'Eusebio, San Michele, Valrovina, Rubbio, Campese, Merlo, Quartiere Pré, Santa Croce.
Rubbio, with an area of 6.835 km2, is the largest quartiere of Bassano, but also the least populated (86 inhabitants in 2009).
Quartiere Prè (an old venetian plural meaning meadows, the modern ven. plural is prai), located in the southern lowland of Bassano, is the second least populated quartiere (299 inhabitants in 2009). Part of it hosts an industrial zone that also falls in the nearby San Lazzaro, but it also contains a considerable rural area which falls within the Parco rurale sovracomunale Civiltà delle Rogge regional park.
San Vito, in the north-eastern part of Bassano, is the most inhabited quartiere (5841 inhabitants in year 2009). It merges with the built-up areas of the bordering comunes Romano d'Ezzelino, San Giuseppe di Cassola and Pove del Grappa.
Territorial variations
Until 1928, the official name of Bassano del Grappa was simply Bassano (as it is still informally called today).
In 1878, the neighbourhood of Campese, previously belonging to the commune of Campolongo sul Brenta is detached from Campolongo and aggregated to Bassano. In 1938, the commune of Valrovina, which also comprised Rubbio, is suppressed and aggregated to Bassano.
Notable people
Luigi Agnolin, football referee
Jacopo Bassano, painter
Jeronimo Bassano, Master of Trumpets and Shawms to the Doge in Venice
Giovanni Battista Brocchi, geologist
Giusto Bellavitis, (1803 -1880), mathematician and senator
Miki Biasion, World Rally Champion
Luisa Vania Campagnolo (born 1968), luthier
Simone Cogo (Sir Bob Cornelius Rifo), Musician and founder of The Bloody Beetroots
Luigi Fabris, sculptor and ceramist, founder of Manifattura Italiana Porcellane Artistiche Fabris
Pietro Fabris, senator
Tommaso Gabrielli, Motorcycle racer
Antonio Gaidon, (1738 -1829), architect, civil engineer, naturalist (Antonio Gaidon)
Tito Gobbi, opera singer
Federico Marchetti, footballer
Francesca Michielin, singer and songwriter
Jacek Pałkiewicz, Polish journalist, traveler and explorer. Fellow (by recommendation from Thor Heyerdahl) of the prestigious London-based Royal Geographical Society and numerous other such societies, he is best known for his discovery of the sources of the Amazon River.
Joseph Pivato, writer and academic in Canada, born in Tezze sul Brenta.
Stefano Rusconi, professional basketball player, who also played in the NBA
Renzo Rosso, Founder and President of Diesel (brand) and the Only The Brave Group
Iacopo Vittorelli, poet
Giovanni Volpato, engraver
International relations
Twin towns – Sister cities
Bassano del Grappa is twinned with:
Main industries in the Bassano del Grappa area
Diesel (brand)
Baxi
Manfrotto
Vimar
ABB
Montegrappa
References
External links
Official website
Personal webpages about the network of tributaries and ditches connected to Brenta in Bassano
Cities and towns in Veneto
Domini di Terraferma
Populated places established in the 2nd century BC
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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| 4,622
|
Q: Spawn object with custom script with Unity Network I have a PlayerPref that spawn more objects in OnStartLocalPlayer().
So in OnStartLocalPlayer() it call Command(assume that called on server) that instantiate GameObject and setup some values of its scripts. At the end it calls SpanWithClientAuthority()...
The thing is that on owner client and on server those script tweeks are correct, but on all other clients it lost all that settings(ex. gameobject ref etc). What do I do wrong?
Once more in nutshell: playerPref GO must have ref list of several other objects, and those objects must have ref to that playerPref GO. (making them part of playerPref GO is not a solution).
A: If I understand your problem correctly, you need the references to be set across all clients who have the same game object. [Command]'s are for client to server. What you need is a [ClientRpc]. Make the OnStartLocalPlayer() call a [ClientRpc] function, in that function (ex: RpcSetRefs()) set the references you need each client to have.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 4,418
|
Dig into the Bible with Logos 5: 15% Off!
You've experienced Scripture with the Faithlife Study Bible. Now you can upgrade to Logos Bible Software' best Bible study tool, Logos 5, for 15% off through May 20. Just use coupon code SPRINGSALE at checkout.
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You'll get a massive digital library, and smart tools that will help you connect to the Word.
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|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,455
|
import { Template } from 'meteor/templating';
import { ReactiveDict } from 'meteor/reactive-dict';
import { FlowRouter } from 'meteor/kadira:flow-router';
import { _ } from 'meteor/underscore';
import { Profiles } from '/imports/api/profile/ProfileCollection';
import { Tastes } from '/imports/api/taste/TasteCollection';
const displaySuccessMessage = 'displaySuccessMessage';
const displayErrorMessages = 'displayErrorMessages';
Template.Edit_Profile_Page.onCreated(function onCreated() {
this.subscribe(Tastes.getPublicationName());
this.subscribe(Profiles.getPublicationName());
this.messageFlags = new ReactiveDict();
this.messageFlags.set(displaySuccessMessage, false);
this.messageFlags.set(displayErrorMessages, false);
this.context = Profiles.getSchema().namedContext('Edit_Profile_Page');
});
Template.Edit_Profile_Page.helpers({
successClass() {
return Template.instance().messageFlags.get(displaySuccessMessage) ? 'success' : '';
},
displaySuccessMessage() {
return Template.instance().messageFlags.get(displaySuccessMessage);
},
errorClass() {
return Template.instance().messageFlags.get(displayErrorMessages) ? 'error' : '';
},
fieldError(fieldName) {
const invalidKeys = Template.instance().context.invalidKeys();
const errorObject = _.find(invalidKeys, (keyObj) => keyObj.name === fieldName);
return errorObject && Template.instance().context.keyErrorMessage(errorObject.name);
},
profile() {
return Profiles.findDoc(FlowRouter.getParam('username'));
},
tastes() {
const profile = Profiles.findDoc(FlowRouter.getParam('username'));
const selectedTastes = profile.tastes;
return profile && _.map(Tastes.findAll(),
function makeTasteObject(taste) {
return { label: taste.name, selected: _.contains(selectedTastes, taste.name) };
});
},
});
Template.Edit_Profile_Page.events({
'submit .profile-data-form'(event, instance) {
event.preventDefault();
const firstName = event.target.First.value;
const lastName = event.target.Last.value;
const title = event.target.Title.value;
const username = FlowRouter.getParam('username'); // schema requires username.
const picture = event.target.Picture.value;
const facebook = event.target.Facebook.value;
const instagram = event.target.Instagram.value;
const bio = event.target.Bio.value;
const selectedTastes = _.filter(event.target.Tastes.selectedOptions, (option) => option.selected);
const tastes = _.map(selectedTastes, (option) => option.value);
const updatedProfileData = { firstName, lastName, title, picture, facebook, instagram, bio, tastes,
username };
// Clear out any old validation errors.
instance.context.resetValidation();
// Invoke clean so that updatedProfileData reflects what will be inserted.
Profiles.getSchema().clean(updatedProfileData);
// Determine validity.
instance.context.validate(updatedProfileData);
if (instance.context.isValid()) {
const docID = Profiles.findDoc(FlowRouter.getParam('username'))._id;
const id = Profiles.update(docID, { $set: updatedProfileData });
instance.messageFlags.set(displaySuccessMessage, id);
instance.messageFlags.set(displayErrorMessages, false);
} else {
instance.messageFlags.set(displaySuccessMessage, false);
instance.messageFlags.set(displayErrorMessages, true);
}
},
});
|
{
"redpajama_set_name": "RedPajamaGithub"
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| 3,379
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Q: Python function for multiplication of arrays by index Can someone help me with this?
A: Have you checked numpy library? it makes working with matrices a lot easier. You can check it out in this links: https://www.tutorialspoint.com/python_data_structure/python_matrix.htm
numpy gives you a .dot method for matrix multiplication. Hope this guides you enough --> https://numpy.org/doc/stable/reference/generated/numpy.dot.html#numpy.dot
A: If you have a matrix like
mat=np.array([[1,2,3,4,5],
[2,3,4,5,6],
[3,4,5,6,7]])
and an array like
arr=np.array([1,2,3])
Then the required function using numpy multiplication can be so simple:
def multiplicate(mat,arr):
mat*arr.reshape((3,1))
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,351
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Q: Why is a slight gap added between rows on HDPI screens, causing this striped effect? I have a perplexing CSS issue, where multiple rows with background colour that should meet have a very slight gap in high DPI screens.
@font-face {
font-family: 'Iosevka';
font-style: normal;
font-weight: 400;
src: url(https://curiousdannii.github.io/parchment-testing/fonts/build/iosevka-custom-extended.woff2) format('woff2');
}
.BufferWindow {
font-family: Georgia;
font-size: 15px;
line-height: 1.4;
overflow-y: auto;
scrollbar-gutter: stable;
}
.BufferLine {
white-space: pre-wrap;
}
span.Style_preformatted {
font-size: 1em;
font-weight: normal;
font-style: normal;
font-family: Iosevka;
}
.reverse {
background: black;
color: white;
display: inline-block;
}
<div class="BufferWindow">
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span><span class="Style_preformatted"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
</div>
Screenshot showing how it's designed to look, on a non-HDPI screen:
Screenshot directly from my phone:
Screenshot from remote dev tools showing the height of one row, which is 21 pixels:
Screenshot from remote dev tools showing the height of all five rows, which is 107 pixels:
So on a non-HDPI screen the height of all five rows is 105 pixels, which is 5 times 21 pixels. Somehow on the HDPI screen it's adding two extra pixels.
So then I saw in dev tools that the height of the BufferLine isn't actually 21 pixels, it's 21.364:
I was wondering if it was that the line-height: 1.4 multiplier was producing something almost but not exactly 21 pixels high, which only becomes relevant on HDPI screens, but 15 * 1.4 = 21 exactly (I must have designed this sensibly in the past ;)). So I'm not sure what's cause that little bit extra in height.
How do I determine which part of my CSS is causing this, and how can I make it more reliable?
A: The arrangement you have is essentially this:
(1/2L = Half Leading)
The characters align by their baselines, but because the ascents and/or descents differ, their tops and bottoms don't align with one another, and the total height is greater than either of the characters.
From this you should be able to see that the problem can be avoided by either using the same font throughout, or by reducing the line height of one of them, so that their half-leadings are so small (or negative) that the total top edge and bottom edge are both determined by the same character.
A: UPDATE: It appears that the below is incorrect - although it worked for me it has not worked for the OP. The following is left here temporarily as a basis for discussion.
The phenomenon seems to occur because there is a difference in the way the two typefaces treat ascenders.
Georgia has a special characteristic that the ascenders come significantly above the caps.
From https://taylorhieber.co/georgia-type-set
If you replace Georgia with a typeface that does not have this difference then the gaps disappear.
This snippet uses another serif font, Times New Roman, but which does not have this 'extra' height on the ascenders:
@font-face {
font-family: 'Iosevka';
font-style: normal;
font-weight: 400;
src: url(https://curiousdannii.github.io/parchment-testing/fonts/build/iosevka-custom-extended.woff2) format('woff2');
}
.BufferWindow {
font-family: 'Times New Roman';
font-size: 15px;
line-height: 1.4;
overflow-y: auto;
scrollbar-gutter: stable;
}
.BufferLine {
white-space: pre-wrap;
}
span.Style_preformatted {
font-size: 1em;
font-weight: normal;
font-style: normal;
font-family: Iosevka;
}
.reverse {
background: black;
color: white;
display: inline-block;
}
<div class="BufferWindow">
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span><span class="Style_preformatted"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
<div class="BufferLine Style_preformatted"><span class="Style_preformatted"> </span><span class="Style_preformatted reverse"> </span></div>
</div>
Footnote: I am not absolutely convinced that it is because the ascenders are higher (relatively) on Georgia than Times New Roman, it may just be that the typefaces differ in how they interpret lineheight overall. Hope someone out there can confirm or deny. In any case it appears that using a different serif font may cure the white lines problem. More experiments with different serif fonts could be interesting.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
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Plaats in Indonesië
Taba Baru (Bermani Ilir), een plaats in het bestuurlijke gebied Kepahiang in de provincie Bengkulu
Taba Baru (Lais), een plaats in het bestuurlijke gebied Bengkulu Utara in de provincie Bengkulu
Taba Baru (Lubuk Linggau Utara I), een plaats in het bestuurlijke gebied Lubuklinggau in de provincie Zuid-Sumatra
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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| 4,533
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Els Alps centrals conformen una porció del sistema montanyós dels Alps, classificació usada en la Partició dels Alps. L'any 1926, després del IX Congrés Geogràfic Italià de 1924, es va adoptar aquest nou sistema de subdivisió del sistema alpí. Aquesta va anar caient en desús d'acord amb la incorporació de la nova classificació SOIUSA, que en lloc de partir els Alps en tres grans sectors (Occidental, Central, Oriental), passa a fer-ho únicament en dos: Alps occidentals i Alps orientals.
En la partició tradicional, els Alps centrals s'extendrien entre el col Ferret i el pas del Brennero, i el cim més elevant en seria el Mont Rosa (4.611 m). Aquesta subdivisió tradicional, molt centrada en una visió italiana, es va superar el 2005 amb la introducció de la Suddivisione Orografica Internazionale Unificata del Sistema Alpino (Subdivisió Orogràfica Internacional Unificada del Sistema Alpí) o SOIUSA.
A més de la classificació de l'any 1926, a vegades també s'ha entès que els Alps centrals són el tram comprès entre el pas de Resia i el pas d'Arlberg, de manresa que els Alps Penins no s'incourien aquí si no als Alps occidentals.
Els Alps centrals se subdivideixen en successives classificacions:
Alps Penins (9)
Alps del Valais (9.a)
Grup de Val Sesia (9.b)
Alps Lepontins (10)
Grup del Monte Leone (10.a)
Grup d'Adula (10.b)
Alps Ticineses (10.c)
Alps Rètics (11)
Grup d'Albula i Silvretta (11.a)
Grup de la Plessur (11.b)
Cadena del Reticone (11.c)
Grup del Fervall (11.d)
Grup del Bernina (11.e)
Grup del Umbraglio (11.f)
Alps Venoste (11.g)
Alps Breonie (11.h)
Alps Sarentins (11.i)
Grup de l'Ortles (11.j)
Cims de la Val di Non (11.k)
Grup de l'Adamello (11.l)
Dolomites de Brenta (11.m)
Alps Bernesos (12)
Massís del Finsteraarhorn (12.a)
Grup del Wildhorn (12.b)
Alps de Uri (12.c)
Alps de Glaris (13)
Grup del Tödi (13.a)
Grup de la Sardona (13.b)
Prealps suïssos (14)
Prealps de Simmental (14.a)
Prealps d'Emmental (14.b)
Prealps de Linth (14.c)
Alps Bavaresos (15)
Alps d'Algovia (15.a)
Alps de la Lechtal (15.b)
Montes d'Achensee (15.c)
Prealps llombards (16)
Prealps de Lugano (16.a)
Alps Orobie (16.b)
Prealps Bergamescos (16.c)
Prealps de Brescia (16.d)
Prealps Giudicarie (16.e)
Grup del Monte Baldo (16.f)
Notes
Referències
Vegeu també
Alps
SOIUSA
Centralss
Geografia física d'Itàlia
Geografia d'Àustria
Geografia física de Suïssa
Geografia física d'Alemanya
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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Q: Sending ASCII Control Characters to Serial Device using libserial on Linux I have a very basic device with which I am trying to interact via a serial connection on Linux. I am on the steep part of the learning curve here, still, so please be gentle!
One of the functions involves sending data to an attached printer. You send a command to the device, which then relays the data you input to the printer attached to the device. The command looks like this:
*
*Send "EXEX*". The device echoes back "EXEX" (the '*' is not echoed yet)
*Send a single byte indicating the length of the data you will send, including a LF at the end.
*Send the data (the device will now echo back the *).
*Send "#". The device will now be ready for another command.
I have a small C++ program to communicate with the device, and I can successfully send single characters and such, but when I try to send this command, I do not get the expected results.
Using Hyperterminal in Windows, it is particularly easy, using alt-key combinations to send ASCII control codes. Just connect and:
*
*Type "EXEX*"
*Type Alt+010 to send a LF character, indicating that you are sending 10 bytes to the printer (nine characters and a LF).
*Type the data you wish to send: "123456789" (nine bytes in length).
*Type Alt+010 again to send a final LF character to the printer.
*Type "#" to finish.
Here is what I cobbled together to try in C++:
#include <SerialStream.h>
#include <string>
#include <iostream>
#include <fstream>
using namespace std;
using namespace LibSerial;
int main(){
char buffer [50];
int n;
n=sprintf (buffer, "EXEX*%c123456789%c#",10,10);
printf("Variable buffer was set to a %d character string: %s\n",n,buffer);
SerialStream my_serial_stream;
my_serial_stream.Open("/dev/ttyS0") ;
my_serial_stream.SetBaudRate( SerialStreamBuf::BAUD_19200 ) ;
my_serial_stream.SetCharSize( SerialStreamBuf::CHAR_SIZE_8 ) ;
my_serial_stream.SetFlowControl( SerialStreamBuf::FLOW_CONTROL_NONE ) ;
my_serial_stream.SetParity( SerialStreamBuf::PARITY_NONE ) ;
my_serial_stream.SetNumOfStopBits(1) ;
my_serial_stream.SetVTime(1);
my_serial_stream.SetVMin(100);
cout<<"Sending Command:\n";
my_serial_stream << buffer;
//my_serial_stream << printf("%s",buffer);
//my_serial_stream << "EXEX*\n123456789\n#";
my_serial_stream.read(next_char,100);
cout<<"Result: "<<next_char<<"\n";
my_serial_stream.Close();
return 0;
}
I also tried both of the commented out lines, and neither worked. The device does not receive the proper characters on the other end.'
I'm certain that this is pretty basic, perhaps something is grabbing the control characters in the middle? If anyone has any ideas on a better way to do this, I would really appreciate it. Specifically, I might need to send a byte with a value anywhere between 1 and 40, depending on the length of the data I wish to send to the printer.
My apologies for being unclear, please let me know if I need to break this down farther.
Many thanks,
Tom
A: The line you send doesn't include the # that you mention in the character sequence.
Have you checked serial comms works on /dev/ttyS0 using gtkterm / cutecom etc?
To test your interface you could read back the serial port. If you have a second port or computer, you could do that by connecting to another port via a null modem. Otherwise you could short pins 2 and 3 of your serial port and check that you are receiving back the characters you send.
You may want to check the return values of the calls to make to the serial library, to see if any errors are returned.
Perhaps there are timing requirements on the printer, and you may need to wait between sending some characters.
I compiled the code and checked the output on another serial port with gtkterm, it does receive the string you would expect:
45 58 45 58 2A 0A 31 32 - 33 34 35 36 37 38 39 0A EXEX*.12 3456789.
It won't affect the sending part of the code, but the receiving looks suspicious. If the read() member function is like the system call and if next_char is a character array, then it won't null terminate the string. Instead you have to look at the return value to get the size, and then null terminate if you are going to use it as a null-terminated C string.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,693
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Ha studiato musica elettronica e composizione con Riccardo Dapelo e pianoforte con Giuseppina Schicchi. È iscritto alla SIAE dal 1995 con le qualifiche di compositore e autore della parte letteraria. Dal 1994 ricopre il ruolo di direttore artistico del gruppo I Giovani Canterini di Sant'Olcese, nel quale canta da contræto (contralto).
Attività
Premio città di Recanati
Nel 1996 risulta vincitore alla VII edizione del Premio città di Recanati - nuove tendenze della canzone d'autore con il brano O trallalero canson de 'na vitta per squadra di canto popolare genovese. Il brano presentato ha come obiettivo principale quello di utilizzare la squadra di canto - così vengono denominati in Liguria i gruppi a cappella che praticano l'arte polivocale del trallalero genovese - come vero e proprio strumento musicale, cercando di proporre una soluzione alternativa rispetto al binomio cantautore-chitarra e il brano, scritto in lingua genovese, si prestava bene all'arrangiamento per squadra di canto trallalero.
Canzone d'autore e ricerca sulla lingua genovese
La partecipazione e la vittoria al Premio città di Recanati sono stati uno sprone per continuare l'attività di autore e cantautore, soprattutto per quanto attiene ai testi in lingua ligure. Besagno ha collaborato con diversi autori, tra i quali ricordiamo Carlo Denei e Mike FC, sia in qualità di arrangiatore che di coautore. Nel 2020 pubblica la raccolta Bambòcce sensa i euggi con testi in genovese, curati da Stefano Lusito.
Musica elettronica
La produzione di Besagno è caratterizzata sostanzialmente dalla musica elettroacustica, studiata e approfondita sotto la guida del M° Riccardo Dapelo. Sono della fine degli anni novanta le sue prime produzioni, tra le quali ricordiamo Allegro Moderabile per suoni di treno, voce di nonno Dria e nastro magnetico, In primo vere, brano sulla circolarità della vita per voci di bimbe e nastro. Nel giugno 2003 presenta il brano in memoria del padre e...tu come saresti? a Synthèse, Festival International des Musiques et Créations Electroniques di Bourges. Il brano viene selezionato ed eseguito al Théâtre Jacques-Cœur di Bourges. Dal 2001 al 2006, periodo durante il quale si tiene annualmente il Festival Synthèse e Bourges é un importante punto di riferimento per la musica elettronica, vengono proposti ed eseguiti i brani: In purissima luce, dedicato all'amico Franco Nocentini prematuramente scomparso; Auto-domestikalia, per suoni di ambiente domestico e nastro; Normalmente, due dedicato a Joseph Marie Jacquard, inventore del telaio; Silvia is so deep, dedicato alla moglie. Nel 2008 il suo brano In primo vere viene selezionato ed eseguito a Emufest, festival internazionale di musica elettronica, curato dal M° Giorgio Nottoli al Conservatorio di Santa Cecilia in Roma. Nel 2015 partecipa a Mascagni Remix- una sezione della rassegna Suoni inauditi curata dal M° Fabio De Sanctis De Benedictis presso L'Istituto Superiore di Studi Musicali "Pietro Mascagni" di Livorno e viene selezionato il suo brano Largo IV per rielaborazione di campioni audio tratti dalle opere di P. Mascagni e nastro magnetico. Il brano viene scritto per essere eseguito sull'impianto di diffusione a 24 canali dell'Auditorium C. Chiti, ubicato nello stesso istituto. Nel 2019 il suo brano Witte Vlam, scritto per rimusicare l'omonimo cortometraggio del periodo del cinema muto, a firma di Charles Dekeukeleire (1930), viene selezionato ed eseguito a Bolzano al festival Rimusicazioni.
Trallalero genovese
Il lavoro su lingua e tradizioni genovesi non poteva prescindere dallo studio del trallalero, praticato in famiglia fin dalla più tenera età. Il nonno materno Andrea "Dria" Cassissa gli fornisce gli elementi necessari per curare la propria preparazione. Nel 1994 fonda, insieme ai canterini Mario Oliva e Andrea Pedemonte, la squadra di canto popolare I Giovani Canterini di Sant'Olcese, oggi più che mai attiva. Nel 2015, con Rinaldo Marti, field recordist e compositore, dá vita al consort vocale Ethnogenova, attualmente formato, oltre che dallo stesso Besagno (contræto - contralto), da Alberto Sacco (primmo - tenore), Alessandro Campora (controbasso - baritono), Fabrizio Parodi (chitàra - voce chitarra) e Alessandro Ghiglino (basso). Il quintetto ha come obiettivo una proposta di ascolto immersivo nel canto popolare genovese grazie all'utilizzo di un microfono quadrifonico che viene posto al centro del cerchio che formano i cantori e a una patch scritta in ambiente MAX/MSP da Rinaldo Marti. Si tratta di un progetto divulgativo e di educazione alla pratica del trallalero genovese, in particolare nelle scuole secondarie di primo grado. Numerose le sue collaborazioni con l'etnomusicologo Mauro Balma e, tra queste, la realizzazione di alcune registrazioni inerenti alla timbrica della voce etnica chitàra (voce chitarra) propria delle formazioni di trallalero, a scopi di studio. Tali registrazioni sono state utilizzate per la stesura di un capitolo, dedicato all'analisi timbrica voce chitarra e al modo particolare con il quale i canterini ottengono quel suono caratteristico, dall'etnomusicologo Giuliano d'Angiolini nel libro Alle origini del trallalero genovese, di M. Balma e G. d'Angiolini, Ed Nota Music, Udine 2018. Nel 2013 rilascia un'intervista a cura di M. Balma, per il volume Local and Global Understandings of Creativities: Multipart Music Making and the Construction of Ideas, Contexts and Contents - edito da Ardian Ahmedaja, nella quale si dialoga sui metodi di resa e istruzione dei cantori in caso di musica tramandata oralmente o di partiture solo pensate e non trascritte.
...ruhe, ruhe!
La ricerca su canzone d'autore, musica elettronica e lingua ligure trova una sintesi nel progetto multimediale ...ruhe, ruhe! Si tratta di un concerto della durata di circa cinquanta minuti dove vengono eseguite otto canzoni in genovese, intervallate da sequenze elettroacustiche e proiezione di immagini e video. Questo progetto, il cui sottotitolo è libero ragionamento sulla sofferenza attraverso le immagini della Passione, è suonato in trio (Sandro Secchi - chitarre e arrangiamenti, Stefano Bosi -fisarmonica, Paolo Besagno - testi, musiche, pianoforte, voce, elettronica.
Collegamenti esterni
Sito ufficiale e canale YouTube dell'autore.
Recensione a Bambòcce sensa i euggi, di Stefano Lusito.
Note
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{
"redpajama_set_name": "RedPajamaWikipedia"
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| 2,407
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Q: How to get guest user sessionid in Yii2? I have used YII's session_id's to track user activity across a website with the following code snippet:
Yii::app()->getSession()->getSessionId()
How do I get same sessionid in Yii2? I tried quite possible way, but its all vain.
Please share the exact code snippet for Yii2.
A: You can try with
Yii::$app->session->getId();
this guide could be useful to you http://www.yiiframework.com/doc-2.0/guide-runtime-sessions-cookies.html
try checking if a session is already active before and ifnot use session open() .
session = Yii::$app->session;
// check if a session is already open
if ($session->isActive) ...
// open a session
$session->open();
// close a session
$session->close();
// destroys all data registered to a session.
$session->destroy();
A: $session = Yii::$app->session;
// if session is not open, open session
if ( !$session->isActive) { $session->open(); }
// get session id
Yii::$app->session->getId();
I got 26 character string in session id. "itddoa1lr247phpds34aemr0v0"
This link can be helpful: http://www.yiiframework.com/doc-2.0/guide-runtime-sessions-cookies.html#sessions
A: Yii::$app->session->Id
And no need to explicitly open session. When you access session data through the session component, a session will be automatically opened if it has not been done so before.
http://www.yiiframework.com/doc-2.0/guide-runtime-sessions-cookies.html#sessions
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
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Mendoncia flagellaris är en akantusväxtart som beskrevs av Raymond Benoist. Mendoncia flagellaris ingår i släktet Mendoncia och familjen akantusväxter. Inga underarter finns listade i Catalogue of Life.
Källor
Akantusväxter
flagellaris
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,258
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using std::string;
using std::vector;
namespace net {
// static
string SpdyUtils::SerializeUncompressedHeaders(const SpdyHeaderBlock& headers) {
SpdyMajorVersion spdy_version = HTTP2;
size_t length = SpdyFramer::GetSerializedLength(spdy_version, &headers);
SpdyFrameBuilder builder(length, spdy_version);
SpdyFramer::WriteHeaderBlock(&builder, spdy_version, &headers);
scoped_ptr<SpdyFrame> block(builder.take());
return string(block->data(), length);
}
// static
bool SpdyUtils::ParseHeaders(const char* data,
uint32_t data_len,
int* content_length,
SpdyHeaderBlock* headers) {
SpdyFramer framer(HTTP2);
if (!framer.ParseHeaderBlockInBuffer(data, data_len, headers) ||
headers->empty()) {
return false; // Headers were invalid.
}
if (ContainsKey(*headers, "content-length")) {
// Check whether multiple values are consistent.
base::StringPiece content_length_header = (*headers)["content-length"];
vector<string> values =
base::SplitString(content_length_header, base::StringPiece("\0", 1),
base::TRIM_WHITESPACE, base::SPLIT_WANT_ALL);
for (const string& value : values) {
int new_value;
if (!base::StringToInt(value, &new_value) || new_value < 0) {
return false;
}
if (*content_length < 0) {
*content_length = new_value;
continue;
}
if (new_value != *content_length) {
return false;
}
}
}
return true;
}
// static
bool SpdyUtils::ParseTrailers(const char* data,
uint32_t data_len,
size_t* final_byte_offset,
SpdyHeaderBlock* trailers) {
SpdyFramer framer(HTTP2);
if (!framer.ParseHeaderBlockInBuffer(data, data_len, trailers) ||
trailers->empty()) {
DVLOG(1) << "Request Trailers are invalid.";
return false; // Trailers were invalid.
}
// Pull out the final offset pseudo header which indicates the number of
// response body bytes expected.
auto it = trailers->find(kFinalOffsetHeaderKey);
if (it == trailers->end() ||
!base::StringToSizeT(it->second, final_byte_offset)) {
DVLOG(1) << "Required key '" << kFinalOffsetHeaderKey << "' not present";
return false;
}
// The final offset header is no longer needed.
trailers->erase(it->first);
// Trailers must not have empty keys, and must not contain pseudo headers.
for (const auto& trailer : *trailers) {
base::StringPiece key = trailer.first;
base::StringPiece value = trailer.second;
if (key.starts_with(":")) {
DVLOG(1) << "Trailers must not contain pseudo-header: '" << key << "','"
<< value << "'.";
return false;
}
// TODO(rjshade): Check for other forbidden keys, following the HTTP/2 spec.
}
DVLOG(1) << "Successfully parsed Trailers.";
return true;
}
// static
string SpdyUtils::GetUrlFromHeaderBlock(const SpdyHeaderBlock& headers) {
SpdyHeaderBlock::const_iterator it = headers.find(":scheme");
if (it == headers.end())
return "";
std::string url = it->second.as_string();
url.append("://");
it = headers.find(":authority");
if (it == headers.end())
return "";
url.append(it->second.as_string());
it = headers.find(":path");
if (it == headers.end())
return "";
url.append(it->second.as_string());
return url;
}
// static
string SpdyUtils::GetHostNameFromHeaderBlock(const SpdyHeaderBlock& headers) {
return GURL(GetUrlFromHeaderBlock(headers)).host();
}
// static
bool SpdyUtils::UrlIsValid(const SpdyHeaderBlock& headers) {
string url(GetUrlFromHeaderBlock(headers));
return url != "" && GURL(url).is_valid();
}
} // namespace net
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{
"redpajama_set_name": "RedPajamaGithub"
}
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Enjoy an evening showcasing some of Nelson's finest singers and musicians.
Whether a cappella harmony, traditional choral or solo, these performers have the passion and dedication that turns talent into excellence.
Welcome to Sounds Excellent 2018!
Nelson Bays Harmony, directed by Kathy Jamieson, has been part of the Nelson music scene for over 20 years. With a focus on performance excellence and a great sense of humour, this wonderful team of women sings four-part a cappella harmonies that are a joy to listen to. The group's repertoire spans popular songs from the 1920s through to modern pop, moving ballads, and upbeat songs that make you want to dance in the aisles.
Bryce Wastney is one of the region's finest singer-songwriters. He has performed extensively across NZ and Australia. Inspired by the music of Cat Stevens, Neil Finn and The Beatles, Bryce's music traverses folk, country, gospel, rock and pop – always soulful whatever the genre. It's the ideal soundtrack for the traveller and free spirit in us all.
The Bachelorettes (Joy Lee, Jasmine Jessen and Hannah Johns) are a string trio from Nelson College for Girls. They were Highly Commended in this year's NZCT Regional Chamber Music Contest. Be prepared to tap your feet along to their rendition of two tango-inspired pieces by Astor Piazzolla.
NCMA Young Voices Choir was set up in November 2017 following a brilliant performance by the original Children's Choir of Karl Jenkin's Cantata Memoria. Tutored by Zoe Bennett, the choir currently has 14 singers, from 8 to 13 years old, and performs all types of musical styles from classical to contemporary music, musical theatre, world music and Aotearoa New Zealand repertoire.
Nelson's oldest world music choir, Mosaic has been bringing a cappella pleasure to audiences for 21 years with their high energy and vibrancy. The choir is directed by Shannel Courtney and has been described as one of Nelson hidden treasures. Their repertoire covers most continents of the world, including songs from the Pacific and a focus on African music with its uplifting harmonies and rhythms. The inspiration that drives Mosaic is the sense of community that singing fosters and the privilege of experiencing the diversity of musical cultures through song.
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{
"redpajama_set_name": "RedPajamaC4"
}
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{"url":"http:\/\/condensedconcepts.blogspot.com\/2016\/11\/","text":"Wednesday, November 30, 2016\n\nPhotosynthesis is incoherent\n\nBeginning in 2007 luxury journals published some experimental papers making claims that quantum coherence was essential to photosynthesis. This was followed by a lot of theoretical papers claiming support. I was skeptical about these claims and in the first few years of this blog wrote several posts highlighting problems with the experiments, theory, interpretation, and hype.\n\nHere is a recent paper that repeats one of the first experiments.\n\nNature does not rely on long-lived electronic quantum coherence for photosynthetic energy transfer Hong-Guang Duan, Valentyn I. Prokhorenko, Richard Cogdell, Khuram Ashraf, Amy L. Stevens, Michael Thorwart, R. J. Dwayne Miller\nDuring the first steps of photosynthesis, the energy of impinging solar photons is transformed into electronic excitation energy of the light-harvesting biomolecular complexes. The subsequent energy transfer to the reaction center is understood in terms of exciton quasiparticles which move on a grid of biomolecular sites on typical time scales less than 100 femtoseconds (fs). Since the early days of quantum mechanics, this energy transfer is described as an incoherent Forster hopping with classical site occupation probabilities, but with quantum mechanically determined rate constants. This orthodox picture has been challenged by ultrafast optical spectroscopy experiments with the Fenna-Matthews-Olson protein in which interference oscillatory signals up to 1.5 picoseconds were reported and interpreted as direct evidence of exceptionally long-lived electronic quantum coherence. Here, we show that the optical 2D photon echo spectra of this complex at ambient temperature in aqueous solution do not provide evidence of any long-lived electronic quantum coherence, but confirm the orthodox view of rapidly decaying electronic quantum coherence on a time scale of 60 fs. Our results give no hint that electronic quantum coherence plays any biofunctional role in real photoactive biomolecular complexes. Since this natural energy transfer complex is rather small and has a structurally well defined protein with the distances between bacteriochlorophylls being comparable to other light-harvesting complexes, we anticipate that this finding is general and directly applies to even larger photoactive biomolecular complexes.\nI do not find the 60 fsec timescale surprising. In 2008, Joel Gilmore and I published a review of experiment and theory on a wide range of biomolecules (in a warm wet environment) that suggested that tens of femtoseconds is the relevant time scale for decoherence.\n\nI found the following section of the paper (page 7) interesting and troubling.\nThe results shown in Figs. 3 (a) and (b) prove that any electronic coherence vanishes within a dephasing time window of 60 fs. It is important to emphasize that the dephasing time determined like this is consistent with the dephasing time of \u03c4hom = 60 fs independently derived from the experiment (see above). It is important to realize that this cross-check constitutes the simplest and most direct test for the electronic dephasing time in 2D spectra. In fact, the only unique observable in 2D pho- ton echo spectroscopy is the homogeneous lineshape. The use of rephasing processes in echo spectroscopies removes the inhomogeneous broadening and this can be directly inferred by the projection of the spectrum on the antidiagonal that shows the correlation between the excitation and probe fields. This check of self-consistency has not been made earlier and is in complete contradiction to the assertion made in earlier works. Moreover, our direct observation of the homogeneous line width is in agreement with independent FMO data of Ref. 12. This study finds an \u223c 100 cm\u22121 homogeneous line width estimated from the low-temperature data taken at 77 K, which corresponds to an electronic coherence time of \u223c 110 fs, in line with our result given the difference in temperature. In fact, if any long lived electronic coherences were operating on the 1 ps timescale as claimed previously (1), the antidiagonal line width would have to be on the order of 10 cm\u22121, and would appear as an extremely sharp ridge in the 2D inhomogeneously broadened spectrum (see Supplementary Materials). The lack of this feature conspicuously points to the misassignment of the long lived features to long lived electronic coherences where as now established in the present work is due to weak vibrational coherences. The frequencies of these oscillations, their lifetimes, and amplitudes all match those expected for molecular modes (41, 42) and not long-lived electronic coherences.\n\nMonday, November 28, 2016\n\nPolanyi and Emergence before \"More is Different\"\n\nThe common narrative in physics is that the limitations of reductionism, the importance of emergence, and the stratification of scientific fields and concepts were first highlighted in 1972, \u00a0by P.W. Anderson in a classic article, \"More is Different\" published in Science. Anderson nicely used broken symmetry as an example of an organising principle that occurs at one strata and as a result of the thermodynamic limit.\n\nThe article was based on lectures Anderson gave in 1967.\nThe article actually does not seem to contain the word \"emergence\". He talks about new properties \"arising\".\n\nI recently learned how similar ideas about emergence and the stratification of fields was enunciated earlier by Michael Polanyi, in\u00a0The Tacit Dimension, published in 1966, based on his 1962 Terry Lectures at Yale.\nThe book contains a chapter entitled \"Emergence\".\n\nHere is a quote:\nyou cannot derive a vocabulary from phonetics; you cannot derive the grammar of language from its vocabulary; a correct use of grammar does not account for good style; and a good style does not provide the content of a piece of prose. ... it is impossible to represent the organizing principles of a higher level by the laws governing its isolated particulars.\nMuch of the chapter focuses on biology and the inadequacy of genetic reductionism. These ideas were expanded in a paper, \"Life's irreducible structure,\" published in Science in 1968.\n\nI recently learned about Polanyi's contribution from\nThe concept of emergence in social sciences: its history and importance\nG.M. Hodgson\n\nHere is a bit of random background.\n\nBefore turning to philosophy, Polanyi worked very successfully in Physical Chemistry. Some readers will know him for his contributions to reaction rate theory, the transition state, a diabatic state description of proton transfer, the LEPS potential energy surface based on valence bond theory, ...\n\nPolanyi was the Ph.D. advisor of Eugene Wigner. Melvin Calvin, a postdoc with Polanyi, and his son, John Polanyi, went on to win Nobel Prizes in Chemistry.\n\nGoogle Scholar lists \"The Tacit Dimension\" with almost 25,000 citations.\nThe book was recently republished with a new foreword by Amartya Sen, Nobel Laureate in Economics.\n\nFriday, November 25, 2016\n\nShould you quit social media?\n\nThe New York Times has an interesting Op-ed. piece\u00a0Quit Social Media. Your Career May Depend on It, by Cal Newport, a faculty member in computer science at Georgetown University.\n\nWhen I saw the headline I thought the point was going to be an important one that has been made many times before; people sometimes post stupid stuff on social media and get fired as a result. Don't do it!\nHowever, that is not his point.\nRather, he says social media is bad for two reasons:\n\n1. It is a distraction that prevents deep thinking and sustained \u00a0\"deep\" work. Because you are constantly looking at your phone, tablet, or laptop or posting on it, you don't have the long periods of \"quiet\" time that are needed for substantial achievement.\n\n2. Real substantial contributions will get noticed and recognised\u00a0without you constantly \"tweeting\" or posting about what you are doing or have done. Cut back on the self-promotion.\n\nOverall, I agree.\n\nWhen I discussed this and my post about 13 hour days with two young scientists at an elite institution they said: \"you really have no idea how much time some people are wasting on social media while in the lab.\" Ph.D students and postdocs may be physically present but not necessarily mentally or meaningfully engaged.\n\nA similar argument for restraint, but with different motivations, is being advocated by Sherry Turkle, a psychologist at MIT. Here is a recent interview.\n\nI welcome discussion.\n\nThursday, November 24, 2016\n\nThe many scales of emergence in the Haldane spin chain\n\nThe spin-1 antiferromagnetic Heisenberg chain provides a nice example of emergence in a quantum many-body system. Specifically, there are three distinct phenomena that emerge that were difficult to anticipate: the energy gap conjectured by Haldane, topological order, and the edge excitations with spin-1\/2.\n\nAn interesting question is whether anyone could have ever predicted these from just knowing the atomic and crystal structure of a specific material. I suspect Laughlin and Pines would say no.\n\nTo understand the emergent properties one needs to derive effective Hamiltonians at several different length and energy scales. I have tried to capture this in the diagram below. In the vertical direction, the length scales get longer and the energy scales get smaller.\n\nIt is interesting that one can get the Haldane gap from the non-linear sigma model. However, it coarse grains too much and won't give the\u00a0topological order or the edge excitations.\n\nIt seems to me that the profundity of the emergence that occurs at the different strata (length scales) is different. At the lower levels, the emergence is perhaps more \"straightforward\" and less surprising or less\u00a0singular (in the sense of Berry).\n\nAside. I spend too much time making this figure in PowerPoint. Any suggestions on a quick and easy way to make such figures?\n\nAny comments on the diagram would be appreciated.\n\nWednesday, November 23, 2016\n\nIt has dubious origins.\n\nI find this a bit embarrassing because there are many scientists, more distinguished than I, who do not have pages.\nWhen people tell me how impressed they are I tell them the story.\n\nAlmost ten years ago some enthusiasts of \"quantum biology\" invited me to contribute a chapter to a book on the subject. The chapter I wrote, together with two students, was different from most of the other chapters because we focussed on realistic models and estimates for quantum decoherence in biomolecules. (Some of the material is here.) This leads one to be very skeptical about the whole notion that quantum coherence can play a significant role in biomolecular function, let alone biological processes. Most other authors are true believers.\n\nI believe that to promote the book one of the editors had one of his Ph.D. students [who appeared to also do a some of the grunt work of the book editing] create a Wikipedia page for the book and for all of the senior authors. These pages emphasised the contribution to the book and the connection to quantum biology.\n\nThe \"history\" of my page states it was created by an account that\nAn editor has expressed a concern that this account may be a\u00a0sock puppet\u00a0of\u00a0Bunzil\u00a0(talk\u00a0\u00b7\u00a0contribs\u00a0\u00b7\u00a0logs).\nI have since edited my page to remove links and references to the book since it is not something I want to be defined by.\n\nAn aside. Today I updated the page because when giving talks I got tired of sometimes being introduced based on outdated information on the page.\n\nHardly, a distinguished history....\n\nThe xkcd cartoon is from here.\n\nMonday, November 21, 2016\n\nThe \"twin\" excited electronic state in strong hydrogen bonds\n\nOne of the key predictions of the\u00a0diabatic\u00a0state picture of hydrogen bonding\u00a0is that there should be an excited electronic state (a twin state) which is the \"anti-bonding\" combination of the two diabatic states associated with the ground state H-bond.\nRecently, I\u00a0posted\u00a0about\u00a0a possible identification of this state in malonaldehyde.\n\nThe following recent paper is relevant.\n\nSymmetry breaking in the axial symmetrical configurations of enolic propanedial, propanedithial, and propanediselenal: pseudo Jahn\u2013Teller effect versus the resonance-assisted hydrogen bond theory\nElahe Jalali, Davood Nori-Shargh\n\nThe key figure is below. The lowest B2 state is the twin state.\nIn the diabatic state picture, Delta is half of the off-diagonal matrix element that couples the two diabatic states.\nSimilar diagrams occur when O is replaced with S or Se.\n\nThe paper does not discuss twin states, but interprets everything in terms of the framework of the\n(A1\u00a0+\u00a0B2) \u2297\u00a0bpseudo-Jahn-Teller effect.\n\nIt uses TD-DFT (Time-dependent Density Functional Theory). It is contentious how reliable that is for excited states in organic molecules.\nThe diabatic states are not explicitly constructed.\nThese issues could be addressed by using higher level quantum chemistry and constructing the diabatic states by a systematic procedure, as was done by Seth Olsen for a family of methine dye molecules.\n\nA video illustrating the length scales of the universe\n\nSometimes when I speak about science to church groups I show the old (1977) video Powers of Ten which nicely illustrates the immense scale of the universe and orders of magnitude.\nI often wished there was a more polished modern version.\nYesterday it was pointed out to me there is,\u00a0Cosmic Eye.\n\nThe phone app can be purchased here\u00a0for $1. Friday, November 18, 2016 Desperately seeking Weyl semi-metals In 2011 it was proposed that pyrochlore iridates (such as Y2Ir2O7) could exhibit the properties of a Weyl semi-metal, the three-dimensional analog of the Dirac cone found in graphene. Since the sociology of condensed matter research is driven by exotica this paper stimulated numerous theoretical and experimental studies. However, as often is the case, things turn out to be more complicated and it seems unlikely that these materials exhibit a Weyl semi-metal. This past week I have read several nice papers that address the issue. Variation of optical conductivity spectra in the course of bandwidth-controlled metal-insulator transitions in pyrochlore iridates K. Ueda, J. Fujioka, and Y. Tokura There is a very nice phase diagram which shows systematic trends as a function of the ionic radius of the rare earth element R=Y, Dy, Gd, ... Most of the materials are antiferromagnetic insulators. The colour shading describes the low energy spectral weight in the optical conductivity up to 0.3 eV. Blue is an insulator and red actually means a very small low energy spectral weight. N can be thought of as the number of charge carriers per unit cell. Specifically, if this was a simple weakly interacting Fermi liquid N=1. Thus, the value of 0.05 for Pr signifies strong electron correlations. [Unfortunately, the paper talks about this as \"weak correlations\"]. In fact, as shown below even in the metallic phase at T=50 K one cannot see the Drude peak down to 10 meV. This presents a theoretical challenge to explain this massive redistribution of spectral weight. Slater to Mott Crossover in the Metal to Insulator Transition of Nd2Ir2O7 M. Nakayama, Takeshi Kondo, Z. Tian, J. J. Ishikawa, M. Halim, C. Bareille, W. Malaeb, K. Kuroda, T. Tomita, S. Ideta, K. Tanaka, M. Matsunami, S. Kimura, N. Inami, K. Ono, H. Kumigashira, L. Balents, S. Nakatsuji, and S. Shin This ARPES study does find band touching at the magnetic metal-insulator transition temperature but as the temperature is lowered the spectral weight is suppressed and there is no sign of Weyl points. Phase Diagram of Pyrochlore Iridates: All-in\u2013All-out Magnetic Ordering and Non-Fermi-Liquid Properties H Shinaoka, S Hoshino, M Troyer, P Werner This LDA+DMFT study shows that a three-band description is important for the R=Y compound. This sets the stage for describing the phase diagram above. I thank Prachi Telang for discussions at IISER Pune about these materials and bad semi-metals that stimulated this post. Wednesday, November 16, 2016 Many reasons why you should NOT work 13 hours per day I am very disturbed at how I encounter people, particularly young people, who work ridiculously long hours. Furthermore, it worries me that some are deluded about what they might achieve by doing this. Due to a variety of cultural pressures I think Ph.D. students from the Majority World are particularly prone to this. First let's not debate exactly how many hours is too many or exceptions to the generalisations below. At the end I will give some caveats. Here are some reasons why very long hours are not a good idea. Something may snap. And, when it does it will be very costly. It may be your mental or physical health, or your spouse, or your children, ... Don't think it won't happen. It does. Long hours may be making you quite inefficient and unproductive. You become tired and can't think as clearly and so make more mistakes, have less ideas, and find it harder to prioritise. It is a myth that long hours is mostly what you need to do to survive or prosper in science. I claim dumb luck is the biggest determining factor in getting a faculty position. Furthermore, when I look at people [students, postdocs, facutly] I don't observe a lot of correlation between real productivity and the hours they work. There are other things that are much more important than long hours. Some of these I have covered in posts about basic but important skills. Others include knowing the big picture, giving good talks, ... These are necessary but not sufficient conditions for survival. Yet many who are \"lab slaves\" seem oblivious to do this. They may have unrealistic expectations about what the long hours will lead to. Some even think long hours are a sufficient condition for survival. You may be wasting a lot of time. Because you can't think clearly and\/or just do whatever your boss or manager tells you to, you may spend a lot of time on tasks that have almost no chance of succeeding: poorly formulated experiments or calculations, applying for grants or jobs out of your league, submitting papers to luxury journals, ... There are also all those papers that you or your boss did not finish. You worked long hours in the lab to get the data and then the paper was never brought to completion because you and\/or your boss had moved on to the next crisis\/opportunity\/hot topic. It may rob you of your joy of doing science. It may be an addiction. Workaholism is as dangerous and as costly as alcoholism, drug and sexual addictions. The only difference is that workaholism is often seen as a virtue. You DO have a choice. One of the great lies of life in the affluent modern West is that people do not have many choices. This is exactly what employers and governments want us to believe. A problem is that people make choices [e.g. I have to get a permanent job in a research university, I have to have a big house, I have to send my kids to a private school, ...] that then severely constrain other choices. You may be being exploited. Universities and many PI's love cheap and compliant labour, whether it is grad students, \"adjunct faculty\" [teaching staff on short term contracts], or \"visiting scholars\" from the Majority World. A few years from now you may regret it. You may have left academia and realise you could have got your current job without working 3 extra hours a day. Why did you do it? Your spouse [if they are still around] sure wishes you hadn't. How many hours is too many? I don't know. There is significant variability in people's stamina and makeup. There are also differences in personal circumstances [e.g. a single person versus someone with two young children at home]. Different tasks in science [analytical calculations, writing, discussing, device fabrication, computer coding, babysitting experiments, ...] differ significantly in how taxing they are intellectually, physically, or emotionally. Also, there may be certain deadlines or tasks that require long hours for a short period of time [a visit to a synchrotron, monitoring a chemical reaction that takes 18 hours, the last week of finishing a thesis, ...] . This is not what I am talking about. I am talking about an unhealthy lifestyle that does not deliver what it claims to. How do you get out of this? First take a break so you can see more clearly the problem. Set some boundaries. Just say NO! Talk to others about the issue. Aim to work smarter not longer. I welcome comments. Monday, November 14, 2016 Why are the macroscopic and microscopic related? Through a nice blog post by Anshul Kogar, I became aware of a beautiful Physics Today Reference Frame (just 2 pages!) from 1998 by Frank Wilczek Why are there Analogies between Condensed Matter and Particle Theory? It is worth reading in full and slowly. But here a few of the profound ideas that I found new and stimulating. A central result of Newton's Principia was \"to prove the theorem that the gravitational force exerted by a spherically symmetric body is the same as that due to an ideal point of equal total mass at the body's center. This theorem provides quite a rigorous and precise example of how macroscopic bodies can be replaced by microscopic ones, without altering the consequent behavior. \" More generally, we find that nowhere in the equations of classical mechanics [or electromagnetism] is there any quantity that fixes a definite scale of distance. Only with quantum mechanics do fundamental length scales appear: the Planck length, Compton wavelength, and Bohr radius. Planck's treatment of blackbody radiation [macroscopic phenomena] linked it to microscopic energy levels. Einstein then performed a similar link between the specific heat of a crystal and the existence of phonons: the first example of a quasi-particle. Aside: I need to think of how these two examples do or do not fit into the arguments and examples I give in my emergent quantum matter talk. Wilczek says it is certainly not logically necessary for there to be any deep resemblance between the laws of a macroworld and those of the microworld that produces it an important clue is that the laws must be\" upwardly heritable\" [This is Wilczek's own phrase which does not seem to have been picked up by anyone later, including himself.] the most basic conceptual principles governing physics as we know it - the principle of locality and the principle of symmetry .... - are upwardly inheritable. He then adds the \"quasi material nature of apparently empty space.\" Overall, I think my take might be a little different. I think the reason for the analogies in the title are that there are certain organising principles for emergence [renormalisation, quasi-particles, effective Hamiltonians, spontaneous symmetry breaking] that transcend energy and length scales. The latter are just parameters in the theory. Depending on the system they can vary over twenty orders of orders of magnitude (e.g., from cold atoms to quark-gluon plasmas). But, perhaps Wilczek would say that once you have symmetry and locality you get quantum field theory and the rest follows.... What do you think? Friday, November 11, 2016 Telling students my personal teaching goals and philosophy It is strange that I have never done this. Furthermore, I don't know anyone who does. Why do this? First, it is helpful for me to think about and decide what my goals actually are, particular relating to the big picture. Second, it will be helpful for students to know. Too often they are guessing. Even worst, I fear that most just assume that my goals are theirs. Then they get frustrated if\/when they discover their goals and\/or values are different. So here are some goals I could think of. They are listed in order of decreasing importance to me. To help you learn to THINK. To inspire you to learn. To help you see this is a beautiful subject. To help you learn skills that are useful in other endeavors (including outside physics). The help you put this subject in the context of others. To help you learn the technical details of the subject. To be your ally not your adversary . My goals are NOT the following. (Listed in no particular order). To make you happy. To spoon feed you. To make life difficult for you. To get high scores on your evaluations of my teaching. To recruit you as a Ph.D. student to work with me. Have you ever done anything like this? Have you ever been in a class where it was done? Do you know anyone who does it? Are there benefits? Thursday, November 10, 2016 Irreversibility is an emergent property Time has a direction. Macroscopic processes are irreversible. Mixing is a simple example. The second law of thermodynamics encodes universal property of nature. Yet the microscopic laws of nature [Newton's equations or Schrodinger's equation] are time reversal invariant. There is no arrow of time in these equations. So, where does macroscopic irreversibility come from? It is helpful to think of irreversibility [broken time-reversal symmetry] as an emergent property. It only exists in the thermodynamic limit. Strictly speaking for a finite number of particles there is a \"recurrence time\" [whereby the system can return to close to its initial state]. However, for even as few as a thousand particles this becomes much longer than any experimental time scale. There is a nice analogy to spontaneously broken symmetry in phase transitions. Strictly speaking for a finite number of particles there is no broken symmetry as the system can tunnel backwards and forwards between different states. However, in reality for even a small macroscopic system the time scale for this is ridiculously long. Deriving irreversibility from microscopic equations is a major theoretical challenge. The first substantial contribution was that of Boltzmann's H-theorem. There are many subtleties associated with why it is not the final answer, but my understanding is superficial... This post was stimulated by some questions from students when I recently visited Vidyasagar University. Monday, November 7, 2016 A concrete example of a quantum critical metal I welcome comments on this preprint. Quantum critical local spin dynamics near the Mott metal-insulator transition in infinite dimensions Nagamalleswararao Dasari, N. S. Vidhyadhiraja, Mark Jarrell, and Ross H. McKenzie Finding microscopic models for metallic states that exhibit quantum critical properties such as$\\omega\/T$scaling is a major theoretical challenge. We calculate the local dynamical spin susceptibility$\\chi(T,\\omega)$for a Hubbard model at half filling using Dynamical Mean-Field Theory, which is exact in infinite dimensions. Qualitatively distinct behavior is found in the different regions of the phase diagram: Mott insulator, Fermi liquid metal, bad metal, and a quantum critical region above the finite temperature critical point. The signature of the latter is$\\omega\/T$scaling where$T\\$ is the temperature. Our results are consistent with previous results showing scaling of the dc electrical conductivity and are relevant to experiments on organic charge transfer salts.\nHere is the omega\/T scaling, which I think is quite impressive.\n\nSaturday, November 5, 2016\n\nThe role of simple models and concepts in computational materials science\n\nToday I am giving the first talk in a session on Computational materials science at the\u00a04th International Conference on Advances in Materials and Materials Processing.\n\nHere are the slides for my talk \"The role of simple models and concepts in computational materials science\".\n\nI will be referring the audience to the article such as those mentioned here, here and here that give a critical assessment of computer simulations and stress the importance of concepts.\n\nI welcome comments, particularly as I think the talk could be stronger and clearer.\n\nThursday, November 3, 2016\n\nVisit to a state university in India\n\nLike everything in India, higher education is incredibly diverse, both in quality, resources, and culture. These statistics give some of the flavour. There are about 800 universities. A significant distinction is between state and central universities. The former are funded and controlled by state governments. The latter (and IITs, IISERs, IISc, TIFR...) \u00a0are funded and controlled by the central (i.e. national\/federal) government. Broadly, the quality, resources, and autonomy (i.e. freedom from political interference) of the latter is much greater. On my many trips to India I have only visited these centrally funded institutes and universities.\n\nThis afternoon I looking forward to visiting the Physics Department of Vidyasagar University. It is funded by the West Bengal state government, and was started in 1981. It is named in honour of Ishwar Chandra Vidyasagar, a significant social reformer from the 19th century.\n\nI am giving my talk on \"Emergent Quantum Matter\".\nHere are the slides.\n\nUpdate. I enjoyed my visit and interacting with the faculty and students. On the positive side, people were enthusiastic and there were some excellent questions from the students. I want to write a blog post about one question. On the negative side, it is sad to see how poorly places like this are resourced: whether infrastructure, lab equipment, lab supplies, library, faculty, or salaries. For example, there are 5 physics faculty members and they teach a full M.Sc. [2 years course work] to about 100 students. This is 2 courses per faculty per semester and obviously, their expertise is stretched to cover all courses. The Ph.D. students mostly have full-time jobs elsewhere and come in the afternoons and evenings to work on their projects. One travels 2 hours each way on public transport.\n\nWednesday, November 2, 2016\n\nHydrogen bonding talk at IIT-Kgp\n\nToday I am giving a seminar, \"Effect of quantum nuclear motion on hydrogen bonding\" in the Chemistry Department at IIT Kharagpur. My host is Srabani Taraphder.\n\nHere are the slides. The talk is mostly based on this paper.\n\nTuesday, November 1, 2016\n\nOrganic spin liquid talk at IIT-Kgp\n\nToday I am giving a seminar in the Physics Department at Indian Institute of Technology (IIT) Kharagpur,\n\"Frustrated organic Mott insulators: from quantum spin liquids to superconductors.\"\nSlides are\u00a0here.\n\nDue to the recent Nobel Prize to Haldane, I included one slide about quantum spin liquids in one dimension.\n\nThe talk material is covered in great detail in a\u00a0review article, written together with Ben Powell.","date":"2017-08-24 05:02:35","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4426438808441162, \"perplexity\": 1739.8041711332226}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-34\/segments\/1502886133032.51\/warc\/CC-MAIN-20170824043524-20170824063524-00094.warc.gz\"}"}
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\section{Introduction}
The mean and median are basic methods for calculating central prototypes from a probability distribution. The median is commonly more robust to outliers than the mean.
Generalizations of such prototypes to Euclidean space can be formulated as a solution to an optimization problem. Suppose we have a set of points in Euclidean space, $\mathcal{X} = \{\mathbf{x}_i\}_{i=1}^p \subset \R^n$ which we would like to
represent as a prototype $\mathbf{y}$ via
solving
\begin{equation}\label{eq:euclopt}
\arg \min_{\mathbf{y} \in A} \sum_{i=1}^p \|\mathbf{x}_i - \mathbf{y}\|^q_2.
\end{equation}
The solution to~\eqref{eq:euclopt} for $A=\R^n$ and $q=2$ is called the centroid, which may be viewed as the generalization of the mean. In fact, the centroid is the component-wise mean of the vectors in $\mathcal{X}$, $\sum_{i=1}^p \mathbf{x}_i/p$. Generalizations of the median involve solving~\eqref{eq:euclopt} when $q=1$. When $A=\mathcal{X}$, the solution is called the medoid, while when $A=\R^n$, the solution is called the geometric median. The geometric median inherits the robustness to outliers from the median without being required to be a point in the dataset; however, it is not as straightforward to compute as a centroid since that calculation is not simply a least squares problem. An iterative algorithm for approximating a geometric median is the Weiszfeld algorithm~\cite{weiszfeld1937point}; each iteration of this algorithm is a weighted centroid problem. Thus, the Weiszfeld algorithm falls into a class known as Iteratively Reweighted Least Squares algorithms (IRLS). These Euclidean prototypes are used as a statistic for a dataset and in common machine learning algorithms like $k$-means and nearest centroid classification.
Not all datasets are best represented using points in Euclidean space. Specifically, image or video datasets are sometimes better represented using subspaces, i.e., as points on a Grassmannian. For example, the smallest principal angle between two subspaces
has proven powerful for modeling illumination
spaces~\cite{beveridge2008principal}.
Hyperspectral data may fail to be linearly separable in Euclidean space but separate
linearly on the Grassmannian~\cite{chepushtanova2017sparse}.
Hence, it is potentially useful to find versions of the Euclidean prototypes on the Grassmannian. A logical generalization of prototypes from Euclidean space to the Grassmannian is to replace the Euclidean $2$-norm in the optimization problems for the centroid, medoid and geometric median with a distance or dissimilarity between subspaces. The centroid is generalized using the geodesic distance in \cite{karcher1977riemannian} and the chordal distance in \cite{draper2014flag}. To the extent of our research, we have have not found a generalization of the medoid. However, the geometric median has been generalized using the geodesic distance and is called the $\ell_2$-median in \cite{aftab2014generalized, fletcher2009geometric}. Prototypes like these have been used alone as a method to classify emotion in images \cite{zhang2018grassmannian}, as a step in a k-means type algorithm \cite{marrinan2014finding} and in feature extraction \cite{fletcher2009geometric}.
Popular machine learning techniques, like dictionary learning, have been adapted to Riemannian manifolds \cite{xie2012dictionary}. Jayasumana et.\ al.\ consider learning on the Grassmannian (and Riemannian manifolds in general) with RBF kernels and advocate for chordal distance (sometimes referred to as the projection norm) kernels on the Grassmannian because chordal distance generates a positive definite Gaussian kernel \cite{jayasumana2015kernel}. More recently, Cherian et.\ al.\ use kernalized Grassmannian pooling for activity recognition \cite{cherian2018non}. Methods for Riemannian optimization like Riemannian SVRG have gained popularity alongside this surge of interest in Riemannian learning \cite{zhang2016riemannian}. Even more uses for subspaces in computer vision and machine learning can be found in, e.g., \cite{aftab2014generalized,bates2015max,draper2014flag,edelmangeometry,marks2012mean,marrinan2014finding,zhang2018grassmannian}.
In this paper we propose the flag median, a prototype which is a generalization of the geometric median to the Grassmannian using the chordal distance. We solve the flag median optimization problem using the novel FlagIRLS algorithm. The FlagIRLS is an IRLS algorithm on the Grassmannian that solves a weighted flag mean problem at each iteration similar to the way an iteration of the Weiszfeld algorithm solves a weighted centroid problem. We conduct experiments with the flag median, $\ell_2$-median and the flag mean on synthetic datasets, the MNIST handwritten digits dataset \cite{deng2012mnist}, the DARPA (Defense Advanced Research Projects Agency) Mind's Eye dataset used in \cite{marrinan2014finding} and the UCF YouTube action dataset \cite{liu2009recognizing}. In these examples we find that the FlagIRLS algorithm tends to converge quickly. We show that the flag
median appears to be more robust
to outliers than the flag mean and
$\ell_2$-median,
and produces the highest cluster purities in the LBG algorithm \cite{linde1980algorithm}.
\section{Background}\label{sec:background}
\subsection{Introduction to the Grassmannian}
For the purposes of this paper, the Grassmannian manifold (a.k.a. "the Grassmannian"), denoted $\Gr(k,n)$, is the manifold whose points correspond to the $k$ dimensional subspaces of $\R^n$. We will represent a point in $\Gr(k,n)$ using a tall $n \times k$ real matrix $\mathbf{X}$ with orthonormal columns. The point on $\Gr(k,n)$ determined by $\mathbf{X}$ is the column space of $\mathbf{X}$ and is denoted $[\mathbf{X}]$. Thus if $\mathbf{X}$ and $\mathbf{Y}$ have the same column space then they determine the same point $[\mathbf{X}]=[\mathbf{Y}]$ on $\Gr(k,n)$.
In order to allow more flexibility in our generalization of optimization problems to subspaces, we work with points that are not all necessarily on the same Grassmannian manifold but are in the same ambient space. Suppose we have a set of subspaces of $n$-dimensional space, $\{[\mathbf{X}_1],[\mathbf{X}_2], \dots, [\mathbf{X}_p]\}$, where $[\mathbf{X}_i] \in \Gr(k_i, n)$. We want to find an $r$-dimensional subspace of $\R^n$, $[\mathbf{Y}^*] \in \Gr(r,n)$, that is in some sense the center of these points, i.e., that $[\mathbf{Y}^*]$ is a solution to
\begin{equation}
\arg \min_{[\mathbf{Y}] \in \Gr(r,n)} \sum_{i=1}^p d([\mathbf{X}_i],[\mathbf{Y}])
\end{equation}
where $d$ measures dissimilarity between its arguments.
Principal angles between subspaces are a common dissimilarity measure that is invariant to orthogonal transformations \cite{bjorck1973numerical,draper2014flag}. Take $[\mathbf{X}], [\mathbf{Y}] \in \Gr(k,n)$. The $i$th smallest principal angle between $[\mathbf{X}]$ and $[\mathbf{Y}]$, $\theta_i ([\mathbf{X}], [\mathbf{Y}]) \in [0, \pi/2]$ is defined as the solution to~\eqref{eq: principal angle problem} \cite{bjorck1973numerical}.
\begin{align}\label{eq: principal angle problem}
\begin{aligned}
\cos \theta_i ([\mathbf{X}], [\mathbf{Y}]) &= \max_{\mathbf{x} \in [\mathbf{X}]} \: \: \max_{\mathbf{y} \in [\mathbf{Y}]} \: \: \mathbf{x}^T \mathbf{y} = \mathbf{x}_i^T \mathbf{y}_i\\
\text{Subject to } &\mathbf{x}^T\mathbf{x} = \mathbf{y}^T \mathbf{y} = 1\\
& \mathbf{x}^T \mathbf{x}_j = \mathbf{y}^T \mathbf{y}_j = 0 \text{ for } j=1,2,\dots, i-1
\end{aligned}
\end{align}
Now let $\theta([\mathbf{X}], [\mathbf{Y}]) \in \R^k$ be the vector of principal angles between $[\mathbf{X}]$ and $[\mathbf{Y}]$. The geodesic distance on $\Gr(k,n)$ is $\| \theta([\mathbf{X}], [\mathbf{Y}]) \|_2$ and the chordal distance on $\Gr(k,n)$ is $\| \sin( \theta([\mathbf{X}], [\mathbf{Y}]) ) \|_2$ \cite{conway1996packing}. We can calculate these quantities when $[\mathbf{X}] \in \Gr(k,n)$, $[\mathbf{Y}] \in \Gr(r,n)$ where $k \neq r$ by setting the last $\max(k,r) - \min(k,r)$ entries of $\theta([\mathbf{X}], [\mathbf{Y}]) \in \R^{\max(k,r)}$ to $0$.
\subsection{Geodesic Distance Prototypes}
The Euclidean mean and geometric median have been translated to the Grassmannian using the geodesic distances. The mean on the Grassmannian using geodesic distance (the solution to~\eqref{eq:geodes} for $q=2$) is called the Karcher mean and the geometric median using geodesic distance (the solution to~\eqref{eq:geodes} for $q=1$) is called the $\ell_2$-median.
\begin{equation}\label{eq:geodes}
\arg \min_{[\mathbf{Y}] \in \Gr(r,n)} \sum_{i=1}^p \| \theta ([\mathbf{X}_i],[\mathbf{Y}])\|_2^q.
\end{equation}
The Karcher mean and the $\ell_2$-median are only computable in the case where all subspaces are of equal dimensions (e.g., $r = k_1 = k_2 \cdots = k_p$) and \cite{marrinan2014finding} use examples to show that the available algorithms to compute these prototypes are slow.
The most common algorithm for finding the solution to the Karcher mean was discovered by Karcher \cite{karcher1977riemannian} and Fletcher et.\ al.\ \cite{fletcher2009geometric} show we can find the $\ell_2$-median using a Weiszfeld-type algorithm. \cite{marrinan2014finding} show that the Karcher mean is not only slow to compute, but also produces lower cluster purities than the $\ell_2$-median in their LBG clustering example so we choose not to use the Karcher mean as a prototype in our experiments (Section \ref{sec:experiments}).
For context, the Weiszfeld algorithm for vectors in $\R^n$ is stated in Algorithm~\ref{alg:weis_algorithm}.
\\
\begin{algorithm}[ht]
\SetAlgoLined
\KwData{ $\{\mathbf{x}_i\}_{i=1}^p \subset \R^n$}
\KwResult{ The geometric median $\mathbf{y} \in \R^n$}
\While{not converged}{
$w_i = \frac{p}{\| \mathbf{x}_i - \mathbf{y} \|_2} \left( \sum_{k=1}^p \frac{1}{\| \mathbf{x}_k - \mathbf{y} \|_2} \right)^{-1}$;\\
$\mathbf{y} \leftarrow \sum_{i=1}^p\frac{w_i \mathbf{x}_i}{p}$;
}
\caption{Weiszfeld Algorithm in $\R^n$}\label{alg:weis_algorithm}
\end{algorithm}
\\
Note that each iteration of the Weiszfeld algorithm (Algorithm \ref{alg:weis_algorithm}) is the solution to the least squares problem~\eqref{eq:euclopt} (with $A=\R^n$ and $q=2$) for the weighted vectors $w_i \mathbf{x}_i$.
The weights, $w_i$, come from the fact that the geometric median $\mathbf{y}$ satisfies (presuming $\mathbf{y} \notin \mathcal{X}$)
\[
\mathbf{y} = \left(\sum_{i=1}^p \frac{\mathbf{x}_i}{\| \mathbf{x}_i - \mathbf{y} \|_2} \right) \Big/ \left( \sum_{k=1}^p \frac{1}{\| \mathbf{x}_k - \mathbf{y} \|_2} \right).
\]
Fletcher et.\ al.\ solve~\eqref{eq:geodes} for $q=1$ by generalizing this approach to Riemannian manifolds.
In Section \ref{sec:experiments}, we use the unweighted Weiszfeld-type algorithm from Fletcher et.\ al.\ with geodesic distance to calculate the $\ell_2$-median on the Grassmannian. Let $d$ be the maximum distance between points in the dataset and let $\delta$ be the convergence parameter. We define $N_{d,\delta}$ as the number of iterations of one run of our implementation. The complexity of our implementation of this algorithm in Section \ref{sec:experiments} is ${O}\left(n p k^2N_{d,\delta}\right)$.
\subsection{The Flag Mean Prototype}
Draper et.\ al.\ \cite{draper2014flag} present the flag mean as an average of subspaces of different dimensions using the squared chordal distance. The optimization problem for the flag mean is
\begin{equation}\label{eq:flag_mean}
\arg \min_{[\mathbf{Y}] \in \Gr(r,n)} \sum_{i=1}^p \| \sin(\theta ([\mathbf{X}_i],[\mathbf{Y}]))\|_2^2.
\end{equation}
This flag mean determines not only a point on $\Gr(r,n)$, it determines a point on various flag manifolds. A flag manifold is a manifold whose points represent a flag of subspaces $[\mathbf{S}_1] \subset [\mathbf{S}_2] \subset \dots \subset [\mathbf{S}_r] = \mathbb R^n$. If we let $s_i = \text{dim}([\mathbf{S}_i])$, then we say the flag is of type ${s_1,s_2,\dots,s_r}$. For more details on flag manifolds, see \cite{monk1959geometry}.
We will refer to a flag mean in this paper as the point on $Gr(r,n)$ determined by the flag. Let $[\mathbf{Y}]$ be the flag mean of $\{[\mathbf{X}_i]\}_{i=1}^k$. Let $\mathbf{y}_i$ be the $i$th column of $\mathbf{Y}$, the orthonormal matrix representation of $[\mathbf{Y}]$. Then the $r^{th}$ ``real'' flag mean is the point on the flag manifold of type $\{1,2,\dots, r, n\}$ defined as in~\eqref{eq:point_on_flag}.
\begin{equation}\label{eq:point_on_flag}
\llbracket \mathbf{Y} \rrbracket = \text{span}\{\mathbf{y}_1\} \subset \text{span}\{\mathbf{y}_1,\mathbf{y}_2\} \subset \dots \subset \text{span}\{\mathbf{y}_1,\mathbf{y}_2, \dots , \mathbf{y}_r\} \subset \mathbb R^n
\end{equation}
The point on $Gr(r,n)$ determined by the flag is $\text{span}\{\mathbf{y}_1,\mathbf{y}_2, \dots , \mathbf{y}_r\} \subset \mathbb R^n$.
Draper et.\ al.\ ~\cite{draper2014flag} show that we can calculate the flag mean by utilizing the singular value decomposition (SVD) of the matrix $[ \mathbf{X}_1, \mathbf{X}_2, \dots , \mathbf{X}_p ]$. The flag mean, as a point on $Gr(r,n)$, is the span of the $r$ left singular vectors of the corresponding to the $r$ largest singular values.
The complexity of this algorithm is $O\left(n \left(\sum_{i=1}^{p}k_i\right)^2\right)$. Marks \cite{marks2012mean} suggests calculating weighted flag means in his dissertation. This weighted flag mean calculation will be used as an iteration of the FlagIRLS algorithm introduced in Section \ref{sec:flagirls}.
\section{Flag Median}
The translation of the geometric median to the Grassmannian using chordal distance is called the flag median. The optimization problem for this novel prototype is in~\eqref{eq:Flag Median}.
\begin{equation}\label{eq:Flag Median}
\arg \underset{[\mathbf{Y}] \in Gr(r,n)}{\min} \sum_{i=1}^p \|\sin \theta([\mathbf{X}_i], [\mathbf{Y}])\|_2
\end{equation}
We call this the flag median since, using FlagIRLS, $[\mathbf{Y}]$ actually is a flag of subspaces rather than a single $r$ dimensional subspace of $n$ dimensional space. This flag median is indeed a median (similar to the geometric median) because it minimizes the chordal distance rather than the squared chordal distance problem in~\eqref{eq:flag_mean} that is solved by the flag mean.
\subsection{Derivation}\label{sec:derivations}
In this section we show that the FlagIRLS algorithm can be used to approximate the flag median. The algorithm derived in this section revolves around weighted flag means of $\{[\mathbf{X}_i]\}_{i=1}^p$. For the rest of this paper we will denote the weight of the subspace $[\mathbf{X}_i]$ as $w_i$.
Notice that the flag median optimization problem in~\eqref{eq:Flag Median} involves a sum of two norms of the vector of sines of principal angles and the flag mean optimization problem in~\eqref{eq:flag_mean} involves squared two norms of the same vector. So, in other words, we are deriving an algorithm, analogous to Weiszfeld and IRLS in the Euclidean setting, that approximates solutions to the $2$-norm problem by iteratively solving squared $2$-norm problems. So the FlagIRLS algorithm \ref{alg:main_algorithm} provides an ``iterative reweighted least squares'' method for approximating the flag median.
Let us begin by translating the flag median problem from~\eqref{eq:Flag Median} to an optimization problem over matrices with orthonormal columns. The eigenvalues of $\mathbf{Y}^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{Y}$ are the entries in the vector $\cos^2( \theta([\mathbf{X}_i], [\mathbf{Y}]))$. Using properties of trace we can show $\text{tr}(\mathbf{Y}^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{Y}) = \sum_{j=1}^{m_i} \cos^2 \theta_j([\mathbf{X}_i], [\mathbf{Y}])$ \cite{bjorck1973numerical}. This allows us to rewrite the flag median problem from~\eqref{eq:Flag Median} as the matrix optimization problem in~\eqref{eq:sin_matrix} where $m_i = \min(r,k_i)$.
\begin{equation}\label{eq:sin_matrix}
\min_{\substack{\mathbf{Y} \in \R^{n \times r} \\ \mathbf{Y}^T\mathbf{Y} = I}} \sum_{i=1}^p \left( m_i-\text{tr}(\mathbf{Y}^T \mathbf{X}_i \mathbf{X}_i^T\mathbf{Y})\right)^{1/2}
\end{equation}
We formulate a Lagrangian from this problem using $\Lambda$ as a symmetric matrix of Lagrange multipliers with entries $\lambda_{ij}$ in~\eqref{eq:sin_lagrange}.
\begin{align}\label{eq:sin_lagrange}
\begin{aligned}
\mathcal{L}(\mathbf{Y},\Lambda) = &\sum_{i=1}^p \left(m_i-\text{tr}(\mathbf{Y}^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{Y})\right)^{1/2} \\
&- \langle \Lambda, \mathbf{Y}^T\mathbf{Y} - I \rangle
\end{aligned}
\end{align}
We then calculate~\eqref{eq:sin_algebra} the Lagrangian with respect to the $j$th column of $\mathbf{Y}$, namely $\mathbf{y}_j$, and set it equal to $0$.
\begin{equation}\label{eq:sin_algebra}
\mathbf{y}_j^T \sum_{i=1}^p \frac{-1}{\left(m_i-\text{tr}(\mathbf{Y}^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{Y})\right)^{1/2}} \mathbf{X}_i \mathbf{X}_i^T \mathbf{y}_j = 2\lambda_{jj}
\end{equation}
Now define the matrix $\mathbf{X}$
\begin{equation}
\label{eq:big mat}
\mathbf{X} = \left[ w_1 \mathbf{X}_1 , w_2 \mathbf{X}_2 , \cdots , w_p \mathbf{X}_p \right]
\end{equation}
where $ w_i = \left(\frac{1}{m_i-\text{tr}(\mathbf{Y}^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{Y})}\right)^{1/4}.$
Combining the information in \eqref{eq:sin_algebra} and the matrix $\mathbf{X}$ in \eqref{eq:big mat}, we see that $\mathbf{Y}$ must be $r$ left singular vectors of $\mathbf{X}$ when $[\mathbf{Y}]$ to be the flag median of $\{[\mathbf{X}_i]\}_{i=1}^p$.
Now let us consider an iterative algorithm with the $j$th iteration of the form $\mathbf{Y}_{j+1} = \text{Flag Mean}\left( \left \{w_i^{(j)} \mathbf{X}_i\right \}_{i=1}^p \right)$ where
$w_i^{(j)} = \left(\frac{1}{m_i-\text{tr}(\mathbf{Y}_j^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{Y}_j)}\right)^{1/4}$.
This algorithm will be formalized in Section \ref{sec:flagirls}. For this type of algorithm, we desire $\mathbf{Y}_{j+1}$ to be an approximation of the flag median of the dataset $\{\mathbf{X}_i\}_{i=1}^p$. We will now show that the columns of $\mathbf{Y}_{j+1}$ should be chosen as the left singular vectors of $\mathbf{X}$ associated with the $r$ largest singular values and therefore the update using the flag mean of $\left \{w_i^{(j)} \mathbf{X}_i \right \}_{i=1}^p$ is the correct update choice.
Let $\mathbf{U}^*$ be some matrix consisting of $r$ left singular vectors of $\mathbf{X}$. To determine $\mathbf{U}^*$ where
\begin{equation}\label{eq:sine_reiterated}
\mathbf{U}^* = \arg \min \sum_{i=1}^p \left( q - \text{tr} (\mathbf{U}^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{U}) \right)^{1/2}.
\end{equation}
we solve the optimization problem
\begin{equation}\label{eq:sine_opt}
\begin{aligned}
&\max_{\substack{ \mathbf{U} \in \R^{n \times r} \\ \mathbf{U}^T \mathbf{U} = \mathbf{I}}} \text{tr} \left( \mathbf{U}^T \mathbf{X} \mathbf{X}^T \mathbf{U} \right) \\
\end{aligned}
\end{equation}
which requires the columns of $\mathbf{U}^*$ to be the left singular vectors of
$\mathbf{X}$
associated with the largest singular values. So we take our update to be $\mathbf{Y}_{j+1} = \mathbf{U}^*$.
\subsection{The FlagIRLS Algorithm}\label{sec:flagirls}
We use an iteratively reweighted least squared flag mean algorithm structure to solve~\eqref{eq:Flag Median}. We will call the weight for subspace $\mathbf{X}_i$, $w_i$. A concern with these weights arises when the denominator of a weight is zero. For the flag median objective function, the denominator is zero when $[\mathbf{Y}]$ is a subspace of $[\mathbf{X}]$ or vise versa.
To avoid singularities, we added a small quantity $\epsilon$ to the denominator. The $w_i$ for the flag median problem is in~\eqref{eq:weights}
\begin{equation}\label{eq:weights}
w_i = \left( \frac{1}{m_i-\text{tr}(\mathbf{Y}^T \mathbf{X}_i \mathbf{X}_i^T \mathbf{Y})+\epsilon}\right)^{1/4}
\end{equation}
We use these weights, along with the flag mean, in the FlagIRLS algorithm as described in Algorithm \ref{alg:main_algorithm}.
\begin{algorithm}[ht]
\SetAlgoLined
\KwIn{ A set of orthonormal subspace representatives $\{\mathbf{X}_i\}_{i=1}^p$ for $\{[\mathbf{X}_i] \in \Gr(k_i, n)\}_{i=1}^p$}
\KwOut{ An orthonormal subspace representative $\mathbf{Y}$ for the flag median $[\mathbf{Y}] \in \Gr(r,n)$}
\While{not converged}{
assign each $w_i$\;
$\mathbf{X} \leftarrow \left [ w_1 \mathbf{X}_1 | w_2 \mathbf{X}_2 | \cdots | w_p \mathbf{X}_p \right]$\;
$\mathbf{U} \Sigma \mathbf{V} ^T = \mathbf{X}$ \%calculate the SVD \;
$\mathbf{Y} \leftarrow \mathbf{U} [:,1:r]$ \%first $r$ columns of $\mathbf{U}$;
}
\caption{The FlagIRLS algorithm. See~\eqref{eq:weights} for the algorithm weights. We assume the columns of $\mathbf{U}$ are sorted from the left singular vector associated with the smallest to the largest singular values of $\mathbf{X}$.} \label{alg:main_algorithm}
\end{algorithm}
An important note is that FlagIRLS is an iterative weighted flag mean algorithm so the outputs of this algorithm come from the left singular vectors of $\mathbf{X}$. We take $r$ singular vectors associated with the $r$ largest singular values of $\mathbf{X}$, i.e., the first $r$ columns of $\mathbf{U}$. However, there are $n$ columns of $\mathbf{U}$, so FlagIRLS actually outputs a flag of subspaces
$ [\mathbf{U} [:,1]] \subset [\mathbf{U} [:,:2]] \subset \cdots \subset [\mathbf{U} [:,:n]].$
This flag is used to distinguish between different prototypes in Section \ref{sec:mnist_experiments} with MNIST digits.
\section{Limitations}
The main limitation of calculating the flag median is the speed of FlagIRLS. This requires that we take the thin SVD of $\mathbf{X} \in \R^{n \times pk}$ every iteration in FlagIRLS. The complexity of the FlagIRLS algorithm is the complexity of the flag mean times the number of iterations of the algorithm, i.e., $O\left(n N_{\delta} \left(\sum_{i=1}^{p}k_i\right)^2\right)$ where $N_{\delta}$ is the number of iterations and $\delta$ is the convergence parameter.
Another current limitation of this work is the lack of proven mathematical guarantees for the flag median and the FlagIRLS algorithm. Although, for all our examples, FlagIRLS converges to a local minimum of the flag median problem, we have not worked out the mathematical theory to find the conditions where FlagIRLS converges. We also still need to determine the conditions where an iteration of FlagIRLS is a contraction mapping. On a larger scale, given a dataset of subspaces of $\R^n$, we have yet to determine where the flag median problem is convex. Section \ref{sec:derivations} shows that FlagIRLS is a logical algorithm for finding the flag median, but further development of the mathematical theory would give us more intuition about which datasets are good for FlagIRLS, how to initialize FlagIRLS and overall provide the user with
a better understand of rates of convergence.
Currently, the FlagIRLS algorithm is run with a number of different initializations to verify convergence.
\section{Experiments}\label{sec:experiments}
In this section we carry out experiments with synthetic data, the MNIST handwritten digits dataset \cite{deng2012mnist}, the Mind's Eye dataset \cite{marrinan2014finding} and the UCF YouTube action dataset \cite{liu2009recognizing}. The goal is to compare the flag median to the flag mean and the $\ell_2$-median and establish the efficiency of FlagIRLS. For all of this section we use FlagIRLS to compute the flag median and the Weiszfeld-type algorithm from \cite{fletcher2009geometric} for the $\ell_2$-median.
The convergence criteria for our implementation of FlagIRLS is as follows. We terminate the algorithm when objective function values of consecutive iterates of FlagIRLS are less than $\delta = 10^{-11}$, or if the $i$th iteration resulted in an increasing objective function value. In the former case, we output the $(i-1)$st iterate. For our weights in all examples we run the FlagIRLS algorithm with $\epsilon = 10^{-7}$.
The convergence criteria of our implementation of the Weiszfeld-type algorithm from \cite{fletcher2009geometric} to calculate the $\ell_2$-median is similar to the Flag IRLS convergence criteria.
We terminate the algorithm when when objective function values of consecutive iterates are less than $\delta =10^{-11} $.
Both FlagIRLS and the Weiszfeld-type algorithm are terminated when we have exceeded $1000$ iterations. FlagIRLS never exceeds $1000$ iterations in our examples.
\subsection{Synthetic Data}
We begin with two experiments on a dataset consisting of $10$ points from $\Gr(3,20)$ and $10$ points from $\Gr(5,20)$. A representative for a point on $\Gr(k,n)$ is sampled in two steps. The first step is to sample an $n \times k$ matrix from a uniform distribution on $[-.5, .5)$, $\mathcal{U} [-.5, .5)$. We then do the QR decomposition of this matrix to get a point on $\Gr(k,n)$. We perform two experiments on this dataset: The first experiment verifies convergence of FlagIRLS, and the second experiment compares the convergence rate of FlagIRLS to Grassmannian gradient descent.
For the first experiment, we run 100 trials of FlagIRLS with different random initializations. For each of these trials, we verify that we have converged by checking 100 points near the FlagIRLS algorithm output. Given one algorithm output, $[\mathbf{X}] \in \Gr(3,20)$, we sample the entries of $\mathbf{Y} \in \mathbb{R}^{20 \times 3}$ from $\mathcal{U} [-0.5, 0.5) $ and check the objective function value at the first $3$ columns of $\mathbf{Q}$ where $\mathbf{Q}$ comes from the QR decomposition of the matrix $\mathbf{X} + 0.00001 \mathbf{Y}$. We call these points ``test points'' for the algorithm output. We say the FlagIRLS algorithm for flag median converged when all the objective function values of the test points are less than or equal to the objective function value for the algorithm output. In this experiment, we find that $100 \%$ of the FlagIRLS trials converge.
We now show an example with the same dataset where we run FlagIRLS and Grassmannian gradient with $100$ random intializations to compute the flag median. The results of this experiment are in Figure \ref{fig:convergence examples}. For this example, Grassmannian gradient descent is implemented with a step size of $0.01$. We find that FlagIRLS converges in fewer iterations than Grassmannian gradient descent for the flag median problem.
\begin{figure}[ht]
\centering
\includegraphics[width=.7\textwidth]{FlagIRLS_convergence.png}
\caption[]{The mean objective function values over $100$ trials with different random initializations. FlagIRLS converges in fewer iterations than gradient descent for the flag median problem for the synthetic dataset.}
\label{fig:convergence examples}
\end{figure}
For our next example, we use a dataset of $200$ points on $\Gr(6,100)$. The points are sampled by first fixing a ``center'' point for the dataset, $[\mathbf{X}_*]$. We do this by taking a random $100 \times 6$ matrix with entries from $\mathcal{U} [-.5, .5)$. We then take $\mathbf{X}_*$ as the first $6$ columns of $\mathbf{Q}$ from the QR decomposition of this random matrix. The $200$ points in the dataset are now calculated via the following steps. For each point, we generate $\mathbf{Z}$ by sampling a random $100 \times 6$ matrix with entries sampled from $\mathcal{U} [-.5, .5)$ and scaling it by $0.01$. We then take the point determined by the first $6$ columns of $\mathbf{Q}$ from the QR decomposition of $\mathbf{X}_* + \mathbf{Z}$.
We then run our FlagIRLS and Weiszfeld-type algorithm implementations with
$20$ random initializations to calculate the flag median and the $\ell_2$-median respectively. For the random initializations, we initialize FlagIRLS and the Weiszfeld-type algorithm at the same point. The results of this experiment are in Table \ref{tab:FlagIRLS_vs_Weiszfeld-type}. We terminate the Weiszfeld-type algorithm after $1000$ iterations regardless of convergence. So perhaps, many of the high iteration runs of Weiszfeld still did not converge even after $1000$ iterations.
\begin{table}[]
\centering
\begin{tabular}{c||c}
Algorithm/ Initialization & Mean Iterations\\
\hline
\hline
FlagIRLS/random& $4.55 \pm 0.50$\\
\hline
Weiszfeld-Type/ datapoint & $795.50 \pm 227.40$\\
\hline
Weiszfeld-Type/ randomly & $968.80 \pm 94.49$\\
\end{tabular}
\caption{The mean number of iterations until convergence of $20$ random initalizations of FlagIRLS and the Weiszfeld-type algorithm on a dataset of $200$ points on $\Gr(6,100)$. FlagIRLS is converges in far fewer iterations than the Weiszfeld-type algorithm and also sports a much lower standard deviation in the number of iterations.}
\label{tab:FlagIRLS_vs_Weiszfeld-type}
\end{table}
Now we will use a dataset that consists of $200$ points on $\Gr(3,20)$. These points consist of a cluster of $180$ points centered around the subspace $[\mathbf{X}_*]$ and $20$ outlier points. The points from the $180$-point cluster are sampled by the following process. We calculate a fixed ``center'' point for the dataset, $[\mathbf{X}_*]$, by taking a $20 \times 3$ matrix with entries from $\mathcal{U} [-.5, .5)$. We then take $\mathbf{X}_*$ as the first $3$ columns of $\mathbf{Q}$ from the QR decomposition of this random matrix. We then calculate the points in the cluster via the following steps. The first step is to generate $\mathbf{Z}$ by sampling a random $20 \times 3$ matrix with entries sampled from $\mathcal{U} [-.5, .5)$ and scaling it by $0.01$. The second step generates one point in the 180 point cluster as the first 3 columns of $\mathbf{Q}$ from the QR decomposition of $\mathbf{X}_* + \mathbf{Z}$. A point from the set of outlier 20 points is the first $3$ columns of the QR decomposition of a random $20 \times 3$ matrix with entries sampled from $\mathcal{U} [-.5, .5)$.
Table \ref{tab:c_dists_center} shows the results of calculating the flag median, $\ell_2$-median and flag mean of this dataset and then computing the chordal distance between $[\mathbf{X}_*]$ and the three different prototypes. Notice the flag median is the least affected by the outliers, the $\ell_2$-median is twice as affected and the flag mean is ten times more affected by the outliers.
\begin{table}[ht]
\centering
\begin{tabular}{c|c}
Algorithm & Chordal Distance\\
\hline
Flag Median & 0.0017\\
$\ell_2$-median & 0.0022\\
Flag Mean & 0.0128\\
\end{tabular}
\caption{The chordal distance between the algorithm result and $[\mathbf{X}^*]$.}
\label{tab:c_dists_center}
\end{table}
\textit{Note: for Table \ref{tab:c_dists_center}, FlagIRLS converges to the flag median in one iteration.}
\subsection{MNIST Handwritten Digits dataset}\label{sec:mnist_experiments}
The MNIST digits dataset is a set of $28 \times 28$ single band images of handwritten digits \cite{deng2012mnist}. We represent an MNIST handwritten digit using an element of $\Gr(1,784)$ to by taking one image, vectorizing it, then dividing the resulting vector by its norm.
For our first example, we see how the flag median, $\ell_2$-median and flag mean prototypes are classified by a MNIST-trained 3-layer neural network. This trained neural network classifier has a $97\%$ test accuracy on the MNIST test dataset. We generate our datasets for this experiment by to taking 20 examples of the digit $1$ and $i$ examples of the digit $9$ from the MNIST training dataset. We let $i=0,1,2,3, \dots, 19$ and this results in $20$ datasets. For each of these datasets, we calculate the flag median, $\ell_2$-median and flag mean, then predict the class of each of these prototypes by passing each through the neural network classifier.
In Figure \ref{fig:MNIST_classifier} we plot the predicted class of each prototype for each dataset by the trained neural network. For this figure, we choose to use the random initialization that resulted in the best predictions of the $\ell_2$-median.
\begin{figure}[ht]
\centering
\includegraphics[width=.7\textwidth]{Classifier_Predictions.png}
\caption[]{The neural network predicted class of the prototype for the dataset with $20$ examples of $1$'s and $i$ examples of $9$'s with $i=0,1,2,\dots,19$.}
\label{fig:MNIST_classifier}
\end{figure}
The $\ell_2$-median and flag mean are misclassified with $i=9$ added examples of $9$'s whereas the flag median is still classified correctly for $i=10$ and $i=11$ added examples $9$'s. Therefore the flag median is the most robust prototype to outliers in this experiment. The common misclassification as $8$ is likely due to the fact that the $1$'s tend to be at an angle, so when averaged, they tend to look like fuzzy $8$'s, especially when some $9$'s have been introduced to the dataset.
Also, the $\ell_2$-median of a dataset of $20$ $1$'s with $15$ to $19$ $9$ digits is misclassified as a $7$. This is likely a result of the different angled $1$'s and the introduction of the examples of $9$'s adding the top of the digit $7$.
Now we use Multi Dimensional Scaling (MDS) \cite{kruskal1978multidimensional} to visualize the movement of the prototypes of an MNIST dataset that is poisoned with outliers. For this experiment, we use $20$ examples of $7$'s and $i=0,2,4,6,8$ examples of $6$'s. This results in $5$ different subspace datsets formed from examples from the MNIST training dataset. We then calculate the flag median, $\ell_2$-median and the flag mean. We generate a distance matrix for all the examples of $6$'s and $7$'s along with the exemplars from each dataset using the geodesic distance and pass the distance matrix through a MDS algorithm to visualize relationships between these subspaces in two dimensions. This example is in Figure \ref{fig:1MDS}.
\begin{figure}[ht]
\centering
\begin{minipage}[b]{.45\textwidth}
\includegraphics[width=\textwidth]{gr1_8.png}
\end{minipage}\hfill
\begin{minipage}[b]{.45\textwidth}
\includegraphics[width=\textwidth]{gr1_8_just_prototypes.png}
\end{minipage}
\caption[]{MDS embedding of flag median, $\ell_2$-median and flag mean with points as one dimensional subspaces. We have $20$ examples of $7$'s and $i$ examples of $6$'s. Each triangle represents a prototype for $i=0,4,8$. The furthest left triangle is the prototype for the dataset with $i=0$ examples of $6$'s and the furthest right triangle is the prototype for the dataset with $i=8$ examples of $6$'s. The lower image is a zoomed in version of the interior of the red box in the upper image to clarify the difference between the exemplars.}
\label{fig:1MDS}
\end{figure}
Notice that the flag mean is moving the most as we add examples of $6$'s and the $\ell_2$-median is moving similarly to the flag mean. The flag median moves substantially less than the other prototypes and therefore is the least affected prototype by the added examples of $6$'s.
We now compute the $r=5$-dimensional flag median and flag mean of a dataset with 20 examples of $7$'s with $i=8$ $6$'s. We plot each of the reshaped columns of the matrix representative of these prototypes in Figure \ref{fig:high_dim}. Notice that the flag mean is more affected by examples of $6$'s than the flag median. This is particularly noticeable in the final column (dimension $5$) where there is a clear $6$ in the image for flag mean whereas the $6$ is not clear in the flag median.
\begin{figure}[ht]
\centering
\includegraphics[width=.9\textwidth]{added_dims_7_6.png}
\caption[]{Each column of the matrix representative for flag median and flag mean on the dataset with $20$ examples of $7$'s and $i=8$ examples of $6$'s.}
\label{fig:high_dim}
\end{figure}
\subsection{Mind's Eye dataset}\label{sec:mindseye_experiments}
The Mind's Eye dataset is a set of grey-scale outdoor video clips that are centered on moving objects (mainly humans) and have a subtracted background. Each video clip consists of 48 frames, each rescaled to a size of $32 \times 32$ pixels. We use the preprocessed data from the k-means experiment from Marrinan et.\ al.\ \cite{marrinan2014finding}. These data and the scripts for the preprocessing can be accessed at \url{https://www.cs.colostate.edu/~vision/summet}. There are 77 labels of the video clips for the action of the centered object in the video. A video clip is represented on $\Gr(48,1024)$ by the span of the $1024 \times 48$ matrix formed by vectorizing and horizontally stacking each frame.
For this example, we use subspaces (points in $\Gr(48,1024)$) that represent clips with action labels bend, follow, pickup, ride-bike and run. There are 27 examples of bend, 32 of follow, 27 of pickup, 17 of ride-bike and 24 of run. We run the Linde-Buzo-Grey (LBG) algorithm \cite{linde1980algorithm,stiverson2019subspace} to cluster these data with different sized codebooks (numbers of centers) and prototype calculation using the flag median, $\ell_2$-median and flag mean. In the LBG algorithm, we calculate distance using chordal distance. For each number of centers, we run 10 trials with different LBG initializations. The results are in Figure \ref{fig:LBG}.
We note that the flag median produces the highest cluster purities for 8, 12, 16 and 20 clusters. In all of the previous experiments we found that the flag median is more robust to outliers which may be the key factor in the success of the flag median prototype LBG implementation. We also note that the $\ell_2$-median and the flag mean LBG implementations have similar cluster purities for each of the codebook sizes. Again, this is consistent with the similar behavior the $\ell_2$-median and the flag mean MNIST experiments (see Figures \ref{fig:MNIST_classifier} and \ref{fig:1MDS}).
\begin{figure}[ht]
\centering
\includegraphics[width=.8\textwidth]{lbg_20trials.png}
\caption[]{An LBG implementation on the Mind's Eye dataset. The results of 3 different implementations of LBG for codebook sizes $4, 8, 12, 16 $ and $20$. The flag median is competitive with the $\ell_2$-median and flag mean for a size $4$ codebook and outperforms $\ell_2$-median and flag mean for codebook sizes $8, 12, 16, 20$.}
\label{fig:LBG}
\end{figure}
\subsection{UCF YouTube dataset}
Our final dataset is a subset UCF YouTube Action dataset \cite{liu2009recognizing}. This dataset contains 11 categories of actions. For each category, the videos are grouped into groups with common features. For this expeeriment, we take approximately one example from each group within an action category. Specifically our dataset consists of 23 examples of basketball shooting, 22 of biking/cycling, 25 of diving, 24 of golf swinging, 24 of horse back riding, 24 of soccer juggling, 23 of swinging, 24 of tennis swinging, 24 of trampoline jumping, 22 of volleyball spiking, and 24 of walking with a dog. Since these RGB videos are quite large, we convert them to greyscale. Then we generate a matrix for each video whose columns are vectorizations of each frame. Finally, we perform the QR decomposition of each video and take the first $10$ columns of $\mathbf{Q}$ to be it's representative on the Grassmannian.
We then run subspace LBG with $48$ dimensional flag mean and the flag median. The results are in Figure \ref{fig:LBG_youtube}. We choose to omit the $\ell_2$-median LBG implementation since the Weiszfeld-type algorithm since it can only compute a $10$ dimensional prototype. We run our LBG implementations with $10$ trials for each of the following codebook sizes: $4,8,12,16$ and $20$. We see the flag median LBG implementation out preform the flag mean LBG implementation in all trials.
\begin{figure}[ht]
\centering
\includegraphics[width=.8\textwidth]{youtube_lbg_10trials.png}
\caption[]{An LBG implementation on the YouTube dataset. The results of 2 different implementations of LBG for codebook sizes $4, 8, 12, 16 $ and $20$. The flag median outperforms flag mean for all codebook sizes.}
\label{fig:LBG_youtube}
\end{figure}
\section{Conclusion}
In this paper we presented a new prototype, the flag median, for clusters of points on the Grassmannian. We propose the FlagIRLS algorithm to approximate solutions to the flag median optimization problem. We run experiments comparing the flag median, flag mean, and the $\ell_2$-median. In our experiments, we find the FlagIRLS generally converges faster than gradient descent. In addition, we discover that the flag median is the most robust to outliers and produces higher cluster purities than the flag mean and $\ell_2$-median algorithms.
Future work with the flag median and FlagIRLS could involve machine learning or add details to the mathematical theory. For machine learning, the flag median can be used as a step in a subspace $k$-means algorithm, Grassmannian $n$-shot learning or any other machine learning algorithm in which calculating an ``average'' is a step. Most likely these types of algorithms will be useful for classifying images and videos. In terms of mathematics, we would like to find domain on which the flag median problem is convex and proofs for the convergence rates of FlagIRLS is an open problem. There are potential connections between this flavor of optimization problem and frame theory; so further investigation in this direction could prove useful. Finally, the flag median could be generalized to other spaces such as Stiefel manifolds.
\newline
\\
\textbf{Acknowledgement:} This work was partially supported by National Science Foundation award NSF-ATD 1830676.
\bibliographystyle{unsrt}
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{
"redpajama_set_name": "RedPajamaArXiv"
}
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package boa.test.datagen.php;
import java.io.IOException;
import org.junit.Test;
public class TestPHPArrayAccessNode extends PHPBaseTest {
@Test
public void arrayAccess() throws IOException, Exception{
nodeTest(load("test/datagen/PHP/ArrayAccessNode.boa"),
load("test/datagen/PHP/ArrayAccessNode.php"));
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,672
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Chris Drury (* 20. srpna 1976, Trumbull, Connecticut, USA) je generální manažer týmu New York Rangers a bývalý americký hokejový útočník.
Profil
Chris Drury je hokejista, který je vítězný typ hráče. Velká část jeho gólů byla právě vítězných. Drury v mládí vynikal nejen v ledním hokeji, ale také v baseballu. Nakonec ale dal přednost lednímu hokeji. Chris v mládí studoval se svým bratrem Tedem na Fairfield College Preparatory School a s hokejovým týmem Fairfieldu vyhrál státní šampionát.
Chris poté odešel na Bostonskou univerzitu, kde se stal hvězdou. S týmem vyhrál národní šampionát NCAA a v roce 1998 získal Hobey Baker Award pro nejlepšího hráče soutěže.
V roce 1994 ho do NHL draftoval tým Quebec Nordiques, který se ale před sezonou 1995/96 stěhoval do Denveru a přejmenoval se na Colorado Avalanche. Za Colorado Avalanche hrál poprvé v sezoně 1998/99 a hned ve svém prvním ročníku v NHL ukázal, co v něm je a vyhrál cenu pro nejlepšího nováčka soutěže Calder Trophy a byl jmenován do 1. All-Rookie Týmu. Je jediným hokejistou, který dokázal vyhrát Calder Trophy i Hobey Baker Award. V sezoně 2000/01 získal s Coloradem Avalanche Stanley Cup.
V roce 2002 reprezentoval USA na ZOH 2002 v Salt Lake City a vybojoval s týmem stříbrnou medaili.
1. října 2002 byl Chris vyměněn společně se Stephanem Yellem do Calgary Flames za Dereka Morrise, Jeffa Shantze a Deana McAmmonda. V Calgary hrál pouze jednu sezónu. 3. července 2003 byl vyměněn se Stevem Béginem do Buffala Sabres za Stevea Reinprechta a Rhetta Warrenera. S Buffalem Sabres se dostali v sezónách 2005/06 a 2006/07 až do finále Východní konference. V Buffalu Sabres mu po sezóně 2006–07 uplynula smlouva a stal se tzv. volným agentem.
V roce 2004 reprezentoval USA na Světovém poháru, kde byl tým USA vyřazený v semifinále.
V roce 2006 reprezentoval USA na ZOH 2006 v Turíně, kde byl tým USA vyřazený ve čtvrtfinále.
1. července 2007 podepsal jako volný agent smlouvu s týmem New York Rangers, kde si od něho slibovali vytvoření dua s Jaromírem Jágrem, toto spojení ale moc nefungovalo. V klubu New York Rangers hrál až do konce své kariéry.
Úspěchy a trofeje
Individuální trofeje
1996 a 1997 jmenován do East Hockey 2.All-star týmu
1996 jmenován do NCAA East 2.All-american týmu
1997 a 1998 jmenován do NCAA East 1.All-american týmu
1997 a 1998 Hráč roku ligy Hockey East
1997 jmenován do NCAA Championship All-tournament týmu
1998 jmenován do Hockey East 1. All-star týmu
1998 Hobey Baker Memorial Award (Tofej pro nejlepšího hráče v americkém univerzitním hokeji)
1999 Calder Memorial Trophy
1999 jmenován do 1. NHL All-rookie týmu
Klubové trofeje
2001 Stanley Cup s Coloradem Avalanche
2001 Clarence S. Campbell Bowl s Coloradem Avalanche
2001 Presidents' Trophy s Coloradem Avalanche
2002 Stříbrná medaile s USA na Zimních Olympijských Hrách
2004 Bronzová medaile s USA na Mistrovství Světa
2010 Stříbrná medaile s USA na Zimních Olympijských Hrách
Smlouva a plat
Smlouva do roku 2012 s New York Rangers, poté volný agent.
2008–09 * 7100000 $
2009–10 * 8050000 $
2010–11 * 8000000 $
2011–12 * 5000000 $
Klubové statistiky
Reprezentační statistiky
Externí odkazy
Američtí lední hokejisté
Američtí hokejoví útočníci
Američtí stříbrní olympijští medailisté
Hráči Buffalo Sabres
Hráči Colorado Avalanche
Hráči Calgary Flames
Hráči New York Rangers
Hokejisté NCAA
Vítězové Stanley Cupu
Stříbrní olympijští medailisté v ledním hokeji
Narození v roce 1976
Žijící lidé
Muži
Narození 20. srpna
Narození v Connecticutu
Hráči draftovaní Quebecem Nordiques
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,347
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Akindale Thoroughbred Rescue
Location: Pawling, NY
Website: akindalehorserescue.org
Average Number of Thoroughbreds: 150
"I never met a horse I didn't like."
Those words were spoken by our founder, the late businessman and philanthropist John Hettinger, a noted racehorse owner and breeder, winner of the Eclipse Award of Merit, and an activist against equine slaughter, fighting successfully for legislation that banned U.S. slaughterhouses.
"He is the modern father of Thoroughbred aftercare," said D.G. Van Clief, president of Akindale Thoroughbred Aftercare. "We owe its existence to John for making this a top-of-consciousness issue. He said he was going to do something and he put his money where his mouth was. He was a champion for retired racehorses."
In 2019, Hettinger was inducted into the National Museum of Racing Hall of Fame for his contributions to the industry and to horses. Hettinger founded Akindale Thoroughbred Aftercare in 2006, two years before his death. He left 300 acres of the family farm in Pawling, N.Y., to carry on his mission and care for the animals he called his "best friends." The property is deeded and dedicated in perpetuity to the care of retired racehorses and continues to successfully transition Thoroughbreds to second careers.
Akindale operates a "Reinventing Racehorses" program that has rehabilitated, retrained, and re-homed hundreds of Thoroughbreds. The program's emphasis is on working directly with the racing industry to be sure Thoroughbreds get the best opportunity for a second career. Once a horse enters the Akindale program, they are guaranteed a safe haven for life. Horses are prepared for a second career and the right home is found for them. If anything goes wrong in the horse's lifetime, they can come back and retire in
our sanctuary and live out their years peacefully.
Akindale is a non-profit organization that accepts tax-deductible donations to help cover the costs of care for the 160-plus Thoroughbreds in its care. Every dollar makes a difference in the lives of our horses, whether it is a one-time cash donation, material donation, or a long-term sponsorship of one of the 130 retirees living in our sanctuary.
Success Stories from Akindale Thoroughbred Rescue
Pardner and Wire Me Rockin – Two OTTBs who faced challenges before being assisted by MidAtlantic Horse Rescue and Akindale Horse Rescue but are now living the good life. Read more >>
Callmetony – This warhorse closed out his racing career in the claiming ranks at age 11 and began a new life in the show ring. Read more >>
Evening Attire – This multiple graded stakes winner racked up earnings of more than $2.9 million in his 69 career starts before becoming a resident of Akindale. Read more >>
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,238
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{"url":"http:\/\/mathoverflow.net\/questions\/58122\/slick-proof-of-kudo-transgression-theorem","text":"# Slick Proof of Kudo Transgression Theorem\n\nThe Kudo Trangression Theorem has to do with the transgression in the Leray-Serre spectral sequence for cohomology in $\\mathbb{Z}\/p$ ($p$ odd). It can be proved by the method of the universal example, once it is shown that in the path-loop fibration sequence $K(\\mathbb{Z}\/p,2n) \\to P(K(\\mathbb{Z}\/p,2n+1)) \\to K(\\mathbb{Z}\/p,2n+1)$\n\n1. the fundamental class $v$ of the fiber transgresses to $u$, that of the base\n\n2. this forces a zig-zag of cancellation, up to $v^{p-1}\\mapsto u \\otimes v^{p-2}$\n\n3. also $v^p$ transgresses to $P^n(u)$, and\n\n4. $u\\otimes v^{p-1}$ \"transgresses\" to $\\beta P^n(u)$.\n\nParts (1), (2) and (3) are easy, but part (4) seems difficult. There is a proof along these lines in a paper of Browder from the mid 1960's (he attributes the proof to Milgram), but the proof of (4) is actually quite hard and leans heavily on algebraic mucking around in the spectral sequence.\n\nDoes anyone know of a clever way to prove (4)?\n\nEdit: Let's say we know by induction that the cohomology of the fiber is what it has to be. Then I think the behavior of the spectral sequence is forced in dimensions below that of $u\\otimes v^{p-1}$. Does this show that $u\\otimes v^{p-1}$ \"transgresses\"? Suppose it does; then its image is $Q(u)$, where $Q$ is a cohomology operation that vanishes when looped (since it is not the transgression of a class in the fiber). Perhaps we can argue that $Q$ must be $\\beta P^n$, up to sign?\n\n-\nI didn't think this was all that hard when I wrote A general algebraic approach to Steenrod operations'' which appeared in 1970. It is available on my web page.","date":"2013-12-07 03:14:32","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8884251713752747, \"perplexity\": 254.64407443504447}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2013-48\/segments\/1386163053174\/warc\/CC-MAIN-20131204131733-00089-ip-10-33-133-15.ec2.internal.warc.gz\"}"}
| null | null |
Behav Brain Sci. 2016 Jan;39:e5. doi: 10.1017/S0140525X15000357.
Memes and the evolution of religion: We need memetics, too.
Department of Psychology,University of Plymouth,Portland Square,Plymouth PL4 8AA,United Kingdom.susan.blackmore@virgin.nethttp://www.susanblackmore.uk.
In their analysis, Norenzayan et al. completely ignore memetics, which, unlike other theories, treats memes as replicators and looks to memetic as well as genetic advantage. Now that memes are evolving ever faster, genetic advantage is less relevant. So when religious and secular values are at odds, we need a memetic analysis to understand what is going on.
|
{
"redpajama_set_name": "RedPajamaC4"
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| 7,864
|
We offer Bike Rentals to both CSU Departments and to CSU Students, Faculty, & Staff. Please choose the appropriate option below for more information.
Choose this option if you are interested in renting one or more bikes for your department.
A CSU Kuali Account number is required for payment and bikes must be stored at a CSU owned/leased facility.
Choose this option if you are interested in renting a bike for your personal use as a CSU Student, Faculty, or Staff member.
We accept Cash, Major Credit Cards, and RAMCash.
A special Thank You! goes out to FC Bikes for providing the bike fleet for this program!
|
{
"redpajama_set_name": "RedPajamaC4"
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| 4,049
|
Q: how to get saved text into edit text How could save the text as a text edit so that after leaving the activity and re-enter, It appear in the same edit text?
switchbutton.setOnClickListener(new OnClickListener() {
public void onClick(View v) {
Switch sw = (Switch) v;
if (sw.isChecked()) {
SharedPreferences settings = getPreferences(0);
SharedPreferences.Editor editor1 = settings.edit();
editor1.putString("nombre", ed1.getText().toString());
SharedPreferences.Editor editor2 = settings.edit();
editor2.putString("apellido", ed2.getText().toString());
SharedPreferences.Editor editor3 = settings.edit();
editor3.putString("ciudad", ed3.getText().toString());
SharedPreferences.Editor editor4 = settings.edit();
editor4.putString("calle", ed4.getText().toString());
SharedPreferences.Editor editor5 = settings.edit();
editor5.putString("numero", ed5.getText().toString());
editor1.commit();
editor2.commit();
editor3.commit();
editor4.commit();
editor5.commit();
} else {
Toast.makeText(OtraActivity.this, "Datos no guardados",
Toast.LENGTH_LONG).show();
}
}});
A: As cloudymusic suggested, use SharedPreferences, and your activity's onPause() store the data, and in the onResume() load it again, and you'll have your EditText's data persistent even when the App is closed
A: In onCreate, onStart, or wherever you have loaded your views, you can load the data from prefs and insert it into your views:
@Override
protected void onCreate(Bundle savedInstanceState) {
SharedPreferences settings = getPreferences(0);
ed1.setText(settings.getString("nombre", ""));
ed2.setText(settings.getString("apellido", ""));
...
}
In onPause, you can save it just as you are when the button is clicked.
A: Use the following onCreate
SharedPreferences settings = getPreferences(0);
ed1.setText(settings.getString("nombre",""));
do the same for the other edittext..
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 9,546
|
{"url":"https:\/\/kluedo.ub.uni-kl.de\/frontdoor\/index\/index\/docId\/1018","text":"## 6 DOF path planning in dynamic environments - A parallel on-line approach\n\n\u2022 This paper presents a new approach to parallel path planning for industrial robot arms with six degrees of freedom in an on-line given 3D environment. The method is based a best-first search algorithm and needs no essential off-line computations. The algorithm works in an implicitly discrete configuration space. Collisions are detected in the Cartesian workspace by hierarchical distance computation based on polyhedral models of the robot and the obstacles. By decomposing the 6D configuration space into hypercubes and cyclically mapping them onto multiple processing units, a good load distribution can be achieved. We have implemented the parallel path planner on a workstation cluster with 9 PCs and tested the planner for several benchmark environments. With optimal discretisation, the new approach usually shows very good speedups. In on-line provided environments with static obstacles, the parallel planning times are only a few seconds.\n\n### Additional Services\n\nAuthor: Dominik Henrich, Christian Wurll, Heinz W\u00f6rn urn:nbn:de:hbz:386-kluedo-9723 Article English 1998 1998 Technische Universit\u00e4t Kaiserslautern 2000\/04\/04 AG-RESY ; PARO ; SKALP Fachbereich Informatik 0 Informatik, Informationswissenschaft, allgemeine Werke \/ 00 Informatik, Wissen, Systeme \/ 004 Datenverarbeitung; Informatik AG RESY Standard gem\u00e4\u00df KLUEDO-Leitlinien vor dem 27.05.2011\n\n$Rev: 13581$","date":"2016-02-10 22:26:59","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.2446846216917038, \"perplexity\": 5339.541512292957}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-07\/segments\/1454701160822.87\/warc\/CC-MAIN-20160205193920-00200-ip-10-236-182-209.ec2.internal.warc.gz\"}"}
| null | null |
Q: Is it correct to pronounce the letter N as "ain" when spelling out words letter by letter? I live in a non-English-speaking country. A lot of people around me pronounce the letter N as "ain" (/eɪn/ in IPA). I am very confused because in dictionaries the letter N can be only pronounced as /en/.
Is this a mistake? Do any native speakers also pronounce the letter N as "ain"?
A: Pronounce the letter N /ɛn/, not /en/
To answer your first question, the OED entry for the letter N (paywalled link) gives only /ɛn/ as the pronunciation in both American and British dialects alike. If you for whatever reason choose to view that dictionary's pronunciation as prescription not description, then the letter N "should" be pronounced to rhyme with men and ten, not with main and feign.
That means that native speakers of English say the name of the letter N as /ɛn/ using the lax DRESS vowel, not /en/ using the tense FACE vowel the way you've described hearing it said in that non-English country you've so far only alluded to.
In some native-speaker accents these two contrasting phonemes do move around a bit compared with how they work out in other dialects, but they should still belong to distinct lexical sets no matter how they are pronounced. The same would hold true with words like gem with an open/lax vowel and name with a close/tense one.
To answer your second question, although I've heard native-speaker accents where pin and pen merge, I've never heard any native-speaker accents where pen and pain merge.
I have heard non-native-speaker accents where this happens, however...
Non-English phonologies
I bet the people wherever you're talking about also pronounce the letter L to rhyme with nail not with bell the way native speakers do. Unlike English, some languages like Spanish do not distinguish lax vowels from tense ones. All five vowels there are considered close, and while some speakers may in some utterances say some words with a vowel that's a little more open there due to its phonological environment, they cannot "hear" this as a separate vowel because they lack minimal pairs.
But because you can, you do, and this ends up being confusing.
That's why in a Spanish accent, the English word bit sounds the same as English word beat, and why English met in their accent would sound like mate to you. Maybe that's what you're hearing happen here: the original language doesn't distinguish an open/lax e from a close/tense one.
English dialectal variations
It turns out that name is a pretty good example for demonstrating variation in this phoneme across various English dialects. Follow that link to see those, and even hear them.
Notice how differently that works out in practice. Sometimes it's one diphthong, sometimes it's another, and at other times it's not a diphthong at all.
*
*You'll find that native-speaker name variants with tense [e] include monophthongs in [neːm], [ne̞ːm] [ne̝ːm], [neˑm], [ne̝ˑm], [ne̝m], and diphthongs in [neˑəm], [nëˑəm], [ne̝ˑəm], [neˑɪm], [ne̞ˑɪm], [neɪm], [ne̝ɪm].
*There are also native-speaker name variants with lax [ɛ] that include a monophthong in [nɛ̝ːm] and diphthongs in [nɛɪm], [nɛˑɪm], [nɛ̝ˑɪm] [nɛ̞ˑɪm].
*There's even [njɛːm] with a rising diphthong and [niˑəm] with a falling one.
Every single one of those is a "correct" but different pronunciation of that same word by native speakers from around the world. It's therefore "correct" in that accent but "incorrect" in others. What sounds normal in one dialect would necessarily sound abnormal in another that uses different rules for realizing its phonemes.
Phonemically those are all /e/, but phonetically they are certainly not always [e]! But this doesn't matter to our ear. Whether the vowel from name is ever some sort of diphthong there is a different matter altogether, one not especially important since that's merely a minor unconscious phonological effect, an offglide made by some speakers and not others. The contrasting feature distinguishing the two phonemes is whether the vowel is the open one as in DRESS or the close one as in FACE.
English has no minimal pairs differing only in whether there's a glide there in a falling diphthong versus a monophthong without a glide. So we have no minimal pair that has /eɪ/ in one word and /e/ in the other, with all else held equal. The same is true with /ɛɪ/ versus /ɛ/.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,652
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We are recruiting for a well-established, international client based in the Medway area.
We have an excellent opportunity to join their manufacturing team, initially on a temporary basis with the potential to become permanent.
You will be working in a busy assembly environment where over a period of time you will learn a variety of roles. Initially, you will be carrying out hand assembly of basic electronic products. In time, you will be trained to use various hand tools and to solder.
Hours of work are initially on a day shift of 8.30am-5pm, Mon-Thurs and 8.30am-4pm on a Friday; moving to a rotating shift pattern of 6am-2pm and 2pm-10pm.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 3,038
|
Drtivá porážka může být:
Drtivá porážka (americká verze), původní americká verze televizní soutěže
Drtivá porážka (britská verze), britská verze televizní soutěže
Drtivá porážka (kanadská verze), kanadská verze televizní soutěže
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,167
|
The wall decor image displayed right above, the Contemporary Home Decoration Ideas, based on the many view counts this wall decor image has gain, it clearly say that this contemporary home decoration ideas is one of readers' most-loved wall decor inspiration. This special wall decor shows many awesome stuff, including dual palm tree island wall decal sticker, pretty looks wall in yellow painted, and few more such as the dark brown finished wooden door and seamless off white floor lustrous, etc.
This wall decor photo, which was uploaded here, was of course not just the only one we would want to recommend to you. There are tons of wall decor photos related to this wall decor gallery set. In the next page, we have Traditional Style Interior Design featuring plain bamboo beaded curtain for doorways and 75 inch drop beaded drapes; labeled with beaded curtains for doorways topic. While in previous page — labeled under home decorative galleries — is the Elegant Living Room featuring ethan allen chadwick sofa and white tufted leather sofa.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,968
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Kalifornijska Południowa Konwencja Baptystyczna (ang. California Southern Baptist Convention) – jedna z 39 stanowych unii baptystycznych wchodzących w skład Południowej Konwencji Baptystycznej.
Początki kościoła baptystycznego w Kalifornii sięgają roku 1849, kiedy to założono pierwszy zbór w San Francisco.
W roku 1851 przybył do Kalifornii pastor z Alabamy, który założył zbór w Santa Clara.
Południowa Konwencja Baptystyczna została powołana w Kalifornii w roku 1940 jako Generalna Konwencja Południowych Baptystów w Kalifornii. W roku 1988 przyjęła ona obecną nazwę.
Jest to największy baptystyczny związek wyznaniowy w Kalifornii. Liczy około 500 tysięcy ochrzczonych członków oraz ponad 2200 zborów. Poza zborami anglojęzycznymi istnieje kilkaset hiszpańskich, koreańskich, chińskich i innych.
Konwencja prowadzi Kalifornijski uniwersytet baptystyczny.
Przypisy
Baptystyczne związki wyznaniowe
Religia w Stanach Zjednoczonych
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,818
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\section{Keywords:} social insects, swarm behavior, automatic tracking method, object detection, online tracking}
\end{abstract}
\section{Introduction}
Swarm behavior is one of the most important features of social insects, which has important significance for the study of embodied intelligence ~\cite{tiacharoen2012design}. Specifically, social insects often tend to cluster into a colony~\cite{vandermeer2008clusters}, which forms a complex dynamical system together with the surrounding environment~\cite{balch2001automatically}.
So far, we do not know enough about the mechanisms behind swarm behaviors of social insects.
The mainly reason is that the key requirement of this research is the ability to track the motions and interactions of each individual robustly and accurately.
However, until the late $20^{th}$ century, biologists still manually marked the motion trajectories on the video to guarantee the quality. They have to track each individual at one time, which might mean watching the entire video 50 times or more in a crowded scene~\cite{poff2012efficient}.
It becomes an inhibiting factor in obtaining the complete and accurate dataset required to analyze the evolution of complex dynamical system.
Therefore, in the past two decades, attempts have been made to automate the tracking process for social insects utilizing computer vision (CV) techniques~\cite{khan2005mcmc, khan2006mcmc, oh2006parameterized, veeraraghavan2008shape, fletcher2011multiple}.
Traditional CV techniques free researchers from manual work through approaches such as foreground segmentation algorithm~\cite{li2008estimating}, temporal difference method~\cite{khan2005mcmc} and hungarian algorithm~\cite{li2009learning}.
Such approaches, however, have failed to address the noise in the image~\cite{zhao2015improved}, resulting in the limitation that a laboratory environment with a clean background is needed.
Nevertheless, many scientifically valuable results are obtained in nature rather than laboratory environment~\cite{schmelzer2009special, kastberger2013social, tan2016honey, dong2018olfactory}.
In recent years, with the popularity of computer vision, many advanced object detection and tracking methods have emerged.
\subsection{Object detection}
Existing methods in object detection are categorized as one-stage or two-stage, according to whether there is a separate stage of region proposal.
One-stage frameworks (e.g., YOLO~\cite{redmon2016you}) are fast, but their accuracy is typically slightly inferior compared with that of two-stage detection.
The popularity of two-stage detection frameworks is enhanced by R-CNN~\cite{girshick2014rich}, which proposes candidate regions via a selective search (SS) algorithm~\cite{uijlings2013selective}, thereby the detector focuses on these RoIs.
However, using the SS algorithm~\cite{uijlings2013selective} to generate region proposals is the main reason causing slow inference.
Fast R-CNN~\cite{girshick2015fast} reduces the computational complexity of region proposals by downsampling the original image, while Faster R-CNN~\cite{ren2015faster} proposes an RPN, which further improves the speed of training and inference.
Given the success of deep learning in general tasks of object detection, researchers also applied to detect specific groups of animals, such as a single mouse~\cite{geuther2019robust}, fruit flies~\cite{murali2019classification}.
These methods are either limited to track a single object, or a fixed number of objects.
General tools~\cite{romero2019idtracker,sridhar2019tracktor} also offer the functionality to detect and track unmarked animals in the image.
However, most of existing methods focus on the condition of ideal lab set-up and none of existing works reported the detection of ants in outdoor environments which contain diverse backgrounds and arbitrary terrains.
\subsection{Multi-object tracking (MOT)}
In the last two decades, vision-based detection and tracking models have been widely used to study social insects~\cite{khan2005mcmc,veeraraghavan2008shape}.
Appearance (particularly color) and motion information are the main metrics used in this category of method.
Due to high similarity of ants' appearance, researchers either use the technique of pigmenting to create more distinct appearance features~\cite{fletcher2011multiple}, or limit the observation to a laboratory setup~\cite{branson2009high, perez2014idTracker}.
State-of-the-art methods, such as Ctrax~\cite{branson2009high} and idTracker~\cite{perez2014idTracker}, for insect tracking are tested in a laboratory setup and use background subtraction for foreground segmentation.
Notably, the operations of background modeling and foreground extraction are time-consuming.
The tracking-by-detection (TBD) paradigm is to match trajectories and detections in two consecutive frames, a process that requires metrics.
The global nearest neighbor model measures motion state to achieve Drosophila tracking~\cite{chiron2013detecting}.
The global nearest neighbor model assumes that the motion state obeys the linear observation model, which commonly uses a constant velocity model - the Kalman filter (KF).
However, changes in ants' speed and direction are difficult to predict, thus appearance information is integrated as a metric.
The DAT method is a mainstream method for ant colony tracking~\cite{fasciano2014ant}.
It allows a combination of multiple metrics, and uses Hungarian algorithm~\cite{li2009learning} to assign detections for trajectories.
The PF method is suitable for solving nonlinear problems~\cite{fasciano2013tracking}, but the growth in the number of particles leads to an exponential increase in the computational cost, preventing the effective multi-object tracking. Using Markov Chain Monte Carlo sampling can reduce computational complexity~\cite{khan2006mcmc}.
A GPU-accelerated semi-supervised framework can further improve tracking accuracy and performance~\cite{poff2012efficient}.
When applying the methods above for tracking ant colonies, they are greatly disturbed by background noise and difficult to overcome the serious occlusion problem in dense scenes.
Long short-term memory~\cite{kim2018multi} and spatial-temporal attention mechanisms~\cite{zhu2018online} have been developed to tackle the problem of long-term occlusion.
A bilinear Long short-term memory structure that couples a linear predictor with input detection features, thereby modeling long-term appearance features~\cite{kim2018multi}.
The spatial-temporal attention mechanism is also suitable for the MOT task.
The spatial attention module makes the network focus on the pattern of matching. Meanwhile the temporal attention module assigns different levels of attention to the sample sequence of the trajectory~\cite{zhu2018online}.
The TBD paradigm-based framework is dependent on detection results. Therefore, severe occlusion is likely to cause tracking failures.
To prevent this situation, a detector with automatic bounding box repairing and adjustment is introduced by a cyclic structure classifier~\cite{dong2016occlusion}.
In this paper, we use a deep learning method to build a detection and tracking framework.
Our method is based on the TBD paradigm and accomplishes the goal of online multi-ant tracking.
To the best of authors' knowledge, this is the first work to achieve robust detection and tracking of ant colony in both indoor and outdoor environments (Figure~\ref{fig:tracking_results}).
Our method is robust in tackling the challenge of visual similarity among colony individuals, handling diverse terrain backgrounds and achieving long-period of tracking.
Our main contributions are as follows:
\begin{itemize}
\item We adopt a two-stage object detection framework, using ResNet-50 as the backbone and position sensitive score maps to encode regions of interest (RoIs).
During the tracking stage, we use a ResNet network to obtain the appearance descriptors of ants and then combine them with motion information to achieve online association.
\item Our method proves to be robust in both indoor and outdoor scenes. Furthermore, only a small amount of training data are required to achieve the goal in our pipeline, which are 50 images chosen for each scene in the detection framework and 50 labels randomly chosen for the tracking framework respectively.
\item We construct an ant database with labeled image sequences, including five indoor videos (laboratory setup) and five outdoor videos, with 4983 frames and 115,433 labels in total.
The database is made publicly available, which is hoped that it will contribute to a deeper exploration on swarm behavior of ant colony.
\end{itemize}
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{tracking_results}
\caption{\label{fig:tracking_results} Tracking results by our method in both indoor and outdoor environments.}
\end{figure}
\section{Materials and methods}
\subsection{Overview}
\label{sec:overview}
Following the TBD paradigm, we propose a uniform framework for detection and tracking to efficiently and accurately track the ant colony in both indoor and outdoor scenes (Figure~\ref{fig:pipeline}).
In the detection phase, we adopt a two-stage object detection framework, using ResNet-50 as the backbone, and encoding RoIs proposed by regional proposal network via position-sensitive score maps.
Then we implement classification and regression through downsampling and voting mechanisms.
(see details in Section~\nameref{sec:two-stage}).
In the tracking stage, we first use ResNet to train the appearance descriptors of ants and measure the appearance similarity between two objects.
Next, the tracking is accomplished by combining appearance and motion information for online association metric. (Section~\nameref{sec:MOT_framework}).
\begin{figure}[!htbp]
\centering
\includegraphics[width=0.65\linewidth, height=0.7\linewidth]{pipeline}
\caption{\label{fig:pipeline} Architecture for detection and tracking.}
\end{figure}
\subsection{Two-stage object detection}
\label{sec:two-stage}
\subsubsection{Regional proposal network}
Regional proposal network (RPN) is proposed in Faster R-CNN \cite{ren2015faster} to generate RoIs.
Compared to SS \cite{uijlings2013selective}, RPN is based on the CNN network structure and can connect the backbone with shared weight, significantly improving detection speed.
We use ResNet-50 as the backbone and replace the fully connected layer with a 1$\times$1 convolution to reduce the dimensions of feature maps.
Considering that ResNet-50 conducts downsampling 32 times, we get 256-d feature maps via a 3$\times$3 Atrous convolution to maintain translation variability.
For each sliding position, we predict k region proposal boxes of different sizes and ratios; these boxes are called anchors.
After the 256-d vector, we connect classification and regression branches through two parallel 1$\times$1 convolution layers.
The classification branch uses softmax to determine whether there is an object in anchor so that this branch has 2$\times$k outputs.
The regression branch will perform a regression on the 4D position parameters of anchors (i.e., center coordinates, width and height) so that there are 4$\times$k outputs.
RPN will propose k$\times$w$\times$h anchors with a w$\times$h feature map, called RoIs.
We use the Non-maximum suppression algorithm \cite{neubeck2006efficient} to filter duplicate anchors and set the IOU threshold to 0.7.
\subsubsection{Position sensitive region of interest}
On the basis of RoIs, the two-stage detection framework classifies and fine-tunes the location of bounding boxes.
In Faster R-CNN, RoIs are scaled to the last feature maps and focusing on these areas through ROIPooling.
Next, each RoI is classified and regressed through two fully connected layers, causing high computational complexity.
In order to reduce the number of parameters, we use RPN-FCN \cite{dai2016r} to generate position-sensitive score maps via a convolutional layer, which is connected to the backbone.
Both classification and regression tasks have independent position-sensitive score maps, forming three parallel branches with RPN.
For the classification task, since we only need to classify ants and background, we use k$\times$k$\times$2 convolution kernels to generate score maps.
k$\times$k indicates that each RoI is divided into k$\times$k regions to encode position information.
Each region is encoded by a specific feature map with two dimensions.
Similarly, we use k$\times$k$\times$4 convolution kernels for fine-tuning the position of RoIs in the regression task.
To focus on RoIs, we perform average pooling on each region to get feature maps, called position sensitive region of interest (PSRoI) pooling, as the following formula shows:
\begin{equation}
r_{c}(i, j | \Theta)=\sum_{(x, y) \in \text {region}(i, j)} z_{i, j, c}\left(x+x_{0}, y+y_{0} | \Theta\right) / n.
\end{equation}
$r_{c}(i, j | \Theta)$ is the result of downsampling in $(i,j)^{th}$ for $c^{th}$ category, and $z_{i, j, c}$ is one score map in the k$\times$k$\times$2 position-sensitive score maps. $(x0, y0)$ represents the left-top corner of RoI. $\Theta$ is the set of parameters of the network, and $n$ is the number of pixels in the region.
For the feature maps, we vote on k$\times$k regions, getting the overall score of RoI on the classification or regression task, as the following formula shows:
\begin{equation}
r_{c}(\Theta)=\sum_{i, j} r_{c}(i, j | \Theta).
\end{equation}
In the formula, $r_{c}(\Theta)$ represents the overall scores of all regions.
Next, we use softmax to implement binary classification, as the following formula shows:
\begin{equation}
s_{c}(\Theta)=e^{r_{c}(\Theta)} / \sum_{c=0}^{C} e^{r_{c}(\Theta)}.
\end{equation}
Here, $s_{c}(\Theta)$ is the probability of $c^{th}$ category.
Finally, we use the Non-maximum suppression algorithm to filter the bounding box.
Since object detection includes classification and regression, we require a multitask loss function.
In this paper, we weight the loss functions of the two tasks. Because softmax is used for the binary classification task, it is natural to adopt cross-entropy loss for the classification task. For the regression task, we calculate the matching degree between the four position parameters and ground truth:
\begin{equation}
L\left(s, t_{x, y, w, h}\right)=L_{c l s}\left(s_{c^{*}}\right)+\lambda\left[c^{*}=1\right] L_{r e g}\left(t, t^{*}\right).
\end{equation}
where $c^{*}$ is the ground truth category label of RoI, and $c^{*} = 1$ represents ants.
$L_{c l s}\left(s_{c^{*}}\right)$ represents cross-entropy loss:
\begin{equation}
L_{c l s}\left(s_{c}\right)=-\log \left(s_{c}\right).
\end{equation}
$L_{r e g}\left(t, t^{*}\right)$ represents the loss of the regression task, including 4 dimensions:
\begin{equation}
L_{r e g}\left(t, t^{*}\right)=\Sigma_{i=1}^{4} g\left(t_{i}^{*}-t_{i}\right).
\end{equation}
In the formula, $t^{*}$ is the predicted position, and $t$ is ground truth after translation and scaling.
\subsection{MOT framework}
\label{sec:MOT_framework}
\subsubsection{Offline ResNet network architecture}
We adopt a 15-layer ResNet network architecture to extract the appearance descriptors of objects, as Figure~\ref{fig:pipeline} shows.
After downsampling eight times, the network will eventually obtain a 128-dimensional feature vector through a fully connected layer. The specific parameters are consistent with \cite{CAO2020107233}.
\subsubsection{Cosine similarity metric classifier}
We modify the parameters of softmax to get a cosine similarity measurement classifier, which can measure the similarity of the same category or different categories. First, the output of a fully connected layer is normalized by batch normalization, ensuring that it is expressed as a unit length $\left\|f_{\Theta}(x)\right\|_{2}=1$, $\forall x \in R^{D}$. Second, we normalize the weights, that is, $\varpi_{k}=\omega /\left\|\omega_{k}\right\|_{2}$, $\quad\forall k=1, \ldots C$.
Cosine similarity metric classifier is constructed as follows:
\begin{equation}
p\left(y_{i}=k | r_{i}\right)=\frac{\exp \left(\kappa \cdot \varpi_{k}^{T} r_{i}\right)}{\sum_{n=1}^{C} \exp \left(\kappa \cdot \varpi_{n}^{T}\right)}.
\end{equation}
Here, $\kappa$ is the free scaling parameter.
Because the cosine similarity classifier follows the structure of softmax, we use the cross-entropy loss for training:
\begin{equation}
L(D)=-\sum_{i=1}^{N} \sum_{k=1}^{C} \mathrm{gt}_{y_{i}-k} \cdot \log p\left(y_{i}=k | r_{i}\right).
\end{equation}
Here, $L(D)$ represents the sum of the cross-entropy loss of $N$ images, $p\left(y_{i}=k | r_{i}\right)$ is the prediction result of $i^{th}$ image in $k^{th}$ label, and $\mathrm{gt}_{y_{i}-k}$ is ground truth.
\subsubsection{Motion matching}
We use the KF model to predict the position of trajectories in the current frame.
Then, we calculate the square of the Mahalanobis distance between the predicted position and the detected bounding box position by measuring the degree of motion matching \cite{wojke2017simple} as follows:
\begin{equation}
d^{(1)}(i, j)=\left(d_{j}-y_{i}\right)^{T} S_{i}^{-1}\left(d_{j}-y_{i}\right).
\end{equation}
Here, $d_{j}$ is the position of the $j^{th}$ detection box, $y_{i}$ is the position of the $i^{th}$ trajectory predicted by the KF, and $S_{i}$ is the covariance matrix between the $i^{th}$ trajectory and the detected bounding box.
We use a 0-1 variable to indicate whether trajectory and detection meet the association conditions. If the Mahalanobis distance meets $t^{\left(1\right)}$, $\left(i,j\right)$ will be added to the association set. The formula can be expressed as:
\begin{equation}
b_{i j}^{(1)}=\left\{\begin{array}{ll}
{1,} & {d^{(1)}(i, j)<t^{(1)}} \\
{0,} & {\text{ otherwise }}
\end{array}\right..
\end{equation}
Here, $b_{i j}^{(1)}$ is the motion association signal.
\subsubsection{Appearance matching}
\label{sec:appearance_matching}
We use the appearance descriptors to measure the appearance similarity between ants. Furthermore, we create a gallery for each trajectory, and each gallery stores the latest 100 appearance descriptors.
Then, we calculate the cosine distance of appearance descriptors between gallery and candidate bounding boxes.
The smallest distance is used as an appearance matching degree as follows:
\begin{equation}
d^{(2)}(i, j)=\min \left\{1-r_{j}^{T} r_{k}^{(i)} | r_{k}^{(i)} \in \mathrm{K}_{i}\right\}.
\end{equation}
where $r_{j}$ is the appearance descriptor of the $j^{th}$ detection box, $r_{k}^{(i)}$ is the $k^{th}$ appearance descriptor of the $i^{th}$ trajectory, $d^{(2)}(i, j)$ represents the appearance matching degree between the $i^{th}$ trajectory and the $j^{th}$ bounding box.
Similarly, we introduce a 0-1 variable as an association signal. If the appearance matching degree from a pair of trajectory and detection boxes meets the threshold, we add it to the association set:
\begin{equation}
b_{i j}^{(2)}=\left\{\begin{array}{ll}
{1,} & {d^{(2)}(i, j)<t^{(2)}} \\
{0,} & {\text { otherwise }}
\end{array}\right..
\end{equation}
where $b_{i j}^{(2)}$ represents the appearance association signal. In this paper, $t^{(2)}$ is set to 0.2.
\subsubsection{Comprehensive matching}
To combine motion and appearance information, we set a comprehensive association signal $b_{i j}$. Only when both motion and appearance matching degree meet the threshold, the $\left(i, j\right)$ pair will be considered for matching. The formula expression is denoted as follows:
\begin{equation}
b_{i j}=\prod_{m=1}^{2} b_{i, j}^{(m)}.
\end{equation}
However, the KF is scarcely possible to track accurately for long periods, because of the motion of ants is complicated.
Therefore, we use the appearance matching degree (Section~\nameref{sec:appearance_matching})
as the association cost.
\subsubsection{Track update}
First, we use matching cascade to match in priority for the most recently associated trajectories, avoiding the trajectory drift caused by long-term occlusion \cite{wojke2017simple}.
During the matching, we use the Hungarian algorithm to find the minimum cost matches in the association cost matrix.
For unmatched trajectories and detection boxes, we calculate the IOU.
If they meet the threshold, they are associated.
After that, trajectories need to be updated. They have three states: unconfirmed, confirmed, and deleted. We assign a new trajectory for each unmatched detection.
Furthermore, if duration of trajectory is less than three, it will be set to an unconfirmed state.
The unconfirmed trajectories need to be successfully associated for three consecutive frames before being converted into confirmed state; otherwise, they will be deleted.
For the unmatched confirmed trajectories, if they are successfully matched in the previous frame, the KF will to estimate and update their motion state in the current frame; otherwise, we will suspend tracking.
Moreover, if the number of consecutively lost frames of confirmed trajectories exceeds the threshold (Amax=30), they will be deleted.
\section{Results}
\label{sec:experiment}
\subsection{Ant colony database}
\label{sec:database}
We establish an video database of ant colony, which contains a total of 10 videos.
Five videos are from an existing published work \cite{CAO2020107233} and captured in the indoor (laboratory) environment.
The remaining five outdoor videos are captured in different backgrounds and are obtained from the online website \emph{DepositPhotos} (http://www.depositphotos.com).
Table~\ref{tab:Statistics} shows detailed video information, where $\mathcal{I}$ represents an indoor video, $\mathcal{O}$ represents an outdoor video.
The resolutions of indoor and outdoor videos are 1920$\times$1080 and 1280$\times$720, respectively.
\begin{table}[tbhp]
\centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\textbf{Sequence} & \textbf{FPS} & \textbf{Resolution} & \textbf{Length} & \textbf{Ants} & \textbf{Annotations} \\ \hline
$\mathcal{I}_1$ & \multirow{5}{*}{25} & \multirow{5}{*}{1920$\times$1080} & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_2$ & & & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_3$ & & & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_4$ & & & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_5$ & & & 1001 (00:40) & 10 & 3510 \\ \hline
$\mathcal{O}_1$ & \multirow{5}{*}{30} & \multirow{5}{*}{1280$\times$720} & 600 (00:20) & 73 & 11178 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_2$ & & & 677 (00:23) & 162 & 25158 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_3$ & & & 577 (00:19) & 133 & 10280 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_4$ & & & 526 (00:18) & 193 & 27902 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_5$ & & & 569 (00:19) & 101 & 22044 \\ \hline
\end{tabular}
\caption{Statistics of ant videos with annotations in indoor and outdoor scenes.}
\label{tab:Statistics}
\end{table}
The videos in our database have a total of 4983 frames.
There are 10 ants per frame in the indoor videos.
The number of ants in each frame is 18-53 in the outdoor videos.
The number of objects in this scenario is significant, considering the fact that the popular COCO benchmark dataset contains only on average 7.7 instances per image.
Some video characteristics present challenges for detection and tracking algorithms, for example over-exposure for indoor videos and diverse background for outdoor ones.
There are caves or rugged terrains in outdoor scenes, and ants may enter or leave the scene.
Different from multi-human tracking, ants are visually similar and this causes significant challenges for tracking.
We manually mark the video frame by frame.
To facilitate training and reduce labeling cost, the aspect ratio of each bounding box is 1:1.
Considering the posture and scale of ants, we set the size of the bounding box to 96$\times$96 for indoor videos and 64$\times$64 for outdoor videos. The database and code will be made publicly available.
\subsection{Evaluation index}
\label{sec:evaluation}
In this paper, the evaluation indicators of detection and tracking performance are as follows:
\begin{itemize}
\item Mean Average Precision (MAP): the weighted sum of the average precision of all videos. The weight value is the proportion of frames.
\item False Positive (FP): the total number of false alarms.
\item False Negative (FN): the total number of objects that do not match successfully.
\item Identity Switch (IDS): the total number of identity switches during the tracking process.
\item Fragments (FM): the total number of incidents where the tracking result interrupts the real trajectory.
\item mean Multi-object Tracking Accuracy (mMOTA): the weighted sum of the average tracking accuracy of all videos. The equation to compute mMOTA is: mMOTA = 1 - (FP + FN + IDS)/NUM\_LABELED\_SAMPLES, where NUM\_LABELED\_SAMPLES is the total number of labeled samples.
\item mean Multi-object Tracking Precision (mMOTP): the weighted sum of the average tracking precision of all videos. Tracking precision measures the intersection over union (IOU) between labeled and predicted bounding boxes.
\item Frame Rate (FR): the number of frames being tracked per second.
\end{itemize}
\subsection{Results of multi-ant detection}
In our ant database, we set up five groups of training sets (Table~\ref{tab:det_results}) and compare their performance with that of the remaining datasets.
The naming conventions are:
\begin{itemize}
\item
$\mathcal{I}_5$+$\mathcal{O}_4$ represents a union of the $5^{th}$ indoor video and the $4^{th}$ outdoor video.
\item
$\mathcal{I}_{1-4}$ represents a union of indoor videos with their IDs of [1,2,3,4].
\item
$\mathcal{I}_5\left(50\right)$ represents the last 50 frames selected from the $5^{th}$ indoor video. This partition strategy ensures the frame continuity for the subsequent tracking task.
\item
$\mathcal{O}_{1-5}\left(-50\right)$ represent the union of 5 subsets, the last 50 frames de-selected from the outdoor videos with their IDs of [1,2,3,4,5].
\end{itemize}
In all scenarios, the detection accuracy of indoor videos is higher than that of outdoor videos, and MAP reaches over 90\%.
We also noticed that the test result for outdoor videos was only 49.7\% on $\mathcal{I}_5$+$\mathcal{O}_4$.
This is because we used only $\mathcal{O}_4$ as the outdoor training set, which is insufficient to cover the wide range of diversity in terms of environmental backgrounds and ant appearances.
\begin{table}[!h]
\centering
\begin{tabular}{@{}llccc@{}}
\toprule
\multicolumn{1}{c}{Training Data} & \multicolumn{1}{c}{Testing Data} & Objects & MAP$\uparrow$ & FR$\uparrow$ \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_4$} & $\mathcal{I}_{1-4}$ & 10 & 90.4 & \textbf{12.5} \\
& $\mathcal{O}_{1-3,5}$ & 28 & 49.7 & 16.0 \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5\left(50\right)$+$\mathcal{O}_{1-5}\left(50\right)$} & $\mathcal{I}_{1-4}$
& 10 & 90.4 & 12.2 \\
& $\mathcal{O}_{1-5}\left(-50\right)$ & 33 & 81.9 & \textbf{17.1} \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_{1-5}\left(100\right)$} & $\mathcal{I}_{1-4}$
& 10 & 90.4 & 12.3 \\
& $\mathcal{O}_{1-5}\left(-100\right)$ & 33 & 82.4 & 16.6 \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_{1-5}\left(200\right)$} & $\mathcal{I}_{1-4}$
& 10 & 90.5 & 12.3 \\
& $\mathcal{O}_{1-5}\left(-200\right)$ & 33 & 85.1 & 16.6 \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_{1-5}\left(300\right)$} & $\mathcal{I}_{1-4}$
& \textbf{10} & \textbf{90.5} & 11.8 \\
& $\mathcal{O}_{1-5}\left(-300\right)$ & \textbf{33} & \textbf{85.8} & 16.2 \\ \bottomrule
\end{tabular}
\caption{\label{tab:det_results} Detection results of different training sets.}
\end{table}
\begin{figure}[!htbp]
\centering
\includegraphics[width=0.6\linewidth]{different_sets}
\caption{\label{fig:different_sets} Detection accuracy of different training sets.}
\end{figure}
In the subsequent experiments, we integrate the images of all outdoor scenes into the outdoor training set and dramatically improve the accuracy of outdoor testing.
Figure~\ref{fig:different_sets} clearly shows the effects of using different training sets.
By further increasing in the number of images in outdoor videos, the detection accuracy of outdoor scenes improves slightly.
For indoor environments, the detection accuracy is impervious to different training sets.
Moreover, reducing the number of images to 50 ($\mathcal{I}_5$ has a total of 351 frames) does not reduce the detection accuracy.
This shows that we need only a small number of training samples to achieve satisfactory results when the training and testing scenarios are the same.
The frame rate is around 12 FR for indoor videos and 16 FR for outdoor ones.
The factor of different image resolution should be accountable for this performance gap.
In practical applications, if accuracy is guaranteed, we tend to use smaller training sets to reduce labeling costs.
Therefore, we use the model trained in "$\mathcal{I}_5\left(50\right)$+$\mathcal{O}_{1-5}\left(50\right)$" for comparison with the other methods in the comparative experiments.
\subsection{Results of multi-ant tracking}
Based on the TBD paradigm, we use detection results as the input to the tracking framework. For offline training, we randomly select 50 labeled samples from $\mathcal{I}_5$ as the training set.
We visualized the tracking results in Figure~\ref{fig:trajectories}.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth, height=1\linewidth]{trajectories}
\caption{\label{fig:trajectories} Tracking trajectories in test videos. Horizontal axes indicate the pixel coordinates in an image. (a-d) indoor scenes. (e-i) outdoor scenes.}
\end{figure}
Table~\ref{tab:tracking_performance} shows the performance of online tracking.
After integrating the images of each outdoor video in the detection training set, our method gets 95\% mMOTA for indoor videos and over 80\% for outdoor videos.
Additionally, mMOTP is around 80\% for both indoor and outdoor videos.
Notably, since the tracking performance depends on the detection result, the tracking task in $\mathcal{O}_{1-3,5}$ fails due to the low-quality detection (the second row in Table~\ref{tab:tracking_performance}).
Except for this failure case, the tracking performance is generally satisfactory considering that we only use 50 labeled samples from one indoor video.
\begin{table}[h]
\centering
{
\begin{tabular}{@{}lcccccc@{}}
\toprule
\multicolumn{1}{c}{Testing Data} & FP$\downarrow$ & FN$\downarrow$ & IDS$\downarrow$ & mMOTA$\uparrow$ & mMOTP$\uparrow$ & FR$\uparrow$ \\ \midrule
$\mathcal{I}_{1-4}$ & 236 & 621 & 21 & 89.9 & 79.9 & 36.5 \\
$\mathcal{O}_{1-3,5}$ & \multicolumn{6}{c}{detection failure} \\ \midrule
$\mathcal{I}_{1-4}$ & 239 & 628 & 22 & \textbf{95.7} & 81 & \textbf{38.9} \\
$\mathcal{O}_{1-5}\left(-50\right)$ & 6078 & 7122 & 625 & 81.8 & 81.9 & 26.2 \\ \midrule
$\mathcal{I}_{1-4}$ & 260 & \textbf{617} & \textbf{14} & 95.7 & 81 & 38.7 \\
$\mathcal{O}_{1-5}\left(-100\right)$ & 4867 & 6018 & 526 & 83.3 & 82.7 & \textbf{28.0} \\ \midrule
$\mathcal{I}_{1-4}$ & 228 & 709 & 17 & 95.4 & 81 & 36.3 \\
$\mathcal{O}_{1-5}\left(-200\right)$ & 4289 & 3421 & 394 & 85.3 & 83 & 24.9 \\ \midrule
$\mathcal{I}_{1-4}$ & \textbf{224} & 820 & 24 & 94.8 & \textbf{81.7} & 35.4 \\
$\mathcal{O}_{1-5}\left(-300\right)$ & \textbf{2644} & \textbf{3007} & \textbf{266} & \textbf{85.8} & \textbf{83.3} & 26.4 \\ \midrule \hline
$\mathcal{I}_{1-4}$/GT & 22 & 23 & 8 & 99.6 & 92.4 & 35.2 \\
$\mathcal{O}_{1-5}\left(-50\right)$/GT & 1697 & 458 & 1064 & 96.2 & 92.4 & 25.9 \\ \bottomrule
\end{tabular}}
\caption{\label{tab:tracking_performance} Tracking performance evaluation. The last two rows indicate that we use the ground truth of detection for tracking, which leads to a boost in tracking performance.}
\end{table}
The time cost of the tracking model is mainly incurred by generating 128-d feature vectors for each detection box.
The average number of objects in outdoor videos is more than three times that in indoor videos.
As for runtime time, FR reaches over 35 in indoor videos and more than 24 in outdoor videos.
We add a set of comparative experiments in the last two rows of Table~\ref{tab:tracking_performance}.
We directly use manually-labeled detection boxes for tracking and compare the detection results on the $\mathcal{I}_{1-4}$ and $\mathcal{O}_{1-3,5}$.
Both mMOTA and mMOTP have been dramatically improved.
This implies that an increase in detection accuracy could further boost the tracking performance of our framework.
\subsection{Comparative experiments}
\label{sec:comparative}
There are two widely used insect tracking software: idTracker \cite{perez2014idTracker} and Ctrax \cite{chiron2013detecting}.
idTracker needs to specify the number of objects before tracking, to create a reference image set for each object.
Meanwhile, Ctrax assumes that objects will rarely enter and leave the arena.
Thus, they are both not capable of tracking in outdoor scenes because of the variable number of ants.
Therefore, we compare these two methods only in videos depicting indoor scenes.
idTracker needs to specify the number of objects before tracking, in order to create a reference image set for each object.
To compare them with our method, we convert their representations into square boxes as our ground truth.
Table~\ref{tab:comparison} and Figure~\ref{fig:spatial_temporal} shows the tracking results.
In addition to a significant improvement of tracking accuracy, our method is 6 and 10 times faster than idTracker and Ctrax (see the column of FR).
\begin{table}[h]
\centering
{
\begin{tabular}{cccccccc}
\toprule
Method & FP$\downarrow$ & FN$\downarrow$ & IDs$\downarrow$ & FM$\downarrow$ & mMOTA$\uparrow$ & mMOTP$\uparrow$ & FR$\uparrow$ \\ \hline
idTracker & 881 & 8479 & 83 & 432 & 54 & 77.4 & 1.3 \\
Ctrax & 2832 & 5646 & 110 & 349 & 58.2 & 79.7 & 0.8 \\
Ours & \textbf{239} & \textbf{628} & \textbf{22} & \textbf{189} & \textbf{95.7} & \textbf{81.1} & \textbf{8.7} \\ \bottomrule
\end{tabular}}
\caption{\label{tab:comparison} Comparison of tracking results on videos $\mathcal{I}_{1-4}$.}
\end{table}
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{spatial_temporal}
\caption{\label{fig:spatial_temporal} Comparison of tracking performance in spatial-temporal dimension ($\mathcal{I}_1$). Horizontal axes indicate the pixel coordinates in an image.}
\end{figure}
We further compare tracking accuracy of idTracker and Ctrax across different indoor videos, as Figure~\ref{fig:comparison} shows.
The large variance of idTracker's performance is affected by the number of static ants, which will cause missing tracking.
Ctrax proves to be robust but with a lower accuracy compared with our method.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{comparison}
\caption{\label{fig:comparison} Comparison of tracking results for indoor scenes.}
\end{figure}
\section{Discussion}
\subsection{Comparison of methods}
In the previous section, we compared two methods of insect tracking, idTracker and Ctrax.
idTracker uses the intensity and contrast of the foreground segmented area to extract appearance features and construct a reference image set for each individual.
However, it can not track motionless individuals.
Figure~\ref{fig:spatial_temporal}(a) shows that only a minority group of ants are successfully tracked over the period of video.
Further, there are some trajectory fragments due to the limitations of the foreground segmentation model for multiple objects.
Compared to our results, the trajectories of Ctrax are incomplete.
This indicates that there are more FN, as Figure~\ref{fig:spatial_temporal}(b) shows.
Ctrax requires a sharp contrast between object and background.
The ants passing through the overexposed areas in the scene will be ignored.
Additionally, Ctrax assumes that the motion of the object obeys the linear distribution.
However, the ants' movement is nonlinear, and their speed and direction might change abruptly, causing IDS in Ctrax.
Our method classifies and regresses twice to locate ants accurately.
During the tracking stage, we use the historical appearance sequence as a reference and update it frame by frame.
Compared with idTracker, our method effectively solves the long-term and short-term dependence of motion states, thereby reducing FM.
Despite that we also assume the linear distribution of motion states, they are used only to filter impossible associations, and have nothing to do with association cost.
We take the appearance distance between trajectories and detection boxes as association cost, thus the model is robust even when the ant movement is complicated.
We take the appearance measure between trajectories and detection boxes as the association cost, thus the model is robust even when the ant movement is complicated.
\subsection{Failure cases}
\label{sec:failure}
\subsubsection{Limitations of detection framework}
The number of ants in outdoor scenes is on average 33 per frame.
It is also typical for ants to involve close body contact with each other for the purpose of information sharing.
Naturally, their extremely-close interactions are highly likely to cause mis-detection (Figure~\ref{fig:failed_detection}(a)).
Additionally, entrances and exits of ants in outdoor scenes are more prone to mis-detection (Figure~\ref{fig:failed_detection}(b)).
Moreover, the dramatically non-rigid deformation of ants is also a factor causing the detection failure (Figure~\ref{fig:failed_detection}(c)).
These three scenarios are all challenging cases that deserve our future efforts.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{failed_detection}
\caption{\label{fig:failed_detection} Examples of failed detection in outdoor scenes.}
\end{figure}
\subsubsection{Limitations of tracking framework}
According to Figure~\ref{fig:drift}, Ant No.41 entered the scene at Frame No.88.
Coincidentally Ant No.32 left the scene at an adjacent region, but its trajectory was not deleted.
At Frame No.93, Ant No.41 drifted to Trajectory No.32.
This defect is caused by insufficient appearance descriptors stored in the gallery of Ant No.41, and it moved near the exit location of another ant.
This kind of mis-association occurs at the image boundary and accounts for the majority of IDS and FM in our experiments.
However, when ants move inside the scope of both indoor and outdoor scenes, our method can accurately track multiple ants simultaneously for a long time, as Figure~\ref{fig:tracking_results} shows.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{drift}
\caption{\label{fig:drift} Drift at the scene boundary. A newly-entered Ant No.41 is mis-associated with an existing Trajectory No.32.}
\end{figure}
\section{Conclusion}
We proposed a complete detection and tracking framework based on deep learning for ant colony tracking.
In the detection stage, we adopted a two-stage object detection framework for the detection task.
We also use a ResNet model to obtain ant appearance descriptors for online associations.
Next, we combined appearance and motion information for the tracking task.
The experimental results demonstrated that our method outperformed two mainstream insect tracking models in terms of accuracy, precision, and speed.
Particularly, our work shows its advantage in robustly detecting and tracking ant colonies in outdoor scenes, which is rarely reported in existing literature.
We believe our method could serve as an effective tool for high-throughput swarm behavior analysis of ant colonies, leading to the development of embodied intelligence.
In future research, we aim to achieve more robust detection.
For example, by exploring additional information of ants' skeletal structure, we can potentially solve the aforementioned failure case of close interaction and nonrigid deformation problem.
We also plan to improve the generalization ability of our detection and tracking frameworks so that it is applicable to a wide range of outdoor environments.
\section*{Conflict of Interest Statement}
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
\section*{Author Contributions}
All authors listed have made substantial, direct, and intellectual contribution to the work; they have also approved it for publication.
In particular, MW, SG and XC contributed to the design of this work; MW, SG and XC contributed to the writing of the manuscript; XC designed and implemented the multi-ant tracking framework; XC conducted the experiments, and analyzed the results.
\section*{Acknowledgments}
This work was supported by the Natural Science Foundation of Fujian Province of China (No. 2019J01002).
\section*{Data Availability Statement}
The training and testing datasets used for this study can be found in the \href{https://data.mendeley.com/datasets/9ws98g4npw/3}{ANTS--ant detection and tracking}.
\bibliographystyle{Frontiers-Harvard}
\section{Keywords:} social insects, swarm behavior, automatic tracking method, object detection, online tracking}
\end{abstract}
\section{Introduction}
Swarm behavior is one of the most important features of social insects, which has important significance for the study of embodied intelligence ~\cite{tiacharoen2012design}. Specifically, social insects often tend to cluster into a colony~\cite{vandermeer2008clusters}, which forms a complex dynamical system together with the surrounding environment~\cite{balch2001automatically}.
So far, we do not know enough about the mechanisms behind swarm behaviors of social insects.
The mainly reason is that the key requirement of this research is the ability to track the motions and interactions of each individual robustly and accurately.
However, until the late $20^{th}$ century, biologists still manually marked the motion trajectories on the video to guarantee the quality. They have to track each individual at one time, which might mean watching the entire video 50 times or more in a crowded scene~\cite{poff2012efficient}.
It becomes an inhibiting factor in obtaining the complete and accurate dataset required to analyze the evolution of complex dynamical system.
Therefore, in the past two decades, attempts have been made to automate the tracking process for social insects utilizing computer vision (CV) techniques~\cite{khan2005mcmc, khan2006mcmc, oh2006parameterized, veeraraghavan2008shape, fletcher2011multiple}.
Traditional CV techniques free researchers from manual work through approaches such as foreground segmentation algorithm~\cite{li2008estimating}, temporal difference method~\cite{khan2005mcmc} and hungarian algorithm~\cite{li2009learning}.
Such approaches, however, have failed to address the noise in the image~\cite{zhao2015improved}, resulting in the limitation that a laboratory environment with a clean background is needed.
Nevertheless, many scientifically valuable results are obtained in nature rather than laboratory environment~\cite{schmelzer2009special, kastberger2013social, tan2016honey, dong2018olfactory}.
In recent years, with the popularity of computer vision, many advanced object detection and tracking methods have emerged.
\subsection{Object detection}
Existing methods in object detection are categorized as one-stage or two-stage, according to whether there is a separate stage of region proposal.
One-stage frameworks (e.g., YOLO~\cite{redmon2016you}) are fast, but their accuracy is typically slightly inferior compared with that of two-stage detection.
The popularity of two-stage detection frameworks is enhanced by R-CNN~\cite{girshick2014rich}, which proposes candidate regions via a selective search (SS) algorithm~\cite{uijlings2013selective}, thereby the detector focuses on these RoIs.
However, using the SS algorithm~\cite{uijlings2013selective} to generate region proposals is the main reason causing slow inference.
Fast R-CNN~\cite{girshick2015fast} reduces the computational complexity of region proposals by downsampling the original image, while Faster R-CNN~\cite{ren2015faster} proposes an RPN, which further improves the speed of training and inference.
Given the success of deep learning in general tasks of object detection, researchers also applied to detect specific groups of animals, such as a single mouse~\cite{geuther2019robust}, fruit flies~\cite{murali2019classification}.
These methods are either limited to track a single object, or a fixed number of objects.
General tools~\cite{romero2019idtracker,sridhar2019tracktor} also offer the functionality to detect and track unmarked animals in the image.
However, most of existing methods focus on the condition of ideal lab set-up and none of existing works reported the detection of ants in outdoor environments which contain diverse backgrounds and arbitrary terrains.
\subsection{Multi-object tracking (MOT)}
In the last two decades, vision-based detection and tracking models have been widely used to study social insects~\cite{khan2005mcmc,veeraraghavan2008shape}.
Appearance (particularly color) and motion information are the main metrics used in this category of method.
Due to high similarity of ants' appearance, researchers either use the technique of pigmenting to create more distinct appearance features~\cite{fletcher2011multiple}, or limit the observation to a laboratory setup~\cite{branson2009high, perez2014idTracker}.
State-of-the-art methods, such as Ctrax~\cite{branson2009high} and idTracker~\cite{perez2014idTracker}, for insect tracking are tested in a laboratory setup and use background subtraction for foreground segmentation.
Notably, the operations of background modeling and foreground extraction are time-consuming.
The tracking-by-detection (TBD) paradigm is to match trajectories and detections in two consecutive frames, a process that requires metrics.
The global nearest neighbor model measures motion state to achieve Drosophila tracking~\cite{chiron2013detecting}.
The global nearest neighbor model assumes that the motion state obeys the linear observation model, which commonly uses a constant velocity model - the Kalman filter (KF).
However, changes in ants' speed and direction are difficult to predict, thus appearance information is integrated as a metric.
The DAT method is a mainstream method for ant colony tracking~\cite{fasciano2014ant}.
It allows a combination of multiple metrics, and uses Hungarian algorithm~\cite{li2009learning} to assign detections for trajectories.
The PF method is suitable for solving nonlinear problems~\cite{fasciano2013tracking}, but the growth in the number of particles leads to an exponential increase in the computational cost, preventing the effective multi-object tracking. Using Markov Chain Monte Carlo sampling can reduce computational complexity~\cite{khan2006mcmc}.
A GPU-accelerated semi-supervised framework can further improve tracking accuracy and performance~\cite{poff2012efficient}.
When applying the methods above for tracking ant colonies, they are greatly disturbed by background noise and difficult to overcome the serious occlusion problem in dense scenes.
Long short-term memory~\cite{kim2018multi} and spatial-temporal attention mechanisms~\cite{zhu2018online} have been developed to tackle the problem of long-term occlusion.
A bilinear Long short-term memory structure that couples a linear predictor with input detection features, thereby modeling long-term appearance features~\cite{kim2018multi}.
The spatial-temporal attention mechanism is also suitable for the MOT task.
The spatial attention module makes the network focus on the pattern of matching. Meanwhile the temporal attention module assigns different levels of attention to the sample sequence of the trajectory~\cite{zhu2018online}.
The TBD paradigm-based framework is dependent on detection results. Therefore, severe occlusion is likely to cause tracking failures.
To prevent this situation, a detector with automatic bounding box repairing and adjustment is introduced by a cyclic structure classifier~\cite{dong2016occlusion}.
In this paper, we use a deep learning method to build a detection and tracking framework.
Our method is based on the TBD paradigm and accomplishes the goal of online multi-ant tracking.
To the best of authors' knowledge, this is the first work to achieve robust detection and tracking of ant colony in both indoor and outdoor environments (Figure~\ref{fig:tracking_results}).
Our method is robust in tackling the challenge of visual similarity among colony individuals, handling diverse terrain backgrounds and achieving long-period of tracking.
Our main contributions are as follows:
\begin{itemize}
\item We adopt a two-stage object detection framework, using ResNet-50 as the backbone and position sensitive score maps to encode regions of interest (RoIs).
During the tracking stage, we use a ResNet network to obtain the appearance descriptors of ants and then combine them with motion information to achieve online association.
\item Our method proves to be robust in both indoor and outdoor scenes. Furthermore, only a small amount of training data are required to achieve the goal in our pipeline, which are 50 images chosen for each scene in the detection framework and 50 labels randomly chosen for the tracking framework respectively.
\item We construct an ant database with labeled image sequences, including five indoor videos (laboratory setup) and five outdoor videos, with 4983 frames and 115,433 labels in total.
The database is made publicly available, which is hoped that it will contribute to a deeper exploration on swarm behavior of ant colony.
\end{itemize}
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{tracking_results}
\caption{\label{fig:tracking_results} Tracking results by our method in both indoor and outdoor environments.}
\end{figure}
\section{Materials and methods}
\subsection{Overview}
\label{sec:overview}
Following the TBD paradigm, we propose a uniform framework for detection and tracking to efficiently and accurately track the ant colony in both indoor and outdoor scenes (Figure~\ref{fig:pipeline}).
In the detection phase, we adopt a two-stage object detection framework, using ResNet-50 as the backbone, and encoding RoIs proposed by regional proposal network via position-sensitive score maps.
Then we implement classification and regression through downsampling and voting mechanisms.
(see details in Section~\nameref{sec:two-stage}).
In the tracking stage, we first use ResNet to train the appearance descriptors of ants and measure the appearance similarity between two objects.
Next, the tracking is accomplished by combining appearance and motion information for online association metric. (Section~\nameref{sec:MOT_framework}).
\begin{figure}[!htbp]
\centering
\includegraphics[width=0.65\linewidth, height=0.7\linewidth]{pipeline}
\caption{\label{fig:pipeline} Architecture for detection and tracking.}
\end{figure}
\subsection{Two-stage object detection}
\label{sec:two-stage}
\subsubsection{Regional proposal network}
Regional proposal network (RPN) is proposed in Faster R-CNN \cite{ren2015faster} to generate RoIs.
Compared to SS \cite{uijlings2013selective}, RPN is based on the CNN network structure and can connect the backbone with shared weight, significantly improving detection speed.
We use ResNet-50 as the backbone and replace the fully connected layer with a 1$\times$1 convolution to reduce the dimensions of feature maps.
Considering that ResNet-50 conducts downsampling 32 times, we get 256-d feature maps via a 3$\times$3 Atrous convolution to maintain translation variability.
For each sliding position, we predict k region proposal boxes of different sizes and ratios; these boxes are called anchors.
After the 256-d vector, we connect classification and regression branches through two parallel 1$\times$1 convolution layers.
The classification branch uses softmax to determine whether there is an object in anchor so that this branch has 2$\times$k outputs.
The regression branch will perform a regression on the 4D position parameters of anchors (i.e., center coordinates, width and height) so that there are 4$\times$k outputs.
RPN will propose k$\times$w$\times$h anchors with a w$\times$h feature map, called RoIs.
We use the Non-maximum suppression algorithm \cite{neubeck2006efficient} to filter duplicate anchors and set the IOU threshold to 0.7.
\subsubsection{Position sensitive region of interest}
On the basis of RoIs, the two-stage detection framework classifies and fine-tunes the location of bounding boxes.
In Faster R-CNN, RoIs are scaled to the last feature maps and focusing on these areas through ROIPooling.
Next, each RoI is classified and regressed through two fully connected layers, causing high computational complexity.
In order to reduce the number of parameters, we use RPN-FCN \cite{dai2016r} to generate position-sensitive score maps via a convolutional layer, which is connected to the backbone.
Both classification and regression tasks have independent position-sensitive score maps, forming three parallel branches with RPN.
For the classification task, since we only need to classify ants and background, we use k$\times$k$\times$2 convolution kernels to generate score maps.
k$\times$k indicates that each RoI is divided into k$\times$k regions to encode position information.
Each region is encoded by a specific feature map with two dimensions.
Similarly, we use k$\times$k$\times$4 convolution kernels for fine-tuning the position of RoIs in the regression task.
To focus on RoIs, we perform average pooling on each region to get feature maps, called position sensitive region of interest (PSRoI) pooling, as the following formula shows:
\begin{equation}
r_{c}(i, j | \Theta)=\sum_{(x, y) \in \text {region}(i, j)} z_{i, j, c}\left(x+x_{0}, y+y_{0} | \Theta\right) / n.
\end{equation}
$r_{c}(i, j | \Theta)$ is the result of downsampling in $(i,j)^{th}$ for $c^{th}$ category, and $z_{i, j, c}$ is one score map in the k$\times$k$\times$2 position-sensitive score maps. $(x0, y0)$ represents the left-top corner of RoI. $\Theta$ is the set of parameters of the network, and $n$ is the number of pixels in the region.
For the feature maps, we vote on k$\times$k regions, getting the overall score of RoI on the classification or regression task, as the following formula shows:
\begin{equation}
r_{c}(\Theta)=\sum_{i, j} r_{c}(i, j | \Theta).
\end{equation}
In the formula, $r_{c}(\Theta)$ represents the overall scores of all regions.
Next, we use softmax to implement binary classification, as the following formula shows:
\begin{equation}
s_{c}(\Theta)=e^{r_{c}(\Theta)} / \sum_{c=0}^{C} e^{r_{c}(\Theta)}.
\end{equation}
Here, $s_{c}(\Theta)$ is the probability of $c^{th}$ category.
Finally, we use the Non-maximum suppression algorithm to filter the bounding box.
Since object detection includes classification and regression, we require a multitask loss function.
In this paper, we weight the loss functions of the two tasks. Because softmax is used for the binary classification task, it is natural to adopt cross-entropy loss for the classification task. For the regression task, we calculate the matching degree between the four position parameters and ground truth:
\begin{equation}
L\left(s, t_{x, y, w, h}\right)=L_{c l s}\left(s_{c^{*}}\right)+\lambda\left[c^{*}=1\right] L_{r e g}\left(t, t^{*}\right).
\end{equation}
where $c^{*}$ is the ground truth category label of RoI, and $c^{*} = 1$ represents ants.
$L_{c l s}\left(s_{c^{*}}\right)$ represents cross-entropy loss:
\begin{equation}
L_{c l s}\left(s_{c}\right)=-\log \left(s_{c}\right).
\end{equation}
$L_{r e g}\left(t, t^{*}\right)$ represents the loss of the regression task, including 4 dimensions:
\begin{equation}
L_{r e g}\left(t, t^{*}\right)=\Sigma_{i=1}^{4} g\left(t_{i}^{*}-t_{i}\right).
\end{equation}
In the formula, $t^{*}$ is the predicted position, and $t$ is ground truth after translation and scaling.
\subsection{MOT framework}
\label{sec:MOT_framework}
\subsubsection{Offline ResNet network architecture}
We adopt a 15-layer ResNet network architecture to extract the appearance descriptors of objects, as Figure~\ref{fig:pipeline} shows.
After downsampling eight times, the network will eventually obtain a 128-dimensional feature vector through a fully connected layer. The specific parameters are consistent with \cite{CAO2020107233}.
\subsubsection{Cosine similarity metric classifier}
We modify the parameters of softmax to get a cosine similarity measurement classifier, which can measure the similarity of the same category or different categories. First, the output of a fully connected layer is normalized by batch normalization, ensuring that it is expressed as a unit length $\left\|f_{\Theta}(x)\right\|_{2}=1$, $\forall x \in R^{D}$. Second, we normalize the weights, that is, $\varpi_{k}=\omega /\left\|\omega_{k}\right\|_{2}$, $\quad\forall k=1, \ldots C$.
Cosine similarity metric classifier is constructed as follows:
\begin{equation}
p\left(y_{i}=k | r_{i}\right)=\frac{\exp \left(\kappa \cdot \varpi_{k}^{T} r_{i}\right)}{\sum_{n=1}^{C} \exp \left(\kappa \cdot \varpi_{n}^{T}\right)}.
\end{equation}
Here, $\kappa$ is the free scaling parameter.
Because the cosine similarity classifier follows the structure of softmax, we use the cross-entropy loss for training:
\begin{equation}
L(D)=-\sum_{i=1}^{N} \sum_{k=1}^{C} \mathrm{gt}_{y_{i}-k} \cdot \log p\left(y_{i}=k | r_{i}\right).
\end{equation}
Here, $L(D)$ represents the sum of the cross-entropy loss of $N$ images, $p\left(y_{i}=k | r_{i}\right)$ is the prediction result of $i^{th}$ image in $k^{th}$ label, and $\mathrm{gt}_{y_{i}-k}$ is ground truth.
\subsubsection{Motion matching}
We use the KF model to predict the position of trajectories in the current frame.
Then, we calculate the square of the Mahalanobis distance between the predicted position and the detected bounding box position by measuring the degree of motion matching \cite{wojke2017simple} as follows:
\begin{equation}
d^{(1)}(i, j)=\left(d_{j}-y_{i}\right)^{T} S_{i}^{-1}\left(d_{j}-y_{i}\right).
\end{equation}
Here, $d_{j}$ is the position of the $j^{th}$ detection box, $y_{i}$ is the position of the $i^{th}$ trajectory predicted by the KF, and $S_{i}$ is the covariance matrix between the $i^{th}$ trajectory and the detected bounding box.
We use a 0-1 variable to indicate whether trajectory and detection meet the association conditions. If the Mahalanobis distance meets $t^{\left(1\right)}$, $\left(i,j\right)$ will be added to the association set. The formula can be expressed as:
\begin{equation}
b_{i j}^{(1)}=\left\{\begin{array}{ll}
{1,} & {d^{(1)}(i, j)<t^{(1)}} \\
{0,} & {\text{ otherwise }}
\end{array}\right..
\end{equation}
Here, $b_{i j}^{(1)}$ is the motion association signal.
\subsubsection{Appearance matching}
\label{sec:appearance_matching}
We use the appearance descriptors to measure the appearance similarity between ants. Furthermore, we create a gallery for each trajectory, and each gallery stores the latest 100 appearance descriptors.
Then, we calculate the cosine distance of appearance descriptors between gallery and candidate bounding boxes.
The smallest distance is used as an appearance matching degree as follows:
\begin{equation}
d^{(2)}(i, j)=\min \left\{1-r_{j}^{T} r_{k}^{(i)} | r_{k}^{(i)} \in \mathrm{K}_{i}\right\}.
\end{equation}
where $r_{j}$ is the appearance descriptor of the $j^{th}$ detection box, $r_{k}^{(i)}$ is the $k^{th}$ appearance descriptor of the $i^{th}$ trajectory, $d^{(2)}(i, j)$ represents the appearance matching degree between the $i^{th}$ trajectory and the $j^{th}$ bounding box.
Similarly, we introduce a 0-1 variable as an association signal. If the appearance matching degree from a pair of trajectory and detection boxes meets the threshold, we add it to the association set:
\begin{equation}
b_{i j}^{(2)}=\left\{\begin{array}{ll}
{1,} & {d^{(2)}(i, j)<t^{(2)}} \\
{0,} & {\text { otherwise }}
\end{array}\right..
\end{equation}
where $b_{i j}^{(2)}$ represents the appearance association signal. In this paper, $t^{(2)}$ is set to 0.2.
\subsubsection{Comprehensive matching}
To combine motion and appearance information, we set a comprehensive association signal $b_{i j}$. Only when both motion and appearance matching degree meet the threshold, the $\left(i, j\right)$ pair will be considered for matching. The formula expression is denoted as follows:
\begin{equation}
b_{i j}=\prod_{m=1}^{2} b_{i, j}^{(m)}.
\end{equation}
However, the KF is scarcely possible to track accurately for long periods, because of the motion of ants is complicated.
Therefore, we use the appearance matching degree (Section~\nameref{sec:appearance_matching})
as the association cost.
\subsubsection{Track update}
First, we use matching cascade to match in priority for the most recently associated trajectories, avoiding the trajectory drift caused by long-term occlusion \cite{wojke2017simple}.
During the matching, we use the Hungarian algorithm to find the minimum cost matches in the association cost matrix.
For unmatched trajectories and detection boxes, we calculate the IOU.
If they meet the threshold, they are associated.
After that, trajectories need to be updated. They have three states: unconfirmed, confirmed, and deleted. We assign a new trajectory for each unmatched detection.
Furthermore, if duration of trajectory is less than three, it will be set to an unconfirmed state.
The unconfirmed trajectories need to be successfully associated for three consecutive frames before being converted into confirmed state; otherwise, they will be deleted.
For the unmatched confirmed trajectories, if they are successfully matched in the previous frame, the KF will to estimate and update their motion state in the current frame; otherwise, we will suspend tracking.
Moreover, if the number of consecutively lost frames of confirmed trajectories exceeds the threshold (Amax=30), they will be deleted.
\section{Results}
\label{sec:experiment}
\subsection{Ant colony database}
\label{sec:database}
We establish an video database of ant colony, which contains a total of 10 videos.
Five videos are from an existing published work \cite{CAO2020107233} and captured in the indoor (laboratory) environment.
The remaining five outdoor videos are captured in different backgrounds and are obtained from the online website \emph{DepositPhotos} (http://www.depositphotos.com).
Table~\ref{tab:Statistics} shows detailed video information, where $\mathcal{I}$ represents an indoor video, $\mathcal{O}$ represents an outdoor video.
The resolutions of indoor and outdoor videos are 1920$\times$1080 and 1280$\times$720, respectively.
\begin{table}[tbhp]
\centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\textbf{Sequence} & \textbf{FPS} & \textbf{Resolution} & \textbf{Length} & \textbf{Ants} & \textbf{Annotations} \\ \hline
$\mathcal{I}_1$ & \multirow{5}{*}{25} & \multirow{5}{*}{1920$\times$1080} & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_2$ & & & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_3$ & & & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_4$ & & & 351 (00:14) & 10 & 3510 \\ \cline{1-1} \cline{4-6}
$\mathcal{I}_5$ & & & 1001 (00:40) & 10 & 3510 \\ \hline
$\mathcal{O}_1$ & \multirow{5}{*}{30} & \multirow{5}{*}{1280$\times$720} & 600 (00:20) & 73 & 11178 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_2$ & & & 677 (00:23) & 162 & 25158 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_3$ & & & 577 (00:19) & 133 & 10280 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_4$ & & & 526 (00:18) & 193 & 27902 \\ \cline{1-1} \cline{4-6}
$\mathcal{O}_5$ & & & 569 (00:19) & 101 & 22044 \\ \hline
\end{tabular}
\caption{Statistics of ant videos with annotations in indoor and outdoor scenes.}
\label{tab:Statistics}
\end{table}
The videos in our database have a total of 4983 frames.
There are 10 ants per frame in the indoor videos.
The number of ants in each frame is 18-53 in the outdoor videos.
The number of objects in this scenario is significant, considering the fact that the popular COCO benchmark dataset contains only on average 7.7 instances per image.
Some video characteristics present challenges for detection and tracking algorithms, for example over-exposure for indoor videos and diverse background for outdoor ones.
There are caves or rugged terrains in outdoor scenes, and ants may enter or leave the scene.
Different from multi-human tracking, ants are visually similar and this causes significant challenges for tracking.
We manually mark the video frame by frame.
To facilitate training and reduce labeling cost, the aspect ratio of each bounding box is 1:1.
Considering the posture and scale of ants, we set the size of the bounding box to 96$\times$96 for indoor videos and 64$\times$64 for outdoor videos. The database and code will be made publicly available.
\subsection{Evaluation index}
\label{sec:evaluation}
In this paper, the evaluation indicators of detection and tracking performance are as follows:
\begin{itemize}
\item Mean Average Precision (MAP): the weighted sum of the average precision of all videos. The weight value is the proportion of frames.
\item False Positive (FP): the total number of false alarms.
\item False Negative (FN): the total number of objects that do not match successfully.
\item Identity Switch (IDS): the total number of identity switches during the tracking process.
\item Fragments (FM): the total number of incidents where the tracking result interrupts the real trajectory.
\item mean Multi-object Tracking Accuracy (mMOTA): the weighted sum of the average tracking accuracy of all videos. The equation to compute mMOTA is: mMOTA = 1 - (FP + FN + IDS)/NUM\_LABELED\_SAMPLES, where NUM\_LABELED\_SAMPLES is the total number of labeled samples.
\item mean Multi-object Tracking Precision (mMOTP): the weighted sum of the average tracking precision of all videos. Tracking precision measures the intersection over union (IOU) between labeled and predicted bounding boxes.
\item Frame Rate (FR): the number of frames being tracked per second.
\end{itemize}
\subsection{Results of multi-ant detection}
In our ant database, we set up five groups of training sets (Table~\ref{tab:det_results}) and compare their performance with that of the remaining datasets.
The naming conventions are:
\begin{itemize}
\item
$\mathcal{I}_5$+$\mathcal{O}_4$ represents a union of the $5^{th}$ indoor video and the $4^{th}$ outdoor video.
\item
$\mathcal{I}_{1-4}$ represents a union of indoor videos with their IDs of [1,2,3,4].
\item
$\mathcal{I}_5\left(50\right)$ represents the last 50 frames selected from the $5^{th}$ indoor video. This partition strategy ensures the frame continuity for the subsequent tracking task.
\item
$\mathcal{O}_{1-5}\left(-50\right)$ represent the union of 5 subsets, the last 50 frames de-selected from the outdoor videos with their IDs of [1,2,3,4,5].
\end{itemize}
In all scenarios, the detection accuracy of indoor videos is higher than that of outdoor videos, and MAP reaches over 90\%.
We also noticed that the test result for outdoor videos was only 49.7\% on $\mathcal{I}_5$+$\mathcal{O}_4$.
This is because we used only $\mathcal{O}_4$ as the outdoor training set, which is insufficient to cover the wide range of diversity in terms of environmental backgrounds and ant appearances.
\begin{table}[!h]
\centering
\begin{tabular}{@{}llccc@{}}
\toprule
\multicolumn{1}{c}{Training Data} & \multicolumn{1}{c}{Testing Data} & Objects & MAP$\uparrow$ & FR$\uparrow$ \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_4$} & $\mathcal{I}_{1-4}$ & 10 & 90.4 & \textbf{12.5} \\
& $\mathcal{O}_{1-3,5}$ & 28 & 49.7 & 16.0 \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5\left(50\right)$+$\mathcal{O}_{1-5}\left(50\right)$} & $\mathcal{I}_{1-4}$
& 10 & 90.4 & 12.2 \\
& $\mathcal{O}_{1-5}\left(-50\right)$ & 33 & 81.9 & \textbf{17.1} \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_{1-5}\left(100\right)$} & $\mathcal{I}_{1-4}$
& 10 & 90.4 & 12.3 \\
& $\mathcal{O}_{1-5}\left(-100\right)$ & 33 & 82.4 & 16.6 \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_{1-5}\left(200\right)$} & $\mathcal{I}_{1-4}$
& 10 & 90.5 & 12.3 \\
& $\mathcal{O}_{1-5}\left(-200\right)$ & 33 & 85.1 & 16.6 \\ \midrule
\multirow{2}{*}{$\mathcal{I}_5$+$\mathcal{O}_{1-5}\left(300\right)$} & $\mathcal{I}_{1-4}$
& \textbf{10} & \textbf{90.5} & 11.8 \\
& $\mathcal{O}_{1-5}\left(-300\right)$ & \textbf{33} & \textbf{85.8} & 16.2 \\ \bottomrule
\end{tabular}
\caption{\label{tab:det_results} Detection results of different training sets.}
\end{table}
\begin{figure}[!htbp]
\centering
\includegraphics[width=0.6\linewidth]{different_sets}
\caption{\label{fig:different_sets} Detection accuracy of different training sets.}
\end{figure}
In the subsequent experiments, we integrate the images of all outdoor scenes into the outdoor training set and dramatically improve the accuracy of outdoor testing.
Figure~\ref{fig:different_sets} clearly shows the effects of using different training sets.
By further increasing in the number of images in outdoor videos, the detection accuracy of outdoor scenes improves slightly.
For indoor environments, the detection accuracy is impervious to different training sets.
Moreover, reducing the number of images to 50 ($\mathcal{I}_5$ has a total of 351 frames) does not reduce the detection accuracy.
This shows that we need only a small number of training samples to achieve satisfactory results when the training and testing scenarios are the same.
The frame rate is around 12 FR for indoor videos and 16 FR for outdoor ones.
The factor of different image resolution should be accountable for this performance gap.
In practical applications, if accuracy is guaranteed, we tend to use smaller training sets to reduce labeling costs.
Therefore, we use the model trained in "$\mathcal{I}_5\left(50\right)$+$\mathcal{O}_{1-5}\left(50\right)$" for comparison with the other methods in the comparative experiments.
\subsection{Results of multi-ant tracking}
Based on the TBD paradigm, we use detection results as the input to the tracking framework. For offline training, we randomly select 50 labeled samples from $\mathcal{I}_5$ as the training set.
We visualized the tracking results in Figure~\ref{fig:trajectories}.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth, height=1\linewidth]{trajectories}
\caption{\label{fig:trajectories} Tracking trajectories in test videos. Horizontal axes indicate the pixel coordinates in an image. (a-d) indoor scenes. (e-i) outdoor scenes.}
\end{figure}
Table~\ref{tab:tracking_performance} shows the performance of online tracking.
After integrating the images of each outdoor video in the detection training set, our method gets 95\% mMOTA for indoor videos and over 80\% for outdoor videos.
Additionally, mMOTP is around 80\% for both indoor and outdoor videos.
Notably, since the tracking performance depends on the detection result, the tracking task in $\mathcal{O}_{1-3,5}$ fails due to the low-quality detection (the second row in Table~\ref{tab:tracking_performance}).
Except for this failure case, the tracking performance is generally satisfactory considering that we only use 50 labeled samples from one indoor video.
\begin{table}[h]
\centering
{
\begin{tabular}{@{}lcccccc@{}}
\toprule
\multicolumn{1}{c}{Testing Data} & FP$\downarrow$ & FN$\downarrow$ & IDS$\downarrow$ & mMOTA$\uparrow$ & mMOTP$\uparrow$ & FR$\uparrow$ \\ \midrule
$\mathcal{I}_{1-4}$ & 236 & 621 & 21 & 89.9 & 79.9 & 36.5 \\
$\mathcal{O}_{1-3,5}$ & \multicolumn{6}{c}{detection failure} \\ \midrule
$\mathcal{I}_{1-4}$ & 239 & 628 & 22 & \textbf{95.7} & 81 & \textbf{38.9} \\
$\mathcal{O}_{1-5}\left(-50\right)$ & 6078 & 7122 & 625 & 81.8 & 81.9 & 26.2 \\ \midrule
$\mathcal{I}_{1-4}$ & 260 & \textbf{617} & \textbf{14} & 95.7 & 81 & 38.7 \\
$\mathcal{O}_{1-5}\left(-100\right)$ & 4867 & 6018 & 526 & 83.3 & 82.7 & \textbf{28.0} \\ \midrule
$\mathcal{I}_{1-4}$ & 228 & 709 & 17 & 95.4 & 81 & 36.3 \\
$\mathcal{O}_{1-5}\left(-200\right)$ & 4289 & 3421 & 394 & 85.3 & 83 & 24.9 \\ \midrule
$\mathcal{I}_{1-4}$ & \textbf{224} & 820 & 24 & 94.8 & \textbf{81.7} & 35.4 \\
$\mathcal{O}_{1-5}\left(-300\right)$ & \textbf{2644} & \textbf{3007} & \textbf{266} & \textbf{85.8} & \textbf{83.3} & 26.4 \\ \midrule \hline
$\mathcal{I}_{1-4}$/GT & 22 & 23 & 8 & 99.6 & 92.4 & 35.2 \\
$\mathcal{O}_{1-5}\left(-50\right)$/GT & 1697 & 458 & 1064 & 96.2 & 92.4 & 25.9 \\ \bottomrule
\end{tabular}}
\caption{\label{tab:tracking_performance} Tracking performance evaluation. The last two rows indicate that we use the ground truth of detection for tracking, which leads to a boost in tracking performance.}
\end{table}
The time cost of the tracking model is mainly incurred by generating 128-d feature vectors for each detection box.
The average number of objects in outdoor videos is more than three times that in indoor videos.
As for runtime time, FR reaches over 35 in indoor videos and more than 24 in outdoor videos.
We add a set of comparative experiments in the last two rows of Table~\ref{tab:tracking_performance}.
We directly use manually-labeled detection boxes for tracking and compare the detection results on the $\mathcal{I}_{1-4}$ and $\mathcal{O}_{1-3,5}$.
Both mMOTA and mMOTP have been dramatically improved.
This implies that an increase in detection accuracy could further boost the tracking performance of our framework.
\subsection{Comparative experiments}
\label{sec:comparative}
There are two widely used insect tracking software: idTracker \cite{perez2014idTracker} and Ctrax \cite{chiron2013detecting}.
idTracker needs to specify the number of objects before tracking, to create a reference image set for each object.
Meanwhile, Ctrax assumes that objects will rarely enter and leave the arena.
Thus, they are both not capable of tracking in outdoor scenes because of the variable number of ants.
Therefore, we compare these two methods only in videos depicting indoor scenes.
idTracker needs to specify the number of objects before tracking, in order to create a reference image set for each object.
To compare them with our method, we convert their representations into square boxes as our ground truth.
Table~\ref{tab:comparison} and Figure~\ref{fig:spatial_temporal} shows the tracking results.
In addition to a significant improvement of tracking accuracy, our method is 6 and 10 times faster than idTracker and Ctrax (see the column of FR).
\begin{table}[h]
\centering
{
\begin{tabular}{cccccccc}
\toprule
Method & FP$\downarrow$ & FN$\downarrow$ & IDs$\downarrow$ & FM$\downarrow$ & mMOTA$\uparrow$ & mMOTP$\uparrow$ & FR$\uparrow$ \\ \hline
idTracker & 881 & 8479 & 83 & 432 & 54 & 77.4 & 1.3 \\
Ctrax & 2832 & 5646 & 110 & 349 & 58.2 & 79.7 & 0.8 \\
Ours & \textbf{239} & \textbf{628} & \textbf{22} & \textbf{189} & \textbf{95.7} & \textbf{81.1} & \textbf{8.7} \\ \bottomrule
\end{tabular}}
\caption{\label{tab:comparison} Comparison of tracking results on videos $\mathcal{I}_{1-4}$.}
\end{table}
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{spatial_temporal}
\caption{\label{fig:spatial_temporal} Comparison of tracking performance in spatial-temporal dimension ($\mathcal{I}_1$). Horizontal axes indicate the pixel coordinates in an image.}
\end{figure}
We further compare tracking accuracy of idTracker and Ctrax across different indoor videos, as Figure~\ref{fig:comparison} shows.
The large variance of idTracker's performance is affected by the number of static ants, which will cause missing tracking.
Ctrax proves to be robust but with a lower accuracy compared with our method.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{comparison}
\caption{\label{fig:comparison} Comparison of tracking results for indoor scenes.}
\end{figure}
\section{Discussion}
\subsection{Comparison of methods}
In the previous section, we compared two methods of insect tracking, idTracker and Ctrax.
idTracker uses the intensity and contrast of the foreground segmented area to extract appearance features and construct a reference image set for each individual.
However, it can not track motionless individuals.
Figure~\ref{fig:spatial_temporal}(a) shows that only a minority group of ants are successfully tracked over the period of video.
Further, there are some trajectory fragments due to the limitations of the foreground segmentation model for multiple objects.
Compared to our results, the trajectories of Ctrax are incomplete.
This indicates that there are more FN, as Figure~\ref{fig:spatial_temporal}(b) shows.
Ctrax requires a sharp contrast between object and background.
The ants passing through the overexposed areas in the scene will be ignored.
Additionally, Ctrax assumes that the motion of the object obeys the linear distribution.
However, the ants' movement is nonlinear, and their speed and direction might change abruptly, causing IDS in Ctrax.
Our method classifies and regresses twice to locate ants accurately.
During the tracking stage, we use the historical appearance sequence as a reference and update it frame by frame.
Compared with idTracker, our method effectively solves the long-term and short-term dependence of motion states, thereby reducing FM.
Despite that we also assume the linear distribution of motion states, they are used only to filter impossible associations, and have nothing to do with association cost.
We take the appearance distance between trajectories and detection boxes as association cost, thus the model is robust even when the ant movement is complicated.
We take the appearance measure between trajectories and detection boxes as the association cost, thus the model is robust even when the ant movement is complicated.
\subsection{Failure cases}
\label{sec:failure}
\subsubsection{Limitations of detection framework}
The number of ants in outdoor scenes is on average 33 per frame.
It is also typical for ants to involve close body contact with each other for the purpose of information sharing.
Naturally, their extremely-close interactions are highly likely to cause mis-detection (Figure~\ref{fig:failed_detection}(a)).
Additionally, entrances and exits of ants in outdoor scenes are more prone to mis-detection (Figure~\ref{fig:failed_detection}(b)).
Moreover, the dramatically non-rigid deformation of ants is also a factor causing the detection failure (Figure~\ref{fig:failed_detection}(c)).
These three scenarios are all challenging cases that deserve our future efforts.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{failed_detection}
\caption{\label{fig:failed_detection} Examples of failed detection in outdoor scenes.}
\end{figure}
\subsubsection{Limitations of tracking framework}
According to Figure~\ref{fig:drift}, Ant No.41 entered the scene at Frame No.88.
Coincidentally Ant No.32 left the scene at an adjacent region, but its trajectory was not deleted.
At Frame No.93, Ant No.41 drifted to Trajectory No.32.
This defect is caused by insufficient appearance descriptors stored in the gallery of Ant No.41, and it moved near the exit location of another ant.
This kind of mis-association occurs at the image boundary and accounts for the majority of IDS and FM in our experiments.
However, when ants move inside the scope of both indoor and outdoor scenes, our method can accurately track multiple ants simultaneously for a long time, as Figure~\ref{fig:tracking_results} shows.
\begin{figure}[!htbp]
\centering
\includegraphics[width=\linewidth]{drift}
\caption{\label{fig:drift} Drift at the scene boundary. A newly-entered Ant No.41 is mis-associated with an existing Trajectory No.32.}
\end{figure}
\section{Conclusion}
We proposed a complete detection and tracking framework based on deep learning for ant colony tracking.
In the detection stage, we adopted a two-stage object detection framework for the detection task.
We also use a ResNet model to obtain ant appearance descriptors for online associations.
Next, we combined appearance and motion information for the tracking task.
The experimental results demonstrated that our method outperformed two mainstream insect tracking models in terms of accuracy, precision, and speed.
Particularly, our work shows its advantage in robustly detecting and tracking ant colonies in outdoor scenes, which is rarely reported in existing literature.
We believe our method could serve as an effective tool for high-throughput swarm behavior analysis of ant colonies, leading to the development of embodied intelligence.
In future research, we aim to achieve more robust detection.
For example, by exploring additional information of ants' skeletal structure, we can potentially solve the aforementioned failure case of close interaction and nonrigid deformation problem.
We also plan to improve the generalization ability of our detection and tracking frameworks so that it is applicable to a wide range of outdoor environments.
\section*{Conflict of Interest Statement}
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
\section*{Author Contributions}
All authors listed have made substantial, direct, and intellectual contribution to the work; they have also approved it for publication.
In particular, MW, SG and XC contributed to the design of this work; MW, SG and XC contributed to the writing of the manuscript; XC designed and implemented the multi-ant tracking framework; XC conducted the experiments, and analyzed the results.
\section*{Acknowledgments}
This work was supported by the Natural Science Foundation of Fujian Province of China (No. 2019J01002).
\section*{Data Availability Statement}
The training and testing datasets used for this study can be found in the \href{https://data.mendeley.com/datasets/9ws98g4npw/3}{ANTS--ant detection and tracking}.
\bibliographystyle{Frontiers-Harvard}
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Our planet is an immense playground for fans of the outdoors. Outdoorsy travelers and courageous adventures, all love this realm blessed with various landscapes and natural wonders. If we could spend our entire lives close to nature, we would probably be a lot happier and healthier. However, since civilization has taken us away from nature's simplicity, planning outdoor adventures from time to time is all we have. Let's make the most of them!
Woman taking a break from skiing on Mont Tremblant, Quebec
Winter fun in Quebec
Winter in Quebec with its glistening snow and fairytale scenery creates wonderlands for skiers and winter sports aficionados. Situated close to the "urban jungle" of Montreal, Mount Royal delights locals and tourists alike with cross-country skiing, snow tubing, and snowshoeing, while the frozen Beaver Lake invites ice skaters to show their gracious skills. Weekends bring people on the slopes of the Laurentian Mountains situated only one hour away from Montreal, many of them flocking to Ville de Mont-Tremblant, known as one of the best ski resorts in North America. The city of Quebec lures people on the fabulous slopes of Le Relais, located only a stone throw away from the urban scenery. Moreover, 15 miles north of Quebec, the slopes of Stoneham add more skiing options, including off-piste slopes, night skiing, and terrain parks for freestylers. The highest vertical drop in Canada (excluding The Rockies) awaits at the Le Massif.
Surfer relaxing in Portugal
Surfing extravaganza in Portugal
Known for its charming and picturesque capital, as well as for its fabulous beaches, Portugal surprises surfers with world-class surf experiences and waves for everyone. While Lisbon's coast is the most popular destination for surfing and home to an active and experienced surfing community, for super tubes you should choose the Peniche area. One of the best-surfing destinations in Europe, Peniche combines amazing surfing with fabulous views and an active nightlife. The Algarve is renowned for its glorious beaches, but it also is a trendy surfing destination with surfing conditions similar to those found in southern California.
Hippopotamus in the Zambezi
Safari experiences in Zimbabwe
Not many safari destinations boast such diversity and beauty as Zimbabwe. Welcoming animal and nature lovers in an authentic and raw setting, the safaris in Zimbabwe will often take you away from tourist crowds and into the spectacular wilderness of Africa. While you may not find luxurious safari packages, you will definitely enjoy a comfortable stay and amazing safari tours. Hwange National Park is usually the best destination for tourists who want to be pampered while exploring the jungle and for travelers who want to visit the most popular park in Zimbabwe. Situated close by to the majestic Victoria Falls, the park is home to lions, zebras, lopards, elephants, giraffes, and many more. Hippos and crocodiles await at the stunning Mana Pools National Park, an ideal destination for both game drives and canoe adventures.
Female climbing trails in the mountains with glaciers in the distance.
Unusual hikes in Kazakhstan
Many hikers are familiar with the paths of the Lake District and the outdoor adventures in Sedona, but how many of them can say that they've hiked on the peaks of Kazakhstan? A land relatively unknown to the world but rich in natural attractions and dramatic scenery, Kazakhstan takes hikers through thick forests and on spectacular glaciers while rewarding them with breathtaking views of endless lakes and an infinite desert. The country is vast and seems disarming at first, but once you get to know its past and present, its legends and people and figure out the visa process, you'll discover a fascinating destination. Whether you choose to trek in the Tien Shan Mountains and climb on mighty peaks like the Podeba Peak or to hike around Lake Balkhash, one of the largest lakes on the planet, you'll discover a land of surreal purity and beauty, a secret corner of the world that awaits to be discovered.
A woman bungy jumping off Kawarau Bridge in Queenstown, New Zealand.
Thrilling adventures in New Zealand
New Zealand, a land of phenomenal beauty and spectacular wilderness invites adrenaline junkies to follow their passions and feed their appetite for the extraordinary. Since the Kiwis invented the "art" of bungee jumping, nothing is too dangerous for them! They love the thrills of living on the edge and have access to the perfect setting for various crazy adventures. Try jumping out of planes…just for the sake of it over Queenstown or Lake Taupo, depending on what views you want to take your breath away! Experience bungee jumping from the Kawarau Bridge, in Queenstown, or off the Sky Tower, in Auckland! Don't worry, you'll live to tell the story! Have fun jet boating on Lake Taupo or the Waitoto River or canyoning in remote locations around Auckland, Canterbury or Nelson. Go tramping (hiking) in the Tongariro Crossing to visit the realm of Mordor or follow the Mueller Hut track in Mount Cook National Park.
New Hapag-Lloyd ship enters service
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Weekend Exploring London
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3 Natural Remedies For Travel Anxiety
Beach Holiday in Bulgaria
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Beach Vacations in Sri Lanka
New Forest, England
Discover Dominica
Safari in Botswana
Scuba Diving in the Bahamas
Safari in Tanzania
Safari Experience in Rwanda
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pandas\_profiling.config.MissingPlot
====================================
.. currentmodule:: pandas_profiling.config
.. autopydantic_model:: MissingPlot
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Archive | lawsuits RSS for this section
in ICE, illegal immigration, lawsuits
Judge allows illegal aliens to sue ICE agents
A Connecticut federal judge ruled against the government and will allow illegal immigrants to sue ICE agents for civil rights violations.
U.S. District Court Judge Stefan Underhill ruled Immigration and Customs Enforcement (ICE) officials are not immune from lawsuits if they are anchored on Constitutional grounds.
The civil rights lawsuit, brought by 11 alleged illegal immigrants claim they were unlawfully apprehended by ICE agents in New Haven, Conn, June 2007.
The plaintiff's attorneys told The New Haven Register "the decision has wide-ranging implications."
Judge Underhill refused the government's plea to dismiss the charges against ICE agents and their supervisors, including Julie Myers ICE's former agency head.
A group of law school interns and their bosses at Jerome N. Frank Legal Services Organization at the Yale Law School brought the lawsuit on behalf of the illegal aliens. The interns argued that the 11 suspected illegal aliens were unfairly treated based on race.
Recently retired ICE agent, John Sakelarides had plenty to say about another lawsuit benefitting illegal aliens.
"This suit is nothing more than a transparent and feeble attempt to scare ICE agents into refusing to enforce the immigration laws of this country. Can you spell intimidation? When it goes to trial, my guess is the agents will be exonerated and what should happen is that the government should seek fees for wasting taxpayer money in defending a baseless lawsuit," Sakelarides said. "Extortion is still illegal in this country. The far reaching implications claim is a clear indication of what the intent of this lawsuit is, namely intimidation."
Sakelarides didn't know if this lawsuit was being brought under the Federal Torts Claim Act (FTCA) or under a Bivens Action.
"I do know, from the past and unfortunately present, experience that suits are often filed under both but the law is clear that they cannot co-exist, there can only be either an FTCA claim or a Bivens Action. It centers on whether or not the federal employee was or was not acting within the scope of their authority," he explained.
"This is borderline criminal. If the agents did something wrong, so be it. But if they didn't and this suit is being brought simply to induce a chilling effect on the enforcement of the immigration law, then these idealistic and naive law students and Yale Law School ought to be held accountable," Sakelarides finished.
ICE officials declined to comment on the lawsuit.
in defamation, illegal immigration, lawsuits
Illegal alien wins defamation case for being called a 'criminal' – set back for 1st Amendment
An illegal-alien day laborer who attacked a U.S. photographer at a notorious San Diego day labor site in 2006, was awarded $2,500 in damages for "defamation per se" by Judge Ronald Styn in a non-jury trial in San Diego Superior Court.
The Mexican national plaintiff, Alberto Jimenez, who was illegally in the country at the time of the attack of Los Angeles photographer John Monti, sued San Diego Minutemen founder Jeff Schwilk for defamation for calling the illegal immigrant attackers "criminals" when he forwarded an email with Monti's pictures of Jimenez and six other suspects who were at the scene of the crime.
Initially the lawsuit was filed in October 2007 and all seven men shown on the flyer sued Schwilk, Monti, and Fox News Corporation for defamation. However, Fox News and Monti were eventually dismissed from the case in 2008 and 2009 and six of the seven plaintiffs dropped their lawsuits against Schwilk in February, leaving only Jimenez vs. Schwilk for the one-day judge-only trial.
When the trial began, Jimenez was not in the courtroom to meet his accuser. Schwilk immediately asked Judge Styn for a directed verdict to dismiss the case, but the judge opted to start the trial without him and gave Jimenez additional time to appear in court as his attorney said he was running late. The plaintiff's attorney, Dan Gilleon, claimed his client was trying to get across the border and needed more time because he was a Tijuana, Mexico resident.
As testimony began Schwilk admitted that he had forwarded Monti's 'wanted' flyer by email the day after the attack. The email was sent to a dozen law enforcement and trusted community leaders so they could be on the lookout for the suspects who were still at large. He explained anyone with knowledge of these pictured men needed to notify San Diego Police Department.
The victim, Mr. Monti then testified that Jimenez was indeed one of the seven men who attacked him from "behind and knocked him into the busy boulevard that Saturday morning in 2006." Monti explained he had gone to the day labor site to investigate the connection between the street-side illegal alien hiring site and child prostitution in the nearby the migrant camps where many of the day laborers live.
Monti also said under oath that Jimenez had previously testified in his criminal trial in 2007 that he was an illegal alien and had no legal papers to gain employment legally in the U.S.
It is worth pointing out that Gilleon wanted to bar television video coverage of the one-day trial, however, the defendant, Schwilk and the judge stated it was an open court and cameras would be permitted in the courtroom.
After a lunch break the defense called Jimenez to the witness stand, but Jimenez still had not arrived in court 4 1/2 hours after the start of the trial.
Gilleon then claimed that "Jimenez' wife was having surgery and he could not attend the trial that day." Judge Styn was clearly tired of the excuses, however he allowed the trial to continue. It became clear the real reason Jimenez could not come to the trial was because he was not eligible to enter the country legally from his home in Tijuana, Mexico.
Schwilk again asked Judge Styn for a directed verdict, but again the judge declined, he wanted to proceed with closing statements. Schwilk pointed out to the judge that he had a right to question his accuser and the judge seemed to agree, but he also wanted to conclude the trial without the plaintiff.
Schwilk said, "I knew at that point the fix was in. This was so obviously a charade by Dan Gilleon and his La Raza employers who brought this frivolous case against us. There was no plaintiff in this case. This was a sham trial with no real accuser and Judge Styn played right along with La Raza's devious plan."
During the closing arguments, Gilleon accused Schwilk of conspiring with Sean Hannity, Alan Colmes and Fox News to spread Jimenez' face around the world in an effort to defame the day laborer by characterizing him as a wanted criminal.
The case against Fox News was dismissed earlier in what was a fishing expedition because Hannity and Colmes aired a story in late November 2006 regarding the assault featuring Monti and his flyer with the seven men's grainy pictures on it. See video here: http://www.youtube.com/watch?v=OqscavZ4tyY
Gilleon asked the judge to award Jimenez $85,000 in damages for four counts of defamation, including $50,000 for providing NewFox s with a copy of the flyer.
Schwilk reminded the judge in his closing arguments that his private emails with Monti's pictures had been sent only to a small group of law enforcement contacts, local community leaders and two local politicians who are leaders in the fight against illegal immigration.
Schwilk says Gilleon and his La Raza employers most likely got the email months after the fact from a disgruntled former member of SDMM who was kicked out of the group in February 2007 for bad behavior. He also pointed out again that Jimenez was an illegal alien and had obviously broken laws by entering the U.S. illegally.
Finally, Schwilk reminded the court that although Jimenez was never formally charged with the assault on Monti, he did participate in the attack and then fled the scene of the crime. "In law enforcement, flight implies guilt," he said.
After a 15-minute recess, Judge Styn returned with his verdict finding Schwilk liable for one count of defamation for "carelessly" calling Jimenez a "criminal." Schwilk plans to appeal.
Asked about the verdict, Schwilk said, "Judge Styn not only ruled against our protection of free speech, he is attempting to de-criminalize illegal aliens and the crimes they commit on American soil. Styn is yet another example of a bad judge attempting to legislate from the bench."
Fellow San Diego Minuteman Barry Shipley observed the entire 5-hour trial and agreed.
"The judge completely ignored the court testimony that Jimenez was a criminal illegal alien and that the truth is a defense in a defamation case. Judge Styn should be removed from the bench," Shipley explained. "He is a danger to our American Constitutional rights".
The plaintiff's attorney Daniel Gilleon was obviously pleased with the outcome of the trial. "We are very pleased with Judge Ronald Styn's ruling. Once again, the Court system has shown that defamation is Schwilk's chief weapon in his 'war' against illegal immigration. The judgments against Schwilk are starting to pile up."
Schwilk saw it differently; "Of course La Raza and agents of the Mexican Government (Gilleon's employers) are pleased that they have been able to chip away at the U.S. First Amendment. Their greatest fear in the border and immigration debate is truth tellers like myself and other leaders fighting daily to secure America from the mass criminal invasion from Mexico. Unfortunately for them, they will never shut down our free speech".
Schwilk said the San Diego Minutemen would not be deterred by these malicious politically motivated lawsuits by the open border groups. "Our corrupt court system is all they have to try to shut us down," he said. "Honestly, we don't get distracted by all this court drama. We're fighting to save San Diego from the invasion of criminals and drug smugglers from Mexico."
SDMM have continued to expose outdoor prostitution rings in nearby McGonigle Canyon, including a highly publicized incident last October in which they caught another field brothel in action with trafficked Mexican girls and numerous Latino "Johns" paying for cheap sex in the bushes.
SDPD again refused to prosecute the suspected pimp and prostitute that were detained, but Mayor Sanders and SDPD were forced to clear out the canyons of migrant squatters yet again.
Schwilk said it is a constant cat and mouse game with the Mexican migrant camps and field prostitution rings. "We've been monitoring the canyons for four years now. We get the camps cleared out, but some of them eventually come back due to SDPD looking the other way at the illegal Mexican squatters. But there are far fewer migrant camps in San Diego County than when we started investigating these crime zones in 2006. We expect our success to continue to drive the Mexican Consulate and its operatives to further desperate measures to try and stop us."
For the first time in more than three years, there are no lawsuits pending against SDMM members. Perhaps the illegal alien activists in San Diego are finally giving up on that failed and costly tactic.
With the latest push for amnesty heating up again, it appears that both sides in the illegal immigration debate are gearing up for some major battles.
Schwilk, who is a retired Marine, says he will fight until America's borders are fully secured and the country returns to an orderly system of legal immigration. He believes most Americans strongly oppose amnesty for the approximately 20 million illegal aliens currently in the U.S.
"We've made huge gains in the past five years, but we still have a ways to go. We're up against some very rich and powerful ethno-centric special interest groups, but no one should underestimate the spirit and determination of the American citizens", he concluded.
For more stories; http://www.examiner.com/examiner/x-10317-San-Diego-County-Political-Buzz-Examiner
in Amnesty, E-Verify, illegal immigration, La Raza, lawsuits, MALDEF
Illegal alien lawsuits continue to clog the courts in California
Well-funded illegal alien activists in Southern California have found a new way to attack Americans fighting for secure borders and enforcement of current immigration laws. The fight has moved from the streets where they wave the their Mexican flag to America's civil courtrooms.
"Allow me to understand this correctly. Illegal aliens, people who have committed a crime by entering this country illegally, and who continue to commit additional crimes by using counterfeit documents to project a status they are not entitled to, are suing cities and citizens for "disrupting their RIGHT to work in the US, even though they have no such right? If any immigrant "rights" organization or other advocacy group is responsible for the filing of such suits, either directly or indirectly, they should be counter claimed against for abuse of process and malicious prosecution. It's time for the good citizens of this country to fight back through the courts," said retired ICE Agent John Sampson who now runs CSI Consulting and Investigations.
The beef about an upcoming lawsuit in San Diego against Jeff Schwilk, founder of the San Diego Minutemen, stems from a violent attack on Los Angeles anti-illegal immigration activist John Monti in November 2006 at the Rancho Penasquitos day-labor site in San Diego.
"Controversial San Diego attorney Daniel Gilleon was hired by La Raza operatives more than three years ago to go after San Diego Minutemen and other pro-security activists," says Schwilk.
Monti had gone to the infamous makeshift hiring site to photograph the day laborers and the law-breaking employers hiring illegal workers. "It is still a felony to hire illegal aliens in the U.S. and studies have shown that almost all day laborers are illegal aliens from Latin America, Schwilk explains.
While photographing the street-side hiring process, Monti was suddenly jumped from behind by at least seven Hispanic men. "They punched him, tried to steal his professional camera, and pushed him into the busy boulevard. Several passersby's witnessed the attack and called 911. When police arrived two minutes later, they found a bloody and shaken Mr. Monti," witnesses reported.
Once the police were called to the scene, all of the attackers had made a run for it. Luckily, Monti photographed many of the laborers prior to the attack and was able to show San Diego Police Department. The next day the victim, Monti, sent the same pictures via email to local San Diego activists so they could call the police if the suspects returned to the day labor site where they usually look for employment every day. At this point Schwilk received the pictures and forwarded them to his local law enforcement contacts and other concerned residents in the area.
Their claim was defendants were disseminating pictures of the suspects with Monti's statement and pointed out that they were wanted for questioning by SDPD regarding day-labor site scuffle. The flyer indicated if anyone had any information about the incident to phone the police.
Police records show the suspects were being sought for questioning and most of them were eventually found and questioned. No charges against the suspects were ever filed, as they all, not surprisingly, claimed that Monti attacked them first. According to Schwilk, local illegal alien activists were seen speaking with the suspects soon after the attacks.
"Witnesses who saw Monti being attacked and beaten were ignored by investigators and the city attorney, who were under extreme pressure from the so-called 'Mexican mafia' to protect the suspects (most of them Mexican citizens) from prosecution," Schwilk contends. Fox News' and Monti's lawsuits were eventually dismissed or settled, but Schwilk demanded a trial to prove that this was just another unfounded, frivolous lawsuit and malicious prosecution meant to harass and silence those who oppose illegal immigration.
This defamation suit also accuses Schwilk of putting up Monti's "wanted posters" in the area around the day labor site – a charge Schwilk denies. Schwilk and a few other concerned citizens did use Monti's pictures to identify two of the suspects standing at the sidewalk hiring area three days after the attack.
"When we saw two of the suspects back at their sidewalk loitering area, we immediately called SDPD. The lead detective of the nearby division came to the scene and explained to us that they had already questioned and released those two men and that an arrest had been made," Schwilk said. "The officers refused to elaborate further, but they told us they were fully investigating the assault on Monti and hoped to bring all of the guilty day laborers to justice."
The lead attorney in this so-called lawsuit is Dan Gilleon of Del Mar(??). In emails from Gilleon obtained by the Examiner, Gilleon repeatedly seems to be asking Schwilk to settle the case over the past year. In the most recent email sent on Feb. 6, Gilleon again offers to Schwilk, "If you want to settle, we'll take $1,000 for each plaintiff, cash now, or $10,000 each in stipulated judgments."
Schwilk has repeatedly told the plaintiff's attorney Gilleon that he has no intention of settling this case because he is the victim not the other way around. In addition to last minute attempts to avoid taking this case to trial, Gilleon failed to depose Schwilk for 2 1/2 years and recently convinced the judge to allow a last-minute deposition just days before the trial is set to commence.
In a voicemail message from Gilleon, he declined to comment on this pending case. However he claims there is a new lawsuit pending against Schwilk. However, court records show no new lawsuits have been filed.
Schwilk, who is now defending himself, says he strongly opposes a deposition at this late date because he believes Gilleon lied to the judge about trying to depose him in October. Evidence submitted to the court last week clearly shows that Gilleon claimed to have served Schwilk a notice of a deposition at a long-abandoned store in Oceanside. Schwilk says he has asked Judge Styn to rescind his recent deposition order and the motion is pending.
The trial begins on Thursday. It is worth pointing out, the attorney for the plaintiffs admitted that all his clients were living in Mexico and may not be able to enter the country legally to attend the trial, according to Schwilk.
It is a sad commentary that the civil court system is filled with well-funded foreign interests who hire activist lawyers to gain control of the country for their illegal clients.
The Minutemen vow to keep fighting no matter how many lawsuits La Raza activists file against them. "Our cause to make our cities and our country safe and secure is too great to be deterred by these people who make a mockery of our legal system," Schwilk finishes.
In another lawsuit filed in Costa Mesa, California, MALDEF the Mexican American Legal Defense Fund charges the city's anti-solicitation ordinance is unconstitutional. Joining in on the February 2, 2010 lawsuit is the ACLU of Southern California and the National Day Laborer's Organizing Network (NDLON).
The civil rights groups filed the lawsuit against the City of Costa Mesa on behalf of the Association de Jornaleros de Costa Mesa and the Colectivo Tonantzin, whose members have been restricted from peaceably expressing their need and availability for employment in the city's public areas due to the ordinance.
Claiming his client's first Amendment right, MALDEF President and General Counsel Thomas A. Saenz said, "Free speech, one of our most cherished rights, belongs to everyone in society. Day laborers seeking work have as much right to express themselves as the largest corporation employing hundreds of thousands. Costa Mesa's anti-solicitation ordinance violates this vital and longstanding constitutional principle."
"The city's anti-solicitation ordinance prohibits any person standing on a sidewalk or other public area from soliciting employment, business or contributions in any manner deemed to be intended to attract the attention of traveling vehicles. The ordinance subjects day laborers and other solicitors to a fine of $1,000 and imprisonment up to six months. The ordinance violates the day laborers' First and Fourteenth amendments rights under the United States Constitution," according Saenz.
"Day laborers have contributed to the Costa Mesa economy for decades," says Pablo Alvarado, director of the National Day Laborer Organizing Network. "Particularly during these tough times, the hard work they provide the community should be rewarded and not the target of destructive law enforcement practices."
The ruling of federal courts throughout the country in the past have ruled in favor of preserving the free speech rights of day laborers, which allows them to continue to solicit work.
"The contention that the civil rights of illegal day laborers are being violated is pure malarkey and if anything at all, the advocates can be criminally charged and prosecuted for aiding and abetting illegal alien immigration," says Vince Johnson in a letter to the city of Costa Mesa.
He goes on to explain this frivolous lawsuit should be recognized as such by any federal judge who may incorporate immigration law regarding employment by undocumented immigrants.
"I suggest that you utilize the resources of Homeland Security/ICE, the US Department of Labor and the US Justice Department to promptly stop this court action and affirm that this country can not be controlled by people who do not even have the right to be in this country. Any meddling by the Mexican government is a clear-cut violation of American law and sovereignty," Johnson finished.
The meddling of Mexico continues to play out on both sides of the border. It was reported by the M3 report (a publication put out by the National Association of Former Border Patrol Officers) that all political parties of the Mexican Senate must reproached the policy change of direction of President, Barack Obama, "who has decided to go back on the promise he made to all the Hispanic groups that supported him, and now he insists on closing the border. The government of Barack Obama seeks to increase the funds to reinforce border security with $4.6 billion to support 20,000 agents of the Border Patrol, as well as to finish the first portion of the 'virtual fence,'" the report stated.
The Mexican government continues by "announcing that he (Obama) will build the missing portion of the 'virtual fence' by means of which everyone who crosses, undocumented or illegally, will be detected and can be immediately jailed and later expelled, (and also) reinforcing the number of agents for customs, as well as for the border. From now on we Mexicans will not know what to believe when we speak with the President of all the Americans, because he is a President who fails to keep his word," according to a member of the Mexican Senate.
One thing is certain Mexico continues to demand the rule of law be tweaked in their favor when it comes to illegal immigration. This will surely set up a very contentious immigration reform debate, one the American people have clearly stated Amnesty will not be a part of the legislation.
For more stories; http://www.examiner.com/x-10317-San-Diego-County-Political-Buzz-Examiner
|
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"redpajama_set_name": "RedPajamaCommonCrawl"
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| 8,682
|
The Berlin Blockade Essay Example
Questions Related to The Berlin Blockade Essay Example
The Berlin Blockade began in 1948. It was an attempt by the Soviet Union to stop France, Great Britain and the United States from being able to travel to their sectors of Berlin. This caused the Western powers to organize an airlift to West Berlin. The blockade was the first major clash of the cold war and because of the actions of the Union of Soviet Socialist Republics, American efforts, and the lasting aftermath, the Berlin Blockade is a key part of the cold war and history itself. The Berlin Blockade first developed due to the actions of the Union of Soviet Socialist Republics. Discussion on what to do with Germany broke down over Soviet accusations that the Western Allies were violating the Potsdam Agreement. The Allied powers decided to unite their occupation zones of Germany and in protest the Soviet union withdrew from the allied control council. "The Western allies announced the introduction of a new currency in Western Germany and West Berlin. Stalin, who favored keeping Germany weak, so that it could never again invade his country, was angered and quickly set up a blockade of West Berlin." (Bodden, 20) France, Britain, and the United States created a new currency, the Deutsche Mark, for their occupation zones. The Soviets feared this would devalue the Reichsmarks that they used in the East. For the Soviets, this was the last straw and marked the beginning of the Berlin Blockade. After the announcement of the Deutsche Mark, Soviet guards halted trains and traffic to Berlin and delayed freight shipments. Then, Soviets severed land and water connections and stopped all rail traffic. Eventually, the Soviet Union stopped supplying food to the non-Soviet sectors of Berlin."He knew that West Berlin had enough food and fuel only for six weeks, and he expected the three Western allies to let the Soviet union have its way. As far as he could see, their only alternative was to use tanks to smash through the road and rail blocks and bring supplies. Such an aggressive action was bound to cause war, and he doubted they would do that." (Kelly, 12)This proves that Stalin was fully aware of what the blockade would lead to for the West Berliners. He wanted his way, and had no concern for the innocent people at risk. He was also confident that the West wouldn't use aggressive tactics to stop him. In 1949, Moscow proposed negotiation to end the blockade. After eleven months the Soviets realized that the blockade had failed, and reopened the borders. "The airlifts continued until May 12, 1949, when the Soviets reopened the East German borders for transportation, realizing that the blockade had failed." (Mur, 13-14) The Americans were successful. West Berlin was freed from the blockade and transportation was allowed through. For the Soviets, this event was an absolute disaster. The world came to see the Russians as international bullies while the United States, France and Britain hardened their resolve on issues related to Germany. It can be seen that the Berlin Blockade played an important role in world history in the actions of the Union of Soviet Socialist Republics and in the response of the Americans. Further proving that the Berlin Blockade is important in history are the efforts the Americans made to help West Berlin. Soviet had halted all traffic from the West to East. Finding a way to supply the city seemed to be the only reasonable response. General Clay, the American commander in Berlin, predicted that: "When Berlin falls, Western Germany will be next. If we withdraw our position in Berlin, Europe is threatened… Communism will run rampant."(Berlin Airlift, schoolshistory.org.uk) The Western allies were put in a tough situation. The Truman administration was convinced losing Berlin would mean losing all of Germany, but using military forces to strike back against the blockade wasn't an option: the risk of turning the Cold War into an actual war was just too great. Truman ordered a massive airlift of supplies into West Berlin. It was a difficult task to provide food, clothing, water, medicine, and other necessities for over two million citizens. The West also introduced a counter-blockade, stopping all rail traffic into East Germany. "You peoples of the world, you, peoples of America, of England, of France. Look at this city, and recognize that this city, this people, must not be abandoned – cannot be abandoned." (Knopp, 22:14) During the airlift Berlin Mayor Ernst Reuter pleads for the West to not abandon them. It showed the desperation of West Berliners and how fearful they were of being left alone to suffer Stalin's abuse. This speech caused a lot of people to stand up and riot against Stalin. In 1949, the blockade ended. The United States had gained widespread public sympathy and Berlin became a symbol of the allies willingness to oppose further Soviet expansion. "With the opening of the gates, a new chapter in postwar history begins to unroll down German highways. Just 10 months and 23 days after the capital was sealed off from the ground, traffic is rolling towards Europe's number one trouble spot. It's a day of triumph for a band of men in the airlift who kept Berliners eating while they were held in an iron ring" (Knopp, 46:34) The United States were able to keep the airlift going long enough for Stalin to realize he wasn't going to win this battle. They showed that they were able to fight past Stalin's walls and that they were not willing to let communism spread. The importance of the blockade was shown not only in the actions of the Union of Soviet Socialist Republics and within the response of the Americans, but through the lasting impact on Berlin as a city. The aftermath of the blockade and airlift also make this an important part in history. The blockade solidified the separation between West and East. After the blockade was lifted, East Berlin became a separate communist republic.. "By the middle of 1961 as many as 30,000 East Germans a month were moving to the West. The East German government had to act to stem this flow as these were the very people they couldn't afford to lose." (Berlin Airlift – Consequences, nationalcoldwarexhibition.org) Stalin was losing support at an extremely fast rate and lots of people were leaving East Berlin for the West. Stalin saw it as the survival of East Germany being threatened. This caused Stalin to bring more drastic countermeasures to keep the West and East separate and as many people under communism as possible. Like the East, after the end of the blockade, the three Western allies combined their sectors of Berlin into one and the state of West Berlin was made. West Berlin was a separate capitalist republic."On May 12, 1949 the Soviet Union officially ended the blockade allowing other mechanisms for transportation of goods into West Berlin. Though the blockade of Berlin had ended, the Allies continued the airlift until September 30. This was to ensure that the people of Berlin had three-month surplus of supplies and food." (The Berlin Airlift-Operation Vittles, wigglywings.weebly) For the West Berliners the threat of Stalin was still real. The West had to make sure that West Berlin had enough supplies if Stalin tried to set up another blockade. This also bought added pro America, pro West ideals. The people of West Berlin were also affected by the end of as a rethe blockade – economically and from the world's opinions. West Berlin continued to prosper as a result of the Marshall Aid and the economic gap between East and West grew. With what the West had to offer in terms of employment, wages and the standard of living many East Berliners decided to move to West Berlin. "It also transformed Berlin, once equated with Prussian militarism and Nazism, into a symbol of democracy and freedom in the fight against Communism." (The Berlin Airlift, 1948–1949, history.state.gov) World War Two caused Berlin and the rest of Germany to be seen in a negative light, but by the end of the blockade, Berlin was transformed into a symbol of freedom and democracy. The lasting effect on Berlin is one of the reasons why the blockade is so important in global history.The Berlin Blockade, through the measures taken by the Union of Soviet Socialist Republics, the United States' efforts and the results of the event, is an important part of history. Though the blockade had ended cold war tension remained high. The blockade foreshadowed the future conflict over the city of Berlin, and the creation of the Berlin Wall.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 9,554
|
{"url":"http:\/\/nrich.maths.org\/6742\/solution","text":"### Times\n\nWhich times on a digital clock have a line of symmetry? Which look the same upside-down? You might like to try this investigation and find out!\n\n### Clock Hands\n\nThis investigation explores using different shapes as the hands of the clock. What things occur as the the hands move.\n\n### Ten Green Bottles\n\nDo you know the rhyme about ten green bottles hanging on a wall? If the first bottle fell at ten past five and the others fell down at 5 minute intervals, what time would the last bottle fall down?\n\n# Weekly Problem 42 - 2009\n\n##### Stage: 2 and 3 Short Challenge Level:\n\nThe only digits which appear the same when reflected are $0$, $1$, $3$ and $8$, so we want to find the number of times the display on the clock is made up of these digits.\n\nThe first digit can be $0$ or $1$, the second digit can be $0$, $1$, $3$ or $8$, the third digit can be $0$, $1$ or $3$ and the fourth digit can be $0$, $1$, $3$ or $8$. Therefore there are $2\\times 4\\times 3\\times 4=96$ different possible displays that are the same when reflected.\n\nThis problem is taken from the UKMT Mathematical Challenges.\n\nView the previous week's solution\nView the current weekly problem","date":"2015-01-28 22:22:36","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.39154621958732605, \"perplexity\": 635.6149631047856}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2015-06\/segments\/1422122034298.3\/warc\/CC-MAIN-20150124175354-00239-ip-10-180-212-252.ec2.internal.warc.gz\"}"}
| null | null |
{"url":"http:\/\/clay6.com\/qa\/60085\/a-charged-particle-of-charge-e-and-mass-m-is-moving-in-an-electric-field-ov","text":"Comment\nShare\nQ)\n\n# A charged particle of charge e and mass m is moving in an electric field $\\overrightarrow{E}$ and magnetic field $\\overrightarrow{B}$. Construct dimension less quantities and quantities of dimension $[T]^{-1}$\n\n$\\begin{array}{1 1} \\text{There cannot be a dimension less quantity. quantity of dimension$[T^{-1}]=w$} \\\\ \\text{There cannot be a dimension less quantity. quantity of dimension$[T^{-1}]=\\alpha$} \\\\ \\text{There cannot be a dimension less quantity. quantity of dimension$[T^{-1}]=F$} \\\\ none\\;of\\; the \\;above \\end{array}$\n\nThere cannot be dimensionless quantity with $e,m,\\overrightarrow{E}$ and $\\overrightarrow{B}$\nAnswer : There cannot be a dimension less quantity. quantity of dimension $[T^{-1}]=w$","date":"2019-08-24 20:14:06","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9512497186660767, \"perplexity\": 383.3972441157836}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-35\/segments\/1566027321696.96\/warc\/CC-MAIN-20190824194521-20190824220521-00217.warc.gz\"}"}
| null | null |
\section{\label{Sec:Intro}Introduction}
After the discovery of the Higgs boson in 2012 by the ATLAS and CMS experiments \cite{ATLAS:2012yve,CMS:2012qbp}, the Standard Model was complete and potentially valid all the way up to the Planck scale. However it is known that there are problems with the Standard Model and it cannot be a complete description of reality. One of these problems is the so-called hierarchy problem, the fact that the Higgs mass is unstable against quantum corrections; there is also the issue that we require dark matter in order to explain the observed density of the universe and new physics is also needed to explain the baryon asymmetry of the universe. The model proposed in this paper goes some way to addressing each of these concerns and would also have experimentally observable consequences.
In the Standard Model, the Higgs vev is introduced at tree level, but in the 1970s Coleman and Weinberg showed that it was possible for the tree level potential to have its minimum at the origin and still develop a minimum away from the origin at loop level \cite{Coleman:1973jx}. This idea was known as dimensional transmutation as we trade the dimensionful parameter of the higgs quadratic term for the dimensionless parameter of the scalar coupling. As this also removes the dimension-2 operator in the standard model, it removes the source of corrections which depend quadratically on the UV cutoff and hence
solves the Hierarchy problem \cite{Bardeen:1995kv,Hempfling:1996ht}. However due to the relationship induced between couplings in the dimensional transmutation it has been known since the 90s that the Standard Model Higgs is too heavy to come from a Coleman-Weinberg theory and one then applies the Coleman-Weinberg mechanism in a hidden sector~\cite{Hempfling:1996ht,Meissner:2006zh,Chang:2007ki,Foot:2007iy,Englert:2013gz}
coupled to the Standard Model for example via a Higgs portal interaction.
It has also been known for many years that some additional matter to that observed in the universe is needed to explain many observations. The first piece of evidence was seen in galactic rotation curves \cite{Rubin:1970zza} and there have since been many other pieces of evidence, such as observations of galactic collisions \cite{Markevitch:2003at} and data from the CMB experiment \cite{Planck:2015fie}, which support the hypothesis of particle dark matter \cite{Bertone:2004pz}. A popular model of dark matter is that the dark matter particle is charged under some new gauge group while being a singlet under the standard model gauge group and conversely all Standard Model particles are singlets under the dark gauge group. The number of particles in the dark sector (those charged under the new gauge group) varies heavily from model to model, some are very minimal including only one particle, while some contain multiple vectors, scalars and fermions. The dark sectors then 'communicate' with the Standard Model through either a Higgs portal or kinetic gauge mixing. See \cite{
Cheung:2007ut,
Hambye:2008bq,
Lindner:2011it,
Baek:2012se,
Hambye:2013dgv,
Carone:2013wla,
Khoze:2014xha,
Khoze:2014woa,
Harris:2014hga,
Alves:2015mua,
Rodejohann:2015lca,
Karam:2016bhq,
Karam:2016rsz,
Khoze:2016zfi,
Bauer:2018egk,
Foldenauer:2018zrz,
Nomura:2020zlm,
Baouche:2021wwa,
Tapadar:2021kgw,
Dasgupta:2021dnl} for examples.
Finally, it was demonstrated by Sakharov that in order to generate the required baryon asymmetry of the universe, thermal equilibrium must be violated in the early universe (amongst other requirements)~\cite{Sakharov:1967dj}. This requirement can be satisfied if there is a first order phase transition in the early universe. However it is known that in the Standard Model the electroweak phase transition is second-order/crossover, but this can be modified with the addition of new physics to create a first-order phase transition and go some way to generating the baryon asymmetry of the universe e.g. \cite{Croon:2018kqn,Li:2020eun,Athron:2019teq}. It is also well known that a first order phase transition can lead to the production of gravitational waves. There were several papers in the late 1980s - early 1990s which first calculated this possibility \cite{Hogan:1986qda,Kosowsky:1992rz,Kosowsky:1991ua,Kamionkowski:1993fg} and it is now a common consideration in dark matter models \cite{Schwaller:2015tja,Jaeckel:2016jlh,Mohamadnejad:2019vzg,Addazi:2017gpt,Breitbach:2018ddu,
Croon:2018erz,Croon:2019iuh,Ellis:2018mja,Ellis:2019oqb,Chala:2019rfk,Ghosh:2020ipy}.
In this paper, we attempt to simultaneously alleviate these various shortcomings by augmenting the Standard Model with a dark sector and imposing classical scale invariance. In this way we can generate the required dark matter abundance whilst simultaneously solving the hierarchy problem through the demand of conformal symmetry at the the classical level. With this new model it will also be possible to generate a strongly first order electroweak phase transition, thus aiding in the generation of the baryon asymmetry (although it should be noted that we will not seek to satisfy all of Sakharov's conditions).
The structure of the paper is as follows; in Section~\ref{sec:model} we introduce the model and derive the symmetry breaking. In Section~\ref{sec:expCon} we calculate the relic abundance of the model and examine in which areas of parameter space we can reproduce the observed relic abundance as well as imposing constraints arising from collider searches for new particles and direct detection experiments looking for dark matter. In Section~\ref{sec:theCon} we examine the theoretical constraints on the model such as ensuring that the vacuum is stable, we respect perturbative unitarity and that the model is perturbative. Finally, in Section~\ref{sec:gravWaves} we calculate the potential in the early universe before looking at which areas of phase space give rise to a strongly first order phase transition and the associated gravitational wave signal before we conclude in Section~\ref{sec:conc}.
\medskip
\section{\label{sec:model}Model and the Coleman-Weinberg Mechanism}
We introduce a classically scale invariant model with a dark sector charged under a new $U(1)$ symmetry and coupled to the standard model through a Higgs portal coupling. Our model is similar to that considered in \cite{Kim:2019ogz}, although we extend their model by allowing our fermions to have different masses (as well as in later sections looking at the model in the early universe by investigating the phase transition and a production of gravitational waves). The model is given by
\begin{equation}
\mathcal{L}=\mathcal{L}_{\rm{SM}}+\mathcal{L}_{kin}+\mathcal{L}_{Y}-V_0\left(H,S\right)
\end{equation}
where $\mathcal{L}_{SM}$ is the Standard Model lagrangian without the Higgs potential, $\mathcal{L}_{kin}$ is the kinetic terms for the new fields\footnote{Note that the term $\epsilon F_{\mu\nu}F'^{\mu\nu}$, where $F_{\mu\nu}$ is the U(1)$_{Y}$ field strength tensor, is also allowed by gauge invariance but we neglect this term in light of strong collider constraints\cite{Hook:2010tw,Curtin:2014cca} and leave it for future work.} :
\begin{equation}
\mathcal{L}_{kin}=\lvert D_{\mu} S\rvert^{2}-\frac{1}{4}F'_{\mu\nu}F'^{\mu\nu}+\overline{\chi}^{a}_{L}\slashed{D}\chi^{a}_{L}+\overline{\chi}^{a}_{R}\slashed{D}_{\mu}\chi^{a}_{R}.
\end{equation}
All new particles, $S, \chi, A'_{\mu}$ are singlets under the standard model gauge group, $G_{\rm{SM}}=SU(3)_{C}\times SU(2)_{L}\times U(1)_{Y}$ and all SM particles are singlets under the new gauge group. The charges of the new particles are given in Table~\ref{table:charges}. $F'_{\mu\nu}$ is the field strength tensor associated with the gauge boson of the new $U\left(1\right)$ and there is implied summation over repeated indices.
\begin{table}
\centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Field & $S$ & $\chi^{1}_{L}$ & $\chi^{1}_{R}$ & $\chi^{2}_{L}$ & $\chi^{2}_{R}$ \\
\hline
$U(1)_{D}$ & 1 & $\frac{1}{2}$ & $-\frac{1}{2}$ & $-\frac{1}{2}$ & $\frac{1}{2}$ \\
\hline
\end{tabular}
\caption{The charges of the dark sector particles under the new $U(1)_{D}$ symmetry. Note that this assignment of charges renders the theory anomaly-free.}
\label{table:charges}
\end{table}
$\mathcal{L}_{Y}$ is the Yukawa coupling of the dark sector:
\begin{equation}
\mathcal{L}_{Y}=y_{1,D}\overline{\chi}^{1}_{L}S\chi^{1}_{R}+y_{2,D}\overline{\chi}^{2}_{L}S\chi^{2}_{R}+h.c.
\end{equation}
The tree-level potential for scalar fields of the new theory is given by:
\begin{equation}
\label{eq:potential}
V_0\left(H,\,S\right)=\lambda_{H}\left(H^{\dagger}H\right)^{2}+\lambda_{S}\left(S^{*}S\right)^{2}-\lambda_{P}\left(H^{\dagger}H\right)\left(S^{*}S\right).
\end{equation}
Note that we require $\lambda_{P}>0$ to create a true minimum away from the origin. Working in the unitary gauge where we can write
\begin{align}
H &=\frac{1}{\sqrt{2}} \begin{bmatrix}
0\\
h
\end{bmatrix}
\quad
S =\frac{s}{\sqrt{2}}
\end{align}
where $h$ and $s$ are real; then the classical scalar potential may be written as
\begin{equation}
V_0\left(h,\,s\right)=\frac{\lambda_{H}}{4}h^{4}+\frac{\lambda_{S}}{4}s^{4}-\frac{\lambda_{P}}{4}h^{2}s^{2}.
\end{equation}
Symmetry breaking in a classically scale invariant model was first considered in \cite{Coleman:1973jx}. In models such as ours many authors consider $\lambda_{P}$ to be small so that the backreaction of the Standard Model on the dark sector is negligible and one can treat the dark sector in the original Coleman-Weinberg formalism (see e.g. \cite{Hempfling:1996ht, Englert:2013gz,Oda:2015gna}) however a more general formalism was later developed by Gildener and S. Weinberg in \cite{Gildener:1976ih} to deal with theories of multiple scalars and it is this formalism we shall follow here in order not be restricted in our choice of parameters.
We write our fields as
\begin{equation}
h=N_{1}\varphi \quad s=N_{2}\varphi
\end{equation}
where $\vec{N}$ is a 2D unit vector. Then our tree-level potential becomes
\begin{equation}
V_{0}\left(h,\,s\right)=\varphi^{4}\left(\frac{\lambda_{H}}{4}N_{1}^{4}+\frac{\lambda_{S}}{4}N_{2}^{4}-\frac{\lambda_{P}}{4}N_{1}^{2}N_{2}^{2}\right).
\end{equation}
To find the minimum we then require $\frac{\partial V_{0}}{\partial N_{i}}=0$ and $V_{0}\left(\phi\textbf{n}\right)=0$. This leads to the constraints:
\begin{gather}
\label{eq:GWConditions}
\lambda_{H}n_{1}^{2}-\frac{\lambda_{P}}{2}n_{2}^{2}=0\\
\lambda_{S}n_{2}^{2}-\frac{\lambda_{P}}{2}n_{1}^{2}=0\\
\frac{\lambda_{H}}{4}n_{1}^{4}+\frac{\lambda_{S}}{4}n_{2}^{4}-\frac{\lambda_{P}}{4}n_{1}^{2}n_{2}^{2}=0
\end{gather}
where $n_{i}$ are the particular values for which the potential reaches its minimum. These equations are satisfied at some renormalisation scale $\mu=\Lambda_{GW}$. These can be solved to yield
\begin{gather}
\label{eq:flatDirection}
n_{1}^{2}=\frac{\lambda_{P}}{\lambda_{P}+2\lambda_{H}}\\
n_{2}^{2}=\frac{2\lambda_{H}}{\lambda_{P}+2\lambda_{H}}
\end{gather}
We now expand the fields about their minima, writing $h=wn_{1}+\tilde{h},\,s=wn_{2}+\tilde{s}$ where $w=\langle \varphi\rangle$ is a classically flat direction.\footnote{The value of $w$ will be stabilised below by the inclusion of quantum effects.} This leads to the mass matrix:
\begin{equation}
\label{eq:massMatrix}
M^{2}=w^{2}\begin{bmatrix}
2\lambda_{H}n_{1}^{2} &-\lambda_{P}n_{1}n_{2} \\
-\lambda_{P}n_{1}n_{2} & 2\lambda_{s}n_{2}^{2}
\end{bmatrix}
\end{equation}
after using the relations in \eqref{eq:GWConditions}. By standard results of linear algebra, this matrix can be diagonalised by a rotation matrix of the form:
\begin{equation}
O=\begin{bmatrix}
\cos\theta &-\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
\end{equation}
where
\begin{equation}
\tan\left(2\theta\right)=\frac{\lambda_{P}n_{1}n_{2}}{\lambda_{s}n_{2}^{2}-\lambda_{H}n_{1}^{2}}.
\end{equation}
We can now write the mass eigenstates:
\begin{equation}
\begin{bmatrix}
h_{1}\\
h_{2}
\end{bmatrix}
=O
\begin{bmatrix}
h\\
s
\end{bmatrix}
\end{equation}
where we identify $h_{1}$ with the SM higgs. The mass eigenvalues are given by
\begin{equation}
M_{h_{1},h_{2}}^{2}=w^{2}\left(\lambda_{H}n_{1}^{2}+\lambda_{s}n_{2}^{2}\pm\sqrt{\left(\lambda_{H}n_{1}^{2}-\lambda_{s}n_{2}^{2}\right)^{2}+\lambda_{P}^{2}n_{1}^{2}n_{2}^{2}}\right).
\end{equation}
After using the relations in \eqref{eq:flatDirection} to simplify this we obtain
\begin{gather}
\label{eq:h1Mass}
M_{h_{1}}^{2}=\lambda_{P}w^{2}\\
M_{h_{2}}^{2}=0.
\label{eq:massless2}
\end{gather}
We recall that $w$ is a classically flat direction that will be stabilised in \eqref{eq:rewrittenPotential}
and also note that \eqref{eq:massless2} is true only at tree level ({\it cf.} \eqref{eq:h2Mass} below).
We shall take the tree level mass for $h_{1}$ but although $h_{2}$ is massless at tree level it receives sizeble corrections at the one-loop level which we shall calculate at the end of this section.
\begin{figure}
\centerline{\includegraphics[width=0.7\textwidth,keepaspectratio]{Figures/ScalarLoop.png}}
\caption{The infinite series of diagrams contributing to the one-loop effective potential arising from a scalar running in the loop.}
\label{fig:scalarLoop}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=0.9\textwidth,keepaspectratio]{Figures/FermionLoop.png}}
\caption{The infinite series of diagrams contributing to the one-loop effective potential arising from a fermion running in the loop. Fermionic diagrams with an odd number of external legs have been excluded as the trace of an odd number of gamma matrices is zero and so such diagrams do not contribute.}
\label{fig:fermionLoop}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=0.8\textwidth,keepaspectratio]{Figures/BosonLoop.png}}
\caption{The infinite series of diagrams contributing to the one-loop effective potential arising from a boson running in the loop.}
\label{fig:bosonLoop}
\end{figure}
To find the new minimum of the theory we must calculate the infinite series of diagrams shown in Figs.~\ref{fig:scalarLoop},\ref{fig:fermionLoop},\ref{fig:bosonLoop} with external legs being either $h$ or $s$ and all possible scalars, fermions and bosons running in the loop. Calculating these diagrams (and counterterms in the $\overline{MS}$ scheme) leads to the 1-loop effective potential (see e.g. \cite{Quiros:1999jp} for a review),
\begin{gather} \nonumber
V_1\left(h,\,s\right)=\frac{1}{64\pi^{2}}\left(\, \sum_{\rm bosons}n_{i} M_{i}^{4}\left(\phi\right)\left(\log\left(\frac{M_{i}^{2}\left(\phi\right)}{\Lambda_{GW}^{2}}\right)-\frac{3}{2}\right)\right.\\
-\,\left.\sum_{\rm fermions}n_{i} M_{i}^{4}\left(\phi\right)\left(\log\left(\frac{M_{i}^{2}\left(\phi\right)}{\Lambda_{GW}^{2}}\right)-\frac{3}{2}\right)\right)
\end{gather}
Note that since $h_{2}$ is massless at tree level it does not contribute to the effective potential at one loop so our sum runs over $h_{1},\,W,\,Z,\,Z',\,t,\chi_{1},\,\chi_{2}$ with degrees of freedom $n_{i}=1,\,6,\,3,\,3,\,12,\,4,\,4$ respectively. Note that all SM fermions contribute to the potential but, as is standard in the literature, we account only for the contribution of the top quark (with a factor of three due to colour) as this is the most significant. Since we are working in a theory with no intrinsic masses we can write for all particles: $M^{2}\left(\phi\right)=\frac{M^{2}\phi^{2}}{w^{2}}$ where $M^{2}$ is the observed mass matrix evaluated at $\phi=w$, so we may rewrite the above equation as
\begin{gather}
V_{1}\left(\varphi\right)=A\varphi^{4}+B\varphi^{4}\log\left(\frac{\varphi^{2}}{\Lambda_{GW}^{2}}\right)
\end{gather}
where
\begin{gather}
A=\frac{1}{64\pi^{2}w^{4}}\left(\,\sum_{\rm bosons}n_{i} M_{i}^{4}\left(\log\left(\frac{M_{i}^{2}}{w^{2}}\right)-\frac{3}{2}\right)-\sum_{\rm fermions}n_{i} M_{i}^{4}\left(\log\left(\frac{M_{i}^{2}}{w^{2}}\right)-\frac{3}{2}\right)\right)\\
B=\frac{1}{64\pi^{2}w^{4}}\left(\,\sum_{\rm bosons}n_{i} M_{i}^{4}-\sum_{\rm fermions}n_{i} M_{i}^{4}\right).
\end{gather}
By definition $V'\left(w\right)=0$ so we obtain the relationship:
\begin{equation}
\label{eq:GWScaleDef}
\log\left(\frac{w}{\Lambda_{GW}}\right)=-\frac{1}{4}-\frac{A}{2B}.
\end{equation}
which allows us to rewrite our potential as
\begin{equation}
\label{eq:rewrittenPotential}
V_{1}\left(\varphi\right)=B\varphi^{4}\left(\log\left(\frac{\varphi^{2}}{w^{2}}\right)-\frac{1}{2}\right).
\end{equation}
At the one loop level the mass of $h_{2}$ is given by\footnote{By examining previous relations one can show that $h_{2}$ and $\phi$ turn out to be the same field}
\begin{equation}
\label{eq:h2Mass}
M_{h_{2}}^{2}=\frac{\partial^{2}V}{\partial\phi^{2}}\bigg\rvert_{\phi=w}=8Bw^{2}.
\end{equation}
Finally we end this section with a summary of which parameters are free and which others are determined by the constraints previously listed. Firstly $v_{h}=246$ GeV and $M_{h_{1}}=125$ GeV are known from experiment. $\lambda_{h}$ has a certain value within the Standard Model but it has not been experimentally measured so we shall regard this as undetermined. We have only one remaining degree of freedom in the scalar sector, once we have picked a value of e.g. $w$ then $\lambda_{P}$ is determined by \eqref{eq:h1Mass} and once $\lambda_{P}$ is determined then the remaining scalar couplings must take their values to satisfy \eqref{eq:GWConditions} (with $n_{1}$ and $n_{2}$ already being determined by the vevs). For our purposes it shall be more convenient to take $\sin\theta$, the mixing angle as our free parameter and determine the scalar couplings and vevs from here.
We shall also take $M_{h_{2}}$ as a free parameter and then $M_{Z_{D}}$ is determined by \eqref{eq:h2Mass}, which in turn determines $g_{D}$ as $M_{Z'}=g_{D}v_{S}$. Finally we have complete freedom in choosing the mass of our fermions, $M_{\chi_{1}},\, M_{\chi_{2}}$ and these in turn shall determine the yukawa couplings $y_{D, 1/2}$. For later convenience we also define $\Delta M_{\chi}$ as the mass splitting between the two fermions and without loss of generality we shall always take $\chi_{1}$ to be the lighter of the two.
\medskip
In summary, the free parameters of our model are $\sin\theta$, $M_{h_{2}}$, $M_{\chi_{1}}$ and $M_{\chi_{2}}$ .
\section{Relic density and Experimental Constraints \label{sec:expCon}}
We can consider our model as a model of dark matter, with $\chi_{1}$ and $\chi_{2}$ serving as the dark matter candidates. To calculate the relic density provided by our mode we use MicrOMEGAs \cite{Belanger:2018ccd} with FeynRules \cite{Alloul:2013bka} being used to generate the model file. At the same time we also use MicrOMEGAs to implement several experimental constraints on our model.
One of the primary constraints on dark sector models comes from direct detection experiments where dark sector particles can scatter off standard model nuclei. This happens in our model due to the mixing between the two scalars.
This constraint can be done within MicrOMEGAs.
We also have constraints on the scalar sector of our model. There have been many searches at the LHC for additional light and heavy scalars. So far all such searches have produced null results and so these analyses constrain the valid parameter space of our model. We implement these constraints using the HiggsBounds and HiggsSignals codes \cite{Bechtle:2020pkv,Bechtle:2020uwn}.
Below we plot the relic density as a function of some of our free parameters and also show some of the constraints coming from direct detection and collider searches. The relic density of the universe has been measured as $\Omega_{DM}h^{2}=0.1200\pm0.0020$ \cite{Workman:2022ynf}. As can be seen in Fig.~\ref{fig:DMresonance}, in order to obtain the correct relic density, we need the mass of our dark fermions to lie in the region around a resonance i.e. $M_{{\chi_{1,2}}}\approx M_{h_{1,2}}/2$, although it should be noted that the allowed region is not particularly narrow. This near-resonance regime is necessary in order for the dark matter to annihilate sufficiently quickly to not produce an overabundance. An alternative is to have the dark matter sufficiently heavy that the annihilation rate is enhanced by the phase space, as shown in Fig.~\ref{fig:DMheavy}. Note that the smaller the value of $\sin\theta_{D}$ the narrower the resonance, or the larger the mass of the dark fermions should be as the annihilation rate is additionally suppressed.
Here $\Delta M_{\chi}=|M_{\chi_1}-M_{\chi_2}|$ and we always choose $\chi_{1}$ to be the lighter fermion.
\begin{figure}
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[width=0.95\linewidth,keepaspectratio]{Figures/relicAbundanceSinTheta025DeltaM0.png}
\caption{The points which correctly produce an acceptable relic abundance for $\sin\theta_{D}=0.25,\,\Delta M_{\chi}=0$ GeV.}
\label{fig:DMresonance}
\end{subfigure}%
\hskip 0.3truecm
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[width=0.95\linewidth,keepaspectratio]{Figures/relicAbundanceSinTheta03DeltaM100.png}
\caption{The points which produce an acceptable relic abundance for $\sin\theta_{D}=0.30,\,\Delta M_{\chi}=100$ GeV.}
\label{fig:DMheavy}
\end{subfigure}
\caption{Areas of our parameter space which do not produce an over-abundance of dark matter.}
\end{figure}
\begin{figure}
\begin{subfigure}{.5\textwidth}
\includegraphics[width=0.99\linewidth,keepaspectratio]{Figures/scalarConstraintsSinTheta03DeltaM5.png}
\caption{Scatter plot of $M_{\chi_{1}}$ against $M_{h_{2}}$ for $\sin\theta_{D}=0.30, \Delta M_{\chi}=5$ GeV with points allowed by constraints from the scalar sector in green and forbidden points in red.}
\label{fig:scalarCon}
\end{subfigure}
\hskip 0.3truecm
\begin{subfigure}{.5\textwidth}
\includegraphics[width=0.99\linewidth,keepaspectratio]{Figures/DDConstraintsSinTheta02DeltaM0.png}
\caption{Scatter plot of $M_{h_{2}}$ against $M_{\chi_{1}}$ for $\sin\theta_{D}=0.20, \Delta M_{\chi}=0$ GeV with points allowed by constraints from the direct detection experiments in green and forbidden points in red.}
\label{fig:DDCon}
\end{subfigure}
\caption{Constraints from collider and direct detection experiments.}
\end{figure}
Hence non-observation of dark matter at the LHC corresponds to an upper bound on the value of $\sin\theta$. Experimental evidence requires $\sin\theta<0.44$ independent of the mass of $h_{2}$. There is also a mass-dependent constraint, which requires $\sin\theta\lesssim0.32$ for $M_{h_{2}}\gtrsim200$ GeV and $\sin\theta\lesssim0.2$ for $M_{h_{2}}\gtrsim400$ GeV, mostly coming from restrictions on the NLO corrections to the mass of the W boson (obviously other constraints exist but none as severe as those coming from the W boson mass in our considered parameter range) \cite{Robens:2015gla}. We also in general require $M_{h_{2}}>M_{h_{1}}/2$ to respect bounds coming from the decays of the SM Higgs to invisibles. We show an example plot of the allowed region of parameter space in Fig.~\ref{fig:scalarCon}. The constraints are largely independent of the fermion mass splitting (although there is some effect).
There are also constraints on the masses of our dark fermions coming from direct detection experiments. Although the fermions do not interact directly with SM quarks/hadrons, they can still interact through the exchange of a mixed scalar, although such diagrams are suppressed by a factor of $\sin\theta$. Such interactions are proportional to $\frac{1}{M_{h_{1}}^{2}}-\frac{1}{M_{h_{2}}^{2}}$ \cite{Bell:2016ekl} and so we require $M_{h_{1}}\approx M_{h_{2}}$ to avoid direct detection constraints. Alternatively we can suppress these diagrams by taking the yukawa coupling of the dark fermions to the scalars to be small i.e. our dark fermions will be light. As one would expect, these constraints become more relaxed for smaller values of $\sin\theta_{D}$. These constraints are shown in Fig.~\ref{fig:DDCon} and as for the scalar sector, the constraints are mostly independent of $\Delta M_{\chi}$.
\section{Theoretical Constraints \label{sec:theCon}}
We shall now examine constraints on the coupling constants coming from vacuum stability, perturbativity and unitarity. From \eqref{eq:rewrittenPotential}, we see that the potential is bounded from below and hence the vacuum is stable if and only if $B\geq0$ (which is automatically satisfied for any positive choice of $M_{h_{2}}^{2}$ by Eq.~\ref{eq:h2Mass}). We also require that the vacuum be stable (bounded from below) at tree level which implies
\begin{equation}
\lambda_{P}^{2}<4\lambda_{S}\lambda_{H}\,, \quad \lambda_{H}>0.
\end{equation}
The requirement of perturbativity simply imples that we have $\lvert g_{i}\rvert<\textrm{const.}$ for all couplings $g_{i}$, i.e. $g_{i}=\lambda_{P},\,\lambda_{H},\,y_{D}\ldots$ \cite{Robens:2015gla}. We choose a constant of $2\pi$ in agreement with \cite{Karam:2016bhq}. We derive and numerically solve the RG equations using SARAH \cite{Staub:2013tta}, and list them in Appendix \ref{appendix:RG}.
Checking the resulting constraints involves evolving the various coupling constants up to high scales using numerical solutions to the RG equations. Rather than doing it for the entire parameter space we will check these constraints for a selection of bench mark points which we define in the next section. Also it is not necessary for these conditions to hold at arbitrarily high scales (perturbativity and vacuum stability are not absolute requirements in any case) as there may be new physics which arises at some higher scale which then contributes in such a way to e.g. stabilise the vacuum. Hence when we numerically check these constraints we only require them to hold up to $\Lambda_{GW}$ (defined by Eq.~\ref{eq:GWScaleDef}) and then give the higher scale at which they are violated, see Table~\ref{Table:benchPoints}.
We now consider constraints from perturbative unitarity for our theory. A partial wave expansion for a scattering amplitude gives
\begin{equation}
\mathcal{M}\left(s,\theta\right)=16\pi\sum_{J=0}^{\infty}\left(2J+1\right)A_{J}\left(s\right)P_{K}\left(\cos\theta\right)
\end{equation}
where $P_{J}$ are the Legendre polynomials and $A_{J}$ are the partial wave amplitudes. Unitarity then imposes the bound that $\lvert Re A_{0}\rvert<\frac{1}{2}$. We consider the tree level amplitudes for the processes: $Z'_{L}Z'_{L}\rightarrow Z'_{L}Z'_{L}, h_{1}h_{1}\rightarrow h_{1}h_{1}, h_{2}h_{2}\rightarrow h_{2}h_{2}$. We use FeynArts \cite{Hahn:2000kx} and FeynCalc \cite{Shtabovenko:2020gxv} to aid in the computation of the amplitudes. For the scalar process, the only relevant diagram at high energy is the four-point interaction (all others are suppressed by $\sim\frac{1}{s}$ due to internal propagators) and so the demand for perturbative unitarity simply imposes
\begin{gather}
\label{eq:sc_un_1}
\frac{6}{16\pi}\left(\lambda_{H}\cos^{4}\theta_{D}-\lambda_{P}\sin^{2}\theta_{D}\cos^{2}\theta_{D}+\lambda_{S}\sin^{4}\theta_{D}\right)<\frac{1}{2} \\
\label{eq:sc_un_2}
\frac{6}{16\pi}\left(\lambda_{H}\sin^{4}\theta_{D}-\lambda_{P}\sin^{2}\theta_{D}\cos^{2}\theta_{D}+\lambda_{S}\cos^{4}\theta_{D}\right)<\frac{1}{2}.
\end{gather}
For the vector boson scattering we obtain
\begin{gather}
\mathcal{M}=-4\frac{g_{D}^{2}v_{s}^{2}\sin^{2}\theta_{D}}{M_{Z'}^{4}}\left(\frac{\left(s-2M_{Z'}^{2}\right)^{2}}{s-M_{h_{1}}^{2}}+\frac{\left(t-2M_{Z'}^{2}\right)^{2}}{t-M_{h_{1}}^{2}}+\frac{\left(u-2M_{Z'}^{2}\right)^{2}}{u-M_{h_{1}}^{2}}\right)-\\
4\frac{g_{D}^{2}v_{s}^{2}}{M_{Z'}^{4}}\cos^{2}\theta_{D}\left(\frac{\left(s-2M_{Z'}^{2}\right)^{2}}{s-M_{h_{2}}^{2}}+\frac{\left(t-2M_{Z'}^{2}\right)^{2}}{t-M_{h_{2}}^{2}}+\frac{\left(u-2M_{Z'}^{2}\right)^{2}}{u-M_{h_{2}}^{2}}\right)\\
\approx -4\frac{\sin^{2}\theta_{D}}{M_{Z'}^{2}}\left(s+t+u+3M_{h_{1}}^{2}\right)-4\frac{\cos^{2}\theta_{D}}{M_{Z'}^{2}}\left(s+t+u+3M_{h_{2}}^{2}\right)\\
\approx -4\sin^{2}\theta_{D}\left(4+3\frac{M_{h_{1}}^{2}}{M_{Z'}^{2}}\right)-4\cos^{2}\theta_{D}\left(4+3\frac{M_{h_{2}}^{2}}{M_{Z'}^{2}}\right).
\end{gather}
Hence we require
\begin{equation}
\label{eq:vec_un}
4\sin^{2}\theta_{D}\left(4+3\frac{M_{h_{1}}^{2}}{M_{Z'}^{2}}\right)+4\cos^{2}\theta_{D}\left(4+3\frac{M_{h_{2}}^{2}}{M_{Z'}^{2}}\right)<8\pi.
\end{equation}
Equations \eqref{eq:sc_un_1}, \eqref{eq:sc_un_2}, \eqref{eq:vec_un}
summarise the unitarity constraints that we require to hold for our model.
\medskip
\section{Phase transition and Gravitational Wave signal\label{sec:gravWaves}}
To discuss the phase transition and gravitational waves we must first compute the one-loop effective potential at finite temperature. It is known that at one loop level, the potential factorises into the zero temperature potential (which we have already calculated) plus thermal corrections. The thermal corrections are given by\footnote{This can be calculated using exactly the same diagrams as the zero-temperature potential but now using the Feynman rules for a theory at finite temperature, see \cite{Quiros:1999jp} for a review}
\begin{gather}
V_{T}=\frac{T^{4}}{2\pi^{2}}\left(\sum_{bosons} n_{i}J_{B}\left(\frac{M_{i}^{2}\left(\phi\right)}{T^{2}}\right)-\sum_{fermions} n_{i}J_{F}\left(\frac{M_{i}^{2}\left(\phi\right)}{T^{2}}\right)\right)
\end{gather}
where the functions $J_{B/F}$ are defined by
\begin{equation}
J_{B/F}\left(x^{2}\right)=\int_{0}^{\infty}\,dy\,y^{2}\log\left(1\mp e^{-\sqrt{x^{2}+y^{2}}}\right).
\end{equation}
\begin{figure}
\centerline{\includegraphics[width=0.45\textwidth,keepaspectratio]{Figures/DaisyDiagram.png}}
\caption{An example of a daisy diagram \cite{Quiros:1999jp} with a scalar field appearing in the outside bubbles. This is then resummed to all orders (the outside series of bubbles) to obtain the thermal mass correction.}
\label{fig:daisyDiagram}
\end{figure}
Finally to go beyond the simple one-loop expressions for the effective potential we were using until now, we now add the resummed contributions of the so-called `daisy diagrams' (shown in Fig.~\ref{fig:daisyDiagram}) to improve the validity of perturbation theory.
This was first done for the Standard Model in \cite{Carrington:1991hz,Arnold:1992rz}, and for our model we have
\begin{gather}
\label{eq:daisy}
V_{daisy}=\frac{T}{12\pi}\sum_{bosons} n_{i} \left(M_{i}^{3}\left(\phi\right)-\left(M_{i}^{2}+\Pi_{i}\left(\phi,T\right)\right)^{\frac{3}{2}}\right)
\end{gather}
where $\Pi_{i}\left(\phi,T\right)$ is the thermal mass correction. Note that, to leading order, fermions do not receive a thermal mass and so do not contribute to the daisy potential \eqref{eq:daisy} and also that it is only the longitudinal mode of the gauge bosons which receives a thermal mass, so the relevant degrees of freedom should be divided by three. The thermal masses are given by:
\begin{gather}
\label{eq:scalarThermalMass}
\Pi_{h/s}=\begin{bmatrix}
T^{2}\left(\frac{\lambda_{H}}{2}+\frac{\lambda_{P}}{12}+\frac{g'^{2}}{16}+\frac{3g_{W}^{2}}{16}+\frac{y_{t}^{2}}{4}\right) & 0 \\
0 & T^{2}\left(\frac{\lambda_{S}}{3}+\frac{\lambda_{P}}{6}+\frac{g_{D}^{2}}{4}+\frac{y_{\chi_{1}}^{2}}{12}+\frac{y_{\chi_{2}}^{2}}{12}\right)
\end{bmatrix} \\
\Pi_{Z'}=\frac{g_{D}^{2}T^{2}}{3} \\
\Pi_{W}=\frac{11}{6}g_{W}^{2}T^{2} \\
\Pi_{Z}=\frac{11}{6} g_{W}^{2}T^{2}
\end{gather}
where the results for the $W$ and $Z$ bosons were taken from \cite{Carrington:1991hz}. Hence the full one-loop effective potential is given by
\begin{equation}
V\left(\phi,T\right)=V_{1}\left(\phi\right)+V_{T}\left(\phi,T\right)+V_{daisy}\left(\phi,T\right)
\end{equation}
Note that to determine the mass of the scalars at finite temperature one must add Eqs.~\ref{eq:massMatrix} and \ref{eq:scalarThermalMass} before finding the eigenvalues. It is known that in order to generate matter-antimatter asymmetry we must have a strongly first order electroweak phase transition. The order parameter for these transitions is given by the ratio ${\phi_{c}}{/T_{c}}$, where the critical temperature, $T_{c}$ , and the critical field strength, $\phi_{c}$ are defined by
\begin{gather}
V\left(\phi_{c},T_{c}\right)=0\\
\partial_\phi V\left(\phi_{c},T_{c}\right)=0,
\end{gather}
i.e. $\phi_{c}$ is a local minimum of the potential that is degenerate with the minimum at the origin at $T_{c}$. To have a strongly first order phase transition we then require ${\phi_{c}}/{T_{c}}\gtrsim1$. The numerical calculation of the order parameters and of various parameters associated with the gravitational wave signal becomes quite slow and so in this section rather than completing a full exploration of the phase space we choose several benchmark points consistent with the constraints from Sections \ref{sec:expCon},\ref{sec:theCon} and compute the order parameters and gravitational wave signal. Our benchmark points are listed in Table~\ref{Table:benchPoints}.
\begin{table}
\centering
\scalebox{0.8}{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
& $\sin\theta_{D}$ &$ M_{h_{2}} $ & $M_{\chi_{1}} $ & $M_{\chi_{2}}$ & $\Omega h^{2}$ & $\Lambda_{\textrm{unit}} $ & $ \Lambda_{\textrm{pert.}}$ &$ \Lambda_{\textrm{stab.}} $ \\
\hline
BP1 & 0.30 & 151\, \textrm{GeV} & 59.5\, \textrm{GeV} & 59.5\, \textrm{GeV} & 0.070 & $ 2.2\times10^{9}\, \textrm{GeV} $ & $ >10^{16}\, \textrm{GeV} $ & $ 9.3\times10^{4}\, \textrm{GeV} $ \\
\hline
BP2 & 0.10 & 320\, \textrm{GeV} & 150\, \textrm{GeV} & 155\, \textrm{GeV} & 0.078 & $ 3.0\times10^{15}\, \textrm{GeV} $ & $ >10^{19}\, \textrm{GeV} $ & $ 3.2\times10^{5}\, \textrm{GeV} $ \\
\hline
BP3 & 0.40 & 121\, \textrm{GeV} & 591\, \textrm{GeV} & 592\, \textrm{GeV} & 0.118 & $ 6.3\times10^{8}\, \textrm{GeV} $ & $ >10^{10}\, \textrm{GeV} $ & $ 8.9\times10^{4}\, \textrm{GeV} $ \\
\hline
BP4 & 0.20 & 331\, \textrm{GeV} & 61\, \textrm{GeV} & 161\, \textrm{GeV} & 0.077 & $ 1.8\times10^{6}\, \textrm{GeV} $ & $ >10^{12}\, \textrm{GeV} $ & $ 2.0\times10^{5}\, \textrm{GeV} $ \\
\hline
BP5 & 0.30 & 120\, \textrm{GeV} & 901\, \textrm{GeV} & 1001\, \textrm{GeV} & 0.118 & $ 5.8\times10^{6}\, \textrm{GeV} $ & $ 3.0\times10^{8}\, \textrm{GeV} $ & $ 1.2\times10^{5}\, \textrm{GeV} $ \\
\hline
\end{tabular}
}
\caption{Table showing our selection of benchmark points. The $\Lambda$ show the scale at which we violate perturbativity, perturbative unitarity and vacuum stability respectively. All chosen points also obey the experimental constraints coming from collider searches and direct detection experiments. Note that due to numerical issues in the software we were unable to determine the exact scale at which perturbativity is violated for most of our benchmark points and so we indicate the maximum scale we were able to check. }
\label{Table:benchPoints}
\end{table}
It is well known that a strongly first order phase transition will produce a gravitational wave signal. Here we calculate this signal and examine the possibility of detection at both present detectors (LIGO, VIRGO etc.) and future detectors (LISA, DECIGO etc.).
A first order phase transition occurs when there is a potential barrier between a false minimum (usually at $\phi=0$) and a true minimum. When this occurs the transition happens as bubbles of true vacuum nucleate in the false vacuum. A gravitational wave signal is produced by three different mechanisms, as reviewed in \cite{Caprini:2015zlo}): collisions between bubbles, sound waves in the plasma, and magnetohydrodynamic turbulence.
Before going on to calculate the signal we briefly outline some bubble nucleation theory necessary for our calculation. The vacuum at $\phi=0$ only becomes metastable at temperatures $T<T_{c}$, however if the barrier is sufficiently high then the tunnelling rate may remain very small even for temperatures much below the critical temperature. Hence it is conventional to also define the \textit{nucleation temperature} at which the probability of one bubble nucleating in one horizon volume is approximately one. The theory of such transitions and bubble nucleation was first addressed in \cite{Coleman:1977py,Callan:1977pt} where it was shown that the decay rate was given by
\begin{equation}
\frac{\Gamma}{V}=Ae^{-S_{4}}
\end{equation}
where the left-hand side is the decay rate per unit volume and on the right-hand side we have $A$ which is a ratio of determinants of quadratic fluctuation operators around the bubble solution that is usually taken as $A\approx T^{4}$ in the literature on dimensional grounds. $S_{4}$ is the action computed on the field profile, $\phi$, satisfying the differential equation
\begin{equation}
\frac{d^{2}\phi}{d\rho^{2}}+\frac{3}{\rho}\frac{d\phi}{d\rho}=V'\left(\phi\right)
\end{equation}
which is the Euler-Lagrange equation for a field in four dimensions with an $O\left(4\right)$ symmetry, $\phi({\bf x}, t)=\phi(\rho)$,
and the boundary conditions,
\begin{equation}
\label{eq:bcb}
\lim_{\rho\to\infty} \phi(\rho) = 0\,, \quad \partial_\rho \phi(0) =0.
\end{equation}
The solution to this classical problem corresponds to the four-dimensional bubble or bounce configuration.
It was shown in \cite{Linde:1981zj} that when working in a theory at finite temperature this four-dimensional approach should be modified to the effectively three-dimensional set-up,
\begin{equation}
\frac{\Gamma}{V}=Ae^{-S_{3}/T}
\end{equation}
and the field profile $\phi$ should satisfy
\begin{equation}
\frac{d^{2}\phi}{d\rho^{2}}+\frac{2}{\rho}\frac{d\phi}{d\rho}=V'\left(\phi\right)
\end{equation}
with the same boundary conditions \eqref{eq:bcb}.
At finite temperature, due to the periodicity of the imaginary time-dimension $0\le \tau \le 1/T$ in the Matsubara formalism, we essentially work in a three-dimensional theory\footnote{The D-dimensional action is given by $S_{D}=\int d\rho d\Omega_{D} \, \rho^{D-1}\left[\left(\frac{d\phi}{d\rho}\right)^{2}+V\left(\phi\right)\right]$ where $\Omega_{D}$ is an integral over the surface of a D-dimensional sphere.}.
It was shown in \cite{Apreda:2001us} that the nucleation temperature, $T_{N}$, is given by solving the equation $S_{3}\left(T_{N}\right)/T_{N}\approx140$.
To describe the gravitational wave spectrum resulting from the first-order phase transition detailed above,
it is conventional to define two more parameters $\alpha$ and $\beta$, in addition to the nucleation temperature $T_N$,
that characterise the phase transition:
\begin{gather}
\alpha=\frac{1}{\rho_{rad}(T_{N})}\left(\Delta V(T_{N})-T_{N}\frac{d \Delta V}{dT}\bigg\rvert_{T=T_{N}}\right) \\
\frac{\beta}{H_{*}}=T_{N}\frac{d(S_{3}/T)}{dT}\bigg\rvert_{T=T_{N}}
\end{gather}
where $H_{*}$ is the Hubble constant at the time of nucleation, $\rho_{rad}$ is the radiation energy density\footnote{$\rho_{rad}(T_{N})=g_{*}\pi^{2}T_{N}^{4}/30$ where $g_{*}$(=117.75 for our model) is the number of relativistic degrees in the plasma at $T_{N}$.} and $\Delta V(T)=V(0,T)-V(v(T),T)$ where $v(T)$ is the global minimum of the potential at temperature $T$. The gravitational wave energy density as the function of their frequency $f$ is then given by the sum of the three production modes \cite{Caprini:2015zlo}
\begin{gather}
\Omega_{Coll}\,h^{2}=1.67\times10^{-5}\left(\frac{H_{*}}{\beta}\right)^{2}\left(\frac{\kappa\alpha}{1+\alpha}\right)^{2}\left(\frac{100}{g_{*}}\right)^{\frac{1}{3}}\left(\frac{0.11v_{w}^{3}}{0.42+v_{w}^{2}}\right)S_{Coll}\left(f\right)\\
\Omega_{SW}\,h^{2}=2.65\times10^{-6}\left(\frac{H_{*}}{\beta}\right)\left(\frac{\kappa_{v}\alpha}{1+\alpha}\right)^{2}\left(\frac{100}{g_{*}}\right)^{\frac{1}{3}}v_{w}\,S_{SW}\left(f\right)\\
\Omega_{MHD}\,h^{2}=3.35\times10^{-4}\left(\frac{H_{*}}{\beta}\right)\left(\frac{\kappa_{MHD}\alpha}{1+\alpha}\right)^{\frac{3}{2}}\left(\frac{100}{g_{*}}\right)^{\frac{1}{3}}v_{w}\,S_{MHD}\left(f\right)
\end{gather}
where $S\left(f\right)$ are the known functions parametrising the dependence on frequency (i.e. determining the shape of the curve as a function of frequency), $v_{w}$ is the velocity of the bubble walls and the $\kappa$-parameters denote the fraction of latent heat that is transformed into sources relevant to each production mode. The precise contribution of the different sources of gravitational waves and formulae for $\kappa$ depend on the dynamics of the bubble walls, see \cite{Caprini:2015zlo} for more details. To determine which regime we lie in we must determine whether the bubble walls are relativistic and whether they 'runaway'($\gamma\rightarrow\infty$).
We do not expect runaway walls as our Z' bosons become massive during the transition and it is known that one should not expect runaway bubbles for a transition where gauge bosons gain a mass \cite{Bodeker:2017cim,Croon:2018erz}. A strongly first order phase transition is expected to give highly relativistic bubble walls and so we take $v_{w}=1$. The exact formulae for the $S(f)$ and $\kappa$ (for our regime) are given in Appendix~\ref{appendix:GW}. In this regime the contribution from collision of bubble walls is negligible so we do not include this in our calculations.
We calculate the bubble profile and the action on the profile using BubbleProfiler \cite{Athron:2019nbd}. The nucleation temperatures and parameters $\alpha,\beta$ are shown in Table~\ref{Table:gravPars} for the benchmark points. The gravitational wave profiles are then plotted in Fig.~\ref{fig:gravWaves} along with the sensitivities of current and planned gravitational wave detectors.
As can be seen from the figure, the gravitational waves produced by our model have too low a frequency to probed by aLIGO but would be probed by the next generation of space-based gravitational wave detectors such as LISA, DECIGO and BBO.
\begin{table}
\centering
\scalebox{0.9}{
\begin{tabular}{|c|c|c|c|c|c|}
\hline
&$ T_{\textrm{c}} $ & $\phi_{\textrm{c}} $ & $T_{\textrm{N}}$ & $\alpha$ & $\frac{\beta}{H_{*}} $ \\
\hline
BP1 & 221\, \textrm{GeV} & 750\, \textrm{GeV} & 84.2\,\textrm{GeV} & 0.547 & 129 \\
\hline
BP2 & 622\, \textrm{GeV} & 2313\, \textrm{GeV} & 115.2\,\textrm{GeV} & 5.70 & 63.5 \\
\hline
BP3 & 129\, \textrm{GeV} & 586\, \textrm{GeV} & 30.8\,\textrm{GeV} & 10.1 & 85.4 \\
\hline
BP4 & 449\, \textrm{GeV} & 1150\, \textrm{GeV} & 273.9\,\textrm{GeV} & 0.0698 & 290 \\
\hline
BP5 & 152\, \textrm{GeV} & 802\, \textrm{GeV} & <10 \textrm{GeV} & - & - \\
\hline
\end{tabular}
}
\caption{Table showing the value of various parameters which are relevant to gravitational waves for our benchmark points. as can be seen from our values of $\phi_{c}, T_{c}$, all of our benchmark points leads to a strongly first order phase transition. Note that for the fifth benchmark point the nucleation temperature is very low and our software encounters problems in this area. Hence we were unable to determine the exact nucleation temperature (it may be that the model does not nucleate in this region of parameter space) and so we do not determine the gravitational wave spectrum for this point.}
\label{Table:gravPars}
\end{table}
\begin{figure}
\centerline{\includegraphics[width=0.9\textwidth,keepaspectratio]{Figures/gravWavesNew.pdf}}
\caption{A plot showing the energy density of gravitational waves for the first four benchmark points. The dashed lines represent the sensitivities of current and future gravitational wave detectors: LISA (blue), eLISA (red), BBO (green), DECIGO (black), Einstein Telescope (pink) and aLIGO (brown).}
\label{fig:gravWaves}
\end{figure}
\section{Conclusion \label{sec:conc}}
We have shown in this paper that a classically scale invariant model can evade all current theoretical and experimental constraints and still account for some or all of the observed dark matter abundance of the universe. Such a model is quite an attractive prospect as it is a relatively minimal model which can solve several problems of the Standard model, primarily dark matter and the hierarchy problem, while also producing a gravitational wave signal which would be observable at the next generation of detectors.
In the context of a minimal model presented here, we have not addressed the question of matter-anti-matter asymmetry. One scenario considered in the literature in classically scale-invariant settings~\cite{Khoze:2013oga,Khoze:2016zfi} is to use a version of leptogenesis via sterile neutrino oscillations~\cite{Akhmedov:1998qx,Drewes:2012ma}, though this would require an extension of our minimal model.
On the other hand, it is reasonably straightforward to provide a realisation of the cosmological inflation in the context of these types of models, following the approach of Ref.~\cite{Khoze:2013uia}.
To achieve this we can extend the Higgs portal interactions of the theory \eqref{eq:potential}
to include an additional real-valued singlet field which is also non-minimally coupled to gravity. In the Einstein frame these interactions generate an exponentially flat potential for the canonically normalised singlet field. This, as explained~\cite{Khoze:2013uia}, provides a successful implementation of a slow-roll cosmological inflation, preserving the classical scale invariance of the model and without the need of perturbative unitarization below the Planck scale. In addition, the singlet can be used as a viable scalar-field Dark Matter component, in addition to fermionic Dark Matter discussed above.
\section*{Acknowledgements}
Research of VVK is supported by the UK Science and Technology Facilities Council (STFC) under grant ST/P001246/1 and DLM acknowledges an STFC studentship.
\newpage
\Appendix{RG Equations \label{appendix:RG}}
{\allowdisplaybreaks \begin{align}
\beta_{g_{X}}^{(1)} & =
g_{X}^{3}\\
\end{align}
{\allowdisplaybreaks \begin{align}
\beta_{\lambda_H}^{(1)} & =
+\frac{27}{200} g_{1}^{4} +\frac{9}{20} g_{1}^{2} g_{2}^{2} +\frac{9}{8} g_{2}^{4} -\frac{9}{5} g_{1}^{2} \lambda_H-9 g_{2}^{2} \lambda_H +24 \lambda_{H}^{2} +\lambda_{HX}^{2}+12 \lambda_H \mbox{Tr}\Big({Y_d Y_{d}^{\dagger}}\Big) \nonumber \\
&+4 \lambda_H \mbox{Tr}\Big({Y_e Y_{e}^{\dagger}}\Big) +12 \lambda_H \mbox{Tr}\Big({Y_u Y_{u}^{\dagger}}\Big) -6 \mbox{Tr}\Big({Y_d Y_{d}^{\dagger} Y_d Y_{d}^{\dagger}}\Big) -2 \mbox{Tr}\Big({Y_e Y_{e}^{\dagger} Y_e Y_{e}^{\dagger}}\Big) -6 \mbox{Tr}\Big({Y_u Y_{u}^{\dagger} Y_u Y_{u}^{\dagger}}\Big) \\
\beta_{\lambda_{HX}}^{(1)} & =
\frac{1}{10} \lambda_{HX} \Big(-9 g_{1}^{2} -45 g_{2}^{2} -60 g_{X}^{2}+120 \lambda_H -40 \lambda_{HX} +80 \lambda_{X} +20 |Y_{XD}|^2 +20 |Y_{XU}|^2 +60 \mbox{Tr}\Big({Y_d Y_{d}^{\dagger}}\Big) \nonumber \\
&+20 \mbox{Tr}\Big({Y_e Y_{e}^{\dagger}}\Big) +60 \mbox{Tr}\Big({Y_u Y_{u}^{\dagger}}\Big) \Big)\\
\beta_{\lambda_{X}}^{(1)} & =
2 \Big(10 \lambda_{X}^{2} + 2 \lambda_{X} |Y_{XD}|^2 + 2 \lambda_{X} |Y_{XU}|^2 + 3 g_{X}^{4} -6 g_{X}^{2} \lambda_{X}- |Y_{XD}|^4 - |Y_{XU}|^4 + \lambda_{HX}^{2}\Big)\\
\end{align}}
{\allowdisplaybreaks \begin{align}
\beta_{Y_{XU}}^{(1)} & =
\frac{1}{2} Y_{XU} \Big(2 |Y_{XD}|^2 -3 g_{X}^{2}+ 4 |Y_{XU}|^2 \Big)\\
\beta_{Y_{XD}}^{(1)} & =
\frac{1}{2} Y_{XD} \Big(2 |Y_{XU}|^2 -3 g_{X}^{2} + 4 |Y_{XD}|^2 \Big)\\
\end{align}}
\Appendix{Gravitational wave formulae \label{appendix:GW}}
All formulae here are taken from \cite{Caprini:2015zlo}. The remaining parameters required for calculation of the gravitational wave spectrum are given below. Firstly we begin with the frequency dependence of the sound wave production.
\begin{equation}
S_{SW}(f)=\left(\frac{f}{f_{SW}}\right)^{3}\left(\frac{7}{4+3\left(\frac{f}{f_{SW}}\right)^{2}}\right)^{\frac{7}{2}}
\end{equation}
where
\begin{equation}
f_{SW}=1.9\times10^{-2}\,\textrm{mHz}\,\left(\frac{1}{v_{w}}\right)\left(\frac{\beta}{H_{*}}\right)\left(\frac{T_{*}}{100\,\textrm{GeV}}\right)\left(\frac{g_{*}}{100}\right)^{\frac{1}{6}}
\end{equation}
The frequency dependence of the gravitational wave production by turbulence is given by
\begin{equation}
S_{turb}(f)=\frac{\left(\frac{f}{f_{turb}}\right)^{3}}{\left(1+\frac{f}{f_{turb}}\right)^{\frac{11}{3}}\left(1+\frac{8\pi f}{h_{*}}\right)}
\end{equation}
where
\begin{gather}
f_{turb}=2.7\times10^{-2}\,\textrm{mHz}\,\left(\frac{1}{v_{w}}\right)\left(\frac{\beta}{H_{*}}\right)\left(\frac{T_{*}}{100\,\textrm{GeV}}\right)\left(\frac{g_{*}}{100}\right)^{\frac{1}{6}} \\
h_{*}=16.5\times10^{-3} \,\textrm{mHz}\, \left(\frac{T_{*}}{100\,\textrm{GeV}}\right)\left(\frac{g_{*}}{100}\right)^{\frac{1}{6}}
\end{gather}
The efficiencies of the two processes are given by
\begin{gather}
\kappa_{v}=\frac{\alpha}{0.73+0.083\sqrt{\alpha}+\alpha} \\
\kappa_{turb}=0.05\kappa_{v}
\end{gather}
\newpage
\bibliographystyle{inspire}
|
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In older houses, a trove of lead risks
A worker in protective gear vacuums lead-contaminated paint chips that had been scraped from an exterior wall of a home in Richmond. (Timothy C. Wright for The Washington Post)
By Dima Williams • Photos by Timothy C. Wright
September 26, 2019 at 10:30 a.m. UTC
Several specks, so few and minuscule that millions can cram onto a penny. That's approximately how much lead-laced dust rested on a window well in the master bedroom of Tiffany Dragos's newly purchased farm house in Ashton, Md.
That amount — 440 micrograms per square foot — slightly exceeds what Maryland and many other states consider safe.
Ingested or breathed in, lead — a toxic metal once widely used in paint, pipes and finishes — accumulates in bones and harms the brain and kidneys, among other organs. Lead poisoning especially affects pregnant women and children, permanently impairing their physical, behavioral and mental health.
According to the U.S. Environmental Protection Agency, nearly 90 percent of abodes constructed before 1940 nationwide are likely to have lead. That probability declines for the subsequent decades, but today any home built before 1978, when lead-based paint succumbed to a national ban, is presumed to contain the element, which can also permeate water and soil.
Dragos knew that the 1703 farm house, where she envisions raising kids, could be a trove of lead risks.
"There could be tons of lead in the house or maybe there is only a little," Dragos said. "It was very hard for me to make a decision [to buy] because I was feeling very confused and emotional."
Kitchen remodels are a big (and expensive) undertaking. Make sure you pick the right approach for you.
Under federal statute, home buyers of pre-1978 residences must receive any known information about lead and be granted the opportunity to conduct an inspection. The District of Columbia, where nearly 90 percent of the housing stock emerged before 1978, also decrees that home sellers disclose any orders from local authorities to mitigate lead hazards.
More than a matter when properties switch hands or welcome new renters, lead exposure is triggering growing concerns as science progressively uncovers its deleterious potency and the nation's homes age. Thus, when extant, lead threats necessitate abatement or interim controls.
Lead inspections
In cities around the country, government agencies — some with federal funding — are implementing lead safety initiatives that often kick in when a child is poisoned. But aside from government-mandated efforts, owners of pre-1978 homes wary about lead should inspect for it.
"The rule of thumb that we propose is in an older home, if the finishes look good, test for lead," said Erik Listou, industry instructor and co-founder of the Living In Place Institute. "Because lead is what kept them intact. It is metal; it doesn't degrade."
An EPA-certified lead inspector should carry out the examination, which utilizes an X-ray fluorescence (XRF) tool. Resembling a police radar gun, it detects lead in multiple layers of interior and exterior paint and on any stained features such as cabinets, doors and windows.
"I have inspected houses built in 1900 that still have the original lead-based paint," said John Burnside of Burnside Enterprises in Colorado. "It might be under two or three layers of paint, but it is still there after 120 years."
A check of 150 spots throughout Dragos's three-century-old property in Montgomery County produced 19 lead-positive results.
While states may postulate disparate cutoff quantities, the Department of Housing and Urban Development denotes paint as lead-based if, per square centimeter, it bears 1 milligram or more of the metal.
But this divulges little about hazards. If chipping, flaking or peeling do not plague the paint, it is safe. If they do, harmful lead dust may settle in.
"We worry about friction points creating lead-based dust," said John Overholser, building consultant and ancillary services manager with Top to Bottom Services in Gaithersburg, Md., which worked with Dragos. "The wood-framed windows that are painted. Doors that don't fit right and rub when you shut them. Railings that are getting rubbed down. Things like that can create that lead dust."
Lead dust requires a risk assessment with wipe samples, which Dragos opted for even though that type of analysis often pertains to documented cases of poisoning.
"I decided for my own peace of mind and having the true picture and reality of what I was dealing with, I would rather invest" in it, she said, adding that the expense, at $565, depended on the number of swipes taken.
There are various lead-dust safety thresholds, taking into account the ways different spaces diverge in how susceptible they are to thorough cleaning and how accessible they are to kids. In June, the EPA and HUD tightened those, slashing contamination standards from 40 to 10 micrograms and from 250 to 100 micrograms of lead per square foot for floors and window sills, respectively.
Lead inspections are costly, so home kits offer a substitute. While some contractors favor them to gauge the need for safeguards against lead, Michael Winn, owner of Winn Design + Build in Falls Church, Va., does not recommend home tests for lay use as they require the baring of every coat of paint.
"Most homeowners don't know to do that," Winn said. "So, they may feel like they don't have lead when they actually do."
Even if done right, home kits only reveal the presence of lead, not its amount.
Lead remediation
Most remediation work calls for EPA-licensed professionals and special permits — not only because of health perils but also because of the possibility of unintentionally spreading lead.
Yet if dust holds lead in lower amounts than the postulated levels, thorough and regular cleaning with wet towels and a high-efficiency particulate air (HEPA) vacuum usually tampers any danger.
Making the case for cottage neighborhoods
This is the approach Mari Fontaine, president of New Hampshire-based Community Builders Construction, took when her family lived in a rental with lead. "I just knew that we need to make sure that we didn't create any dust," she said. "We really kept it very, very clean."
Specialized cleaning represents one of the interim controls and "is generally the most common, easiest and cheapest" method to address lead hazards, said Brandon Colunga, senior project environmental specialist with the District's DMY Capitol.
Interim controls are meant as temporary measures that demand periodic monitoring. They also include wet scraping — so that it doesn't create dust — of deteriorating paint and the blanketing of lead-tainted soil from an abraded exterior with mulch or gravel.
Total removal is what the Living In Place Institute's Listou prefers. Anything short of it is a Band-Aid, he said. But it could be quite a costly solution.
Lead projects can range from $3,000 to $10,000 depending on the amount of lead, the size of the affected area and the kind of remedy, among other variables. And, expenses may swell further with any renovations — new doors or a new lick of paint — warranted after the job.
"Our clients have always been a bit shocked by the cost of that," said Susan Isaacs, real estate agent with Compass's Isaacs Team in the District. "I think for a lot of people, the removal and replacement is prohibitive."
Despite the cost, Dragos plans to replace the lead-laced window. "That is going to be the biggest investment," she said, accenting her intention to also gradually take out the rest of the lead-bearing components throughout the Ashton property.
Short of removal
For reasons beyond expense, removal is not always feasible. Historic neighborhoods, for example, may impose rules on what antique features are indispensable. In such instances, other abatement practices allow for what is deemed long-term prevention of hazardous exposure while keeping lead in place.
For some in 55-plus communities, townhouses with elevators meet all their needs
Encapsulation is one option. Sealing any withering paint, a lead encapsulant often also discourages kids from coming in contact with it. "The encapsulant a lot of times contains Bitrex, which is a bitter chemical that prevents the child from wanting to lick on the paint or chew on the paint," said Aaron Whitmore, owner of Richmond-based Blake Contracting.
Another method is enclosure, or the covering of lead-bearing surfaces, including the construction of dry walls and the installation of carpets.
Despite its scope, any lead abatement activity heeds stringent rules. A containment area — a plastic tent over the house when working on the exterior — needs to be constructed. Rigid removal procedures — sealing any waste and disposing of it in regulated landfills — must be followed. Meticulous cleaning should oust any gathered dust — a process that contractors repeat until a follow-up lead inspection indicates no lead hazards.
Hazards, though, can also lurk in a house's plumbing as pipes and fixtures used to be made of lead.
"Lately, there has been a lot of focus on lead pipes and water contamination in lead pipes," Isaacs said.
Of the 105,000 private water service pipes in the District, about 7,500 are made of lead, according to a 2018 fiscal memorandum on the replacement of lead service lines. While the material of roughly 82,000 pipes is unknown, it is assumed to be lead for half of them.
Any mitigation, however, may pose a challenge, especially so for service pipes, or those that carry water from a city's main to a property and lie on both public and private premises. Thus, the District has recently enacted a mandate that coordinates pipe replacements among owners.
Pipes might be hard to inspect and replace but lead-loaded water can be easily tested and filtered through a system affixed to the kitchen's faucet or at the start of an unleaded private service pipe.
Dragos's water proved free of lead. After a lab analysis, so did the soil.
Content with the outcome, she said, "my advice is to just go ahead and invest in the [lead] tests so you know exactly what you are dealing with."
|
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Charitable Trusts and Foundations choose to partner with Toybox because we are a flexible and innovative organisation striving to deal with a massive global issue: the needs of street children.
Partnering with Toybox links your support to some of the most vulnerable children in the world – street children. Your donation will go to the projects and children that need it most and will be put to life-changing work straight away.
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|
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| 9,127
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/* globals getRegistration */
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global.registration = getRegistration();
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"redpajama_set_name": "RedPajamaGithub"
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| 8,652
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Home English Electric Lightning
Supersonic interceptor and jet fighter capable of unrivalled performance and capability during the Cold War era.
Lightning P1A (WG760) taking off at Warton during the Press Day 1955
The English Electric supersonic interceptor (WG760), piloted by Roland Beamont, first flew at Boscombe Down, Wiltshire as the English Electric P1 on 4th August 1954.
Initial designs were led by WEW 'Teddy' Petter although the aircraft is mostly credited to his successor Freddie Page (later Sir Frederick Page and Chairman of the Aircraft Group of BAC).
A unique feature of the design was its vertically staggered engine configuration of two Rolls-Royce Avon turbojets, housed within the fuselage. The aircraft was initially conceived as an interceptor, designed to defend airfields housing Britain's V Force of bombers, comprised of the Avro Vulcan, Handley Page Victor and the Vickers Valiant. It was thought that during the Cold War of the 1960's these could be vulnerable to attack from the air in any future nuclear conflict.
Petter's initial design was for an aircraft capable of Mach 1.5 and he determined that as a consequence a conventional 40° swept wing would be required. A proposal was submitted November 1948, and after the project was provisionally accepted by English Electric, it was given the designation P.1 in January 1949.
On 29th March 1949, the Ministry of Supply granted their approval for work to begin on a more detailed design, as well as the creation of wind tunnel models and a full-size mock-up. The design developed at quite a pace and in the latter part of 1949, the target speed was broadened to Mach 2. This meant that in Petter's opinion, the required wing sweep needed to be increased to 60° with the ailerons moved to the wingtips. Low-speed wind tunnel tests showed vortex issues would be generated by the wing, creating a large downwash on the tailplane. This was quickly resolved by lowering the height of the tailplane.
The project suffered a major blow however, when Petter suddenly resigned in December 1949 after his demands for greater autonomy for his Lightning Design Team were not met. Freddie Page took over as Design Team Leader for the English Electric P.1, for which the Ministry of Supply had issued Specification F23/49 and subsequently expanded the scope of ER103, to include fighter-level manoeuvring.
On 1st April 1950, English Electric then received an official contract for two flying and one static airframe.
The aerodynamicists at the government led Royal Aircraft Establishment (RAE) were deeply sceptical of swept wing concepts and so Short Brothers in Belfast were contracted to produce the Short SB.5. Built between 1950 - 52, the Short SB.5 was to prove the design of both the wing and the tailplane, and to fully assess overall flight handling. It was effectively a low-speed highly unorthodox, adjustable wing research aircraft, designed so that different wing sweep angles could be tested by a single aircraft. After testing a range of wing and tail configurations, on 2nd December 1952 it was agreed that Petter's 60-degree wing sweep specification was indeed the most effective.
Shorts SB5, 2nd aircraft for Lightning P1 high speed stability and control 1953
The English Electric P.1 wing design, combined with 2 x Rolls-Royce Avon engines configured in a unique stack-staggered arrangement, produced an aircraft with a speed capability of Mach 2. Additionally, it also gave the aircraft an unrivalled rate of climb which was often described as being 'a pilot sitting on two rockets'. It should be noted however that the initial flights of the P.1 utilised un-reheated Armstrong Siddeley Sapphire engines due to delays in the development program of the Avon engine.
The first three prototype aircraft were hand-built at the English Electric factory at Salmesbury, Lancashire from 1953 onwards.
Work started on a second group of prototypes, designated English Electric P1.B, and these represented a significant improvement, incorporating extensive alterations to the forward fuselage and the much-awaited reheated Rolls Royce Avon R24R engines. The P1.B aircraft (XA847, XA853 & XA856) also saw the introduction of the iconic and instantly recognisable conical inlet cone as well as variable nozzle reheat, provision for weapons systems that were integrated with the ADC and AI.23 radar. The three earlier prototypes meanwhile were updated and reclassified from P.1 to P1.A.
The two flying prototypes (WG760 & WG763) and the static aircraft (WG765) were fitted with the un-reheated Armstrong Siddeley Sapphire turbojets. This limited the internal fuselage space and meant that fuel capacity was relatively small, resulting in a limited endurance during testing. This also had a knock-on effect that the narrow tyres wore out much faster because of the greater number of take-off and landing phases. Visually, the prototypes looked similar to the production aircraft however, apart from rounded-triangular intakes, short fins and a reduced level of normal operational equipment.
In May 1954, one of the initial prototypes (WG760) was relocated to Boscombe Down for pre-flight ground taxi trials and it flew for the first time on the morning of 4th August 1954, piloted by Roland 'Bee' Beamont.
Not only were the initial flight tests a success, just a week later it officially achieved supersonic flight for the first time, exceeding the speed of sound during only its third flight. In fact, it was later discovered during the data analysis of its first flight some days later that it had already exceeded Mach 1 (1,225 km/h). However, these flights had proven the English Electric P.1 design to be viable although they were restricted to Mach 1.51 (1,850 km/h) due to directional stability limits.
In May 1956, the English Electric P.1B received the official name 'Lightning' said to have been derived to reflect the aircraft's supersonic capabilities.
English Electric P.1B with name for ceremony
The second P1.B prototype (XA847) then made its first flight on 4th April 1957, again with Beamont taking the aircraft to a speed in excess of Mach 1. After a summer of proving and flight tests, he then 'pushed the envelope further' on 25th November 1957, exceeding Mach 2, becoming the first British aircraft to achieve such a speed.
During the early flight trials of the P.1B speeds in excess of 1,000 mph were achieved daily. Whilst the English Electric P.1B was potentially faster than the Fairey Delta FD2 (which broke the air speed record at 1,132 mph on 10th March 1956) it lacked the fuel capacity to provide the 'one run in each direction at maximum speed' in order to qualify for the record.
The first operational English Electric Lightning variant, designated Lightning F.1, was designed as an interceptor and to best perform this mission, emphasis was placed on rate-of-climb, acceleration, and speed, rather than range and so a radius of operation of 150 miles (240 km) from the V bomber airfields was the initial requirement. Aden guns and interchangeable weapon pack were developed alongside onboard radar for missile guidance, search, and range capability.The next two English Electric Lightning variants (the F.1A and F.2) contained only minor changes whilst the English Electric Lightning F.3 was a different story.
It had higher thrust Rolls-Royce Avon 301R engines, a larger more squared-off fin and strengthened inlet cone. This raised the service clearance from Mach 1.7 (2,083 km/h) to Mach 2.0 (2,450 km/h). Changes and deletions on the weapons front were also included and the English Electric Lightning F.3 was the highest performance variant at that point. However, additional performance resulted in higher fuel consumption which in turn reduced the range which proved an issue for some air forces around the world. Although the next variant, the English Electric Lightning F.6, was already in development, there was an urgent need to find an interim solution.
This was found in the English Electric Lightning F.3A which introduced 2 simple solutions. The first and most obvious was a new, non-jettisonable 610-imperial-gallon ventral fuel tank as well as a new, kinked, conically cambered wing leading edge. This second change accommodated a slightly larger leading edge fuel tank, raising the total usable internal fuel by 716 imperial gallons. This increased fuel only solved one of the major customer concerns as the lack of cannon armament was also thought to be a major deficiency as most pilots thought that cannons were most important to fire warning shots during any intercept mission.
Flying for the first time on 6th June 1965, the English Lightning F.6 was the ultimate variant for the RAF and was almost identical to the F.3A apart from its ability to be fitted with 2 x 260-imperial-gallon ferry tanks, mounted on pylons above the wings. In an emergency these tanks could be jettisoned. The modified ventral tank allowed for two ADEN cannons, mounted in the front. This did have a negative effect on the fuel capacity from 610 to 535 imperial gallons although this was accepted by pilots who announced that the English Electric Lightning F.6 was once again a 'real fighter' once more.
3 x English Electric Lightning F6 in formation
The final British Lightning was the F.2A, developed in 1966. Effectively, this was an F.2 upgraded with the cambered wing, the squared fin, and the same ventral tank as the Lightning F.6. It retained the A.I.23 Radar, Firestreak missile, nose cannon and the Avon 211R engines. Although the F.2A lacked the overall thrust of the later variants, it had a much larger range and was used to great effect in low-altitude interception over West Germany.
Often referred to as the 'Export Lightning' the English Electric F.53 Lightning was developed by BAC as a private venture. From the outset, this variant was designed to have a multi-role capability changing between interception, reconnaissance, and ground-attack duties. Based upon the airframe and avionics of the F6, it incorporated large ventral fuel tank, cambered wing, and over-wing pylon mountings for drop tanks. It boasted a capability to carry a large menu of weaponry as well as the almost essential defensive Aden Cannons.
English Electric Lightning F Mk.53 RSAF
A further development was the English Electric Lightning T.55 2-seater, equipped for combat duties. With a fuselage broadly similar to the T.5 variant, it also utilised the wings and large ventral tank from the F.6. Unfortunately, whilst retaining all of the English Electric Lightning's exceptional climb rate and manoeuvrability, it also suffered the same difficulty of maintenance.
By 1963, the newly formed British Aircraft Corporation (which had subsumed English Electric) were working on a design for a two-seat Lightning development with a variable-geometry wing.
In addition to a variable-sweep wing (sweeping between 25 degrees and 60 degrees), the design featured an extended ventral pack, an enlarged dorsal fin fairing, an arrester hook, and a revised inward-retracting undercarriage. Initially conceived as a carrier-based aircraft, the 'VG Lightning' concept was also revised into a land-based interceptor, intended for RAF use. Various alternative engines were proposed, such as the newer Rolls-Royce Spey engine. With air inlets relocated to the sides of the fuselage, this would have seen the end of the famous Lighting nose cone.
The official ceiling of the BAC Lightning was a closely guarded secret although it is said to be in excess of 60,000 ft and it is well renowned for its exceptional rate of climb at 20,000 ft per minute.
Although never credited with an official 'kill' the English Electric / BAC Lightning did shoot down one aircraft when it was called upon to deal with a Hawker Harrier which unintentionally continued to head for the East German border after the pilot had ejected following apparent engine failure.
The BAC Lightning served in 16 variants within the Air Forces of Kuwait and Saudi Arabia as well as continuing in service with the RAF from 1959 until 1988, and some 337 aircraft were built in its 34-year history. Although proposed, the Sea Lightning FAW1 was never built.
Today, it remains one of the most loved and revered of British Fighter aircraft and represents an era of unrestricted noise and power. Although there are a number of preserved examples around the world, sadly there are no longer any English Electric Lightnings in the sky.
Click here for the memories of Lightning Test Pilot Craig Penrice
EE Lightning Image Gallery
EE Lightning F.1 with Roland Beamont Flight Test Crew
EE Lightning F.1 RAF (XM145) ground engine runs at night
EE Lightning F.1 (XM149) landing with chute deployed
EE Lightning F.2 RAF (XN776 & x3 others) fuselage assembly
EE Lightning F.2 Formation
EE Lightning F.2a (XN783), F.3 (XP739), F.3 (XP746) & F.2 (XN779) in formation
EE Lightning F.2a RAF (XN793 & x1 other) in close formation
EE Lightning F.6 RAF (XP697) air to air
EE Lightning F.6 with Beamont at controls 1964
EE Lightning F.6 (XS919) air to air
EE Lightning F.63 Royal Saudi Air Force (53-683 & 53-689) on approach to Dhahran
English Electric P.1A
3 built Single-seat supersonic research aircraft.
English ElectricP.1B
3 built Single-seat operational prototypes to meet Specification F23/49, 20 development aircraft ordered in February 1954 - Type was officially named 'Lightning' in October 1958.
English Electric Lightning F.1
20 built Development batch aircraft, single-seat fighters delivered from 1959. Nose-mounted twin 30 mm ADEN cannon, two Firestreak missiles, VHF Radio and Ferranti AI-23 "AIRPASS" radar.
BAC Lightning F.1A
28 built Single-seat fighter, delivered in 1961. Featured Avon 210R engines, an inflight refuelling probe and UHF Radio.
BAC Lightning F.2
44 built Single-seat fighter (an improved variant of the F.1), delivered in 1962 - 31 later modified to F.2A standard, five later modified to F.52 for export to Saudi Arabia.
31 conversions Single-seat fighter (F.2s upgraded to near F.6 standard); featuring Avon 211R engines, retained ADEN cannon and Firestreak (replaceable Firestreak pack swappable with ADEN Cannon Pack for a total of four ADEN Cannon), arrestor hook and enlarged Ventral Tank for two hours flight endurance.
70 built Single-seat fighter with upgraded AI-23B radar, Avon 301R engines, new Red Top missiles, enlarged and clipped tail fin due to aerodynamics of carriage of Red Top, and deletion of ADEN cannon.
16 built Single-seat fighter with extended range of 800 miles due to large ventral tank and new cambered wings.
BAC Lightning T.4
22 built Two-seat side-by-side training version, based on the F.1A with two aircraft converted to T.5 prototypes and two aircraft later converted to T.54.
22 built Two-seat side-by-side training version, based on the F.3. One former RAF aircraft later converted to T.55 for Saudi Arabia.
39 built Featured new wings with better efficiency and subsonic performance, overwing fuel tanks and a larger ventral fuel tank, reintroduction of 30 mm cannon (initially no cannon but later in the forward part of the ventral pack rather than in the nose), use of Red Top missiles. Nine aircraft were converted from F.3 and 15 from F.3A.
BAC Lightning F.7 Proposed single-seat interceptor featuring variable geometry wings, extended fuselage, relocated undercarriage, underwing hardpoints, cheek-mounted intakes, new radar and use of the Sparrow/Skyflash AAMs. Never built.
25 conversions Slightly modified ex-RAF F.2 single-seat fighters for export to Saudi Arabia.
346 built / 1 conversion Export version of F.6 with pylons for underwing bombs or rocket pods.
42 conversions Ex-RAF T.4 two-seat trainers supplied to Saudi Arabia (two converted).
58 built / 1 conversion Two-seat side-by-side training aircraft (export version of the T.5), 6 for the Royal Saudi Arabian Air Force, two T.55Ks for the Kuwaiti Air Force and one converted from T.5 that crashed before delivery).
BAC Sea Lightning FAW.1 Proposed two-seat Royal Navy Fleet Air Arm carrier capable variant with variable-geometry wing; not built.
Specifications (F.6)
Powerplant 2 x Rolls-Royce Avon 301R after-burning turbojets (16,000lb with AB)
Span 34 ft 10 in (10.6m)
Maximum Weight 45,750lb 20,752kg) take-off weight
Capacity 1 Pilot (1 student & 1 instructor in training variants)
Armament 2 x 30mm Aden cannons with hard point mountings for Air to Air missiles.
Maximum Speed Mach 2.0 (1,300mph / 2,100 kph) at 36,000 ft
Maximum range 850miles (1,370km)
Number built
337 All variants and prototypes
Variant & Serial No
Location and website (If available)
BAC Lightning P.1A
(WG7601st Prototype)
RAF Museum Cosford, England
www.rafmuseum.org.uk/cosford
(WG763 2nd Prototype)
Museum of Science and Industry, Manchester, UK
www.msimanchester.org.uk
P.1B/F.1 (XG329)
Norfolk & Suffolk Aviation Museum, Flixton, UK
www.aviationmuseum.net
P.1B / F.1 (XG337)
F.1A(XM178)
Savigny-les-Beaune, France
www.aviationmuseum.eu
F.1A (XM135)
Imperial War Museum Duxford, UK
www.iwm.org.uk
Thorpe Camp Museum,Tattershall Thorpe, Lincolnshire
www.thorpecamp.wixsite.com
F.2A (XN730)
Luftwaffe Museum, Gatow, Germany
www.kommando.streitkraeftebasis.de
PS Aero Museum, Baarlo, Netherlands
www.psaero.com
Flugausstellung Hermeskeil, Germany
www.flugausstellung.de
National Museum of Flight, East Fortune, UK
www.nms.ac.uk
F.3(XP706)
AeroVenture, Doncaster, UK
www.southyorkshireaircraftmuseum.org.uk
F.3 (XR713)
Bruntingthorpe Aerodrome, Leicestershire, UK
www.bruntingthorpeaviation.com
F.3(XR749)
Score Group Integrated Valve and Gas Turbine Plant, Peterhead, UK
www.score-group.com
RAF Coningsby, Lincolnshire, UK
www.raf.mod.uk/rafconingsby
F.6(XS897 / XP765)
F.6(XS903)
Castle Motors, Liskeard, Cornwall, UK
www.castlemotors.com
BAE Systems, Warton Aerodrome, Lancashire
www.baesystems.com
RAF Museum, London, UK
RAF Manston History Museum, Manston, Kent, UK
www.rafmanston.co.uk
F.6 (S929)
RAF Akrotiri, Cyprus
www.rafakrotiri.co.uk
T.4(XL629)
MoD Boscombe Down, Wiltshire, UK
T.5(XS417)
Newark Air Museum, Newark, UK
Farnborough Air Sciences Trust, Farnborough, UK
www.airsciences.org.uk
Skegness Water Leisure Park, Lincolnshire, UK
www.skegnesswaterleisurepark.co.uk
Cranfield Airport, Bedfordshire, UK
www.cranfieldairport.com
Fenland and West Norfolk Aviation Museum, Wisbech, UK
www.museumsnorfolk.org.uk
F.52(XN770)
Royal Saudi Air Force Museum, Riyadh, Saudi Arabia
www.mod.gov.sa/Services/SaqrAlJazira/Pages/default.aspx
Anglo American Lightning
www.lightning422.com
F.53(53-418)
Kuwait Science and Natural History Museum, Kuwait City
www.ksnhm.weebly.com
F.53(ZF578 / XR753)
Tangmere Military Aviation Museum, Tangmere, UK
F.53(ZF579)
Gatwick Aviation Museum, Charlwood, UK
www.gatwick-aviation-museum.co.uk
BAE Systems, Samlesbury, UK
Bentwaters Cold War Museum, Suffolk, UK
www.bcwm.org.uk
.53(ZF583)
Solway Aviation Museum, Carlisle Airport Cumbria, UK
www.solway-aviation-museum.co.uk
Dumfries and Galloway Aviation Museum, Dumfries, UK
www.dumfriesaviationmuseum.com
East Midlands Airport Aeropark, Castle Donington, UK
www.eastmidlandsaeropark.org
F.53(ZF592 as 53-686)
City of Norwich Aviation Museum, Norwich, UK
www.cnam.org.uk
F.53(ZF594 / XS733)
North East Aircraft Museum, Sunderland, UK
www.nelsam.org.uk
F.53(? )
Abdullah Al-Mubarak Air Base, Kuwait
T.54(XN989)
Main entrance to King Abdul-Aziz Air Base, Dhahran, Saudi Arabia
T.55(55-716)
T.55(ZF598 / 55-713)
Airworthy aircraft
The following aircraft were listed for sale in 2020 and their current location is unknown.
(ZU-BBD / XS452)
Offered for Sale 2020
(ZU-BEW / XR773)
(ZU-BEY / XR693)
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,480
|
08 6311 7080 info@hammondlegal.com.au
Industrial Manslaughter – What you need to know as an employer or employee
In June 2017, the WA Parliament introduced an amendment to the Criminal Code which recognises the offence of industrial manslaughter.
Companies will now be held liable where a worker dies as a result of the company's negligence.
The change is based on the principle that "every worker has the right to a safe and healthy workplace, and all workers are entitled to expect to be able to return home from work safe and well".
An employer, if found guilty, could face imprisonment for life.
Currently the law only holds individual persons responsible for manslaughter.
This change removes the obstacle of holding companies themselves liable for the death of an employee in the workplace.
What is Industrial Manslaughter and Negligence?
Industrial manslaughter will occur when:
A worker dies in the course of employment, or is injured and later dies;
The company's conduct or failure to do an act causes the death;
The company has been negligent;
The company has contributed to the death; and
The company must have known that the conduct would be likely to cause the death but failed to act.
What happens if the employer is part of a corporation?
The new law can result in the company being required to:
Publicise its offence including any penalties imposed;
Undertake a project for improvement of safety and health; and
Pay fines.
The Court can only impose these measures if the cost incurred by the company is less than $5,000,000.00.
What happens if a Corporation fails to comply with the Orders?
A Commissioner can carry out any action required to make a company comply with any orders.
The Commissioner may publicise the failure of a company to carry out the orders of the Court.
Furthermore, the legislation proposes that the costs incurred by the Commissioner in enforcing the Court's Orders will be a debt owed by the corporation to the State.
The new law allows for a tougher stance on safety.
This isn't a new concept either.
The United Kingdom and Australian Capital Territory have similar laws.
It is a tough law.
However, it is fair to expect criminal sanctions to be imposed in a workplace where a company's negligent acts lead to the death of a worker.
This law brings WA into line with current community standards, ensuring companies take a more proactive stance on Health & Safety in the workplace.
We expect to come home from work, our families expect it, and now if you don't, your company may be held liable.
If you have any questions about these changes to the Criminal Code or how it affects you, contact us on (08) 6311 7080 or click here to make an enquiry, today.
Hammond Legal
Level 2, 82 Beaufort Street
Perth, Western Australia 6000
Telephone: +61 (0)8 6311 7080
Facsimile: +61 (0)8 6311 7081
Email: info@hammondlegal.com.au
© 2017 Hammond Legal | All rights reserved
Carolyn Smiddy-Brown
Carolyn graduated from James Cook University in 1996 with an honours degree in law. Since then has worked in North Queensland, Hong Kong, China and Albany, Western Australia.
Carolyn practices in criminal law, dispute resolution and civil, commercial and estate litigation. She has represented clients in a diverse range of matters from murder to complex interstate litigation. She regularly appears in the Supreme Court, District Court and Magistrates Court of Western Australia.
Carolyn is experienced in negotiating and resolving disputes. She strives to provide clients with independent, expert, professional, objective legal advice and representation in order to achieve the best possible outcome.
Janet Burke
Janet joined Hammond Legal in June 2018 as the Practice Manager.
She has worked in the legal industry in medium sized, multi-disciplined law firms for over 10 years. Prior to this Janet worked for Aldi Store UK for 16 years at their head office as part of their finance team.
Janet is enthusiastic, positive and thorough in everything she does and is great at multi-tasking.
Joshua Clarey
In 2017, Josh graduated from the University of Notre Dame with a Bachelor of Laws. He completed his graduate year with barrister Simon Watters assisting on a number of high-profile criminal law jury trials and appeals. Prior to this, he also completed a Bachelor of Commerce at the University of Western Australia with majors in management, marketing and a minor in business law.
Josh completed his Graduate Diploma of Legal Practice through the College of Law and was admitted to practice as a lawyer in 2018.
He is passionate about criminal law and commercial litigation, and has experience in representing clients in drug and traffic offences, domestic violence, and serious assaults.
Greg Boland
Greg Boland was admitted to legal practice in 1980, after graduating from the University of Western Australia with a Bachelor of Jurisprudence and a Bachelor of Laws.
He has spent time in private practice in both city and country firms, has had an extensive career in the public sector and has held many senior roles whilst at Legal Aid WA.
Greg brings a broad range of knowledge to Hammond Legal in the areas of criminal, civil administrative and family law, and also vast experience, appearing in numerous courts all over WA and the Commonwealth territories of Christmas Island and Cocos Islands.
Greg has also served as an elected local government Councillor in the Town of Cottesloe.
Anthony Dique
In 2014, Anthony graduated from Murdoch University with a Bachelor of Laws and since 2016 has been a solicitor in Western Australia.
He practices primarily in civil litigation with experience in appeals, family law, wills and estate planning, employment law, commercial drafting and property law.
Anthony has managed matters in most Courts in Western Australia including the Supreme Court, State Administrative Tribunal and Family Court.
Arfa Shoukat
In 2016, Arfa graduated from Curtin University with a Bachelor of Laws. During her time at Curtin University, she participated in a Summer Exchange Programme at the University of Ghent, Belgium, where she studied three International Law units, namely Human Rights, Public International Law and International Commercial Law. She also has a Graduate Diploma of Legal Practice from the College of Law.
Arfa has a particular interest in the area of personal injuries and has extensive experience in motor vehicle accident claims and criminal injuries compensation. Other than the area of personal injuries Arfa also has experience in a variety of areas including wills and estates, property law, commercial law, and family law. She is also keenly interested in International Law, social justice principles, and pro bono legal work. Arfa has previously volunteered for a community legal centre and prepared wills and other legal documents for the Salvation Army Community Wills Day.
In 2017, Arfa was admitted to the Supreme Court of Western Australia as a lawyer.
Miriam Corbould
Miriam advises in the area of commercial disputes and litigation.
She has advised individuals and companies in respect of contractual claims, debt recovery, negligence and consumer law matters and has extensive experience in negotiating commercial settlements. Miriam has also represented clients in various courts and tribunals in relation to personal injury, employment law and inheritance claims.
In 2014, Miriam graduated from the University of Western Australia with a Bachelor of Laws and a Bachelor of Arts. Upon obtaining a Graduate Diploma of Legal Practice, Miriam was admitted as a lawyer of the WA Supreme Court in 2015.
Admitted to practice in 1985, John has practiced as a barrister and solicitor in Western Australia for over 30 years.
John has extensive experience in civil litigation, commercial disputes, crime, employment law and local government law. He has a vast knowledge of different practice areas and has advised individuals, companies and partnerships in a wide range of commercial litigious and non-litigious matters.
John is an excellent advocate and has appeared as Counsel in most Courts and Tribunals in Western Australia. He has also appeared at the Royal Commissions, local government inquiries and before disciplinary boards.
John is also able to assist with media and public relations, a vital element in the mitigation of damage to personal reputation and corporate risk management.
As Mayor of the Town of Cottesloe (1997 – 2003) and Chairman of the Town's Town Planning Scheme Review Committee (1999 – 2003), John has worked at length with local government officers and planning consultants.
In 2002, John was awarded the Centenary Medal for service to local government.
John is well known for his role as a commentator on the radio station 6PR and on Channel 9.
John is also a Notary Public.
John is a graduate of the University of Western Australia and has a Bachelor of Arts, Bachelor of Laws and Master of Laws degree.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,638
|
using System;
using System.Collections.Generic;
namespace ElevatorSharp.Console
{
internal class FloorArrivalRate
{
internal int Floor { get; set; }
internal TimeSpan StartTime { get; set; }
internal TimeSpan EndTime { get; set; }
public int msPerArrival { get; internal set; }
public TimeSpan LastArrival { get; internal set; }
internal List<int> DestinationWeights = new List<int>();
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,377
|
{"url":"https:\/\/zbmath.org\/?q=an%3A1181.91326","text":"# zbMATH \u2014 the first resource for mathematics\n\nA hidden Markov model of credit quality. (English) Zbl\u00a01181.91326\nSummary: This paper presents a hidden Markov model of credit quality dynamics, and highlights the use of filtering-based estimation methods for models of this kind. We suppose that the Markov chain governing the \u201ctrue\u201d credit quality evolution is hidden in \u201cnoisy\u201d or incomplete observations represented by posted credit ratings. Parameters of the model, namely credit transition probabilities, are estimated using the EM algorithm. Filtering methods provide recursive updates of optimal estimates so the model is \u201cself-calibrating\u201d. The estimation procedure is illustrated with an application to a data set of Standard & Poor\u2019s credit ratings.\n\n##### MSC:\n 91G40 Credit risk 91G70 Statistical methods; risk measures 62P05 Applications of statistics to actuarial sciences and financial mathematics\n##### Keywords:\ncredit quality; filtering; hidden Markov models; EM algorithm\nastsa\nFull Text:\n##### References:\n [1] Altman, E.I., The importance and subtlety of credit rating migration, Journal of banking and finance, 22, 1231-1247, (1998) [2] Altman, E.I.; Kao, D.L., Rating drift in high-yield bonds, Journal of fixed income, 2, 15-20, (1992) [3] Bangia, A.; Diebold, F.X.; Schuermann, T., Rating migration and the business cycle, with application to credit portfolio stress testing, Journal of banking and finance, 26, 445-474, (2002) [4] Basel Committee on Banking Supervision, 2000. Credit ratings and complementary sources of credit quality information. BCBS Publications No. 3, Bank for International Settlements. [5] Basel Committee on Banking Supervision, 2001. A new capital adequacy framework, Basel. [6] Basel Committee on Banking Supervision, 2005. Studies on validation of internal rating systems. BCBS Publications No. 14, Bank for International Settlements. [7] Carey, M.; Hrycay, M., Parametrizing credit risk models with rating data, Journal of banking and finance, 25, 197-270, (2001) [8] Delianedis, G., Geske, R., 2003. Credit risk and risk neutral default probabilities: information about rating migrations and defaults. In: EFA Annual Conference Paper No. 962. Available at SSRN $$\\langle$$http:\/\/ssrn.com\/abstract=424301\u27e9. [9] Elliott, R.J.; Krishnamurthy, V., New finite-dimensional filters for parameter estimation of discrete-time linear Gaussian models, IEEE transactions on automatic control, 44, 938-951, (1999) \u00b7 Zbl\u00a00959.93055 [10] Elliott, R.J.; Aggoun, L.; Moore, J.B., Hidden Markov models. estimation and control, (1995), Springer Berlin \u00b7 Zbl\u00a00819.60045 [11] Krishnamurthy, V.; Chung, S.H., Signal processing based on hidden Markov models for extracting small channel currents, () [12] Lando, D.; Sk\u00f8deberg, T., Analyzing rating transitions and rating drift with continuous observations, Journal of banking and finance, 26, 423-444, (2002) [13] L\u00f6ffler, G., An anatomy of rating through the cycle, Journal of banking and finance, 28, 695-720, (2004) [14] L\u00f6ffler, G., Avoiding the rating bounce: why rating agencies are slow to react to new information, Journal of economic behavior and organization, 56, 365-381, (2004) [15] Shumway, R.H.; Stoffer, D.S., Time series analysis and its applications with R examples, (2006), Springer Berlin \u00b7 Zbl\u00a00502.62085 [16] Standard & Poor\u2019s, 2000. Standard & Poor\u2019s COMPUSTAT (North America) User\u2019s Guide. Standard & Poor\u2019s, Englewood, CO. [17] Wendin, J.; McNeil, A.J., Dependent credit migrations, Journal of credit risk, 2, 87-114, (2006)\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.","date":"2022-01-18 19:38:49","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.46547386050224304, \"perplexity\": 11479.781787358193}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-05\/segments\/1642320300997.67\/warc\/CC-MAIN-20220118182855-20220118212855-00030.warc.gz\"}"}
| null | null |
{"url":"https:\/\/www.sanfoundry.com\/physics-questions-answers-current-electricity-temperature-dependence-resistivity\/","text":"# Physics Questions and Answers \u2013 Current Electricity \u2013 Temperature Dependence of Resistivity\n\n\u00ab\n\u00bb\n\nThis set of Physics Multiple Choice Questions & Answers (MCQs) focuses on \u201cCurrent Electricity \u2013 Temperature Dependence of Resistivity\u201d.\n\n1. Identify the material which is suitable for making standard resistors.\na) Silver\nb) Copper\nc) Constantan\nd) Germanium\n\nExplanation: Alloys like constantan or manganin are used for making standard resistance coils due to their high resistivity values and very small temperature coefficient.\n\n2. What is the unit of the temperature coefficient of resistance (\u03b1)?\na) oC\nb) oC-1\nc) $$\\frac {\\Omega}{ ^{\\circ} C}$$\nd) $$\\frac { ^{\\circ} C}{\\Omega}$$\n\nExplanation: The temperature coefficient of resistivity is defined as the increase in resistivity per unit resistivity per degree rise in temperature.\nThe unit of the temperature coefficient of resistance is oC-1.\n\n3. Which of the following relation is significant for metals when the temperature increases?\na) Resistivity increases and conductivity decreases\nb) Resistivity decreases and conductivity decreases\nc) Resistivity and conductivity do not change with temperature\nd) Temperature dependence is non-linear\n\nExplanation: The resistivity of a metal increases and the conductivity decreases with the increase in temperature. With an increase in temperature, the free electrons collide more frequently with the metal ions. The mean collision time also decreases.\n\n4. Identify the type of material based on the T-\u03c1 graph given below.\n\na) Silicon\nb) Polymer\nc) Nichrome\nd) Copper\n\nExplanation: For metals, the temperature coefficient of resistivity is positive. At lower temperatures, the resistivity of a pure metal increases as a higher power of temperature. So, the answer is copper, which is a metal.\n\n5. Which among the following has weak temperature dependence values with resistivity?\na) Silver\nb) Copper\nc) Nichrome\nd) Germanium\n\nExplanation: Alloys have high resistivity. The resistivity of nichrome has weak temperature dependence. At absolute zero, a pure metal has negligibly small resistivity while an alloy like nichrome has some residual resistivity.\n\n6. The resistivity of semiconductors and insulators decreases linearly with the increase of temperature.\na) True\nb) False\n\nExplanation: The resistivity of semiconductors and insulators decreases exponentially with the increase in temperature. This is because the number density of free electrons increases exponentially with the increase in temperature.\n\n7. The resistivity of \u2018X\u2019 decreases with temperature and its coefficient of resistivity is negative. Identify X.\na) Silver\nb) Silicon\nc) Copper\nd) Nichrome\n\nExplanation: The coefficient of resistivity is negative for semiconductors and their resistivity decreases with temperature. The relaxation time does not change with temperature but the number density of free electrons increases exponentially with the increase in temperature. Consequently, the resistivity decreases exponentially with the increase in temperature.\n\n8. A wire has a resistance of 5.5 \u03a9 at 19oC and 21.5 \u03a9 at 200oC. Find the temperature coefficient of resistivity(\u03b1) of the material.\na) 0.016 oC-1\nb) 0.160 oC-1\nc) 1.600 oC-1\nd) 16.00 oC-1\n\nExplanation: Temperature coefficient \u03b1=$$\\frac {(R_2-R_1)}{R_2(T_2-T_1)}$$.\n\u03b1 = $$\\frac {(21.5-5.5)}{(5.5(200-19))}$$\n= 0.01607 oC-1.\n\n9. Which of the following is not a valid reason for using alloys to make standard resistors?\na) Alloys have a high value of resistivity\nb) They are least affected by air and moisture\nc) Alloys have a large temperature coefficient\nd) Their contact potential with copper is small\n\nExplanation: Alloys have a high value of resistivity. They have a very small temperature coefficient. So their resistance does not change appreciably even for several degrees rise of temperature. That leaves the answer \u2013 alloys have a large temperature coefficient.\n\n10. Identify the temperature at which the resistance of copper would be double of its resistance at oC. Given \u03b1 (temperature coefficient of resistivity) for copper=3.9 x 10-3 oC-1.\na) 125oC\nb) 256oC\nc) 1080oC\nd) 273oC\n\nExplanation: \u03b1=$$\\frac {(R_2-R_1)}{R_2(T_2-T_1)}$$.\n\u03b1=$$\\frac {(2R_0-R_0)}{R_0(T-0)} = \\frac {1}{T}$$.\nT=$$\\frac {1}{\\alpha}$$\n= $$\\frac {1}{3.9}$$ x 10-3 oC-1\n= 256oC.\nTherefore, the required temperature is 256oC.\n\nSanfoundry Global Education & Learning Series \u2013 Physics \u2013 Class 12.\n\nTo practice all areas of Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.\n\nParticipate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!","date":"2021-11-28 01:48:33","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5903176069259644, \"perplexity\": 2198.1755241709516}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964358443.87\/warc\/CC-MAIN-20211128013650-20211128043650-00315.warc.gz\"}"}
| null | null |
Q: How to execute the find command in a shell script without getting an error message "missing argument to `-execdir'"? This works fine:
$ find /home/me/folder -type f -iname "*.rar" -execdir unrar x -o- {} \;
However for this script it does not work:
#!/bin/bash
find /home/me/folder -type f -iname "*.rar" -execdir unrar x -o- {} \;
Can you give me a hint why this error message appears?
$ ./unrar_script.sh
find: missing argument to `-execdir'
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 1,247
|
{"url":"https:\/\/www.ssccglapex.com\/the-average-of-6-observations-is-45-5-if-one-new-observation-is-added-to-the-previous-observations-then-the-new-average-becomes-47-the-new-observation-is\/","text":"### The average of 6 observations is 45.5. If one new observation is added to the previous observations, then the new average becomes 47. The new observation is :\n\nA. 58 B. 56 C. 50 D. 46 Answer: Option B\n\n### Solution(By Apex Team)\n\n$\\begin{array}{l}\\text{Let the new observation be = x}\\\\ \\text{According to the question,}\\\\ \\text{6 \u00d7 45.5 + x = 7 \u00d7 47}\\\\ \\text{273.0 + x = 329}\\\\ \\text{x = 329 \u2013 273}\\\\ \\text{x = 56}\\end{array}$\n\nA. 20\nB. 21\nC. 28\nD. 32\n\nA. 18\nB. 20\nC. 24\nD. 30\n\nA. 10 years\nB. 10.5 years\nC. 11 years\nD. 12 years","date":"2021-06-20 19:32:18","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8941103219985962, \"perplexity\": 4233.184080654811}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-25\/segments\/1623488253106.51\/warc\/CC-MAIN-20210620175043-20210620205043-00028.warc.gz\"}"}
| null | null |
<?php
namespace CachetHQ\Cachet\Commands\User;
final class InviteTeamMemberCommand
{
/**
* The invte emails.
*
* @var string
*/
public $email;
/**
* The validation rules.
*
* @var string[]
*/
public $rules = [
'emails' => 'required|array|email',
];
/**
* Create a new invite team member command instance.
*
* @param array $email
*
* @return void
*/
public function __construct($emails)
{
$this->emails = $emails;
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,105
|
\section{Introduction}
\label{sec:intro}
Modern collider experiments such as the Large Hadron Collider (LHC) at CERN depend heavily on the modelling of particle collisions and simulations of detector response to examine physics processes within the experiments. This modelling is used to construct different possible outcomes from particle collisions, used both for the identification of certain physical processes, and for the construction of event backgrounds. Consequently, such simulations play a crucial role in modern high energy physics, and are usually carried out by Monte Carlo event generators such as \textsc{Pythia}~\cite{Pythia}, \textsc{Herwig}~\cite{Herwig} and \textsc{Sherpa}~\cite{Gleisberg:2003xi}.
The theoretical description of LHC events can be highly complex. In a typical event, hundreds of particles are produced as a result of the evolution of an event from the collision of two protons to the formation of long-lived hadrons, leptons and photons. The collision process can be separated into several stages. The protons consist of many partons, each carrying a fraction of the total proton energy. When protons collide, two of their partons can interact with each other via a large momentum transfer, thereby giving rise to the so-called hard interaction. In this part of the collision, large interaction scales are probed, possibly accessing new physics. However, if color-charged particles are produced during the hard interaction process, they are likely to emit further partons. This results in a parton shower, providing a mechanism that evolves the process from the hard interaction scale down to the hadronisation scale $\mathcal{O}(\Lambda_{\mathrm{QCD}})$, where non-perturbative processes rearrange the partons into colour-neutral hadrons.
The hard interaction and the parton shower are the two parts of the event evolution that can be described perturbatively and largely independently of non-perturbative processes, as a result of the factorisation theorem \cite{Collins:1989gx}. In addition, they are by far the most time-consuming parts of the event simulation and pose, therefore, the bottleneck in the generation of pseudo-data for ongoing measurements at the LHC.
While a speed improvement in calculating the hard process and the parton shower is crucial for the interpretation of high-energy collision experiments, the conceptual methods used to calculate either of these two parts are distinctly different. For a mathematical description of the hard interaction, scattering matrix elements are calculated, which nowadays rely on helicity amplitude methods to cope with the ever-increasing complexity of the partonic scattering process \cite{Parke:1986gb,Berends:1987me}.
Instead, the parton shower is technically implemented through a Markov chain algorithm ordered in some measure of showering time $t$, where splitting functions define the probability for a parton to branch into two partons and Sudakov factors \cite{Sudakov:1954sw} determine the probability for the system not to change between two shower times\footnote{For more details see \cite{Buckley:2011ms} and references therein.} $t_\mathrm{in}$ and $t_\mathrm{end}$.
Recent developments in combining helicity amplitudes with the parton shower have shown to improve the theoretical description of scattering events including multiple jets \cite{Catani:2001cc,Mangano:2001xp,Lonnblad:2001iq,Hoche:2015sya, Fischer:2017yja,Fischer:2017htu}, in hypothesis testing \cite{Prestel:2019neg,Soper:2014rya} and in particular in the construction of spin-dependent parton showers \cite{Richardson:2018pvo}.
With practical quantum computers becoming available, there has been growing interest in harnessing the power and advantages that these machines may provide. This interest extends to applying the abilities of quantum computers to describe processes in field theories, with the hope of exploiting the intrinsic `quantumness' of these novel machines to calculate quantum phenomena efficiently. Current quantum computers are divided into two classes: quantum annealers and universal gate quantum computers (GQC). The former is based on the adiabatic theorem of quantum mechanics to find the ground state of a complex system. Quantum annealers perform continuous-time quantum computations and are therefore well-suited to study the dynamics of quantum systems, even quantum field theories \cite{Abel:2020ebj,Abel:2020qzm}, and in solving optimisation problems, e.g. applied to Higgs phenomenology \cite{MottQuantum}. However, they are not universal.
Despite their severe limitation due to the relatively small number of qubits of current machines, GQC are a popular choice for the implementation of algorithms to calculate multi-particle processes \cite{Jordan:2011ci,Garcia-Alvarez:2014uda,Jordan:2014tma,Jordan:2017lea,Preskill:2018fag,bauer2019quantum,Moosavian:2019rxg,Alexandru:2019ozf, Alexandru:2019nsa, Lamm:2019uyc, Lamm:2020jwv}, often with field theories mapped onto a discrete quantum walk \cite{Marque-Martin:2018PRA,Arrighi:2018PRA,Jay:2019PRA,DiMolfetta:2020QIP} or a combined hybrid classical/quantum approach \cite{Lamm:2018siq,Harmalkar:2020mpd,Wei:2019rqy,Matchev:2020wwx}.
Here, we aim to provide a first step towards a generic implementation of quantum algorithms, applicable to QGC devices, for the most time-consuming parts of the event generation in high-energy collisions, i.e. the calculation of the hard process in terms of helicity amplitudes and the simulation of the parton shower\footnote{A first implementation of a parton shower algorithm was provided in \cite{bauer2019quantum}, where interference effects in the parton shower evolution were studied.}.
As depicted in Fig.~\ref{fig:encoding}, QC calculations proceed in general in three stages: i) encoding of the initial state, i.e. an initial wavefunction, using a specific representation of the problem, ii) applying unitary operations on this state, which on a GQC is realised through circuits, and iii) measuring a specific property of interest, i.e. a projection onto the final state vector.
\begin{figure}[!h]
\centering
\includegraphics[width=0.8\textwidth]{Figures/encode.pdf}
\caption{Schematic setup of generic quantum computing calculations with the following steps: (i) encoding of the initial state, (ii) the application of (unitary) operations and (iii) the measurement of the transformed state.}
\label{fig:encoding}
\end{figure}
Following this structure, we will elucidate how the calculation of a hard process in terms of helicity amplitudes or the parton shower can be performed using a GQC. Specifically, we use the IBM Q Experience~\cite{IBMQ}, which provides access to a range of public access quantum computers and a 32-qubit Quantum Simulator~\cite{32_sim}. We have designed the circuits with a focus on limiting the number of qubits needed to perform the calculations. While our code can be run on a real quantum device, the current quantum machines cannot outperform classical computers. The quantum circuits presented here, therefore, serve as a template and nucleus for future developments.
This paper is organised as follows: in Sec.~\ref{sec:helicity}, we motivate and detail the implementation of our QC algorithm for helicity amplitudes for a 1$\rightarrow$2 process and a 2$\rightarrow$2 scattering process, Sec.~\ref{sec:parton} contains the description of our 2-step parton shower algorithm and Sec.~\ref{sec:conclusions} offers a summary and conclusions.
\section{Helicity amplitude algorithm}
\label{sec:helicity}
Scattering processes are calculated using conventional techniques by squaring the scattering amplitude and then performing a sum of all possible helicity processes using trace techniques. For a process with $N$ possible Feynman diagrams, this results in $N^2$ terms in the squared amplitude. Therefore, for processes with a large number of Feynman diagrams, such calculations become extremely complicated. In contrast, helicity amplitude calculations provide a more efficient way of calculating such processes, as one calculates the amplitude for a specific helicity setup. The different helicity combinations do not interfere, and therefore the full amplitude can be obtained by summing the squares of all possible helicity amplitudes.
Helicity amplitude calculations are based on the manipulation of helicity spinors. As the Lorentz group Lie algebra can be written as the direct sum of two $SU(2)$ sub-algebras, i.e. $so(3,1)= su(2) \oplus su(2)$, there are two specific complex representations each specified by two degrees of freedom which solve the massless Weyl equation: a right-handed Weyl spinor, associated with the representation $(\frac{1}{2},0)$, and a left-handed Weyl spinor, associated with the representation $(0,\frac{1}{2})$.
Consequently and for concreteness, the helicity spinor $\vert p \rangle^{\dot{a}}$ for a massless state can be chosen to be expressed as
\begin{align}
\vert p \rangle^{\dot{a}}=\sqrt{2E}\begin{pmatrix}
\cos\frac{\theta }{2}\\
\sin\frac{\theta }{2}e^{i\phi }
\end{pmatrix},
\label{eq:spinor}
\end{align}
associated with momentum $p^\mu$ and energy $E$, such that $p^\mu p_\mu=-m^2$ using the $\eta _{\mu \nu }$=diag(-1, +1, +1, +1) metric convention.
This spinor is parametrised by the angles $\theta$ and $\phi$, where the other spinors $\langle p \vert_{\dot{a}}$, $| p ]_a$ and $[ p |^a$ are related by $p_{a\dot{b}}=- | p ]_a \langle p \vert_{\dot{b}}$ and $p^{\dot{a}b} = - \vert p \rangle^{\dot{a}} [ p |^b$. The correspondence between the two-dimensional helicity spinors and four-component Dirac spinors associated with Feynman rules is demonstrated in Appendix~\ref{sec: DiracHelicity}.
To facilitate and implement such calculations on a GQC, we use \emph{qubits}, the quantum analogue of the \emph{bit} for classical computation. The state of the qubit is defined on a two-dimensional complex vector space with states $\vert 0 \rangle$ and $\vert 1 \rangle$ forming the orthonormal basis for this space. A qubit can thus be formed by a linear superposition of these orthonormal basis states. By considering a general qubit parametrized by two angles
\begin{align}
\vert \psi \rangle=\cos\frac{\theta }{2}\vert 0 \rangle+e^{i\varphi }\sin\frac{\theta }{2}\vert 1 \rangle=\begin{pmatrix}
\cos\frac{\theta }{2}\\
\sin\frac{\theta }{2}e^{i\phi }
\end{pmatrix},
\label{eq:qubit}
\end{align}
we can represent the qubit on a three-dimensional unit sphere called the Bloch sphere. Performing unitary operations on qubit states corresponds to rotating states in the Bloch sphere.
Remarkably, comparing Eqs.~(\ref{eq:spinor}) and (\ref{eq:qubit}), helicity spinors can be represented through a qubit, modulo an overall normalisation factor $\sqrt{2E}$, and the calculation of helicity amplitudes follows the identical structure shown in Fig.~\ref{fig:encoding}, i.e. quantum operators act on an initial state to eventually perform the projection onto a final state.
In contrast to classical computers, where all numerical quantities are converted into a binary system representation, on which an algorithm is applied, and then transformed back into quantities that can be understood in terms of a numerical result, in a quantum computing algorithm, the helicity spinor is a faithful representation of the object the circuit directly operates on. The spinors can be directly represented as vectors on the Bloch sphere, which provides the most efficient encoding of the state on which the algorithm operates.
This indicates that GQC provide an ideal framework for the calculation of helicity amplitudes.
Consequently, we will exploit that the spinors used to calculate helicity amplitudes naturally live in the same representation space as qubits. This motivates the manipulation of the direct correspondence of the $\theta$ and $\phi$ variables of the qubit states and helicity spinors to represent the spinors on a quantum circuit. We further encode operators acting on spinors as quantum circuits of unitary operations. These can be applied to qubits (rotating vectors on the Bloch sphere) to calculate helicity amplitudes. The helicity spinors $\vert p \rangle^{\dot{a}}$,$(\langle p \vert_{\dot{a}})^\textrm{T}$, $| p ]_a$ and $([ p |^a)^\textrm{T}$ are visualised for $\theta=\pi/4$, $\phi=\pi/2$, $E=1/2$, as vectors on the Bloch sphere in Fig.~\ref{fig: bloch}, in direct analogy to their respective qubit representation.
This study aims to create the basic building blocks to encode spinor helicity calculations on a quantum circuit.
These basic building blocks are then used to construct quantum algorithms for two simple examples of helicity calculations: i) the contraction of an external polarisation vector corresponding to a $g\rightarrow q\bar{q}$ vertex and ii) the construction of s and t-channel amplitudes for a $q\bar{q}\rightarrow q\bar{q}$ process with identical initial and final quark flavours. `Helicity registers' are crucially introduced into these circuits to control the helicity of each particle involved. In addition, we introduce a superposition state between the helicity qubits of $\vert + \rangle =\vert 1\rangle$ and $\vert - \rangle =\vert 0 \rangle$ by applying Hadamard gates to the helicity registers. In doing so, we can calculate both helicities of each particle involved simultaneously, thus fully utilising the quantum nature of the computation. This advantage is further exploited by the simultaneous computation of s and t-channel amplitudes for the $q\bar{q}\rightarrow q\bar{q}$ process.
This section is organised as follows: a description of the quantum circuit for the 1$\rightarrow$2 process of $g\rightarrow q\bar{q}$ is given in Sec.~\ref{sec:vertex}, together with a comparison of the results of the algorithm as run on a real machine and a simulator, the quantum circuit and the results for the 2$\rightarrow$2 process of $q\bar{q}\rightarrow q\bar{q}$ are given in Sec.~\ref{sec:qqqq}, and a brief discussion of the generalisation of the algorithm to 2$\rightarrow n$ processes follows in Sec.~\ref{sec: helicityFuture}.
\begin{figure}[ht!]
\centering
\begin{subfigure}{0.24\textwidth}
\centering
\includegraphics[scale = 0.4]{Figures/Uqarr_Bloch.png}
\subcaption{$\vert p \rangle^{\dot{a}}$}
\end{subfigure}
\begin{subfigure}{0.24\textwidth}
\centering
\includegraphics[scale = 0.4]{Figures/Uqsq_Bloch.png}
\subcaption{$\vert p ]_a$}
\end{subfigure}
\begin{subfigure}{0.24\textwidth}
\centering
\includegraphics[scale = 0.4]{Figures/Uparr_Bloch.png}
\subcaption{$(\langle p \vert_{\dot{a}})^\textrm{T}$}
\end{subfigure}
\begin{subfigure}{0.24\textwidth}
\centering
\includegraphics[scale = 0.4]{Figures/Upsq_Bloch.png}
\subcaption{$([p \vert^{a})^\textrm{T}$}
\end{subfigure}
\caption{A visualisation of the helicity spinors $\vert p \rangle^{\dot{a}}$,$\langle p \vert_{\dot{a}}$, $(| p ]_a)^\textrm{T}$ and $([ p |^a)^\textrm{T}$ for $\theta=\pi/4$, $\phi=\pi/2$, $E=1/2$ on the Bloch sphere, following the choice of representation of Eq.~(\ref{eq:spinor}).}
\label{fig: bloch}
\end{figure}
\subsection{Constructing helicity spinors and scalar products on the Bloch sphere}
The helicity spinors have been implemented on the quantum circuit by constructing Bloch sphere representations, like the ones shown in Fig.~\ref{fig: bloch}. The helicity spinor decompositions are outlined in detail in Appendix~\ref{sec: helicityAmpGates}. They utilise the Qiskit $U_3 (\theta, \phi, \lambda)$ gate, which applies a rotation to a single qubit. The rotation is defined by,
\begin{equation}
U_3 (\theta, \phi, \lambda) = \begin{pmatrix} \cos \big( \frac{\theta}{2} \big) & - e^{i\lambda} \sin \big( \frac{\theta}{2} \big) \\ e^{i\phi} \sin \big( \frac{\theta}{2} \big) & e^{i(\phi + \lambda)} \cos \big( \frac{\theta}{2} \big) \end{pmatrix}.
\end{equation}
A simple $U_3$ gate acting on a $\vert 0 \rangle$ state has been used to create the $\vert q \rangle^{\dot{a}}$ spinor, where $\theta $ and $\phi $ variables of the $U_3$ gate corresponded to the $\theta $ and $\phi $ variables of the helicity spinor. The $\vert q ]_a$ spinor has been created by sequentially applying a $U_3^\dagger$ rotation and a $NOT$ gate, where here the $\theta $ and $ \lambda $ variables of the $U_3$ gate corresponded to the $\theta $ and $\phi $ variables of the $\vert q ]_a$ spinor.
To construct the scalar products $\langle pq \rangle$ or $[pq]$ on a quantum computer, $2\times2$ unitary gates $U_{\langle p}$ and $U_{[p}$ were created such that, when they act on the $\vert q \rangle^{\dot{a}}$ and $\vert q]_a$ spinors respectively, the scalar product values correspond to the first component of the final qubit state, i.e. the complex coefficient associated with the $\vert 0 \rangle$ state. It should be noted that the factors of $\sqrt{2E}$ in the definition of the helicity spinors have not been accounted for such that the spinor-qubit states are normalized to one on the quantum register. As a consequence, these factors must be added after the results have been obtained from the quantum computer.
\subsection{1$\rightarrow$2 amplitude calculation}
\label{sec:vertex}
A simple application of the helicity amplitude approach is the calculation of a 1$\rightarrow$2 process. Here we will consider the process of $q \rightarrow g\overline{q}$ by calculating the $gq\overline{q}$ vertex,
\begin{align}
\mathcal{M}_{gq\overline{q}} &= \langle p_f \vert \bar{\sigma }_{\mu } \vert p_{\overline{f}} ] \epsilon^\mu_\pm,
\end{align}
where $p_f$ and $p_{\overline{f}}$ are the momenta associated with the fermon and anti-fermion respectively. The gluon polarisation vectors are defined as~\cite{elvang_huang_2015},
\begin{align}
\epsilon^\mu_+ &= - \frac{\langle q \vert \bar{\sigma }^{\mu } \vert p ]}{\sqrt{2} \langle q p \rangle}, &\epsilon^\mu_- = - \frac{\langle p \vert \bar{\sigma }^{\mu } \vert q ]}{\sqrt{2} [q p]}.
\label{eq:polarisation}
\end{align}
From this, it is possible to create a circuit where each four-vector present in the amplitude, i.e. the fermion anti-fermion vertex and polarisation vector, is calculated individually on a series of 4 qubits. This is done by using the corresponding Pauli gates for each four-vector component on each qubit. However, this will lead to a large circuit depth due to the number of gates required to do such a calculation. Therefore it is useful to simplify the expression for the amplitude using the Fierz identity,
\begin{equation}\label{eqn: Fierz}
\langle p \vert \bar{\sigma }^{\mu } \vert q ] \langle k \vert \bar{\sigma }_{\mu }\vert l] = 2 \langle p k \rangle [q l ].
\end{equation}
With this, the amplitude for the $gq\overline{q}$ vertex becomes
\begin{align}\label{eqn: gqqbarAmp}
\mathcal{M}_+ &= -\sqrt{2} \frac{\langle p_{f} q \rangle [ p_{\overline{f}} p ]}{ \langle q p \rangle}, &\mathcal{M}_- = - \sqrt{2} \frac{\langle p_f p \rangle [p_{\overline{f}} q]}{ [q p]}.
\end{align}
As a consequence of this simplification, the number of qubits needed to calculate the amplitude on the quantum computer can be reduced from 10 to 4. The circuit for calculating this amplitude is shown in Fig.~\ref{fig: gqqbarCircuit}. The three $q_i$ qubits calculate the three scalar products from Eq.~(\ref{eqn: gqqbarAmp}) using the gate decompositions outlined in Appendix~\ref{sec: helicityAmpGates}. These rotation gates are controlled from the helicity register, $h$. If $h$ is in the $\vert 1 \rangle$ state, then the helicity is positive and the $\mathcal{M}_+$ amplitude is calculated; if $h$ is in the $\vert 0 \rangle$ state, then the helicity is negative and the $\mathcal{M}_-$ amplitude is calculated. The three calculation qubits, $q_i$, are then measured by the quantum machine.
\begin{figure}[ht!]
\centering
\includegraphics[scale = 0.5]{Figures/gqqbar1.pdf}
\caption{$gq\overline{q}$ vertex circuit. The amplitude for the process is calculated on the $q_i$ qubits, which are controlled from the helicity register. The $q_i$ qubits are then measured by the quantum computer.}
\label{fig: gqqbarCircuit}
\end{figure}
Figure~\ref{fig:gqqbarResults} shows the results of the algorithm for a random selection of small scattering angles, with runs on the IBM Q 32-qubit Quantum Simulator~\cite{32_sim} and the IBM Q 5-qubit Santiago Quantum Computer~\cite{ibmq_santiago}; both of which have been compared to theoretical predictions of the probability distributions extrapolated directly from analytic calculations of the helicity amplitude, calculated using the S@M software~\cite{zbMATH05804587}. The simulator has been run without a noise profile for 10,000 shots. The results agree well with theoretically predicted values, to within 1$\sigma$. From these distributions, one can determine the helicity setup of the process and consequently reconstruct the helicity amplitudes.
The Santiago machine has been run on the maximum shot setting of 8192 for 100 runs, leading to a total of 819,200 shots of the algorithm. Figure~\ref{fig:gqqbarResults} shows that the quantum computer's performance does not match that of a perfect machine, as expected. Therefore, the simulator is rerun with the noise profile of the Santiago device and a comparison between this and the quantum computer is shown and discussed in Appendix~\ref{app:helicityAmpCalc}.
The results from the quantum computer, shown in Fig.~\ref{fig:gqqbarResults}, have been achieved by isolating the individual helicity processes on the quantum circuit, and removing the superposition between the positive and negative processes. The full amplitude is achieved through the implementation of a Hadamard gate on the helicity qubit, which puts the system into a superposition state of the positive and negative processes. The helicity of the process is then determined by measuring the helicity register. The qubit setup chosen here has been used in order to best reduce the \textit{CNOT} qubit errors and limits the number of \textit{SWAP} operations needed in the algorithm. The Santiago machine is a 5-qubit quantum computer, with all qubits connected inline to their adjacent qubit. The helicity qubit, $h$, from Fig.~\ref{fig: gqqbarCircuit} has been assigned to qubit 4 on the Santiago machine, with the $q_i$ qubits on the 2nd, 3rd and 5th qubits of the Santiago machine. The optimum qubit setup would have the $h$ qubit fully connected to the $q_i$ qubits, thus fully minimising the \textit{SWAP} operation errors. However, the available machines with such a qubit mapping on the public IBM Q experience have a lower quantum volume than the Santiago machine, which reports a quantum volume of 32. Consequently, the trade of ideal qubit mapping for a better quantum volume has been made.
\begin{figure}[ht!]
\centering
\begin{subfigure}{\textwidth}
\centering
\includegraphics[scale = 0.4]{Figures/gqqbar/gqqbarPos.pdf}
\end{subfigure}
\begin{subfigure}{\textwidth}
\centering
\includegraphics[scale=0.4]{Figures/gqqbar/gqqbarNeg.pdf}
\end{subfigure}
\caption{Results for the $q\rightarrow g\overline{q}$ helicity amplitude calculation. Comparison between theoretically calculated probability distribution, quantum simulator and real quantum computer.}
\label{fig:gqqbarResults}
\end{figure}
One of the key sources of error in the quantum computer is readout noise. Error mitigation methods have been used to optimise the output from the quantum computer and reduce readout noise effects. This has been done using the Qiskit Ignis software~\cite{IBMQ}, which provides tools for noise characterisation and error correction based on noise models of the quantum machines. The method involves testing simple qubit states on a series of calibration circuits, which are run using the quantum simulator with the noise profile of the Santiago machine. The response matrix created from this is shown in Fig.~\ref{fig:responseMatrix}. This response matrix is calculated immediately before running the algorithm and then applied to the machine results to obtain the error corrected results, as shown in Fig.~\ref{fig:gqqbarResults}.
\begin{figure}[ht!]
\centering
\includegraphics[width = 0.5\textwidth]{Figures/gqqbar/Response_Matrix_Santiago.pdf}
\caption{IBM Q Santiago 5-qubit Quantum Computer Response Matrix for measurement error correction on the 4 qubit helicity amplitude calculation algorithm.}
\label{fig:responseMatrix}
\end{figure}
\subsection{2$\rightarrow$2 amplitude calculation}
\label{sec:qqqq}
Extending from the $1\rightarrow 2$ case in Sec.~\ref{sec:vertex}, the implementation of a full helicity amplitude calculation for the s and t-channels of a $2 \rightarrow 2$ scattering process is presented here\footnote{Note, for the calculation of the $1 \to 3$ case only minor modifications are needed.}. As an example, we consider a $q\overline{q} \rightarrow q\overline{q}$ process. The initial state quark and antiquark are labelled as particles 1 and 2 respectively and the final state quark and antiquark as 3 and 4. In total, there are only 4 non-zero helicity configurations possible for each s and t-channel process. The relevant amplitudes are,
\begin{align}\label{eqn: 2to2ampsfull}
\mathcal{M}_{s{(+-+-)}} &= -\langle 2 \vert \bar{\sigma }^{\mu } \vert 1] \frac{1}{s_{12}}[ 3 \vert \sigma _{\mu } \vert 4 \rangle, &\mathcal{M}_{s{(+--+)}} = -\langle 2 \vert \bar{\sigma }^{\mu } \vert 1] \frac{1}{s_{12}}\langle 3 \vert \bar{\sigma }_{\mu } \vert 4 ]
\end{align}
and
\begin{align}\label{eqn: 2to2amptfull}
\mathcal{M}_{t{(++--)}} &= -\langle 3 \vert \bar{\sigma }^{\mu } \vert 1] \frac{1}{s_{13}}[ 2 \vert \sigma _{\mu } \vert 4 \rangle, &\mathcal{M}_{t{(+--+)}} = -\langle 3 \vert \bar{\sigma }^{\mu } \vert 1] \frac{1}{s_{13}}\langle 2 \vert \bar{\sigma }_{\mu }\vert 4 ]
\end{align}
where the +/- signs denote the helicity of the outgoing-particles 1, 2, 3 and 4 and
\begin{equation}
s_{ij} = -(p_{i}+p_{j})^{2}=\langle i j \rangle [j i ].
\end{equation}
The other non-zero amplitudes are obtained by complex conjugation.
The calculation is performed in the Centre-of-Mass (CM) frame and the momenta of individual particles is defined such that the only dependent input variable is the angle, $\theta$, through which the quark (and antiquark) is scattered. In the CM frame, the overall magnitude of energy, $E$, associated with the momenta of each particle also drops out of the final helicity amplitude and is therefore not considered in this example.
In the `all-outgoing' convention of spinor-helicity formalism~\cite{elvang_huang_2015}, the momenta of incoming particles are flipped so that the incoming quark (1) (antiquark (2)) is mapped to an outgoing antiquark (quark) with opposite helicity. In the quantum algorithm, each quark-antiquark vertex is calculated on a 4-qubit quantum register, $q_i$. The outgoing antifermion spinor, $ q \rangle/q]$, is implemented on the vertex quantum register, $q_i^j$, followed by the two dimensional representation of the gamma matrices, $\sigma ^{\mu }/\bar{\sigma }^{\mu }$, and then finally the vertex is closed with the opposite helicity outgoing fermion spinor, $[q/\langle q$. A single qubit, $s$, is used to calculate the denominator of the gluon propagator. The calculation is controlled both from the helicity registers, $h_i$, which determine what helicity configuration the particles are in, and the amplitude qubit, $p$, which controls whether the s or t-channel process is calculated. A schematic of the quantum circuit is shown in Fig.~\ref{fig: 2to2}. Through this implementation, each component of the helicity amplitude can be calculated and extracted from the machine.
\begin{figure}[ht!]
\centering
\includegraphics[scale = 0.5]{Figures/2to2.pdf}
\caption{Circuit for the $q\overline{q}\rightarrow q\overline{q}$ process helicity amplitude calculation. The $q^j_i$ registers are used to calculate the $q\overline{q}$ vertices, and these are controlled from the helicity registers, $h_i$, which dictate the helicity configuration of the process.}
\label{fig: 2to2}
\end{figure}
This method is powerful as it allows for each component of the calculation to be extracted, however it leads to a complicated circuit, especially if one implements a method of dealing with incorrect helicity setups. As in Sec.~\ref{sec:vertex}, the circuit can be simplified by directly calculating the scalar products required for the final amplitudes. The amplitudes given in Eqs.~(\ref{eqn: 2to2ampsfull}) and~(\ref{eqn: 2to2amptfull}) can be simplified using Eq.~(\ref{eqn: Fierz}) (and that $ [p \vert \sigma ^{\mu } \vert q \rangle = \langle q \vert \bar{\sigma }^{\mu } \vert p])$ to give the final forms,
\begin{align}\label{eqn: 2to2amps}
\mathcal{M}_{s_{(+-+-)}} &= 2\frac{\langle 2 4 \rangle [3 1]}{\langle 1 2 \rangle [2 1]}, &\mathcal{M}_{s_{(+--+)}} = 2\frac{\langle 2 3 \rangle [4 1]}{\langle 1 2 \rangle [2 1]}
\end{align}
and
\begin{align}\label{eqn: 2to2ampt}
\mathcal{M}_{t_{(++--)}} &= 2\frac{\langle 3 4 \rangle [2 1]}{\langle 1 3 \rangle [3 1]}, &\mathcal{M}_{t_{(+--+)}} = 2\frac{\langle 3 2 \rangle [4 1]}{\langle 1 3 \rangle [3 1]}.
\end{align}
Using these expressions, the number of qubits needed for the circuit is reduced from 17 to 12 qubits. Another advantage is that the machine now only has to read out 3 qubits, where previously 8 qubits were read out per run. On these three qubits, each of the scalar products is calculated. The quark-antiquark vertex scalar products from the numerator are calculated on the first two qubits, and the denominator of the gluon propagator is calculated on the third qubit. Only one scalar product needs to be calculated for the denominator since~\cite{elvang_huang_2015},
\begin{equation}
\langle i j \rangle = [ j i ] ^*,
\end{equation}
therefore the second scalar product can be determined from the same qubit.
This simplified circuit is run on the IBM Q 32-qubit Quantum Simulator~\cite{32_sim} for 10,000 runs and compared to theoretically calculated probability distributions, extrapolated directly from analytic calculations of the helicity amplitude, calculated using the S@M software~\cite{zbMATH05804587}. Using the equivalence between helicity spinors and orthogonal pure state qubits, these theoretical predictions have been obtained from the probabilities of each of the qubits to be in the $\vert 0 \rangle$ or $\vert 1 \rangle$ state, which correspond to the magnitude squared of the upper and lower components of the helicity spinor respectively. The results from the quantum simulator show that the output of the quantum circuit lies within 1$\sigma$ of the theoretically predicted probability distribution and are shown in Fig.~\ref{fig: 2to2Comparison} for both the s and t-channel in a specific helicity configuration.
\begin{figure}[!h]
\centering
\includegraphics[width=\textwidth]{Figures/HelicityAmp_2to2/st+-+-.pdf}
\caption{Comparison between theoretically predicted qubit final state probabilities and 32-qubit quantum simulator output for the s and t-channel $q\overline{q} \rightarrow q\overline{q}$ process in the (+,-,+,-) helicity configuration. The quark (antiquark) scattering angle has been chosen as $\theta_3$ = $\frac{\pi}{4}$.}
\label{fig: 2to2Comparison}
\end{figure}
\subsection{Generalisation to $2 \to n$ amplitude calculations}\label{sec: helicityFuture}
It can be shown, using the BCFW recursion formula \cite{Cachazo:2004kj, Britto:2005fq} and the relations in Eq.~(\ref{eq:polarisation}), that scattering amplitudes for massless partons can be reduced to a combination of scalar products between helicity spinors\footnote{A well-known example is the Parke-Taylor formula for a $2\to n$ gluon scattering process, where the gluons i and j have helicity (-) and all other gluons have helicity (+). Then the formula provides the following expression for the amplitude $\mathcal{A}_n$,
\begin{equation}
\mathcal{A}_n[1^+ \cdots i^- \cdots j^- \cdots n^+]= (-g_s)^{n-2} \frac{\left < ij \right >^4 }{\left < 12 \right > \left < 23\right > \cdots \left < n1 \right >}.
\end{equation}
}. Consequently, the algorithm presented in Secs.~\ref{sec:vertex} and \ref{sec:qqqq} can be generalised to multi-particle amplitudes straightforwardly, as the tools are already created, namely the circuit decompositions of the helicity spinors from Appendix~\ref{sec: helicityAmpGates}. The number of calculation qubits, $q_i$, and the number of helicity qubits, $h_i$, needed in the algorithm both scale linearly with the number of final state particles, $n$. As the number of helicity qubits, $h_i$, scales linearly, then so does the number of work qubits needed in the algorithm. Each scalar product calculation requires two spinor operations, and so the algorithm can be easily extended without adding disproportionate complexity. The circuit depth scales linearly with an increase in the number of scalar products, calculated on the $q_i$ qubits, and the number of helicity qubits, $h_i$, added to the circuit.
It is interesting and practical to consider the extension of the simple helicity amplitude algorithms presented here to more complicated processes that are likely to be present in high energy collisions, such as those studied at the LHC. As we have seen in Sec.~\ref{sec:vertex}, modern public access quantum computers do not perform to a standard where one could extrapolate accurate calculations of helicity amplitudes, even for a single vertex. However, the performance of public access computers is well below that of state of the art machines, such as the IBM 53-qubit machine and the Honeywell machine. The latter, in unpublished work, claims to have the world's best Quantum Volume of 64~\cite{yirka_2020}. Such computers do not have the same restrictions as the smaller, less capable public access machines. The more powerful machines offer more choice for qubit setup and mapping, and the ability to perform more operations before decoherence in the machine starts to affect the circuit output. We can speculate that the algorithms presented here would be very accurate on these machines, especially the vertex calculation, which comprises a maximum of only 33 operations across 4 qubits.
The main difficulty of extending such algorithms for helicity amplitude calculations on quantum computers comes not only from limitations due to the number of qubits, but also the machine's fault tolerance. The more complicated the helicity amplitude calculation, the more operations are needed to calculate it. Therefore, a machine needs not only sufficient qubits but also the ability to implement many operations without excess noise. For the algorithm proposed, the immediate challenge is not the number of qubits available, but the number of operations that can be reliably implemented on the circuit. With advancements in the Quantum Volume of quantum computers~\cite{Jurcevic2020DemonstrationOQ}, this limitation will likely be overcome on current hardware. It is possible that near-future computers will have the ability to perform accurate and precise calculations and also have a large number of qubits. IBM recently announced their roadmap for the future and the goal of having machines with the number of qubits exceeding 1,000 by 2023~\cite{ibm_future}. Therefore, it is highly likely that these near-future devices will be able to perform precise helicity amplitude calculations for processes with a large number of particles.
\section{Parton shower algorithm}
\label{sec:parton}
After the hard process is calculated, the next step in simulating a scattering event at a high-energy collider experiment is the parton shower stage. The parton shower evolves the scattering process from the hard interaction scale down to the hadronisation scale. We propose an algorithm for simulating a QCD parton shower using IBM Quantum Experience \cite{IBMQ} software and hardware. The quantum circuit has been implemented to simulate a 2-step QCD parton shower with collinear splittings only. Section~\ref{sec: showertheory} provides the theoretical outline for the shower algorithm and discusses the splitting functions and probability calculations implemented in the quantum circuit. A brief overview of the quantum circuit is given in Sec.~\ref{sec: showerimplementation}, and a comparison between the results of the algorithm and theoretically calculated probability distributions are discussed in Sec.~\ref{sec: showerresults}. A glossary of quantum logic gates is given in Appendix~\ref{app: definitions} and a detailed overview of the quantum circuit for the algorithm in Appendix~\ref{app: showercircuit}.
\subsection{Theoretical outline of shower algorithm}\label{sec: showertheory}
We present a parton shower algorithm with the ability to simulate a general, discrete QCD parton shower, harnessing the quantum computer's ability to remain in a quantum state throughout the algorithm. In contrast to classical methods, the algorithm does not need to explicitly keep track of individual shower histories. Instead, our algorithm constructs and maintains a wavefunction that consists of a superposition of all possible shower histories, with the final measurement projecting out a specific quantity of the final state. Consequently, the algorithm presented inherently simulates the quantum interference between all possible final states, without the need for extensive computational logic present in current classical algorithms. In a classical algorithm, a physically meaningful quantity can only be extracted from a parton shower calculation after summing over all possible shower histories, requiring them to be stored on a physical memory device. Our quantum algorithm avoids the need for such an intermediate step, as the measurement is performed on the superposition of all shower histories directly.
The goal is to create the foundation for constructing a general quantum algorithm that can simulate a full QCD parton shower. To comply with the current capabilities of public access quantum computers and simulators provided by IBM Quantum Experience \cite{IBMQ}, the algorithm presented here uses a simplified model consisting of one flavour of quark and a gluon. This reduces the number of qubits needed, and the algorithm can be run on the IBM Q 32-qubit Quantum Simulator \cite{32_sim}. To further reduce the number of required qubits, only collinear splittings are considered within the model. By neglecting the soft-limit, there is no need to keep track of the detailed kinematics of the particles in the shower history.
Collinear emission occurs when a parton splits into two massless particles which have parallel 4-momenta, such that the total momentum, $P$, is distributed between the particles as
\begin{align}\label{Eqn. CollinearEmission}
p_i &= x P, &p_j = (1- x) P,
\end{align}
thus, $(p_i + p_j)^2 = P^2 = 0$ \cite{Taylor_2017}.
The emission probabilities in the algorithm are calculated using the collinear splitting functions outlined in~\cite{Dokshitzer:1977sg,Gribov:1972ri,ALTARELLI1977298,Marzani_2019}. A consequence of the collinear limit being a semi-classical interpretation with 1-to-2 splittings leads to the presence of a diagonal colour charge in the splitting functions, $C_{ii}$. The splitting for a quark to a gluon and a quark, with momentum fractions $z$ and $1-z$ respectively, is described at Leading Order (LO) by
\begin{equation}\label{eqn: quarkSplittings}
P_{q\rightarrow qg}(z) = C_F \frac{1 + (1 - z)^2}{z},
\end{equation}
with $C_F = 4/3$. The gluon splitting can be divided into two parts, with the first describing the splitting of a gluon to a quark-antiquark pair and the second describing the splitting of a gluon to two gluons,
\begin{align}\label{eqn: gluonSplittings}
P_{g \rightarrow q\overline{q}}(z) = n_f T_R (z^2+ (1-z)^2), & &P_{g\rightarrow gg}(z) = C_A \Big[ 2 \frac{1 - z}{z} + z(1-z) \Big],
\end{align}
where $C_A=3$ and $T_R=1/2$.
Here, $n_f$ is the number of massless quark flavours, and $T_R$ is the colour factor. It should be noted that both splitting functions have a soft singularity at $z$ = 0; the hard-collinear limit only takes into account finite $z$.
Further to calculating the splitting functions, the Sudakov factors have been used to determine whether an emission occurred in the step. The Sudakov factors for a QCD process are given by \cite{Sudakov:1954sw}
\begin{equation}
\Delta_{i, k} (z_1 , z_2)= \exp \Big[ - \alpha_s^2 \int^{z_2}_{z_1} P_k(z^\prime) dz^\prime \Big],
\end{equation}
and are used to calculate the non-emission probability. The running of the strong coupling, $\alpha_s$, is not simulated in this algorithm and for ease has been set to 1. For any given step $N$, there are $N$ possible particles present, and so the probability that none of the particles split is given by
\begin{equation}\label{eqn: sudakovs}
\Delta_{\textrm{tot}} (z_1, z_2) = \Delta_g^{n_g} (z_1, z_2) \Delta_q^{n_q}(z_1, z_2) \Delta_{\overline{q}}^{n_{\overline{q}}}(z_1, z_2).
\end{equation}
Finally, the probability of a certain splitting is therefore obtained from
\begin{equation}\label{Eqn. probCalc}
\textrm{Prob}_{k \rightarrow ij} = \big( 1 - \Delta_k \big) \times P_{k \rightarrow ij} (z).
\end{equation}
To implement the algorithm efficiently, preference has been given to gluons splitting to a quark-antiquark pair. This splitting preference implementation is explained in depth in Appendix~\ref{app: showercircuit}, but, for definiteness, the probability of a gluon splitting to two gluons is calculated as
\begin{equation}\label{eqn: probCalc}
\textrm{Prob}_{g \rightarrow gg} = \big( 1 - \Delta_g \big) \times \big( 1 - P_{g \rightarrow q\overline{q}} (z) \big) \times P_{g \rightarrow gg} (z).
\end{equation}
For the energy scale considered here, this should have a small affect on the results as $P_{g \rightarrow q\overline{q}}(z)~\ll ~P_{g \rightarrow gg}(z)$.
\subsection{Implementation on quantum circuit}\label{sec: showerimplementation}
A quantum circuit has been constructed to simulate a parton shower with collinear splittings. The circuit comprises of particle registers, emission registers and history registers and uses a total of 31 qubits. The algorithm is discretised into individual steps. An emission can occur in each step, and the probabilities are calculated from the splitting functions and Sudakov factors. To meet the 32 qubit limit of the IBM Q Quantum Simulator \cite{32_sim}, the algorithm has been limited to two steps, but it is generally extendable. Figure~\ref{fig: 1stepcircuit} shows the circuit diagram for a single step.
\begin{figure}[!h]
\centering
\includegraphics[scale = 0.5]{Figures/partonshower.pdf}
\caption{Circuit diagram for one step of the algorithm. The circuit comprises particle registers, emission registers and history registers.}
\label{fig: 1stepcircuit}
\end{figure}
The algorithm follows a similar method to that described in~\cite{bauer2019quantum}, first counting the particles present in the simulation, determining whether an emission has occurred and if so, assessing which splitting did occur, then finally updating the particle content of the simulation. In contrast to the method shown by \cite{bauer2019quantum}, the algorithm presented here has the ability to simulate a QCD process with splittings for both gluons and quarks implemented using the splitting functions outlined in Eqs.~(\ref{eqn: quarkSplittings}) and (\ref{eqn: gluonSplittings}). The addition of such splitting functions leads to significant changes to the algorithm compared to that presented in~\cite{bauer2019quantum}, specifically in the History and Update gates of the algorithm, shown in Fig.~\ref{fig: 1stepcircuit}. The implementation of these gates is outlined in detail in Appendix~\ref{app: showercircuit}. Unlike the algorithm presented in~\cite{bauer2019quantum}, we have chosen not to introduce flavour mixing at the start of the algorithm. Instead, the superposition and interference between the possible output states are introduced in the tailored History and Update gates. With the ability to simulate gluon and quark splittings, the algorithm is thus well suited to hadronic parton shower simulation and provides the foundations for a general parton shower algorithm for use on a GQC.
The parton shower algorithm is designed to operate on the public access IBM~Q 32-qubit Quantum Simulator~\cite{32_sim}, which allows for a total of two steps to be simulated on the machine. As the machine is a simulator, it does not suffer from noise or a limit on the number of operations due to qubit decoherence effects, therefore giving a simulation of a perfect machine. As a consequence, error checking is easily done with direct comparison to theoretically predicted probability distributions, and this is discussed in Sec.~\ref{sec: showerresults}.
One of the main benefits of using a quantum computing (QC) algorithm for the simulation of QCD parton showers over classical methods is the computational simplicity of the algorithm. When dealing with interference of different splittings in the shower process, the algorithm presented here offers a much less computationally complex approach than that provided by modern Monte Carlo event generators. This is achieved by utilising the unique ability to maintain the quantum computer in a fully quantum state throughout the algorithm, and only collapse to a classical circuit by measurement at the end of the process. This allows for the system to account for all possible parton shower histories simultaneously. In contrast, modern Monte Carlo methods must manually keep track of the particle splitting histories to consider all possible contributions to a specific final state. For a two-step, discrete parton shower, this is a relatively easy task for a modern Monte Carlo generator. However, the quantum computing field is still in its infancy; the true potential of quantum computing for simulating QCD parton showers will become apparent with the advancement of quantum technologies. With more available qubits and machines with improved hardware, the algorithm presented here will have the ability to simulate quantum effects, without the extensive and complex computational logic that a classical computer would need. Therefore, quantum computers offer an avenue to explore processes that contain a large number of shower particles, thus requiring complicated parton histories and computing power, not currently achievable with modern classical techniques. Beyond QCD parton showers, this feature of a quantum computing algorithm can be of particular interest for cosmic-ray air showers, where millions of long-lived particles are simulated \cite{Wentz:2003bp,Schichtel:2019hfn}.
\subsection{Results of parton shower}\label{sec: showerresults}
A comparison of the output from the parton shower algorithm and theoretical predictions of the splitting probabilities is made, and the results are shown in Fig.~\ref{fig:2step}. The algorithm was run for 10,000 shots using the IBM Q 32-qubit Quantum Simulator~\cite{32_sim}, with a momentum interval of $z_{\textrm{lower}}$~=~0.3 to $z_{\textrm{upper}}$~=~0.5, and no noise simulation. Here the theoretical predictions have been calculated using the collinear splitting functions from Eqs.~(\ref{eqn: quarkSplittings}) and (\ref{eqn: gluonSplittings}), using the method outlined in Sec.~\ref{sec: showertheory}. The $z$ value used for the particle splitting probabilities from Eq.~(\ref{eqn: probCalc}) is the mid-point of the momentum interval used in the algorithm. The results are in agreement with the theoretically calculated probabilities to within 1$\sigma$.
A consequence of running the algorithm on a quantum simulator is that there will be no noise in the results, unlike a real quantum computer. Therefore, problems with the algorithm can be identified through direct comparison with the theoretical calculations.
In the future, if the algorithm can be run on a real quantum computer with enough qubits, then IBM Q offers a range of noise reducing schemes for its devices through the Qiskit software \cite{IBMQ}. Another advantage of using the quantum simulator is that it automatically chooses an optimum qubit setup. In a real quantum computer, the user has to select a qubit mapping in order to optimise the operation of the computer. For future use of the algorithm, this can be done using the calibration data provided by IBM Q.
\begin{figure}[ht!]
\centering
\begin{subfigure}{0.89\textwidth}
\centering
\includegraphics[width=1\textwidth]{Figures/2Step_Gluon.pdf}
\subcaption{Initial particle a gluon.}
\end{subfigure}
\begin{subfigure}{0.89\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/2Step_Quark.pdf}
\subcaption{Initial particle a quark.}
\end{subfigure}
\begin{subfigure}{0.89\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/2Step_AntiQuark.pdf}
\subcaption{Initial particle an antiquark.}
\end{subfigure}
\caption{Results from the quantum circuit compared to theoretical predictions for two steps of the parton shower with momentum interval of $z_{\textrm{lower}}$~=~0.3 to $z_{\textrm{upper}}$~=~0.5 and the initial state particle of (a) gluon, (b) quark and (c) antiquark. }
\label{fig:2step}
\end{figure}
\section{Summary and conclusions}
\label{sec:conclusions}
The accurate modelling of the complexity of collisions at experiments, such as the Large Hadron Collider, relies on theoretical calculations of multi-particle processes. Such calculations can be factorised into: the hard interaction, which models the large momentum transfer, and the parton shower, which models the evolution from the hard interaction scale down to the hadronisation scale.
We present general and extendable quantum computing algorithms that provide calculations of the hard interaction process and the parton shower, as a first step towards a quantum computing algorithm to describe the full collision event at the LHC.
The hard interaction calculation uses helicity amplitudes by exploiting the equivalence of spinors and qubits, and encoding operators as a series of unitary operations in the quantum circuit, thus demonstrating an excellent use case of quantum computers to model the intrinsic quantum behaviour of the system. A quantum algorithm is constructed for two simple examples of helicity calculations; a $gq\bar{q}$ vertex and the $q\overline{q}\rightarrow q\overline{q}$ process. By applying Hadamard gates to helicity registers, we introduce a superposition state between the helicity qubits and can therefore calculate the positive and negative helicities of each particle involved simultaneously. This is further exploited in the simultaneous computation of s and t-channel amplitudes for the $q\bar{q}\rightarrow q\bar{q}$ process, thus fully utilising the quantum nature of the computation. A comparison between the theoretical predictions and the output of the quantum algorithm shows very good agreement. Furthermore, the successful implementation of the $gq\bar{q}$ vertex algorithm on a real machine is also demonstrated by comparing results from the machine with a simulator.
We also present a quantum algorithm for simulating collinear emission in a two-step, discrete parton shower with a maximum of three final state particles, utilising the quantum computer's ability to remain in a quantum state throughout the simulation. In contrast to classical implementations of parton showers, where individual shower histories have to be stored on a physical memory device, our quantum computing algorithm constructs a wavefunction for the whole parton shower process, which contains a superposition of all shower histories. As a result, we do not need to keep track of individual shower histories explicitly, and a physical quantity of the shower process can be obtained through a measurement of the wavefunction. The results from the algorithm, as performed on the IBM Q 32-qubit Quantum Simulator~\cite{32_sim}, show good agreement with theoretical predictions. The algorithm builds on previous work~\cite{bauer2019quantum} by including a vector boson and boson splittings, which leads to significant changes in its implementation. The ability to simulate gluon and quark splittings makes the algorithm presented here well suited to hadronic parton shower simulation and provides the foundations for developing a general parton shower algorithm. With advancements in quantum technologies, this algorithm can be extended to include all flavours of quarks without adding disproportionate computational complexity.
With IBM recently setting their goal of exceeding 1,000 qubits by 2023~\cite{ibm_future} and advancements in the development of devices with better Quantum Volume~\cite{Jurcevic2020DemonstrationOQ}, we are on the brink of a quantum revolution. These developments would allow the algorithms presented in this paper to be extended to reflect the processes seen in experiments such as the LHC. The consequence of such advancements would be algorithms that can fully model the dynamics of quantum field theories to provide accurate and precise helicity amplitude calculations and simulations of parton showers.
\vskip 2 \baselineskip
\noindent {\it{{\bf Acknowledgements:}~~We would like to acknowledge the use of the IBM Q for this work.
We are grateful to the authors of \cite{bauer2019quantum} for answering questions on the circuit presented in their work and for sharing their preliminary codes. M.S would like to thank Steve Abel and Daniel Maitre for helpful discussions. K.B and M.S are supported by the STFC under grant ST/P001246/1. S.M and S.W are supported by a grant from the Royal Society.}}
\vspace{1.0cm}
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{"url":"https:\/\/mathematica.stackexchange.com\/questions\/160048\/mapping-a-function-over-a-graphs-edges-or-vertices?noredirect=1","text":"# Mapping a function over a graph's edges or vertices\n\nTake a graph with VertexWeights and\/or EdgeWeights like\n\ng = Graph[{1, 2, 3}, {1 <-> 2, 2 <-> 3, 3 <-> 1},\nEdgeWeight -> {1 <-> 2 -> \"edge1\", 2 <-> 3 -> \"edge2\", 3 <-> 1 -> \"edge3\"},\nVertexWeight -> {1 -> \"vertex1\", 2 -> \"vertex2\", 3 -> \"vertex3\"},\nVertexLabels -> \"VertexWeight\", EdgeLabels -> \"EdgeWeight\"]\n\n\nHow can one map a function over vertices and\/or edges?\n\nUsing PropertyValue and VertexList\/EdgeList one can define a version of Map that works on graphs.\n\nFor edges:\n\ngraphEdgeMap[f_, g_] :=\nWith[{weights = PropertyValue[{g, #}, EdgeWeight] & \/@ EdgeList[g]},\nGraph[g, EdgeWeight -> Thread[EdgeList[g] -> (f \/@ weights)] ] ]\n\n(* Operator form *)\ngraphEdgeMap[f_][g_] := graphEdgeMap[f, g]\n\n\nand analogous for vertices:\n\ngraphVertexMap[f_, g_] :=\nWith[{weights = PropertyValue[{g, #}, VertexWeight] & \/@ VertexList[g]},\nGraph[g, VertexWeight -> Thread[VertexList[g] -> (f \/@ weights)] ] ]\n\n(* Operator form *)\ngraphVertexMap[f_][g_] := graphVertexMap[f, g]\n\n\n## Usage example\n\nh \/\/ graphEdgeMap[ToUpperCase] \/\/ graphVertexMap[Style[#, Red] &]\n\n\nAlthough used for styling\/string manipulation in this example, these functions can of course be used for computation as well.\n\n\u2022 Related: mathematica.stackexchange.com\/q\/95935\/12 and mathematica.stackexchange.com\/q\/143601\/12 The performance of the method you use leaves something to be desired. (But then why is doing this so difficult in the first place? In other non-Mathematica graph manipulation packages this is much easier.) Nov 16, 2017 at 10:28\n\u2022 I don't remember all the various bugs I hit when implementing this for IGraph\/M, but here's one that comes up fairly often: try using graphEdgeMap on g = KaryTree[5, EdgeWeight -> {1., 2., 3., 4.}]. I should have commented the code better and explain why things are done the exact way they are\u2014unfortunately this is something that can't be understood by reading only the code (you'd think you have an opportunity to do something more compactly than it currently is in IGraph\/M but then it triggers bugs in edge cases). Nov 16, 2017 at 10:35\n\u2022 @Szabolcs I agree with your sentiment that this is more difficult than it should be. There should be a build-in for this really. Nov 16, 2017 at 10:44\n\u2022 I suggested a builtin to WRI more than once, including in the W Community thread I linked below my post. Nothing happened so far. I think first they'd have to fix the mess the property system is. Nov 16, 2017 at 10:45\n\nIGraph\/M has functions precisely for this. They are implemented purely on Mathematica (they don't use the igraph C core), so you can look at the implementation.\n\nSadly, making this work reliable and with reasonable performance is not at all easy. It requires distinguishing between the various kinds of properties and crossing the minefield of property-related bugs. So I won't go into details, instead, I'll show how to use the functionality in IGraph\/M. For many more example, see its documentation (evaluate IGDocumentation[]).\n\nNeeds[\"IGraphM\"]\n\ng = ExampleData[{\"NetworkGraph\", \"Friendship\"}]\n\n\nLook at the labels with the vertex property extractor:\n\nIGVertexProp[VertexLabels][g]\n(* {Placed[\"Anna\", After], Placed[\"Rose\", Above],\nPlaced[\"Nora\", Above], Placed[\"Ben\", Before], Placed[\"Larry\", Above],\nPlaced[\"Carol\", Below], Placed[\"Rudy\", Below],\nPlaced[\"Linda\", Above], Placed[\"James\", Below]} *)\n\n\nMake them uppercase by mapping a function over the VertexLabels property\n\nIGVertexMap[MapAt[ToUpperCase, {1}], VertexLabels, g]\n\n\nWe needed MapAt, because, as you can see above, all vertex labels are given as Placed expressions.\n\nCompute edge betweenness values for each edge and store them in the EdgeWeight property:\n\nwg = g \/\/ IGEdgeMap[# &, EdgeWeight -> EdgeBetweennessCentrality]\n\n\nWe are mapping the identity function #& here, as we don't want to transform the values in any way.\n\nNotice that we used the operator form of IGEdgeMap.\n\nCopy the value of the EdgeWeight property to the \"betweenness\" edge property:\n\nwg = IGEdgeMap[# &, \"betweenness\" -> IGEdgeProp[EdgeWeight], wg]\n\n\nThe right-hand-side of -> must always contain a function that returns a value for each edge in order when applied to the graph. The result will be transformed with the function given as the first argument, and then stored in the property specified in the left-hand-side of ->.\n\nThe IGEdgeProp and IGVertexProp property extractors are useful for constructing the RHS of ->, and are often used alone there. But they can also be combined with other functions. Here's how to colour each edge based on the \"betweenness\" edge property. We Rescale the value to fit the domain of the colour function.\n\nIGEdgeMap[ColorData[\"Rainbow\"],\nEdgeStyle -> Rescale@*IGEdgeProp[\"betweenness\"], wg]\n\n\nHere's a more complex styling example from the documentation, showing operator forms, and mapping a function that takes multiple arguments.\n\ng = Graph[ExampleData[{\"NetworkGraph\", \"EastAfricaEmbassyAttacks\"}],\nImageSize -> Medium];\n\ng \/\/\nIGEdgeMap[ (* save original weight in \"weight\" property *)\nIdentity, \"weight\" -> IGEdgeProp[EdgeWeight]\n] \/*\nIGEdgeMap[ (* invert edge weights for betweenness calculation; for this purpose we need large weight = weak connection *)\n1\/# &, EdgeWeight\n] \/*\nIGEdgeMap[ (* thickness by original weight, colour by betweenness based on inverse weight; map two-argument function *)\nDirective[AbsoluteThickness[9 #1], ColorData[\"Rainbow\"][#2]] &,\nEdgeStyle -> {IGEdgeProp[\"weight\"], Rescale@*EdgeBetweennessCentrality}\n]\n\n\nFinally, here are the usage messages of the functions related to this functionality area:\n\nThere's also the question of what to do when a requested property value does not exist. For custom properties, values may be assigned only to some edges\/vertices. The current behaviour of IGraph\/M is to return Missing[...]` in these cases.\n\n\u2022 Technically, the version of IGraph\/M containing this is still a prerelease, so I am still taking suggestions for improvements; but at this point I'll only make breaking changes if the improvements are significant. Nov 16, 2017 at 10:43\n\u2022 This framework was mostly developed in this Wolfram Community thread, which also doubles as a feature request for something similar to be built in. Nov 16, 2017 at 10:44","date":"2022-09-25 17:05:52","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.19563071429729462, \"perplexity\": 3565.492635831836}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-40\/segments\/1664030334591.19\/warc\/CC-MAIN-20220925162915-20220925192915-00504.warc.gz\"}"}
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Carriera
In carriera ha vinto un titolo di doppio, il Bell Challenge nel 1999, in coppia con Amy Frazier. Nei tornei del Grande Slam ha ottenuto il suo miglior risultato raggiungendo i quarti di finale di doppio agli US Open nel 1998 e di doppio misto a Wimbledon nel 2000.
Statistiche
Doppio
Vittorie (1)
Doppio
Finali perse (1)
Collegamenti esterni
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"redpajama_set_name": "RedPajamaWikipedia"
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700 million Euros spent for players after the Sir Alex Ferguson's era in Manchester United and the team doesn't look like anything. Most probably this was the feeling of every Red Devil fan when they watched the first game of the season in the Premier League against Leicester. Despite winning by 2:1 over the champions from three seasons ago, the United of Jose Mourinho in no way looks like a team that can win the championship this year.
The chaos in the team is full. The defender Matteo Darmian, who announced that he wanted to leave the team two weeks ago was in the starting eleven. In attack Marcus Rashford was lost somewhere between Alexis Sanchez and Juan Mata. In the midfield, Paul Pogba is doing his best to get a transfer to Barcelona. The word that summarizes everything in Manchester United at this point is a confusion.
Betting prediction for the match between Burnley and Watford could you find here.
That's why the match against Brighton away doesn't seems to be a great one. Of course it is not because the hosts in the match started the season well. The loss against Watford away was certainly not expected, but now the hosts know they have all the chances to get something out of the match against United.
The odds offered by the bookmakers are in favor of the guests, with bet365 offering 1.72 for an United's win. The draw is estimated at 3.5 and exactly 5 is the odds for the hosts to win the match. This can surely be used by those who like to bet on the Premier League matches, as this odds score for an away win seems a bit overpriced.
So, if we look at the odds offered for Asian Handicap, we will see that the offer of a home win in this match with a predetermined advantage of one goal is 1.6. This, though low coefficient is certainly a value bet.
For those who would like to gamble with better odds, the bet for a draw in the match certainly seems to be reasonably well grounded. However we will stick to the Asian lines options.
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{
"redpajama_set_name": "RedPajamaC4"
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Qualifying Requirement : BIDDER MUST PROVIDE SAMPLE. SAMPLE SHALL BE SENT TO MRPU RECEIPT GODOWN, DAE Nodal Facility Centre, Dr.Ambedkar Road, Pallavaram,CHENNAI-600043, AND SCANNED COPY OF DOCUMENTARY PROOF/DELIVERY CHALLAN SHALL BE UPLOADED ALONG WITH THE OFFER, FAILING WHICH YOUR OFFER WILL BE LIABLE FOR REJECTION. ACCEPTANCE OF OFFER IS SUBJECT TO SATISFACTORY FEEDBACK OF THE SAMPLE PROVIDED. THE INDENTED QUANTITY MAY BE PROCURED IN PARTS, IF REQUIRED. OFFERS FROM SUPPLIERS IN AND AROUND CHENNAI/PUDUCHERRY ONLY WILL BE CONSIDERED FOR EVALUATION.
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{
"redpajama_set_name": "RedPajamaC4"
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\section{Introduction}
\subsection{Variational principle}
A crucial
requirement for any theoretical model of coronal structures
is to give account of the stability and evolution of far--from--equilibrium states which are responsible of the characteristic rich topology and dynamics of the solar corona.
This implies to
consider the coupling of thermal and mechanical equations.
Different stability analysis of
solar structures can be found in the literature, generally
restricted to special types of perturbations and specific
equilibrium models. These includes, models that consider adiabatic
configuration such as the ones analyzed via the classical
criterion of \citet{ber} or those that
presuppose static equilibrium and analyze thermal stability. In
the application of Bernstein's
criterion, the adiabatic assumption implies that the energy balance equation is not required
and thus dissipation is
impossible. Also the assumption of static models is a strong, and
often unjustified, restriction for open systems.
In this paper we apply an energy
principle to analyze the stability of solar coronal loops when helical modes are present.
The principle was obtained in previous papers (Paper I: \citep{cos1};
\citep{cos2}; see also \citep{us0}) using a general
procedure of irreversible thermodynamics -based on firmly
established thermodynamic laws- that can be understood as an
extension of Bernstein's MHD principle to situations far from
thermodynamic equilibrium.
In Paper I and in \citep{cos2} we showed how to obtain the variational principle for solar
coronal structures
from
the equations that describe the dynamics of the system. The method consists of obtaining a
Lyapunov function,
also known as generalized potential, that represents the
mathematical expression of the stability conditions. The principle is subject to physically reasonable
requirements of hermiticity and antihermiticity over the matrices.
For a more detailed presentation see Paper I and the references therein.
\subsection{Solar coronal loops}
MHD loop oscillations in the corona are known to be
strongly damped, mostly having decaying times of few periods
$N_{p}\approx 2 - 7 \ periods.$ While thermal conduction, with the
contribution of
radiative cooling mechanisms, could be the main
cause of the damping of pure
MHD slow magnetoacoustic mode oscillations they are unimportant for the MHD fast
modes.
Resonant absorption and phase mixing seem more promising
in giving account of the rapid decay (\citep{gh94}; hereafter HG,
\citep{go02}) of the ideal fast
oscillations of these strongly inhomogeneous and structured plasma systems.
Inhomogeneous
equilibrium distributions of plasma density
and temperature varying continuously across the magnetic field led to plasma waves with
continuous intervals of eigenfrequencies. The
occurrence of the Alfv\'en ideal MHD continuum in a thin edge layer
is derived from the highly
anisotropic character of the fast magnetoacoustic waves
giving rise to a peak of the amplitudes where the perturbation
develops large gradients and the absorption has maxims. However, there is another type of continuum commonly known as slow magnetosonic continuum associated to the inhomogeneity of the equilibrium parameters along the axis of the loop (see Paper I). This inhomogeneities are associated, for example, to changes in the
density concentration at the loop basis. If the magnetic field is twisted the inhomogeneities led to the coupling of Alfv\'en and slow magnetosonic continuum modes (\citep{bel}).
The resonant absorption mechanism of
wave heating consists on the non--dissipative transference of
wave energy from the collective line-tied wave with fast discrete eigenvalues (kinetic energy of the fast
radial component) to a local resonant mode in the Alfv\'en continuum, (kinetic energy of the
azimuthal component), which is then dissipated in an enhanced
manner. Then, the continuum oscillations are converted
into heat by dissipative processes; as the medium has large
gradients in the Alfv\'en speed, the oscillations of neighboring
field lines become out of phase and shear Alfv\'en waves lead to
enhanced viscous and ohmic dissipation (see \citep{pr83} for the
linear regime and \citep{nak97} for the nonlinear one). The mode
conversion from the collective to the local mode occurs in a time
that is non--dissipative and generally much shorter than the
second time scale which is related to the dissipative damping of the
small--scale perturbations of the local mode in the resonance
layer (\citep{rob00}; \citep{van04}).
The whole temporal pattern description of
modes that exhibit a combination of global (discrete line--tied
fast eigenmode) and localize (Alfv\'en continuum mode) behaviour
is known as quasi--mode. Moreover, the mixed nature of the modes
is not only due to the temporal behaviour but also to the
boundary value problem giving rise to a spatial behaviour which is
also of a mixed nature, i.e. coronal loops with line--tying
constraints cannot support pure waves: Alfv\'en, slow or fast
magnetoacoustic modes.
HG
studied the mixed spectral description of coronal loops (i.e. the
resulting superposition of basic waves which adjust the line--tied
condition) without assuming a straight magnetic field and forcing
the loop to follow the photospheric velocity perturbations. They
found that pure Alfv\'en and pure slow modes are obtained as
singular limiting cases of cluster spectra of Alfv\'en--fast or
slow--fast modes, where the fast components are localized in a
photospheric boundary associated to the line--tied condition:
the coronal part of the
loop acting as a resonant cavity of large Alfv\'en
components and fast components, with a small but
rapidly varying amplitude, located in the photospheric boundary layer.
They found that heating of coronal loops by resonant absorption is
due to the line--tied Alfv\'en continuum which no longer depends
on the poloidal magnetic field and that the corresponding
eigenmodes have a global ballooning feature which is characterized
by an accumulation point given by the Alfv\'en frequency. In \citep{gh93} (hereafter
GH), a variational principle, based in Bernstein's
principle, was obtained to derive the Alfv\'en and slow continuum
frequencies in a line--tied inhomogeneous cylinder. Stability
considerations led them to conclude the global stability of
coronal loops.
In this paper, following results of Paper I we apply our energy principle to consider the stability and mode structure of loop inhomogeneous coronal models with non--vanishing helicity. Our principle has the advantages that it does not require a WKB approximation and that, as was mentioned, it allows the consideration of the coupling of the thermal and mechanical equations that are necessary to analyze far from equilibrium states.
\section{The MHD stability criterion for coronal structures}
Solar coronal conditions with large Reynolds numbers are well fitted by
ideal MHD plasma models (i.e. infinite electrical
conductivity $\sigma \gg 1 $ leading to vanishing viscosity and
ohmic dissipation). Thus, the fundamental equations considered are the
mass conservation equation,
the perfect gas law or state equation for a fully
ionized $H$ plasma and the induction equation, with vanishing magnetic diffusivity due to the conductivity properties. The energy balance equation takes the form:
\begin{equation}
\frac{\rho^{\gamma}}{(\gamma-1)}\frac{D}{D
t}(\frac{p}{\rho^{\gamma}})=-\nabla\cdot\vec{F_{c}}-L_{r}+H\label{1}
\end{equation}
$\vec{F_{c}}$ is the heat flux due to particle conduction along
the loop, $L_{r}$ is the net radiation flux and $H$ the heating function
which was chosen as in Paper I: $H = h \rho + H_{0}$.
Eq.~\ref{1} expresses the fact that the gain in particle
energy (internal plus kinetic) is due to the
external heating sources represented by the heating function, heat
flow and radiation losses;
all other heating sources were considered as vanishing terms
implying that the optically thin assumption holds. Note that the non--ideal contribution in the energy equation ($L$) is associated to the open character of the loop system.
Once the linearization around a nonlinear equilibrium or stationary state is performed, and after a straightforward manipulation procedure where the hermiticity requirements are fulfilled the generalized energy principle and the respective frequencies are obtained (Paper I and
\citep{cos2}) as:
\begin{equation}
\delta^{2} W_{p} =\frac{1}{ 2}\int ( \vec{\xi}^{*} \beta F \vec{\xi}+T_{1}^{*} AT_{1}
+T_{1}^{*}B\vec{\xi} -\vec{\xi}^{*}BT_{1})d^{3}x\geq 0.
\label{2}
\end{equation}
\begin{equation}
\omega^{2} =- \frac{\int ( \vec{\xi}^{*} \beta F\vec{\xi} +
T_{1}^{*}AT_{1}+T_{1}^{*}B\vec{\xi}-\vec{\xi}^{*}BT_{1}^{*} )
d^{3}x}{\int (\vec{\xi^{*}} \beta \rho_{0}\vec{\xi} )d^{3}x}
\label{3}
\end{equation}
with the same normalization condition as in Paper I.
$ F$ is the known Bernstein operator for the system, $\xi$ and $T_{1}$ are the motion and temperature perturbations and operators $A$ and $B$ are as in Paper I.
For the non-dissipative cases ($L=0$ or equivalently $T_{1}=0$), last expressions (discarding the presence of factor $\beta$ which appears in the equations to fit the Hermitian and anti--Hermitian conditions) are reduced to the
well--known Bernstein MHD energy principle and its respective frequencies.
\section{Application to an inhomogeneous loop model with non--vanishing helicity}
On one hand, the azimuthal component of the loop perturbation is
believed to be one of the principle responsible of resonant
absorption and damping of ideal oscillations; on
the other, this component is associated to the storage of
magnetic energy in systems with non--vanishing helicity which
eventually is released by instabilities. Thus, we are interested in
analyzing the changes produced in the stability of
non--homogeneous loops subject to helical perturbations. This is, loops
with inhomogeneous distributions of plasma density and
temperatures subject to body modes and with non--vanishing helicity. In this case, the
Alfv\'en, slow and fast magnetoacoustic cylinder
modes cannot longer be associated to the azimuthal, longitudinal
and radial components respectively. The
observational importance of helical modes cannot be neglected
and it is poorly known how helicity affects important physical
features of mode oscillations (e.g., damping mechanisms,
stability and periods). However, a mode classification can be accomplished
via the analysis of the mode variations, described in an orthogonal basis, while helicity is varied. The basis is formed by the orthogonal displacements:
parallel and perpendicular to the magnetic field and the radial
(and perpendicular to the surface of the tube) one, of
observational interest.
The fundamental modes are generally observationally and
energetically more important than their harmonics. For these
global modes the inhomogeneous nature of the medium cannot be
ignored and it determines the structure of the disturbance which
cannot be taken as sinusoidal, making the traditional normal mode
analysis useless for this treatment (sinusoidal dependence with constant coefficients), i.e. at least a WKB approximation, of weakly varying parameters compared to a typical wavelength, is required. Moreover, the occurrence of
either an infinitely degenerate eigenvalue or an accumulation
point giving rise to a continuous spectrum are associated to inhomogeneities.
We consider two types of inhomogeneities: the inhomogeneity of the equilibrium parameters along the loop axis, and the inhomogeneity across the loop axis when the radius is varied. As a first order approximation we neglect the effect
of gravitational stratification and thus confine the analysis to
characteristic spatial scales lower than the pressure scale height
in the solar corona. In order to analyze the stability and to
obtain the frequencies and modes the physical quantities in
eq.~\ref{2} and eq.~\ref{3} must be calculated along the loop
structure.
\subsection{Mechanical equilibrium}
To determine an equilibrium configuration we assume
force--free equations. This assumption is justified for coronal conditions due to the fact that in plasmas with low
$\mathcal{\beta}$ (gas pressure over the magnetic pressure) the
pressure gradient can be neglected in comparison to the Lorentz
force. For the chromosphere and the photosphere the force--free approximation may not be a good one. However, it is a widespread supposition \citep{rud}:
perturbed systems are believed to relax to new force-free, minimum energy states and chromospheric conditions seem to be well fitted to force--free models from $4. \ 10^{5} \ m$ \citep{asc4} (Chapter 5).
Coronal loops are generally modeled as thin cylindrical
fluxtubes where the curvature and related forces can be neglected so
the cylindrical geometry can be applied. The fluxtube is assumed
as line--tied to the photospheric plasma through its footpoints
which are forced to follow the photospheric velocity
perturbations. The random velocity field creates vorticity
generally twisting the coronal fluxtubes. Thus, a relation between
the helical twist and the force--free parameter can be derived as
follows (e.g. \citep{stu94}). The coronal loop model is obtained
from the equations
\begin{equation}
\nabla \times \mathbf{B_{0}}=\alpha(r) \mathbf{B_{0}} \ \ \ \mathbf{j} \times \mathbf{B_{0}}=0. \label{3}
\end{equation}
Also, since $\mathbf{B_{0}}$ is force--free, $\nabla p=0$ everywhere and thus has
a constant value along the loop. We consider a straight cylinder with a nonuniform distribution of density and temperature and a resulting uniform
twist over an initially non--rotated field $ \textbf{B}=(0,0,B_{z})$ yielding
the unperturbed
magnetic field
$$\mathbf{{B_{0}}}= (B_{r},B_{\phi},B_{z})=B_{0}(0, \frac{br}{\Delta}\frac{1}{\Delta})$$
\noindent with $\Delta=1+b^{2}r^{2}$ and $ b=2 \Pi N_{t}/L$ ($
N_{t}$ number of turns over the cylinder length $ L$). Then,
\begin{equation}
\frac{B_{\phi}}{B_{z}}=\frac{r \partial \phi}{\partial z}=\frac{r 2 \pi N_{t}}{L}=br \\
\alpha(r) =\frac{2b}{\Delta}\label{4}
\end{equation}
We assume a given value of
the cylinder radius $r=R$, thus the line element results a function of
the coordinate
$z$: $s=s(z)$. The dependence with the radial component will be taken into account by considering different values of the radius $R$.
\begin{equation}
ds^{2}=R^{2}d \phi^{2}+dz^{2}=\left( 1+R^{2} b^{2}\right) dz^{2} =\Delta dz^{2}\label{5}
\end{equation}
\subsection{Thermal equilibrium}
The thermal equilibrium is obtained, as in Paper I, assuming $L=0$ in the balance energy equation
(eq.~\ref{1}) . The procedure
developed consists in obtaining the function of the temperature along the
arc element $s$ by integrating eq.~\ref{1} with the constraint $L=0$ and
replacing border conditions: the temperature at the bottom $T_{b}=10^{4}K$ and the temperature at the top $T_{t}=10^{6}K$. The known expression (see chapter 6 of
\citet{pri}) is obtained
\begin{equation}
\left[\frac{dT}{ds}\right]^{2}=\frac{p^{2}\chi}{2k_{B}^{2}k_{0}(\alpha
+\frac{3}{2})}T^{\alpha-\frac{7}{2}} \left[1 - (
\frac{T}{T_{t}})^{2-\alpha}\right]\label{6}
\end{equation}
which has to be inverted to obtain
$T=f^{-1}(s)$ \citep{ar95} as
\begin{equation}
\frac{dT}{ds}=\mathcal{A}\left[\frac{d\mathbb{B}_{v}}{dv}\frac{dv}{dT}\right]^{-1} \ \ \ where \ \ \ \mathbb{B}_{v}(\frac{1}{2},q)=\int_{0}^{v}t^{p-1}(1-t)^{q-1}dt \label{7}
\end{equation}
\noindent
with
$$p= \frac{1}{2}; v=1-(\frac{T}{T_{t}})^{2-\alpha}; q=(\frac{\alpha}{2}+\frac{3}{4}) (2-\alpha)+1 $$
$$ \mathcal{A}=(2-\alpha)T_{t}^{\frac{\alpha}{2}-\frac{11}{4}}((p^{2}\chi)/(2k_{0}
(\alpha+\frac{3}{2})k_{B}^{2}))^{\frac{1}{2}}.$$
We use $\alpha=-\frac{1}{2}$ so $q=\frac{6}{5}$
to numerically calculate
the modes, \\
$ s=\frac{1}{\mathcal{A}}\mathbb{B}_{v}(\frac{1}{2}, \frac{6} {5})\rightarrow\mathcal{A}=\frac{5}{2}T_{t}^{3}(\frac{p^{2}\chi} {2k_{0}k_{B}^{2}})^{1/2}.$\\
Also, from boundary conditions $\upsilon =0$, thus the constant value of the heating function results
$H_{0}= 7
p^{2} \chi T^{\alpha -2}_{t}/\left(8 k_{B}^{2}(\alpha
+\frac{3}{2})\right).$
\subsection{The perturbation}
To calculate the stability and the structure of the modes the
general perturbation along the equilibrium magnetic field is
written
\begin{equation}
\vec{\xi}=[\zeta_{r}(r,z)
\mathbf{e}_{t}+i \zeta_{\phi}(r,z) \mathbf{e}_{\phi}+\zeta_{z}(r,z) \mathbf{e}_{z}]e^{im\phi} \ \ T_{1}=T_{1}(r,z)e^{im\phi}\label{9}
\end{equation}
with $r=R$.
The $\phi$ dependence only appears in the exponents that multiply the perturbation; the integration with respect to this coordinate is straightforward.
Then, representing the equilibrium functions of the different
quantities with a 0 sub-index, defining
\noindent
$$ \mathbf{e}_{t}=(Rb \mathbf{e}_{\phi}+ \mathbf{e}_{z} )/ \sqrt{\Delta} \ \ \ \nabla_{\parallel}=\mathbf{e}_{t}(\mathbf{e}_{t}\cdot\nabla)\nonumber \ \ \ \rho_{t}=\frac{m_{p}}{k_{B}T_{t}}$$
\noindent
with $\mathbf{e}_{\phi}, \mathbf{e}_{z}$ the cylindrical versors and $\mathbf{e}_{t}$ the tangential versor, we obtain a non--dimensional expression for the energy principle
of eq.~\ref{2}:
\begin{equation}
\delta^{2} W_{p}= \delta^{2} W_{c}+\delta^{2} W_{m}+\delta^{2} W_{hc}+\delta^{2} W_{r}
\label{10}
\end{equation}
where $\delta^{2} W_{c}$ is the generalized potential energy associated to compressional terms, $\delta^{2} W_{m}$ corresponds to the magnetic contributions, $\delta^{2} W_{hc}$ corresponds to the heat conduction terms and $\delta^{2} W_{r}$ to the radiative contributions. The explicit form of these functions are given in the Appendix. The Bernstein's generalized potential energy corresponds to the magnetic contribution and part of the compressional one. In the generalized version of the energy principle additional terms appear in the $\delta^{2} W_{c}$ term and also $\delta^{2} W_{hc}$ and $\delta^{2} W_{r}$ are entirely new terms.
\section{Results and discussion}
Convective motion of the photosphere is believed to provide the
energy that is storage in twisted magnetic coronal fields allowing
the presence of long--lived coronal structures until it is
released by instabilities (\citep{raa}; \citep{vrs}). On the
other hand, continuous spectra are generally associated to stability.
An accepted conjecture establishes that unstable
modes have a discrete spectrum (see
\citep{fre} or \citep{pri}). There are two types of possible continuous spectra in this problem. The inhomogeneous character of the equilibrium parameters along the loop axis can lead to a continuum that couples to the Alfv\'en continuum
\citep{bel}; e.g, when the disturbances considered are comparable to the inhomogeneous characteristic wavelength stable eigenvalues can give rise to a continuous
spectrum ($L/2$, the equilibrium structure in the $z$ component). This is the case studied in Paper I. On the other hand, GH
established, for non vanishing helicity systems, that
there is a continuous spectrum associated with the line-tied Alfv\'en
resonance leading to the damping and heating by the resonant absorption mechanism and thus, directly relate to
the stability of loops. They also pointed out how to obtain the
resonant singular limit $\omega$, from the class of physically
permissible solutions,
\begin{equation}
\omega(r)=\frac{nB_{z}(r)}{\int_{-L}^{L}\sqrt{\rho(z)}dz}.
\label{11}
\end{equation}
This resonance results because of the absence
of an explicit dependence on the azimuthal magnetic field component ($B_{\varphi}$).
Thus, in order to understand in which conditions which mechanism can dominate
and give account of the different
scenarios i.e., the driving of the
instability or the damping of mode oscillations, it is critical to gain
knowledge about the dynamics and energetic contribution of twisted structures.
Yet, the implications of the twisting in theoretical and
observational descriptions are poorly known; e.g., there is no
clarity about the modification of the dispersion relation and observational data
are indirectly inferred.
In this paper we focused our attention to describe the changes in
periods, stability and mode structure of coronal loops when the
helicity, the magnetic field intensity and the radius are varied. For loops
with vanishing helicity it is well established that the Alfv\'en
line--tied resonance continuum is responsible of the damping of
kink ($m=1$) quasi--modes via the transfer of energy from the
radial component into the azimuthal one, i.e., from discrete
global modes into the local continuum modes where phase mixing
can take place. Still, the twisting of the magnetic field leads to
the coupling of MHD cylindrical modes making difficult to provide
a classification in terms of the behavior of pure--like modes.
In order to calculate modes and frequencies we followed the
schematic procedure described in Paper I and in
\citep{gal1}. We used a symbolic manipulation program to integrate
the equations. $\delta^{2}W_{p}$ and the perturbations were
expanded in a six dimensional--Fourier basis on the independent
coordinate $z$ that adjusts to
border conditions, i.e., the four perturbated components (eq.~\ref{9})
were expanded in a six mode basis to
obtain $24$ eigenvalues and eigenvectors for each of the helicity
and the magnetic field values.
Only the first
eighteen eigenvalues were considered (the others are more than two order of magnitude smaller
and accumulate at zero; the eigenvectors are also
vanishingly small).
Thus, a quadratic form for $\delta^{2}W_{p}$
was obtained and was minimized with the Ritz variational
procedure. A matrix discrete eigenvalue problem subject to a
normalization constraint was obtained. From the resulting modes and the
generalized potential energy (eq.~\ref{2}): $\delta^{2}W_{p}\geq0$ the stability of each mode was determined.
The coronal loop parameters used were: $L=10^{10}cm$ (or $L=100Mm$),
$T_{b}=10^{4}K$ $T_{t}=10^{6}K$ $n_{e}=10^{8}cm^{-3}$ electron
number density $p_{t} =2k_{B}T_{t} \;$;
$\rho_{t}=m_{p}p_{t}/k_{B}T_{t}$.
Frequencies and modes were calculated for
two different values of the magnetic field: $B_{0}=10G$ and
$B_{0}=100G$, and for three different values of the helicity $b=(3.1 \ 10^{-8} ; \ 3.1
\ 10^{-7}; \ 1.9 \ 10^{-6})$ which correspond to the adimentional values:
$b_{a}=(2.8; \ 28; \ 170)$ with $N_{t}=(0.45; \ 4.3; \ 13.7)$, $N_{t}:$ the
number of turns
over the cylinder length. These helicity values defined as weak,
moderated and strong helicity respectively correspond to the
classification given in \citep{asc4} (Chapter 5).
The adimentional radius
was initially chosen as $R=0.01$.
In what
follows we summarize the conclusions obtained from the data analysis which are displayed in three
tables.
Table 1 shows the periods (in minutes) for weak, moderate and strong helicity
for two values of the magnetic field intensity ($B_{0}=10G$ and $B_{0}=100G$
(left and right panel respectively). S and U
letters indicate the stable--unstable character of the modes.
From the table we see that:
\noindent
I) Weak helicity modes are stable. This is in accordance with the analytic results by \citet{rud} who studied nonaxisymmetric oscillations of a thin twisted magnetic tube with fixed ends in a zero-beta plasma.
\noindent
II) Higher modes have an accumulation point at zero, indicating the presence of
a continuum spectra of stable modes (as in Paper I). Note that, calculus performed
via discrete basis, as in our case, give spectra that are necessarily discrete.
Thus, an
accumulation of discrete eigenvalues
suggests a stable continuum spectrum.
\noindent
III) The $B_{0}=10G$ case has larger periods than the $B_{0}=100G$ one.
For moderate and strong helicity the eigenvalues of the first panel follow a scaling
law with that of the second one i.e., they scale with the magnetic field intensity exactly as the Alfv\'en speed does $P_{10G}\simeq 10 P_{100G}$.
\noindent
IV) As HG and GH, we note a clustering of the spectra associated to the
change from real to imaginary eigenvalues (and viceversa). There is a pronounced change (in the spacing of the periods or/and in the stability) from the sixth mode to the seventh mode. This is noted by a double line in Table 1 and related to the importance of the parallel component with respect to the perpendicular component (see Table 2).
Up to period number ten real -- imaginary eigenvalues of the first panel
($B_{0}=10G$) correspond to real--imaginary ones of the second panel ($B_{0}=100G$). Also,
excepting large order periods $n>10$, when
the helicity is increased from weak to moderate the imaginary stable
eigenvalues turn to imaginary unstable ones. For $B_{0}=10G$ and weak
and moderate helicity cases there are five different groups of periods
($P_{1}-P_{6};P_{7}-P_{10};P_{11}-P_{12};P_{13}-P_{14};P_{15}-
P_{18}$) (see also Table 2). The clustering is more difficult to establish i.e., the
differences are less pronounced, with increasing magnetic field
intensity and larger order periods.
\noindent
In order to compare our results with those given by these authors
we calculated the expression eq.~\ref{11} for our modes. We found that, all periods excepting $P_{1}-P_{6}$ weak helicity modes satisfy
the relation and thus, they belong to the Alfv\'en continuum spectrum
justifying the scaling law described in III.
As HG, we conclude that the change in the real--complex character of the
$P_{6}-P_{7}$ eigenvalues is associated to the existence of an accumulation
point of the resonant Alfv\'en continuum, however we find that this change is not necessarily related to a change
in the stability as they claimed. Note that all modes with weak
helicity are stable (even the imaginary ones); in all the other cases the imaginary character of the eigenvalues is associated to instability.
Yet, the continuum stable eigenvalue conjecture
is here still valid \citep{fre}, \citep{pri}; it applies to a spectrum with an accumulation point in zero; we found stable modes for all the helicity values and for the two magnetic field values with $P_{n>14}$. Note that the analysis of stable modes is still of interest because depending on the relative characteristic times of stable and unstable modes the stable ones could be active
and accessible to observations.
\noindent
The presence of at least one unstable mode means that the equilibrium
state is unstable. Thus, taking into account the whole range of stable modes, we confirm previous results
leading to conclude that field configurations with some
degree of twisting give a stabilizing effect allowing the storage
of magnetic energy \citep{raa2}, i.e., when the helicity is augmented the stable weak case turns to an unstable one suggesting a critical value.
\bigskip
In Paper I, we obtained only one unstable mode classified as slow
magnetoacoustic mode due to the almost longitudinal character (parallel to the
magnetic field) of the wavevector perturbation
and to the fact that the period did not changed with the
intensity of the magnetic field, resembling acoustic waves with sound speeds,
$v_{s}$, independent of the magnetic field. The characteristic
unstable time obtained in Paper I was $\tau_{u}=36 \ min$, corresponding
to a typical slow magnetoacoustic fundamental period with a characteristic
wavelength of the order of the loop length $L/2$. Also,
we obtained a continuous set of stable modes classified as
fast magnetoacoustic modes due to their large value component orthogonal
to the magnetic field and to the fact that the eigenvalues scale with
the intensity of
the magnetic field as $P_{11G}\simeq 10 P_{100G}$; thus resembling
the dependence of the
Alfv\'en waves $v_{A} \sim B_{0}$.
Table 2 (First Panel) displays the resulting features associated to the
relative intensity of the parallel and perpendicular to the field components
($(\xi_{\parallel},\xi_{\perp},\xi_{r}) $ is an orthogonal basis)
and their classification as slow--like (S) or fast--like (F). The relative
phase between the components is also indicated in the table
by P (in phase) and IP (inverted phase). Table 2 (Second Panel) also shows the
intensity relationship between the cylindrical components.
In order to classify the modes and to compare
with the slow and fast
magnetoacoustic modes obtained in Paper I, we calculated the cylindrical
mode components and also the tangential and normal to the field
components
($\xi_{\parallel}=
(Rb\xi_{\phi}+\xi_{z})/\Delta; \xi_{\perp}=
(\xi_{\phi}-Rb\xi_{z})/\Delta$).
Our interest in the $\xi_{\parallel}$,
$\xi_{z}$, $\xi_{r}$, $\xi_{\perp}$
and $\xi_{\phi}$ components resides in that: First, when the helicity is weak, the $\xi_{\parallel}$
component is expected to play the slow-mode role of $\xi_{z}$
in Paper I.
Second, the $\xi_{r}$ component is related to the
fast modes and determines
the resonant absorption mechanism when uniform cylindrical
flux tubes are considered by
the transferring of energy
to the $\xi_{\phi}$ component. When helicity and inhomogeneous distribution of equilibrium parameters are present it is worth investigating the transferring
of energy from the $\xi_{r}$ component to the others. In this
case the resonant damping of global oscillations
will occur by conversion of kinetic energy of the
radial component into kinetic energy of the
$\xi_{\parallel}$ and $\xi_{\perp}$ components;
both components forming the plane orthogonal to
$\xi_{r}$, and equal to the plane formed by $\xi_{\phi}$ and $\xi_{z}$.
From the analysis of the amplitude of the components of the $P_{1}-P_{6}$ modes with respect to the $P_{7}-P_{18}$ ones in the weak helicity case i.e., real and imaginary eigenvector respectively, we could classify the first ones as slow-like modes because: I) their tangential components $\xi_{\parallel}$
are at least an order of magnitude larger
than the normal ones $\xi_{\perp}$; II) as the helicity is
weak $\xi_{\parallel}\approx \xi_{z}$ and
$\xi_{r}\rightarrow 0; \xi_{\phi}\rightarrow 0$, the
wavevector is almost tangential to the magnetic field; III) they have
a larger characteristic time and a shorter characteristic speed than the imaginary eigenvectors. On the contrary, imaginary eigenvalues are associated to large values of
the $\xi_{r}$ component and $\xi_{\perp}$ component (due to large values of $\xi_{\phi}$ (see Table 2 Second Panel)) , and small values of
the $\xi_{\parallel}$ and $\xi_{z}$ components.
As in Paper I, when the eigenvalues change form real to imaginary
the period strongly diminishes and a change in
the type of mode from the slow to fast magnetoacoustic type occurs. In opposition to Paper I where the
acoustic mode has the same eigenvalue for both magnetic field
intensities, here the modes are affected by the strengthening of
the magnetic field leading to an-order-of-magnitude shorter period than in the non--helicity case.
The $\xi_{\parallel}$ and $\xi_{\perp}$ components are in an inverted phase for real eigenvector modes and in phase for imaginary eigenvector modes.
For moderate helicity the overall description is similar but all the cases having non vanishing
$\xi_{\phi}$ component and all the periods in the resonant
line-tied continuum. As was mentioned, real--imaginary eigenvalues
correspond to stable--unstable behavior.
In the strong helicity case, as the weak and moderate ones, we
note for $P_{1}-P_{6}$ larger, but comparable, values of the
$\xi_{\parallel}$ component with respect to the $\xi_{\perp}$
component. In this case the two components of the mode are in phase.
This relationship between the $\xi_{\parallel}$ and $\xi_{\perp}$
components of Table 2 (FP), and their associated phases is found again in the
modes with $P_{15} - P_{18}$. In spite that these features are associated to the slow
magnetoacoustic characterization, Table 2 (SP) shows that as $\xi_{z}$ is
vanishingly small, the strong helicity case cannot be classify as a
slow mode.
\bigskip
When helicity is present the mixed character of the modes manifests itself making difficult to identify the components that are involved in the damping mechanism. However, taking into account the resonant frequency of eq.~\ref{11}, we noted that (HG) all the modes, except those with $P_{1}-P_{6}$
periods of the weak helicity case, have resonant frequencies suggesting
that resonant
absorption in helical modes is associated to modes with significant values of
$\xi_{\perp}$ component. If this argument is correct we can affirm
that the damping mechanism of body helical modes
is associated to the transfer of kinetic energy of the radial component into
kinetic
energy of the $\xi_{\perp}$ component which is not only related to the $\xi_{\phi}$
cylindrical contribution but also to the $\xi_{z}$ one
by the expression
$ \xi_{\perp}=(\xi_{\phi}-Rb\xi_{z}) \Delta$.
\bigskip
We also analyzed the change of the period as a function of the radius for different values of the helicity.
We found that, for weak helicity, the increasing of the radius leads to a decrease of the periods. This is in accordance with observations, e.g., observed sausage modes are associated with thicker and denser loop structures and lower periods; while in other case (unstable cases) the increasing of the radius leads to an increase of the period.
\bigskip
Table 3 -First and Second Panel- shows the variation of the radius $R$ with the twist
$bR$ for weak and moderate helicity respectively.
\citep{rud} has conjectured that the line-tying condition at the tube ends should stabilize the tube and has suggested a critical value ($\sim L b<q$; with $q$ a positive constant and $L$ the loop length) for the onset of instability. Also,
\citep{lin} found that when the helicity grows beyond a critical value, the kink isolated twisted magnetic flux tubes below the photosphere become unstable. In fact, Table 3 can be seen as the variation of $R$ with the twist value: $bR$, for constant values of the helicity $b$ in the two cases: weak and moderate respectively. Stability is guaranteed when the loop radius is varied between $R=0.01$ and $R=0.1$ and the helicity is weak $b=0.05$ (for almost the same value of the length of the loop, $L$). However, when the helicity is incremented to
$b=0.5$ even for the radius of $R=0.01$ the loop structure is unstable, thus, instability can be associated with the presence of helicity values larger than a critical one.
\bigskip
Figure~\ref{fig:tres} shows the general potential energy for $P_{6}$ and
$P_{7}$ in the weak and moderate cases. Note the change
of this function when the system turns from stable to unstable, as
helicity is augmented i.e., from $\delta^{2} W_{p}>0$ to
$\delta^{2} W_{p}<0$. Figure~\ref{fig:tres}a and Figure~\ref{fig:tres}c
display the total energy composed by the compressional,
radiative, thermal and magnetic energy contributions of $P_{6}$ mode in the weak and moderate case respectively.
The same features but for the $P_{7}$ mode are shown in Figure~\ref{fig:tres}e and Figure~\ref{fig:tres}g. Figure~\ref{fig:tres}b and Figure~\ref{fig:tres}d show the
magnetic energy content alone for $P_{6}$ mode in the weak and moderate case respectively. Figure~\ref{fig:tres}f and Figure~\ref{fig:tres}h show the magnetic energy content for $P_{7}$ mode and for the weak and moderate case respectively. It can be seen, in this and in all
the other cases, that the magnetic energy content has a determinant role on
the stability--instability of the system, i.e., the stability changes when the magnetic generalized potential energy changes sign. Thus, a result of
this analysis is that the stability of twisted coronal loops is
fundamentally determined by the storing of magnetic energy, being the
other contributions less significant. Meanwhile, when the helicity is weak or vanishingly small and the magnetic contribution has a stabilizing effect the other non-dominant contributions, as the non-adiabatic ones, can play an important role. This makes possible, for example, the damping of fast excitations due to resonant absorption. Yet, even when one of these contributions is unstable, stable modes could be active for a while if their characteristic periods are shorter than the characteristic time of the instability. This is the case of Paper I, where we obtained a slow mode with an unstable characteristic times of $\tau\sim 36 \ min$ coexisting with stable fast modes with periods about $P\sim 1 \ min$; moreover, we showed that the instability can be nonlinearly saturated giving rise to a limit-cycle solutions, i.e., an oscillation between
parallel plasma kinetic energy and plasma internal energy where
the magnetic energy plays no relevant role.
Thus, the
contribution to the stable--unstable character of the modes is
mostly due to the magnetic energy content and not to other
energetic contributions.
Note that as the balance energy equation takes into account non--adiabatic contributions, i.e., radiation, heat flow and heat function (with $L=0$ at the equilibrium), the resulting perturbations are not constrained to the force--free condition. So, one result of the analysis is that the pertubation energy contribution is mainly due to magnetic forces.
Thus, for these type of twisted magnetic field
models, non--adiabatic perturbations (e.g. thermal perturbations) and resonant absorption seem unimportant to guarantee stability; a loop system with weak
storage of magnetic energy (low values of the
helicity) could be released if the helicity is suddenly
increased, e.g., by footpoint motions. Meanwhile all the "zoo" of the coronal seismology can be active and accessible to observations.
\begin{figure*}
\includegraphics[width=4.3cm]{P6ewp.eps}
\includegraphics[width=4.3cm]{P6emwp.eps}
\includegraphics[width=4.3cm]{P6emp.eps}
\includegraphics[width=4.3cm]{P6emmp.eps}
\includegraphics[width=4.3cm]{P7ewp.eps}
\includegraphics[width=4.3cm]{P7emwp.eps}
\includegraphics[width=4.3cm]{P7emp.eps}
\includegraphics[width=4.3cm]{P7emmp.eps}
\caption{Energy content of the sixth and seventh mode for $B_{0}=10G$. a) Total potential energy and b) magnetic potential energy respectively for the sixth mode $P_{6}=1.23 \ min$ and for weak helicity. c) Total potential energy and d) magnetic potential energy respectively for the sixth mode $P_{6}=1.23 \ min$ and for moderate helicity.
e) Total potential energy and f) magnetic potential energy respectively for the seventh mode $P_{7}=0.07 \ min$ and for weak helicity. g) Total potential energy and h) magnetic potential energy respectively for the sixth mode $P_{7}=0.07 \ min$ and for moderate helicity.}
\label{fig:tres}
\end{figure*}
\begin{table*}
\begin{tabular}{cccccccc}
\hline
$P_{i}$&$weak $&$moderate$&$strong$&$ \ \ \ $& $weak$&$moderate$&$strong$ \\ \hline
$P_{1}$&$1.921 \ S $&$0.209 \ S$&$0.525 \ i \ U$&$ \ \ \ $&$0.159 \ S$&$0.021 \ S$&$0.052 \ i \ U$\\ \hline
$P_{2}$&$1.869 \ S $&$0.204 \ S$&$0.450 \ S$&$ \ \ \ $& $0.158 \ S$&$0.020 \ S$&$0.044 \ S$\\ \hline
$P_{3}$&$1.535 \ S $&$0.169 \ S$&$0.430 \ S$&$ \ \ \ $& $0.154 \ S$&$0.017 \ S$& $0.042 \ S$ \\ \hline
$P_{4}$&$1.533 \ S $&$0.168 \ S$&$0.424 \ i \ U$&$ \ \ \ $&$0.153 \ S$&$0.0167 \ S$&$0.042 \ i \ U$\\ \hline
$P_{5}$&$1.306 \ S $&$0.143 \ S$&$0.206 \ i \ U$& $ \ \ \ $&$0.151 \ S$&$0.014 \ S$&$0.020 \ i \ U$\\ \hline
$P_{6}$&$1.228 \ S $&$0.135 \ S$&$0.177 \ i \ U$&$ \ \ \ $& $0.15 \ S $&$0.013 \ S$&$0.017 \ i \ U$\\ \hline \hline
$P_{7}$&$0.068 \ i \ S$&$0.070 \ i \ U$&$0.125 \ S$&$ \ \ \ $& $0.0047 \ i \ S$&$0.007 \ i \ U$&$0.0125 \ S$\\ \hline
$P_{8}$&$0.064 \ i \ S$&$0.066 \ i \ U$&$0.122 \ S$&$ \ \ \ $& $0.0046 \ i \ S$&$0.006 \ i \ U$&$0.012 \ S$\\ \hline
$P_{9}$&$0.042 \ i \ S $&$0.044 \ i \ U$&$0.101 \ S$&$ \ \ \ $& $0.0044 \ i \ S$&$0.0043 \ i \ U$&$0.0101 \ S$\\ \hline
$P_{10}$&$0.041 \ i \ S$&$0.043 \ i \ U $&$0.100 \ S$&$ \ \ \ $& $0.0043 \ i \ S$&$0.0042 \ i \ U$&$0.01 \ S$\\ \hline
$P_{11}$&$0.033 \ S $&$0.036 \ S$&$0.989 \ S$&$ \ \ \ $& $0.0042 \ i \ S$& $0.0036 \ S$&$0.099 \ S$\\ \hline
$P_{12}$&$0.032 \ S $&$0.035 \ S$ &$0.096 \ S$&$ \ \ \ $& $0.0041 \ i \ S$&$0.0035 \ S$ &$0.0096 \ S$\\ \hline
$P_{13}$&$0.030 \ i \ S $&$0.031 \ i \ U$ &$0.085 \ S$&$ \ \ \ $&$0.003 \ S$&$0.0031 \ i \ U$ &$0.0086 \ S $\\ \hline
$P_{14}$ & $0.027 \ i \ S $&$0.029 \ i \ U$&$0.081 \ S $&$ \ \ \ $&$0.0026 \ S$&$0.003 \ i \ U$&$0.0081 \ S $\\ \hline
$P_{15}$&$0.025 \ S $&$0.027 \ S $&$0.077 \ S $&$ \ \ \ $&$0.0025 \ S $&$0.0027 \ S $&$0.0077 \ S $\\ \hline
$P_{16}$&$0.024 \ S $&$0.026 \ S $&$0.076 \ S $&$ \ \ \ $&$0.002 \ S $&$0.003 \ S $&$0.0076 \ S $\\ \hline
$P_{17}$&$0.02 \ S $&$0.02 \ S $&$0.063 \ S $&$ \ \ \ $&$0.0024 \ S $&$0.0021 \ S $&$0.0063 \ S $\\ \hline
$P_{18}$&$0.018 \ S $&$0.02 \ S $&$0.059 \ S $&$ \ \ \ $&$0.0024 \ S $&$0.0025 \ S $&$0.006 \ S $\\ \hline
\end{tabular}
\caption{\label{tab:table1} Eighteen first periods associated to stable (S) and unstable (U) eigenvalues (minutes) for A) Left panel: $B_{0}=10G$ with A1) left column: weak helicity, A2) middle column: moderate helicity, A3) right column: strong helicity and B) Right panel: $B_{0}=100G$ with B1, B2, B3 the same as in A. Larger order modes were discarded.}
\end{table*}
\begin{table*}
\begin{tabular}{ccccccc}
\hline
$P_{i}$&$weak $&$moderate$&$strong \ \ \ \ $&$weak $&$moderate$&$strong$\\ \hline
$P_{1}$&$\xi_{\parallel} \gg \xi_{\perp} \mapsto 0 \ S; \ IP $&$\xi_{\parallel} > \xi_{\perp} \ S; \ IP $&$\xi_{\parallel} \geq \xi_{\perp}
\ P \ \ \ \ $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r}\mapsto 0 $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r} $&$ \xi_{r} \leq \xi_{\phi}; \xi_{z} \mapsto 0 $\\ \hline
$P_{2}$&$\xi_{\parallel} \gg \xi_{\perp} \mapsto 0 \ S; \ IP $&$\xi_{\parallel} > \xi_{\perp} \ S; \ IP $&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r}\mapsto 0 $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r} $&$ \xi_{r} \leq \xi_{\phi}; \xi_{z} \mapsto 0 $\\ \hline
$P_{3}$&$\xi_{\parallel} \gg \xi_{\perp} \mapsto 0 \ S; \ IP $&$\xi_{\parallel} > \xi_{\perp} \ S; \ IP $&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r}\mapsto 0 $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r} $&$ \xi_{r} \leq \xi_{\phi}; \xi_{z} \mapsto 0 \ $\\ \hline
$P_{4}$&$\xi_{\parallel} \gg \xi_{\perp} \mapsto 0 \ S; \ IP $&$\xi_{\parallel} > \xi_{\perp} \ S; \ IP $&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r}\mapsto 0 $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r} $&$ \xi_{r} \leq \xi_{\phi}; \xi_{z} \mapsto 0 $\\ \hline
$P_{5}$&$\xi_{\parallel} \gg \xi_{\perp} \mapsto 0 \ S; \ IP $&$\xi_{\parallel} > \xi_{\perp} \ S; \ IP $&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r}\mapsto 0 $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r} $&$ \xi_{r} \leq \xi_{\phi}; \xi_{z} \mapsto 0 $\\ \hline
$P_{6}$&$\xi_{\parallel} \gg \xi_{\perp} \mapsto 0 \ S; \ IP $&$\xi_{\parallel} > \xi_{\perp} \ S; \ IP $&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r}\mapsto 0 $&$ \xi_{z}\gg \xi_{\phi} \sim \xi_{r} $&$ \xi_{r} \leq \xi_{\phi}; \xi_{z} \mapsto 0 $\\ \hline \hline
$P_{7}$&$\xi_{\perp} \gg \xi_{\parallel} \mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$ &$\xi_{\perp} \geq \xi_{\parallel} \ IP \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0$&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z} \mapsto 0 $&$\xi_{z} > \xi_{r} > \xi_{\phi} $\\ \hline
$P_{8}$&$\xi_{\perp} \gg \xi_{\parallel}\mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\perp} \geq \xi_{\parallel} \ IP \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{z} > \xi_{r}> \xi_{\phi} $ \\ \hline
$P_{9}$&$\xi_{\perp} \gg \xi_{\parallel}\mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\perp} \geq \xi_{\parallel} \ IP \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{z} > \xi_{r} > \xi_{\phi} $\\ \hline
$P_{10}$&$\xi_{\perp} \gg \xi_{\parallel}\mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\perp} \geq \xi_{\parallel} \ IP \ \ \ \ $& $\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{z}> \xi_{r} > \xi_{\phi} $\\ \hline
$P_{11}$&$\xi_{\perp} \gg \xi_{\parallel}\mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\perp} \geq \xi_{\parallel} \ IP \ \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi} \gg \xi_{z}\mapsto 0 $&$\xi_{z} > \xi_{r} > \xi_{\phi} $ \\ \hline
$P_{12}$&$\xi_{\perp} \gg \xi_{\parallel}\mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\parallel} \geq \xi_{\perp} \ IP \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $ &$ \xi_{r}> \xi_{\phi} > \xi_{z} $ \\ \hline
$P_{13}$&$\xi_{\perp} \gg \xi_{\parallel} \mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\perp} \geq \xi_{\parallel} \ IP \ \ \ \ $&$\xi_{r}\sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{z} > \xi_{r} > \xi_{\phi} $\\ \hline
$P_{14}$&$\xi_{\perp} \gg \xi_{\parallel}\mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$ &$\xi_{\perp} \geq \xi_{\parallel} \ IP \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{z} > \xi_{r} > \xi_{\phi} $\\ \hline
$P_{15}$&$\xi_{\perp} \gg \xi_{\parallel} \mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0$&$ \xi_{r}> \xi_{\phi} > \xi_{z} $ \\ \hline
$P_{16}$&$\xi_{\perp} \gg \xi_{\parallel} \mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$ \xi_{r}> \xi_{\phi} > \xi_{z} $ \\ \hline
$P_{17}$&$\xi_{\perp} \gg \xi_{\parallel} \mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$&$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0$&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$ \xi_{r}> \xi_{\phi} > \xi_{z} $ \\ \hline
$P_{18}$&$\xi_{\perp} \gg \xi_{\parallel} \mapsto 0 \ F; \ P $&$\xi_{\perp} > \xi_{\parallel} \ F; \ P$ &$\xi_{\parallel} \geq \xi_{\perp} \ P \ \ \ \ $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$\xi_{r} \sim \xi_{\phi}\gg \xi_{z}\mapsto 0 $&$ \xi_{r}> \xi_{\phi} > \xi_{z} $\\ \hline
\end{tabular}
\caption{\label{tab:table2} First Panel: Intensity relationship between the tangential and normal to the field components of the eighteen first periods for $B_{0}=10G$ and for weak (first column), moderate (second column) and strong helicity (third column) cases. The (P) indicates in phase and (IP) indicates inverted phase. Second Panel: Intensity relationship between the cylindrical components of the eighteen first periods for $B_{0}=10G$ and for weak (first column), moderate (second column) and strong helicity (third column) cases.}
\end{table*}
\begin{table*}
\begin{tabular}{cccccccc}
\hline
$R$&$L $&$R/2L$&$Twist=bR \ \ \ \ \ \ \ \ \ $&$R$&$L $&$R/2L$&$Twist=bR$ \\ \hline
$0.01$& $ 9.05 \ 10^{7} $&$0.005 $&$ 0.028\ \ \ \ \ \ \ \ \ $&$0.01$& $ 8.07 \ 10^{7} $&$0.005 $&$ 0.28$\\ \hline
$0.02$&$ 9.04 \ 10^{7} $&$ 0.01 $&$ 0.057 \ \ \ \ \ \ \ \ \ $&$0.015$&$ 8.32 \ 10^{7} $&$ 0.008 $&$ 0.43 $\\ \hline
$0.03$&$ 9.02 \ 10^{7} $&$ 0.015 $&$ 0.085 \ \ \ \ \ \ \ \ \ $&$0.02$&$ 7.86 \ 10^{7} $&$ 0.011 $&$ 0.57$\\ \hline
$0.04 $&$ 8.99 \ 10^{7} $&$ 0.02 $&$ 0.11\ \ \ \ \ \ \ \ \ $&$0.025 $&$ 7.38 \ 10^{7} $&$ 0.015 $&$ 0.71$\\ \hline
$0.05 $&$ 8.96 \ 10^{7} $&$ 0.025 $&$ 0.14\ \ \ \ \ \ \ \ \ $&$0.03 $&$ 6.88 \ 10^{7} $&$ 0.02 $&$ 0.85$\\ \hline
$0.06 $&$ 8.9 \ 10^{7} $&$ 0.03 $&$ 0.17\ \ \ \ \ \ \ \ \ $&$0.04 $&$ 5.97 \ 10^{7} $&$ 0.03 $&$ 1.13$\\ \hline
$0.1 $&$ 8.7 \ 10^{7} $&$ 0.05 $&$ 0.28\ \ \ \ \ \ \ \ \ $&$0.05 $&$ 5.2 \ 10^{7} $&$ 0.04 $&$ 1.42$\\ \hline
\hline
\end{tabular}
\caption{\label{tab:table5} First Panel - Stable case: Variation of the Radius with the Twist for weak helicity $b=0.05$ and $B_{0}=10G$. Second Panel - Unstable case: Variation of the Radius with the Twist for moderate helicity $b=0.5$ and $B_{0}=10G$.}
\end{table*}
\section{Appendix: Generalized potential energy terms}
From the procedure described above and extensively exemplified in Paper I we
can obtain -laboriously but in a straightforward way- the explicit terms for the energy principle given in eq.~\ref{10}:
$$\delta^{2} W_{p}= \delta^{2} W_{c}+\delta^{2} W_{m}+\delta^{2} W_{hc}+\delta^{2} W_{r} $$
where the right side of the equation corresponds to the compressional, magnetic, heat conduction and radiative contributions respectively.
The compressional term $ \delta^{2} W_{c}=\delta^{2} W_{c1}+\delta^{2} W_{c2}$ has an additional contribution ($\delta^{2} W_{c2}$) with respect to Bernsteins principle:
$$ \delta^{2} W_{B}= \delta^{2} W_{c1}+ \delta^{2} W_{m}$$
$$
\delta^{2} W_{c1}=\frac{1}{2}\int_{-1}^{1}dz \beta\left\{ T_{0} \rho_{0} (1-m) \frac{\xi_{r}^{2}}{R^{2}}-\frac{m}{R}T_{0}\left(\Delta\frac{d\rho_{0}}{ds}\xi_{z}\xi_{\phi}+\rho_{0}\left(\frac{\xi_{r}\xi_{\phi}}{R} -\right.\right.\right.$$
$$ \left. \left. -\frac{m}{R}\xi_{\phi}^{2}+\frac{d\xi_{\phi}}{dz}\right)\right)+\Delta \frac{dT_{0}}{ds}\left(\Delta \frac{d\rho_{0}}{ds}\xi_{z}^{2}+\rho_{0}(\frac{\xi_{r}\xi_{\phi}}{R} -\frac{m}{R}\xi_{\phi}\xi_{z}+\xi_{z}\frac{d\xi_{z}}{dz}) \right) +
$$
$$
+T_{0}
\left(\Delta^{2}\frac{d^{2}\rho_{0}}{ds^{2}}\xi_{z}^{2}+\Delta\frac{d\rho_{0}}{ds}
\xi_{z}\frac{d\xi_{z}}{dz}+\rho_{0}(\frac{\xi_{z}}{R}\frac{d\xi_{r}}{dz}-
\frac{m}{R}\xi_{z}\frac{d\xi_{\phi}}{dz}
+\xi_{z}\frac{d^{2}\xi_{z}}{dz^{2}})+\right.
$$
$$\left. \Delta\frac{d\rho_{0}}{ds}(\frac{\xi_{r}\xi_{\phi}}{R} -\frac{m}{R}
\xi_{\phi}\xi_{z}+\xi_{z}\frac{d\xi_{z}}{dz}) \right)
$$
The magnetic contribution is:
$$ \delta^{2} W_{m}=C_{1}\left\{ \beta \Delta \left (\frac{m}{R}B_{\phi}B_{z}\xi_{r}
\xi_{\phi}-B_{\phi}B_{z}\frac{d\xi_{r}}{dz}\xi_{z}\right. \right. $$
$$
-\left. (\frac{B_{\phi}B_{z}}{R}+B_{\phi}\frac{dB_{z}}{dr})
\xi_{r}^{2}+(\frac{m}{R}B_{\phi}B_{z}\xi_{r}\xi_{\phi}+B_{\phi}
\frac{dB_{\phi}}{dr} \xi_{r}^{2})\right) -$$$$
-\beta \left( (B_{z}^{2}\frac{d\xi_{r}^{2}}{dz}+\frac{m^{2}}{R^{2}}
B_{\phi}^{2}\xi_{r}^{2})+(B_{\phi}^{2}\frac{d\xi_{z}^{2}}{dz}+2B_{\phi}+
\frac{dB_{\phi}}{dr}\xi_{r}\frac{d^{2}\xi_{z}}{dz}+
\frac{dB_{\phi}^{2}}{dr}\xi_{r}^{2}+B_{z}^{2}\frac{d^{2}
\xi_{\phi}}{dz})+\right.
$$$$
\left. \left. \left ( (B_{z}+R\frac{dB_{z}}{dr})^{2}\frac{\xi_{r}^{2}}{R^{2}}-
2\frac{m}{R^{2}}B_{z}(B_{z}+R\frac{dB_{z}}{dr})
\xi_{r}\xi_{\phi}+(\frac{m}{R}B_{z})^{2}\xi_{\phi}^{2}+
(\frac{m}{R}B_{\phi})^{2}
\xi_{z}^{2}\right) \right) \right\}
$$
The heat conduction term results:
$$\delta^{2} W_{hc}=-C_{2}\left\{5\frac{T_{0}^{3/2}}{\Delta}
\frac{dT_{0}}{ds}T_{1}\frac{dT_{1}}{dz}+T_{1}^{2}
\left( -T_{0}^{5/2}(\frac{mb}{\Delta})^{2}\frac{15}{4}T_{0}^{1/2}
\frac{dT_{0}^{2}}{ds}+\right.\right.
$$
$$
\left. \left. +\frac{5}{2}T_{0}^{3/2}\frac{d^{2}T_{0}}{ds^{2}}\right)
+\frac{1}{\Delta^{2}}T_{0}^{5/2}T_{1}\frac{d^{2}T_{1}}{dz^{2}}\right\}
$$
The new compressional contribution is expressed as:
$$\delta^{2} W_{c2}=-\beta \left(\frac{m}{R} \rho_{0}\xi_{\phi}T_{1}+\Delta\frac{d \rho_{0}}{ds}
\xi_{z}T_{1}+\rho_{0}\xi_{z}\frac{dT_{1}}{dz}\right)
$$
and the term associated to radiation results:
$$
\delta^{2} W_{r}=-\alpha T_{1}^{2}\rho_{0}^{2}T_{0}^{\alpha-1}-\beta \left(\frac{m}{R} \rho_{0}\xi_{\phi}T_{1}+\Delta\frac{d \rho_{0}}{ds}
\xi_{z}T_{1}+\rho_{0}T_{1}\frac{d\xi_{z}}{dz}+
\frac{\rho_{0}}{R} \xi_{r}T_{1}\right)$$
\noindent
where the following changes were made:
$$\rho\rightarrow\frac{\rho}{\rho_{t}}; \ \
T\rightarrow\frac{T}{T_{t}}; \ \
B_{\phi,z}\rightarrow\frac{B_{\phi,z}}{B_{0}}; \ \ b\rightarrow b S$$
$$ r,z\rightarrow\frac{r,z}{S}; \ \ \delta^{2}
W_{p}\rightarrow\ \delta^{2} W_{p}/\left(\chi
T_{t}^{\alpha+1}\rho_{t}^{2}L/m_{p}^{2}\right),$$
$S=\Delta L$ and the non--dimensional constants: $$C_{1}=
\rho_{t}^{2}T_{t}^{\alpha+1}B_{0}^{2}/(\mu_{0}k_{B} T_{t}n_{e}); \ \
C_{2}=c T_{t}^{\frac{7}{2}-\alpha}/(S^{2} n_{e}^{2}).$$
were used. All the quantities were defined as in Paper I.
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"redpajama_set_name": "RedPajamaArXiv"
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These are the moments when a misanthrope, like myself, appreciates the nameless, faceless people.
Based on your application, your new, very important profession under the Green New Deal will be Atrazine Water Chemist and your new salary will be $120,735/yr.
You know what's funny about all this unwilling to work crap?
I truly am unwilling to work. I pulled my Socialism Support Card back when Zero was presidents. Heck, everyone else was getting freebies, it was my turn! Unfortunately, I've been too successful with that plan, I still refuse to work.
In fact, my wife who still holds a job, just for the insurance. I pester her constantly about finding something simpler, with less pay. While we will always take more money, it's just not worth it to work hard doing so, especially just to give a good percentage of it to someone else.
I still technically pull an income, I just don't really work for it. The best part about it, since I'm not getting a W2, I no longer pay Social Insecurity.
Wouldn't it be cheaper if the Billionaires made AOC disappear Stalin Style?
I can't wait! I'm gonna be a Senior Buzzfeed Proofreader at $140,370.00 a year.
Three steps before "Profit!"? THREE! That's one more step, than when I was "willing" to work! What kind of capitalist conspiracy is this!?
Spirit Cooking Chef: $90,598/yr…crap! I want to be a Buzzfeed Proofreader!
Atrazine Water Chemist: $123,000/yr!! wOO hOO!
Fast Food Kiosk Screen Smasher $173,567/year–Wow! I am ready to be destructive!
The People's Cube…for Gen Z!
Good thing I didn't state that I wanted to meet Homer.
That would've been awkward, me being a budding Epic Poem Composer whose starting salary is $135,572 !
Black Market Toilet Paper Investigator…$100,451/yr.
I legit LOL'd at the gender options: 'Attack Helicopter'. I picked that, natch.
Thawed ice salesman, Government Accounts, $128,000 per year plus commission.
Move over Amy Schumer, I'm gonna be like a Plus Sized Hand Model pullin down $127,356.41 a year.
I'm a certified JB Welder. How much will that be worth?
Years ago a local freebie newspaper had a column written by a fortune teller of some sort. A young woman wrote in asking if she and her boyfriend would ever get married. The woman said her boyfriend never had a job because working would "infringe on his liberty." Funny how he didn't care about the liberty of people who did work and had to give up some of their money to support someone who refused to work.
Way better than engineering! count me in … or out…. or whatever.
My work will never be done given the rate of growth in this field lately.
Job security AND I get to be cruel ?!
You! Yeah, you by the water cooler. You're a fluffnutter today. Don't ask. Just check in with me after work to see if you've evolved into another gender by then. Wait. make it 6pm. Want to warm up with a few shots of tequila first.
Why have objective standards when we can continue to move to a purely tribal-based system of justice? It's been tribal enough for years, but there has unfortunately been some attempt to maintain objectivity and judge the individual on their merits instead of their membership in a class. That won't work much longer. As Selective Outrage Project Manager, you'll get to pick and choose whose malfeasances are ignored- and whose are magnified- simply by their superficial qualities!
So does this mean I would be a journalist?
Based on your application, your new, very important profession under the Green New Deal will be Paint-drying Watcher and your new salary will be $96,294/yr.
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"redpajama_set_name": "RedPajamaC4"
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\section{Introduction and Summary}\label{first}
In 2009 Petr Ho\v{r}ava formulated new
proposal of quantum theory of gravity
(now known as
Ho\v{r}ava-Lifshitz gravity (HL
gravity))
that is power counting renormalizable
\cite{Horava:2009uw,Horava:2008ih,Horava:2008jf}
that is also expected that it
reduces do General Relativity in the
infrared (IR) limit
\footnote{For review and
extensive list of references, see
\cite{Horava:2011gd,Padilla:2010ge,Mukohyama:2010xz,Weinfurtner:2010hz}.}.
The HL gravity is based on
an idea that
the Lorentz symmetry is restored in IR
limit of given theory while it is absent
in its high energy regime. For that reason
Ho\v{r}ava considered
systems whose scaling at short
distances exhibits a strong anisotropy
between space and time,
\begin{equation}
\bx' =l \bx \ , \quad t' =l^{z} t \ .
\end{equation}
In order to have power counting
renormalizable theory we have to demand that $z\geq 3$
in $(3+1)$ dimensional space-time. It turns out however
that the symmetry group of given theory is reduced from the full
diffeomorphism invariance of General Relativity to the foliation
preserving diffeomorphism
\begin{equation}\label{fpdi}
x'^i=x^i+\zeta^ i(t,\bx) \ , \quad
t'=t+f(t) \ .
\end{equation}
Due to the fact that the diffeomorphism is restricted (\ref{fpdi})
one more degree of freedom appears that is a spin$-0$ graviton. The
existence of this mode could have very significant consequences
either for the consistency of given theory or for the
phenomenological applications of HL gravity.
For that reason it
would be desirable to formulate HL gravity where the number of the
physical degrees of freedom is the same as in case of General
Relativity. Such a proposal was formulated by
Ho\v{r}ava and Malby-Thompson
in \cite{Horava:2010zj} in the context of the projectable HL gravity
\footnote{See also \cite{Huang:2010ay} and
\cite{Das:2011tx}}.
Their construction is
based on an
extension of
the foliation preserving diffeomorphism in
such a way that the theory is invariant
under additional
local $U(1)$ symmetry. The
resulting theory is known as non-relativistic covariant theory of
gravity.
It was shown in \cite{Horava:2010zj,daSilva:2010bm} that the
presence of this new symmetry implies that the spin-0 graviton
becomes non-propagating and the spectrum of the linear fluctuations
around the background solution coincides with the fluctuation
spectrum of General Relativity.
It is also well known that General Relativity contains large number
of symmetries. Fixing all these symmetries we find that there are
only two physical degrees of freedom left. Then we can ask the
question whether it is possible to formulate the action for these
physical degrees of freedom that is not based on the principle of
covariance of the action under general diffeomorphism. The
construction of such an action was proposed recently in two very
interesting papers \cite{Khoury:2011ay,Khoury:2013oqa}.
The basic idea presented there was to perform the
conformal traceless decomposition of the gravitational field
\cite{Brown:2005aq} so that we have one degrees of freedom
corresponding to the scale factor of the metric while we have five
degrees of freedom of the metric that is restricted to have
unit determinant. Then it was shown in \cite{Khoury:2011ay} that by
gauge fixing of the Hamiltonian constraint one can eliminate the
scale factor together with the conjugate momenta. As a result we
obtain the action for five degrees of freedom that is invariant
under spatial diffeomorphism where now Hamiltonian is determined by
the solving of the Hamiltonian constraint of the General Relativity
for the momentum conjugate to the scale factor. This analysis was
further generalized in a very nice paper in \cite{Khoury:2013oqa}
where the starting point was the action for the five physical
degrees of freedom where it is required that given theory is
invariant under spatial diffeomorphism. In other words we demand
that the constraints that are generators of the spatial
diffeomorphism are the first class constraints. We also have to
require that these generators are preserved during the time
evolution of the system. Then the requirement of the closure of the
algebra of the Poisson brackets of these constraints together with
the requirement of their time preservation determines the form of
the Hamiltonian and the form of these constraints. When it is
presumed that these constraints depend on partial derivatives of
$g_{ij}$ trough the scalar curvature we find that the original
General Relativity action is reproduced.
The goal of this paper is to formulate HL gravity for the
gravitational physical degrees of freedom only in the similar way
as in \cite{Khoury:2011ay}. To do this we start
with another version of HL gravity that has
the correct
number of physical degrees of freedom and which
is known as Lagrange multiplier modified HL gravity
\cite{Kluson:2011xx}. This model is based on the formulation
of the HL gravity with reduced symmetry
group known as
\emph{restricted-foliation-preserving
Diff} (RFDiff)
HL gravity
\cite{Blas:2010hb,Kluson:2011xx}. This is the theory that is
invariant under following symmetries
\begin{equation}\label{rfdtr}
t'=t+\delta t \ , \delta
t=\mathrm{const} \ , \quad x'^i=
x^i+\zeta^i(\bx,t) \ .
\end{equation}
The characteristic property of Lagrange multiplier modified HL
gravity is an absence of the Hamiltonian constraint
\cite{Kluson:2010na} and also presence of the additional constraint
which changes the constraint structure of given theory so that
the number of physical degrees of freedom is the same as in the case
of General Relativity. Then in order to separate physical degrees of
freedom of HL gravity we perform conformal traceless decomposition
of the gravitational field, following
\cite{Brown:2005aq,Kluson:2012tw}. In this procedure we introduce
new additional scalar field with additional symmetry so that the
number of physical degrees of freedom is the same. Performing
Hamiltonian analysis we also identify two second class constraints
that, together with the gauge fixing scaling symmetry allow us to
find Hamiltonian for the physical degrees of freedom, at least in
principle. These physical degrees of freedom are
metric with unit determinant and conjugate traceless
momenta so that the number of physical degrees of freedom is the
same as in the case of General Relativity. On the other hand there
are also important differences. Since this theory arises from the
theory with the complicated second class constraints we find that
there is a very complicated symplectic structure on the phase space
of the physical degrees of freedom. Secondly, even if we can claim
that these second class constraints can be solved in principle we
find that their solutions have the form of the non-local
perturbative expansions. In other words it is hard to see how such a
theory could be useful for some practical computations or even for
its path integral formulation.
However we mean that the analysis performed here suggests very
interesting direction in further research. The starting point would
be the general form of the action for the physical degrees of
freedom as was analyzed in \cite{Khoury:2013oqa} where we now
presume that the additional term in the diffeomorphism constraint
depends either on higher order of scalar curvature as for example
$R_{ij}R^{ij}$ or it depends on $R_{ij}$ non-locally. Then we should
proceed as in \cite{Khoury:2013oqa} where we demand that the Poisson
brackets of the spatial diffeomorphism constraints close on the
constraint surface. Then from the requirement of the preservation of
these constraints during the time evolution of the system we could
determine corresponding Hamiltonian density. We hope to return to
this problem in future.
\section{Brief Review of Lagrange Multiplier
Modified HL Gravity}\label{second} We begin this section
with the brief review of the Lagrange multiplier modified RFDiff
invariant HL gravity, for more detailed treatment see
\cite{Kluson:2011xx}.
RFDiff invariant Ho\v{r}ava-Lifshitz
gravity was introduced in
\cite{Blas:2010hb}, see also \cite{Kluson:2010na}. In
\cite{Kluson:2011xx} this action was extended by introducing
Lagrange multiplier term that ensures that the spatial curvature is
constant. Explicitly,Lagrange multiplier modified RFDiff HL gravity
has the form
\begin{equation}\label{RFDaction}
S=\frac{1}{\kappa^2} \int dt d^3\bx \sqrt{h}(\tK_{ij}
\mG^{ijkl}\tK_{kl}-\mV(h)+\mG[R]\mA) \ ,
\end{equation}
where $\mG[R]=R-\Omega$, where $\Omega$ is constant, $\mA$ is
Lagrange multiplier that transforms as scalar
\begin{equation}
\mA'(t',\bx')= \mA(t,\bx) \
\end{equation}
under (\ref{rfdtr}). Further, $\tK_{ij}$ introduced in
(\ref{RFDaction}) is
modified extrinsic
curvature
\begin{equation}
\tK_{ij}=\frac{1}{2}(\partial_t h_{ij} -\nabla_i N_j-\nabla_j N_i) \
\end{equation}
that differs from the standard extrinsic
curvature by absence of the lapse $N(t)$.
Further the generalized De Witt metric
$\mG^{ijkl}$ is defined as
\begin{equation}
\mG^{ijkl}=\frac{1}{2}(h^{ik}h^{jl}+ h^{il}h^{jk})-\lambda
h^{ij}h^{kl} \ ,
\end{equation}
where $\lambda$ is a real constant that in case of General
Relativity is equal to one. Finally $\mV(h)$ is a general function
of $h_{ij}$ and its covariant derivative. The analysis performed in
\cite{Kluson:2011xx} showed that this theory possesses the same
number of physical degrees of freedom as General Relativity. For
that reason we mean that this action is a good candidate for the
conformal traceless decomposition of the gravitational field and
possible identification of the physical degrees of freedom of HL
gravity.
\section{Conformal Traceless Decomposition}
The conformal-traceless decomposition of the gravitational field was
firstly performed in \cite{York:1998hy} in its initial value problem
\footnote{For review and extensive list of references, see
\cite{Gourgoulhon:2007ue}.}. In order to implement
conformal-traceless decomposition we follow \cite{Brown:2005aq} and
define $h_{ij}$ and $\tK_{ij}$ as
\begin{equation}\label{defcon}
h_{ij}=\phi^4 g_{ij} \ , \quad
\tK_{ij}=\phi^{-2}A_{ij}+\frac{1}{3}\phi^4 g_{ij}\tau \ .
\end{equation}
We see that this definition is redundant since the multiple of the
fields $g_{ij},\phi,A_{ij},\tau$ give the same physical metric
$h_{ij}$ and modified extrinsic curvature $\tK_{ij}$. In fact, we
see that the decomposition (\ref{defcon}) is invariant under the
conformal transformation
\begin{eqnarray}\label{gaugecon}
g'_{ij}(\bx,t)&=&\Omega^4(\bx,t)g_{ij}(\bx,t) \ , \quad
\phi'(\bx,t)=\Omega^{-1}(\bx,t)
\phi(\bx,t) \ , \nonumber \\
A'_{ij}(\bx,t)&=&\Omega^{-2}(\bx,t)A_{ij}(\bx,t), \quad
\tau'(\bx,t)=\tau(\bx,t) \ .
\nonumber \\
\end{eqnarray}
We also see that (\ref{defcon}) is
invariant under following transformation
\begin{equation}\label{secondsym}
\tau'(\bx,t)=\tau(\bx,t)+\zeta(\bx,t) \ , \quad A'_{ij}(\bx,t)=
A_{ij}(\bx,t) -\frac{1}{3}\zeta(\bx,t) \phi^6 g_{ij}(\bx,t) \ .
\end{equation}
Clearly the gauge fixing of these symmetries we can eliminate $\tau$
and $\phi$.
In order to perform the Hamiltonian analysis of the conformal
decomposition of the action (\ref{RFDaction}) we firstly
rewrite the
action (\ref{RFDaction}) to its Hamiltonian form. To do this we
introduce the conjugate momenta
\begin{eqnarray}\label{defmom}
P^{ij}=\frac{\delta S}{\delta
\partial_t h_{ij}}=
\frac{1}{\kappa^2}\sqrt{h}\mG^{ijkl}\tK_{kl} \ , \quad
P_i=\frac{\delta S}{\delta \partial_t N^i}=
0 \ , \quad P_{\mA}=\frac{\delta S}{\delta \partial_t\mA}\approx 0 \ . \nonumber \\
\end{eqnarray}
Then we easily determine corresponding
Hamiltonian
\begin{equation}
H=\int d^3\bx (\partial_t h_{ij}P^{ij}-\mL)= \int d^3\bx
(\mH'_T+N^i\mH'_i) \ ,
\end{equation}
where
\begin{equation}
\mH'_T=\frac{\kappa^2}{\sqrt{h}}
P^{ij}\mG_{ijkl}P^{kl}+\sqrt{g}\mV(h)-\sqrt{h}\mA\mG(R) \ ,
\quad
\mH'_i=-2h_{ij}\nabla_k P^{jk} \ .
\end{equation}
Using the Hamiltonian and the corresponding canonical variables we
write the action (\ref{RFDaction}) as
\begin{equation} S=\int dt L=
\int dt d^3 \bx (P^{ij}\partial_t h_{ij}-\mH)= \int dt d^3\bx
(P^{ij}\partial_t h_{ij}-N \mH'_T- N^i\mH'_i) \ .
\end{equation}
Then we insert the decomposition (\ref{defcon}) into the definition
of the canonical momenta $P^{ij}$
\begin{equation}\label{Pabh}
P^{ij}=\frac{1}{\kappa^2} \sqrt{g}(\phi^{-4}\tmG^{ijkl}A_{kl}
+\frac{1}{3}\phi^2\tau \tmG^{ijkl}g_{kl}) \ ,
\end{equation}
where the metric $\tmG^{ijkl}$ is defined as
\begin{equation}
\tmG^{ijkl}=\frac{1}{2}(g^{ik}g^{jl}+g^{il}g^{jk})-\lambda
g^{ij}g^{kl} \ , \quad \mG^{ijkl}=\phi^{-8}\tmG^{ijkl} \ .
\end{equation}
Note that $\tmG^{ijkl}$ has the inverse
\begin{equation}
\tmG_{ijkl}=\frac{1}{2}(g_{ik}g_{jl}+
g_{il}g_{jk})-\frac{\lambda}{3\lambda-1} g_{ij}g_{kl} \ , \quad
\tmG_{ijkl}= \phi^8 \mG_{ijkl} \ .
\end{equation}
Using (\ref{Pabh}) and (\ref{defcon}) we rewrite $P^{ij}\partial_t
h_{ij}$ into the form
\begin{eqnarray}
P^{ij}\partial_t h_{ij}&=&
\left(\frac{1}{\kappa^2}\sqrt{g}\tmG^{ijkl}A_{kl} +
\frac{\sqrt{g}}{3\kappa^2} \phi^6(1-3\lambda)\tau
g^{ij}\right)\partial_t g_{ji}+ \nonumber \\
&+& \left(\frac{4}{\kappa^2}\sqrt{g}\phi^{-1}
A_{kl}g^{kl}(1-3\lambda)+\frac{4\sqrt{g}}{\kappa^2}
(1-3\lambda)\phi^5\tau\right)\partial_t\phi
\ . \nonumber \\
\end{eqnarray}
We see that it is natural to identify the expression in the
parenthesis with momentum $\pi^{ij}$ conjugate to $g_{ij}$ and
$p_\phi$ conjugate to $\phi$ respectively
\begin{eqnarray}\label{canmom}
\pi^{ij}&=&\frac{1}{\kappa^2}
\sqrt{g}\tmG^{ijkl}A_{kl}+\frac{\sqrt{g}}{3\kappa^2}(1-3\lambda)
\phi^6\tau g^{ij} \ , \nonumber \\
p_\phi&=&\frac{4}{\kappa^2} \sqrt{g}\phi^{-1}
A_{ij}g^{ji}(1-3\lambda)+ \frac{4\sqrt{g}}{\kappa^2}
(1-3\lambda)\phi^5 \tau \ . \nonumber
\\
\end{eqnarray}
Then using (\ref{canmom}) we obtain following primary constraint
\begin{equation}
\Sigma_D: p_\phi \phi-4\pi^{ij}g_{ji}=0 \ .
\end{equation}
As we will see below this is the constraint that generates conformal
transformation of the dynamical fields.
Further, using
(\ref{canmom}) we find the relation between $P^{ij}$ and $\pi^{ij}$
in the form
\begin{equation}
P^{ij}=\phi^{-4}\pi^{ij} \ .
\end{equation}
Then we find that the kinetic term in $\mH_T$ takes the form
\begin{equation} \frac{\kappa^2}{\sqrt{h}}
P^{ij}\mG_{ijkl}P^{kl}= \frac{\kappa^2\phi^{-6}}{\sqrt{g}}
\pi^{ij}\tmG_{ijkl}\pi^{kl} \ .
\end{equation}
As the next step we introduce the
decomposition (\ref{defcon}) into the contribution $ \int d^3\bx N^i
\mH'_i$. Using the relation between Levi-Civita connections
evaluated with the metric components $h_{ij}$ and
$g_{ij}$
\begin{equation}
\Gamma_{ij}^k(h)=\Gamma_{ij}^k(g)+ 2\frac{1}{\phi} (\partial_i\phi
\delta^k_j+\partial_j \phi\delta_i^k-\partial_l\phi g^{kl}g_{ij}) \
\end{equation}
and also if we define
$n_i$ through the relation
$N_i=\phi^4n_i$
we obtain
\begin{eqnarray}
\int d^3\bx N^i\mH'_i&=&
\int d^3\bx n^i \mH''_i \ , \nonumber \\
\end{eqnarray}
where
\begin{equation}
\mH''_i=-2g_{ik}D_j \pi^{jk}+
4\phi^{-1}\partial_i \phi g_{kl}\pi^{kl} \ ,
\nonumber \\
\end{equation}
where the covariant derivative $D_i$ is defined using the
Levi-Civita connection $\Gamma^k_{ij}(g)$.
Observe that with the help of the constraint $\Sigma_D $
we can write the constraint $\mH''_i$ as
\begin{eqnarray}
\mH''_i=
-2g_{ik}D_j \pi^{jk}+ \partial_i\phi p_\phi-
4\phi^{-1}\partial_i\phi \Sigma_D\equiv \hat{\mH}_i -
4\phi^{-1}\partial_i\phi \Sigma_D
\end{eqnarray}
so that we see that it is natural to
identify $\hat{\mH}_i$ as an independent constraint. In
fact, we will see that the smeared form of this constraint
generates the spatial diffeomorphism.
Finally we should proceed to the analysis of the spatial curvature
and generally the whole potential term $\mV$. Note that this is the
function of the covariant derivative, $R$ and $R_{ij}$. Using the
following formulas
\begin{eqnarray}
R_{ij}[h]&=& R_{ij}[g]+\frac{6}{\phi^2}D_i\phi D_j\phi-
\frac{2}{\phi} D_iD_j\phi-2\frac{g_{ij}}{\phi}D_k[g^{kl}D_l\phi]
-\frac{2}{\phi^2}
g_{ij}D_k\phi g^{kl}D_l\phi \ , \nonumber \\
R[h]&=&\phi^{-4}[ R[g] -\frac{8}{\phi}
g^{ij}D_iD_j\phi
] \ , \nonumber \\
\end{eqnarray}
Then using also the relation between Levi-Civita connections
evaluated on $h$ and $g$ we find that the potential term is
generally function of $\phi$ and $g$ whose explicit form is not
needed here. As a result we find the action in the form
\begin{eqnarray}\label{actiontraceless}
S&=&\int dt d^3\bx(\pi^{ij}\partial_t g_{ij}+ p_\phi\partial_t\phi
-n^i\hat{\mH}_i-\frac{\kappa^2\phi^{-6}}{\sqrt{g}}
\pi^{ij}\tmG_{ijkl} \pi^{kl}-\sqrt{g}\phi^6\mV(\phi,h)+\nonumber \\
&+&\sqrt{g}\phi^6 \mA
\mG(\phi^{-4}R[g]-\frac{8}{\phi^5}g^{ij}D_iD_j\phi)
-\lambda
\Sigma_D) \ ,
\nonumber \\
\end{eqnarray}
where we included the primary constraint $\Sigma_D$ multiplied by
the Lagrange multiplier $\lambda$.
Now we can proceed to the Hamiltonian analysis of the conformal
decomposition of the gravitational field given by the action
(\ref{actiontraceless}).
Clearly we have following primary
constraints
\begin{equation}
\pi_i\approx 0 \ , \pi_{\mA}\approx 0 \ ,
\quad \Sigma_D\approx 0 \ ,
\end{equation}
where $\pi_i$ and $\pi_{\mA}$ are momenta conjugate to $n^i$ and
$\mA$
with following
non-zero Poisson brackets
\begin{equation}
\pb{n^i(\bx),\pi_j(\by)}= \delta^i_j\delta(\bx-\by) \ , \quad
\pb{\mA(\bx),\pi_{\mA}(\by)}=\delta(\bx-\by) \ .
\end{equation}
Further, the preservation of the primary constraints $\pi_i$ and
$\pi_{\mA}$ implies following secondary ones
\begin{equation}
\hat{\mH}_i\approx 0 \ , \Phi_1 \equiv
\frac{1}{\kappa^2}\sqrt{g}\phi^6 \mG\approx 0 \ .
\end{equation}
Now we should analyze the requirement of the preservation of the
primary constraint $\Sigma_D$ during the time evolution of the
system. First of all the explicit calculations give
\begin{eqnarray}\label{SigmaDg}
\pb{\Sigma_D(\bx),g_{ij}(\by)}&=&
4g_{ij}(\bx)\delta(\bx-\by) \ , \nonumber \\
\pb{\Sigma_D(\bx),\pi^{ij}(\by)}&=& -4\pi^{ij}(\bx)\delta(\bx-\by) \
, \nonumber
\\
\pb{\Sigma_D(\bx),\phi(\by)}&=& -\phi(\bx)\delta(\bx-\by) \ ,
\nonumber
\\
\pb{\Sigma_D(\bx),p_\phi(\by)}&=& \phi(\bx)\delta(\bx-\by) \
\nonumber
\\
\end{eqnarray}
using the canonical Poisson brackets
\begin{equation}
\pb{g_{ij}(\bx),\pi^{kl}(\by)}= \frac{1}{2} (\delta_i^k\delta_j^l+
\delta_i^l\delta_j^k)\delta(\bx-\by) \ , \quad
\pb{\phi(\bx),p_\phi(\by)}=\delta(\bx-\by) \ .
\end{equation}
It turns out that it is useful
to introduce the smeared forms of the constraints
$\hat{\mH}_i,\Sigma_D$
\begin{equation}
\bT_S(N^i)= \int d^3\bx N^i \hat{\mH}_i \ , \quad \bD(M)=\int d^3\bx
M \Sigma_D \ ,
\end{equation}
where $N^i$ and $M$ are smooth functions on $\mathbf{R}^3$. Then
using (\ref{SigmaDg}) and also
\begin{eqnarray}
\pb{\Sigma_D(\bx),\Gamma_{ij}^k(\by)}=2
\delta_j^k\partial_{y^i}\delta(\bx-\by)+2\delta_i^k\partial_{y^j}
\delta(\bx-\by)-2g^{kl}(\by)\partial_{y^l}\delta(\bx-\by)g_{ij}
(\by) \
\nonumber \\
\end{eqnarray}
we easily find that
\begin{equation}\label{bDmHT}
\pb{\bD(M),\mH'_T(\by)}=0 \ ,
\end{equation}
where
\begin{equation}
\mH'_T=\frac{\kappa^2\phi^{-6}}{\sqrt{g}} \pi^{ij}\tmG_{ijkl}
\pi^{kl}+\sqrt{g}\phi^6\mV(\phi,g) \ .
\end{equation}
To proceed further we use following Poisson brackets
\begin{eqnarray}\label{pbbtSgp}
\pb{\bT_S(N^i),g_{ij}(\bx)}&=&-N^k\partial_k g_{ij}(\bx)-
\partial_i N^kg_{kj}(\bx)-g_{ik}\partial_j N^k(\bx) \ , \nonumber \\
\pb{\bT_S(N^i),\pi^{ij}(\bx)}&=&-\partial_k (N^k\pi^{ij})(\bx) +
\partial_k N^i\pi^{kj}(\bx)+\pi^{ik}\partial_k N^j(\bx) \ , \nonumber \\
\pb{\bT_S(N^i),\phi(\bx)}&=&-N^i\partial_i\phi(\bx) \ , \nonumber \\
\pb{\bT_S(N^i),p_\phi(\bx)}&=&-\partial_i (N^i p_\phi)(\bx) \
\nonumber \\
\end{eqnarray}
and hence it is easy to see that
\begin{equation}
\pb{\bT_S(N^i),\Sigma_D(\bx)}=-N^i\partial_i \Sigma_D(\bx)-
\partial_i N^i\Sigma_D(\bx) \
\end{equation}
that together with (\ref{bDmHT}) implies that $\Sigma_D \approx 0$
is the first class constraint.
Now we proceed to the analysis of the preservation of the secondary
constraints $ \hat{\mH}_i\approx 0$ and $\Phi_1\approx 0$. Note that
the total Hamiltonian takes the form
\begin{eqnarray}
H_T=\int d^3\bx (\mH_T'
+\lambda \Sigma_D
+n^i\hat{\mH}_i+\gamma p_{\mA}+\Gamma_{I}\Phi_1 ) \ , \nonumber \\
\end{eqnarray}
where $\gamma$ is the Lagrange multiplier corresponding to the
constraint $p_{\mA}$ while $\Gamma_{I}$ is the Lagrange multiplier
corresponding to the constraint $\Phi_1\approx 0$.
In case
of $\hat{\mH}_i$ we find following Poisson brackets
\begin{equation}
\pb{\hat{\mH}_i(\bx),\hat{\mH}_j(\by)}=
\hat{\mH}_j(\bx)\frac{\partial} {\partial
x^i}\delta(\bx-\by)-\hat{\mH}_i(\by)\frac{\partial}{\partial y^j}
\delta(\bx-\by) \
\end{equation}
which implies that the smeared form of the diffeomorphism
constraints takes the familiar form
\begin{equation}
\pb{\bT_S(N^i),\bT_S(M^j)}= \bT_S(N^j\partial_j M^i- M^j\partial_j
N^i) \ .
\end{equation}
Further using (\ref{pbbtSgp}) we easily find
\begin{eqnarray}
\pb{\bT_S(N^i),\mH'_T(\bx)}= -\partial_i N^i\mH'_T(\bx) -
N^i\partial_i \mH'_T(\bx) \ , \nonumber \\
\pb{\bT_S(N^i),\Phi_1(\bx)}=-\partial_i N^i\Phi_1(\bx)-
N^i\partial_i\Phi_1(\bx) \ \nonumber \\
\end{eqnarray}
that implies that $\hat{\mH}_i$ are the first class constraints
that are preserved during the time evolution of the system.
Finally we analyze the time evolution of the constraint
$\Phi_1\approx
0$
. Using following formulas
\begin{eqnarray}
\pb{R(\bx),\pi^{ij}(\by)}&=& -R^{ij}(\bx)\delta(\bx-\by)+D^i D^j
\delta(\bx-\by)-g^{ij} D_k D^k\delta(\bx-\by) \ ,
\nonumber \\
\pb{\Gamma_{ij}^k(\bx),\pi^{mn}(\by)}&=& \frac{1}{4}g^{kp} [
D_i\delta (\delta_j^m\delta_p^n+
\delta_j^n\delta_p^m)\delta(\bx-\by)+\nonumber \\
&+& D_j (\delta_p^m\delta_i^n+\delta_p^n\delta_i^m)\delta(\bx-\by)-
D_p (\delta_i^m\delta_j^n+\delta_i^n\delta_j^m)\delta(\bx-\by)]
\nonumber \\
\end{eqnarray}
we find that the time derivative of $\Phi_1$ is equal to
\begin{eqnarray}
\partial_t\Phi_1&=&\pb{\Phi_1,H_T}\approx
\nonumber \\
&-&\frac{2\kappa^2 }{\phi^4\sqrt{g}} (R_{ij}\pi^{ij}-
\frac{\lambda}{3\lambda-1}R\pi)+\nonumber \\
&+&\frac{2\kappa^2\phi^2}{\sqrt{g}}D_kD_l[\phi^{-6}\pi^{kl}] +
\frac{2\kappa^2\phi^2}{\sqrt{g}}\frac{(1-\lambda)}{3\lambda-1}D_k
D^k[\phi^{-6}\pi]
-\nonumber \\
&-&16\frac{\kappa^2}{\phi^5\sqrt{g}}
(\pi^{ij}-\frac{\lambda}{3\lambda-1}g^{ij}\pi)D_iD_j\phi+\nonumber
\\
&+&\frac{16 \phi\kappa^2}{\sqrt{g}}D_i\phi D_j[\phi^{-6}\pi^{ij}]-
\frac{8\kappa^2\phi}{\sqrt{g}}\frac{2\lambda-1}{3\lambda-1}D_i\phi
g^{ij} D_j[\phi^{-6}\pi]\equiv \Phi_2 \ , \nonumber \\
\end{eqnarray}
where $\Phi_2$ is an additional constraint that has to be imposed
on the system. Following
\cite{Henneaux:1992ig,Govaerts:2002fq,Govaerts:1991gd}
we include the constraint $\Phi_2$ into
the definition of the total Hamiltonian that now has the form
\begin{eqnarray}
H_T&=&\int d^3\bx (\mH_T'-\sqrt{g}\phi^6\mA (\phi^{-4}R[g]-
8\phi^{-5}g^{ij}D_iD_j\phi-\Omega)+\lambda \Sigma_D+\nonumber
\\
&+&n^i\hat{\mH}_i+\gamma p_{\mA}+\Gamma_{I}\Phi_1 +\Gamma_{II}\Phi_2
) \ . \nonumber \\
\end{eqnarray}
Now we should again check the stability of all constraints. It is
easy to see that the primary constraints together with $\bT_S(N^i)$
are preserved while the time evolution of the constraint
$\Phi_1\approx 0$ is equal to
\begin{eqnarray}\label{parttPhi2}
\partial_t \Phi_1&=&\pb{\Phi_1,H_T}
\approx \int d^3\bx \left(\Gamma^{II}(\bx)
\pb{\Phi_1,\Phi_{2}(\bx)}\right)
\approx \nonumber \\
&\approx& \int d^3\bx \Gamma^{II}(\bx)\pb{\Phi_1,\Phi_{2}(\bx)}=0 \
.
\nonumber \\
\end{eqnarray}
As follows from the explicit form of the constraints $\Phi_{1,2}$ we
have
\begin{eqnarray}\label{DBphi12}
\pb{\Phi_1(\bx),\Phi_{2}(\by)}\neq 0 \ .
\end{eqnarray}
Then we find that the equation (\ref{parttPhi2}) gives
$\Gamma^{II}=0$. In the same way the requirement of the preservation
of the constraint $\Phi_{2}$ implies
\begin{eqnarray}\label{partPhi2}
\partial_t\Phi_{2}\approx
\int d^D\bx (\pb{\Phi_{2},\mH_T(\bx)} +
\Gamma^I(\bx)\pb{\Phi_{2},\Phi_{1}(\bx)})=0 \ .
\nonumber \\
\end{eqnarray}
Using the fact that $\pb{\Phi_{2},\mH_T(\bx)}\neq 0$ and also the
equation (\ref{DBphi12}) we see that (\ref{partPhi2}) can be solved
for $\Gamma^I$. In fact, (\ref{DBphi12}) shows that
$\Phi_1$ and
$\Phi_{2}$ are the second class constraints. We also see from the
previous analysis that no additional constraints have to be imposed
on the system.
In order to find the action for the physical degrees of freedom we
have to finally fix the gauge symmetry generated by $\Sigma_D$. To
do this we introduce the gauge fixing function
\begin{equation}
\Phi_{G.F.}=\sqrt{g}-1 \ .
\end{equation}
It is easy to see that there is non-zero Poisson bracket between
$\Phi_{G.F.}$ and $\Sigma_D$ so that they are the second class
constraints. In summary we have following collection of the second
class constraints
\begin{equation}
\Phi_1=0 \ , \Phi_2=0 \ , \Sigma_D=0 \ , \Phi_{G.F.}=0 \ .
\end{equation}
The goal is to eliminate some degrees of freedom from these
constraints, at least in principle. In fact, from $\Phi_1$, which
is version of Lichnerowitz-York equation
\cite{York:1971hw}, we
express $\phi$ as
\begin{equation}
\phi=\frac{1}{8}\nabla^{-1}(\phi R[g]-\phi^5\Omega) \ .
\end{equation}
where $\nabla^{-1}$ is inverse operator to $g^{ij}D_iD_j$. We can
solve the equation above perturbatively around some constant
$\phi_0$. Further, from $\Sigma_D$ we express $p_\phi$. Finally,
$\Phi_{G.F.}$ reduces number of degrees of freedom in $g$ to be
equal to five and from $\Phi_2$ we express $\pi$ as the function of
remaining degrees of freedom. In summary, the physical degrees of
freedom of Lagrange multiplier modified HL gravity are
\begin{equation}
g_{ij}, \quad \sqrt{g}=1 \ , \quad \tilde{\pi}^{ij} \ ,
g_{\ij}\tilde{\pi}^{ji}=0 \ .
\end{equation}
Note that there are three first class constraints $\hat{\mH}_i$ so
that by gauge fixing these constraints we should eliminate remaining
three degrees of freedom in $g_{ij}$. In other words the physical
content of given theory is the same as in General Relativity.
From the previous analysis we see that $\phi$ is a non-local
function of $R$. The same situation also occurs in case of $\pi$ so
that in principle $\hat{\mH}_i$ has the form
\begin{equation}
\hat{\mH}_i=-2g_{ik}\nabla_j
\tilde{\pi}^{kj}+p_\phi(\nabla^{-1}f(g,\pi))\nabla_i
\phi(\nabla^{-1}R(g))-\frac{1}{3}\nabla_i
\pi(\nabla^{-1},\tilde{\pi},g) \ .
\end{equation}
By definition they are the first class constraints which however
depends non-locally on the canonical variables.
We would like to stress
that this result can be considered as the generalization of
the analysis performed
in \cite{Khoury:2013oqa,Khoury:2011ay} to the case of HL gravity. In
more details, papers cited above were devoted to the construction
of the action for the physical modes of the gravitational fields
only that are the metric that obeys the condition $\sqrt{g}=1$ and
the conjugate momenta $\tilde{\pi}^{ij}$ that are traceless. In
order to have the right number of physical degrees of freedom this
action should be invariant under spatial diffeomorphism.
Explicitly, the action studied there has the form
\begin{equation}
S=\int dt d^3\bx
(\dot{g}_{ij}\tilde{\pi}^{ij}-\pi_H-N^i\tilde{\mH}_i) \ ,
\end{equation}
where
\begin{equation}\label{KhourimH}
\tilde{\mH}_i=-2g_{ij}\nabla_k \tilde{\pi}^{jk}-\nabla_i \pi_K \ ,
\end{equation}
where $\pi_K$ is arbitrary function that has to be determined in
such a way that $\tilde{\mH}_i$ are the first class constraints.
However when we restrict to the case when this function depends on
the partial derivatives of $g_{ij}$ through the scalar curvature $R$
we find that the functions $\pi_N$ and $\pi_K$ are uniquely
determined and leads to the General Relativity action. Note also
that the symplectic structure used in given paper has the standard
form
\begin{equation}\label{stsymp}
\pb{g_{ij},\tilde{\pi}^{kl}}=\frac{1}{2}(\delta_i^k
\delta_j^l+\delta_i^l \delta_j^k)-\frac{1}{3}g_{ij}g^{kl} \ .
\end{equation}
In case of Lagrange multiplier modified HL gravity the situation is
more involved. The symplectic structure is determined by the Poisson
brackets of the second class constraints $\Phi_1,\Phi_{2}$ which is
rather complicated. More precisely, let us denote all second class
constraints as $\Phi_A=(\Sigma_D,\Phi_{G.F.},\Phi_1,\Phi_2)$. Then
the Poisson bracket between the constraints $\Phi_A$ can be written
as
\begin{equation}
\pb{\Phi_A(\bx),\Phi_B(\by)}= \triangle_{AB}(\bx,\by) \ ,
\end{equation}
where the matrix $\triangle_{AB}$ has following structure
\begin{equation}\label{triangleAB}
\triangle_{AB}(\bx,\by)= \left(\begin{array}{cccc}
0 & * & 0 & 0 \\
{*} & 0 & 0 & * \\
0 & 0 & 0 & * \\
0 & * & * & * \\
\end{array}\right) \ ,
\end{equation}
where $*$ denotes non-zero elements that depend on the phase space
variables and their derivatives.
Now it is easy to see that the Dirac brackets between the canonical
variables that are defined as
\begin{eqnarray}
& &\pb{g_{ij}(\bx),g_{kl}(\by)}_D= -\int d\bz d\bz'
\pb{g_{ij}(\bx),\Phi_A(\bz)} (\triangle^{-1})^{AB}(\bz,\bz')
\pb{\Phi_B(\bz'),g_{kl}(\by)}\ ,
\nonumber \\
& &\pb{\pi^{ij}(\bx),\pi^{kl}(\by)}_D= -\int d\bz d\bz'
\pb{\pi^{ij}(\bx),\Phi_A(\bz)} (\triangle^{-1})^{AB}(\bz,\bz')
\pb{\Phi_B(\bz'),\pi^{kl}(\by)}
\ ,
\nonumber \\
& &\pb{g_{ij}(\bx),\pi^{kl}(\by)}_D= \pb{g_{ij}(\bx),\pi^{kl}(\by)}
-\nonumber \\
&-&\int d\bz d\bz' \pb{g_{ij}(\bx),\Phi_A(\bz)}
(\triangle^{-1})^{AB}(\bz,\bz') \pb{\Phi_B(\bz'),\pi^{kl}(\by)} \
\nonumber \\
\end{eqnarray}
depend on the phase-space variables.
Secondly, the resulting
action is non-local due the presence of the inverse operator
$\nabla^{-1}$.
Even if the action for the physical degrees of freedom of Lagrange
multiplier modified HL gravity is rather involved we mean that the
results derived here should be considered as the starting point for
further research of HL gravity when we generalize the analysis
performed in \cite{Khoury:2013oqa,Khoury:2011ay} in several
different ways. We could start with the action for the physical
modes with the symplectic structure given by the equation
(\ref{stsymp}) and presume that $\pi_K$ depends on $R$ non-locally
and try to determine the original action. Another possibility is to
consider the case when $\pi_K$ in (\ref{KhourimH}) depends on
$R_{ij}R^{ij}$ and covariant derivatives of $R_{ij}$. We hope to
return to these problems in future.
\\
\noindent {\bf Acknowledgement}
This work was supported by the
Grant Agency of the Czech Republic under the grant P201/12/G028.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,012
|
Matt Lewis and Renato Poliafito's first book, _Baked_ , was published to major critical acclaim and raved about across the blogosphere. Since then, Lewis and Poliafito's Brooklyn—based bakery has continued to garner national recognition: They've gone on to open a second Baked in Charleston, South Carolina, and their products are now found in stores across the country. And yet, while their hearts remain close to home, in this new book the authors present a tribute to the most beloved desserts from across the country.
From Mississippi Mud Pie to Black & White Cookies, from Devil's Food Cake with Angel Frosting to Red Velvet Whoopie Pies, Lewis and Poliafito take on our country's treasured treats. These are desserts that have been passed down through generations, tucked away in Grandma's cookbooks, and shared at church suppers and small town gatherings.
Of course, Lewis and Poliafito reinvent these recipes with their signature tongue-in-cheek style, introducing a new air of sophistication to these favorites. Think Grasshopper Bars with a fudgy brownie base, light mint filling, and a dark ganache glaze; Salt-n-Pepper Sandwich Cookies that are an homage to the Oreo; and Caramel Apple Cake covered in a slightly sweet caramel frosting. They even include the recipe for their most in-demand creation, the Sweet & Salty Brownie. Including favorite recipes for classic desserts to sadly neglected sweets, the recipes in this collection—from breakfast treats to late-night confections and everything in-between—will delight readers.
## _**Contents**_
INTRODUCTION: GETTING BAKED, AGAIN
EVERYTHING YOU NEED TO KNOW TO GET BAKED
[CHAPTER 1
BREAKFAST](ch01.html#ch01)
[CHAPTER 2
TARTS AND PIES](ch02.html#ch02)
[CHAPTER 3
COOKIES AND BARS](ch03.html#ch03)
[CHAPTER 4
CAKES](ch04.html#ch04)
[CHAPTER 5
CONFECTIONS AND PASTRY](ch05.html#ch05)
ACKNOWLEDGMENTS
SOURCES
CONVERSION CHARTS
INDEX
## **_Introduction_
GETTING BAKED, AGAIN**
### **OBSESSIVE-COMPULSIVE**
I no longer vacation like a normal person. Or, rather, I no longer _plan_ vacations like a normal person. Now, when cobbling together and researching the traditional parts of a holiday (destination, arrival time, hotel, and length of stay), I also include at least one dessert or pastry excursion of note. This complicates matters for both my traveling companions and myself. Whereas before, a simple ski vacation included the basic elements of planning, I now add a layer of complexity that most of my friends would rather skip. In addition to choosing the mountain, the condo, and the best month to ski, I might also insist that we spend a day off-slope in search of a much-written-up, much-blogged-about creamy bourbon milkshake. The kind made with homemade vanilla bean ice cream, homemade butterscotch sauce, and top-shelf bourbon and blended at the same location by the same family for many, many years.
Early on, I referred to my dessert jaunts as "research." I would tell myself (and my very tired and very full traveling companions) that I needed to "test" every chocolate chip cookie, every brownie, all manner of cakes and cupcakes, and every type of breakfast pastry to bring a deeper understanding to my development work at Baked or on the Baked books or for possible future concepts. Truth is, I would be eating my way through the United States and beyond, regardless of my occupation. It is my obsession and compulsion.
People often ask me if I ever tire of eating brownies (a signature Baked product). Absolutely not. What about chocolate chip cookies—don't I want to take a break from chocolate chip cookies? Never. I enjoy finding the subtle differences from bakery to bakery and state to state. I want to know why a certain bakery in North Carolina always makes a crunchy chocolate chip with classic, tiny chips, and another bakery in Oregon always makes chewy ones with large chunks of chocolate. Is the disparity regional? Did the cookie in North Carolina originate from a hand-me-down heirloom recipe? Is the Oregon cookie an amalgam, a collaboration by a series of pastry chefs who rotated in and out the bakery door?
My dessert "research" is neither conventional nor scientific. The format is loose, and the results are beside the point. I enjoy what I do, and I will always choose Pierre Hermé over Mona Lisa. My worldview is formed by desserts and the people who make them, and my true satisfaction is derived from finding pockets of regionalism in an increasingly homogenous America.
### **I'LL SHOW YOU MINE, IF YOU SHOW ME YOURS**
There are obvious regional dessert specialties. The black and white cookie is nothing short of a phenomenon in and around New York City, while remaining distinctly off the radar in much of the rest of the country. Banana cream pie is a staple of the Los Angeles area restaurant and bakery circuit, yet it makes only brief and lackluster appearances elsewhere. Booze, especially the likes of Kentucky bourbon and Tennessee rye, is prominent in Southern desserts. Local fruits and produce are often directly tied to a regional specialty (i.e., blueberries in Maine, peaches in Atlanta). These observations are interesting, but it is the lesser-known regional subtleties that make my head explode. Dig a little deeper and unearth the treasures.
During Renato's and my travels for various food-related research (a great benefit to owning a bakery and writing a cookbook), we met people who introduced us to favorite desserts that had been passed down through generations and at church suppers and small town gatherings. These recipes, carefully archived, are perhaps hyper-regional. These are recipes that are typically neatly handwritten on brittle, yellowing paper and tucked away in Grandma's favorite cookbooks. An anomaly in a digital world.
We are always honored when people share these recipes with us (even, surprisingly, the Jell-O pretzel salad), and we are happy to share in return. It's an "I'll show you mine if you show me yours" scenario that has been deeply influential in our American baking repertoire, and we work daily to re-create or restore these truly great baking principles. Give us your vintage recipes, and we'll—ever so lovingly—turn them on their head.
### **ABANDONMENT ISSUES**
Another subset of American desserts that Renato and I pursue with glee could be classified as Abandoned Desserts. Boston cream pie, Mississippi mud, all things grasshopper, and their equivalents are desserts that never quite endured like the mainstays of American baking: chocolate chip cookies, apple pie, and brownies. The reasons for their gradual decline are varied, though still explainable.
Desserts, like fashion, are highly influenced by cycles and trends. If you were afloat in a sea of lava cake (aka molten chocolate cake) during the nineties, you were not alone. The dessert was on every restaurant menu (regardless of cuisine and price point) throughout the decade. As of this writing, spiking desserts with bacon is de rigueur. These fashions are part of a natural cycle. Lava cake will slowly fade into misty, dew-covered nostalgia, and bacon-flavored chocolate will become a fleeting trend, like parachute pants. When Renato and I dig through our pile of neglected desserts, we like to focus on investigating those beauties that lived through a few heady trend cycles and then were unjustly tossed to the gutter, like grasshopper pie.
We also look for baked goods that almost achieved classic status, like chiffon cake. For some reason, these desserts never quite reached their potential, and we place the blame squarely on the terrible versions people have encountered. After all, eating a substandard chiffon cake is like eating a kitchen sponge and sawdust sandwich.
Regardless, the Abandoned Desserts are something of a preoccupation of ours. We feel that it is our duty to provide some support to these old friends.
### **ABOUT THE BOOK**
This book, _Baked Explorations_ , is a tribute to beloved American desserts—treats and baked goods that are regional gems, fading beauties, or family secrets. It is a compilation of some of the more interesting items we found in our travels.
Renato and I set about looking for the roadways and history to all things sweet and realized that there is no direct route. Most recipes we found had been altered and tweaked by so many hands that one person's simple chocolate cake recipe was completely different from the next-door neighbor's, not to mention the recipe from the person two states over.
In some instances, we left recipes virtually unchanged from the version we were given. In other cases we performed a Baked makeover. This is not a compendium, nor a voluminous history of baked goods. Think of it as an easygoing road trip with stop-offs at the quirky and unusual monuments. Of course we included a few Baked customer favorites as well.
I hope you enjoy making and eating every recipe, and we encourage you to share a few with us.
##
**KITCHEN TOOLS AND EQUIPMENT**
MY KITCHEN OVERFLOWETH WITH USELESS, DISCARDED, AND UNNECESSARY, KITCHEN TOOLS AND UTENSILS. It is an archive of the superfluous, acquired many years ago during a brief, but painful, addiction to all forms of gadgetry. In an effort to help you avoid this brutal form of bread-machine and garlic press sadism, I have compiled a list of the most basic tools you will need for your baking kitchen. You can make anything in this book with just the tools and equipment listed below.
_**Baking sheets:**_ I recommend buying heavy-duty, inexpensive, light-colored, rimmed, baking sheets. I almost always bake on basic aluminum half-sheet pans (18 by 13 inches) found at restaurant supply stores and various retailers (see Sources). Generally speaking, the newfangled insulated and cookie-specific sheet pans are overpriced and not as effective.
_**Bench knife:**_ A bench knife, or dough scraper, is an extraordinarily useful tool. Generally a 3 inch by 5 inch sheet of metal attached to a wooden or plastic handle, it is used to cut, portion, and turn dough. And it is extremely effective for scraping down and cleaning surfaces.
_**Brownie and bar pans:**_ I always use a light metal or glass baking dish for bars and brownies. Dark metal pans produce unsavory, extra-crispy edges.
_**Bundt pan:**_ Every home baker's equipment collection should include at least one Bundt pan. I use the basic 10- and 12-cup versions made by Nordic Ware. They are heavy, easy to use, and should last forever. There are also many decorative Bundt pans on the market (that turn out shapes like rosettes or castles), and you can absolutely use them for our recipes. Just make sure to grease all the nooks and crannies of the specialty pans to prevent unsightly surface breakage.
_**Cake pans:**_ Keep it simple, straightforward, and economical. Use the professional aluminum cake pans available from almost any kitchen supply store (see Sources). Stay away from dark pans to prevent crispy cake edges.
_**Cake turntable:**_ It is much easier to decorate a cake (or sugar cookies, or brownies) on a rotating cake turntable. Heavy-duty turntables (which we recommend) are not inexpensive, but they are well worth the cost if you are an avid decorator. I would avoid the plastic versions altogether; they tend be flimsy and fall apart easily.
_**Candy and chocolate thermometers:**_ Candy and chocolate thermometers come in many shapes, price points, and styles (including an incredibly cool laser version). For the beginning candy maker, I recommend the old-school inexpensive clip-on candy thermometer. Make sure it has gradations of 2 to 5 degrees and a range of 100 to 400 degrees F. Also, many basic candy thermometers mark all the stages of candy making (hard-ball, soft-ball, etc.)—this makes things all the easier for you. This type of thermometer should cost no more than fifteen dollars.
_**Cooling racks:**_ I generally use two cooling racks when baking cookies and cakes so I have enough room for everything that comes out of the oven. I prefer the basic collapsible version (color and materials do not matter) that fits perfectly over a half-sheet pan.
_**Cupcake pans:**_ The cupcakes in this book were tested in the familiar 12-cup cupcake/muffin pan made of light-colored metal. If you elect to use a different cup size, you will have to change the baking time accordingly. The basic rule of thumb: Mini muffins or cupcakes usually bake in half the suggested baking time or less, while the larger pans usually require time and half in the oven.
_**Food processor:**_ If you bake and cook more than occasionally, I wholeheartedly recommend purchasing a large (9- to 12-cup) food processor. I know they are oversized, heavy, not exactly beautiful, and often viewed as difficult to clean. However, once you own one, you will never give it up. It chops nuts and graham crackers with ease. It makes some batters, some icing, and pie dough with the flick of a switch. It's like an assistant, only better. Go get one. It has many uses in the savory kitchen as well.
_**Ice cream maker:**_ Ice cream makers are fun to have around and experiment with. I own a fifty-dollar Cuisinart ice cream machine, and it works just fine (as long as you freeze the bowl ahead of time). For the more serious ice cream maker, there are more serious machines with larger capacities and built-in compressors (which eliminate the whole frozen-bowl business).
_**Ice cream scoop:**_ The ice cream scoop with a release mechanism is a very important tool, if not a necessity. I use several different sizes of scoops to form perfect and uniform cookie dough balls and portion cake batters—not to mention scoop ice cream.
_**Loaf pans:**_ As with all pans, simpler equals better. Loaf pans (9 by 5 by 3 inches or thereabouts) should be inexpensive and made of light-colored metal. I have three very old, very inexpensive, very effective, very battered Chicago Metallic versions, and I have a feeling they will outlast me.
_**Measuring cups and spoons:**_ For liquid measurements, I recommend Pyrex (glass) 2-cup and 4-cup sizes. They're handy for melting butter in a microwave, too.
For dry measurements, I recommend a basic set of metal measuring cups from ¼- to 2-cup sizes.
All the recipes in the book were tested by scooping dry ingredients into the measuring cup, then leveling the top of the cup with a straight-edged knife (often referred to as the "spoon and sweep" method). All light and dark brown sugars should be packed tightly and leveled to the top of the cup.
For measuring spoons, use the most basic set of metal spoons you can find. They usually come locked together by a metal ring, starting with ¼ teaspoon and going up to 1 tablespoon. Measure all ingredients level with the rim of the spoon.
_**Microplane:**_ Microplane actually refers to a brand. They make the long, thin graters most commonly found in commercial and home kitchens. At home, I use one grater strictly for spices and another strictly for zesting fruit and, if need be, grating cheese. Do not be tempted to purchase a specialized "zesting" tool; the Microplane is more practical and has many more uses.
_**Microwave oven:**_ I know the microwave oven is frowned upon in many foodie kitchens; however, I am not ashamed to say that I use mine constantly. It melts chocolate, reheats coffee, and makes boiling water for tea a cinch. To melt chocolate (or butter) in the microwave, use short bursts of low power, regularly removing the ingredient to stir it, then repeating the process until it is fully melted. If you don't have a microwave, no worries; you can melt butter and chocolate in a double boiler.
_**Mixing bowls:**_ You should own a three-to-five-bowl set of spouted nesting mixing bowls, preferably made of melamine. Melamine bowls are lightweight, super cheap, and easy to clean (I hardly use my old ceramic bowls, but I just can't stand to part with them).
_**Parchment paper:**_ Parchment paper is essential to the home baker. I use it to line cookie sheets, cake pans, and loaf pans. It keeps things from sticking, and it is a much less messy option than cooking sprays and other grease-containing items (in particular that horrible flour spray). If you are baking cookies, you can reuse the parchment paper at least once. I find that silicone baking sheet liners don't produce the same kind of browning as parchment does.
_**Pie plates or tins:**_ You can bake a pie in almost any pie plate or tin, but I am partial to metal and Pyrex glass. Disposable tins generally produce a soggier crust, and ceramic pie plates, while by far the most attractive, tend to conduct heat unevenly (often due to the make and age of the pan).
_**Pie weights:**_ Pie weights help the dough hold its shape and prevent it from shrinking while baking. You can buy specially made pie weights from most kitchen stores or save a few dollars and use dried beans.
_**Skillets:**_ I highly recommend getting a set (about three sizes) of cast-iron skillets. I found mine at a garage sale. They were inexpensive and already seasoned (a huge plus to purchasing used cast-iron ware), and I swear they are easier to clean than many more expensive pans. Additionally, I like the chewy edges they produce on both the cheese grits and the chocolate skillet cake in this book.
_**Spatulas:**_ You should own several high-heat spatulas in a variety of shapes and sizes. They are essential for scraping down bowls, mixing light batters, and folding egg whites.
I also have offset spatulas—several small metal ones—for detail-oriented jobs like smoothing batters into baking pans, loosening cakes from the sides of pans, removing the first brownie, and swirling or marbling batters.
_**Standing mixer:**_ Even though a standing electric mixer is big and a bit expensive, every home baker or baking enthusiast should have one with at least the three basic attachments: whisk, paddle, and dough hook. I have had my KitchenAid mixer for thirteen years, and it has never given me one problem.
_**Tart pans:**_ I use several sizes of tart pans in the book for variety; however, you do not have to own this many to use the recipes (all the tart recipes herein offer pan substitutions where applicable). Tart pans have a removable bottom that makes it easier to remove the tart from its pan. When storing mini tart pans, we suggest layering them between sheets of paper towels or parchment paper to make sure they do not stick together while nesting.
_**Whisk:**_ Home chefs should not worry themselves about owning the many varied types of whisks on the market. I used a basic wire whisk with a wooden handle (medium to large size will do) for every recipe that calls for whisking. These are great tools for combining dry ingredients (mixing flour, baking soda, and salt together, for instance), but do not use your whisk as an all-purpose stirring device (a silicone spatula works better); you can accidentally whisk too much air into your batters.
**TERMS AND BRAND RECOMMENDATIONS**
### **CHOCOLATE**
It is imperative when making a chocolate-based dessert to use the best possible chocolate. If you plan on doing a lot of baking, it may be more cost-effective to purchase larger blocks of chocolate online or at local specialty stores. Luckily, in recent years, it has gotten easier to find good-quality chocolate at even the supermarket. The recipes in this book were tested using Callebaut Divine and Scharffen Berger chocolate, and I highly recommend both.
Of course, there are many other wonderful brands of chocolate to choose from, and as you get more familiar with them, you will start to align yourself with a few favorites. See Sources section for a complete list of where to buy them.
_**Chocolate percentages:**_ The percentage label on a bar of chocolate is confusing. One brand's 64 percent chocolate bar is often completely different from another brand's 64 percent bar. The percentage is really referring to the cocoa mass in the bar itself, but the proportions of sugar, milk solids, and any other ingredients can be wildly different. While I could write an entire chapter on this subject, it is perhaps easier to recommend the following bars for this book's recipes (see chart below).
_**Cocoa powder:**_ Every recipe in this book is made with Valrhona cocoa powder. It is deep, dark, and delicious. Don't get too caught up in the Dutched (cocoa powder treated with alkali) versus natural debate. Instead, pay more attention to the color and smell of the cocoa powder. Some mass-produced cocoas are almost gray—avoid them. If you can't find Valrhona, look for a dark-colored cocoa. I never use sweetened cocoa in this book (or for that matter, any other time).
_**Melting chocolate in a double boiler:**_ A double boiler is a great way to melt chocolate or chocolate and butter together. To create one, you need a mediumsized pan or saucepan filled with water, and a (preferably metal) bowl that will sit partway inside the pan without touching the water. The chocolate (or delicate sauce) goes in the top pan, and the idea is that you are less apt to burn chocolate in this manner. Most double-boilers are warmed over low to low-medium heat.
_**Coffee Extract:**_ Pure coffee extract is essentially coffee in concentrated form, and it is usually found in the baking section of most supermarkets. Nielsen Massey makes a wonderful version that imparts a smooth, never bitter, coffee taste. I use it in both the Chocolate Coffee Cake and the Coffee Ice Cream, though you will find many other uses for it, too (like whipped cream, marshmallows, or hot fudge, for example).
_**Vanilla Paste:**_ I am a big fan of Neilsen Massey's Madagascar Bourbon Pure Vanilla Bean Paste. The paste is thick and fragrant and contains real vanilla bean seeds that give light-colored frostings, fillings, and icings a wonderful speckled appearance. Generally speaking, vanilla bean paste is slightly more concentrated than extract but it can be substituted evenly with pure vanilla extract. You could also use slightly less paste.
_**Salt:**_ In recent years there has been an explosion in salt sophistication. Salt now comes in many forms, sizes, and colors. In order to streamline the recipes in this book, I broke down salt into two categories.
Salt: When I refer to salt in the book's recipes, I mean kosher salt. However, you can easily substitute table or iodized salt without a fear of ruining any recipe.
Sea Salt/Fleur de Sel: In the instances when I refer to fleur de sel or sea salt, I suggest using a fine-grained (or if it is your want, slightly chunky) fleur de sel. Lately, I have been partial to Le Saunier de Camargue brand (easily found on the Internet or at gourmet markets).
* * *
**IF A RECIPE CALLS FOR A DARK CHOCOLATE OF 60 TO 72 PERCENT, USE ANY OF THE FOLLOWING:**
_Scharffen Berger's Home Baking Bar 62% (found in most supermarkets)_
_Scharffen Berger's Home Baking Bar 70% (found in most supermarkets)_
_Callebaut Chocolate Block 60% (found in specialty markets—often chopped and repackaged by the market)_
_Callebaut Chocolate Block 70% (found in specialty markets—often chopped and repackaged by the market)_
_Divine 70% Dark Chocolate (Fair Trade)_
**IF A RECIPE CALLS FOR A MILK CHOCOLATE, USE ANY OF THE FOLLOWING:**
_Jacques Torres Milk Chocolate Bar (found in specialty markets and online)_
_Scharffen Berger Milk Chocolate Bar 41% (found in most supermarkets)_
* * *
### **TWO POWDERED FLAVORINGS**
Though I have been playing with various flavored powders of late (I was given a few jars of flavored powders that look much better than they taste), I am still uniquely partial to my tried-and-true standbys: instant espresso powder and malted milk powder.
_**Instant espresso powder:**_ This is not interchangeable with ground espresso beans. Instant espresso powder is intended to dissolve easily and is great for most baking applications. It can be is used to cut sweetness, accentuate chocolate flavor, and heighten the coffee-like taste of your pastries. Ground espresso beans will not dissolve and can give your baked goods a grainy texture. I used Medaglia d'Oro brand espresso powder in testing all these recipes.
_**Malted milk powder:**_ Primarily still used to make soda-fountain drinks, this is one of our favorite ingredients. We think its tangy, nutty flavor enhances both vanilla- and chocolate-based desserts. My favorite brand is Carnation malt, which is carried in most grocery stores, but if you can't find that, Ovaltine chocolate malt drink mix can be used as well.
### **TWO IMPORTANT TECHNIQUES**
The world of pastry relies heavily on a vast array of proper "techniques." My pastry world, and this book, are dominated by these two: folding and sifting.
_**Folding:**_ When a recipe calls for folding, this means the act of gently mixing two parts of a batter together—no fast stirring, no whisking. The best way to do it is to use a rubber or silicone spatula and concentrate on turning the bottom part of the batter (often the heavier part) into the top part of the batter (often the lighter part) by scraping the sides of the bowl, then sweeping and twisting inward.
_**Sifting:**_ Only a few recipes in this book require dry ingredients to be sifted. Sifting is the act of adding air to the dry ingredients to produce lighter cakes and baked goods. To do it, I recommend shaking the ingredients through a large sieve. Sieves are less expensive and easier to clean than the special sifting knickknacks on the market today. Besides, they have many other uses, while a sifter has only one.
##
**MONKEY BUBBLE BREAD**
**PUMPKIN CHEDDAR MUFFINS**
**FARM STAND BUTTERMILK DOUGHNUTS THREE WAYS**
**NUTELLA SCONES**
**CARROT COCONUT SCONES WITH CITRUS GLAZE**
**BAKED CHEESE GRITS**
**MOM'S OLIVE OIL ORANGE BUNDT**
**OATMEAL CHOCOLATE CHIP CAKE WITH CREAM CHEESE FROSTING**
**HONEY CORN MUFFINS**
**NEW YORK-STYLE CRUMB CAKE**
**CORNMEAL GRIDDLE CAKES**
**MALTED WAFFLES**
**BAKED FRENCH TOAST**
**DOUBLE-CHOCOLATE LOAF WITH PEANUT BUTTER CREAM CHEESE SPREAD**
Breakfast, or the concept of a "proper breakfast," can be unpredictable. I like it that way. When I feel inspired, I like rummaging around the pantry and refrigerator for unexpected muffin, scone, or pancake ingredients. I might use up some fresh fruit, chop some chocolate, stir in a bit of brandy, or break apart a stale baguette. When I am feeling less ambitious, I might just reheat leftover macaroni and cheese, or grab a bagel from the local deli, or both. I leave myself open to either option—I consider myself a breakfast optimist, and I never plan in advance.
It's not that I am blasé about breakfast. Actually, I am quite a breakfast advocate; I just never structure the meal like I might a lunch for friends or a large dinner party. I have never "dressed" for breakfast (a frightening idea!), and I don't enjoy the idea of sitting formally at a table in the morning. I prefer to fly solo for my first meal of the day, and most likely I am hunched over the morning news, be it on my laptop or the daily paper.
My carefree roll-out-of-bed-and-grab-your-own-breakfast attitude is largely a part of my upbringing. Mom encouraged the scour-and-devour breakfast scenario that still is part my daily routine. On occasion we were treated to last-minute innovations like a spruced-up muffin mix (usually loaded with butterscotch or chocolate chips) or a pancake burdened with more toppings than a tricked-out ice cream sundae. Other times, it was a simple store-bought, and probably not very good, coffee cake. My breakfast never looked like the hearty abundance of a tweaked-and-Photoshopped Denny's picture menu.
While digging for this book, I unearthed more recipes for breakfast than any other section. People are passionate about their first meal of the day, and the nostalgia runs deep—deeper than with most recipes. I whittled the written and oral submissions down, keeping to the sweeter side of things, and edited them down again by preserving the items that felt the most homey without being too kitsch. I can honestly say that I had the hardest time regulating myself with breakfast during the book's testing phase. One time I lost self-control, nearly consuming half a loaf of Monkey Bubble Bread all by my lonesome. The other recipes in this chapter are equally delicious. I still daydream about the Double-Chocolate Loaf with Peanut Butter Cream Cheese Spread. It is a rewarding and handsome breakfast loaf with a sinful flair. Mom's Olive Oil Orange Bundt is coffee-klatch heaven, and the Malted Waffles are a great excuse to use your waffle iron. If you are one of those rare anti-sweet breakfast people, I recommend the Baked Cheese Grits. Actually, I recommend the cheese grits no matter what. Have a great breakfast.
### **MONKEY BUBBLE BREAD**
I SUGGEST ONLY MAKING THIS FROM-SCRATCH BREAD IF YOU ARE HAVING A LARGE GATHERING. Otherwise, you could end up (like me) eating more than you should. Simply put, this is addictive stuff. I liken these warm, gooey bread balls to the most amazing glazed doughnut hole you have ever had. There are several recipes floating about for monkey bread that use canned biscuit dough, but I ask you to kindly refrain from this expedient fix because the result won't be as tasty, and it is more expensive. The origin of the name monkey bread or bubble bread is quite hard to pinpoint, and while many dubious answers exist (the bread resembles a monkey puzzle tree or monkeys love to pull things apart), none of them are definitive, and some are cloyingly cute. I hate cloyingly cute. Suffice it to say that the source of the name is just one of life's great mysteries, and we should leave it at that.
**YIELD: ONE 10-INCH BUNDT**
**_Ingredients_**
FOR THE MONKEY BUBBLE BREAD
_1¼ cups whole milk_
_2 teaspoons instant yeast_
_4 cups all-purpose flour_
_5 tablespoons sugar_
_1 teaspoon salt_
_1 egg_
_5 tablespoons unsalted butter, melted_
FOR THE CINNAMON SUGAR COATING
_1¼ cups firmly packed dark brown sugar_
_2 teaspoons cinnamon_
_½ cup (1 stick) unsalted butter, melted and cooled_
#### **MAKE THE MONKEY BUBBLE BREAD**
Generously spray the inside of a 10-inch Bundt pan with nonstick cooking spray.
In a small saucepan, warm your milk to slightly above room temperature, then remove it from the heat, add the yeast, and whisk to dissolve. (Do not warm it beyond 110 degrees F or you will kill the yeast).
In the bowl of a standing mixer fitted with the paddle attachment, beat the flour, sugar, and salt until combined.
In a small bowl, beat the egg with a fork and add it to the dry ingredients. Mix on low speed until combined.
Keeping the mixer on low, slowly stream in the milk until combined. Add the melted butter and mix until the dough comes together. Replace the paddle attachment with the dough hook attachment. Continue to mix on medium speed until the dough becomes silky and tacky, but not sticky, 8 to 10 minutes. The dough should mound together and easily come off the bottom of the mixing bowl. (If the dough is too wet, add some flour. If it is too dry, add a tiny bit of water.)
Spray the bottom and sides of a large bowl with cooking spray. Place the dough in the bowl and roll it around to make sure it is completely covered in oil. Cover the bowl with plastic wrap or a dish towel and let it rest in a warm area until the dough has doubled in size, approximately 1 hour.
Line a sheet pan with parchment paper.
Use your clean hands to push down and deflate the dough. Remove it from the bowl and pat it into a rough circle approximately 8 inches diameter. Use a bench knife or serrated knife to cut dough into 1- to 1½-inch pieces (about ½ ounce each)—alternatively, use your hands to pinch apart the dough. Roll the pieces into balls (they don't have to be perfectly round). Place the balls on the sheet pan (you will get about 60 pieces in all). Cover the balls lightly with plastic wrap.
#### **MAKE THE CINNAMON SUGAR COATING**
In a small bowl, stir together the sugar and cinnamon. Place the melted butter in a separate bowl.
#### **ASSEMBLE THE BREAD**
Remove the plastic wrap from the dough balls and dip one ball in the melted butter. Let the excess butter drip back into the bowl, roll the ball in the brown sugar mixture, and place it in the Bundt pan. Continue this process with each ball, until you have several layers, arranging them as if you are building a brick wall.
Wrap the Bundt pan tightly in plastic wrap. Set it in a warm area of the house for about 1 hour, or until the dough balls have doubled in size and appear puffy.
Preheat the oven to 350 degrees F. Remove the plastic and bake the Bundt until the top layer is deep brown and the caramel coating begins to bubble around the edges, about 30 minutes.
Cool the bread for 5 minutes, then turn it out directly onto a platter and serve warm. Should you have any leftovers (this is rare, I promise you), simply reheat them in a 300-degree oven until warm to the touch.
**_Baked Note_**
There are a lot of monkey bread misconceptions, and I will do my darnedest to dispel them. First, you do not need an icing or topping for this bread—too sweet. Second, you can make the dough ahead of time. Once the dipped dough has been placed in the pan, wrap it tightly, refrigerate it, and bring it back to room temperature to proof the dough before baking. Lastly, this is one of those breads that exists to be eaten warm, straight from the oven. Once the caramel begins to cool, reheat the bread in the oven before serving.
### **PUMPKIN CHEDDAR MUFFINS**
IF THERE WERE A PLACE ON EARTH WHERE YOU COULD EXPERIENCE A NEW ENGLAND FALL FOR TEN MONTHS OUT OF THE YEAR, I WOULD PROBABLY MOVE THERE. I would pursue leaf peeping like a sport, build a crackling fire nightly, and indulge in every hearty autumn recipe at my whim. Until I find this utopia, I will make do with my annual three months of fall. I will churn through umpteen pumpkins (pumpkin bread, pumpkin soup, toasted pumpkin seeds) and hundreds of pounds of Vermont cheddar (grilled cheese, cheese and crackers, fondue), and on a few mornings, I will combine the two in this very autumnal muffin. Like all good muffins, this one is quick to put together. The pumpkin base is moist but spiced with cayenne and black pepper so the sharpness of the cheddar has a chance to shine. I also like to top the muffin with a little extra cheese, so you get a savory-sweet morning experience.
**YIELD: 12 MUFFINS**
**_Ingredients_**
_1 cup canned solid-pack pumpkin puree_
_3 tablespoons sour cream_
_2 large eggs_
_½ cup (1 stick) unsalted butter, melted and cooled_
_2 cups all-purpose flour_
_1½ teaspoons baking powder_
_¼ teaspoon cayenne pepper_
_1½ teaspoons salt_
_1½ teaspoons freshly ground black pepper_
_½ cup firmly packed dark brown sugar_
_1¼ cups (about 4 ounces) grated sharp cheddar_
_2 tablespoons pumpkin seeds, optional_
Preheat the oven to 400 degrees F. Lightly spray each cup of a standard 12-cup muffin pan with a little bit of vegetable spray and use a paper towel to spread the oil evenly along the bottom and up the sides of each cup.
In a large bowl, whisk together the pumpkin and sour cream. Add the eggs and butter and whisk until combined.
In another large bowl, whisk together the flour, baking powder, cayenne pepper, salt, black pepper, and brown sugar. Make a well in the middle of the dry ingredients. Pour the wet ingredients into the well, and fold until just combined. Fold in three-quarters of the cheese.
Divide the batter among the muffin cups. Sprinkle the remaining cheddar and the pumpkin seeds on top of the muffins. Bake them for 20 minutes, or until golden brown. Let the muffin pan cool on a rack for 10 minutes before turning out the muffins. Serve them warm.
Muffins taste best when eaten fresh, but they can be made ahead of time and reheated in a 200-degree oven.
**_Baked Note_**
I am addicted to the raw-milk cheddar offered by several farms in Vermont. The flavors are more dimensional than ordinary cheddar (though, yes, I am still a fan of the pasteurized version); raw-milk cheeses are nutty and chocolatey and earthy—and different from farm to farm. This recipe works well with any cheddar, the sharper the better, but make sure you try a raw-milk one if the opportunity presents itself (in or out of this muffin recipe).
### **FARM STAND BUTTERMILK DOUGHNUTS THREE WAYS**
IF I WERE A BETTER PERSON, I WOULD MAKE THESE MORE OFTEN. I would avoid the supermarket or mass-produced doughnut. I would take a stand and refuse to eat a doughnut that was not prepared by hand and eaten fresh from the fryer. These delicious doughnuts are what a doughnut should be, the type you might pick up from the side of the road at a local farm or farm stand. And though I'm often too lazy and lethargic to fire up the fryer, they really aren't that difficult to make.
Farm stand doughnuts are usually sold coated with cinnamon sugar and tucked inside a paper bag. Sometimes they are made with cider, and sometimes they are made with buttermilk, and they are always worth stopping for. I prefer the buttermilk variety (it produces a cakier doughnut), and I prefer mine dipped in chocolate, but they taste great au naturel as well.
Each topping makes enough for one batch of doughnuts. If you want to use more than one topping for your batch, reduce the amounts by half or by two-thirds, accordingly.
**YIELD: ABOUT 10 LARGE DOUGHNUTS PLUS DOUGHNUT HOLES**
**_Ingredients_**
FOR THE DOUGHNUTS
_3½ cups all-purpose flour_
_¾ cup granulated sugar_
_½ teaspoon baking soda_
_2 teaspoons baking powder_
_1 teaspoon salt_
_1 teaspoon freshly grated nutmeg_
_1 teaspoon cinnamon_
_2 large eggs_
_¾ cup buttermilk_
_¼ cup sour cream_
_¼ cup (½ stick) unsalted butter, melted and slightly browned and cooled_
_Vegetable oil for frying_
FOR THE CHOCOLATE DIP
_4 ounces good-quality dark chocolate (60 to 70%), coarsely chopped_
_½ cup heavy cream_
_2 tablespoons unsalted butter_
_Sprinkles to decorate (optional)_
FOR THE VANILLA GLAZE
_2 cups confectioners' sugar_
_¼ cup whole milk_
_1 teaspoon vanilla paste or 1½ teaspoons pure vanilla extract_
_Sprinkles to decorate (optional)_
FOR THE CINNAMON SUGAR
_1¼ cups granulated sugar_
_3 tablespoons cinnamon_
#### **MAKE THE DOUGHNUTS**
Line one baking sheet with parchment paper and another baking sheet with two layers of paper towels.
In a large bowl, whisk together the flour, sugar, baking soda, baking powder, salt, nutmeg, and cinnamon.
In a medium bowl, whisk the eggs, buttermilk, and sour cream until combined. Add the melted, cooled butter and whisk again.
Make a well in the center of the flour mixture and pour the liquid ingredients into the well. With a rubber spatula, slowly fold the flour into the liquid center until the mixture forms a sticky dough.
Turn the dough out onto a work surface lightly dusted with flour. Sprinkle the top of the dough with flour and pat it out until it is about ½ inch thick.
Use two round cutters (3¼ inch and 1½ inch for large doughnuts; 2½ inch and 1 inch for smaller doughnuts). Dip the large cutter in flour and press out the rounds. Dip the smaller cutter in the flour and cut out the center of each dough round. Arrange both doughnuts and doughnut holes on the parchment-lined baking sheet, pat the dough scraps back together, and use them to make as many more doughnuts and doughnut holes as possible. Chill the dough while you heat the oil.
Pour enough oil into a deep skillet to make a layer approximately 1 inch to 1½ inches deep. Slowly heat the oil over medium-high heat until it is 365 to 370 degrees F.
While you are waiting for the oil to reach temperature, make the assorted toppings.
#### **MAKE THE CHOCOLATE DIP**
Place the chopped chocolate in a medium wide-mouthed bowl. In a small saucepan, heat the cream until it is just about to boil. Pour the cream over the chocolate and wait 1 minute. Whisk until smooth. Whisk in the butter. Keep the mixture warm.
#### **MAKE THE VANILLA GLAZE**
In a medium wide-mouthed bowl, whisk together the sugar, the milk, and the vanilla paste.
#### **MAKE THE CINNAMON SUGAR**
In a medium wide-mouthed bowl, whisk together the sugar and cinnamon.
#### **TO FRY THE DOUGHNUTS**
Once the oil reaches temperature, gently lift the large doughnuts off the baking sheet and place them in the hot oil. Do not crowd the skillet—make no more than 3 doughnuts at a time. Once they have browned on one side (this takes 2 to 3 minutes), turn them over with tongs or a slotted spoon and continue to cook for another minute or just until browned (they can overcook or burn rather quickly). Using a slotted spoon, transfer the doughnuts to the paper towel–lined baking sheet and continue to fry the rest of the dough until finished. The doughnut holes will cook faster and can be made in two or three batches after the doughnuts are done.
#### **ASSEMBLE THE DOUGHNUTS**
Once you have finished frying, work quickly to dip the doughnuts in the chocolate or vanilla glaze, or the cinnamon sugar. If you like, decorate the chocolate or vanilla doughnuts with sprinkles. Serve immediately.
**_Baked Note_**
When you fry the doughnuts, make sure you maintain the correct oil temperature throughout the process. Generally speaking, doughnuts taste best served immediately after they've emerged from the fryer (and taken a quick dip in sugar or chocolate or vanilla); however, I have managed to find a few uses that play to the strengths of leftover (or day-old) doughnuts. Chop them into big coarse crumbs, toast them lightly, and add them to vanilla ice cream as a mix-in (if you are making it from scratch) or a topping (if you are serving store-bought). Doughnuts also work wonders (very rich wonders) when aused as the base of a bread pudding.
### **NUTELLA SCONES**
YES, RENATO AND I LOVE NUTELLA, THE LITTLE (OR BIG) JAR OF HAZELNUT AND CHOCOLATE BLISS FROM ITALY. Once stocked only by specialty stores, Nutella can now be found virtually everywhere. If you have not tried it, I beg you to stop everything, go to the nearest grocery store, and buy at least two jars: one for baking and one for a daily midday boost directly from said jar to your mouth.
These scones (secretly my favorite scones) have a decent-size dollop of Nutella folded into a cocoa-based dough. They aren't overly sweet, and the hazelnuts provide a great texture. Technically, they are still a breakfast treat, but they tend to make a bigger splash at brunch when people feel better about eating indulgently. Scones, no matter the ingredients, are still technically more difficult to put together than a muffin or quick bread. They require a little practice to perfect (i.e., getting a feel for the texture you want as you work in the butter and making sure you don't overwork the dough). However, once you master the scone, it will take you just a few moments to put together, bake, and serve a comfy little breakfast or tea snack. If you prefer to make your own, all-natural "Nutella," see the Homemade Nutella recipe.
**YIELD: 6 TO 8 SCONES**
**_Ingredients_**
_2 cups unbleached all-purpose flour_
_¼ cup granulated sugar_
_¼ cup dark unsweetened cocoa powder (like Valrhona)_
_1 tablespoon baking powder_
_½ teaspoon salt_
_6 tablespoons (¾ stick) cold unsalted butter, cut into chunks_
_1 large egg_
_½ cup heavy cream_
_¾ cup toasted hazelnuts, coarsely chopped_
_½ cup Nutella_
Preheat the oven to 375 degrees F and place the rack in the center. Line a baking sheet with parchment paper.
In a large bowl, whisk the flour, sugar, cocoa powder, baking powder, and salt until combined.
Add the butter. Use your fingertips to rub it into the flour until the butter is pea size and the mixture is coarse.
In a separate bowl, whisk together the egg and cream. Slowly pour the wet ingredients into the dry ingredients and stir until the dough just comes together. Gently and briefly knead the dough with your hands. Add the toasted hazelnuts and knead gently to incorporate. Flatten the dough into a rectangle approximately 6 by 12 inches (it does not need to be precise) and spread ¼ cup of the Nutella on top in a crisscross pattern. Roll the dough up to make a cylinder about 6 inches long, turn it on its end, and gently flatten it into a disk about 1¾ inches high. Do not overwork the dough.
Cut the dough into 6 or 8 wedges and place them on the prepared baking sheet. Bake the scones for 18 to 20 minutes, rotating the baking sheet halfway through, or until a toothpick inserted into the center of a scone comes out clean. Do not overbake.
Transfer the scones to a wire rack to cool completely. Place the baking sheet with the parchment still on it underneath the rack.
#### **ASSEMBLE THE NUTELLA SCONES**
Heat the remaining ¼ cup Nutella in a microwave until pourable, about 10 seconds on high. Pierce the tops of the scones a few times with a fork. Use a spoon (or two spoons—one to scoop, one to scrape) to drip the warm Nutella in a zigzag pattern over the tops of the hot scones. Transfer them to a refrigerator to set for 5 minutes, then serve immediately.
Most scones have a lifespan of 24 hours or less; however, these scones taste pretty darn good on day two provided you wrap them tightly and store them at room temperature.
**_Baked Note_**
My number-one piece of advice for those new to scone making: Do not knead too much. I think we must be born with an urge to create perfect, smooth, pliable, pillowy dough. Resist the temptation. Stop working the dough the minute it comes together. Do not worry if there are a few dry bits scattered throughout.
### **CARROT COCONUT SCONES WITH CITRUS GLAZE**
CARROTS HAVE LONG BELONGED TO THE SWEET BAKING SPECTRUM. In fact, it is likely that cakes sweetened with carrot have been around for many centuries and that carrot cake–type recipes like those we are familiar with today started appearing as early as 1914. Carrots have a high sugar content and subtle taste, which makes them easy to sneak into all manner of baked goods. Additionally, I have succumbed to deep self-delusion and equate all carrot baked things with health and nutrition. True, carrots contain vast amounts of vitamin A and dietary fiber, but a scone is still a scone (and for that matter, a cake is still a cake). These carrot scones are delectable whatever their nutritional value. (Like all scones, they are at their very best the day they are made.) They are dense without being heavy, sweet without being cloying, and the coconut adds some simple texture.
**YIELD: 6 TO 8 SCONES**
**_Ingredients_**
FOR THE SCONES
_2¾ cups all-purpose flour_
_½ cup granulated sugar_
_½ cup rolled oats_
_1 tablespoon baking powder_
_¼ teaspoon salt_
_1 cup shredded sweetened coconut_
_½ cup (1 stick) cold unsalted butter, cut into ½-inch chunks_
_1 large egg_
_¾ cup buttermilk_
_1 tablespoon pure vanilla extract_
_¼ cup carrot puree_
_1 egg white, beaten_
FOR THE CITRUS GLAZE
_1 tablespoon fresh lemon juice_
_2 tablespoons fresh orange juice_
_1 cup confectioners' sugar_
#### **MAKE THE CARROT COCONUT SCONES**
Preheat the oven to 400 degrees F and position the rack in the center. Line a baking sheet with parchment paper.
In a large bowl, whisk together the flour, sugar, oats, baking powder, salt, and shredded sweetened coconut.
Add the butter. Use your fingertips to rub the butter into the flour until the butter is pea size and the mixture is coarse.
In a separate bowl, whisk together the egg, buttermilk, vanilla and carrot puree. Slowly pour the wet ingredients into the dry ingredients and stir until the dough just comes together. Gently and briefly knead the dough with your hands. The dough will be sticky and may need to be sprinkled with flour.
Roll the dough up, turn it on its end, and gently flatten it into a disk about 1¾ inches high. Do not overwork the dough.
Whisk the egg white with 1 tablespoon water. Set aside.
Cut the dough into 6 or 8 wedges and place the scones on the prepared baking sheet. Brush the tops with the egg white wash. Bake for 18 to 20 minutes, rotating the baking sheet halfway through, or until a toothpick inserted into the center of a scone comes out clean. Do not overbake.
Transfer the scones to a wire rack to cool completely. Place the baking sheet, with the parchment still on it, underneath the rack.
#### **MAKE THE CITRUS GLAZE**
Whisk all ingredients together in a medium bowl. The glaze should be loose enough to drizzle. If it is too thick, add a little more orange juice. If it is too loose, add a little more confectioners' sugar.
Drizzle the glaze over the scones and allow it to set before serving.
**_Baked Note_**
I can assume that if you are attempting this recipe, you are not expecting a classic English scone. I feel obliged to tell everyone that this is more of a "morning cookie" with a scone ego. Make no mistake —this scone is less sweet than a typical cookie, and the oats and carrot lend an air of nutrition, but deep down this is a perfect treat for morning, noon, and night. Queen Elizabeth might scoff, but no one else will.
* * *
Truthfully, I prefer carrots in the "puree" form. I prefer carrot soup to carrot sticks and a carrot–mashed potato concoction to a side of steamed carrots. I use this puree in various recipes (cakes, pancakes, lasagna, and soup), and it is quick, easy, and perfect for our Carrot Coconut Scones. Double it, triple it, and freeze for later use.
_1 medium carrot
¼ cup orange juice_
Place the carrot and orange juice in a medium glass microwaveable bowl. Cover or wrap tightly in plastic wrap.
Microwave on high for about 5 minutes. If the carrot is fork-tender, it is ready. If it is not fork-tender, continue to microwave in 30-second bursts until it is.
Blend (in blender or food processer) the carrot and the orange juice until smooth or, alternatively, mash with a potato masher until lump free.
* * *
### **BAKED CHEESE GRITS**
I HAD TO SNEAK THIS SAVORY RECIPE INTO THE BOOK. Though, thematically, it is quite different from everything else in this book, grits—specifically these baked cheese grits—make up a large portion of my cooking and baking DNA. I was first introduced to grits—real stone-ground grits swimming in butter and cheese—while nursing a fairly nasty hangover during my first year of college at the University of Alabama, Tuscaloosa. It was the beginning of a beautiful relationship that has outlasted many others. Grits, a staple of Southern cuisine, are as homey, comfortable, variable, and delicious as America's beloved macaroni and cheese. I hope everyone will have a chance to try this neglected delicacy, as it is more than just a Southern caricature and maligned stereotype. Honestly, grits in any form please me. They are wonderful without adornment, and even better with copious amounts of butter and cheese. This basic recipe for baked cheese grits (adapted from _The Lee Bros. Southern Cookbook_ ) is a breakfast crowd-pleaser and the perfect accompaniment to bacon or sausages.
**YIELD: 4 DECENT-SIZE SERVINGS**
**_Ingredients_**
_2 cups whole milk_
_1 cup stone-ground grits_
_1 teaspoon kosher salt_
_1 teaspoon freshly ground black pepper_
_1 cup, packed, grated extra-sharp cheddar cheese, about 4 ounces_
_1 cup, packed, grated Monterey Jack cheese, about 4 ounces_
_1 tablespoon unsalted butter_
Lightly butter the bottom and sides of a 10-inch cast-iron skillet or, alternatively, a similar size baking dish.
Pour the milk and 2 cups water into a medium saucepan, cover, and heat on medium-high until the mixture boils, about 5 minutes. Uncover the pot, add the grits, salt, and pepper, and reduce the heat to medium. Stir constantly until grits are the consistency of thick soup, about 8 minutes. Reduce to a simmer, stirring every 2 minutes, and cook for about 15 to 20 minutes to let the grits thicken further. Cook for 10 to 15 minutes longer, stirring constantly to prevent the grits from sticking to the bottom of the pan. The grits will be really thick at this point.
Remove the pan from the heat and stir in ¾ cup of the cheddar cheese, ¾ cup of the Monterey Jack cheese, and the butter. Pour the grits into the prepared skillet and top them with the remaining cheeses. Turn the oven to broil.
Place the skillet directly under the heating element for 2 to 3 minutes, or until the cheese topping is melted and starts to brown. Serve immediately.
**_Baked Note_**
What is the difference between stoneground and instant grits? Well, technically, instant grits are grits with the germ (i.e., the healthy part) removed. Tastewise, I liken instant grits to instant mashed potatoes: flavorless, manufactured, and without depth. I may be terribly lazy in the kitchen, and you won't find me making my own peanut butter anytime soon, but I beg of you, always use stone-ground grits.
### **MOM'S OLIVE OIL ORANGE BUNDT**
TECHNICALLY SPEAKING, THIS LIGHT AND PLEASING ORANGE CAKE IS NOT RENATO'S MOM'S. Renato's mom would be the first to admit this. It belonged to his mother's neighbor, a lovely French woman named Annette, who arrived and left their neighborhood in Queens before he was born. So, yes, this is really Annette's Olive Oil Orange Bundt (Annette from Marseille, France, to give proper attribution), but his mom adapted and baked it so many times, he truly associates it only with her. Mom's Olive Oil Orange Bundt is great for breakfast with tea and coffee, or sliced and served in the afternoon with a tart dessert wine.
**YIELD: ONE 10-INCH BUNDT**
**_Ingredients_**
_3 cups all-purpose flour_
_1 tablespoon baking powder_
_½ teaspoon salt_
_4 large eggs, separated_
_2 cups granulated sugar_
_1 cup plain yogurt_
_¾ cup good-quality extra-virgin olive oil_
_Freshly grated zest of 2 oranges_
_1 teaspoon vanilla paste, or 1½ teaspoons pure vanilla extract_
_¼ cup confectioners' sugar, sifted, for dusting_
Preheat the oven to 350 degrees F. Generously spray the inside of a 10-inch Bundt pan with nonstick cooking spray; alternatively, butter it well, dust it with flour, and knock out the excess flour.
In a large bowl, whisk together the flour, baking powder, and salt. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the egg yolks until they are pale and light; slowly pour in the sugar until it is completely incorporated. Add the yogurt and olive oil and mix until thoroughly combined. Add the orange zest and vanilla, and mix until just incorporated.
Add the flour mixture to the wet ingredients in two parts, beating after each addition until just combined (this will take about 10 seconds). Scrape down the bowl and beat again for 5 seconds.
In another large bowl, beat the egg whites until stiff peaks form. Scoop 1 cup of the egg whites into the batter. Use a rubber spatula to gently fold them in. After about 30 seconds of folding, add the remaining egg whites and gently fold until they are almost completely combined. Do not rush the folding process.
Pour the batter into the prepared pan and bake for 40 to 50 minutes, rotating the pan halfway through the baking time, or until a small sharp knife inserted into the cake comes out clean. Transfer the pan to a wire rack to cool completely. Gently loosen the sides of the cake from the pan (I sometimes use an offset spatula for this) and turn it out onto the rack.
Just before serving, dust the cake with the confectioners' sugar.
The cake can be stored at room temperature, covered tightly, for up to 3 days.
**_Baked Note_**
When making this cake, try to use a really good, fruity olive oil (Renato likes Paesano extra-virgin olive oil) to bring out the citrusy tones in the cake. Also, you'll notice I skipped glazing this cake altogether (a rarity at Baked) because the simple dusting of confectioners' sugar is really all it needs. If you insist on a glaze, though, I won't stop you (see the sidebar).
* * *
Throughout Renato's entire childhood, he enjoyed this cake sans glaze. I can only assume that his mom thought a glaze was unnecessary and too sweet. However, I will admit that this quick-and-easy orange glaze makes a great visual and is a great way to use up the oranges you zested for the cake.
_2 cups confectioners' sugar
¼ cup fresh orange juice_
In a large bowl, whisk together the confectioners' sugar and orange juice until the glaze is pourable. (If it is too thick, add a few more drops of orange juice. If it is too thin, add a few more tablespoons of confectioners' sugar.) Drizzle the glaze along the crown of the Bundt, allowing it to drip down the sides. Allow the glaze to set before serving.
* * *
### **OATMEAL CHOCOLATE CHIP CAKE WITH CREAM CHEESE FROSTING**
IT MAY SOUND DEVILISH, BUT THIS CAKE—WHICH IS LIKE A DENSE BANANA BREAD WITH THE FLAVOR OF A REALLY GOOD OATMEAL CHOCOLATE CHIP COOKIE—IS PERFECT FOR BREAKFAST. It is like coffee cake for the cookie enthusiast. Of course, it is also ideal for either an afternoon or a midnight snack, but it was conceived as a breakfast treat. The original recipe was buried deep within a Time-Life cookbook, and it revealed a split personality: Half the cake felt like an ambrosia, while the other half felt like a fancy coffee-klatch dessert. It was the South (ambrosia) meets North (coffee klatch) aspect of the dessert that originally caught my eye, and I have been making it ever since. Though I am hesitant to mention it, the delicious cream cheese frosting is not necessary if you are attempting a strict coffee cake interpretation. If you are serving it as an afternoon snack, however, the frosting adds a sweet little sugar rush.
**YIELD: ONE 9-BY-13-INCH CAKE**
**_Ingredients_**
FOR THE CAKE
_8 ounces chocolate chips_
_½ teaspoon bourbon, Scotch, or favorite liquor_
_1½ cups plus 2 tablespoons all-purpose flour_
_1 cup rolled oats_
_½ cup (1 stick) unsalted butter, cut into small cubes, at room temperature_
_2 eggs, slightly beaten_
_¾ cup granulated sugar_
_1¼ cups firmly packed dark brown sugar_
_½ teaspoon salt_
_1 teaspoon baking soda_
_1 teaspoon baking powder_
_1½ teaspoons cinnamon_
FOR THE CREAM CHEESE FROSTING
_5 tablespoons unsalted butter, softened_
_5½ ounces cream cheese, softened_
_2 cups confectioners' sugar, sifted_
_¾ teaspoon pure vanilla extract_
#### **MAKE THE CAKE**
Preheat the oven to 375 degrees F and position the rack in the center. Butter the sides and bottom of a 9-by-13-inch glass or light-colored metal baking pan. Heat 1¼ cups water to boiling.
Place the chocolate chips in a small bowl and toss them with the bourbon until covered. Sprinkle 2 tablespoons of the flour over the chips and toss until coated. This will keep them from settling at the bottom during baking. Set aside.
Place the oats and cubed butter in a large bowl. Pour the boiling water over the oat mixture, wait 30 seconds, and stir to moisten all the oats and melt the butter. Set the mixture aside for 25 to 30 minutes.
In a separate bowl, whisk together the eggs, both sugars, salt, baking soda, baking powder, and cinnamon until combined. Fold in the cooled oatmeal and stir until well combined. Gently fold in the remaining flour and then the chocolate chips. Pour the batter into the prepared pan.
Bake the cake for 40 to 45 minutes, or until a toothpick inserted in the center comes out clean.
Let the cake cool in the pan on a wire rack for at least 30 minutes.
#### **MAKE THE CREAM CHEESE FROSTING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter until it is completely smooth. Add the cream cheese and beat until combined.
Add the confectioners' sugar and vanilla and beat until smooth, about 1 minute. Cover the bowl tightly and refrigerate for at least 30 minutes. (The frosting can be made 1 day ahead. Let it soften at room temperature before using.)
#### **ASSEMBLE THE CAKE**
Spread a thin, even layer of frosting over the cooled cake. Chill it for 15 minutes so that it can set. Slice and serve. The frosted cake can be kept, refrigerated and tightly covered, for up to 3 days. Bring the cake back to room temperature before serving. (An unfrosted one will keep for 3 days, tightly covered, at room temperature.)
**_Baked Note_**
Unlike most of my favorite cakes, this one does not require a standing mixer. You do not have to cream any butter or whip any egg whites. You just dump in the ingredients, stir, and fold. It is quick, easy, and satisfying. The cake tastes great right out of the oven, but most of my tasters and testers think it's best after resting for a day.
### **HONEY CORN MUFFINS**
THIS MUFFIN MOMENT BEGAN FOR RENATO WITH A BOX OF BASIC JIFFY-BRAND CORN MUFFIN MIX. The boxed mix, ever present in his college apartment, produces a simple and straightforward muffin. It is not revelatory. It is sturdy and sound. However, Renato also found that it is extremely versatile as a base recipe, and he set about playing with new additions to zest up the Jiffy muffin. He tried chipotle for a little smoky heat, canned corn for texture, and honey for sweetness. The honey corn muffin was the clear favorite, and he eventually conceived of a recipe that didn't rely on the boxed mix. The honey adds a hint of sweetness without being overt, and the muffin is light (some would say more Northern in origin) without losing any character. Serve these warm and with lashings of butter and more honey.
**YIELD: 12 MUFFINS**
**_Ingredients_**
_2 large eggs_
_1 cup buttermilk_
_¼ cup honey_
_¼ cup (½ stick) unsalted butter, melted and cooled_
_1¼ cups yellow cornmeal_
_¾ cup all-purpose flour_
_1 tablespoon baking powder_
_¼ cup firmly packed light brown sugar_
_2 tablespoons granulated sugar_
_1 teaspoon salt_
_Butter or honey to taste (for serving)_
Preheat the oven to 400 degrees F. Lightly spray each cup of a standard 12-cup muffin pan with a little bit of vegetable spray and use a paper towel to spread the oil evenly along the bottom and up the sides of each cup.
In a medium bowl, lightly whisk the eggs. Add the buttermilk, honey, and butter and whisk again until combined. Set aside.
In a large bowl, whisk together the cornmeal, flour, baking powder, both sugars, and salt. Make a well in the middle of the dry ingredients, pour the wet ingredients into the well, and fold the dry into the wet until just mixed.
Fill each muffin cup about three-quarters full. Tap the bottom of the pan against counter to level the batter. Bake for 12 to 15 minutes, or until the tops are golden brown and a toothpick inserted in the center of a muffin comes out clean. Transfer the pan to a wire rack to cool for 10 to 15 minutes. Pop the muffins out while they're still warm and serve them with a generous helping of butter or honey.
Leftover muffins (should you have any) taste great sliced and toasted in a toaster oven.
**_Baked Note_**
I have spent my entire recipe-writing career advising people to avoid darkcolored baking pans. However, this is a recipe that is improved by baking it in a heavy, dark metal muffin or cupcake tin. The pan will "brown up" the edges and the sides of your muffins perfectly, leaving a tender crumb inside. If you don't have a dark pan, no worries—they will still taste delicious.
### **NEW YORK-STYLE CRUMB CAKE**
I LEARNED THE HARD WAY: NEW YORK-STYLE CRUMB CAKE IS NOT TO BE CONFUSED WITH COFFEE CAKE—EVER. A very passionate born and bred New Yorker (aka Renato Poliafito) informed me, quite brutally, about the not-so-subtle differences between the two. It was a dressing down I won't ever forget. It was as if I'd confused Picasso with Norman Rockwell. First and foremost, New York crumb cake is all about the crumb topping. It is obscenely large in proportion to the cake. In fact, the topping is nearly identical in thickness to—or even thicker than—the cake. Second of all, the crumb should never contain nuts—no crushed nuts, no whole nuts, no hint of a nut whatsoever. Finally, a true New York crumb cake is swirl free. This was the hardest part for me to reconcile, as I love a chocolate nut swirl, and this cake seems like a natural swirl candidate. But I obeyed the New York Crumb Commandments and am now a convert myself.
**YIELD: ONE 9-BY-13-INCH CAKE**
**_Ingredients_**
FOR THE CRUMB TOPPING
_1 cup firmly packed dark brown sugar_
_½ cup granulated sugar_
_½ teaspoon salt_
_1½ tablespoons cinnamon_
_1 cup (2 sticks) unsalted butter, melted and warm_
_2½ cups all-purpose flour_
FOR THE CAKE
_2½ cups all-purpose flour_
_¾ teaspoon baking powder_
_1 teaspoon baking soda_
_½ teaspoon salt_
_12 tablespoons (1½ sticks) unsalted butter_
_1½ cups granulated sugar_
_2 large eggs_
_1¼ cups sour cream_
_1 teaspoon pure vanilla extract_
Preheat the oven to 350 degrees F and position the rack in the center. Butter the sides and bottom of a glass 9-by-13-inch pan. You can use a metal pan, but the edges of the cake may turn crispy (although that is not traditional, it is not an altogether bad thing).
#### **MAKE THE CRUMB TOPPING**
In a medium bowl, stir together both sugars, the salt, and cinnamon. Add the melted butter and whisk until combined. Fold in the flour until it is absorbed and set the mixture aside.
#### **MAKE THE CAKE**
Sift the flour, baking powder, baking soda, and salt together in a medium bowl. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, cream the butter until it is completely smooth and ribbonlike. Scrape down the bowl and add the sugar. Beat the mixture until it starts to look fluffy.
Add the eggs, one at a time, and beat until incorporated. Scrape down the sides of the bowl and mix again for 30 seconds. Add the sour cream and vanilla and beat just until incorporated. Add the dry ingredients in three parts, scraping down the bowl before each addition, beating only until it is just incorporated.
#### **ASSEMBLE THE CAKE**
Pour the batter into the prepared pan. Use your hands to scoop up a handful of the topping and make a fist. The topping should hold together. Break off in chunks and drop them over the cake. Repeat to use all the topping. Remember, the topping layer will look outrageously thick.
Bake the cake for 45 to 55 minutes, or until a toothpick inserted in the middle comes out clean. Rotate the pan two times during the baking process. Cool the entire pan on a wire rack for about 30 minutes before serving.
The cake will last for 3 days, tightly covered, at room temperature.
**_Baked Note_**
Renato likes this cake with really huge crumb chunks. To attain these gargantuan boulders of sugar, make sure you give the crumb time to rest. I sometimes cheat the process and spread the topping mixture on a parchment-lined baking sheet to make it dry a bit faster; however, you don't want it to dry out completely.
### **CORNMEAL GRIDDLE CAKES**
THE CORNMEAL GRIDDLE CAKE IS A HERITAGE RECIPE SO STEEPED IN NOSTALGIA THAT MANY PEOPLE ASSUME IT IS A CREATION OF THEIR FAMILY OR SMALL TOWN, BUT LIKE SO MANY OLD-SCHOOL RECIPES, THE ORIGINS ARE CLOUDY AT BEST. The term "griddle cake" is nearly synonymous with pancakes, flapjacks, and hotcakes. A cornmeal griddle cake is a pancake made with cornmeal, and the end result is hearty without being heavy. While I usually appreciate a good deal of maple syrup on my pancakes, I tend to load up my cornmeal cakes with heaps of fresh butter. It's like a fresh, hot, buttery, flat, crunchy corn muffin.
**YIELD: 10 TO 12 LARGE OR 20 SMALL CAKES**
**_Ingredients_**
_1¼ cups all-purpose flour_
_½ teaspoon salt_
_1 tablespoon baking powder_
_½ teaspoon baking soda_
_1 cup yellow cornmeal_
_2 tablespoons firmly packed light brown sugar_
_2 large eggs_
_1 cup buttermilk_
_2 tablespoons unsalted butter, melted_
_4 to 6 tablespoons unsalted butter for the skillet_
In a medium bowl, sift together the flour, salt, baking powder, and baking soda. Set aside.
Bring 1½ cups water to a boil. Place the cornmeal in a large bowl. Stirring continuously, slowly pour the boiling water over the cornmeal. Keep stirring until the mixture has cooled to lukewarm, almost room temperature. Add the brown sugar and stir until combined.
In a medium bowl, whisk the eggs until pale yellow. Add the buttermilk and whisk until blended. Add the flour mixture, alternating with the buttermilk mixture, to the cornmeal in three parts (beginning and ending with the flour mixture), stirring after each addition until just combined. Stir in the melted butter.
Heat a skillet or griddle pan over medium-low heat.
Add 1 or 2 tablespoons butter to the skillet and make sure it coats the surface. (Note: The first cornmeal griddle cake soaks up a fair amount of the butter and generally speaking, will not be your best handiwork. However, the subsequent griddle cakes will be a thing of beauty.) Drop griddle cakes in ¼-cup batches into the skillet (they will spread—do not crowd the pan). Cook until the bottoms are medium-brown, about 3 minutes, and the tops are bubbly, then flip the griddle cakes over and cook the other side for about 2 minutes and serve immediately. Continue cooking and serving until all the batter is gone. Serve with generous amounts of sweet butter.
**_Baked Note_**
Generally speaking, you can substitute white cornmeal for yellow cornmeal in most recipes, including this one. I believe that yellow cornmeal (which is made from yellow corn) has a "cornier" taste and prefer it for that reason. White cornmeal is made from white corn, obviously, and some people prefer its less aggressive corn taste.
### **MALTED WAFFLES**
I ARRIVED VERY LATE TO THE WAFFLE PARTY. I suppose I had a difficult time justifying getting yet another piece of kitchen equipment (both in terms of cost and its occupation of precious kitchen storage space), and I figured my pancakes were so good no one would ever miss a waffle. I was wrong. Friends, fans, and acquaintances peppered me with waffle recipe suggestions. One Baked fan is purported to own at least six waffle irons (I hope his kitchen is larger than mine). I finally gave in and bought a basic and inexpensive waffle iron. It is not fancy, and it is not a vaunted piece of vintage cooking equipment, but it really does the trick.
I could eat malted waffles all day long, every day, and be quite satisfied. The sweet nuttiness of the malt powder renders the ordinary, simple waffle especially addictive. Personally, I like these waffles drizzled with a bit of melted butter and smattering of chocolate chips, but they taste great with pure maple syrup or confectioners' sugar and whipped cream. And if you don't have a waffle iron, you now know what to ask for on your approaching birthday. Or you could just borrow one from a wacky, waffle-addicted friend.
**YIELD: MAKES ABOUT 10 WAFFLES**
**_Ingredients_**
_2 cups all-purpose flour_
_1 cup malt powder_
_2 tablespoon firmly packed light brown sugar_
_1 teaspoon baking soda_
_1½ teaspoons baking powder_
_½ teaspoon salt_
_2 large eggs, at room temperature_
_2½ cups buttermilk_
_6 tablespoons (¾ stick) unsalted butter, melted and cooled to room temperature_
_Maple syrup, butter and / or chocolate chips for serving_
Preheat the oven to 225 degrees F. Prepare a waffle iron with cooking spray or vegetable oil per the manufacturer's instructions.
In a large bowl, whisk together the flour, malt powder, sugar, baking soda, baking powder, and salt. In a separate bowl, whisk the eggs slightly, add the buttermilk and butter, and whisk again.
Make a well in the center of the dry ingredients and pour the buttermilk mixture into it. Fold the dry ingredients into the wet ever so gently until just combined—there will be some visible lumps. Cook the waffles according to the manufacturer's instructions for your iron. Generally speaking, you will use ¼ to ½ cup batter per waffle (depending on the size of your waffle iron). Cook the waffles until they are golden brown or a little darker (I actually prefer darker ones). Transfer the waffles directly to a rack in your oven to keep them warm while you make the rest. Serve immediately with maple syrup, butter, and chocolate chips.
**_Baked Note_**
Waffles, like most batter cakes, go from light and fluffy to tough and doughy really quickly. All it takes is a bit of overmixing when you combine wet and dry ingredients, and suddenly you have lost your "waffle." While most people are considerate and careful when making a cake, they tend to be less so when making breakfast items. Chalk that up to the heartier mentality that breakfast conjures.
### **BAKED FRENCH TOAST**
I WISH I COULD SAY IT IS MY DEEP LOVE OF COMFORT FOOD (MAC AND CHEESE, ALL MANNER OF GRATINS, AND BAKED CHICKEN DISHES) THAT HAS ME ENTRENCHED FIRMLY IN A CASSEROLE PHASE, BUT I FEAR IT IS JUST A GRADUAL DESCENT INTO LAZINESS. I find them much easier to serve or transport to a gathering. This baked version of French toast is emblematic of this new stage in my life. It is hearty, easy—you put it together the night before your breakfast gathering—delicious, and simple to manipulate (you can swap out fruits and nuts at will).
**YIELD: ONE 8-INCH SQUARE CASSEROLE (ABOUT 4 SERVINGS)**
**_Ingredients_**
FOR THE BAKED FRENCH TOAST
_1 loaf French, Italian or Challah bread, about 11 ounces_
_5 large eggs_
_1 cup half-and-half_
_½ cup whole milk_
_1 teaspoon pure vanilla extract_
_¼ teaspoon cinnamon_
_½ cup whole skinned almonds, coarsely chopped_
FOR THE RASPBERRY SAUCE
_1 cup fresh raspberries_
_2 tablespoons sugar_
_1 teaspoon fresh lemon juice or raspberry liqueur_
#### **MAKE THE BAKED FRENCH TOAST**
Generously butter the sides and bottom of an 8-inch square baking pan. Cut the loaf into 1- or 1½-inch slices and arrange them in the pan.
In a large bowl, whisk the eggs just until they break up. Add the half-and-half, milk, vanilla, and cinnamon. Whisk until combined. Pour the mixture over the bread slices. Cover tightly with plastic wrap and refrigerate for at least 8 hours or overnight.
Preheat the oven to 350 degrees F. Use your fingers to flip each bread slice over, making sure to coat the entire surface in the liquid. Arrange the bread in an overlapping pattern, sprinkle it with almonds, and bake for 35 to 40 minutes, or until the French toast is golden brown and the mixture is puffy. Set the French toast aside to cool in the pan.
#### **MAKE THE RASPBERRY SAUCE**
In a small saucepan, toss the raspberries with the sugar. Cook the mixture over low heat until the berries start to break down, about 10 minutes. Strain the cooked raspberry sauce into a bowl to remove the seeds, then stir in the lemon juice.
#### **ASSEMBLE THE FRENCH TOAST**
Pour the raspberry sauce directly over the warm French toast (or serve it alongside in a gravy boat) and serve immediately. Top with whole fresh raspberries, if you have any left over.
**_Baked Note_**
If you are interested in a creamier bread pudding.type recipe, increase the half and-half to 1 cup and the milk to 3.4 cup. And any hearty, crusty white bread makes a great substitute for the baguette.
### **DOUBLE-CHOCOLATE LOAF WITH PEANUT BUTTER CREAM CHEESE SPREAD**
I HAVE BEEN TOUTING THE CHOCOLATE-AS-BREAKFAST PHILOSOPHY FOR SOME TIME NOW. I am neither a medical doctor nor a nutritionist, but I wholeheartedly believe that a few bites of chocolate throughout the day will save you from an all-out midnight binge (the kind where you devour a pint of ice cream and black out). Though this philosophy has not been tested under the supervision of a scientist, I can assure you that the Double-Chocolate Loaf is a great way to start your own experimentation. It is extremely easy to make—it is a quick bread, after all—and the chocolate flavor is intense without being too sweet. Slice and eat it plain, or toast and spread it with a bit of butter, cream cheese, or our recommended spread: Peanut Butter Cream Cheese. It is wonderful with coffee and the morning newspaper.
**YIELD: ONE 9-BY-5-INCH LOAF**
**_Ingredients_**
FOR THE DOUBLE-CHOCOLATE LOAF
_¾ cup firmly packed dark brown sugar_
_1 cup dark unsweetened cocoa powder, (like Valrhona), sifted_
_1½ cups all-purpose flour_
_¼ cup granulated sugar_
_1½ teaspoons baking soda_
_¾ teaspoon baking powder_
_1 teaspoon salt_
_2 large eggs_
_1 large egg yolk_
_¾ cup buttermilk_
_½ cup vegetable oil_
_1 teaspoon pure vanilla extract_
_8 ounces good-quality dark chocolate (60 to 72%), coarsely chopped_
FOR THE CREAM CHEESE SPREAD
_5 ounces cream cheese, softened_
_2 tablespoons creamy peanut butter_
_⅓ cup sugar_
#### **MAKE THE DOUBLE CHOCOLATE LOAF**
Preheat the oven to 350 degrees F and position the rack in the center. Butter a 9-by-5-inch loaf pan, dust it with flour, and knock out the excess flour.
Place the brown sugar in the bowl of a standing mixer fitted with the paddle attachment. Press out any lumps with the back of a large spoon. Add the cocoa, flour, granulated sugar, baking soda, baking powder, and salt. Scrape down the sides and bottom of the bowl.
In a separate bowl, whisk the eggs and egg yolk until blended, then add the buttermilk, oil, and vanilla; whisk until combined.
Turn the mixer to low and slowly stream the wet ingredients into the dry ones, mixing just until combined. Stir in the dark chocolate chunks by hand.
Pour the batter into the prepared pan and bake for 1 hour to 1 hour and 10 minutes, or until a toothpick inserted in the center of the loaf comes out clean.
Let the cake cool in the pan for 15 minutes, then turn it out onto a wire rack to cool completely.
#### **MAKE THE PEANUT BUTTER SPREAD**
In the bowl of a standing mixer fitted with the paddle attachment, beat together the cream cheese and peanut butter until smooth. Add the sugar and beat until incorporated. (If you are not using the spread immediately, place it in a ramekin, tightly cover it in plastic wrap, and refrigerate it for up to 3 days.)
Serve the loaf plain or toasted, topped with the peanut butter spread.
The loaf will keep, in an airtight container or wrapped tightly, at room temperature for up to 3 days.
**_Baked Note_**
This loaf is the ultimate gift. I double (or triple) the recipe, cool the loaves completely, and remove them from the pans. To give the loaves a nifty bakery look, I like to wrap the loaves in brown parchment paper, and then wrap them in plastic and drop them off to friends along with a card.
* * *
During many bouts of testing several recipes for various books and articles, I noticed something quite peculiar: loaves (or quick breads) actually look and taste better baked in a conventional home oven compared with the large convection-style ovens often found in restaurants and patisseries (mine included). The longer bake of a conventional oven coupled with the absence of a convection fan allows the loaf to bake up from the bottom, creating a perfect dome, crack, and moist crumb. The convection fan, while perfect for cookies and most cakes, often bakes the loaves a tad too quick, resulting in a shorter, tighter loaf—one that is still quite delicious, but not as impressive as those produced in a home oven.
That said, I compiled a list of a few home loaf baking tips:
You can double (or triple) almost any loaf recipe without fail.
Make sure your loaf is baked all the way through. This is important, as many loaves are tricky and appear baked from the outside (including puffy dome and crack) while remaining undercooked. If you pull your loaf out early, it will collapse in on itself, creating a "sunken" loaf.
Loaves freeze extremely well. Let cool entirely to room temperature and just wrap tightly in two layers of plastic wrap and one layer of aluminum foil, and freeze for up two months. Bring the loaves back to room temperature by way of your refrigerator for 8 hours, then let sit at room temperature for 4 hours before unwrapping the loaves.
Mix-ins (chocolate chips, fruits, nuts) are hit-and-miss with loaves, depending on the density of your batter, and may sink to the bottom during baking. Toss your mix-ins in a bit of liquor, then coat them in flour to allow the mix-ins to become suspended in the batter.
* * *
##
**CLASSIC PIE DOUGH**
**BUTTERMILK PIE ( WITH A HINT OF MAPLE SYRUP)**
**PEANUT BUTTER BANANA CREAM PIE**
**SAWDUST PIE**
**ALMOND JOY TART**
**PEACHES AND DREAM PIE**
**MISSISSIPPI MUD PIE (A), AKA COFFEE ICE CREAM TART**
**MALTED CRISP TART**
**ORANGE CREAMSICLE TART**
**BLACKBERRY PIE**
**WHISKEY PEAR TART**
**PECAN TASSIES**
Pies can be heartbreaking. They are burdened with the heavy weight of all-American nostalgia. That is a lot to shoulder. A pie loaded with berries or apples or nuts or custard can be uniquely disappointing—at once soggy, oversweet, tasteless, and, yes, sad. I can think of no other dessert that has been mined so deeply to reflect a grandmotherly aura, and yet so many pies do Grandma wrong.
Pies can also be otherworldly. Transformative. A seasonal treat that celebrates and showcases local ingredients like few other recipes, sweet or savory. However, there are few shortcuts to a great pie. I have yet to meet a grocery-store pie, suffocated in shrink-wrap or a mindless see-through box, that is worth one calorie. (I am okay with store-bought, slice-and-bake cookie dough, though, which mimics the home-baked thing much better.) Frozen pie crusts might save a few minutes in the kitchen, but they are uniformly boring. You lose control over thickness and flakiness. Even more unpleasant are the premade crumb crusts available these days. Homemade versions can be pulled together in less than 5 minutes and at a substantial cost savings.
Pie makers already know all of the above. The home bakers I've met in person or through e-mail never utter the words frozen or premade. Never. That's why it was so gratifying to work on this chapter. Though I deify the classic fruit pie (flaky crust encasing a fresh fruit filling), a great many other American pie and tart recipes made their way to me that were lighthearted, easy to make, and thoroughly delicious.
Our Whiskey Pear Tart is a showstopper. It's a little bit boozy and a little bit serious and very, very tasty. It's classic without being stuffy. The Mississippi Mud Pie (A) was created to satisfy a West Coast contingent that insisted the pie must be made with coffee ice cream—Mississippi Mud Pie (B) contains no ice cream at all. I happen to like both versions a great deal. And of course, the Peanut Butter Banana Cream Pie deserves a cult following. It is a pie stuffed full of all of my favorite ingredients (chocolate, peanut butter, and bananas) yet it manages to taste light, sophisticated and deviously delicious.
### **CLASSIC PIE DOUGH**
THIS PIE DOUGH, OUR BASIC PIE DOUGH, IS EXACTLY THE SAME AS THE ONE FROM OUR FIRST BOOK, BAKED: NEW FRONTIERS IN BAKING. There is little need for variance on this recipe. This dough is flaky, easy to work with, and a bit more forgiving than some.
**YIELD: 2 BALLS OF DOUGH, ENOUGH FOR TWO SINGLE-CRUST
9-INCH PIES OR ONE DOUBLE-CRUST PIE**
**_Ingredients_**
_3 cups all-purpose flour_
_1 tablespoon sugar_
_1 teaspoon fine salt_
_1 cup (2 sticks) cold unsalted butter_
In a medium bowl, whisk the flour, sugar, and salt together. In a measuring cup, stir ¾ cup water with several ice cubes until it is very cold.
Cut the cold butter into cubes and toss them in the flour mixture to coat. Put the mixture in the bowl of a food processor and pulse in short bursts until the butter pieces are the size of hazelnuts.
Pulsing in 4-second bursts, slowly drizzle the ice water into the food processor through the feed tube.
As soon as the dough comes together in a ball, stop adding water. Remove the dough from the food processor and divide it in half. Flatten each piece into a disk and wrap each disk first in parchment paper and then in plastic wrap. Refrigerate the dough until firm, about 1 hour. (The dough can be kept refrigerated for up to 3 days or frozen for up to 3 months. Thaw it in the refrigerator before proceeding with your recipe.)
**_Baked Note_**
The dough will feel and look sticky, or at least stickier than you might be used to. Don't fret. Once it firms up in the fridge, it will be perfect.
### **BUTTERMILK PIE ( WITH A HINT OF MAPLE SYRUP)**
I WOKE UP ONE DAY IN THE MIDDLE OF WINTER WITH AN URGE TO VISIT QUEBEC CITY. It was a quick, fairy-tale trip complete with crackling fireplaces, large and comfortable beds, and plenty of cheese and chocolate. I was also taken with the omnipresent maple pie. Well, perhaps I was more than taken with this dessert, since I tasted many versions as I ate my way through Quebec City. It vaguely reminded me of a too-sweet version of the buttermilk pie a friend's mom used to make. Buttermilk pie is supposedly of Texan origin, though you will not find it on many Texas menus these days. It is essentially a custard thrown into a pie shell; it comes together in mere minutes (not including the baking time). I fused my favorite aspects of Canadian maple pie with that forgotten buttermilk pie and came up with this sweet maple-infused custard hybrid.
**YIELD: ONE 9-INCH PIE**
**_Ingredients_**
_1 ballClassic Pie Dough_
_4 large eggs_
_2 tablespoons plus 1 teaspoon flour_
_¾ cup granulated sugar_
_¼ cup firmly packed dark brown sugar_
_½ cup (1 stick) unsalted butter, melted and cooled_
_1 cup buttermilk_
_¼ cup pure maple syrup_
Dust a work surface with a sprinkling of flour and roll the dough ball out into a 12-inch round. Transfer it to a 9-inch pie plate and carefully work it into the bottom and up the sides, folding it under and crimping the edges as you go. Wrap and freeze the crust until it is firm, about 30 minutes (it will keep this way for up to 3 months).
Preheat the oven to 325 degrees F.
In large bowl, lightly beat the eggs. Whisk in 2 tablespoons of the flour. Don't be alarmed if the mixture appears curdled after the addition of the flour—it will come back together. Add both sugars and whisk until the mixture is well combined and uniform in color. Whisk in the butter, buttermilk, and maple syrup until completely combined.
Sprinkle ½ teaspoon of the remaining flour over the unbaked pie crust. Pour the batter into the shell, sprinkle with the remaining ½ teaspoon flour, and bake for about 1 hour, or until the custard is set.
Set the pie on a wire rack to cool completely. Serve it at room temperature. Refrigerate any leftover pie, tightly covered, for up to 2 days.
**_Baked Note_**
I should warn you ahead of time that this pie, with its pale dough and brownish filling, would not win any beauty pageants. It is not a showstopper or a centerpiece, and it almost repels attempts to gussy it up. On the other hand, if you are not desperate to impress with a visual smorgasbord, this homey little pie packs a lot of unexpected taste and texture. It is sweet and custardy and captivating. Think of it as the dessert equivalent of frogs' legs —ugly, but addictive.
### **PEANUT BUTTER BANANA CREAM PIE**
TRUE STORY: I WAS HAUNTED BY THE BANANA CREAM PIE. For a brief period in my life, it was all I thought about and ate. I tried many variations and made many variations, but I always found something wrong with the recipe. Too sweet. Too grainy. Too slimy. I lay awake at night rethinking the details, reworking the formula in my head.
This recipe is the culmination of all my thoughts and fever dreams, and it is now officially my favorite pie to make and eat. The crust is made with a classic vanilla wafer, which is less assertive than a traditional pie crust, and the banana pudding is light and pleasing. The whole pie is topped with a layer of smooth and dreamy peanut-butter-flavored cream cheese, which brings the dessert together. No, it is not traditional, but it is delicious. Oh, and the bananas are tossed in a bit of orange juice to keep them from turning black. Trust me, you do not want to eat black bananas.
**YIELD: ONE 9-INCH PIE**
**_Ingredients_**
FOR THE VANILLA WAFER CRUST
_6 ounces vanilla wafer cookies_
_6 tablespoons (¾ stick) cold unsalted butter, cut into ½-inch cubes_
_2 tablespoons sugar_
FOR THE BANANA PUDDING FILLING
_⅓ cup sugar_
_1½ tablespoons cornstarch_
_⅛ teaspoon salt_
_1 cup heavy cream_
_½ cup whole milk_
_2 large egg yolks_
_1 vanilla bean_
_1 tablespoon unsalted butter_
_3 ripe bananas, peeled_
_2 tablespoons orange juice_
FOR THE PEANUT BUTTER TOPPING
_3 ounces cream cheese, softened_
_½ cup confectioners' sugar_
_1 teaspoon pure vanilla extract_
_⅓ cup creamy peanut butter (do not use old-fashioned or freshly ground)_
_⅔ cup heavy cream, chilled_
ASSEMBLY
_1 ripe banana, peeled (optional)_
_1 tablespoon orange juice (optional)_
_Chocolate covered peanuts (optional)_
Preheat the oven to 350 degrees F and position the rack in the middle.
#### **MAKE THE VANILLA WAFER CRUST**
Place the vanilla wafer cookies, butter, and sugar in a food processor. Pulse in short bursts until the mixture resembles a moist crumb. Turn the mixture out into your pie plate and press it into the bottom and up the sides. Using the back of a large spoon will help you to create an even crust.
Bake the crust until it is golden brown, 10 to 12 minutes. If it begins to puff while baking, use the back of the spoon to press it gently down. Allow the baked crust to cool completely.
#### **MAKE THE BANANA PUDDING FILLING**
In a medium saucepan, whisk together the sugar, cornstarch, and salt. Slowly, while whisking continuously, stream in the cream, then the milk. Add the egg yolks. Cut the vanilla bean in half lengthwise and, using the tip of the knife or a small teaspoon, scrape the seeds into the saucepan. Whisk until the mixture is combined. Discard the bean.
Turn the heat to medium-high and, whisking occasionally, bring the mixture to a boil, about 5 minutes. Remove the pan from the heat, add the butter, and stir vigorously for about 2 minutes to release excess heat. Spread the warm pudding over the cooled crust and chill the pie until the filling is completely cool, about 1 hour.
After the filling has chilled, thinly slice the bananas on a diagonal. Toss the slices in the orange juice. Transfer the banana slices to a paper towel and pat them dry. Arrange them in a single layer over the pudding to cover it completely. Return the pie to the refrigerator while making the peanut butter topping.
#### **MAKE THE PEANUT BUTTER TOPPING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the cream cheese and confectioners' sugar until smooth. Add the vanilla and peanut butter, and beat until just combined.
In a clean bowl of a standing mixer, use the mixer fitted with the whisk attachment and whip the cream until soft peaks form. Remove the bowl from the mixer and, with a rubber spatula, gently fold the whipped cream into the peanut butter mixture until the topping is uniform in color.
Spread the peanut butter layer evenly over the bananas on the pie. Chill for at least 3 hours or as long as 8 hours.
#### **TO ASSEMBLE THE PIE**
Just before serving, thinly slice the banana on a diagonal and toss it with the orange juice.
Arrange the banana slices around the top edge of the pie, then arrange chocolate-covered peanuts around the edges of the banana slices, if desired. Serve immediately.
The pie tastes best if eaten within 24 hours. On the rare occasion that you have leftovers, wrap the pie in plastic wrap and refrigerate it for up to 3 days.
**_Baked Note_**
This pie works best if made in a glass pie plate. I am a huge fan of decorative (and funky) pie plates, but glass allows the crust to brown properly. If you want to use a metal or ceramic pie plate, adjust the oven temperature to 375 degrees F and be sure to check the crust more often as you bake it.
### **SAWDUST PIE**
I JUST ASSUMED THAT SAWDUST PIE WAS A HERITAGE RECIPE. The name implies a rugged backstory, one with humble or Depression-era beginnings. I imagined the recipe was steeped in great dustbowl prairie lore and passed down to each generation with a bit of history and a sprinkling of legend. I was completely wrong. Sawdust pie was created at Patti's 1880's Settlement Restaurant in Grand Rivers, Kentucky, sometime after 1975 (the exact date is hard to pin down) and took hold in the national conscience after a version of the recipe was published in _Bon Appétit_ in May 1983. The pie is composed of graham cracker crumbs, coconut, and pecans, which, when thrown together, are supposed to resemble sawdust. The original recipe was a little too sweet. I reimagined it with a little less sugar, a tiny bit of white chocolate, and some much-needed salt. The pie is usually topped with whipped cream and banana slices, and I suggest you always serve it warm.
**YIELD: ONE 9-INCH PIE**
**_Ingredients_**
_1 ballClassic Pie Dough_
_1 cup granulated sugar_
_¼ cup firmly packed dark brown sugar_
_1¼ cups unsweetened flaked coconut_
_1¼ cups pecans, coarsely chopped_
_1½ cups graham cracker crumbs, about_
_21 graham crackers_
_2 ounces high-quality white chocolate, coarsely chopped_
_½ teaspoon salt_
_6 egg whites_
_1 teaspoon pure vanilla extract_
_Simple Whipped Cream for serving (optional)_
_Sliced bananas for serving (optional)_
Dust a work surface with a sprinkling of flour. Roll out the dough ball into a 12-inch round. Transfer it to a pie dish and carefully work it into place, folding any overhang under and crimping the edge as you go. Wrap and refrigerate the crust for at least 30 minutes.
Preheat the oven to 350 degrees F.
In a large bowl, use a wooden spoon to combine the sugars, coconut, pecans, graham cracker crumbs, white chocolate, and salt. Add the egg whites and vanilla and stir until just combined—the egg whites should coat all the ingredients.
Transfer the filling to the prepared pie shell. Bake until filling is set to the touch, 30 to 40 minutes. Cool the pie for at least 1 hour before serving it warm, with whipped cream and sliced bananas, if you like.
**_Baked Note_**
I tend to add chocolate to a lot of desserts that didn't start out with it. This is simply my preference (and addiction). In a dramatic act of self-control, I did not add dark chocolate to this recipe . . . though if you share my chocolate addiction, feel free to stir in about 4 ounces of chocolate chips or chopped dark chocolate with the coconut.
### **ALMOND JOY TART**
THOUGH I AM NOT A TRUE COCONUT PERSON, PEOPLE GO BONKERS FOR OUR COCONUT CAKES AND COCONUT MACAROONS, ESPECIALLY AT OUR CHARLESTON LOCATION, WHERE COCONUT IS KING. And I have managed to work my way into the coconut groove with this riff on the Almond Joy. That legendary candy bar was the creation of the Peter Paul Candy Manufacturing Company of Connecticut, now absorbed by Hershey's. Mr. Peter Paul Halajian certainly knew his way around coconut: He developed the Almond Joy's precursors—bars of moist coconut flake covered in chocolate—made them fresh at night, and sold them door-to-door the following day. This recipe is my wistful version of what a fresh Almond Joy must taste like.
**YIELD: SIX 4-INCH INDIVIDUAL TARTS**
**_Ingredients_**
FOR THE ALMOND TART DOUGH
_1 large egg_
_¼ cup whole toasted almonds_
_¼ cup sugar_
_1¼ cups plus 2 tablespoons all-purpose flour_
_¼ teaspoon salt_
_½ cup (1 stick) cold unsalted butter, cut into ½- inch cubes_
FOR THE COCONUT CREAM FILLING
_8 ounces good-quality white chocolate, coarsely chopped_
_1 cup heavy cream_
_2 cups unsweetened shredded coconut (if you're coconut obsessed, go ahead and use 2¼ cups)_
_1 tablespoon light rum_
FOR THE CHOCOLATE GLAZE AND GARNISH
_2 ounces good-quality milk chocolate, coarsely chopped_
_2 ounces good-quality dark chocolate (60 to 72%), coarsely chopped_
_½ cup heavy cream_
_6 whole toasted almonds_
#### **MAKE THE ALMOND TART DOUGH**
In a small bowl, lightly whisk the egg and set it aside.
Put the almonds and sugar in the bowl of a food processor and pulse until the almonds are finely ground. Add flour and salt and pulse again just until mixed. Add the butter and pulse until sandy (about 6 to 10 quick pulses). Pour in the egg and pulse just until the dough begins to cohere into a ball. Form the dough into a disk, wrap it tightly in plastic wrap, and refrigerate it for at least 1 hour or overnight.
#### **MAKE THE COCONUT CREAM FILLING**
Place the white chocolate in a medium heatproof bowl.
In a small saucepan set over medium heat, heat the cream just to a boil. Pour it over the white chocolate and let it stand for 30 seconds. Slowly, starting in the center of the bowl, whisk the cream and white chocolate until smooth. Cover and refrigerate this ganache for 4 hours or overnight before proceeding.
#### **ASSEMBLE THE TART**
Dust a work surface with flour. Place the disk of chilled dough on the work surface and divide it into 6 equal portions. Shape each into a smooth disk. (Note: The dough will be sticky. Make sure to turn it with a bench knife or offset spatula as needed and keep the working surface floured.) Use a rolling pin to roll each piece of dough into a 5½-inch circle just over 1/8 inch thick. Very gently press each dough round into a 4-inch tart pan with removable bottom.
Place the tart pans in the freezer for 30 minutes. Preheat the oven to 375 degrees F.
Line the tart crusts with aluminum foil, and fill each one three-quarters full with pie weights or dried beans. Bake them for 15 minutes, then remove the foil and weights and bake for another 10 minutes, or until lightly browned. Transfer the tart pans to a wire rack to cool.
#### **MAKE THE COCONUT CREAM FILLING**
Meanwhile, in the bowl of a standing mixer fitted with the whisk attachment, beat the white chocolate ganache at medium speed until soft peaks form. Do not overwhip. Gently fold in the coconut and the rum. Divide the filling evenly among the cooled tart shells and place them in the refrigerator while you make the chocolate glaze.
#### **MAKE THE CHOCOLATE GLAZE**
Place the milk and dark chocolates in a medium heatproof bowl.
In a small saucepan, heat the heavy cream until it is just about to boil. Pour it over the chocolates and whisk to combine. Let the mixture set for about 10 minutes. Remove the tarts from the refrigerator and spoon the glaze evenly over each one. Top each tart with one almond and refrigerate again until the glaze sets up, about 10 minutes.
The tarts can be stored, tightly covered, in the refrigerator for up to 2 days.
**_Baked Note_**
It would seem that a chocolate tart crust would be the perfect shell for a riff on the Almond Joy, but it fell short during many tests and trials. The chocolate overwhelmed the coconut, and somehow it just felt entirely contrived.
### **PEACHES AND DREAM PIE**
I SPENT A GOOD PART OF MY YOUTH IN THE MIDDLE OF FLORIDA, FLUSH AGAINST THE GULF OF MEXICO. I lived in a subdivision of a subdivision, as was the nature of the beast, and I was blissfully unaware of the farmers' market revolution stirring in other parts of the country. All our fruit was purchased from a dimly lit grocery store, and often we bought the canned variety—we viewed canned fruit as a much better nutritional alternative to, say, cookies.
This creamy peach pie is a great revision of a recipe that has floated about the country in many forms. It is tasty, as well as quick and easy to put together. On a side note, I did unearth several no-bake peaches-and-cream pie recipes . . . I did everyone a favor and promptly reburied them.
**YIELD: ONE 9-INCH PIE**
**_Ingredients_**
_1 ballClassic Pie Dough_
FOR THE PEACHES AND DREAM FILLING
_10 canned peach halves, or about 2½ cups fresh diced, peeled peaches_
_2 large eggs_
_1 cup sour cream_
_2 tablespoons honey_
_¼ teaspoon salt_
_½ cup firmly packed dark brown sugar_
_2 tablespoons all-purpose flour_
FOR THE PIE TOPPING
_½ cup firmly packed dark brown sugar_
_⅓ cup all-purpose flour_
_4 tablespoons cold unsalted butter, cut into ½-inch pieces_
#### **MAKE THE CRUST**
Dust a work surface with a sprinkling of flour. Roll the dough ball out into a 12-inch round. Transfer the dough to a pie dish and carefully work it into place, folding any overhang under and crimping the edge as you go. Cover the crust in plastic wrap and refrigerate it for at least 30 minutes. Preheat the oven to 375 degrees F.
#### **MAKE THE PEACHES AND DREAM FILLING**
Arrange the peach halves cut side up on the bottom of the pie shell or spread the diced fresh peaches in the pie shell. In a medium bowl, whisk together the eggs, sour cream, and honey until they just come together. Sprinkle the mixture with the salt, brown sugar, and flour, and whisk until just combined. Pour the mixture over the peaches.
#### **MAKE THE PIE TOPPING**
Place the sugar, flour, and butter in a bowl. Use your hand to work the butter into the dry ingredients until the mix looks like coarse sand. Assemble the pie. Sprinkle the pie topping across the filling, and bake for 45 minutes, or until the filling is bubbly (place a sheet pan on the oven rack directly below the pie to catch any filling that bubbles over). Let the pie cool overnight before serving.
**_Baked Note_**
At least 90 percent of the peaches-andcream pie recipes that crossed my path while I researched this book specified the use of canned peaches. Perhaps they were written in that era when fresh peaches were difficult to locate, or perhaps the sugar content of a canned peach was an added plus, but I tested fresh peaches in place of their canned counterparts in several recipes without any problems.
### **MISSISSIPPI MUD PIE (A), AKA COFFEE ICE CREAM TART**
USUALLY THE DIFFERENCES IN A POPULAR RECIPE—SAY RED VELVET CAKE—ARE A MATTER OF SUBTLE VARIATIONS (I.E., LESS FLOUR, MORE SUGAR, CREAM CHEESE FROSTING VERSUS VANILLA FROSTING) AND PERSONAL TWEAKS (SOME PEOPLE TRADE THE RED FOOD DYE FOR BEET JUICE). However, on occasion, a recipe will undergo seismic differences from state to state and person to person. The Mississippi Mud Pie is one of those recipes. Mississippi mud means many things to many people, and we took the liberty of reinterpreting two of the most popular versions (see version B). This recipe is the simpler one. In fact, you can put it together rather effortlessly (though there are stretches of time to account for the chilling and freezing steps), and I find that perfectly charming: chocolate cookie crust covered in a layer of chocolate fudge covered in a layer of coffee ice cream and drizzled with more bourbon fudge. You could swap the coffee ice cream for another flavor, but then you couldn't call it Mississippi mud.
**YIELD: ONE 9-INCH PIE**
**_Ingredients_**
FOR THE CHOCOLATE COOKIE CRUST
_30 chocolate wafer cookies, about 6 ounces_
_1 tablespoon granulated sugar_
_6 tablespoons (¾ stick) unsalted butter, melted_
FOR THE FILLING
_4 ounces good-quality dark chocolate (60 to 72%)_
_¼ cup plus 1 tablespoon heavy cream_
_3 tablespoons unsalted butter_
_2 tablespoons light corn syrup_
_1 cup confectioners' sugar, sifted_
_1 tablespoon Kentucky bourbon_
_1 pint good-quality coffee ice cream_
_½ cup toasted pecans, coarsely chopped_
FOR THE BOURBON FUDGE TOPPING
_2 tablespoons heavy cream_
_2 tablespoons unsalted butter_
_1 tablespoon light corn syrup_
_3 ounces good-quality dark chocolate (60 to 72%)_
_1 teaspoon Kentucky bourbon_
#### **MAKE THE CHOCOLATE COOKIE CRUST**
In a food processor, pulverize the wafer cookies into a very fine crumb. You should have about 1½ cups. Place the crumbs into a bowl, add the sugar, and stir until combined.
Pour the melted butter over the crumbs and mix well. Transfer the crumb mixture to a 9-inch pie plate and press it into the bottom and up the sides. Use the back of a large spoon to get an even crust. Set the crust aside in the refrigerator.
#### **MAKE THE FILLING**
Place the chocolate in a large heatproof bowl.
In a medium saucepan, bring the cream, butter, and corn syrup to a simmer. Remove the mixture from the heat, pour it over the chocolate, and let sit for 1 minute. Then whisk the chocolate mixture until it is completely smooth. Whisk in the confectioners' sugar and bourbon.
Spread the fudge evenly over the bottom of the pie crust, cover it, and refrigerate for 2 hours.
Soften the coffee ice cream by placing the container in the microwave for 10 seconds on high. Put it into a large bowl and use a rubber spatula to beat it until it is slightly malleable. Spread the ice cream over the chilled fudge filling, sprinkle it with pecans, gently pressing them into the ice cream, and freeze the pie for about 1½ hours, or until the ice cream is firm.
#### **MAKE THE BOURBON FUDGE TOPPING**
In a small saucepan over low heat, heat the cream, butter, and corn syrup together until the mixture begins to simmer. Remove the pan from the heat and add the chocolate. Whisk until the fudge is smooth—if you still have a few stray chocolate chunks, reheat the mixture over very low heat until they are completely melted. Stir in the bourbon.
Beat the fudge topping until it reaches room temperature, and pour it over the ice cream and pecan layer in a zigzag pattern. Freeze the Mississippi Mud Pie until it is set, about 20 minutes. To serve the pie, cut it with a warmed sharp knife.
The pie will keep in the freezer, tightly covered, for up to 4 days.
**_Baked Note_**
During our dessert explorations, I noticed the Mississippi mud devotees were among the more fanatical bakers. A few even suggested —no, demanded—that I print their specific ice cream brands or ice cream recipes in the book. In the end, I've decided that might be too restrictive. Use any high-quality store-bought coffee ice cream, with the shortest ingredient list possible, or make your own.
### **MALTED CRISP TART**
TRUTH BE TOLD, MY SEARCH FOR TREASURED AND HERITAGE-INSPIRED TART RECIPES WAS A LITTLE BLEAK. I dug up hundreds of heirloom pie and cookie and cake recipes, but the tart recipes were extremely scarce, or incomplete, or not really tarts. Instead, I tasked pastry chef extraordinaire Melissa Fritz Walters with creating a tart that is a little bit malty and little bit crispy, and she built the tart of my candy-colored dreams. The Malted Crisp Tart is a Melissa-original, and it quickly became my favorite dessert. The brown sugar crust is a near perfect receptacle for the dense malted milk chocolate and light malted diplomat cream, and the layers of caramelized rice crispies provide an unexpected but ultimately desired crunch. The tart is definitely designed for a swanky gathering, but I have been known to eat a slice in my pajamas in front of the television.
**YIELD: ONE 9-INCH TART**
**_Ingredients_**
FOR THE BROWN SUGAR CRUST
_1½ cups all-purpose flour_
_¼ teaspoon salt_
_1 tablespoon malted milk powder_
_10 tablespoons (1¼ sticks) unsalted butter, chilled (cut into ½-inch pieces)_
_½ cup firmly packed light brown sugar_
_½ teaspoon pure vanilla extract_
FOR THE CARAMELIZED CRISPIES
_⅓ cup sugar_
_2 cups crisped rice cereal_
FOR THE MILK CHOCOLATE GANACHE
_8 ounces good-quality milk chocolate, coarsely chopped_
_⅔ cup heavy cream_
_2 teaspoons malted milk powder_
FOR THE MALTED DIPLOMAT CREAM
_1¼ cups whole milk_
_⅓ cup sugar_
_1 large egg yolk_
_1 large egg_
_1 tablespoons plus 1½ teaspoons cornstarch_
_2 tablespoons malted milk powder_
_2 tablespoons unsalted butter_
_2 teaspoons pure vanilla extract_
_5 ounces heavy cream_
FOR THE TART ASSEMBLY
_1 cup crushed malted milk balls_
_Malted Milk Balls, to garnish_
_Caramelized Crispies, to garnish_
#### **MAKE THE BROWN SUGAR CRUST**
Lightly spray a napkin or paper towel with vegetable oil and use the napkin to apply the oil to the sides and bottom (and nooks and crannies) of the tart pan.
Place the flour, salt, malted milk powder, butter, sugar, and pure vanilla in a food processor and pulse until the mixture is crumbly.
Scoop the crumb mixture out into the prepared tart pan and use your hands to press the mixture into bottom and up the sides of the tart pan (the crust should not look too thick).
Preheat the oven to 350 degrees F.
Place the tart pan in the freezer for 20 minutes.
Remove the tart pan from the freezer, place on a baking sheet, and bake until the tart is golden brown, about 20-30 minutes.
Transfer the tart pan to a wire rack and cool completely.
#### **MAKE THE CARAMELIZED CRISPIES**
Line a half-sheet baking pan with a Silpat or, alternatively, a sheet of aluminum foil sprayed with vegetable oil.
In a small saucepan over low heat, stir together 2 tablespoons water with the sugar and bring just to a very low boil for about 1 minute.
Add the crisped rice cereal and stir until the mixture is dry. Keep stirring until sugar begins to caramelize and the pan begins to smoke. Fold the mixture over and over until all the crispies are coated with an amber layer of sugar. Once the crispies are completely coated, turn them out onto the Silpat and cool completely.
Break the candy up into large chunky pieces and set aside.
#### **MAKE THE MILK CHOCOLATE GANACHE**
Place the milk chocolate in a medium-size heatproof bowl.
In a small saucepan over low heat, whisk together the heavy cream and malt powder. Bring the mixture to a simmer (tiny bubbles will form around the edges of the cream; it should not be a rolling boil). Remove from the heat and pour over the milk chocolate. Let the mixture sit for 2 minutes. Starting in the center of the bowl, and working your way out to the edges, whisk the chocolate ganache in a circle until completely smooth.
#### **TO BEGIN ASSEMBLING THE TART**
Gently pour the ganache into the cooled tart shell. Top with crushed malted milk balls and 1 cup caramelized crispies. Very gently press the crushed malt balls and caramelized crispies into the ganache.
Refrigerate the tart while you make the Malted Diplomat Cream.
#### **MAKE THE MALTED DIPLOMAT CREAM**
Set a fine-mesh sieve over a medium bowl.
In a medium saucepan, bring the milk to a simmer and keep warm.
In a medium bowl, whisk the sugar, egg yolk, egg, cornstarch, and malted milk powder together until the mixture is pale, about 1 minute.
Whisk half of the warm milk into the egg yolk mixture, then pour the mixture into the remaining milk in the saucepan and cook over medium heat, whisking constantly, until thickened, about 5 minutes. Remove from the heat and whisk in the butter and vanilla. Strain the pastry cream through the sieve and press a piece of plastic wrap directly onto the surface of the cream to prevent a skin from forming. Put in the refrigerator for about 1 hour until chilled.
Remove the pastry cream from the refrigerator and whip until creamy.
In a separate bowl, whip the heavy cream with a whisk until soft peaks form and fold into the pastry cream.
#### **FINISH ASSEMBLING THE TART**
Remove the tart from the refrigerator and cover the Milk Chocolate Malt layer with the Malted Diplomat Cream. Garnish with a few whole malt balls and caramelized crispies. Refrigerate the tart to set up, about 30 minutes, and serve immediately. The tart can be stored, tightly covered, in the refrigerator for up to 2 days.
**_Baked Note_**
Yes, there are a lot of bowls involved in this recipe, but it is not difficult —so don't be afraid to attempt it. I tried to break out the steps in an orderly manner, but feel free to break it up over two days: make the tart dough (unbaked) and caramelized crispies on the first day and the other parts on day two.
### **ORANGE CREAMSICLE TART**
GENERALLY SPEAKING, I AM NOT A BIG SODA DRINKER. In fact, the only time I ever really want a carbonated beverage is during a sour morning-after hangover. It wasn't always so. During the endless summers of grade school, I craved orange soda pop and only orange soda pop—no grape, no cola (certainly never water). This light, citrusy, summery tart is an elegant nod to my old obsession. It actually features orange soda as a main ingredient, yet the look and texture suggest something decidedly more upscale. It's soda-fountain quaint with a whiff of Le Cordon Bleu, and it is delicious.
**YIELD: ONE 9-INCH TART**
**_Ingredients_**
FOR THE ORANGE CREAM SODA FILLING
_½ cup (1 stick) unsalted butter, cut into_
_½-inch cubes_
_1¼ teaspoons unflavored gelatin_
_Zest and juice of 2 medium lemons
(2 tablespoons zest and ¼ cup juice)_
_Zest and juice of 3 large oranges
(3 tablespoons zest and 1 cup juice)_
_1 cup orange cream soda_
_3 large eggs_
_2 large egg yolks_
_¾ cup sugar_
FOR THE ORANGE TART DOUGH
_½ cup (1 stick) unsalted butter_
_¼ cup sugar Zest of 1 orange (2 tablespoons)_
_¼ teaspoon salt_
_1 large egg_
_1½ cups all-purpose flour_
FOR THE ORANGE WHIPPED TOPPING
_1 cup heavy cream_
_2 tablespoons sugar_
_2 tablespoons orange cream soda_
#### **MAKE THE ORANGE CREAM SODA FILLING**
Place the butter in a large bowl. Set aside.
In a wide bowl, sprinkle the gelatin evenly over the lemon juice (take care that the gelatin does not clump).
In a medium saucepan, stir together the orange juice and soda. Bring the liquid to a boil and cook until it is reduced by half, or 1 cup. Turn the heat to low and whisk to release excess heat.
In a medium bowl, whisk together the lemon and orange zest, eggs, egg yolks, and sugar and pour the mixture into the saucepan. Cook over medium-low heat, whisking constantly, until a candy thermometer reads 180 degrees F, or the curd can easily coats the back of a wooden spoon.
Remove the pan from the heat and add the gelatin mixture. Whisk until the gelatin is completely combined. Pour the liquid through a fine-mesh sieve directly onto the butter. Whisk the mixture furiously until it has increased a bit in volume (the faster you whisk, the more voluminous it will be). Cover the top of the curd with plastic wrap, pressing the plastic directly onto the curd's surface, and refrigerate for at least 4 hours.
#### **MAKE THE ORANGE TART DOUGH**
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter, sugar, zest, and salt until light and fluffy. Add the egg, and beat just until incorporated. Scrape down the bowl, add the flour all at once, and beat just until the dough comes together in a ball. Do not overbeat, or your crust will be hard.
Remove the dough from the bowl, shape it into a disk with your hands, wrap it tightly in plastic wrap, and refrigerate it for at least 30 minutes.
Dust a work surface with a sprinkling of flour. Use a rolling pin to roll the dough into a 10-inch circle about ¼ inch thick. (Note: The dough will be sticky. Make sure to turn it over with a bench knife or offset spatula as needed and to keep the work surface floured.)
Ever so gently, guide the dough, without pulling it, into a 9-inch tart pan with a removable bottom and lightly press it into place. Roll the rolling pin over the pan to trim off excess. Place the tart pan in the freezer for 30 minutes.
Preheat the oven to 375 degrees F.
Line the tart shell with aluminum foil and fill it three-quarters full with pie weights or dried beans. Bake for 15 minutes, then remove the foil and weights and bake for another 10 minutes, or until lightly browned. Transfer the tart pan to a wire rack to cool.
#### **ASSEMBLE THE TART**
In the bowl of a standing mixer fitted with the whisk attachment, beat the curd on high for 5 minutes, then spoon it into the tart and level the filling with an offset spatula. Refrigerate the tart for 1 hour to set completely.
#### **MAKE THE ORANGE WHIPPED TOPPING**
Pour the cream into a chilled metal bowl and beat it with a chilled whisk for about 1 minute. Sprinkle the sugar and orange cream soda on top and continue whisking vigorously until soft peaks form. (The whipped cream can be made in the bowl of a standing mixer fitted with the whisk attachment, but the hand-whisking method burns more calories. Also, if you prefer, you can substitute Simple Whipped Cream, for the flavored topping.)
To serve, gently push up on the tart bottom to release it from the pan. Top the tart with orange whipped cream.
The tart tastes best if eaten within 24 hours but can be kept, covered, in the refrigerator for up to 2 days.
**_Baked Note_**
If you want to add a little more sweetness to your tart (and to keep the crust from getting soggy) brush the bottom of the tart shell with 2 ounces of melted white chocolate. Let the chocolate set for about 5 minutes in the refrigerator, then pour the curd over it.
### **BLACKBERRY PIE**
AT THE AGE OF TWELVE, WHEN I CONSUMED ONE HEARTY PINT, ABOUT A WEEK'S WORTH, OF BLACKBERRIES IN TWENTY MINUTES, I WAS COMPLETELY UNAWARE THAT BERRIES, LIKE ALL FOOD, HAD A DISTINCT POINT OF ORIGIN. I just assumed that food came from a grocery store, that it was plentiful, and that it was universal. It was a revelation when I went to my first pick-your-own farm and I saw actual fruit on bushes and trees. It was also a revelation when I asked for the blackberry section of the farm and was told that I was a few states away from prime blackberry country. This pie is very straightforward and truly great with fresh, hand-picked blackberries (check out pickyourown.org for a comprehensive list of farms); however you will not taste the slightest hint of guilt if you get your blackberries via the grocery store route.
**YIELD: ONE 9-INCH PIE**
**_Ingredients_**
_2 ballsClassic Pie Dough_
_1½ tablespoons fresh lemon juice_
_1½ teaspoons freshly grated lemon zest_
_¾ cup granulated sugar_
_⅓ cup firmly packed dark brown sugar_
_⅓ cup all-purpose flour_
_¼ teaspoon salt_
_7 cups fresh blackberries_
_1½ tablespoon butter, cut into 8 tiny pieces_
_1 large egg, beaten_
_1½ tablespoons raw cane sugar_
Dust a work surface with a sprinkling of flour. Roll out one of the balls of dough into a 12-inch round. Transfer the dough to a 9-inch pie plate and gently work it into place, folding any overhang under and crimping the edge as you go. Wrap and refrigerate the crust for at least 30 minutes.
Preheat the oven to 350 degrees F.
In a small bowl or cup, stir together the lemon juice and zest. In a large bowl, stir together the sugars, flour, and salt. Add the blackberries and gently toss everything together with your hands. Sprinkle the lemon juice mixture over the top of the berries and toss again. Pour the blackberry mixture into the prepared pie shell and scatter the pieces of butter over the top.
Dust a work surface with a sprinkling of flour. Roll the remaining ball of chilled dough into a 12-inch round and place it over the pie filling. Trim the dough, leaving about a ½-inch overhang. Crimp the edges together, brush with the beaten egg, and sprinkle with the raw sugar. Cut six long steam vents into the top crust. Bake the pie until the filling bubbles and the crust is golden, about 1 hour. Cool the pie on a rack for at least 1 hour. Serve warm or at room temperature.
The pie can be stored in the refrigerator, tightly covered, for up to 2 days. Bring it to room temperature or reheat it in a warm oven before serving.
**_Baked Note_**
Blackberries are sometimes labeled "Marionberries" or "Marion blackberries " as this is the most prolific commercial type of blackberry (especially in the western half of the United States). It is, essentially, a crossbred berry notable for both its taste and its production.
### **WHISKEY PEAR TART**
I BASICALLY BAKE WITH MY LIQUOR CABINET OPEN AND IN FULL VIEW. I PREFER FLAVORED LIQUEURS (KAHLÚA, FRAMBOISE, AND SO FORTH) OVER EXTRACTS (VANILLA EXTRACT EXCLUDED), AS THEY TEND TO BE MORE SUBTLE. Extracts tend to attack and take over a dessert. Liqueurs are content to stay in the background. Whiskey, on the other hand, is an entirely different dessert companion; it sexes things up with smoky undertones. I cannot imagine our pear tart sans whiskey. The pears take on a robustness and soul unlike their dainty teetotaler counterparts, and it cuts the sweetness of the almond cream filling just perfectly. Incidentally, I like to serve this tart for dessert alongside a shot of the same whiskey I used for baking.
This is a very easy recipe, but for the sake of making it easier, I broke it up into two parts. Since you have to "poach" the pears overnight, I suggest making the tart dough at the same time. The following day is a cinch: Just roll and bake the dough, and fill and bake the tart.
**YIELD: ONE 14-BY-4-INCH RECTANGULAR TART
OR ONE 11-INCH ROUND TART**
**_Ingredients_**
FOR THE PEARS AND POACHING LIQUID
_1 (15-ounce) can pear halves in heavy syrup, about 6 halves_
_1½ tablespoons fresh lemon juice_
_2 tablespoons whiskey_
_3 tablespoons sugar_
_1 tablespoon pure vanilla extract_
FOR THE BASIC SWEET TART DOUGH
_¼ cup sugar_
_1½ cups all-purpose flour_
_¼ teaspoon salt_
_½ cup (1 stick) cold unsalted butter, cut into ½-inch cubes_
_1 large egg, beaten_
FOR THE ALMOND CREAM FILLING
_¼ cup (½ stick) unsalted butter, cool but not cold_
_4½ ounces almond paste_
_1 large egg_
_1½ tablespoons cornstarch_
_1 tablespoon whiskey_
FOR THE PEAR GLAZE
_Reserved syrup and reserved "poaching" liquid from pears_
_1 teaspoon whiskey_
_¾ teaspoon cornstarch_
#### **MAKE THE PEARS AND POACHING LIQUID**
Strain the pears and reserve the heavy syrup (for the glaze) in a small, covered bowl or cup in the refrigerator.
In a medium, nonreactive bowl, whisk together the lemon juice, whiskey, sugar, and vanilla. Toss the pears with the liquid, cover the bowl tightly with plastic wrap, and refrigerate overnight.
#### **MAKE THE SWEET TART DOUGH**
Put the sugar, flour, and salt in a food processor and pulse until combined. Add the butter and pulse until sandy (about 6 to 10 quick pulses). Add the egg and pulse just until the dough begins to form a mass. Form the dough into a disk, wrap it tightly in plastic, and refrigerate it overnight (or for at least 1 hour).
#### **BAKE THE CRUST**
Dust a work surface with a sprinkling of flour. Use a rolling pin to roll the dough about ¼ inch thick into either a rectangle about 15 inches long or into a round about 12 inches in diameter. (Note: The dough will be sticky. Make sure to turn it with a bench knife or offset spatula as needed and keep the working surface floured. Some people find it easier to roll dough between two layers of plastic wrap. This can ease transfer and be a bit less messy.)
Ever so gently, guide the dough into the tart pan, without pulling it, and lightly press it into place. Roll the rolling pin over the pan to trim off excess dough. Place the tart pan in the freezer for 30 minutes.
Preheat the oven to 375 degrees F.
Line the tart shell with aluminum foil and fill it three-quarters full with pie weights or dried beans. Bake for 15 minutes, then remove the foil and weights and bake for another 10 minutes, or until lightly browned. Transfer the tart pan to a wire rack to cool. Leave the oven on.
#### **MAKE THE ALMOND CREAM FILLING**
In the bowl of a standing mixer fitted with the paddle attachment, cream the butter and almond paste on medium speed until the mixture is light, fluffy, and smooth, 3 to 4 minutes. Add the egg and beat until combined. Sprinkle the cornstarch over the filling and turn the mixer to low. Drizzle in the whiskey and beat until it is combined. Spread the almond cream filling evenly over the cooled tart shell.
Drain the pear halves, reserving the soaking liquid, and arrange them decoratively on top of the almond cream. Bake for 35 to 40 minutes, or until the almond cream puffs up and sets and the crust turns golden brown. Let the tart cool on a wire rack while you make the glaze.
#### **MAKE THE PEAR GLAZE**
Place the syrup and soaking liquid in a medium pan over medium heat and gently boil until the liquid is reduced to about ¾ cup. Remove it from the heat and whisk quickly and continuously for 1 minute to speed cooling. Add the whiskey and cornstarch and whisk to combine. Set the pan over medium-high heat, bring the glaze to a boil, and cook it for 1 minute. Use a pastry brush to apply the glaze gently to the tart.
Remove the tart from the pan and serve it as soon as possible. The tart will keep at room temperature, covered, for up to 3 days, but the crust will turn slightly soggy after the first day.
**_Baked Note_**
At first, I was hesitant to use canned fruit for this tart, but if you find the right brand (with all natural ingredients), you will get a consistent and wonderful tart every time. If you happen to come across excellent fresh pears at a farmers' market, poach away, using the traditional method on the opposite page. This is a two day project so make sure you read through all the steps before getting started.
* * *
Suffice it to say, there are many, many ways to poach your own pears. You can use a variety of liquids (water, wine, half water/half wine, diluted fruit juice), and you can tweak the liquid according to your mood (add spices, other fruits, vanilla, and sugars). It is a recipe with endless possibilities, and I suggest you modify the below ingredients at will. This quick poaching method is only a roadmap, so feel free to throw your personality in the poaching pot:
_4 firm and ripe pears
1 cup sugar
1 bottle of cheap and cheerful sweet dry wine
Zest and juice of 1 orange_
Peel the pears, core them, and cut them in half. Set aside.
In a large saucepan set over low heat, stir together the sugar and the wine until dissolved.
Stir in the orange zest and juice, increase the heat to medium, and wait for the liquid to simmer.
Once the liquid reaches a low boil, add the pears and simmer for 15–30 minutes. During the poaching process it is important to make sure the liquid covers the pears the entire time.
The pears are done when a sharp knife inserted into the bulbous end of the pear slides in and out easily. Check your pears every few minutes after the 15-minute mark, as cooking time is determined by the size and ripeness of your pears.
Remove the pears and let cool if you are using in a recipe, or serve warm with any accompaniment (ice cream, whipped cream, etc.). The poaching liquid can be reused. Store the poaching liquid in an airtight container in the refrigerator for up to 1 week.
* * *
### **PECAN TASSIES**
TRACKING DOWN THE ORIGIN OF A RECIPE IS A TASK BEST SUITED TO THE MOST DOGGED REPORTER. A recipe might begin in one state, be adopted by another, and then manipulated by a third. Following the winding trail might be an insurmountable task, and that is why I will not argue with my Aunt Judy, from upstate New York, about her pecan tassies. She insists, in a very proud way, that it is a local regional recipe. I happen to believe it is from the South. We will never know who is right, because pecan tassie information is scarce. There are virtually no stories or lore (at least none that I can find) surrounding these mini desserts, yet I always assumed the tassie emerged from some heritage holiday collection. A pecan tassie is essentially a miniature, gooey, pecan pie with a hassle-free crust. They are transportable, individual, and picturesque. I doubt they will ever completely satisfy a pecan pie craving (the crust-to-filling ratio of the full-size pie is nearly perfect), but they will certainly tide you over.
**YIELD: ABOUT 40 TASSIES**
**_Ingredients_**
FOR THE TASSIE SHELLS
_1 cup (2 sticks) unsalted butter, softened, cut into ½-inch cubes_
_6 ounces cream cheese, softened, cut into 1-inch cubes_
_1 tablespoon sugar_
_2 cups all-purpose flour_
FOR THE TASSIE FILLING
_2 large eggs_
_1½ cups firmly packed light brown sugar_
_2 tablespoons pure vanilla extract_
_⅛ teaspoon salt_
_1 cup toasted pecans, coarsely chopped_
#### **MAKE THE TASSIE SHELLS**
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and cream cheese together on medium speed until the mixture is lump free. Add the sugar and beat again for 15 seconds. Scrape down the bottom and sides of the bowl. Add the flour in four parts, at low speed, until the mixture is just combined and a dough forms. Pinch off a walnut-size piece of dough, roll it into a ball, and set it aside. Continue pinching and rolling until all the dough is gone. You will end up with approximately 40 pieces. Place each ball into an individual mini muffin pan, then use your fingers to press the dough into the bottom and up the sides of each pan.
Once all the tassie shells have been formed, place the muffin pan in the refrigerator while you make the filling.
Preheat the oven to 350 degrees F.
#### **MAKE THE TASSIE FILLING**
In a medium bowl, whisk the eggs just until they break apart. While whisking, gradually add the brown sugar until completely combined. Add the vanilla and salt and whisk again. Add ½ cup of the chopped pecans and stir until they are completely mixed in.
#### **ASSEMBLE THE TASSIES**
Sprinkle the remaining ½ cup pecans into the tassie shells. Spoon in enough filling into the shells until they are about three-quarters full.
Bake for about 15 minutes. Reduce the oven temperature to 250 degrees F, and bake for another 10 minutes, or until the filling is set.
Allow the tassies to cool for at least 30 minutes. Serve them warm or at room temperature. Pecan tassies taste best the day they are made; however, you can wrap them in plastic and keep them at room temperature for up to 2 days.
**_Baked Note_**
It is worth noting that many heritagetype recipes were transcribed in the days before the standing mixer. This seems obvious, but it never really occurred to me until I started to sort through people's archives. This pecan tassie recipe is adapted directly from my aunt, but I updated it to make use of a standing mixer (and a few other modern items). If you want to do it the authentic way, simply blend the crust ingredients with a fork.
* * *
There are two ways to toast nuts: the oven method and the skillet method. Both are easy.
**Oven Method** Preheat your oven to 300 degrees F. Spread the nuts in an even layer on a rimmed baking sheet and toast until fragrant. (Be sure to toss and flip your nuts half-way through the baking process).
**Skillet Method** (I actually prefer this method.) Place the nuts in a single layer in a large skillet set over medium heat. Stir and flip the nuts frequently until fragrant and almost golden. Usually, the nuts will toast more quickly (about half the time listed above) using this method; however, keep a keen nose at the ready, as toasting times will vary depending on heat source.
* * *
##
**CHOCOLATE MINT THUMBPRINTS**
**COWBOY COOKIES**
**SWEET & SALTY BROWNIE**
**GRASSHOPPER BARS**
**HEARTLAND TURTLE BARS**
**SALT-N-PEPPER SANDWICH COOKIES**
**CLASSIC SHORTBREAD WITH FLEUR DE SEL**
**CHOCOLATE GINGER MOLASSES COOKIE**
**MALTED MILK SANDWICH COOKIES**
**CHOCOLATE WHOOPIE PIES**
**BLACK AND WHITE COOKIES**
**SPECULAAS**
**RED VELVET WHOOPIE PIES**
**PEANUT BUTTER AND JELLY BARS**
**ROSEMARY APRICOT SQUARES**
**AUNT SABRA KING'S PUDDING BARS**
**JOE FROGGERS, OR GINGER RUM MOLASSES COOKIES**
I rarely make use of my two cookie jars. One is a basic, perfectly adequate, glass jar with the latest in airtight technology. The other is heavy ceramic, shaped and painted to resemble an art deco gas pump (file under the strange things we accumulate). Both are gathering dust. For a few reasons, I never make more cookies than are immediately required. First, cookies really do taste best the day they are made or, even better, straight from the oven. I recommend that bakers refrigerate almost any dough and bake it as needed. Second, I am lazy. Most recipes yield several dozen cookies, yet I only like to bake one or two dozen at a time. I do not have the patience to endure more than one bake cycle. Lastly, I love cookies too much. I possess little self-control if a chocolate chip, a black and white, or the much-beloved cowboy cookie is lurking nearby. If I bake twelve cookies, chances are I will eat twelve cookies, unless there is someone around to share them with.
Thankfully, my willpower isn't tested by bar cookies or brownies. For some reason, my brain views a bar cookie (at least when left in its original baking receptacle) as part of a larger whole, something that would have to be consumed en masse if eaten at all. Better to just slice a bit off, remove it from the baking dish, move far away, and forget about the rest.
I really enjoy making bars in the same way that I like making casserole-type dishes for potluck dinner parties. They are easy to tote about, a little bit homey, usually not too time-consuming, and delicious (if you use the right recipe). Regretfully, bar cookies (and, to some degree, cookies of any type), make only brief appearances in the realm of storied five-star desserts. Odd how ice cream is currently celebrated and critically acclaimed on several fine-dining menus, yet brownies are still considered the handiwork of a "fern bar" or "quick casual" restaurant. Life is unfair.
The cookie and bar recipes throughout this chapter are informed by some of the most interesting stories and lore in the entire book. The Joe Frogger, a wonderful ginger molasses cookie, has a backstory that borders on Hollywoodesque, and the Cowboy Cookie, while completely addictive, probably has nothing to do with cowboys. I enjoyed all the research that went into this chapter, and I had an even better time testing (i.e., eating) my way through the recipes. I hope you have as much fun as I did...and don't miss the Grasshopper Bars.
### **CHOCOLATE MINT THUMBPRINTS**
THE PROBLEM WITH CATEGORIZING CERTAIN COOKIES AS "HOLIDAY" OR "CHRISTMAS" TREATS IS THAT IT TENDS TO PREVENT A GREAT RECIPE FROM BEING USED YEARROUND. That is a tragedy, especially if the cookie is as addictive as our Chocolate Mint Thumb-print. I originally made these cookies for a magazine's holiday story (hence the presence of mint), but I now find myself whipping them up for almost any occasion. The chocolate cookie base is perfectly textured—crunchy, crumbly—not too sweet, and the filling is a simple, creamy, white chocolate peppermint ganache (make sure you use a high-quality white chocolate).
**YIELD: ABOUT 40 COOKIES**
**_Ingredients_**
FOR THE CHOCOLATE MINT THUMBPRINTS
_2 ounces good-quality dark chocolate (60 to 72%)_
_2 ounces mint chocolate (or Andes mint chocolate candies)_
_1½ cups all-purpose flour_
_½ cup dark unsweetened cocoa powder (like Valrhona)_
_¾ teaspoon salt_
_1 cup (2 sticks) unsalted butter, cut into_
_1-inch cubes, at room temperature_
_⅓ cup granulated sugar_
_2 tablespoons firmly packed dark brown sugar_
_2 large egg yolks_
_1 teaspoon pure vanilla extract_
_1 cup coarse sugar for rolling_
FOR THE WHITE CHOCOLATE FILLING
_3 ounces good-quality white chocolate, coarsely chopped_
_3 tablespoons heavy cream_
_½ teaspoon pure peppermint extract_
#### **MAKE THE CHOCOLATE MINT THUMBPRINTS**
Melt the dark chocolate and mint chocolate together in a microwave or over a double boiler. Whisk until smooth, then set aside to cool.
In a medium bowl, whisk together the flour, cocoa powder, and salt. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter until creamy. Add the granulated and brown sugars and beat on medium-high speed until light and fluffy, about 3 minutes. Add the egg yolks and vanilla and beat again until combined. Scrape the chocolate into the mixer and beat just until incorporated. Scrape down the sides and bottom of the bowl and add the flour mixture all at once. Beat on low speed, scraping the side of the bowl occasionally, until the dough is smooth. Transfer it to a sheet of plastic wrap and pat it into a disk; wrap and refrigerate it until it is chilled and firm, at least 30 minutes.
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
Pour the coarse sugar into a shallow bowl.
With clean hands, form tablespoon-size dough balls, taking care that they have no lumps or cracks. Roll each ball in the coarse sugar and place it on a prepared baking sheet. Use your thumb or a small dowel to make an indentation in the center of the cookie. Bake for 10 minutes, remove the sheet from the oven, and use your thumb or the dowel to make the indentation more visible. Return the sheets to the oven and bake for another 4 to 5 minutes. (These are the type of cookies that can overbake very quickly—pull them out at the first signs of cracking.) Set the baking sheets on a wire rack to cool for 5 minutes. Use a spatula to transfer the cookies to the rack to cool completely before filling them.
#### **MAKE THE WHITE CHOCOLATE FILLING**
Place the white chocolate in a glass measuring cup with a pour spout. Put the cream in a microwave-safe bowl or cup and microwave it on high power until it boils, about 30 seconds. Pour the hot cream over the white chocolate and let stand for 30 seconds, then whisk until smooth. Stir in the peppermint extract. Fill the thumbprint cookies with the white chocolate ganache and refrigerate them until set, about 30 minutes.
These cookies taste great at room temperature or directly from the refrigerator. They can be stored in an airtight container in the refrigerator for up to 3 days.
**_Baked Note_**
If you want a perfect thumbprint in every cookie (and in order to avoid burning yourself), feel free to use the handle of a wooden spoon to make the initial indentation and subsequent one. It also makes the whole process slightly faster.
### **COWBOY COOKIES**
THE NAME, COWBOY COOKIE, IS A BIT FANTASTICAL. I mean, I am fairly certain that there is only just the faintest hint of connection (if any) between actual cowboys and these cookies. My basic, modest research cannot find a reference anywhere stating that they were invented by cowboys, enjoyed by cowboys, or made in honor of cowboys. It seems the name is simply a sly reference to the rugged manliness implied by the cookie's bigness and embrace-it-all ingredients.
Whatever their origins, they are delicious. The typical components of the cowboy cookie are oatmeal, chocolate chips, walnuts or pecans, crunchy exterior, and chewy interior. Ours is slightly different. I opted to do away with the nuts and added salty pretzels; we also tossed in some instant espresso powder to add a grown-up (maybe cowboyish) flavor.
**YIELD: MAKES ABOUT 36 COOKIES**
**_Ingredients_**
_1¾ cups all-purpose flour_
_1 teaspoon baking soda_
_1 teaspoon baking powder_
_½ teaspoon salt_
_2 cups rolled oats_
_14 tablespoons (1¾ sticks) unsalted butter, cool but not cold, cut into 1-inch cubes_
_¾ cup granulated sugar_
_1 cup firmly packed dark brown sugar_
_1 large egg_
_1 large egg yolk_
_1 teaspoon vanilla extract_
_1 teaspoon instant espresso powder_
_2 cups semisweet chocolate chunks (about 12 ounces)_
_¾ cup thin salty pretzels (about_
_1½ ounces), broken into tiny pieces but not crushed into dust_
In a medium bowl, whisk together the flour, baking soda, baking powder, and salt. Add the oats and stir to combine.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and sugars together until smooth and creamy. Add the egg and egg yolk, beating until the mixture looks light and fluffy. Scrape down the sides and bottom of the bowl, add the vanilla, and beat for 5 seconds. Dissolve the espresso powder in ¼ cup hot water and add it to the bowl, mixing until combined.
Add half of the dry ingredients and mix for 15 seconds. Add the remaining dry ingredients and beat until just incorporated. Scrape down the sides and bottom of the bowl and fold in the chocolate chunks and ¼ cup of the pretzel pieces.
Cover the bowl tightly and refrigerate the dough for at least 4 hours.
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
Use a small ice cream scoop with a release mechanism to scoop out dough in 2 tablespoon–size balls (or use a tablespoon measure) and place the dough balls onto the prepared baking sheet about 1 inch apart. Sprinkle the remaining ¼ cup pretzel pieces over the dough balls. Use the palm of your hand to press the dough down lightly; don't smash the cookie—you just want to slightly flatten the ball and push the pretzel pieces into the dough.
Bake for 11 to 13 minutes, rotating the pans halfway through the baking time, until the edges of the cookies are golden brown or just start to darken.
Set the pan on a wire rack for 10 minutes to cool. Use a spatula to transfer the cookies to the rack to cool completely. They can be stored in an airtight container for up to 3 days (though I doubt they will last that long).
**_Baked Note_**
This recipe (and a few others in this book) calls for cool, but not cold butter. To be specific, I am suggesting you remove the butter from the refrigerator, cut it up into cubes, and plan to use it within 15 to 20 minutes. Using colder (i.e. not room temperature) butter will prevent your cookies from spreading. Conversely, if you prefer thinner, slightly crispier cookies, let your butter come to room temperature.
### **SWEET & SALTY BROWNIE**
THE GENESIS BEHIND THE FORMATION OF OUR BAKERY, BAKED, WAS TO ELEVATE THE CLASSIC AMERICAN BROWNIE FROM UBIQUITOUS MEDIOCRITY TO BONA FIDE LUXURY. It is brownie as fetish. The Sweet & Salty Brownie is the Baked brownie in extreme, a decadence on par with other, more well-known (and perhaps more respected) desserts. The brownie itself is a riff on our famous deep, dark brownie (sans instant espresso powder) filled with a smoky, dark caramel and topped with just a hint of sea salt and sugar. The caramel taste is more of a hint and less of an explosion, befitting this elegant dessert; the result is a fudgey-chocolatey-carameley brownie. I should also reveal that the Sweet & Salty Brownie is our most requested recipe, owing to the brownie's featured moment on the Food Network, where it was lauded with praise as one of the best salty foods in the United States. That is a lot of hype for one brownie, but in this case, I think it is worthy.
**YIELD: 12 LARGE BROWNIES OR 24 SMALL BROWNIES**
**_Ingredients_**
FOR THE FILLING
_1 cup sugar_
_2 tablespoons light corn syrup_
_½ cup heavy cream_
_1 teaspoon fleur de sel_
_¼ cup sour cream_
FOR THE BROWNIE
_1¼ cups all-purpose flour_
_1 teaspoon salt_
_2 tablespoons dark unsweetened cocoa powder (like Valrhona)_
_11 ounces quality dark chocolate (60 to 72%), coarsely chopped_
_1 cup (2 sticks) unsalted butter, cut into_
_1-inch cubes_
_1½ cups sugar_
_½ cup firmly packed light brown sugar_
_5 large eggs, at room temperature_
_2 teaspoons vanilla extract_
FOR THE ASSEMBLY
_1½ teaspoons fleur de sel_
_1 teaspoon coarse sugar_
#### **MAKE THE CARAMEL**
In a medium saucepan, combine the sugar and corn syrup with ¼ cup water, stirring them together carefully so you don't splash the sides of the pan. Cook over high heat until an instant-read thermometer reads 350 degrees F, or until the mixture is dark amber in color (keep a close eye on the caramel at all times, as it goes from golden brown to black and burnt very quickly), 6 to 8 minutes. Remove for the heat, and slowly add the cream (careful, it will bubble up) and then the fleur de sel. Whisk in the sour cream. Set aside to cool.
#### **MAKE THE BROWNIE**
Preheat oven to 350 degrees F.
Butter the sides and bottom of a glass or light-colored metal 9-by-13-inch pan. Line the bottom with a sheet of parchment paper, and butter the parchment.
In a medium bowl, whisk together the flour, salt, and cocoa powder.
Place the chocolate and butter in the bowl of the double boiler set over a pan of simmering water, and stir occasionally until the chocolate and butter are completely melted and combined. Turn off the heat, but keep the bowl over the water of the double boiler, and add both sugars. Whisk until completely combined and remove the bowl from the pan. The mixture should be at room temperature.
Add three eggs to the chocolate mixture and whisk until just combined. Add the remaining eggs and whisk until just combined. Add the vanilla and stir until combined. Do not overbeat the batter at this stage, or your brownies will be cakey.
Sprinkle the flour mixture over the chocolate. Using a spatula, fold the dry ingredients into the wet ingredients until there is just a trace amount of the flour mixture visible.
#### **ASSEMBLE THE SWEET& SALTY BROWNIE**
Pour half of the brownie mixture into the pan and smooth the top with a spatula. Drizzle about ¾ cup of the caramel sauce over the brownie layer in a zigzag pattern, taking care to make sure the caramel does not come in contact with the edges of the pan or it will burn. Use your offset spatula to spread the caramel evenly across the brownie layer. In heaping spoonfuls, scoop the rest of the brownie batter over the caramel layer. Smooth the brownie batter gently to cover the caramel layer.
Bake the brownies for 30 minutes, rotating the pan halfway through the baking time, and check to make sure the brownies are completely done by sticking a toothpick into the center of the pan. The brownies are done when the toothpick comes out with a few moist crumbs.
Remove the brownies from the oven and sprinkle with the fleur de sel and coarse sugar.
Cool the brownies completely before cutting and serving.
The brownies can be stored, tightly wrapped at room temperature, for up to 4 days.
**_Baked Note_**
That old adage "less is more" holds true here. You might be tempted to add more caramel than recommended, but temper your indulgence. If you build too much of a caramel layer, it more than likely will seep out and burn during baking. In fact this recipe will make more than enough caramel for one batch of brownies. If you are a caramel addict, use the extra to drizzle on the brownie post-baking/pre-serving.
* * *
Brownies, unlike many bar cookies, freeze particularly well. In fact, some brownie fanatics claim freezing improves the texture. I am still undecided. Regardless, there is an art to freezing your brownies to avoid the sticky, sweaty surface condensation.
**1.** Allow the brownies to cool to room temperature. Wrap your room temperature brownies in two layers of plastic wrap. It is best to wrap the brownies directly as opposed to wrapping the pan of brownies. Place your brownies in the freezer.
**2.** When you are ready to defrost your brownies, remove them from the freezer and place them in the refrigerator for 8 hours or overnight. Then remove the brownies from the refrigerator and let them sit at room temperature for at least 1 hour.
**3.** Unwrap and eat your brownies.
Generally speaking, brownies will keep in the freezer for up to 1 month, but I know people who have kept them far beyond that threshold and have lived to tell about it.
* * *
### **GRASSHOPPER BARS**
THE BAKED GRASSHOPPER BAR WAS SUPPOSED TO BE A SHORT-LIVED EXPERIMENT, A QUIRKY RIFF ON THE MINTY CLASSIC 1950S COCKTAIL WITH WHICH WE ARE SLIGHTLY OBSESSED. The impulse to make it came from a cookbook my mom had stashed away, an ancient tome with an emphasis on Midwestern cuisine. Unfortunately, the only thing it seemed to have going for it was page after page of hideous gelatin-infused recipes, each more gruesome than the one before. Still, something about the grasshopper pie was intriguing. Even though the original tastes like medicinal Jell-O, we persevered through countless iterations. And the bar, which was supposed to be just a jokey tongue-in-cheek confection, has endured. The layers work in perfect harmony: thick brownie base, light mint filling, and a dark ganache top. Interestingly enough, this sweet treat works equally well as a classy plated dessert and as an after-school snack.
**YIELD: 12 LARGE BROWNIES OR 24 SMALL BROWNIES**
**_Ingredients_**
FOR THE BROWNIE BASE
_¾ cups flour_
_½ teaspoon salt_
_1 tablespoon dark unsweetened cocoa powder (like Valrhona)_
_5 ounces good quality dark chocolate (60 to 72%), coarsely chopped_
_½ cup (1 stick) butter, cut into 1 inch cubes_
_¾ cup sugar_
_¼ cup firmly packed light brown sugar_
_3 large eggs, at room temperature_
_1 teaspoons vanilla extract_
FOR THE BUTTERCREAM
_¾ cup sugar_
_2 tablespoons flour_
_¾ cup milk_
_2 tablespoons heavy cream_
_1½ sticks (¾ cup) butter, softened but still cool, cut into small cubes_
_3 tablespoons crème de menthe_
_1 teaspoon peppermint extract_
FOR THE CHOCOLATE GLAZE
_6 ounces good-quality dark choccolate (60 to 72%) coarsely chopped_
_1 teaspoon light corn syrup_
_½ cup (1 stick) unsalted butter, softened, cut into cubes_
#### **MAKE THE BROWNIE BASE**
Preheat the oven to 325 degrees F.
Butter the sides and bottom of a glass or light-colored metal 9-by-13-inch pan. Line the bottom with a sheet of parchment paper, and butter the parchment. In a medium bowl, whisk together the flour, the salt, and cocoa powder.
Configure a large size double boiler. Place the chocolate and the butter in the bowl of the double boiler and stir occasionally until the chocolate and butter are completely melted and combined. Turn off the heat, but keep the bowl over the water of the double boiler and add both sugars. Whisk the sugars until completely combined. Remove the bowl from the pan. The mixture should be at room temperature.
Add three eggs to the chocolate/butter mixture and whisk until just combined. Add the vanilla and stir until combined. Do not overbeat the batter at this stage or your brownies will be cakey.
Sprinkle the flour/cocoa/salt mix over the chocolate. Using a spatula (do not use a whisk) fold the dry ingredients into the wet until there is just a trace amount of the flour/cocoa mix visible.
Pour the batter into the prepared pan, smooth the top with an offset spatula, and bake for approximately 12 to 15 minutes, rotating halfway through the baking time. The brownies should be just a tad underdone (not too gooey, but ideally, just 1 minute from being cooked through completely). A toothpick inserted into the brownies at an angle should contain a few loose crumbs. Remove the brownies from the oven and let cool completely while you make the creme de menthe filling.
#### **MAKE THE BUTTERCREAM**
In a medium heavy-bottomed saucepan, whisk the sugar and flour together. Add the milk and cream and cook over medium heat, whisking occasionally until mixture comes to a boil and has thickened, 5 to 7 minutes.
Transfer the mixture to the bowl of an electric mixer fitted with the paddle attachment. Beat on high speed until cool. Reduce the speed to low and add the butter and mix until thoroughly incorporated. Increase the speed to medium-high and beat until filling is light and fluffy.
Add the crème de menthe and peppermint extract and mix until combined. If the filling is too soft, chill slightly in the refrigerator and then mix again until it is the proper consistency.
If the filling is too firm, place the bowl over a pot of simmering water and re-mix to proper consistency. Spread the filling evenly across the top of the brownie layer and place the pan in the refrigerator, for a minimum of 45 minutes, while you make the chocolate glaze.
#### **MAKE THE CHOCOLATE GLAZE**
In a large non-reactive metal bowl, combine the chocolate, corn syrup, and butter. Set the bowl over a saucepan of simmering water and cook, stirring with a rubber spatula, until the mixture is completely smooth. Remove the bowl from the pan and stir vigorously for 1 minute to release excess heat.
Pour the mixture over the chilled crème de menthe layer and use an offset spatula to spread it into an even layer. Place the pan back in the refrigerator for 1 hour, or until the glaze hardens.
Remove the pan from the refrigerator, wait about 15 minutes for the glaze to soften slightly, and cut the bars with a warm knife. Cut into squares and serve immediately.
The bars can be stored in the refrigerated, tightly covered, for up to 4 days.
**_Baked Note_**
I rarely advocate underbaking in the quest to develop texture (the raw cookie dough thing was never my cup of tea); however, many of my recipe testers admitted to underbaking the brownie base portion of this recipe by just a few minutes. I tried it, and I admit a fudgier texture is a nice option. The recipe is written for the full baking time, but the final decision is yours.
### **HEARTLAND TURTLE BARS**
THIS TURTLE BAR RECIPE WAS PASSED ON TO ME BY A NEWFOUND ACQUAINTANCE I ENCOUNTERED WHILE TRAVELING ALONG THE BORDER OF SOUTH DAKOTA AND IOWA. She assured me that the bar came from the heartland—she dug up the recipe some time ago from one of those church cookbooks, which she found in Minnesota. These bars are wonderfully addictive snacks. I suppose the oatmeal in the crust is my favorite part, and it's what sets this bar apart from other "turtle" offerings.
**YIELD: 24 BARS**
**_Ingredients_**
FOR THE BAR TOPPING AND BASE
_1½ cups all-purpose flour_
_¼ teaspoon salt_
_¾ teaspoon baking soda_
_1 cup firmly packed dark brown sugar_
_1¾ cups rolled oats_
_1 cup (2 sticks) unsalted butter, melted_
_1 cup toasted pecans, chopped into large pieces_
_1½ cups chocolate chips_
FOR THE CARAMEL FILLING
_½ cup firmly packed light brown sugar_
_10 tablespoons (1¼ sticks) unsalted butter, cut into cubes_
_2 tablespoons heavy cream_
#### **MAKE THE BAR TOPPING AND BASE**
Preheat the oven to 350 degrees F. Butter the sides and bottom of a 9-by-13-inch glass or light-colored metal baking pan. Line the pan with parchment paper so that the paper overhangs the pan on two sides. Butter the parchment. In a medium bowl, whisk together the flour, salt, and baking soda. Use your hands to rub in the brown sugar. Add the oats and stir until the ingredients are evenly combined. Make a well in the center of the dry ingredients, then pour in the melted butter and stir until the entire mixture is wet and combined.
Spread two-thirds of the mixture across the bottom of the prepared pan and bake for about 10 minutes. Remove the pan from the oven to cool (but leave the oven on). Sprinkle the pecans and chocolate chips across the cooled crust.
#### **MAKE THE CARAMEL FILLING**
In a medium saucepan over medium-high heat, melt the sugar and butter together. Bring the mixture to a boil and boil for 1 minute, stirring constantly (the caramel will begin to darken quickly at this point). Remove the pan from the heat, stir in the cream, and pour the caramel directly over the chocolate pecan layer. Use an offset spatula to evenly distribute the caramel. Sprinkle the remaining oatmeal mixture onto the caramel and bake for 10 to 12 minutes, or until the top is golden brown.
Let the bars cool in the pan for about 15 minutes, then place the pan in the refrigerator and chill for 1 hour to firm up. Cut and serve. The bars can be stored, tightly wrapped, in the refrigerator or at room temperature for up to 3 days.
**_Baked Note_**
This is quite a buttery dessert. I, personally, have never found anything too buttery (I would be happy to use an entire stick on one bagel), but if you want to cut back on the butter in this recipe, leave out a tablespoon or two in the oatmeal crust. Just remember, this is a heritage recipe — avoid substituting margarine at all costs.
### **SALT-N-PEPPER SANDWICH COOKIES**
I SUPPOSE THE BAKED OBSESSION WITH COOKIE SANDWICHES IS BUILT ON THE MEMORY AND ENDURANCE OF THE OREO. The Oreo, a cookie only the American palate could love, was probably created in our hometown of New York City (at the Nabisco factory in Chelsea), though it is hardly a regional treat anymore. There is no sense in trying to upgrade, tweak, or improve upon this American icon, and it should be left undisturbed. So, consider this recipe a mere tribute. The chocolate sandwich cookies are crisp and toothsome, with a hint of salt and pepper that works beautifully with the cocoa powder, and we filled the whole thing with a pretty close approximation of the classic Oreo filling, sans trans fats and other chemical no-nos. Unlike the original, these do not cry out to be dunked in milk, and they actually make a rather chic after-dinner treat.
**YIELD: APPROXIMATELY 36 COOKIE SANDWICHES**
**_Ingredients_**
FOR THE COOKIES
_3½ cups all-purpose flour_
_¾ teaspoon salt_
_¼ teaspoon fleur de sel, plus more for decorating_
_2 teaspoons white pepper_
_¼ cup dark unsweetened cocoa powder (like Valrhona)_
_1½ cups (3 sticks) unsalted butter, cut into_
_1-inch cubes, cool, but not cold_
_1¼ cups granulated sugar_
_1¾ cups confectioners' sugar_
_3 large egg yolks_
_1 tablespoon pure vanilla extract_
_3 ounces good-quality dark chocolate (60 to 72%), melted_
FOR THE VANILLA FILLING
_5 ounces vegetable shortening, at room temperature_
_4 tablespoons (½ stick) unsalted butter, cut into small chunks, at room temperature_
_3¼ cups confectioners' sugar, sifted_
_½ teaspoon salt_
_1 tablespoon pure vanilla extract_
_1 teaspoon light rum_
#### **MAKE THE COOKIES**
In a large bowl, sift together the flour, salt, fleur de sel, white pepper, and cocoa powder. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and sugars together until light and fluffy, about 3 minutes. Scrape down the bowl, and add the egg yolks, one at a time, beating until each is incorporated. Add the vanilla and melted chocolate and beat until uniform in color. Scrape down the sides and bottom of the bowl and beat again for 10 seconds.
Add half of the dry ingredients and beat for 15 seconds. Again, scrape down the bowl, add the remaining dry ingredients and beat until just incorporated.
Loosely shape the dough into two balls, wrap them tightly in plastic wrap, and refrigerate them for at least 3 hours.
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
Unwrap one ball of dough and divide it into two equal portions. Place the first portion on a lightly flour-dusted work surface and return the other to the refrigerator.
Use your hands to knead the dough until pliable and form into a small disc. Roll the dough into a ¼-inch-thick round. It will be slightly sticky, so you may have to flip and lightly flour it a few times while you work. Use a 2-inch round cookie cutter to create your sandwich tops and bottoms, and transfer them to the prepared baking sheets, leaving about 1 inch of space around each cookie. Continue the process with the remaining dough. Extra dough scraps can be refrigerated and rerolled, if desired.
Sprinkle the tops of the cookies with a little fleur de sel, then bake them for 10 to 12 minutes, rotating the sheets halfway through the baking time. The tops of the cookies should look a bit dry and possibly cracked. Place the baking sheets on wire racks to cool for 5 minutes. Use a spatula to transfer the cookies to the racks to cool completely before filling them.
#### **MAKE THE VANILLA FILLING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the shortening and butter until lump free and smooth. Add the sugar in three parts, mixing each part until just combined. Add the salt, vanilla, and rum and beat again for 10 seconds. The filling should be thick but spreadable (like the inside of an Oreo). If it is too thick, add a drop or two of water as needed. Keep adding water to reach the desired consistency, but do not add too much water or the filling will be too thin.
Alternatively if the mixture is too thin, add a few tablespoons of confectioners' sugar.
#### **ASSEMBLE THE SALT-N-PEPPER SANDWICH COOKIES**
Use a pastry bag or a small spoon to apply about 2 tablespoons of filling to the flat side of a cookie. Place another cookie, flat side down, on top. Press down slightly so that the filling spreads to the edges of the cookie. Repeat until all the sandwich cookies are made. Let them set up for about 15 minutes before serving. Store the cookies at room temperature in an airtight container for up to 3 days.
**_Baked Note_**
In this recipe, I suggest using a 2-inch round cookie cutter; however, it is only for guidance. Obviously, you can use any size, shape, and type of cookie cutter you fancy or have on hand. Also, note that while I prefer a crisp 1/4-inch cookie, I have plenty of friends that like them a bit thicker and chewier. If you are like them, simply roll out the dough to 1/2 inch thick and bake the cookies for a minute less.
### **CLASSIC SHORTBREAD WITH FLEUR DE SEL**
I HAVE SCOTTISH ROOTS. My paternal grandmother was originally from a small town just outside of Edinburgh, and she was steadfast in defending Scottish cuisine to her American children and grandchildren. It was not an easy battle. Her haggis, the classic Scottish dish made of many interior sheep parts, while probably the best haggis ever made, rarely won support or adulation at our family dinner table. But Grandma was persistent. She cooked through a veritable catalog of Scottish dishes before finally winning us over with mountains of classic, buttery, crispy shortbread. It was her mother's recipe, and she baked it with devotion. I grew to accept shortbread as part of the American cookie spectrum. The original recipe calls for an egg yolk to be spread over the rolled out dough and folded in gently by overlapping the dough into the egg. The idea is very interesting, but I still opted for the standing mixer route. Otherwise, the recipe is as written almost eighty years ago (the fleur de sel is my addition).
**YIELD: ABOUT 48 COOKIES**
**_Ingredients_**
_1 pound (4 sticks) unsalted butter, cut into_
_½-inch cubes, cool but not cold_
_1 cup plus 2 tablespoons superfine sugar, or 1 cup confectioners' sugar_
_½ teaspoon salt_
_3½ cups plus 2 tablespoons all-purpose flour_
_½ cup rice flour_
_2 egg yolks_
_1 tablespoon fleur de sel_
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter on high speed until smooth. Add the sugar and salt and beat again just until incorporated, about 2 minutes. In two additions, using a wooden spoon or the absolute lowest speed on your mixer, stir in 3½ cups of the all-purpose flour and all the rice flour, just until incorporated. Add the egg yolks, one at a time, and stir just until combined. If the dough looks too wet, fold in the remaining 2 tablespoons flour. Turn the dough out onto a lightly floured surface, and knead until it is uniform. Do not overwork it. Divide the dough into eight equal balls, then shape them into disks, wrap them in plastic, and refrigerate until firm, at least 1 hour.
Preheat the oven to 325 degrees F. Line two baking sheets with parchment paper.
Dust a work surface with a sprinkling of flour. Working with one piece of chilled dough at a time while leaving the others in the refrigerator, roll a disk into a slightly less than ½-inch round. Cut the round into wedges like a pizza.
Prick the top of the shortbread with the tines of a fork and sprinkle a bit of fleur de sel over the surface. Transfer the cookies to a prepared baking sheet. Bake the shortbread for 17 to 22 minutes, or until they just begin to brown.
Set the pans on a wire rack to cool for 10 minutes before transferring the shortbreads to the rack to cool completely.
Shortbread will keep in an airtight container, at room temperature, for 5 days.
**_Baked Note_**
Shortbread is fickle, or at least complicated. I baked this recipe about thirty times, and each time the cookies tasted different than before. Finally, I hit shortbread nirvana. My advice: pay careful attention to the starting temperature of the butter. It should be cool, but not directly out of the refrigerator. And when I say "cut the butter into cubes," I mean you should really cut it into cubes. No irregular chunks, rectangles, or other shortcuts.
* * *
I am fairly certain that my grandmother's eyes would pop out of her head at the mere suggestion of adding anything to her classic short-bread recipe. She might sniff at the idea of a chocolate dip as "unnecessarily rich" and introducing flavors into the mix as "missing the point." It is with this knowledge, that I offer a few riffs on the classic shortbread. Apologies to grandma in advance.
**Chocolate Dip** Line a baking sheet with parchment paper. Melt about 8 to 10 ounces of good-quality dark chocolate in a deep bowl. Let cool to room temperature. Dip each cookie, about halfway, into the chocolate and place on the baking sheet. Allow the cookies to set completely, about 45 minutes, before serving.
**Lemon** Make the recipe "as is" but add about 3 teaspoons of fresh lemon zest with the sugar. Orange zest works just as well.
**Vanilla Bean Paste** Make the recipe "as is" but add about 1 tablespoon of vanilla bean paste with the sugar and omit the fleur de sel. Other extracts (i.e., coffee, maple) can be substituted as well.
* * *
### **CHOCOLATE GINGER MOLASSES COOKIE**
THIS VERSION OF THE CLASSIC GINGER MOLASSES COOKIE IS GIVEN A BAKED MAKEOVER, REPLETE WITH CHOCOLATE, LARGER THAN SHOULD BE ALLOWED, AND TOOTHSOME. I assure you it deserves a spot on your list of rotating recipes.
**YIELD: ABOUT 36 COOKIES**
**_Ingredients_**
FOR THE MOLASSES COOKIES
_3¼ cups all-purpose flour, plus more for dusting_
_⅓ cup dark unsweetened cocoa powder (like Valrhona)_
_1 tablespoon ground ginger_
_2 teaspoons ground cinnamon_
_1 teaspoon ground cloves_
_1 tablespoon baking soda_
_½ teaspoon baking powder_
_1 teaspoon salt_
_5 tablespoons unsalted butter, softened_
_⅓ cup vegetable shortening_
_½ cup firmly packed dark brown sugar_
_1 large egg, at room temperature_
_½ cup molasses_
_2 ounces bittersweet chocolate, melted and cooled_
FOR THE ICING
_1¼ cups confectioners' sugar_
_1 large egg white_
_1 teaspoon fresh lemon juice_
#### **MAKE THE MOLASSES COOKIES**
In a medium bowl, whisk the flour with the cocoa powder, ginger, cinnamon, cloves, baking soda, baking powder, and salt. In the bowl of a standing electric mixer fitted with the paddle, beat the softened butter with the shortening at medium speed until the mixture is smooth, about 30 seconds. Add the sugar and beat until fluffy, about 2 minutes.
Add the egg to the cookie batter and beat until incorporated. Beat in the molasses and then the melted chocolate. Add the flour mixture in three batches, beating between additions. Divide the dough into three equal parts. Shape each part into a disk, then wrap each one in plastic wrap and refrigerate the cookie dough until chilled, about 2 hours.
Preheat the oven to 350 degrees F. Line two large baking sheets with parchment paper. On a lightly floured work surface, roll out one disk of dough ¼ inch thick. Using 4- to 5-inch cookie cutters, cut the dough into shapes and transfer to the prepared baking sheets. Reroll the dough scraps and cut out more cookies. Bake the cookies for about 7 minutes, rotating the pans halfway through the baking time until the tops are dry. Let the cookies cool on the pans for 5 minutes, then transfer to wire racks to cool completely. Repeat the process with the remaining dough.
#### **MAKE THE ICING**
In a medium bowl, combine the confectioners' sugar with the egg white and lemon juice and whisk until the icing is completely smooth. Scrape the icing into a piping bag fitted with a small tip. Decorate the cookies as desired. Let stand until the icing dries, about 30 minutes. Store tightly covered for 3 days.
### **MALTED MILK SANDWICH COOKIES**
THERE IS A TREASURE TROVE OF A BOOKSTORE IN NEW YORK CITY CALLED BONNIE SLOT-NICK COOKBOOKS. It is a local bookshop filtered through a West Village lens: small, quaint, very knowledgeable staff of one, odd hours, well stocked. Better yet, it trades largely in vintage cookbooks. Luckily, Bonnie lets customers browse (though I always buy more than I need), and it was during such a session that I made an important discovery. Malt powder, one of my favorite ingredients, was a star in vintage cookbooks and was widely featured in the Pillsbury Bake-Off contenders for years. Malt powder was everywhere, and then it nearly disappeared. I am glad that a malt resurgence has recently taken hold, and I am equally pleased with this spin on the classic Oreo. If you are feeling exceedingly motivated, I suggest you make the Salt-n-Pepper Cookies at the same time you make these. They both contain the same filling (so you can just double it).
**YIELD: ABOUT 30 SANDWICH COOKIES**
**_Ingredients_**
FOR THE COOKIES
_4 cups all-purpose flour_
_¾ cup malt powder_
_2 teaspoons baking powder_
_½ teaspoon baking soda_
_½ teaspoon salt_
_1 cup (2 sticks) unsalted butter, at room temperature_
_1 cup firmly packed dark brown sugar_
_1 cup granulated sugar_
_2 eggs_
_⅓ cup sour cream_
_2 teaspoons pure vanilla extract_
FOR THE VANILLA FILLING
_5 ounces vegetable shortening, at room temperature_
_4 tablespoons (½ stick) unsalted butter, cut into small chunks, at room temperature_
_3¼ cups confectioners' sugar, sifted_
_½ teaspoon salt_
_1 tablespoon pure vanilla extract_
_1 teaspoon light rum_
#### **MAKE THE COOKIES**
In a large bowl, whisk together the flour, malt, baking powder, baking soda, and salt. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and sugars together until light and fluffy. Scrape down the bowl and add the eggs, one at a time, beating until each is incorporated. Add the sour cream and vanilla and beat until just incorporated. Add half of the dry ingredients all at once and beat for 15 seconds. Again, scrape down the bowl, then add the remaining dry ingredients and beat until just incorporated. The mixture should come together almost in a ball.
Loosely shape the dough into two balls, wrap them tightly in plastic wrap, and refrigerate for at least 3 hours.
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
Divide each dough ball in half, to make four portions. Place one portion on a lightly flour-dusted work surface and return the other three to the refrigerator.
Roll out the dough so that it is ¼ inch thick. The dough will be sticky, so you may have to flip and lightly flour it a few times while you work. Use a 2-inch round cookie cutter to create the sandwich tops and bottoms, and transfer them to the prepared baking sheets, leaving about 1 inch of space around each cookie. Extra dough scraps can be refrigerated and rerolled once more, if desired.
Bake the cookies for 10 to 12 minutes, or until they are just slightly browned. Place the baking sheets on wire racks to cool for 5 minutes. Use a spatula to transfer the cookies to the racks to cool completely.
While the cookies cool, prepare the filling.
#### **MAKE THE VANILLA FILLING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the shortening and butter until lump free and smooth. Add the sugar in three parts, mixing each part until just combined. Add the salt, vanilla, and rum and beat again for 10 seconds. The filling should be thick but spreadable (like the inside of an Oreo). If it is too thick, add a drop or two of water as needed. Keep adding water to reach desired consistency, but do not add too much water or the filling will be too thin.
Alternatively if the mixture is too thin, add a few tablespoons of confectioners' sugar.
#### **ASSEMBLE THE MALTED MILK SANDWICH COOKIES**
Use a pastry bag or a small spoon to apply about 2 tablespoons of filling to the flat side of a cookie. Place another cookie, flat side down, on top. Press down slightly so that the filling spreads to the edges of the cookie. Repeat until all the sandwich cookies are made. Let them set up for about 15 minutes before serving. Store the cookies at room remperature in an airtight container for up to 3 days.
**_Baked Note_**
Keep a close eye on these cookies while they're baking.they go from chewy to crispy in a matter of minutes. The recipe instructions are written with a crispier cookie in mind.I prefer the crunch in contrast with the smoothness of the filling. But those who like their cookies on the chewier side can bake the cookies a few minutes shy of the suggested cooking time and pull them out before they start to crack on top.
### **CHOCOLATE WHOOPIE PIES**
LET'S FACE IT, THE CHOCOLATE WHOOPIE PIE IS OFFICIALLY A CLASSIC AMERICAN DESSERT. I am the first person to admit that the term "classic" is overused, abused, and on the verge of becoming meaningless; however, it would be a travesty not to give this dessert its due. The whoopie pie, essentially two cake-like cookies with a generous helping of a creamy filling sandwiched between them, has its roots in the Northeast. Pennsylvanians will tell you they originated in Pennsylvania Dutch country, and the people of Maine will argue that it comes from their state. This version features a very moist, deep-chocolate cookie and a light and fluffy vanilla filling (for good measure, we also provide an option for a peanut butter filling). Enjoy—and spread the whoopie religion. It can become as ubiquitous as the chocolate chip cookie and the all-American brownie with a little help from you.
**YIELD: 10 TO 12 LARGE OR 15 TO 17 SMALL PIES**
**_Ingredients_**
FOR THE CHOCOLATE COOKIES
_3½ cups all-purpose flour_
_¼ teaspoon salt_
_1¼ teaspoons baking powder_
_1¼ teaspoons baking soda_
_¾ cup dark unsweetened cocoa powder (like Valrhona)_
_2 teaspoons instant espresso powder_
_½ cup hot coffee_
_2 cups firmly packed light brown sugar_
_¾ cup canola oil_
_1 large egg_
_1 teaspoon pure vanilla extract_
_½ cup buttermilk, shaken_
FOR THE SWISS VANILLA FILLING
_5 large egg whites_
_1½ cups sugar_
_2 cups (4 sticks) unsalted butter, cut into ½-inch cubes, cool but not cold_
_¼ teaspoon salt_
_1 teaspoon pure vanilla extract_
#### **MAKE THE CHOCOLATE COOKIES**
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
In a large bowl, whisk together the flour, salt, baking powder, and baking soda, and set aside.
In another large bowl, whisk together the cocoa powder and espresso powder. Add the hot coffee and ½ cup hot water and whisk until both powders are completely dissolved.
In a medium bowl, stir the brown sugar and oil together. Add this to the cocoa mixture and whisk until combined. Add the egg, vanilla, and buttermilk and whisk until smooth.
Use a rubber spatula to gently fold the dry ingredients into the wet ingredients. Make sure to scrape down the sides and bottom of the bowl as you fold.
Use a small ice cream scoop with a release mechanism to drop heaping tablespoons of the dough onto the prepared baking sheets about 1 inch apart. Bake for 10 to 15 minutes, until the cookies are just starting to crack on top and a toothpick inserted into the center of a cookie comes out clean. Let the cookies cool completely on the pan while you make the Swiss vanilla filling.
#### **MAKE THE SWISS VANILLA FILLING**
In a medium bowl, whisk the egg whites and sugar together (see note below and remember to substitute the sugar for the peanut butter filling variation). Set the bowl over a pan of simmering water but do not let the water touch the bottom of the bowl. Heat the mixture until the sugar is completely dissolved and the color is a milky white, about 2 to 3 minutes.
Transfer the egg mixture to the bowl of an electric mixer fitted with the whisk attachment and beat on medium-high speed (start slowly at first) until smooth and fluffy, about 5 minutes. Remove the whisk attachment and replace with the paddle attachment. Add the cubed butter and beat on medium-high speed (start slowly at first) until smooth and fluffy, about 5 minutes. If the butter-cream looks like it is breaking, don't worry, it will eventually come together.
Add the salt and vanilla and beat for 5 seconds to combine.
Peanut Butter Filling Variation: Replace the sugar with 1 cup granulated sugar and ½ cup light brown sugar (packed tightly). Fold in ¼ cup unsalted smooth peanut butter after adding the vanilla exract.
#### **ASSEMBLE THE WHOOPIE PIES**
Turn half of the cooled cookies upside down (flat side facing up).
Use an ice cream scoop or a tablespoon to drop a large dollop of filling onto the flat side of the cookie. Place another cookie, flat side down, on top of the filling. Press down slightly so that the filling spreads to the edges of the cookie. Repeat until all the cookies are used. Put the whoopie pies in the refrigerator for about 30 minutes to firm up before serving.
The whoopie pies will keep for up to 3 days, on a parchment-lined baking sheet covered with plastic wrap, in the refrigerator. Bring the whoopies to room temperature before serving.
**_Baked Note_**
I am hesitant to place a typical yield amount on this recipe. I encountered so many large, sandwich-size whoopie pies during my travels that mini whoopies seem disingenuous. Still, the final recipe is written for smaller whoopies —if you aim For the larger (Pennsylvania Dutch —size) version, you will need to increase the cooking time by a few minutes.
### **BLACK AND WHITE COOKIES**
I SPENT THE FIRST TWO YEARS OF MY TIME IN NEW YORK CITY LOOKING FOR THE PERFECT BLACK AND WHITE COOKIE. It was a mission bordering on obsession. The black and white cookie is nearly ubiquitous throughout the five boroughs, and there are many versions to be found in the various delis, bakeries, and even grocery stores. The black and white is, in my book, the official cookie of New York City. Though this honor is probably not recognized by any city agency, it is assumed, and that status is duly protected by the citizens. In fact, the merits of the various versions of the cookie are so heavily debated among devotees that we were nervous about supporting any one recipe. In the end, we went with a fairly straightforward adaptation: cakey vanilla cookie, thin layer of vanilla (white) frosting and thin layer of chocolate (black) frosting.
**YIELD: 12 TO 18 COOKIES**
**_Ingredients_**
FOR THE COOKIE
_3 cups all-purpose flour_
_½ teaspoon baking soda_
_½ teaspoon baking powder_
_½ teaspoon salt_
_14 tablespoons (1¾ sticks) unsalted butter, cool but not cold_
_1¼ cups granulated sugar_
_2 large eggs_
_1 large egg yolk_
_¾ cup buttermilk_
_1 tablespoon pure vanilla extract_
_1 tablespoon freshly grated lemon zest_
FOR THE BLACK AND WHITE FROSTING
_3¾ cups confectioners' sugar_
_4 to 5 tablespoons whole milk_
_3 tablespoons heavy cream_
_2 teaspoons pure vanilla extract_
_½ cup plus 2 tablespoons dark unsweetened cocoa powder (like Valrhona)_
#### **MAKE THE COOKIES**
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
In a large bowl, sift together the flour, baking soda, baking powder, and salt.
In the bowl of a standing mixer fitted with the paddle attachment, cream the butter and sugar until fluffy, about 2 minutes. Scrape down the bowl and add the eggs and egg yolk one at a time, beating briefly after each addition. Scrape down the sides and bottom of the bowl again, and mix on low speed for 10 seconds. Add the flour mixture in three parts, alternating with the buttermilk in two parts (end with the flour mixture). Scrape down the bowl; add the vanilla and lemon zest, and mix on low speed for a few more seconds.
Using a ¼-cup ice cream scoop, drop the dough onto the prepared baking sheets, leaving about 3 inches around each cookie. (You'll be able to fit about 6 cookies onto each sheet.)
Bake the cookies for about 17 minutes, rotating the sheets between the oven racks halfway through the baking time, until edges are golden brown and the tops spring back when gently touched. Place the baking sheets on wire racks to cool for 5 minutes, then transfer the cookies to the racks to cool completely.
#### **MAKE THE BLACK AND WHITE FROSTING**
In a large bowl, whisk together the confectioners' sugar, milk, cream, and vanilla. If the mixture is too thick, add milk by the teaspoon until the desired consistency is reached. Pour half of the frosting (about ¾ cup) into a separate bowl and add the cocoa powder and 1 teaspoon water. Stir to incorporate the cocoa powder. The chocolate frosting should be about the same consistency as the "white" frosting. If it is too thick, keep adding water by the teaspoon until you get the right balance.
Use an offset spatula to spread white frosting on half of the flat side of each cookie. Let stand until almost set, about 20 minutes. Clean the spatula and use it to spread chocolate frosting over the unfrosted half of each cookie. (If the frosting thickens up while you are working, whisk it until it loosens.) Let the frosted cookies set completely, about 1 hour, before serving them.
Black and White Cookies are essentially little cakes, and they taste best eaten the day they are made. However, you can store them in an airtight container for up to 3 days at room temperature.
**_Baked Note_**
Icing a black and white cookie isn't difficult —a slightly messy decoration just screams "homemade," and that's not a terrible thing. However, if you are aiming for a perfectly iced cookie, fill a pastry bag fitted with a medium tip with the vanilla frosting, draw carefully defined outlines on half the cookie, then fill them in. Let the vanilla dry thoroughly and follow with the chocolate.
### **SPECULAAS**
THOSE DAMNED DELTA AIR LINES BISCOFF COOKIES HAUNT ME. I really hate flying, but I have to admit they are the one part of it that I enjoy. There are few cookies that I would pilfer in large quantities from a stewardess's mobile cart, but I would do almost anything to satiate my craving for those. They are actually speculaas, a type of Dutch or Belgian shortcrust biscuit. Traditionally served on St. Nicholas's Eve (December 6), speculaas are crunchy, golden, and chock-full of the traditional holiday spices: cinnamon, ginger, nutmeg, and cloves. They are also highly addictive and pair well with tea or coffee.
**YIELD: ABOUT TWENTY-FOUR 2-INCH ROUND COOKIES**
**_Ingredients_**
_1¾ cups all-purpose flour_
_1 cup firmly packed dark brown sugar_
_½ teaspoon baking soda_
_1½ tablespoons cinnamon_
_½ teaspoon freshly grated nutmeg_
_½ teaspoon ground cloves_
_½ teaspoon ground ginger_
_½ teaspoon salt_
_10 tablespoons (1¼ sticks) butter, cool but not cold, cut into ½-inch cubes_
_1 egg, beaten_
_1 teaspoon freshly grated orange zest_
_Coarse sugar_
In a large bowl, whisk together the flour, brown sugar, baking soda, cinnamon, nutmeg, cloves, cardamom, ginger, and salt
Drop the butter over the flour mixture, and use a large fork or a pastry cutter to cut the butter into the flour until the mixture resembles coarse sand.
Add the beaten egg and orange zest, and cut the mixture again until just combined.
Use your hands to knead the dough (do not overwork it) until it forms a ball. The dough should be slightly sticky and break apart easily, but shouldn't stick to your hands. Cover it in plastic wrap and chill for at least 1 hour.
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
Unwrap and divide the chilled dough into two equal portions. Place one on a lightly flour-dusted work surface and return the other to the refrigerator.
Roll the dough into a ¼-inch-thick round. You may have to flip and lightly flour the dough a few times while rolling it out to keep it from sticking. Use any cookie cutter (a rectangular or oblong shape is the most traditional) to cut out the cookies, and transfer them to the prepared baking sheets, leaving about 1 inch of space around them. Extra dough scraps can be refrigerated and rerolled once more, if desired.
Sprinkle the tops of the cookies with coarse sugar. Bake the cookies for 15 minutes, rotating the baking sheets halfway through the baking time. The tops of the cookie should be just a bit dry and dark brown. Remove from the oven and place the baking sheets on wire racks to cool for 5 minutes. Use a spatula to transfer the cookies to the racks to cool completely.
Speculaas can be stored at room temperature, tightly covered, for up to 5 days.
**_Baked Note_**
During this absurdly fun cookie research, I ran into many cousins or relatives of speculaas. There are fluffier, anisefl avored versions from Germany and mass-produced, less-spiced versions known collectively as Dutch windmill cookies. The recipe printed here is as close to the Biscoff as it was possible to get.
* * *
Thankfully, the sugar section of the local grocery store is growing. Offerings now include many organic, specialty, and alternative sugars and I urge you to experiment with them. In the interest of keeping things simple (and in the interest of saving precious kitchen space) I only use a few sugars in this book.
**Coarse Sugar** Larger grained sugar, perfect for decorating. Also known as pearl sugar.
**Confectioners Sugar** Commonly known as "icing," "10X," or "powdered" sugar, confectioners' sugar is often used to make frostings, icings, and whip cream. Confectioner's sugar contains a small amount of corn starch.
**Dark Brown and Light Brown Sugar** Brown sugar contains molasses (dark brown sugar contains more molasses than light brown sugar) and I use it with abandon.
**Granulated Sugar** This is the regular, omnipresent sugar. Lately, I have been playing with organic granulated sugar and have been extremely pleased with the results.
* * *
### **RED VELVET WHOOPIE PIES**
RED VELVET CAKE, ONCE A MINOR CURIOSITY, IS NOW UBIQUITOUS IN AMERICAN BAKE-SHOPS. It is a trend that outlasted its trendiness. Though most closely identified as a Southern recipe, the red velvet probably originated somewhere in the Northeast as a cocoa powder- and buttermilk-based cake. In theory, the cocoa powder reacts with the baking soda to create a reddish hue, but somewhere along the way, the red quotient began to be upped with an unhealthy dose of red dye. Unfortunately, a rash of basic vanilla cakes containing buckets of red dye suddenly became "red velvet cakes." True red velvet needs at least three ingredients to qualify for the moniker: cocoa powder, buttermilk, and shortening (it lends a great fluffy texture). I include all of these in our Red Velvet Whoopie Pies, and fill them with a delectable cream cheese frosting. It is a Southern/Northern hybrid that looks great and tastes even better.
**YIELD: 10 TO 12 LARGE OR 15 TO 17 SMALL PIES**
**_Ingredients_**
FOR THE WHOOPIE PIES
_2½ cups all-purpose flour_
_3 tablespoons dark unsweetened cocoa powder (like Valrhona)_
_½ teaspoon baking powder_
_½ teaspoon baking soda_
_½ teaspoon salt_
_¼ cup canola oil_
_½ cup plus 2 tablespoons buttermilk_
_½ cup butter (1 stick), softened, cut into pieces_
_1 tablespoon vegetable shortening, at room temperature_
_¾ cup firmly packed dark brown sugar_
_¼ cup granulated sugar_
_1 large egg_
_1 teaspoon pure vanilla extract_
_1 tablespoon red gel food coloring_
FOR THE CREAM CHEESE FILLING
_3 cups confectioners' sugar_
_½ cup (1 stick) unsalted butter, softened_
_8 ounces cream cheese, softened_
_1 teaspoon pure vanilla extract_
_¼ teaspoon salt_
FOR THE ASSEMBLY
_½ cup toasted walnuts, chopped coarsely_
#### **MAKE THE PIES**
Preheat the oven to 350 degrees F. Line two baking sheets with parchment paper.
In a large bowl, sift together the flour, cocoa powder, baking powder, baking soda, and salt.
In a small bowl, whisk together the canola oil and buttermilk.
In the bowl of an electric mixer fitted with the paddle attachment, cream the butter and shortening until smooth. Scrape down the bowl and add the sugars. Beat until the mixture is light and fluffy, about 5 minutes. Add the egg and vanilla extract and beat until combined. Scrape down the sides and bottom of the bowl, add the red gel food coloring, then mix on low speed for a few more seconds to incorporate. Do not overmix.
Turn the mixer to low. Add the flour mixture, alternating with the buttermilk mixture, in three separate additions, beginning and ending with the flour mixture until just combined. Scrape down the sides and bottom of the bowl, then mix on low speed for a few more seconds.
Cover with plastic wrap and chill the batter in the refrigerator for about 15 minutes.
Remove the batter from the refrigerator. Use a small ice cream scoop with a release mechanism to drop heaping tablespoons of the dough onto the prepared baking sheets about 1 inch apart. Bake for 10 to 20 minutes, until the cookies are just starting to crack on top and a toothpick inserted into the center of a cookie comes out clean. Let the cookies cool completely on the pan while you make the filling.
#### **MAKE THE CREAM CHEESE FILLING**
Sift the confectioners' sugar into a medium bowl and set aside.
In the bowl of an electric mixer fitted with the paddle attachment, beat the butter until is completely smooth. Add the cream creese and beat until combined.
Add the confectioners' sugar, vanilla, and salt and beat until smooth. Be careful not to overbeat the filling, or it will lose structure. (The filling can be made 1 day ahead. Cover the bowl tightly and put it in the refrigerator. Let the filling soften at room temperature before using).
#### **ASSEMBLE THE WHOOPIE PIES**
Spread the walnuts in an even layer on a small plate.
Turn half of the cooled cookies upside down (flat side facing up).
Use an ice cream scoop or a tablespoon to drop a large dollop of filling onto the flat side of the cookie. Place another cookie, flat side down, on top of the filling. Press down slightly so that the filling spreads to the edges of the cookie. Turn the whoopie on its side and roll through the walnuts. Repeat until all the cookies are used. Put the whoopie pies in the refrigerator for about 30 minutes to firm up before serving.
The whoopie pies will keep for up to 3 days on a parchment-lined baking sheet covered with plastic wrap, in the refrigerator.
**_Baked Note_**
I would never call myself a proper Southern gentleman (and I think my friends would concur), but I do know that a Southern red velvet cake is always finished with walnuts (whole or crushed). I would never insist on serving this cookie with nuts, as I am aware of the considerable increase of nut allergies. I do, however, insist that, if you use the nuts, you do not substitute them with pecans or some other type of nut. That would just be immoral.
### **PEANUT BUTTER AND JELLY BARS**
THE PEANUT BUTTER AND JELLY BAR IS, OF COURSE, A NOD TO THE MOST FAMOUS OF LUNCH-BOX SANDWICHES. It is a playful take (playful without being silly) on an iconic American flavor pairing. If you are a PB and J fan, I urge you to give these a try. They have been a staple of the Baked menu since opening day, and are among our most popular snack items. The heaps of peanut butter and mounds of jelly are sandwiched between a sweet crust and chunky crumble. It's a bit of a picnic dessert. Or the perfect lunch substitute.
**YIELD: 15 LARGE BARS**
**_Ingredients_**
THE SWEET PASTRY DOUGH
_¼ cup granulated sugar_
_1½ cups all-purpose flour_
_¼ teaspoon salt_
_½ cup (1 stick) cold unsalted butter, cut into ½-inch cubes_
_1 large egg_
FOR THE PEANUT BUTTER FILLING
_1 cup (2 sticks) unsalted butter, at room temperature_
_2 cups smooth peanut butter or 1 cup smooth peanut butter and 1 cup chunky peanut butter_
_1¾ cups confectioners' sugar_
_½ teaspoon pure vanilla extract_
FOR THE CRUMB TOPPING
_¾ cups all-purpose flour_
_¼ teaspoon salt_
_½ teaspoon baking powder_
_¼ teaspoon baking soda_
_¼ teaspoon cinnamon_
_⅓ cup firmly packed dark brown sugar_
_⅔ cup rolled oats_
_6 tablespoons (¾ stick) cold unsalted butter, cut into ½-inch pieces_
FOR THE ASSEMBLY
_2 heaping cups good-quality jelly or preserves_
#### **MAKE THE SWEET PASTRY DOUGH**
Butter the sides and bottom of a glass or light-colored metal 9-by-13-inch pan. Line the bottom with a sheet of parchment paper, and butter the parchment. Put the sugar, flour, and salt in a food processor and pulse until combined. Add the butter and pulse until sandy (about 6 to 10 quick pulses). In a small bowl, whisk the eggs and pour them into the food processor. Pulse just until the dough begins to hold together. Form the dough into a disk, wrap it tightly in plastic, and refrigerate for at least 1 hour or overnight.
Dust a work surface with a sprinkling of flour. Using a rolling pin, roll the dough into a rectangle slightly larger than 9 by 13 inches (the size of the pan) and about ¼ inch thick. (The dough might be sticky. Make sure to turn it with a bench knife or offset spatula as needed and keep the working surface floured. Some people find it easier to roll the dough between two layers of parchment paper. This can make it easier to transfer and be a bit less messy.)
Ever so gently, guide the dough into the pan and lightly press it—without pulling—into the bottom; it is not necessary to bring the dough up the sides of the pan, only to completely cover the bottom of the pan. Trim off any excess. Place the pan in the freezer for 30 minutes.
Preheat the oven to 375 degrees F.
Remove the pan from the freezer, line it with aluminum foil, and fill it three-quarters full with pie weights or dried beans. Bake for 15 minutes, then remove the foil and weights and bake for another 10 minutes, or until the crust is lightly browned. Transfer the pan to a wire rack to cool.
Reduce the oven temperature to 325 degrees F.
#### **MAKE THE PEANUT BUTTER FILLING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter until it is completely smooth. Add the peanut butter and beat until combined. Add the confectioners' sugar and vanilla and beat until smooth. Scrape down the sides of the bowl and beat again. Turn the mixture out onto the crust and, using an offset spatula, spread it into an even layer. Chill the peanut butter layer while you make the crumb topping.
#### **MAKE THE CRUMB TOPPING**
In a large bowl, whisk together the flour, baking powder, salt, baking soda, and cinnamon. Add the brown sugar and use your hands to rub it in until the mixture is uniform in color. Stir in the oats.
Place the dry mix in the bowl of a standing mixer fitted with the paddle attachment. Add the butter and beat on low speed until loose crumbs form.
#### **TO ASSEMBLE THE BARS**
Spread the jelly in an even layer over the peanut butter filling. Sprinkle on the crumb topping until the jelly is no longer visible.
Bake the bars for 20 to 25 minutes, rotating the pan halfway through, or until the top is brown.
Transfer the pan to a wire rack to cool completely, then cut the bars and serve.
The bars can be stored in the refrigerator in an airtight container for up to 2 days.
**_Baked Note_**
This is a rare peanut butter flavored dessert for which you can easily replace the smooth peanut butter with a chunkier variety in equal parts. I almost always make the bars with grape jelly, but feel free to use your favorite preserves.
### **ROSEMARY APRICOT SQUARES**
I AM, AND ALWAYS WILL BE, A DESSERT PURIST. I prefer my desserts fuss free, full of chocolate, and usually accompanied by a side of vanilla ice cream. I shy away from forced savory/sweet combinations and headline grabbers (bacon cupcakes, anyone?)—but I will make and try anything at least once. Usually, I then retreat to a peanut butter–chocolate default. So I was a little bit surprised by how hard I fell for these bars. Our Rosemary Apricot Squares are the creation of Eric Wolitzky and Patrick Panella, two rising talents in the baking world (coincidentally, they put some time in at Baked). The moment they presented them to me, I added them to our regular menu rotation. The rosemary short dough is light and elegant without being overpowering, and it combines well with the not-so-sweet apricot filling. They are great for breakfast, afternoon coffee, and casual dinners. In short, they are my chocolate-alternative dessert. If there has to be such a thing.
**YIELD: 9 LARGE BARS**
**_Ingredients_**
FOR THE ROSEMARY SHORT DOUGH
_1¾ cups all-purpose flour_
_½ teaspoon salt_
_2½ teaspoons fresh rosemary leaves, minced_
_12 tablespoons (1½ sticks) unsalted butter, cut into ½-inch cubes, at room temperature_
_½ cup confectioners' sugar, sifted_
_¾ teaspoon pure vanilla extract_
FOR THE APRICOT FILLING
_2 cups dried California apricots (about 8½ ounces)_
_½ cup granulated sugar_
_3 tablespoons honey_
_2 tablespoons brandy_
_Pinch salt_
FOR THE CRUMB TOPPING
_½ cup all-purpose flour_
_½ cup firmly packed dark brown sugar_
_⅓ cup pecans, coarsely chopped_
_Pinch salt_
_3 tablespoons cold unsalted butter, cut into ½-inch cubes_
#### **MAKE THE ROSEMARY SHORT DOUGH**
Lightly spray a 9-inch square baking pan with nonstick cooking spray and line it with parchment paper, allowing the parchment to just overhang on two sides.
In a medium bowl, whisk together the flour, salt, and rosemary. In the bowl of a standing mixer fitted with the paddle attachment, beat the butter with the confectioners' sugar and vanilla at medium speed until fluffy, approximately 2 minutes. Turn the mixer to the lowest speed and stream in the flour mixture. Scrape the dough into the prepared pan, lightly flour your clean hands, and press it into an even layer. Place the pan in the refrigerator for at least 30 minutes.
Preheat the oven to 350 degrees F.
Bake the short crust until it is golden, 25 to 30 minutes, rotating the pan halfway through the baking time. Cool the pan on a wire rack. Leave the oven on.
#### **MAKE THE APRICOT FILLING**
Place the apricots, sugar, honey, brandy, and salt in a medium saucepan with 1½ cups water and simmer over low heat for 40 to 50 minutes, or until the apricots are fork-tender and most of the liquid has evaporated or thickened. Remove the pan from the heat and stir the mixture to release excess steam. Scrape the apricot mixture into a food processor and puree until smooth.
#### **MAKE THE CRUMB TOPPING**
In the bowl of a standing mixer fitted with the paddle attachment, combine the flour, brown sugar, pecans, and salt. Mix on low speed for 15 seconds. Add the butter and mix until a sandy crumb begins to form, about 1 minute. (At this point, the crumb topping can be stored, covered, in the refrigerator until ready to use).
#### **ASSEMBLE THE ROSEMARY APRICOT SQUARES**
Spread the apricot filling over the shortbread, then sprinkle the crumb topping over the filling. Bake for 20 to 25 minutes, or until the crumb has browned. Let the pastry cool for at least 30 minutes in the pan, then lift it out using the parchment paper overhang and cut it into bars. The bars can be stored in refrigerator, tightly wrapped, for up to 3 days.
**_Baked Note_**
I am hesitant to become your go-to source of apricot info, but here are a few personal notes: First, I always buy unsulfured apricots; the sulfured ones seem one step away from a freaky science experiment. Second, generally speaking, I find California apricots are tangier and have a stronger apricot flavor than some of their Middle Eastern counterparts.
### **AUNT SABRA KING'S PUDDING BARS**
BLAIR VAN SANT, OUR FORMER PASTRY CHEF AND KITCHEN MANAGER EXTRAORDINAIRE, WAS ALSO OUR RESIDENT PUDDING EXPERT. He has since returned to the South, but he remains a close friend and an avid pudding freak. In the service of this book, I asked him to dig up his oldest pudding inspiration. Blair got back to me in twenty-six seconds flat. Aunt Sabra King's Pudding Bar is a messy Sunday supper–style dessert that epitomizes the beauty of Southern baking: not too fussy, very rich, and slightly boozy. Aunt Sabra herself is a gin-drinking, card-playing, silver Cadillac–driving lady. The exact origin of this recipe is unknown, but Aunt Sabra King (through a variety of inspired tweaks and nods) made it her very own.
**YIELD: 24 BARS**
**_Ingredients_**
FOR THE GRAHAM CRUST
_2 cups crushed graham cracker crumbs_
_1 tablespoon firmly packed dark brown sugar_
_½ teaspoon salt_
_10 tablespoons (1¼ sticks) unsalted butter, melted_
FOR THE CHOCOLATE PUDDING FILLING
_4 ounces good-quality dark chocolate (60 to 72%), coarsely chopped_
_3 tablespoons cornstarch_
_¾ cup granulated sugar_
_6 egg yolks_
_1 teaspoon pure vanilla extract_
_2 cups whole milk_
_½ cup heavy cream_
_2 tablespoons whiskey_
_Simple Whipped Cream for serving (optional)_
_Dark cocoa powder (like Valrhona) for serving (optional)_
#### **MAKE THE GRAHAM CRUST**
Preheat the oven to 325 degrees F.
In a medium bowl, stir together the graham cracker crumbs, brown sugar, and salt. Pour the butter over the crumb mixture and stir until well combined. The mixture will feel wet. Turn the crumb mixture out into a 9-by-13-inch pan and press it into the bottom and up the sides. Use the back of a large spoon to get an even layer. Place the crust in the refrigerator for 30 minutes.
Bake the crust for approximately 10 minutes, or until it is golden brown. Set it on a wire rack to cool while you make the filling.
#### **MAKE THE CHOCOLATE PUDDING FILLING**
Place the chocolate in a medium heatproof bowl and set it aside.
In a large, heatproof bowl, whisk together the cornstarch and ½ cup of the sugar. Add the egg yolks and whisk until combined. Add the vanilla and whisk again.
In a medium saucepan over medium heat, whisk together the milk, cream, and the remaining ¼ cup sugar. Bring the mixture just to a boil. Add a third of the hot milk mixture to the egg mixture, whisking constantly. Keep whisking the egg mixture and add another third of the hot milk mixture. Transfer the egg mixture into the saucepan with the milk mixture and, whisking constantly, bring it to a boil over medium-high heat. Boil for 2 to 3 minutes, or until the pudding is very thick.
Remove the pan from the heat and strain the mixture through a fine-mesh sieve directly onto the chocolate. Stir until it is smooth. Add the whiskey and stir again.
#### **ASSEMBLE THE PUDDING BARS**
Let the pudding mixture cool for about 20 minutes. Whisk it one more time until it is smooth and pour it over the cooled graham crust. Spread the pudding into an even layer. Cover it with plastic wrap and refrigerate for at least 4 hours.
Cut into squares and serve with whipped cream and a sprinkling of cocoa, if desired.
The bars can be stored, tightly covered, in the refrigerator for up to 2 days.
**_Baked Note_**
These pudding bars work well as pudding cups too. Simply press the graham crust into the bottom and up the sides of a 12-cup muffin pan, then bake and fill them as directed. Use an offset spatula to remove the individual cups and serve immediately.
### **JOE FROGGERS, OR GINGER RUM MOLASSES COOKIES**
I HAD ACTUALLY NEVER HEARD OF THE JOE FROGGER UNTIL I STUMBLED UPON A BAKERY IN MAINE SELLING THESE COOKIES BY THE DOZEN. They were huge—each one the size of a small pancake—and utterly addictive. Though I found the Joe Frogger in Maine, it is a native of Marblehead, Massachusetts, named after a famous resident, Joe Brown. In short, "Uncle Joe" supposedly baked these molasses-heavy cookies by the dozens for the fishermen and denizens of Marblehead, and he sold them at his tavern on Gingerbread Hill (how fitting!). Regardless of the historical origins of the Joe Frogger, I am a fan. The molasses and dark rum blend perfectly, and it has a likeable chewy snap.
**YIELD: 36 TO 48 COOKIES, DEPENDING ON SIZE OF THE CUTTER**
**_Ingredients_**
_4 cups all-purpose flour_
_1½ teaspoons salt_
_1½ teaspoons ground ginger_
_½ teaspoon freshly grated nutmeg_
_½ teaspoon ground cloves_
_1 teaspoon baking soda_
_4 tablespoons (½ stick) unsalted butter_
_¼ cup vegetable shortening_
_¾ cup firmly packed dark brown sugar_
_¼ cup granulated sugar_
_1¼ cups molasses_
_3 tablespoons dark rum_
_Coarse sugar for decor_
Whisk the flour, salt, ginger, nutmeg, cloves, and baking soda together. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and shortening together until there are no visible lumps. Add both sugars and beat just until incorporated. Scrape down the bowl, add the molasses, and beat until the mixture is uniform in color.
Prepare ⅓ cup very hot water. Add the flour mixture to the butter mixture, alternating with the hot water, in three parts, beginning and ending with the flour mixture. Scrape down the bowl, add the rum, and mix for 15 seconds. Cover the bowl and chill for at least 3 hours or overnight.
Preheat the oven to 375 degrees F. Line two baking sheets with parchment paper. Dust a work surface with a sprinkling of flour. Roll the dough into a ¼-inch thick round. Cut out the cookies with a 2- to 3-inch round cookie cutter, and transfer them to the prepared baking sheets. Sprinkle a tiny bit of sanding sugar onto each cookie.
Bake the cookies for 8 to 12 minutes, until they are set. Place the baking sheet on a wire rack to cool for 5 minutes. Use a spatula to transfer the cookies to the rack to cool completely. Store in an airtight container for up to 3 days.
**_Baked Note_**
The original Joe Frogger uses vegetable shortening exclusively. I adapted the recipe to include some butter, but it would be a mistake to take out the shortening entirely. An all-butter cookie, while tasty, will spread more and won't retain the same bite over time. In other words, an all-butter cookie would not be a Joe Frogger, and that would be a shame. If you prefer a chewy cookie, bake for 8 minutes; if you like a crispier cookie, err on the 12-minute side.
##
**DEVIL'S FOOD CAKE WITH ANGEL FROSTING**
**BOSTON CREAM PIE CAKE**
**CARAMEL APPLE CAKE**
**BURNT SUGAR BUNDT CAKE WITH CARAMEL RUM FROSTING**
**STUMP DE NOËL**
**QUICK SKILLET SNACK CAKE**
**LADY PRALINE CHIFFON CAKE**
**CHOCOLATE COFFEE CAKE WITH DARK CHOCOLATE GANACHE**
**TOMATO SOUP CUPCAKES WITH MASCARPONE FROSTING**
**MISSISSIPPI MUD PIE (B), AKA MUDDY MISSISSIPPI CAKE**
**MAPLE CUPCAKES WITH MAPLE CREAM CHEESE FROSTING**
**SUNDAY NIGHT CAKE**
**AUNT SASSY CAKE**
My cake patience is wearing thin. I can no longer pretend to be enthralled with a cake that is not really a cake. If it's in the shape of a designer shoe or is a lifelike replica of a favorite pet, it is really less about the cake and more about the design. It is form over function, cake as afterthought. This does not happen to other foods. I have yet to attend a party where the chef has sewn together a string of delicious steaks into a golf club or fedora. I have never seen (and hope to never see) a baby rattle composed of salmon fillets. But cake abuse has no limits. I admire the design and architecture involved to create these stunning showstoppers, but they are, too often, not something I would want to eat.
I am still a cake pusher. I adamantly believe that cake is the great centerpiece dessert for any affair, however ordinary—not just birthdays, graduations, and holidays. Too often, we relegate cake (primarily layer cake) to event status, but I like to serve it just for its own sake. I am too impatient to wait for the next wedding, or birthday, or engagement, or holiday.
While rummaging about old recipes and dusty cookbooks, I discovered a surprising thing about cakes. They haven't really changed much. I have a recipe for a chocolate cake that is at least forty years old, and it is probably no different from something you would find in the pages of today's food magazines. The methodology might be slightly different. The newer recipes, including ours, are written for standing mixers and other gadgets like food processors and mini grinders, while the older ones were geared toward elbow grease—centered techniques. Also, instructions are becoming more and more precise and increasingly lengthy. For better or worse, recipes went from being a quick paragraph to many paragraphs to a page or two. I am unsure about how to stop this predicament, as I am part of the problem. In just one instance, I transcribed my grand-mother's favorite apple cake recipe from a small index card. After interpreting her shorthand and adding a few precise instructions, the recipe magically grew to two pages. I can't help myself.
The cakes in this chapter were culled to include a range representing every aspect of American baking. Of course we have the beloved cupcake, America's current reigning dessert champion, smartly symbolized by our wonderful Maple Cupcakes with Maple Cream Cheese Frosting. My personal favorite, Burnt Sugar Bundt Cake, is everything a Bundt cake should be: tasty, hearty, and a great accompaniment to any meal, even breakfast. Then, of course, we included an array of stunning three-layer cakes. You should try them all, especially the Chocolate Coffee Cake with Dark Chocolate Ganache. Actually, we hope you have a chance to work your way through each and every cake. We did, and we survived.
### **DEVIL'S FOOD CAKE WITH ANGEL FROSTING**
THE ORIGINS OF THE TERM "DEVIL'S FOOD" TO DESCRIBE CHOCOLATE CAKE ARE EXTREMELY DIFFICULT, IF NOT IMPOSSIBLE, TO IDENTIFY. However, even a cursory investigation shows that devil's food cake recipes started to appear at least as early as the 1920s. The recipes all vary slightly. There are versions with heaps of cinnamon, versions including a variety of nuts, and quite a few with mashed potatoes. In the end, however, devil's food cake is really just a chocolate cake that's more chocolaty than most. This cake is dense and moist without being heavy, and it's full of a rich chocolate flavor imparted by the combination of dark chocolate, dark cocoa powder, and coffee.
It is in my nature to cover any and every chocolate cake in a thick chocolate frosting, but I rather love the contrasting flavor of this fluffy, white Angel frosting. It is worth noting that this recipe makes a great cupcake. If you are adapting it for cupcake use, be sure to reduce the baking time by 25 minutes, keep a watchful eye on the oven, and only fill the cup holders three-quarters of the way full.
**YIELD: ONE 8-INCH, 2-LAYER CAKE**
**_Ingredients_**
FOR THE DEVIL'S FOOD CAKE
_1 ounce good-quality dark chocolate (60 to 72%), broken into a few pieces_
_½ cup dark unsweetened cocoa powder (like Valrhona)_
_⅔ cup hot coffee_
_⅓ cup whole milk_
_1 ⅓ cups all-purpose flour_
_1 teaspoon baking soda_
_½ teaspoon salt_
_10 tablespoons (1¼ sticks) unsalted butter, cut into ½-inch cubes, softened_
_1 cup firmly packed dark brown sugar_
_½ cup granulated sugar_
_3 large eggs_
_1 teaspoon pure vanilla extract_
FOR THE ANGEL FROSTING
_5 large egg whites, at room temperature_
_1½ cups granulated sugar_
_1 tablespoon light corn syrup_
_1 teaspoon vanilla paste (or 1½ teaspoons pure vanilla extract)_
#### **MAKE THE DEVIL'S FOOD CAKE**
Preheat the oven to 325 degrees F. Butter two 8-inch round cake pans, line the bottoms with parchment paper, and butter the parchment. Dust the parchment with flour and knock out the excess flour.
Place the chocolate and cocoa powder in a medium heatproof bowl. Pour the hot coffee directly over them and whisk until combined. Add the milk and whisk until smooth.
In another bowl, sift together the flour, baking soda, and salt. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and sugars on medium speed until fluffy, about 3 minutes. Add the eggs, one at a time, beating well after each addition, then add the vanilla and beat until incorporated. Scrape down the bowl and mix again for 30 seconds.
Add the flour mixture in three parts, alternating with the chocolate mixture, beginning and ending with the flour mixture.
Divide the batter into the prepared pans and smooth the tops. Bake for 35 to 40 minutes, rotating the pans halfway through the baking time, until a toothpick inserted in the center of the cake comes out clean. Transfer the pans to a wire rack and let cool for 45 minutes. Turn the cakes out onto the rack and let them cool completely. Remove the parchment.
#### **MAKE THE ANGEL FROSTING**
Place the egg whites in the bowl of a standing mixer fitted with the whisk attachment. Set aside.
In a medium saucepan over low heat, stir together 1¼ cups of the sugar, the corn syrup, and ¼ cup water. Once the sugar is dissolved, increase the heat to medium-high, and clip a candy thermometer onto the side of the pot. Heat the mixture, without stirring, to almost soft-ball stage (about 235 degrees F)—do not let it go above 235 degrees F.
While you wait for the syrup to reach the soft-ball stage, whip the egg whites on medium speed until soft peaks form—do not beat beyond this.
As soon as the sugar mixture reaches the soft-ball stage, remove the pan from the heat.
Sprinkle the remaining ¼ cup sugar over the soft peaks of the egg whites and turn the mixer to low. Slowly stream in the hot sugar syrup. Once all the syrup has been added, increase the speed to medium-high and beat the icing for about 7 minutes until it is thick and shiny. Add the vanilla and beat again for 10 seconds.
#### **TO ASSEMBLE THE DEVIL'S FOOD CAKE**
Place one cake layer on a serving platter. Trim the top to create a flat surface and evenly spread about 1 cup frosting on top. Place the next layer on top, then trim and frost it the same way. Frost the sides of the cake with the remaining frosting. Serve immediately.
Angel frosting tastes best if it is served within 4 hours of being made.
**_Baked Note_**
Even though this recipe uses almost an entire cup of hot coffee, the taste is undetectable in the final dessert. The coffee merely enhances the chocolate flavor while cutting a little bit of the sweetness. If you want, you can substitute the coffee for 2 teaspoons of instant espresso powder dissolved in ⅔ cup boiling water.
### **BOSTON CREAM PIE CAKE**
AMAZINGLY, SOME DESSERTS—REALLY GREAT AMERICAN DESSERTS—ENTER THE AMERICAN CONSCIOUSNESS, LINGER FOR YEARS, AND SLOWLY DISAPPEAR. For some reason, they never reach icon or classic status. They become unfashionable or, even worse, caricatures of their former selves. This has been the fate of the once-beloved Boston cream pie. It used to be ubiquitous on dessert menus, in diner cases, and even at the odd birthday party. My theories for its disappearance are many, but mainly I blame the inadequate (or bad) versions that started to appear throughout the country. A poorly executed Boston cream pie is one bad dessert. There is nothing worse than fake pastry cream or an over-spongy sponge.
Our Boston Cream Pie Cake is a tribute and an homage. We tweaked the original to make a four-layer beast with all the important parts intact: a bouncy milk sponge, classic vanilla and chocolate pastry cream, and a gooey chocolate glaze. Messy? Yes. Good? Absolutely.
**YIELD: ONE 8-INCH, 4-LAYER ROUND CAKE**
**_Ingredients_**
FOR THE MILK SPONGE CAKE
_1¾ cups cake flour_
_1½ teaspoons baking powder_
_¾ teaspoon salt_
_6 tablespoons (¾ stick) unsalted butter, cut into small pieces_
_¾ cup whole milk_
_4 large eggs_
_1¼ cups sugar_
_1½ teaspoons pure vanilla extract_
FOR THE PASTRY CREAM FILLING
_7 large egg yolks_
_¾ cup sugar_
_¼ teaspoon salt_
_⅓ cup cornstarch_
_3 cups whole milk_
_3 tablespoons unsalted butter_
_1 tablespoon pure vanilla extract_
_2 tablespoons light rum_
_6 ounces good-quality dark chocolate (60 to 72%), finely chopped_
FOR THE CHOCOLATE GLAZE
_5 ounces good-quality dark chocolate (60 to 72%), coarsely chopped_
_2 tablespoons dark unsweetened cocoa powder (like Valrhona)_
_¼ cup sugar_
_¼ cup heavy cream_
_¼ cup light corn syrup_
_Pinch salt_
_½ teaspoon pure vanilla extract_
_1½ tablespoons unsalted butter_
#### **MAKE THE MILK SPONGE CAKE**
Preheat the oven to 325 degrees F. Butter two 8-inch round cake pans, line the bottoms with parchment paper, and butter the parchment. Dust the parchment with flour and knock out the excess flour.
In a large bowl, sift the cake flour, baking powder, and salt together. Set aside.
In a small saucepan over low heat, stir together the butter and milk until the butter is just melted. Do not overheat. Remove the pan from the heat and set aside.
In the bowl of a standing mixer fitted with the whisk attachment, beat the eggs, sugar, and vanilla on medium speed until the mixture is pale and has tripled in volume, about 5 minutes.
Remove the bowl from the standing mixer and sprinkle a third of the flour mixture over the egg mixture. Gently fold the two together, using a rubber spatula. Add the rest of the flour mixture and fold again.
Add the warm milk mixture (if it has cooled completely, reheat it slightly) to the batter and gently fold until just incorporated.
Divide the batter between the prepared pans and smooth the tops. Bake for 20 to 25 minutes, rotating the pans halfway through the baking time, until a toothpick inserted in the center comes out clean. The cake might appear wobbly if you shake the pan, but if pressed gently in the middle, it should spring back.
Transfer the cake pans to a wire rack and let cool for 20 minutes. Turn the cakes out onto the rack and let them cool completely. Remove the parchment.
#### **MAKE THE PASTRY CREAM FILLING**
In a large bowl, whisk together the egg yolks, half of the sugar, the salt, and cornstarch until the mixture is pale, thick, and smooth.
In a medium saucepan over medium heat, stir together the milk and the remaining half of the sugar. Continue stirring until the mixture boils. Remove the pan from the heat, and pour about a third of the liquid into the egg mixture, whisking constantly. Transfer the tempered egg mixture back to the saucepan with the remaining milk mixture and, whisking constantly, bring to a boil over medium-high heat. Boil for 2 minutes.
Remove the pan from the heat and strain through a fine-mesh sieve into a large bowl. Stir in the butter, vanilla, and rum until combined.
Scoop a third of the pastry cream into a different bowl and set it aside. Add the chocolate to the warm pastry cream remaining in the pan and stir until melted and smooth. Scoop the chocolate cream into a bowl. Once both pastry creams have cooled for about 15 minutes, wrap both bowls in plastic wrap, pressing the plastic directly onto the top of the cream to prevent a skin from forming. Refrigerate until firm, about 4 hours or overnight.
#### **MAKE THE CHOCOLATE GLAZE**
Place the chocolate and cocoa powder in a large bowl and set it over a saucepan of simmering water, stirring occasionally, until completely melted and smooth.
In another small saucepan, combine the sugar, cream, corn syrup, and salt with ¼ cup water. Place the pan over low heat and warm, without stirring, until the sugar dissolves. Increase the heat to medium and simmer for about 4 minutes, stirring frequently. Remove the pan from the heat and whisk in the vanilla and butter. Whisk to release excess heat, then add the reserved chocolate mixture and stir until smooth. Set the sauce aside; it will thicken while you assemble the cake.
#### **ASSEMBLE THE BOSTON CREAM PIE CAKE**
Using a serrated knife, cut the cake layers in half horizontally. Place the first cake layer on a platter and scoop half of the chocolate pastry cream on top. Spread the cream evenly to the cake's edges using an offset spatula. Add a second cake layer and cover that with all the vanilla cream. Add a third cake layer and cover it with the remaining chocolate cream. Place the last cake layer on top and press gently with your palms to help the cake layers adhere to the pastry cream.
Very slowly, pour the thickened chocolate glaze onto the center of the top of the cake. Use an offset spatula to spread it out to the edges so that it drips down the sides of the cake. Let the sauce thicken for about 20 minutes before serving.
The cake tastes best when eaten within 24 hours. If you have leftovers, tent the entire cake in foil and refrigerate. Serve it chilled or at room temperature.
* * *
**_Baked Notes_**
Generally speaking, it is easier (though not necessary) to trim cake layers when they are frozen. If you have the time to bake your cake layers the day before you need to serve the dessert, simply wrap the cooled cake twice in plastic and place it in the freezer. One more thing: I really like this cake with a rum soak. If the idea appeals to you, give it a try: First, use a toothpick and poke a few holes in the sponge after it cools from baking. Then heat 1 part sugar with 2 parts water until the sugar is dissolved. Add some rum (about 1 tablespoon) and brush the syrup over the layers before assembling the cake.
### **CARAMEL APPLE CAKE**
PEOPLE IN UPSTATE NEW YORK ARE ABSOLUTELY WILD ABOUT THEIR APPLES. It's a wildness bordering on religious conviction, and they can rattle off the differences in varieties, which apples work best for which recipes, and how many acres of orchards have been lost to development. They also feel a sense of ownership of all things apple, even though New York does not hold the title of largest apple-producing state (that would go to Washington). Unsurprisingly, there are many variations on apple cake floating about the Northeast, each with a particular family-inflected nuance. Our version is hearty. It is moist, dense, lightly spiced, and covered in a slightly sweet caramel frosting that gives the cake a homey county-fair feel. This cake is pure fall. It is dedicated to the apple lovers of New York, and I can say for certain that I have never made it before August or after January.
**YIELD: ONE 8-INCH, 3-LAYER CAKE**
**_Ingredients_**
FOR THE APPLE CAKE
_4 cups all-purpose flour_
_2 teaspoons baking soda_
_1 teaspoon baking powder_
_1 teaspoon salt_
_2 teaspoons cinnamon_
_1 teaspoon ground allspice_
_1 teaspoon ground cloves_
_1½ cups (3 sticks) unsalted butter, cut into 1-inch cubes, at room temperature_
_2½ cups sugar_
_2 large eggs_
_4 cups homemade applesauce (see Baked Note) or store-bought unsweetened applesauce_
FOR THE CARAMEL BUTTERCREAM
_1½ cups sugar_
_⅓ cup all-purpose flour_
_1½ cups whole milk_
_⅓ cup heavy cream_
_1½ cups (3 sticks) unsalted butter, soft but cool, cut into small pieces_
_1 teaspoon pure vanilla extract_
_⅓ cup plus 2 tablespoons Classic Caramel Sauce, at room temperature_
#### **MAKE THE APPLE CAKE**
Preheat the oven to 325 degrees F. Butter three 8-inch round cake pans, line the bottoms with parchment paper, and butter the parchment. Dust the parchment with flour and knock out the excess flour.
Sift the flour, baking soda, baking powder, salt, cinnamon, allspice, and cloves together into a large bowl. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter until creamy, about 4 minutes. Add the sugar and beat until light and fluffy, about 3 minutes. Add the egg and beat until combined.
Add the flour mixture to the mixer bowl in three parts, alternating with the applesauce, beginning and ending with the flour mixture. Scrape down the bowl, then mix on low speed for a few more seconds.
Divide the batter among the prepared pans and smooth the tops. Bake for 40 to 45 minutes, rotating the pans halfway through the baking time, until a toothpick inserted in the center of the cake comes out clean. Transfer the pans to a wire rack and cool for 20 minutes. Turn the cakes out onto the rack, remove the parchment, and let cool completely.
#### **MAKE THE CARAMEL BUTTERCREAM**
In a medium, heavy-bottomed saucepan, whisk the sugar and flour together. Add the milk and cream and cook over medium heat, whisking occasionally, until the mixture comes to a boil and has thickened, about 10 to 15 minutes.
Transfer the mixture to the bowl of a standing mixer fitted with the paddle attachment. Beat on high speed until cool. Reduce the speed to low and add the butter and vanilla; mix until thoroughly incorporated. Increase the speed to medium-high and beat until the frosting is light and fluffy.
Add ⅓ cup of the caramel and continue mixing until combined. If the frosting is too soft, put the bowl in the refrigerator to chill slightly, then beat again until it is the proper consistency. If the frosting is too firm, set the bowl over a pot of simmering water and beat with a wooden spoon until it is the proper consistency.
#### **ASSEMBLE THE CARAMEL APPLE CAKE**
Place one cake layer on a serving platter. Trim the top to create a flat surface and evenly spread about 1¼ cups of the frosting on top. Add the next layer, trim and frost it, then add the third layer. Spread a very thin layer of frosting over the sides and top of the cake and put it in the refrigerator for about 15 minutes to firm up. (This is known as crumb coating and will help to keep loose cake crumbs under control when you frost the outside of the cake.) Frost the sides and top with the remaining frosting. Drizzle on a few swirls of caramel and refrigerate the finished cake for 15 minutes to firm it up before serving.
This cake will keep beautifully in a cake saver at room temperature for up to 3 days, as long as the weather is cool and humidity free. Otherwise, place the cake in a cake saver and refrigerate it for up to 3 days. Let a chilled cake sit at room temperature for at least 2 hours before serving.
**_Baked Note_**
If you prefer, you can make your own applesauce for this cake. It's quite easy. The basic recipe is to peel and cut your apples, then place them in a medium saucepan with small amount (maybe a ⅓ cup) apple cider, ground cinnamon to taste, and a tiny bit of dark brown sugar (optional). Cover the saucepan and cook for about 30 minutes. Uncover and mash as you would potatoes.
### **BURNT SUGAR BUNDT CAKE WITH CARAMEL RUM FROSTING**
I AM ALL ABOUT BUNDT CAKES. IN FACT, TRUTH BE TOLD, I PREFER MAKING BUNDT CAKES TO JUST ABOUT ANYTHING. Bundts generally have a heartier feel and they are much easier to assemble than the classic three-layer cake. They travel well, and they keep like a dream. This Burnt Sugar Bundt Cake is loosely based on of one of those amazing cakes you find in a church supper cookbook and end up making over and over again. The burnt sugar adds a nice caramel flavor, and the coconut milk gives the cake a terrific moist crumb.
**YIELD: ONE 10-INCH BUNDT CAKE**
**_Ingredients_**
FOR THE BURNT SUGAR LIQUID
_½ cup granulated sugar_
_½ cup heavy cream_
_Approximately ¾ cup coconut milk_
_1½ tablespoons fresh lemon juice_
FOR THE BUNDT CAKE
_3 cups all-purpose flour_
_1 teaspoon baking powder_
_½ teaspoon baking soda_
_1 teaspoon salt_
_1¼ cups unsalted butter (2½ sticks) cut into 1-inch cubes, at room temperature_
_2 cups granulated sugar_
_4 large eggs_
_1 teaspoon pure vanilla extract_
_Burnt Sugar Liquid (see above)_
FOR THE CARAMEL RUM FROSTING
_½ cup (1 stick) unsalted butter_
_2 tablespoons dark rum_
_2 ⅓ cups confectioners' sugar_
_Burnt Sugar Liquid (see above)_
#### **MAKE THE BURNT SUGAR LIQUID**
In a medium saucepan over medium heat, slowly melt the sugar. Use a wooden spoon to stir it continuously to ensure even melting. When the sugar turns a dark caramel color, remove the pan from the heat and slowly stream in the cream while continuing to stir (don't worry if mixture starts to clump). Return the pan to medium heat and stir until completely combined; cook for 2 minutes longer, stirring.
Transfer the burnt caramel mixture to at least a 2-cup heatproof liquid measuring cup (like Pyrex) and add enough coconut milk to make 1¼ cups liquid. Add the lemon juice. Whisk to combine, divide the mixture in half, and set both portions aside.
#### **MAKE THE BUNDT CAKE**
Preheat the oven to 325 degrees F. Generously spray the inside of a 10-inch Bundt pan with nonstick cooking spray; alternatively, butter it thoroughly, dust it with flour, and knock out the excess flour.
In a medium bowl, whisk together the flour, baking powder, baking soda, and salt. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and sugar until pale and fluffy. Scrape down the bowl and add the eggs, one at a time, beating until each is incorporated. Add the vanilla and beat for 5 more seconds.
Retrieve one of the reserved portions of burnt sugar liquid. Add the flour mixture in three parts, alternating with the burnt sugar, beginning and ending with the flour mixture. Scrape down the sides and bottom of the bowl and beat again for 10 seconds. Pour the batter into the prepared pan and bake for 45 to 50 minutes, or until a small sharp knife inserted into the center of the cake comes out clean. Transfer the pan to a wire rack to cool completely. Gently loosen the sides of the cake from the pan and turn it out onto the rack.
#### **MAKE THE CARAMEL RUM FROSTING**
Put the butter, rum, confectioners' sugar, and remaining portion of burnt sugar liquid in a food processor. Pulse in short bursts until the frosting is shiny and smooth.
#### **ASSEMBLE THE BURNT SUGAR BUNDT CAKE**
Use an offset spatula to spread the frosting over the crown of the Bundt in a thick layer. Top with caramel shards, if you like (see sidebar below). Let the frosting set before serving. The cake will keep in an airtight container, at room temperature, for up to 3 days.
**_Baked Note_**
The frosting on this cake is very versatile. If you want to omit the alcohol, substitute 1 teaspoon pure vanilla extract. If you want a thinner frosting, increase the dark rum to 3 or 4 tablespoons. If you want a thicker, richer frosting keep adding 2 tablespoons of confectioners' sugar as you beat until you reach your desired consistency.
* * *
This caramel shard decoration is entirely optional—but it looks spectacular. Consider using it when you bring this cake to a potluck picnic in the park or a brunch.
_½ cup granulated sugar_
Place the sugar in a small saucepan. Add enough water—it takes approximately 1 teaspoon—to make it the texture of wet sand. Cook on high heat until the sugar turns amber. Pour the caramel onto a half-sheet pan lined with a Silpat (or other silicone nonstick) baking mat. Let cool. Break the cooled caramel into small shards and use them to decorate the frosted cake.
* * *
### **STUMP DE NOËL**
THE STUMP DE NOËL, A CHRISTMAS HOLIDAY PIÈCE DE RÉSISTANCE, IS THE BAKED INTERPRETATION OF THE INFAMOUS BÛCHE DE NOËL, UBIQUITOUS IN FRANCE. Our stump is every bit as alluring as a classic bûche, though we played with a few of the elements to make it more distinctly Baked. Our stump is an allusion to our fondness for woodland creatures, and it can be iced and finished to reflect the dinner: tongue-in-cheek, realistic, or slightly menacing. Traditionally, the bûche consists of a springy yellow sponge filled with chocolate buttercream. I made the sponge cake slightly less spongy, and the filling is chock-full of our favorite flavor, malt. I even added a little texture with crushed malt balls. Obviously, this is a slightly showy, purely holidayesque affair, but don't be afraid to make it whenever you damn well please.
**YIELD: 24 SERVINGS**
**_Ingredients_**
FOR THE DARK CHOCOLATE AND MALTED BUTTERCREAMS
_5 large egg whites, at room temperature_
_1½ cups sugar_
_1 teaspoon pure vanilla extract_
_2 cups (4 sticks) unsalted butter, at room temperature_
_4 ounces good-quality dark chocolate (60 to 72%), melted and cooled_
_¼ cup malted milk powder_
_12 malted milk ball candies, crushed_
FOR THE CHOCOLATE CAKE ROLL
_1 cup all-purpose flour_
_¼ cup dark unsweetened cocoa powder (like Valrhona), sifted_
_¼ teaspoon salt_
_2 tablespoons instant espresso powder_
_12 ounces good-quality dark chocolate (60 to 72%), melted and cooled_
_12 large eggs, at room temperature, separated_
_1 ⅓ cups sugar_
_4 teaspoons pure vanilla extract_
_¼ teaspoon cream of tartar_
_6 tablespoons (¼ stick) unsalted butter, melted and cooled_
#### **MAKE THE DARK CHOCOLATE AND MALTED BUTTERCREAMS**
In the bowl of a standing mixer, combine the egg whites and sugar. Set the bowl over a pot of simmering water and whisk until the sugar is dissolved and the egg whites are just warm to the touch. Return the bowl to the mixer and fit it with the whisk attachment. Add the vanilla and beat the egg whites at high speed until firm and glossy, about 5 minutes. With the machine running, whisk in the butter a few tablespoons at a time. If the mixture begins to look curdled, continue to beat until it is smooth before adding more butter.
Transfer 1½ cups of the buttercream to a bowl and whisk in the melted chocolate. Cover the chocolate buttercream and refrigerate.
Dissolve the malt powder in 2 tablespoons hot water, then beat it into the buttercream remaining in the mixer. Beat in the crushed milk balls. Cover the malt buttercream and refrigerate.
#### **MAKE THE CHOCOLATE CAKE ROLL**
Preheat the oven to 350 degrees F. Butter two 17-by-12-inch rimmed baking sheets and line them with parchment paper, leaving a 1-inch overhang on all the short sides. Butter the paper and dust it with flour.
In a small bowl, whisk together the flour, cocoa powder, and salt. In another small bowl, dissolve the espresso powder in ¼ cup hot water, then stir in the chocolate.
In the bowl of a standing mixer, whisk together the egg yolks and ⅔ cup of the sugar. Set the bowl over a pan of simmering water and whisk until the sugar is dissolved. Transfer the bowl to a mixer fitted with the whisk attachment and beat at high speed until the yolks are pale and thick, about 5 minutes. Beat in the melted chocolate mixture along with the vanilla. Transfer the mixture to a large bowl.
Thoroughly wash and dry the mixer bowl and the whisk attachment. In the clean bowl, beat the egg whites with the cream of tartar on medium-high speed until soft peaks form. Gradually add the remaining ⅔ cup sugar and continue beating at high speed until the whites are glossy, about 2 minutes longer. Whisk a quarter of the egg whites into the cake batter, then fold in the remaining whites until no streaks remain.
In a small bowl, whisk the melted butter with ½ cup of the batter; fold this mixture into the batter. In two batches, sift the cocoa powder mixture over the batter and gently fold it in. Divide the batter between the prepared pans and use an offset spatula to spread it evenly.
Bake the cakes for about 18 minutes, until they feel springy and slightly dry; shift the pans from top to bottom and front to back halfway through the baking time. Transfer the pans to wire racks and cool completely. Run the tip of a knife around the edges, cover the cake surface with parchment paper and a baking sheet, and turn it out of the pan; peel off the parchment liner.
#### **TO ASSEMBLE THE CAKE**
Spread the malt buttercream over the cakes. Using a ruler, cut each cake precisely in half lengthwise, cutting through the parchment lining them; you should have four 6 by 17-inch strips of cake. Roll one strip into a tight coil, removing the paper as you roll. Roll the three remaining cake strips around the coil in the same way to form a very wide, short jelly roll. Set the cake on a large plate, spiraled end up. Frost the outside of the cake with the chocolate buttercream. Refrigerate the cake until set, at least 8 hours. If desired, decorate it with the meringue mushrooms, cranberries, and rosemary sprigs, and serve, cutting the cake into wedges or horizontal slices. (We like to cut the stump lengthwise for the full effect. It is thin, but large.)
**_Baked Note_**
This cake is spectacular, but it does require some thoughtful preparation. It is important to make and chill the dark chocolate and malted buttercreams before baking the cake. Also, right before you start assembling the cake, pull the buttercream out of the refrigerator so it reaches room temperature in time for filling the buche. Finally, I suggest making the optional meringue mushrooms, the sugared cranberries, and the sugared rosemary while your filled and frosted buche is setting up in the refrigerator.
### **QUICK SKILLET SNACK CAKE**
A GOOD, SOLID, EASY-TO-PUT-TOGETHER, AND EASY-TO-BAKE SNACKIN' CAKE SHOULD BE PART OF EVERY HOME BAKER'S REPERTOIRE, AND THIS IS OUR GO-TO SOLUTION. It's a springy chocolate cake with a slathering of fudgy frosting and my favorite part: It's baked in a skillet, which gives the sides a fun crunch. This is the cake I make when I have a hankering for something less celebratory and more quick and dirty—the kind of cake I can throw together for an impromptu afternoon gathering. If I am toting the cake to a destination, I put the whole thing back in the skillet for ease in carrying.
**YIELD: ONE 10-INCH SKILLET CAKE**
**_Ingredients_**
FOR THE CHOCOLATE CAKE
_½ cup unsweetened dark cocoa powder (like Valrhona)_
_2 ounces good-quality dark chocolate (60 to 72%), coarsely chopped_
_1 teaspoon instant espresso powder_
_1½ cups all-purpose flour_
_1 teaspoon baking soda_
_1 teaspoon salt_
_½ cup (1 stick) unsalted butter, cut into 1-inch cubes, at room temperature_
_2 tablespoons vegetable shortening, at room temperature_
_1 cup firmly packed dark brown sugar_
_¼ cup granulated sugar_
_1 teaspoon pure vanilla extract_
_3 large eggs_
_¼ cup plus 2 tablespoons buttermilk, shaken vigorously_
FOR THE CHOCOLATE FROSTING
_½ cup (1 stick) unsalted butter, softened_
_1 cup confectioners' sugar, sifted_
_1 tablespoon pure vanilla extract_
_3 ounces good-quality dark chocolate (60 to 72%), melted and cooled_
#### **MAKE THE CHOCOLATE CAKE**
Preheat the oven to 350 degrees F.
Grease a 10-inch, cast-iron skillet or ovenproof stainless-steel skillet with butter. (The heavy, dark-colored cast-iron skillet will make the sides of the cake more crispy than a stainless steel one.) Line the pan with parchment paper and butter the parchment. Dust the parchment with flour and knock out the excess.
In a small, heatproof bowl, whisk together the cocoa powder, chocolate, and espresso powder. Add ¾ cup very hot water, wait 1 minute, and then whisk the mixture until it is melted and smooth. Set aside to cool.
In another small bowl, whisk together the flour, baking soda, and salt.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and shortening together on medium speed until creamy, 2 to 3 minutes. Add the sugars and vanilla and beat until fluffy, about 3 minutes. Scrape down the bowl, add the eggs one at a time, and beat until just combined. Turn the mixer to its lowest setting, and in a slow, steady stream, add the reserved chocolate mixture. Scrape down the bowl again, then turn the mixer to low. Add the flour mixture in three parts, alternating with the buttermilk, beginning and ending with the flour mixture. Scrape down the bowl, then mix for a few more seconds and pour the batter into the prepared skillet. Smooth the surface with a spatula.
Bake for 40 to 45 minutes, rotating the skillet halfway through the baking time, until a toothpick inserted in the center of the cake comes out clean. Transfer the pan to a wire rack to cool for about 15 minutes. Run a paring knife around the sides of the pan and flip the cake out onto a cooling rack. Turn the cake right side up and let it sit on the rack until completely cool.
#### **MAKE THE CHOCOLATE FROSTING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter on high speed until creamy, about 2 minutes. Add the confectioners' sugar all at once and beat until completely blended, about 2 minutes. Add the vanilla and beat for 15 seconds. Scrape down the bowl and add the melted, cooled chocolate. Beat until smooth, continuing to scrape down the sides of the bowl as needed until the frosting is uniform in color.
Transfer the skillet cake to a cake board or serving platter. Use an offset spatula to spread the frosting evenly across the top. Serve it immediately or refrigerate it, if necessary. Bring it back to room temperature before serving.
**_Baked Note_**
I often get asked about the necessity of using shortening in this recipe. Do you have to do it? Not really. Can you replace it with butter? Yes, I suppose. Just note that shortening gives the cake a really great springy texture, and shortening is not necessarily the devil it is made out to be. Crisco, the classic vegetable shortening, makes a trans-fat-free version, and Spectrum Naturals makes an organic trans-fat-free version.
### **LADY PRALINE CHIFFON CAKE**
I FEEL LIKE I WAS BORN AT THE TAIL END OF THE CHIFFON CRAZE. It is as if the chiffon, a slightly denser cousin of angel food cake, was in vogue for a lifetime, then slowly drifted into irrelevance, like coq au vin. I remember chiffon cakes as regal affairs baked by regal ladies of a certain age, the kind of ladies who wore a large brooch pinned to their expensive jacket.
The chiffon cake was invented in California (the aptly named Harry Baker is credited with being its creator), but our Lady Praline Chiffon is dedicated to all stately Southern ladies and to the pecan.
**YIELD: ONE 10-INCH TUBE CAKE**
**_Ingredients_**
_¼ cup firmly packed dark brown sugar_
_½ cup plus 2 tablespoons granulated sugar_
_1¼ cups cake flour_
_2 teaspoons baking powder_
_½ teaspoon salt_
_5 egg yolks plus 1 egg_
_1 tablespoon pecan liqueur (if you can find it —otherwise, use an almond liqueur like Amaretto, or pure almond extract)_
_¾ cup vegetable oil_
_5 egg whites_
_¼ teaspoon cream of tartar_
_zest of 1 small to medium orange_
Preheat the oven to 325 degrees F. In a medium bowl, use your hands to rub the dark brown sugar into the granulated sugar until all lumps are gone and the sugars are combined. In a large bowl, sift the flour, baking powder, and salt together. Whisk the sugar mixture into the flour mixture.
In another bowl, whisk together the egg yolks, the egg, and liqueur until smooth. Stir in the oil and ½ cup water until combined. Make a well in the center of the dry ingredients and pour the wet mixture into it. Use a rubber spatula to fold the wet ingredients into the dry ingredients, add the orange zest, and fold until just combined. Do not overmix.
In a clean bowl, whisk the egg whites, cream of tartar, and orange zest until stiff peaks form. Do not overbeat. Gently fold the egg whites into the batter and pour it into an ungreased 10-inch tube pan, with a removable bottom.
Bake the cake for 50 to 60 minutes, rotating the pan halfway through the baking time, until a toothpick inserted in the center comes out clean. Invert the pan over a wire rack to cool completely, about 2 hours. (If the cake has risen above the top of the pan, invert it onto the neck of a bottle to hold it aloft.) To release the cake, run an offset metal spatula along the inner and outer edges of the pan. Sprinkle the wire rack with a little confectioners' sugar so the cake will not stick, and place the cake on the rack. Sprinkle slices with confectioners' sugar before serving.
**_Baked Note_**
You are either a chiffon person or not. Chiffon, like angel food cake, is butterfree, and the cake has a springy crumb that people tend either to adore or abhor. For those who like it, nothing is better for a summer soiree than a light chiffon.
### **CHOCOLATE COFFEE CAKE WITH DARK CHOCOLATE GANACHE**
THE CHOCOLATE COFFEE CAKE IS PURE EGO. It is a basic Baked creation that does not readily fit into any regional or historical category. It is a perennial customer favorite, and the recipe is requested often enough to demand inclusion in this book. The chocolate cake is dark, moist, and inviting, while the coffee buttercream filling and frosting is a bit adult without leaving an extreme coffee aftertaste. The entire three-layer affair is dressed in a thick, drippy, sexy chocolate ganache and studded with chocolate-covered espresso beans. It looks like fall and tastes like heaven. If you are coffee-averse, I promise you will enjoy this cake nonetheless. The coffee flavor is subtle and smooth, but necessary, in that it contrasts perfectly with the dark chocolate.
**YIELD: ONE 8-INCH, 3-LAYER CAKE**
**_Ingredients_**
FOR THE CLASSIC CHOCOLATE CAKE
_¾ cup dark unsweetened cocoa powder (like Valrhona)_
_⅔ cup sour cream_
_2 ⅔ cups all-purpose flour_
_2 teaspoons baking powder_
_1 teaspoon baking soda_
_½ teaspoon salt_
_¾ cup (1½ sticks) unsalted butter, cut into 1-inch cubes, softened_
_½ cup vegetable shortening_
_1½ cups granulated sugar_
_1 cup firmly packed dark brown sugar_
_3 large eggs, at room temperature_
_1 tablespoon pure vanilla extract_
FOR THE COFFEE BUTTERCREAM
_1½ cups granulated sugar_
_⅓ cup all-purpose flour_
_1½ cups whole milk_
_⅓ cup heavy cream_
_1½ cups (3 sticks) unsalted butter, soft but cool, cut into small pieces_
_1 teaspoon pure vanilla extract_
_3 tablespoons coffee extract_
FOR THE CHOCOLATE GLAZE
_8 ounces good-quality (60 to 72%) dark chocolate, coarsely chopped_
_¾ cup (1½ sticks) unsalted butter, softened and cut into ½-inch pieces_
_1 tablespoon light corn syrup_
ASSEMBLY
_10 to 12 chocolate-covered espresso beans_
#### **MAKE THE CLASSIC CHOCOLATE CAKE**
Preheat the oven to 325 degrees F. Butter three 8-inch round cake pans, line them with parchment paper, and butter the parchment. Dust the parchment with flour and knock out the excess flour.
In a medium bowl, mix the cocoa powder and sour cream with 1¼ cups hot water and set aside to cool.
In a large bowl, sift the flour, baking soda, baking powder, and salt together and set aside.
Using a standing mixer fitted with the paddle attachment, beat the butter and shortening together on medium speed until light and fluffy, about 5 minutes—the mixture will appear to string or ribbon throughout the bowl. Add the sugars and beat on medium speed until light and fluffy, about 5 more minutes. Add the eggs, one at a time, mixing about 10 to 15 seconds after each addition until the egg is incorporated into the mixture. Then turn the mixer to low, add the vanilla, and beat until incorporated. Scrape down the sides of the bowl and mix again for 30 seconds.
Beginning with the dry ingredients, add the dry mixture and the cocoa mixture to the mixer bowl in three alternating parts, ending with dry.
Divide the batter among the prepared pans. Use an offset spatula to level the batter. Bake the cakes for 35 to 40 minutes, rotating the pans halfway through the baking time, until a toothpick inserted in the center comes out clean. Transfer the pans to a wire rack and cool for 30 to 45 minutes. Turn the cakes out onto the rack and let them cool completely. Remove the parchment.
#### **MAKE THE COFFEE BUTTERCREAM**
In a medium, heavy-bottomed saucepan, whisk the sugar and flour together. Add the milk and cream and cook over medium heat, whisking occasionally, until the mixture comes to a boil and has thickened, about 10 to 15 minutes.
Transfer the mixture to the bowl of a standing mixer fitted with the paddle attachment. Beat on high speed until cool (this takes about 7 to 9 minutes of mixing; however, you can speed up the process by pressing bags of frozen berries or frozen corn around the sides and bottom of the mixing bowl). Reduce the speed to low and add the butter; mix until thoroughly incorporated. Increase the speed to medium-high and beat until the frosting is light and fluffy, about another 1 to 2 minutes.
Add the vanilla and coffee extracts and continue mixing until combined. If the frosting is too soft, put the bowl in the refrigerator to chill slightly, then beat again until it is the proper consistency. If the frosting is too firm, set the bowl over a pot of simmering water and beat with a wooden spoon until it is the proper consistency.
#### **TO ASSEMBLE THE CAKE**
Place one cake layer on a serving platter. Trim the top to create a flat surface, and evenly spread about 1¼ cups frosting on top. Add the next layer, trim and frost it, then add the third layer. Spread a very thin layer of frosting over the sides and top of the cake and put it in the refrigerator for about 15 minutes to firm up. (This is known as crumb coating and will help to keep loose cake crumbs under control when you frost the outside of the cake.) Spread the sides and top of the cake with the remaining frosting. Refrigerate it for 15 minutes to it firm up.
#### **MAKE THE CHOCOLATE GLAZE**
Place the chocolate, butter, and corn syrup in the top of a double boiler. Using a rubber spatula, stir the mixture until the chocolate and butter are completely melted and smooth.
Remove the pan from the heat and stir the glaze to release excess heat. Drizzle glaze over the cake. Refrigerate the cake for about 15 minutes to set the glaze before serving.
#### **GLAZE THE CAKE**
Line a rimmed baking sheet with parchment paper. Place your cake on a wire rack over the baking sheet. Slowly pour about ¾ cup of the glaze over the cake. Use a small offset spatula to smooth it out to the edges. Place the cake in the refrigerator for 5 minutes to set the glaze. Remove from the refrigerator and slowly pour the rest of the glaze over the cake. It should run down the edges in thick streams. You should be able to control the size and length of the streams by the pour. Feel free to experiment, and have no fear in playing around. This is the fun part, and there is no right or wrong way. Garnish with chocolate-covered espresso beans. Chill the entire cake for approximately 20 minutes, or until glaze is set, then transfer to cake plate. Serve at room temperature.
The cake can be stored, covered in a cake dome or cake saver, at room temperature for up to 3 days.
**_Baked Note_**
There is no road map to creating the perfect chocolate ganache topping. The design (i.e., the width and number of drips) is entirely up to you. I did a fairly light ganache topping for this particular photo. For heavier ganache coverage, double the ganache recipe, pour half over the cake, and refrigerate it for 15 minutes. Then pour the remaining ganache on top.
### **TOMATO SOUP CUPCAKES WITH MASCARPONE FROSTING**
THE TOMATO SOUP CUPCAKE DESERVES SOME EXPLANATION. Though the unusual main ingredient adds a hint of kitsch to the recipe, it makes for a moist and easy-to-put-together cupcake. Additionally, it brings an interesting flavor to the over-saturated cupcake market (don't worry—the tomato taste is not prominent). If you're tired of the same vanilla-frosted chocolate cake, this recipe merits your attention. Tomato soup cake does not dominate a certain region, though it seems to have touched a nerve with a great many Midwestern grandmothers (according to our nonscientific study). This recipe is based on one that originally served as the foundation for a spice cake. We tinkered with it to tone down the spice so that the tang provided by the tomato soup is not completely overwhelmed. Feeling adventurous? Try this recipe.
**YIELD: 24 CUPCAKES**
**_Ingredients_**
FOR THE TOMATO SOUP CUPCAKES
_2 (10¾-ounce) cans condensed tomato soup, preferably low-sodium_
_1 teaspoon baking soda_
_3½ cups all-purpose flour_
_1½ teaspoons cinnamon_
_½ teaspoon freshly grated nutmeg_
_½ teaspoon ground allspice_
_¼ teaspoon salt_
_1 teaspoon baking powder_
_¾ cup (1½ sticks) unsalted butter, at room temperature_
_1 cup granulated sugar_
_1 cup firmly packed light brown sugar_
_4 large eggs_
FOR THE MASCARPONE FROSTING
_¾ cup (1½ sticks) unsalted butter, softened_
_12 ounces mascarpone cheese, softened_
_4 cups confectioners' sugar, sifted_
_1 teaspoon pure vanilla extract_
#### **MAKE THE TOMATO SOUP CUPCAKES**
Preheat the oven to 325 degrees F. Line two 12-cup cupcake pans with paper liners.
In a large bowl, sprinkle the baking soda over the top of the tomato soup and stir well. Set aside.
In a medium bowl, sift together the flour, cinnamon, nutmeg, allspice, salt, and baking powder.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and both sugars together on medium speed until fluffy, 3 to 4 minutes. Add the eggs, one at a time, and beat until just combined. Scrape down the sides and bottom of the bowl and beat for a few seconds. Turn the mixer to low. Add the flour mixture in three parts, alternating with the tomato soup, beginning and ending with the flour mixture. Scrape down the bowl again, and mix on low speed for a few more seconds.
Fill the prepared cupcake pan about three-quarters full. Bake the cupakes for 25 to 28 minutes, or until a toothpick inserted in the center of a cupcake comes out clean.
Allow the cupcakes to cool for 30 minutes in the pan, then turn them out onto wire racks to cool completely.
#### **MAKE THE MASCARPONE FROSTING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the softened butter until it is completely smooth. Add the mascarpone and beat until combined.
Add the sugar and vanilla and beat until smooth. Be careful not to overbeat; this will cause the frosting to lose structure. (At this point, if you want to, you can tightly cover the frosting and refrigerate it overnight. Let it soften at room temperature before using.)
#### **TO ASSEMBLE THE CUPCAKES**
If you have a pastry bag, simply fit it with the largest tip, fill the bag with frosting, and pipe enough over each cake to cover the cupcake with a big mound. If you do not have a pastry bag, use an ice cream scoop with a release mechanism to scoop the frosting and dispense it onto the top of the cupcake. You can also use an offset spatula to frost the cupcakes.
Refrigerate any leftover cupcakes in an airtight container for up to 3 days. Bring the cupcakes to room temperature before serving.
**_Baked Note_**
Feeling really adventurous? In the mood for something even a little strange? Try making these cupcakes a play on sweet/savory. Omit the cinnamon, nutmeg, and allspice; replace them with 1 teaspoon freshly ground pepper and increase the salt to ½ teaspoon.
### **MISSISSIPPI MUD PIE (B), AKA MUDDY MISSISSIPPI CAKE**
THOUGH MISSISSIPPI MUD PIE WAS A STAPLE OF SOUTHERN MENUS DURING MY COLLEGIATE YOUTH, THERE IS NO REAL INDICATION THAT THIS DESSERT WAS CREATED IN MISSISSIPPI (OR EVEN THE SOUTH). To further complicate matters, it seems that Mississippi mud pie (or cake) varies widely in interpretation and means many things to many people. If I had to identify the characteristics of a typical Mississippi mud, I'd say it is a very dense, very sweet chocolate cake. Fudgy comes to mind. It is also probably covered or made with marshmallows and topped with pecans and chocolate sauce. The usual Mississippi mud is far too sweet and strangely dense for my taste buds these days, so I created a dreamier, more elegant version. I bake a flourless chocolate cake inside a cookie crust and top it with a layer of silky chocolate pudding and whipped cream. It is, by far, the Baked staff favorite.
**YIELD: ONE 9-INCH ROUND CAKE**
**_Ingredients_**
FOR THE CHOCOLATE COOKIE CRUST
_16 ounces chocolate sandwich cookies such as Oreos (35 to 40 cookes), crushed_
_5 tablespoons unsalted butter, melted_
FOR THE FLOURLESS CHOCOLATE CAKE
_4 tablespoons (½ stick) unsalted butter_
_6 ounces good-quality dark chocolate (60 to 70%), chopped_
_2 tablespoons plus 1 teaspoon instant espresso powder_
_¼ cup strong coffee, at room temperature_
_¼ teaspoon salt_
_1 tablespoon pure vanilla extract_
_6 large eggs, separated, at room temperature_
_1 cup sugar_
FOR THE CHOCOLATE PUDDING
_¾ cup sugar_
_½ cup dark unsweetened cocoa powder (like Valrhona)_
_¼ cup cornstarch_
_¼ teaspoon salt_
_4 large egg yolks_
_2½ cups whole milk_
_3 tablespoons unsalted butter_
_2 teaspoons pure vanilla extract_
_3 ounces good-quality dark chocolate (60 to 70%)_
ASSEMBLY
_Simple Whipped Cream_
#### **MAKE THE CHOCOLATE COOKIE CRUST**
Preheat the oven to 300 degrees F. Lightly spray a 9-inch springform pan with nonstick cooking spray. Line the pan with parchment paper and lightly spray the parchment and sides of the pan.
In a food processor, grind the cookies to a very fine crumb. You should have about 3½ cups. Put the crumbs in a small bowl. Pour the melted butter over them and mix with a rubber spatula until well combined.
Turn the crumb mixture into the prepared pan and press it into the bottom and up the sides, leaving about ½ inch between the top of the crust and the top of the pan. Use the back of a large spoon to get an even layer of crust. Place the pan in the freezer and let the crust set for about 10 minutes.
Bake the crust in the oven until it is dry to the touch, about 10 minutes. Transfer the pan to a wire rack and let cool.
#### **MAKE THE FLOURLESS CHOCOLATE CAKE**
Increase the oven temperature to 350 degrees F.
Using a double boiler or microwave, melt the butter and chocolate together. Set aside to cool.
In a small bowl, whisk together the espresso powder, coffee, salt, and vanilla. Set aside.
In the bowl of a standing mixer fitted with the whisk attachment, beat the egg yolks with ½ cup of the sugar until the mixture is light and has almost doubled in volume, about 5 minutes. Add the chocolate mixture and beat until just combined. Scrape down the sides and bottom of the bowl and mix on low speed for 5 seconds. Add the coffee mixture and beat until just combined. Scrape down the sides and bottom of the bowl and mix on low for 5 seconds.
In a clean bowl fitted with the whisk attachment (or you can elect to do this step by hand if you are feeling strong), beat the egg whites until foamy. Gradually increase the speed to high and add the remaining ½ cup sugar, beating until soft peaks form.
Scoop 1 cup of the egg whites into the chocolate mixture. Use a rubber spatula to gently fold in the egg whites. After about 30 seconds of folding, add the remaining egg whites and continue folding until they are almost completely combined. Do not rush the folding process, work gently, and take care not to overmix. Pour the batter onto the cooled cookie crust and bake for 38 to 42 minutes, until the cake is set but still jiggles slightly. It might not appear to be completely cooked. Transfer it to a wire rack and cool completely. (As it cools, the cake will deflate in the center and look sunken. Do not despair, this is just the way it settles.) Tightly wrap and refrigerate the cake for at least three hours or overnight.
#### **MAKE THE CHOCOLATE PUDDING**
In a medium saucepan, whisk together the sugar, cocoa powder, cornstarch, and salt. Add the egg yolks and whisk until combined. The mixture will look like a thick paste. Slowly pour in the milk, whisking constantly.
In a saucepan over medium heat, bring the mixture to a boil, whisking constantly to prevent it from burning on the bottom of the pan. Boil for 30 seconds, then transfer it to a medium bowl. Add the butter, vanilla, and chocolate and whisk until combined. Continue to whisk for a few more minutes to cool the mixture slightly. Let the pudding stand for 15 minutes at room temperature. Press a piece of plastic wrap directly onto the surface of the pudding to prevent a skin from forming, and chill it for at least 3 hours.
#### **TO ASSEMBLE THE MISSISSIPPI MUD PIE**
Stir the pudding to loosen it, then pour it on top of the cake, making sure to stay inside the cookie-crust border. Use an offset spatula to spread the pudding into an even layer. Return the cake to the refrigerator for at least 30 minutes while you prepare the whipped cream topping. Spread whipped cream across the pudding layer, all the way out to the sides, unmold the cake, and serve it immediately.
The cake can be kept, covered, in the refrigerator for up to 2 days.
**_Baked Note_**
This is an easy, though many-stepped, recipe. Don't fear, just break up the parts over the course of two days. Make the cookie crust and cake on day one, and make the pudding on the day you're going to serve the dessert. Keep in mind that the cake requires 3 hours to set before it can be cut. The whipped cream topping can be made 15 minutes before serving.
* * *
_1¼ cups heavy cream
2 tablespoons granulated sugar_
Pour the cream into a chilled metal bowl and beat with a chilled whisk for about 1 minute or until soft peaks form. Sprinkle the sugar on the cream and continue whisking vigorously until stiff peaks form.
Yield: about 2 cups whipped cream
* * *
### **MAPLE CUPCAKES WITH MAPLE CREAM CHEESE FROSTING**
THOUGH IT MAY BE A CLICHÉ IN THIS COUNTRY, VERMONT IS VIRTUALLY SYNONYMOUS WITH MAPLE SYRUP. Maple festivals figure heavily into the Vermont calendar, and all manner of maple merchandise (aprons, magnets, candies, cookbooks . . .) is available for purchase in the retail shops that line many of Vermont's historic towns. Real maple syrup deserves its reputation. It gives these cupcakes a great robust flavor, moist crumb, and golden brown color, while the cream cheese frosting with a touch of maple syrup provides the perfect accompaniment. The cupcakes are truly autumnal, but I encourage you to make them during any season you please.
**YIELD: 24 CUPCAKES**
**_Ingredients_**
FOR THE MAPLE CUPCAKES
_3 cups all-purpose flour_
_3 teaspoons baking powder_
_1 teaspoon salt_
_½ cup (1 stick) butter, slightly softened, cut into chunks_
_2 tablespoons vegetable shortening, at room temperature_
_2 cups pure maple syrup (I use grade B to bake with but any grade will suffice)_
_3 egg yolks_
_1 large egg_
_1¼ cups whole milk_
_1 cup walnuts, toasted and coarsely chopped_
FOR THE CREAM CHEESE MAPLE FROSTING
_¾ cup (1½ sticks) unsalted butter, softened_
_12 ounces cream cheese, softened_
_4 cups confectioners' sugar, sifted_
_2 tablespoons maple syrup_
ASSEMBLY
_Whole toasted walnuts (optional)_
#### **MAKE THE MAPLE CUPCAKES**
Preheat the oven to 325 degrees F. Line two 12-cup cupcake pans with paper liners.
In a medium bowl, sift together the flour, baking powder, and salt.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and shortening until ribbonlike. Turn the mixer to low and stream in the maple syrup. Increase the speed to medium-high and beat until the mixture is nearly uniform in color, about 3 minutes.
Add the egg yolks and egg, one at a time, and beat until just incorporated. Scrape down the sides and bottom of the bowl. Add half of the flour mixture and mix on low speed until incorporated. Stream in the milk. Stop the mixer, add the rest of the flour, then turn the mixer on until just combined. Scrape down the sides and bottom of the bowl and fold in the walnuts.
Fill the prepared cupcake pan about three-quarters full. Bake the cupcakes for 20 to 25 minutes, rotating the pans halfway through the baking time, until a toothpick inserted in the center of a cupcake comes out clean. Note: These cupcakes take longer to bake than traditional cupcakes due to the maple syrup.
Allow the cupcakes to cool for 15 minutes in the cupcake pan, then turn them out onto wire racks to cool completely.
#### **MAKE THE CREAM CHEESE MAPLE FROSTING**
In the bowl of a standing mixer fitted with the paddle attachment, beat the softened butter until it is completely smooth. Add the cream cheese and beat until combined.
Add the sugar and the maple syrup and beat until smooth. Be careful not to overbeat the frosting or it will lose structure. (At this point, if you want to, you can tightly cover the frosting and refrigerate it for a day. Let it soften at room temperature before using.)
#### **ASSEMBLE THE CUPCAKES**
There are many ways to frost a cupcake. If you have a pastry bag, simply fit with the largest tip, fill the bag with frosting, and pipe enough to cover the cupcake in a big mound. If you do not have a pastry bag, use an ice cream scoop with a release mechanism to scoop the frosting and dispense it onto the top of the cupcake. You can also use an offset spatula to frost the cupcakes. Top with whole toasted walnuts.
Refrigerate any leftover cupcakes in an airtight container for up to 3 days. Bring cupcakes to room temperature before serving.
**_Baked Note_**
Do not, I repeat, do not use imitation maple syrup in this recipe. Actually, avoid imitation maple syrup at all times. It is usually composed of corn syrup and food coloring and, sadly, contains very little, if any, real maple syrup. In short, it's hard to think of a more disingenuous grocery store product.
### **SUNDAY NIGHT CAKE**
THE WONDERFUL SUNDAY NIGHT CAKE, AS ITS NAME IMPLIES, IS A GREAT WAY TO END A WEEKEND. It is a no-frills affair—no layers to fill, no curds to set, no egg whites to whip. It is just an exceedingly pleasing, simple, and delicious gently spiced sour cream cake with easy chocolate frosting. The theme behind Sunday Night Cake is ease and comfort. Therefore, if your guests are seeking something more complicated, like a six-layer cake or a roulade, they should be directed elsewhere or asked to return on a different night (say, Friday). Our pastry chef, Eric Wolitzky, stumbled upon this cake while entering his Edna Lewis phase. Although he loves the original recipe and Edna Lewis, the grande dame of Southern cooking, his version is decidedly more Baked. Eric toyed with the crumb to make it springier, and he covered the cake in a puddinglike chocolate frosting. Even so, it is still faithful in spirit to the beloved original.
**YIELD: ONE 9-INCH SQUARE CAKE**
**_Ingredients_**
FOR THE CAKE
_1¾ cups cake flour_
_2 teaspoons baking powder_
_1 teaspoon salt_
_½ teaspoon cinnamon_
_10 tablespoons (1¼ sticks) unsalted butter, at room temperature, cut into ½-inch pieces_
_¾ cup granulated sugar_
_½ cup firmly packed light brown sugar_
_3 large eggs_
_1 cup sour cream_
FOR THE CHOCOLATE FROSTING
_¾ cup plus 2 tablespoons granulated sugar_
_3½ tablespoons cornstarch_
_2 tablespoons dark unsweetened cocoa powder (like Valrhona)_
_3 ounces good-quality unsweetened chocolate, coarsely chopped_
_6 tablespoons (¾ stick) unsalted butter, cut into ½-inch pieces, at room temperature_
#### **MAKE THE CAKE**
Preheat the oven to 350 degrees F. Line a 9-inch square cake pan with parchment paper and butter the sides and bottom of the parchment paper.
In a large bowl, sift together the flour, baking powder, salt, and cinnamon. Set aside.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and both sugars on medium speed until light and fluffy, about 2 minutes. Add the eggs, one at a time, and beat until just incorporated. Scrape down the sides and bottom of the bowl and beat for a few seconds. Turn the mixer to low. Add the flour mixture in three parts, alternating with the sour cream, beginning and ending with the flour mixture. Scrape down the bowl and beat for a few more seconds.
Pour the batter into the prepared pan and bake for 35 to 40 minutes, or until a toothpick inserted in the center comes out clean. Set the pan on a wire rack to cool for at least 20 minutes, loosen the sides of the cake from the pan then turn the cake out onto the rack. Remove the parchment and flip the cake right side up. Let the cake completely cool.
#### **MAKE THE CHOCOLATE FROSTING**
In a medium saucepan, whisk together the sugar, cornstarch, and cocoa powder. Add the chopped chocolate. Pour 1 cup boiling water into the pan, wait 30 seconds, then whisk until the mixture is combined and the chocolate is melted.
Turn the heat to medium-high and whisk continuously for about 5 minutes, or until the mixture begins to thicken. (Once pudding begins to thicken, it will come together very quickly.)
Remove the pan from the heat and pour the mixture into the bowl of a standing mixer fitted with the paddle attachment. Beat on high until the steam escapes and the mixture is room temperature. Add the butter and mix for an additional 2 to 3 minutes, until the frosting is light and puddinglike. If you prefer a fluffier, more spreadable frosting, continue to mix for a few minutes longer.
Frost the top of the cake, allowing a little of the frosting to drip down the edges. Chill for 5 minutes to set the frosting. Serve immediately.
The cake can be stored, tightly covered, in the refrigerator for up to 3 days. Bring it back to room temperature before serving.
**_Baked Note_**
If you want your Sunday Night Cake to be truly hassle free, you can opt out of making the frosting (although we really, really like the frosting) and sprinkle the top of the cake with some confectioners' sugar instead.
### **AUNT SASSY CAKE**
A DESSERT THAT LINGERS IN MY SUBCONSCIOUS ON A NEAR-CONSTANT BASIS IS BAKED'S AUNT SASSY CAKE. I have an unhealthy attachment to this cake. The light cake layers are studded with a heap of crushed pistachios, and the entire thing is filled and frosted with whipped honey vanilla buttercream. It is a dream cake, the kind you will make once every now and then but remember often. It is not a one-bowl, whip-it-up-in-an-hour cake, and it is limited to events where nut allergies are not a problem. Oh, about the name: It shall forever remain cloaked in secrecy. Suffice it to say that neither of us actually has an Aunt Sassy.
**YIELD: ONE 8-INCH, 3-LAYER CAKE**
**_Ingredients_**
CAKE
_1 cup shelled pistachios_
_2½ cups cake flour_
_¾ cup all-purpose flour_
_1 tablespoon baking powder_
_1 teaspoon baking soda_
_¾ teaspoon salt_
_½ cup (1 stick) unsalted butter, softened_
_½ cup vegetable shortening_
_1¾ cups sugar_
_1 tablespoon pure vanilla extract_
_1 large egg_
_3 large egg whites, at room temperature_
_¾ teaspoon cream of tartar_
HONEY VANILLA BUTTERCREAM
_1½ cups sugar_
_⅓ cup all-purpose flour_
_1½ cups whole milk_
_⅓ cup heavy cream_
_1½ cups (3 sticks) unsalted butter, soft but cool, cut into small pieces_
_1 teaspoon pure vanilla extract_
_3 tablespoons honey_
ASSEMBLY
_⅓ cup crushed shelled pistachios_
#### **MAKE THE CAKE**
Preheat the oven to 325 degrees F. Butter three 8-inch round cake pans, line the bottoms with parchment paper, and butter the parchment. Dust the parchment with flour and knock out the excess flour.
In the bowl of a food processor, pulse the pistachios until they are coarsely chopped. Transfer about 2 tablespoons' worth of the coarse pistachios to a large bowl. Continue to process the rest of the pistachios until they are almost powdery—but not a superfine dust. Stir the pistachio powder into the reserved coarse pistachios. Sift the flours, baking powder, baking soda, and salt together over the large bowl containing the pistachio mix. Stir to combine.
In the bowl of a standing mixer fitted with the paddle attachment, beat the butter and shortening on medium speed until creamy, 3 to 4 minutes. Add the sugar and vanilla and beat until fluffy, about 3 minutes. Scrape down the bowl, add the whole egg, and beat until just combined. Turn the mixer to low.
In a measuring cup, make 1½ cups ice water. Add the flour mixture to the mixer in three parts, alternating with the ice water, beginning and ending with the flour mixture. For each addition, turn the mixer to low to add ingredients, then up to medum speed for a few seconds until incorporated. Scrape down the bowl, then mix on low speed for a few more seconds.
In a medium bowl, whisk the egg whites and cream of tartar until soft peaks form (You can do this by hand. Don't be intimidated, it should only take 2 to 3 minutes). Do not overbeat. Gently fold the egg whites into the batter.
Divide the batter among the prepared pans and smooth the tops. Bake for 40 to 45 minutes, rotating the pans halfway through the baking time, until a toothpick inserted in the center of the cake comes out clean. Transfer the pans to a wire rack and let cool for 20 minutes. Turn the cakes out onto the rack and let cool completely. Remove the parchment paper.
#### **MAKE THE HONEY VANILLA BUTTERCREAM**
In a medium, heavy-bottomed saucepan, whisk the sugar and flour together. Add the milk and cream and cook over medium heat, whisking occasionally, until the mixture comes to a boil and has thickened, about 10 to 15 minutes.
Transfer the mixture to the bowl of a standing mixer fitted with the paddle attachment. Beat on high speed until cool (this takes at least 7 to 9 minutes of mixing; you can speed up the process by pressing bags of frozen berries or frozen corn against the sides and bottom of the mixing bowl). Reduce the speed to low and add the butter; mix until thoroughly incorporated. Increase the speed to medium-high and beat until the frosting is light and fluffy, 1 to 2 minutes.
Add the vanilla and honey and continue mixing until combined. If the frosting is too soft, put the bowl in the refrigerator to chill slightly, then beat again until it is the proper consistency. If the frosting is too firm, set the bowl over a pot of simmering water and beat with a wooden spoon until it is the proper consistency.
#### **ASSEMBLE THE CAKE**
Place one cake layer on a serving platter. Trim the top to create a flat surface, and evenly spread about 1¼ cups frosting on top. Add the next layer, trim and frost it, then add the third layer. Spread a very thin layer of frosting over the sides and top of the cake and put it in the refrigerator for about 15 minutes to firm up. (This is known as crumb coating and will help to keep loose cake crumbs under control when you frost the outside of the cake.) Spread the sides and top of the cake with the remaining frosting. Garnish the cake with crushed pistachios and refrigerate it for 15 minutes to it firm up before serving.
This cake will keep beautifully in a cake saver at room temperature for up to 3 days, if the weather is cool and humidity free. Otherwise, put it in a cake saver and refrigerate it for up to 3 days. Let the cake sit at room temperature for at least 2 hours before serving.
**_Baked Note_**
Unfortunately and fortunately, I have never been able to control myself around pistachios. I suppose, left unchecked, I could devour my weight in pistachios in a few hours. That said, I have a few friends who seem unaware that nuts can go stale rather quickly (and this has created some slightly uncomfortable after-dinner moments). Left at room temperature, pistachios should be eaten within a few days to a week at the latest from purchase.check the package for specific freshness information. They can be stored in an airtight container in the refrigerator for up to 2 months.
##
**BUCKEYES**
**CHOCOLATE PEANUT BUTTER FONDUE**
**BANANAS FOSTER FRITTERS**
**MARSHMALLOW CHOCOLATE CUPS**
**COFFEE ICE CREAM**
**STRAWBERRY JELL-O SALAD**
**THE NO-BAKE PEANUT BUTTER COOKIE**
**SIMPLE BLUEBERRY PARFAITS**
**CARAMEL POPCORN WITH PEANUTS AND CHOCOLATE**
**CLASSIC CARAMEL SAUCE ( FOR COFFEE CAKE AND THE LIKE)**
**VANILLA BEAN AND CHOCOLATE BUDINO**
**MERINGUE MUSHROOMS, OR SHANDI'S CANDIES**
**SOFT CANDY CARAMELS**
**CHOCOLATE HAZELNUT SPREAD**
My family had a candy bowl, not a dish. It was nothing extraordinary, just a nondescript and nicked soup bowl filled with candy. At various times, the bowl held Jolly Ranchers, lollipops, M&Ms, individually wrapped caramels, and Hershey's Kisses (by far the most common bowl filler). During the holidays, the ordinary candy bowl was whisked away and replaced with a fanciful red and green holiday dish, but it reappeared on December 26. I loved that candy bowl.
As expected, on a few occasions, I overindulged. I still do. I am a recovering candy addict, and I no longer leave myself open to temptation by setting out extraneous candy bowls, dishes, or multipack candy bar minis. Now, when I have a hankering for candy, I make it instead. The process, like baking, is soothing; plus, lately I have a thing for candy thermometers and chemistry.
At the moment, I am an amateur candy maker with great aspirations. I am happy to report that while I was researching this book, many people submitted recipes that showed a dizzying grasp of the science involved. Like the hand-written, almost philosophical digression on spun sugar. No, that didn't make this book. Quite simply, those recipe suggestions were more ambitious than I had predicted. I intend to attempt a spun sugar cage shortly, but I need to develop my candy-making skills further. In the meantime, I've included a few regional favorites. Buckeyes are known for both their kid-friendly approach and addictive combo of peanut butter, chocolate, and graham, while the Marshmallow Chocolate Cups are a riff on an old-time candy. Making those dark-chocolate cups filled with marshmallow gave me a feeling of accomplishment that I'd imagine is usually reserved for worldly chocolatiers.
This is also the chapter where I stuffed in some of the more interesting—er, stuff that defies a label—items. I can honestly say that Strawberry Jell-O Salad, which is not really a salad, is one of the strangest but most popular items I made. My taste testers still ask when I might be making that "Jell-O strawberry pretzel thing" again. I also had to find a place for the lovely banana fritters, which are easy and should be made more often. Finally, I recommend everyone take a crack at the Caramel Popcorn. It is sinful. And it makes the most amazing gift.
### **BUCKEYES**
I GREW UP WITH THIS DEEPLY REGIONAL CLASSIC CANDY. Fear not, there is no gauzy colored memory attached, it was just the one candy my mom (not exactly a friend of the kitchen) would make repeatedly because it was both quick and easy. And truly tasty. The buckeye is perhaps best known to those in Ohio and the surrounding states. Judging by the reactions of a sampling of friends, it is almost completely unknown on the coasts. I present you with a less sweet version than the one Mom used to make (less sweet is perhaps relative when discussing this candy—hers were quite literally sugar bombs), but I stayed true to the original peanut butter ball dipped in chocolate. Ohioans would be proud. And, yes, leave a little bit of the peanut butter exposed, so that the final candy modestly resembles the horse chestnut from the Buckeye State.
**YIELD: 36 TO 42 BUCKEYES**
**_Ingredients_**
_¼ cup cream cheese, softened_
_1½ cups peanut butter_
_1 cup graham cracker crumbs (about 14 graham crackers)_
_3 cups confectioners' sugar_
_10 tablespoons (1¼ sticks) unsalted butter, melted and cooled_
_12 ounces good-quality dark chocolate (60 to 72%), coarsely chopped_
#### **MAKE THE CANDY**
In the bowl of a standing mixer fitted with the paddle attachment, beat the cream cheese and peanut butter until combined. Add the graham cracker crumbs and beat on medium speed for 10 seconds. Add the confectioners' sugar and butter. Beat at low speed for 20 seconds to prevent the sugar from spilling over, then gradually increase the speed until the mixture is completely combined. Scrape down the sides and bottom of the bowl and beat again. The mixture will feel slightly dry. Set the peanut butter filling aside while you melt the chocolate.
In the top of a double boiler set over hot water, melt the chocolate, stirring frequently until it is completely smooth. Pour the chocolate into a small, deep bowl. Let it cool to tepid (about 100 degrees F, body temperature) while you shape the peanut butter centers.
#### **ASSEMBLE THE BUCKEYES**
Line a sheet pan with parchment paper. Scoop out slightly more than 1 tablespoon's worth of filling and use your hands to form it into a ball. (For uniform balls, use a medium-size melon baller or a very small ice cream scoop with a release mechanism.) Place the ball on the prepared sheet pan and repeat the process until all the filling has been shaped. The balls can sit fairly close to each other on the sheet, just make sure they are not touching.
One by one, using a fork or large skewer, dip each ball into the chocolate. Roll the ball around from side to side to cover almost the entire peanut butter center, leaving a small part uncovered. Manipulate the buckeye so that the dripping chocolate covers the holes made by the fork. Let the excess chocolate drip back into the bowl and return each chocolate-covered buckeye to the pan. Refrigerate the entire sheet pan for about 30 minutes to set the chocolate before serving.
Buckeyes will keep for up to 3 days, tightly covered, in the refrigerator.
**_Baked Note_**
This recipe is fun for large parties and big gatherings. Just pass them around like hors d'oeuvres—they make perfect finger food. Also, feel free to substitute chunky peanut butter for smooth peanut butter; however, be aware that all-natural versions do not work as well.
### **CHOCOLATE PEANUT BUTTER FONDUE**
I HAVE MANY ALLEGIANCES TO MANY CANDY BARS AND THEIR ILK. KIT KATS RANK HIGH. As do both plain and peanut M&Ms. And I never travel without a bag of Whoppers or Maltesers. However, I covet the Reese's Peanut Butter Cup above all else. The timeless combo of milk chocolate shell and peanut butter filling is about as boldly American as it gets (my European friends consider peanut butter a strange and unworthy experience). In the 1920s, Mr. Harry Burnett Reese created the candy. His appropriately named company, the H. B. Reese Candy Company, was eventually sold to Hershey's, but the recipe remains relatively unchanged. Our fondue is just another solid reminder that Baked celebrates chocolate and peanut butter unabashedly. Our preferred dipping accompaniment is a brownie—a Baked brownie—but marshmallows, apples, pears, and celery work as well.
**ABOUT 2 CUPS**
**_Ingredients_**
_8 ounces good-quality milk chocolate, coarsely chopped_
_1½ cups smooth peanut butter_
_2 tablespoons heavy cream_
Place the milk chocolate in the top of a double boiler and stir occasionally until it is completely melted and smooth. Add the peanut butter and heavy cream and stir until combined.
Transfer the mixture to your favorite fondue pot and keep it warm. Serve with your choice of dipping items.
**_Baked Note_**
This recipe must be made with a highquality milk chocolate, like Valrhona, Callebaut, or Scharffen Berger. Some of the mass-market brands lack real cocoa butter and use almost no cacao, thereby affecting the consistency of your fondue.
### **BANANAS FOSTER FRITTERS**
AS WITH LIFE IN GENERAL, I HAVE NOTICED THAT MY COOKING LIFE HAS GONE THROUGH VARIOUS PHASES. There was the cake phase, the cookie phase, the anything chocolate and anything caramel phases, and perhaps inevitably, there was the frying phase. I dipped my toe in the frying waters ever so gently with these addictive banana fritters, and from there I started frying everything I could get my hands on: chicken, doughnuts, and dumplings. These fritters are quick, easy, and surprisingly tasty little snacks. The banana flavor really shines through. I last brought them out for an afternoon book-group/coffee-club meeting, and there were many requests for seconds. Treat the rum sauce as optional if you like, but it's well worth the effort. By the by, the rum sauce is awfully tasty on pancakes, French toast, and drizzled on hearty bread puddings.
**YIELD: 6 LARGE OR 10 SMALL FRITTERS**
**_Ingredients_**
FOR RUM DIPPING SAUCE
_4 tablespoons (½ stick) unsalted butter, cut into 1-inch cubes_
_⅓ cup firmly packed dark brown sugar_
_⅓ cup heavy cream_
_1 teaspoon banana liqueur (or pure vanilla extract)_
_2 tablespoons dark rum_
_Pinch cinnamon (optional)_
FOR THE FRITTER DOUGH
_1¼ cups all-purpose flour, plus more if needed_
_3 tablespoons firmly packed dark brown sugar_
_1 teaspoon baking powder_
_½ teaspoon salt_
_¼ teaspoon ground allspice_
_1 teaspoon cinnamon_
_3 medium ripe bananas_
_1 tablespoon dark rum_
_1 teaspoon banana liqueur or pure vanilla extract_
_1 tablespoon unsalted butter, melted and cooled_
_Vegetable oil for frying_
_¼ cup confectioners' sugar_
#### **MAKE THE RUM DIPPING SAUCE**
In a medium saucepan over medium heat, stir the butter and sugar together until smooth. Add the cream and bring to a boil. Remove the pan from the heat and stir in the liqueur, rum, and cinnamon, if using. Set aside until serving time. (The sauce can be made ahead. Allow it to cool, then cover it tightly in plastic wrap and refrigerate. Rewarm it over low heat in a saucepan or microwave it in short 15-second blasts before serving.)
#### **MAKE THE FRITTER DOUGH**
In a large bowl, whisk together the flour, brown sugar, baking powder, salt, all-spice, and ½ teaspoon of the cinnamon. Use your hands to rub the chunks of sugar into the flour mixture and whisk again (it is okay to have a few chunky sugar pieces remaining).
In another large bowl, mash the bananas with your hands or a heavy spoon and stir in the rum, liqueur, and butter.
Use a rubber spatula to fold the dry ingredients into the banana mixture. The mix should look wet, but it should still hold its shape when scooped into a small ball. If the dough is too thin, keep folding in flour 1 tablespoon at a time until it stiffens up. Refrigerate the dough while the frying oil heats and you prepare the sugar topping.
Pour enough oil into a deep skillet to fill it ¾ inch to 1 inch deep. Slowly heat the oil over medium-high heat until it registers 375 degrees F on a deep-frying thermometer.
While the oil heats, in a small bowl, whisk together the confectioners' sugar and remaining ½ teaspoon cinnamon. Set aside.
#### **FRY THE FRITTERS**
Line a plate with a double layer of paper towels and set it near your work area.
Using a small spatula and a small spoon, two spoons, or an ice cream scoop with a release mechanism, drop heaping spoonfuls of dough into the oil. Do not crowd the skillet. Cook until the fritters have browned on one side, 2 to 3 minutes. Using a slotted spoon or tongs, turn them over and continue to cook for another 2 minutes, or until browned. Do not overcook or burn the fritters. Use the slotted spoon to transfer the fritters to the prepared plate and continue frying dough until finished.
Place the fritters on a serving plate and sift the cinnamon sugar over them. Serve immediately with rum dipping sauce.
**_Baked Note_**
Do not be afraid to fry. It is not as scary as it may initially seem. To keep the mess and time to a minimum, follow these basic rules: 1) Use a high-sided skillet or pan. 2) Do not overcrowd the skillet, or the oil will drop in temperature and your fritters will absorb too much fat. 3) Cool the oil completely before discarding it.
### **MARSHMALLOW CHOCOLATE CUPS**
IT ALL STARTED WITH THE MALLO CUP. Specifically the one manufactured by the Boyer Candy Company in Altoona, Pennsylvania. It was my gateway candy. The fluffy marshmallow filling encased in a dark chocolate candy cup had me hooked from bite number one, and soon I was hoarding and hiding the precious Mallos throughout my room and in the back of the freezer. The Mallo is still decidedly a northeastern treat. It is greatly appreciated within a certain distance from Altoona, and virtually unknown outside of it. I, along with many Mallo enthusiasts, hope to convert the masses and show them the light. In fashioning my own version of the beloved treat, my recipe came to differ from the original in a few ways. It produces a smaller, bite-size candy; the proportion of chocolate to marshmallow is larger; and the shell is made of a mix of dark and milk chocolates. It is easy and fun to make and appropriately delicious. And if you are ever near Altoona, Pennsylvania, it pays to drop by the Boyer factory store for the real thing—I particularly enjoy the discounted "irregulars."
**YIELD: 40 TO 60 CANDIES**
**_Ingredients_**
FOR THE CHOCOLATE CANDY CUPS
_16 ounces good-quality dark chocolate (60 to 72%), coarsely chopped_
_8 ounces good-quality milk chocolate (such as Valrhona or Callebaut), coarsely chopped_
FOR THE MARSHMALLOW FILLING
_1 envelope (about 2½ teaspoons) unflavored gelatin_
_1 cup sugar_
_1 teaspoon pure vanilla extract_
_⅛ teaspoon salt_
#### **MAKE THE CHOCOLATE CANDY CUPS**
Arrange 30 miniature candy cups (approximately 1 inch in diameter) on a baking sheet. For stability's sake, I suggest using a double layer of cups for each candy (so you will need to buy a total of 60 cups to make 30 candies). This helps your chocolate cup to maintain its shape.
In a large nonreactive metal bowl, combine the chocolates. Set the bowl over a saucepan of simmering water and stir with a rubber spatula until the two chocolates have completely melted together and the mixture is smooth.
Remove the bowl from the simmering water and stir for about 15 seconds to release excess heat. Use either a small spoon or a pastry bag fitted with one of the smallest tips to fill the candy cups just under a quarter full with chocolate. Using a pastry brush, brush the chocolate from the bottom of each cup up the sides to completely cover the inside of the cup with chocolate. Place the cups in the refrigerator while you make the marshmallow filling. Set the remaining chocolate aside.
#### **PREPARE THE MARSHMALLOW FILLING**
Attach a small plain tube tip to a clean pastry bag and set aside.
In the bowl of a standing mixer, sprinkle the gelatin over ⅓ cup cold water.
In a small saucepan over medium heat, gently stir together the sugar and ¼ cup water. Stop stirring and put a candy thermometer in the saucepan. Bring the mixture to a boil over medium-high heat and cook, still without stirring, until it reaches the soft ball stage, 235 degrees F.
Remove the pan from the heat and slowly stream it into the gelatin. Whisk vigorously for about 30 seconds to release excess heat, then place the bowl on the standing mixer fitted with the whisk attachment and mix on medium-high speed for 5 minutes. Add the vanilla and salt and continue to whisk for about 2 minutes longer. You do not want to whisk the marshmallow to soft peaks; it should be slightly looser than that. Working quickly, pour the marshmallow filling into the prepared pastry bag.
#### **ASSEMBLE THE MARSHMALLOW CHOCOLATE CANDY CUPS**
Pipe the marshmallow directly into the chocolate cups, filling each one a bit more than three-quarters of the way full. Gently knock the pan to level the filling.
If the reserved chocolate has hardened, set it over simmering water to remelt it. Spoon a top layer of the chocolate onto the marshmallow filling to cover it, gently knock the pan again, and place the cups back in the refrigerator to completely set.
The candy cups will keep, covered, in the refrigerator for up to 4 days. Generally speaking, they can be enjoyed directly from the refrigerator or after a few minutes at room temperature, but they will begin to melt or bloom if left unchilled for too long.
* * *
**_Baked Notes_**
Do not, whatever you do, use supermarket chocolate chips for this recipe. Use highquality chocolate like Valrhona, Callebaut, or Scharffen Berger. These brands are usually available in block or bar form and can be found online or at most gourmet food stores. This recipe makes a lot of candies. You can cut the chocolate candy cup recipe in half, keep the marshmallow filling recipe intact, and use larger cup papers (do not use regular cupcake papers) with the end result being a higher ratio of marshmallow to chocolate.
### **COFFEE ICE CREAM**
THE REFRAIN IS FAMILIAR AND CONSISTENT. When people find out I own a bakery, they comment, "I could never own a bakery, I would be too tempted to eat everything all the time." I feel the exact same way about the owners and managers of ice cream parlors. I could easily eat a gallon of ice cream. It is such a problem for me that I often walk the streets of New York taking nonlinear routes to avoid certain ice cream outposts. As ice cream and coffee are two of my favorite vices, it seems natural that I like them even better together. At Baked, we are advocates of the Portland coffee roasters, Stumptown.
**YIELD: ONE QUART**
**_Ingredients_**
_6 egg yolks_
_1¾ cups heavy cream_
_2 cups whole milk_
_¾ cup sugar plus 2 tablespoons_
_1 teaspoon salt_
_3 tablespoons instant espresso powder_
_1 tablespoon Kahlúa_
#### **MAKE THE COFFEE ICE CREAM**
Put the egg yolks in a large heatproof bowl and set aside.
In a medium saucepan, stir together the heavy cream, milk, sugar, salt and instant espresso powder. Bring the mixture to a slow, consistent simmer (just this side of a boil) and remove from heat.
Whisk the egg yolks until just combined, then slowly stream in half of the hot coffee cream mixture while whisking constantly. Transfer the egg mixture back to the medium saucepan containing the other half of the coffee cream mixture. Heat over medium-low heat, stirring constantly, until the mixture is thick enough to coat the back of a spoon (about 175 degrees on an instant-read thermometer).
Remove from heat and strain the mixture through a fine-mesh sieve into a bowl. Whisk in the Kahlúa, and let mixture cool to room temperature.
Press a piece of plastic wrap directly onto the surface of the mixture to prevent a skin from forming. Refrigerate for 4 hours.
Pour into an ice cream machine and freeze, following the manufacturer's directions.
**_Baked Note_**
I religiously reiterate the following: Ground espresso is not the same as instant espresso. Ground espresso does not dissolve in liquids, and it produces baked goods and ice cream with a gritty texture while instant espresso dissolves completely resulting in smooth textured baked goods and ice cream. I use the Medaglia D'Oro brand of instant espresso powder, and a small jar will last you for many baking cycles.
### **STRAWBERRY JELL-O SALAD**
I WOULD BE REMISS IF I DIDN'T FEATURE THIS RECIPE IN OUR SECOND COOKBOOK OUTING. It was discussed with vigor, passion, and enthusiasm as I made my way around America soliciting regional recipe ideas and thoughts from home bakers. Oddly, Jell-O salad is nearly impossible to pin down to one region. Many locales lay claim to some form of the dessert. The confusion of ownership probably resides in the many twists and turns the recipe has made from generation to generation. Generally speaking, I found the Jell-O salad of the West to be more of a "mix it all together and add pineapple bits" affair, while the Jell-O salad of the East remains a clearly layered, strawberry only, dessert. This version is pretty much a facsimile of an original. A crunchy pretzel crust holds a layer of cream cheese frosting topped with thick strawberry Jell-O. In theory, it shouldn't work, but it does. I did not add any flourishes, and I did not attempt to dress it up for mass consumption. You are either a Jell-O Salad fan or a hater. Decide for yourself.
**YIELD: ONE 9-BY-13-INCH DESSERT OR
ABOUT 15 GENEROUS SERVINGS**
**_Ingredients_**
FOR THE JELL-O LAYER
_1 (6-ounce) package strawberry Jell-O_
_2 cups fresh strawberries (about_
_14 ounces), sliced_
FOR THE PRETZEL CRUST
_8 ounces salty pretzels (about 4 cups)_
_¾ cup (1½ sticks) unsalted butter, cut into ½-inch pieces_
_2 tablespoons firmly packed dark brown sugar_
FOR THE CREAM CHEESE LAYER
_1 (8 ounce package) cream cheese, at room temperature_
_1 cup sugar_
_¾ cup heavy cream_
#### **MAKE THE JELL-O LAYER**
Prepare the Jell-O according to the package directions. Add the sliced strawberries and refrigerate until it is partially set (this takes about 45 minutes to 1 hour depending on the depth and type of vessel you use to chill the Jell-O). Make the other parts of the recipe while you wait for the Jell-O to nearly set.
#### **MAKE THE PRETZEL CRUST**
Preheat the oven to 325 degrees F. Butter the sides and bottom of a 9-by-13-inch baking pan, or spray it with nonstick cooking spray.
In a food processor, pulse the pretzels into coarse crumbs. Do not overprocess into a powder—if anything, err on the side of having a few chunkier pretzel pieces (it's simply my preference). Alternatively, crush the pretzels with the bottom of a metal baking cup.
In a medium saucepan, melt the butter over low heat. Whisk in the brown sugar and remove from the heat. Stir in the pretzel pieces until combined and turn the mixture out into the prepared pan. Use your hands or the back of a large spoon to press the mixture into the bottom of the pan, but do not press it up the sides. Bake the crust for 10 minutes, then set it aside to cool completely.
#### **MAKE THE CREAM CHEESE LAYER**
Put the cream cheese and sugar in the bowl of a standing mixer fitted with the paddle attachment. Beat on medium speed until well combined.
In a clean bowl, whip the cream, either by hand or using the standing mixer with the whip attachment, until soft peaks form. Use a rubber spatula to fold the whipped cream into the cream cheese mixture.
#### **ASSEMBLE THE STRAWBERRY JELL-O SALAD**
Spread the cream cheese mixture over the cooled crust. Chill for 5 minutes in the refrigerator. Pour the almost-set Jell-O over the cream cheese layer to cover. Refrigerate the dessert for 2 about hours or more to set before serving.
The Jell-O salad can be stored, tightly covered and refrigerated, for up to 3 days.
**_Baked Note_**
Jell-O Salad requires you to leave your culinary criticism and classically honed foodie personality at the door. This illnamed dessert requires a completely open mind. Feel free to swap out the strawberry Jell-O for any other fruit flavor, like orange, lemon, or lime. I have yet to try a chocolate gelatin or a chocolate pudding layer in place of the strawberry layer, but I am fairly certain it will work just fine.
### **THE NO-BAKE PEANUT BUTTER COOKIE**
THERE SEEMS TO HAVE BEEN A MOMENT IN AMERICAN HISTORY WHEN EVERY MOTHER MADE A VERSION OF THIS NO-BAKE COOKIE FOR HER CHILDREN. A large number of the people contributing ideas to this book submitted a version of this dessert and they all claimed to have watched (or helped) Mom make it. Perhaps the ease with which this recipe comes together explains its ubiquity. It was something every mother could make, regardless of time constraints or baking ability. It is surprisingly good, and well textured. It is fully capable of inspiring a classic midnight craving—I found myself sneaking them from the fridge along with a glass of milk. I am hesitant to call it a cookie, though, as it isn't baked, but you can call it whatever you like.
**YIELD: ABOUT 36 COOKIES**
**_Ingredients_**
_½ cup whole milk_
_2 cups sugar_
_¼ cup dark unsweetened cocoa powder (like Valrhona)_
_½ cup (1 stick) butter, cut into ½-inch cubes, softened_
_1 cup chunky peanut butter_
_3 cups rolled oats_
_1 teaspoon pure vanilla extract_
Line two baking sheets with parchment paper.
In a heavy saucepan over medium heat, stir together the milk, sugar, cocoa powder, and butter until the butter is melted and the sugar is dissolved. Bring the mixture to a boil, stop stirring, and boil for a full 90 seconds. Remove the pan from the heat and add the peanut butter, oats, and vanilla. Stir until the mixture is combined.
Use a small ice scream scoop with a release mechanism or, alternatively, a tablespoon to drop the no-bake cookies onto the baking sheet (leave some room around them; they will spread). Let the cookies cool, then refrigerate them for at least 1 hour. They can be eaten directly from the refrigerator or at room temperature.
Store the cookies between layers of parchment paper in a tightly sealed container for up to 3 days.
**_Baked Note_**
Yes, you can replace the chunky peanut butter with smooth. And yes, you can use natural peanut butter instead—but if you do, add about ⅓ cup more oats to absorb the extra oil.
### **SIMPLE BLUEBERRY PARFAITS**
MAINE IS THE SINGLE LARGEST PRODUCER OF LOWBUSH BLUEBERRIES IN THE WORLD, AND THAT ALONE IS REASON ENOUGH TO VISIT THE STATE. Prime blueberry season, August through September, also happens to coincide exactly with the perfect time for a road trip to Maine. It is dry and cool and sunny, while the rest of the East Coast is humid, hot, and insufferable. I highly recommend a drive along the coast with many stop-offs at the roadside farm stands filled with blueberries. It's the perfect way to end a long, sticky summer. I buy them by the bagful, and September becomes a month of blueberry muffins, blueberry waffles, and blueberry parfaits.
**YIELD: 4 SERVINGS**
**_Ingredients_**
_4 to 6 ounces vanilla wafers, depending on how thick you want to make your wafer layer_
_2½ cups fresh blueberries_
_¼ cup Grand Marnier_
_1 teaspoon orange juice_
_¾ cup heavy cream_
_1 teaspoon confectioners' sugar, sifted_
_Freshly grated zest of 1 orange for garnish_
Choose four clear, wide-mouthed glasses for serving. Put the vanilla wafers in a food processor and pulse them into coarse crumbs.
In a small saucepan, combine 1¼ cups of the blueberries and the Grand Marnier. Cook over medium heat until the blueberries break down, about 5 minutes. Let them cool to room temperature. (If you're in a hurry, put the mixture in the refrigerator to cool it quickly.)
In the bowl of a standing mixer fitted with the whisk attachment, whip the cream on medium speed for 1 minute. Sprinkle the confectioners' sugar over the cream, turn the mixer to high, and beat until soft peaks form.
Add ¾ cup of the fresh blueberries and the orange juice to the room-temperature blueberry syrup and stir until combined.
Sprinkle about a tablespoon of the crumbled vanilla wafers to cover the bottom of each glass, top with a heaping tablespoonful of the blueberry mixture, then top that with a dollop of whipped cream. Continue to layer until you fill the glasses or until you run out of ingredients. Garnish with the remaining ½ cup blueberries and finish each parfait with a sprinkle of orange zest. Serve immediately.
**_Baked Note_**
Obviously, this book is not written for the diet-minded; however, this recipe can easily be given a healthier spin. Simply replace the whipped heavy cream with about 11/2 to 2 cups of yogurt. The dessert will be equally delicious, though the yogurt is slightly heavier.
### **CARAMEL POPCORN WITH PEANUTS AND CHOCOLATE**
BY AND LARGE, I HAVE NEVER BEEN DRAWN TO POPCORN. It is a snack that I rarely think about and only eat a bit reluctantly, if someone offers me his or her bucket at the movies. It was not until Renato introduced me to the voodoo-like nature of homemade caramel popcorn that I understood the allure. The otherwise average kernels turn into something else entirely. The salty peanuts, crunchy popcorn, and dark chocolate form an otherworldly snack that has the overall effect of encouraging epic consumption and borderline hysteria. Finally, since the yield on this recipe is rather large, we encourage you to get out cellophane bags and decorative bows and use a portion of your caramel popcorn for gifts. It makes a great present and will save you a few otherwise unavoidable calories.
**YIELD: 24 CUPS**
**_Ingredients_**
_1 cup unpopped kernels or 24 cups popped corn_
_1 cup (2 sticks) unsalted butter, cut into chunks_
_2 cups firmly packed dark brown sugar_
_½ cup light corn syrup_
_2 tablespoons unsulfured molasses_
_½ teaspoon salt_
_¾ teaspoon baking soda_
_1½ teaspoons pure vanilla extract_
_1¼ cups salted peanuts_
_8 ounces good-quality milk chocolate, melted and tempered (seethis page)_
_8 ounces good-quality dark chocolate (60 to 72%), melted and tempered (seethis page)_
If you are starting with kernels, pop them using any method you prefer and let them cool. (We like a hot-air popper.) Preheat the oven to 250 degrees F. Place the popcorn in a large roasting pan. Line a sheet pan with parchment paper.
In a medium saucepan over low heat, start to melt the butter. Add the brown sugar, corn syrup, and molasses, and stir gently with a heatproof spatula. Continue to cook over medium heat, stirring only occasionally, until the mixture starts to boil. Clip a candy thermometer to the side of the pan and bring the syrup to the soft-ball stage, approximately 240 degrees F. Remove the pan from the heat and stir in the salt, the baking soda, and vanilla. Pour the caramel over the popcorn in large streams, then sprinkle with the peanuts. Use your spatula to fold the popcorn until it is completely coated with caramel.
Place the roasting pan in the oven and bake for 15 minutes. Use a spatula to lift, flip, and coat the popcorn in the warm caramel, then continue baking for another 20 minutes. Cool the caramel popcorn in the pan for 5 minutes and transfer it to the lined sheet pan. Cool for approximately 15 minutes and drizzle the milk chocolate and dark chocolate in crisscross patterns over the top. Let the chocolate set before breaking the popcorn into serving-size pieces. Store in an airtight container for up to 1 week.
**_Baked Note_**
This caramel popcorn virtually begs for chocolate, but if you don't want to go through the trouble and time of tempering it, we completely understand. You can just leave it out. Additionally, this recipe was adapted from none other than my Aunt Judy, by way of a Land O'Lakes Butter recipe that she has kept for untold years.
* * *
In this book, I use tempered chocolate in the delicious caramel popcorn, but tempering also comes in handy for homemade truffles, dipping macaroons, and making molded chocolate candy bars and novelties.
The process of tempering chocolate is a lot less complicated than explaining the science of tempering chocolate; however, I feel obligated to give you a brief intro:
_**What is going on?**_
There are many methods of tempering chocolate, but each is accomplished through a process of melting, cooling, and agitating. Our favorite method is to seed the chocolate. Here, a portion of already tempered chocolate is added to melted chocolate. This helps stimulate the formation of stable beta crystals—or, to put it simply, reduces the temperature of the melted chocolate and helps it come to temper.
_**Getting started**_
It is much easier to temper in a cool kitchen. Measure out 1 pound of chocolate—dark or milk, depending on the recipe—and chop into small chunks. I prefer a couverture chocolate that contains at least 32 percent cocoa butter, as it is thinner when melted and ideal for dipping.
_**Tempering**_
Place about a quarter of the chocolate in a bowl and set aside. Put the remainder in a large heat-proof bowl and place over a pot of simmering water. Melt the chocolate until an instant-read thermometer placed in the middle of the bowl reads 120 degrees F.
Remove the bowl from the heat and add the reserved chopped chocolate. With a rubber spatula, stir the chocolate vigorously without stopping until it has completely melted and cooled to a temperature of 80 degrees F. The chocolate should thicken considerably.
Place the bowl of chocolate back over the simmering water and stir with a spatula. If you are using dark chocolate, bring it to a temperature between 86 and 90 degrees F. If you are using milk chocolate, a temperature between 84 and 87 degrees F works best.
Test the chocolate to make sure it has reached a full temper. Dip a small metal spatula into the chocolate and place it on the counter. The chocolate should begin to set in 3 to 5 minutes and have a satiny shine, without streaks. If the chocolate has not set after 5 minutes or it looks speckled or streaked, you should continue to agitate the chocolate with the spatula until it is properly tempered.
When your chocolate is tempered, you may begin using the chocolate. Keep in mind that tempered chocolate sets up quickly. If you notice the chocolate in your bowl is beginning to harden, place it back over the simmering water to reheat, but only for a few seconds.
* * *
### **CLASSIC CARAMEL SAUCE ( FOR COFFEE CAKE AND THE LIKE)**
THIS CLASSIC AND EASY CARAMEL SAUCE IS A COMPONENT OF OUR CARAMEL APPLE CAKE. It is also a perfect stand-alone recipe and can be used on all manner of cakes, ice creams, and quick breads. This recipe makes more than you actually need for the Caramel Apple Cake itself, so I encourage you to find a tasty use for the leftovers.
**YIELD: ABOUT 2 CUPS**
**_Ingredients_**
_1½ cups sugar_
_¼ cup corn syrup_
_½ cup (1 stick) of butter, softened, cut into ½-inch cubes_
_1½ cups heavy cream_
In a medium saucepan with high sides, combine the sugar and corn syrup with ½ cup water. Stir the mixture gently so you don't slosh any of it up the sides of the pan. Turn the heat to medium-high and continue stirring until the sugar dissolves. Increase heat to high, stop stirring, and allow the mixture to boil. Once it begins to turn a rich caramel color (if you don't want to eyeball it, take the caramel to 300 degrees F on a candy thermometer), remove it from the heat, add the butter and cream, and stir until combined.
You can save the caramel sauce, tightly covered, in the refrigerator for up to 1 week. Let it come to room temperature before using it on cakes, ice creams, or quick breads.
If you want a warm topping, heat the caramel sauce in short bursts in the microwave or in the top of a double boiler.
### **VANILLA BEAN AND CHOCOLATE BUDINO**
TO TELL YOU THE TRUTH, BUDINO IS JUST A FANCY ITALIAN WORD FOR PUDDING. Good old American pudding is pretty much indistinguishable from the more debonair-sounding budino—but this is a debate I avoid with certain family members. I have named this dessert in deference to them, and it is a really great pudding.
Made-from-scratch pudding is superior to the packaged variety in every imaginable way and in theory, it won't take you much longer to make. I prefer a richer pudding experience, pudding slightly thicker and slightly deeper than average. Though I am partial to dark chocolate, I think the milk chocolate in this dessert works extremely well layered against the creamy vanilla bean pudding. I suggest serving this in small glass cups—a little goes a long way, and you'll probably want to show off the layers. And remember, if anybody asks, this is a budino, not a pudding.
**YIELD: 8 SERVINGS**
**_Ingredients_**
_1 tablespoon bourbon_
_1 vanilla bean_
_4 ounces good-quality milk chocolate_
_2 large egg yolks_
_2 large eggs_
_¼ cup heavy cream_
_½ cup sugar_
_⅓ cup cornstarch, sifted_
_½ teaspoon salt_
_3 cups whole milk_
_1 tablespoon unsalted butter_
_Simple Whipped Cream to serve (optional)_
Choose eight small wide-mouthed glasses for serving.
Put the bourbon in a medium heatproof bowl. Cut the vanilla bean in half lengthwise, and, using the tip of a knife or small spoon, scrape the seeds into the bourbon. Discard the vanilla bean and stir the mixture to combine.
Place the milk chocolate in another medium heatproof bowl.
In a third heatproof bowl, whisk together the egg yolks, whole eggs, and cream.
In a medium saucepan, whisk together the sugar, cornstarch, and salt. Whisk in the milk. Cook the mixture over medium-high heat until it just begins to boil, stirring occasionally. Whisking constantly, pour a third of the milk mixture over the egg mixture, then add another third of the hot milk mixture. Transfer the tempered egg mixture to the saucepan with the milk mixture and, whisking constantly, bring it to a boil. Cook for 2 to 3 minutes, or until the pudding is very thick.
Pour half of the pudding over the vanilla bean mixture and half over the milk chocolate. Add half of the butter to the vanilla mixture and whisk vigorously to cool the pudding slightly. Add the remaining butter to the chocolate mixture and again whisk vigorously to release the heat.
#### **ASSEMBLE THE DESSERT**
Divide the vanilla pudding among the serving glasses. Chill them for 20 minutes. While the vanilla pudding is in the refrigerator, whisk the chocolate pudding every 5 minutes to release excess heat.
Spoon the chocolate layer on top of the vanilla, cover (see Sidebar below), and chill the puddings until firm, about 2 hours, before serving.
Top with whipped cream, if you like, and serve immediately.
The puddings can be stored, tightly covered, in the refrigerator for up to 2 days. Do not add the whipped cream until you are ready to serve them.
**_Baked Note_**
If you prefer to make just vanilla pudding, omit the milk chocolate. If you wish to make just the chocolate pudding, double the milk chocolate and omit the vanilla. bean and bourbon. And, of course, ignore the layering instructions.
* * *
Silly me, I just assumed that nobody wants a skin (you know—that thick, continuous top layer) on pudding. Well, I was dead wrong. In fact, I have met just as many skin pudding people as no-skin pudding people, and each side is quite vocal.
Either version is easy to accomplish. If you want a skin, wrap the pudding vessel tightly before refrigerating it and do not let the plastic wrap come in contact with the top of the pudding. If you do not want a skin, gently press the plastic wrap down onto the surface of the pudding, then refrigerate.
* * *
### **MERINGUE MUSHROOMS, OR SHANDI'S CANDIES**
THEORETICALLY, I AM A YOLK MAN. I CAN WORK THROUGH A HILL OF EGG YOLKS IN A MATTER OF DAYS IF I AM FEELING GAME FOR A RUN OF RICH CHOCOLATE PUDDINGS AND VELVETY ICE CREAMS. Inevitably, I am left with a surplus of egg whites. And when I have extra egg whites, I make meringues. A meringue is a simple cloudlike confection of whipped egg whites and sugar, and it is likely they have been part of the baking pantheon for hundreds of years. They are crunchy, light, and very sweet, a lovely little afternoon snack. Our meringue mushrooms are a featured accompaniment to our Stump de Noël, but I encourage you not to limit your meringue making to holiday cake decor. Make them any time you have leftover egg whites (assuming you won't be using them for Angel frosting). It should be noted that these meringues are dedicated to a kind customer, Shandi, who had a limitless appetite for these candies.
**YIELD: ABOUT 2 DOZEN (SMALL) MUSHROOMS**
**_Ingredients_**
_3 large egg whites, at room temperature_
_¾ cup sugar_
_¼ teaspoon cream of tartar_
_1 ounce good-quality white chocolate, melted and cooled_
_2 ounces dark unsweetened cocoa powder, (like Valrhona) for sifting over assembled mushrooms_
Preheat the oven to 200 degrees F. Line two sheet pans with parchment paper.
Whisk the egg whites and sugar together in the heat proof bowl of a standing mixer. Set the bowl over a saucepan of simmering water (double boiler method). Cook, whisking constantly, until the sugar is completely dissolved and the mixture registers 140 degrees F on an instant-read thermometer, 6 to 8 minutes.
Transfer the bowl to a standing mixer fitted with the whisk attachment. Beat it on high speed until stiff peaks form. Add the cream of tartar when the mixture begins to thicken, or after 3 minutes. Keep beating for another minute or so until stiff peaks hold.
Fill a pastry bag fitted with a large round tip with the meringue. To make the caps, hold the pastry bag close to the parchment paper–lined pan and pipe out a small dome (about a tablespoon) of meringue, pulling up at the very end of piping to give your cap some height.
To form the mushroom stems, hold the bag close to the parchment paper and pipe the meringue, pulling up as you go, into small cone shapes. Make the same number of stems as caps.
Place the pans in the oven and bake the meringue pieces for 90 minutes, rotating the baking sheets halfway through the baking time. Turn off the oven, prop the door slightly open, and leave the meringues in place for at least 2 hours longer, or overnight.
#### **ASSEMBLE THE MERINGUE MUSHROOMS**
Turn the caps over and use a toothpick to make a tiny hole large enough to fit the tip of the stem into. Fill the hole with a tiny bit of white chocolate. Gently press the stem into place and allow the chocolate to set. Sift cocoa powder over the assembled mushrooms.
* * *
**_Baked Notes_**
As with most egg-white-based recipes, it is important that the egg whites be 100 percent egg yolk free. Should any yolk bits fall into the whites, carefully and thoroughly scoop out and discard them. Also, when I say egg whites at room temperature, I mean it. Especially for this recipe. If you want to make a classic meringue instead of the mushrooms, add 1 teaspoon almond extract or liqueur along with the cream of tartar. Pipe out large, puffy clouds of meringue and bake them as per the instructions.
* * *
The Stump de Noël is a thing of beauty regardless of whether it is stark-raving naked or wrapped up in baubles and bling. If you wish to add some adornment to your "Stump" or any other holiday treat, try out this simple recipe for sugared cranberries and sugared rosemary.
Line a half-sheet pan with parchment paper. Place the superfine sugar in a small bowl.
In a medium saucepan, stir together 1 cup of water with the sugar, then add the cinnamon stick. Bring the mixture to a boil over medium-high heat and stir until the sugar is completely dissolved. Remove from the heat and pour into a heatproof, wide-mouthed bowl. Let the liquid cool for a few minutes, then remove the cinnamon stick.
Drop the cranberries in the syrup and stir to coat the cranberries completely. Remove the cranberries, a few at a time, with a slotted spoon (tap the spoon to release excess syrup) and drop them in the superfine sugar. Toss the cranberries in the sugar to coat completely, and place on the parchment paper to dry. Repeat the above process with the rosemary. Decorate "the Stump" at will.
* * *
### **SOFT CANDY CARAMELS**
MOM WAS RESOLUTE. HALLOWEEN WAS HER FAVORITE HOLIDAY, AND SHE WAS UNYIELDING IN THE CANDY SELECTION. The shelves of the grocery store were stocked, literally bulging, with eye-catching displays of my favorite candy bars and confections, wrapped in special Halloween-themed packaging and perfectly bite-size (or two-bite-size) for trick-or-treaters. Mom ignored these, barely acknowledged my pleadings for the Hershey's multipack, and moved quickly to the less-colorful part of the candy aisle. She always made a beeline for the caramel candies, Brach's maybe, the kind wrapped tightly in white wax paper. The kind that stuck in your teeth and tasted overwhelmingly like vanilla. The kind of thing I didn't want Mom to give out on Halloween.
Over the years, I grew to appreciate the caramel candies and even anticipated their presence in our house; however, the homemade version is infinitely better. These soft candy caramels have a pleasingly smooth texture and the smoky, buttery dark caramel taste that is absent in their mass-produced counterparts.
**YIELD: ABOUT 70 CARAMELS**
**_Ingredients_**
_2 cups light corn syrup_
_1 cup granulated sugar_
_1 cup firmly packed light brown sugar_
_2 cups heavy cream_
_⅔ cup condensed milk_
_½ cup (1 stick) unsalted butter, at room temperature, cut into cubes_
_2 teaspoons pure vanilla extract_
_1 teaspoon sea salt or fleur de sel_
Lightly spray a 9-inch square baking pan with nonstick cooking spray and line it with aluminum foil, allowing the foil to just overhang on the sides. Lightly coat the foil with nonstick cooking spray.
In a medium saucepan, gently stir together the corn syrup and sugars along with ¼ cup water. Set the saucepan over low heat and continue to stir gently until the sugars dissolve (avoid sloshing the sides of the pan). Once the sugar has dissolved, clip a candy thermometer to the side of the pan, turn the heat up to medium-high, and wait for the mixture to reach 240 to 245 degrees F, about 7 minutes. Keep a watchful eye on the temperature while you proceed with the next step—you do not want it to exceed 250 degrees F.
Meanwhile, in a small saucepan, stir together the cream and condensed milk and set over medium heat. Gently warm the mixture; do not let it boil.
Once the sugar mixture turns amber, remove it from the heat and stir in the butter and warm milk mixture until completely combined (be careful about splattering; it usually bubbles up when you add the milk mixture). Place the pan back on medium heat, stop stirring, and bring the mixture back to 245 to 250 degrees F.
Remove the pan from the heat, stir in the vanilla and salt, and pour the caramel into the prepared pan. Allow the candy to set for 8 hours, or overnight.
#### **CUT AND WRAP THE SOFT CANDY CARAMELS**
Place a sheet of parchment over the caramel and invert it onto a cutting surface. Remove the aluminum foil.
Spray a chef's knife with nonstick cooking spray or carefully rub a tiny bit of vegetable shortening along the blade. Cut the caramels into 1 by ½-inch rectangles, and immediately wrap them in wax paper or candy papers. Twist the ends of the papers to resemble old-school candies. Distribute with brio to friends and family.
The caramels will keep in an airtight container at room temperature for up to 10 days.
**_Baked Note_**
It is important to make sure your candy thermometer is accurate before making caramels (or, really, any candy). A batch of ruined caramels is annoying and expensive. If you are unsure or if you haven't used your thermometer in some time, boil some water in a medium pot, clip the thermometer to the side of the pot, and wait 5 minutes. The thermometer should read 212 degrees F.
### **CHOCOLATE HAZELNUT SPREAD**
I AM A MAN OF GOOD INTENTIONS. I HAVE GRAND IDEAS ABOUT RAISING MY OWN CHICKENS (FOR EGGS), CHURNING MY OWN BUTTER, AND GRINDING MY OWN SPICES ON AN AS-NEEDED BASIS. I would also like to make homemade Nutella more often. Unfortunately, I am still without my own chickens, and I rarely churn my own butter—but I have reorganized my spice drawer with a tilt toward the fresh, and I have been making this chocolate hazelnut spread repeatedly for a wide swath of friends. I am still addicted to the jarred stuff. I still impulsively place at least two tubs of Nutella in my grocery cart without conscious thought, but I also thoroughly enjoy the homemade version. Adapted from the brilliant blogger Jessica Su of Su Good Sweets, it is thicker and a bit richer than Nutella, and it is heavenly. I am still uncertain whether it is less expensive (after all, hazelnuts are pricey), but it is well worth the effort. Serve it on toast, bananas, graham crackers, pancakes, tucked inside a crepe, warmed and poured on ice cream, or with almost anything else you dream of.
**YIELD: APPROXIMATELY 1½ CUPS**
**_Ingredients_**
_2 cups whole hazelnuts_
_1¼ cups confectioners' sugar, sifted_
_⅓ cup dark unsweetened cocoa powder, (like Valrhona)_
_½ teaspoon pure vanilla extract_
_Dash salt_
_3 to 4 tablespoons hazelnut oil (or alternatively vegetable oil)_
Preheat the oven to 350 degrees F. Line a baking sheet with parchment paper.
Spread the hazelnuts across the prepared pan in a single layer and toast them in the oven for 8 minutes. Toss the nuts, then toast them for another 5 to 6 minutes, until they are fragrant and have turned a dark brown. Let the nuts cool completely.
Remove the skins from the cooled nuts by placing them in a damp towel and rubbing them together. Discard the skins.
Place the nuts in the bowl of a food processor and process for 3 to 5 minutes, until the nuts liquefy and become buttery. Scrape down the bowl and process for 30 seconds longer.
Add the sugar, cocoa powder, vanilla, salt, and 3 tablespoons of the hazelnut oil, and process again for about 1 more minute, or until the mixture is smooth and spreadable. If it is too thick, add more hazelnut oil, a teaspoon at a time, until the right consistency is achieved.
Use the spread immediately or store it in the refrigerator, in a tightly sealed container, for up to 2 weeks. Initially, the spread will be wet and thin (perfect for dipping). However it will thicken considerably in the refrigerator and have a consistency akin to chilled peanut butter. For spreading purposes, remove from the refrigerator 15 minutes before using, it will be more pliable and spreadable.
**_Baked Note_**
This recipe can be used to replace the Nutella in the hazelnut scones. Because the homemade version lacks the spreadable consistency of the original, though, make sure to bring it to room temperature and stir in an extra teaspoon of hazelnut oil beforehand.
## _**Acknowledgments**_
_Baked Explorations_ was a little bit of a beast—a beast that we loved and nurtured, but a monster all the same. Without the help of a great circle of friends and acquaintances, we may have been entirely consumed.
First, a much deserved special thank you to Eric Wolitzky. He is a talented pastry chef, a studious technician, and such a great interpreter of Baked dreams that we thought he might be living in our brains. His version of the Mississippi Mud is to be blamed for our sudden involuntary weight gain.
Also, big love to Alison Fargis, agent extraordinaire. At once, protective sister, kick-ass baker, and avid reader—the book trade is a better world with her in it.
Both of our editors, Luisa Weiss and Natalie Kaire, deserve a quiet place in Heaven for dealing with our endless missed deadlines, our severe misgivings about letting a recipe go, and our haphazard self-editing process. Free cakes to you both for life—we appreciate you both more than you know. And thanks to everyone else at STC for giving Baked a tactile voice especially Alissa Faden for designing the heck out of this book.
A hearty thanks to the small army of testers who took the time to spin an endless array of butter, eggs, sugar and chocolate into gold: Liz Moore, Rachel Boller (forever in my heart as the cake lady), Joann Tamburro, Clay Smith, Lianna Allday, Gretchen Lewis (aka multi-tasker), Larry Lewis (aka swell dad), Nancy Mongiovi, and Jessie Sheehan (aka Brookie Queen).
Matt Holbein deserves special recognition as roadie/crew member/all-around Baked believer. Sven Wiedmann for his persistent patience. Grace (mom) Poliafito for the cake. And Rafi Avaramovitz gets billing for his consistent belief. As always Martha Stewart holds a special place in our hearts for giving Baked its first big break.
The lush photos are served up by the same snazzy crew from the first cookbook, including Tina Rupp who can sex up Jell-O Salad like no one's business, and prop-mistress Leslie Seigel.
We have to thank the coolest set of employees, past and present, to ever grace a bakery. We would like to thank them all, but special props to Melissa Fritz—our southern Bakemistress, Jessica Bacchus, Stephanie Francis (Helga), Stephanie Whitten (Vera), Kristine Moberg (can we come back to SD?), and Lesli Heffler. Lesli was never a proper employee, much to our consternation, but we always wanted her to be and she is a genius food stylist.
If we forget anyone, please hunt us down and ask us for brownies.
Matt and Nato
## _**Sources**_
### CANDY
#### **ECONOMY CANDY**
108 Rivington Street
New York, NY 10002
800. 352-4544
www.economycandy.com
Maltesers, Milk Chocolate-covered Peanuts, and a multitude of other candies and chocolates.
#### **KOPPERS CHOCOLATE**
800.325.0026
www.kopperschocolate.com
Malted Milk Balls (Whoppers), Chocolate-covered Peanuts available in bulk quantities.
### CHOCOLATE AND OTHER SPECIALTY INGREDIENTS
#### **CALLEBAUT CHOCOLATE**
www.callebaut.com
#### **DIVINE CHOCOLATE**
202-332-8913
www.divinechocolateusa.com
Fair Trade quality chocolate and cocoa.
#### **JAQUES TORRES**
212.414.2462
350 Hudson Street
New York, NY 10014
www.mrchocolate.com
#### **SCHARFFEN BERGER**
866.608.6944
Available in most grocery/specialty stores.
www.scharffenberger.com
#### **VALRHONA CHOCOLATE**
www.valrhona-chocolate.com
Also available in many specialty stores.
#### **WHOLE FOODS**
Organic foods as well as a great variety of high-grade chocolates. Go to www.wholefoodsmarkets.com for locations in your area.
#### **NEILSEN MASSEY**
800-525-7873
www.nielsenmassey.com
Great source for Vanilla Bean Paste and Coffee Extract (and of course, Pure Vanilla Extract)
#### **SALTWORKS**
800-353-7258
www.saltworks.us
Wide variety of salts sold in bulk portions.
#### **ANSON MILLS**
803-467-4122
www.ansonmills.com
Stone ground grits. Get some.
#### **BOBS RED MILL**
800-553-2258
www.bobsredmill.com
Grits and specialty flours.
#### **INDIA TREE**
800-369-4848
www.indiatree.com
Great resource for specialty sugars.
### KITCHEN & BAKING EQUIPMENT
#### **CRATE & BARREL**
Call 800.967.6696 for store locations in your area.
www.crateandbarrel.com
#### **JB PRINCE**
800.473.0577
www.jbprince.com
#### **KING ARTHUR FLOUR BAKERS CATALOGUE**
Norwich, VT 05005
800.827.6836
www.kingarthurflour.com
#### **KITCHENAID APPLIANCES**
Call 800.334.6889 for a KitchenAid distributor in your area.
kitchenaid.com
#### **NEW YORK CAKE & BAKE**
800.942.2539
56 West 22nd Street
New York, NY 10010
www.nycake.com
A fantastic resource for the New York-based baker. Pans, tools and decorating equipment.
#### **PFEIL & HOLING**
58-15 Northern Blvd.
Woodside, NY 11377
800.247.7955
www.cakedeco.com
Decorating supplies sold in bulk.
#### **SUR LA TABLE**
Call 800.243.0852 for store locations in your area.
www.surlatable.com
#### **WILLIAMS-SONOMA**
Call 800.541.1262 for store locations in your area.
www.williams-sonoma.com
## _**Conversion Chart**_
Weight Equivalents: The metric weights given in this chart are not exact equivalents, but have been rounded up or down slightly to make measuring easier.
**AVOIRDUPOIS** | **METRIC**
---|---
¼ oz | 7 g
½ oz | 15 g
1 oz | 30 g
2 oz | 60 g
3 oz | 90 g
4 oz | 115 g
5 oz | 150 g
6 oz | 175 g
7 oz | 200 g
8 oz (½ lb) | 225 g
9 oz | 250 g
10 oz | 300 g
11 oz | 325 g
12 oz | 350 g
13 oz | 375 g
14 oz | 400 g
15 oz | 425 g
16 oz (1 lb) | 450 g
1 ½ lb | 750 g
2 lb | 900 g
2 ¼ lb | 1 kg
3 lb | 1.4 kg
4 lb | 1.8 kg
Volume Equivalents: These are not exact equivalents for American cups and spoons, but have been rounded up or down slightly to make measuring easier.
**AMERICAN** | **METRIC** | **IMPERIAL**
---|---|---
¼ tsp | 1.2 ml
|
½ tsp | 2.5 ml
|
1 tsp | 5.0 ml
|
½ Tbsp (1.5 tsp) | 7.5 ml
|
1 Tbsp (3 tsp) | 15 ml
|
¼ cup (4 Tbsp) | 60 ml | 2 fl oz
⅓ cup (5 Tbsp) | 75 ml | 2.5 fl oz
½ cup (8 Tbsp) | 125 ml | 4 fl oz
⅔ cup (10 Tbsp) | 150 ml | 5 fl oz
¾ cup (12 Tbsp) | 175 ml | 6 fl oz
1 cup (16 Tbsp) | 250 ml | 8 fl oz
1¼ cups | 300 ml | 10 fl oz (½ pint)
1½ cups | 350 ml | 12 fl oz
2 cups (1 pint) | 500 ml | 16 fl oz
2½ cups | 625 ml | 20 fl oz (1 pint)
1 quart | 1 liter | 32 fl oz
**OVEN MARK** | **F** | **C** | **GAS**
---|---|---|---
Very cool | 250–275 | 130–140 | ½–1
Cool | 300 | 150 | 2
Warm | 325 | 170 | 3
Moderate | 350 | 180 | 4
Moderately hot | 375 | 190 | 5
|
400 | 200 | 6
Hot | 425 | 220 | 7
|
450 | 230 | 8
Very hot | 475 | 250 | 9
## _**Index of Search Terms**_
### **A**
almond joy tart
angel frosting
apple caramel cake
apricot rosemary squares
### **B**
banana cream pie, peanut butter
bananas Foster fritters
bars
grasshopper
heartland turtle bars
peanut butter and jelly
pudding
rosemary apricot
sweet & salty brownie
blackberry pie
blueberry parfaits
bread, monkey bubble
brownie, sweet & salty
budino, vanilla bean and chocolate
bundt cake
burnt sugar
olive oil orange
buttercream
caramel
coffee
dark chocolate
honey vanilla
malted
mint
buttermilk doughnuts
buttermilk pie
### **C**
cake. _See also_ bundt cake; cupcakes
Aunt Sassy
Boston cream pie
caramel apple
chocolate coffee
crumb, New York-style
devil's food
lady praline chiffon
muddy Mississippi
oatmeal chocolate chip
quick skillet snack
stump de noël
Sunday night
caramel
caramel apple cake
caramelized crispies
caramel rum frosting
heartland turtle bars
popcorn
sauce, classic
soft candy caramels
carrot coconut scones
cheese grits, baked
chiffon cake, lady praline
chocolate. _See also_ white chocolate
buckeyes
buttercream
caramel popcorn with peanuts and
chocolate coffee cake
chocolate ginger molasses cookie
chocolate hazelnut spread
chocolate mint thumbprints
chocolate peanut butter fondue
cowboy cookies
devil's food cake
double-chocolate loaf
frosting
ganache
glaze
heartland turtle bars
marshmallow chocolate cups
Mississippi mud pie
Nutella scones
oatmeal chocolate chip cake
pudding
salt-n-pepper sandwich cookies
shortbread dipped in
skillet snack cake
stump de noël
sweet & salty brownie
and vanilla bean budino
whoopie pies
citrus glaze
coconut
almond joy tart
carrot coconut scones
sawdust pie
coffee
buttercream
ice cream
ice cream tart
cookies. _See also_ bars; sandwich cookies; whoopie pies
black and white
chocolate ginger molasses
chocolate mint thumbprints
cowboy
Joe Froggers
no-bake peanut butter
shortbread with fleur de sel
speculaas
corn
baked cheese grits
caramel popcorn
cornmeal griddle cakes
honey corn muffins
cranberries, sugared
cream, whipped
cream cheese
frosting
peanut butter spread
strawberry Jell-O salad
creamsicle tart
crumb cake, New York-style
cupcakes
maple
tomato soup
### **D**
devil's food cake
doughnuts, buttermilk
### **F**
fleur de sel
salt-n-pepper sandwich cookies
shortbread with
sweet & salty brownie
fondue, chocolate peanut butter
French toast, baked
fritters, bananas Foster
frosting. _See also_ buttercream; ganache
angel
black and white
caramel rum
chocolate
confectioner's sugar
cream cheese
mascarpone
### **G**
ganache
chocolate
white chocolate peppermint
ginger cookies
chocolate ginger molasses
ginger rum molasses
glaze. _See also_ ganache
chocolate
citrus
orange
pear
vanilla
griddle cakes, cornmeal
grits, cheese, baked
### **H**
hazelnuts
chocolate hazelnut spread
Nutella scones
honey corn muffins
honey vanilla buttercream
### **I**
ice cream, coffee
coffee ice cream tart
### **J**
Jell-O salad, strawberry
jelly and peanut butter bars
### **L**
lemon shortbread
loaf, double-chocolate
### **M**
malt
buttercream
malted crisp tart
sandwich cookies
waffles
maple cupcakes
maple syrup, buttermilk pie with
marshmallow chocolate cups
mascarpone frosting
meringue mushrooms
mint
chocolate mint thumbprints
grasshopper bars
molasses cookies
chocolate ginger
ginger rum
muffins
honey corn
pumpkin cheddar
### **N**
Nutella
homemade
Nutella scones
### **O**
oatmeal chocolate chip cake with cream cheese frosting
olive oil orange bundt
orange
creamsicle tart
glaze
olive oil orange bundt
shortbread
### **Q**
parfaits, blueberry
pastry cream
peaches and dream pie
peanut butter
banana cream pie with
buckeyes
chocolate fondue with
cookie, no-bake
cream cheese spread with
and jelly bars
whoopie pie filling
peanuts, caramel popcorn with chocolate and
pear whiskey tart
pecans
heartland turtle bars
Mississippi mud pie
pecan tassies
sawdust pie
pepper-n-salt sandwich cookies
pie. _See also_ whoopie pies
blackberry
buttermilk
Mississippi mud pie (A)
peaches and dream
peanut butter banana cream
sawdust
pie dough, classic
pistachio cake
popcorn, caramel
pretzels
cowboy cookies
strawberry Jell-O salad
pudding. _See also_ budino
banana
chocolate
pudding bars, Aunt Sabra King's
pumpkin cheddar muffins
### **R**
rosemary, sugared
rosemary apricot squares
rum
caramel rum frosting
ginger rum molasses cookies
### **S**
salt. _See also_ fleur de sel
salt-n-pepper sandwich cookies
sandwich cookies
malted milk
salt-n-pepper
sauce, classic caramel
scones
carrot coconut
Nutella
shortbread with fleur de sel
spread
chocolate hazelnut
peanut butter cream cheese
strawberry Jell-O salad
sugar
sugared cranberries and rosemary
### **T**
tarts
almond joy
coffee ice cream tart
malted crisp tart
pecan tassies
whiskey pear
thumbprints, chocolate mint
tomato soup cupcakes
turtle bars, heartland
### **V**
vanilla
and chocolate budino
glaze
honey vanilla buttercream
shortbread
### **W**
waffles, malted
whipped cream, simple
whiskey pear tart
white chocolate
almond joy tart
chocolate mint thumbprints
sawdust pie
whoopie pies
chocolate
red velvet
### _**Recipe Index**_
### **A**
Almond Joy Tart
Aunt Sabra King's Pudding Bars
Aunt Sassy Cake
### **B**
Baked Cheese Grits
Baked French Toast
Bananas Foster Fritters
Black and White Cookies
Blackberry Pie
Boston Cream Pie Cake
Buckeyes
Burnt Sugar Bundt Cake with Caramel Rum Frosting
Buttermilk Pie with a Hint of Maple Syrup
### **C**
Caramel Apple Cake
Caramel Popcorn with Peanuts and Chocolate
Carrot Coconut Scones with Citrus Glaze
Chocolate Coffee Cake with Dark Chocolate Ganache
Chocolate Ginger Molasses Cookie
Chocolate Hazelnut Spread
Chocolate Mint Thumbprints
Chocolate Peanut Butter Fondue
Chocolate Whoopie Pies
Classic Caramel Sauce
Classic Pie Dough
Classic Shortbread with Fleur de Sel
Coffee Ice Cream
Coffee Ice Cream Tart
Cornmeal Griddle Cakes
Cowboy Cookies
### **D**
Devil's Food Cake with Angel Frosting
Double-Chocolate Loaf with Peanut Butter Cream Cheese Spread
### **F**
Farm Stand Buttermilk Doughnuts
### **G**
Ginger Rum Molasses Cookies
Grasshopper Bars
### **H**
Heartland Turtle Bars
Honey Corn Muffins
### **I**
Joe Froggers
### **J**
Lady Praline Chiffon Cake
### **K**
Malted Crisp Tart
Malted Milk Sandwich Cookies
Malted Waffles
Maple Cupcakes with Maple Cream Cheese Frosting
Marshmallow Chocolate Cups
Meringue Mushrooms
Mississippi Mud Pie (A), aka Coffee Ice Cream Tart
Mississippi Mud Pie (B), aka Muddy Mississippi Cake
Mom's Olive Oil Orange Bundt
Monkey Bubble Bread
### **N**
New York-Style Crumb Cake
The No-Bake Peanut Butter Cookie
Nutella Scones
### **O**
Oatmeal Chocolate Chip Cake with Cream Cheese Frosting
Orange Creamsicle Tart
### **P**
Peaches and Dream Pie
Peanut Butter and Jelly Bars
Peanut Butter Banana Cream Pie
Pecan Tassies
Pumpkin Cheddar Muffins
### **Q**
Quick Skillet Snack Cake
### **R**
Red Velvet Whoopie Pies
Rosemary Apricot Squares
### **S**
Salt-N-Pepper Sandwich Cookies
Sawdust Pie
Shandi's Candies
Simple Blueberry Parfaits
Soft Candy Caramels
Speculaas
Strawberry Jell-O Salad
Stump de Noël
Sunday Night Cake
Sweet & Salty Brownie
### **T**
Tomato Soup Cupcakes with Mascarpone Frosting
### **V**
Vanilla Bean and Chocolate Budino
### **W**
Whiskey Pear Tart
Published in 2010 by Stewart, Tabori & Chang
An imprint of ABRAMS
Text copyright © 2010 Matt Lewis and Renato Poliafito
Illustrations/photographs copyright © 2010 Tina Rupp
All rights reserved. No portion of this book may be reproduced, stored in a retrieval system, or transmitted
in any form or by any means, mechanical, electronic, photocopying, recording, or otherwise,
without written permission from the publisher.
Library of Congress Cataloging-in-Publication Data
Lewis, Matt.
Baked explorations : classic American desserts revisited / Matt Lewis,
Renato Poliafito; photography by Tina Rupp.
p. cm.
ISBN 978-1-58479-850-7 (alk. paper)
1. Baking. 2. Desserts. 3. Baked (Bakery) I. Poliafito, Renato. II.
Title.
TX765.L673 2010
641.8'6—dc22
2010016610
Editor: Natalie Kaire
Designer: Alissa Faden
Production Manager: Tina Cameron
Stewart, Tabori & Chang books are available at special discounts when purchased in quantity for
premiums and promotions as well as fundraising or educational use. Special editions can also be created
to specification. For details, contact specialsales@abramsbooks.com or the address below.
www.abramsbooks.com
**About the Authors**
MATT LEWIS AND RENATO POLIAFITO left their day jobs in advertising to open their bakery, Baked, in Brooklyn to immediate praise from fans across the country. They have since opened another Baked in Charleston, South Carolina. The authors have been featured on Oprah, the Today Show, the Food Network, and The Martha Stewart Show. Their first book was an IACP award nominee. Lewis and Poliafito live in New York City.
TINA RUPP a New York-based photographer who specializes in food photography. Her work can be found regularly in Food & Wine and Food Network Magazine, various national advertisements, and cookbooks.
JACKET PHOTOGRAPHS © 2010 TINA RUPP
JACKET DESIGN BY ALISSA FADEN
115 WEST 18TH STREET
NEW YORK, NY 10011
WWW.STCBOOKS.COM
|
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| 8,444
|
Q: iOS MVC Design Pattern - Layer between REST resource models and view controllers What do you think about adding a layer with logic between the Models and View Controllers in the MVC pattern?
We have models for REST resources that consist of readonly properties and methods for updating/fetching data on server. These models are placed in a framework that is used by multiple iOS / OS X apps, so we don't want to add any app specific code to them.
The view controllers access these REST resource models through a layer where frequently used logic are placed, for example filtering and cacheing large collections. The layer consists of classes where each class corresponds to a REST resource and each instance of the class has an instance of a REST resource model. The REST resource is public so the resource and its values can be read directly by view controllers.
What should the classes in this layer be called?
Are they controllers, helpers, perhaps utils or something else? Or is this an invalid Cocoa design pattern?
For example, if the resource is named "Activity" could its class be named "ActivityController"?
A: From my understanding, it seems that what you are using is generally speaking a Wrapper on a resource model. More specifically given that you are placing filtering logic inside it, maybe Proxy is the correct name. I don't think that this is an invalid Cocoa design pattern. Anyway
The REST resource is public so the resource and its values can be read directly by view controllers.
sounds odd. The resource should be private and the proxy should provide functions to access to model's values.
|
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| 4,516
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With the help of our quality application, you can hack Facebook Messenger and check out profile pictures of friends and see who they are chatting with. Besides text messages, you can easily check out audio and video messages, as well as exchanged photos and stickers.... You can personalise Skype for Business by adding a profile photo and posting a customised comment in the What's Happening Today field. Both of these customisations are displayed to your Skype for Business contacts.
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Take a look at the fanciest treehouse in Somerset - and you can stay there
Unparalleled luxury in a tree near Taunton
THE BIGGEST STORIES ACROSS TAUNTON IN YOUR INBOX
It's a world away from the forts you might have built when you were a kid.
Nestling between the Quantock Hills and Exmoor National Park in the village of Halse, the Tree House has it's own hot tub and sleeps a family of up to four.
All about The Tree House
Wrapped around a lime tree on a working farm, The Treehouse is built from wood and extending over two floors, it's a mix of luxury and cosiness.
The outside of the treehouse
Boasting a master bedroom with a six feet bed, spiral staircase, bathroom with a roll top bath, octagonal living room, plasma TV and wifi, woodburner, hot-tub, use of an indoor swimming pool under-floor heating and a washer, drier and dishwasher.
Prices in low season from £595 (short break) and £895 (per week); high season from £945 (short break) and £1,865 (per week). Visit premiercottages.co.uk or call the owner on 01984 656622.
The sitting room
How popular is it?
Very. While there are bookings available, it's few and far between, so if you fancy treating yourself to a stay there, don't hang around too long.
|
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New CEO Scott Egan addresses criticism that RSA is tough to trade with and commits to making it easier for regional brokers to work with the insurer as the provider reports an underwriting loss of £106m in 2018.
The Group lamented "poor results" in its commercial lines division as the company reported further underwriting losses in its UK business.
She takes over from Scott Egan, who moved into the UK & international CEO role following Steve Lewis' departure.
Emmanuel Kenning and Ida Axling give their thoughts on this week's biggest stories.
UK and International CEO Steve Lewis left the provider this week and was replaced by Scott Egan.
Dual pricing – the "crack cocaine" of insurance - was the headline article of the month for brokers.
As the sweltering heat continued into September brokers saw opportunities in Marsh's JLT deal and were also interested in Ardonagh buying Swinton and RSA's shock profit warning.
Forget having a summer playlist! July 2018 saw brokers more interested with who topped Allianz's "hit list" and Axa's replacement for Amanda Blanc, along with finally having a date for the senior managers' regime.
As we moved into spring brokers were interested in the names of the top five most complained about insurers, Axa inking a $15bn deal with XL, the collapse of CBL and Alpha as well as mergers and acquisitions.
Brokers warned to prepare for prices to increase by 10%-15% as the provider's departure causes "ripple effect".
Geoff Jones, who helped turnaround RSA's Ireland business, will lead the division which recently pulled out of a number of business lines.
|
{
"redpajama_set_name": "RedPajamaC4"
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| 5,234
|
Q: How to handle rapid like/unlike at updating UI ios I have an app. In this app, a user can like/unlike something. If the user tap the button like, it would change text to unlike, vice versa.
To make the like/unlike event run seamlessly to user, i change the text first, and then make request to my API, saying that the user like/unlike this. The API decides whether the action is liking or unliking depends on value at database. The API then returns a message stating the action made ("liked"/"unliked"). The app receives it, and then update the UI according to the result, in case the action intended by user fails.
If all runs smoothly, the user won't detect the changes made by API result.
This is the flow of liking something
user like -> button text changes to "unlike"
-> app make a request -> request is queued to operation queue
-> request run -> API decides whether that something is liked/unliked by the user
-> API returns action (in this case, "liked") -> app updates button text ("unlike")
Now my question are:
*
*How do I handle rapid button tap by user?
*How do I handle failed requests (internet disconnected, or no signal) while handling the problem no. 1?
nb: I don't want to disable the button (the app has to run seamlessly. Facebook app don't do it either, i just checked). Oh, and I use AFHTTPRequestOperationManager and set its maxConcurrentOperationCount to 1.
A: I resolved my problem using this answer
with a bit modification. The first time user click the button, I set a request flag to false, so that even if the user click the button many times, request won't be made before the first request is done.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 637
|
{"url":"http:\/\/www.lofoya.com\/Solved\/1435\/two-friends-a-and-b-run-around-a-circular-track-of-length-510","text":"# Difficult Time, Speed & Distance Solved QuestionAptitude Discussion\n\n Q. Two friends $A$ and $B$ run around a circular track of length 510 metres, starting from the same point, simultaneously and in the same direction. $A$ who runs faster laps $B$ in the middle of the $5^{th}$ round. If $A$ and $B$ were to run a 3 km race long race, how much start, in terms of distance, should $A$ give $B$ so that they finish the race in a dead heat?\n \u2716\u00a0A. 545.45 metres \u2714\u00a0B. 666.67 metres \u2716\u00a0C. 857.14 metres \u2716\u00a0D. Cannot be determined\n\nSolution:\nOption(B) is correct\n\n$A$ and $B$ run around a circular track. $A$ laps $B$ in the middle of the $5^{th}$ lap. i.e. when $A$ has run four and a half laps he has covered a distance which is 1 lap greater than that covered by $B$'s.\n\nTherefore, when $A$ runs \u00a0$\\dfrac{9}{2}$\u00a0laps, $B$ runs $\\dfrac{7}{2}$\u00a0laps\n\nThis is same as saying when $A$ runs 9 laps, $B$ runs 7 laps.\n\ni.e in a race that is 9 laps long, $A$ can give $B$ a start of 2 laps.\n\nSo, if the race is of 3000 metres long, then $A$ can give $B$ a start of\n\n$\\left(\\dfrac{2}{9}\\right)\\times 3000=$666.67 metres.\n\nThe information with regard to the length of the circular track is redundant information.","date":"2016-12-10 10:44:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 2, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5402317643165588, \"perplexity\": 903.2703286595756}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-50\/segments\/1480698543035.87\/warc\/CC-MAIN-20161202170903-00311-ip-10-31-129-80.ec2.internal.warc.gz\"}"}
| null | null |
Q: MKMapView calloutview height I want to increase height of calloutview but i tried lot of code but still height is not increasing ,
Please tell me how can i increase calloutbubble height
I am using this code
- (MKAnnotationView *)mapView:(MKMapView *)map viewForAnnotation:(id <MKAnnotation>)annotation
{
static NSString *AnnotationViewID = @"annotationViewID";
MKPinAnnotationView *annotationView = (MKPinAnnotationView *)[map dequeueReusableAnnotationViewWithIdentifier:AnnotationViewID];
if (annotationView == nil)
{
annotationView =[[MKPinAnnotationView alloc] initWithAnnotation:annotation reuseIdentifier:AnnotationViewID];
}
NSLog(@"array count %lu ",(unsigned long)[arrayOfAnnotation count]);
if ([arrayOfAnnotation containsObject:annotation])
{
for (OSAnnotation *anootation in arrayOfAnnotation)
{
if (anootation==annotation)
{
//NSLog(@"UserName : %@ ",model.fullName);
annotationView.image = [UIImage imageNamed:@"pinGold24.png"];
annotationView.frame=CGRectMake(0, 0,32,32);
annotationView.canShowCallout = YES;
annotationView.draggable=NO;
annotationView.userInteractionEnabled=YES;
// NSLog(@"latitude = %f logintude = %f",lati,longi);
UIView *calloutView = [[UIView alloc]initWithFrame:CGRectMake(0, 0, 300, 100)];
UIImageView *imgListingImage = [[UIImageView alloc]initWithFrame:CGRectMake(10, 10, 60, 60)];
imgListingImage.backgroundColor = [UIColor redColor];
UILabel *lbl_title = [[UILabel alloc] initWithFrame:CGRectMake(80, 10, 200, 21)];
lbl_title.text = map_title;
[lbl_title setTextColor:[UIColor darkGrayColor]];
[calloutView addSubview:lbl_title];
[calloutView addSubview:imgListingImage];
annotationView.leftCalloutAccessoryView=calloutView;
annotationView.rightCalloutAccessoryView = nil;
}
}
}
else
{
annotationView.canShowCallout=YES;
annotationView.draggable=NO;
annotationView.image=[UIImage imageNamed:@"pinGold24.png"];
}
return annotationView;
}
my calloutview is looking like this
|
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"redpajama_set_name": "RedPajamaStackExchange"
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| 5,924
|
Pimelodella mucosa är en fiskart som beskrevs av Eigenmann och Ward, 1907. Pimelodella mucosa ingår i släktet Pimelodella och familjen Heptapteridae. Inga underarter finns listade i Catalogue of Life.
Källor
Malartade fiskar
mucosa
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 144
|
{"url":"https:\/\/3dprinting.stackexchange.com\/questions\/10980\/reasons-for-a-pla-print-not-sticking-to-bed-all-the-sudden","text":"# Reasons for a PLA print not sticking to bed all the sudden?\n\n\u2022 Plastic: Same Matterhacker PLA (filament I use every day)\n\u2022 Printer: Anycubic i3 Mega (the one I use every day)\n\u2022 Slicer: Ultimaker Cura 4.2.1\n\nI don't know what's causing it, I haven't changed any slicer settings to my knowledge, I haven't changed anything on the printers end, and I'm using the same filament I've always used. But for some reason, the first layer is simply not sticking. At first I noticed when doing a print the nozzle seemed a little higher than normal for the first layer, but then it started having problems where 0\u00a0% of the filament would stick to the bed and it would all just come off and turn into a mess. I've checked and checked, but I see no reason the printer would just start doing this now all the sudden when it's worked perfectly for a year now.\n\nEDIT: Something I've noticed since posting this is that older sliced models seem to print just fine, which means there's something about the newser slicer settings that's causing it. I don't know what I would have changed though and\/or how to restore to my original settings.\n\n\u2022 Did you re-level your bed? Sep 6, 2019 at 22:35\n\u2022 Sure sounds like the nozzle height has been increased a little, or the bed has lowered a little. Sep 7, 2019 at 1:38\n\u2022 @P\u1d00\u1d1c\u029fs\u1d1b\u1d07\u02802 I did not, however after trying to do so I found lowering it made things worse, and lifting it up had no effect until it got high enough to where the extruder just knocked the filament off the bed. Sep 7, 2019 at 2:44\n\u2022 @CrossRoads I didn't change anything, I don't know how that would have happened, but after trying to re-align the bed to a better position it did not help (See my response to P\u1d00\u1d1c\u029fs\u1d1b\u1d07\u02802 for more information on that). I don't know how my nozzle height would have changed, but if it did how do I re-calibrate that? Sep 7, 2019 at 2:46\n\u2022 Please add an image and optionally a link to a video recording for us to see what is actually happening. Is this a nozzle to bed distance problem, or an adhesion problem. Do you use an adhesive agent between bed and filament?\n\u2013\u00a00scar\nSep 7, 2019 at 7:43\n\nThere are 3 general factors about print adhesion you always have to keep in mind:\n\n\u2022 Have a sufficient surface for the print to stick. A pyramid printed on the tip can't print properly.\n\u2022 Check the leveling of your bed occasionally and relevel the bed. By removing prints, one can easily unlevel it over time without noticing it.\n\u2022 Clean your print bed from fingerprints and grease every so often. Fats are good separators between the print and the bed. Getting them off with Isopropyl alcohol or other solvents can restore print surfaces in an instant.\n\nIn this specific case, there are some hints that make the general things less of an issue though: Old sliced items print fine, newer not. This hints that you changed something in the print settings. Among the settings that are good for adhesion, check your old G-code for the following three:\n\n\u2022 Bed temperature. I use 60\u00a0\u00b0C bed temperature for PLA and have good results on bed adhesion. Others print with 50\u00a0\u00b0C. However, going too low can make the plastic not stick well anymore.\n\u2022 Extrusion temperature. When the plastic extrudes, it has to be molten enough to push out enough and cold enough to solidify within moments and stick to the surface of the bed. If it is too hot, it would be dragged along, if it's too cold it doesn't get to stick either. I use 190-200\u00a0\u00b0C for PLA.\n\u2022 The first layer height might be different. I usually use 0.2\u00a0mm for this setting, no matter what the actual layer height is, and get good adhesion and not too much trouble with tiny unevenness.\n\u2022 The reason might be a mechanical issue, in that the Z-endstops (in an Anycubic i3, there are two, hidden in the frame sides) might have bent, moved or misaligned over time. Check its positioning. If the mount is broken, there are replacement part designs.\n\u2022 1. This is nothing new, I follow that all the time and I guarantee that's not the problem. 2. I've tried that already as noted in my original post 3. Ok, I can try that, I don't think it would happen so suddenly and everywhere on the bed at once though. These are good general tips, and I will be looking into them more to see the cause but I'm 99% sure that's not the problem due to it being so sudden, what other reasons do you know of? Sep 8, 2019 at 7:14\n\u2022 @UltraGamer With painters tape, it can be a problem that the tape is slick.The last addition makes me thinkng though... Sep 8, 2019 at 7:51\n\u2022 I'm not using painters tape though, because the bed is glass, usually, there are no problems with it sticking to the bed at all. You did add another 3 reasons though, I've checked the first 2 and they are fine, I don't think I changed any slicer settings but I am going to try to reset it to see if that improves it. What is Z-endstop though and how can I re-position it? Sep 8, 2019 at 18:53\n\u2022 @UltraGamer the z endstop is the physical limit switch that triggers \"here is Z=0\" Sep 8, 2019 at 19:09\n\u2022 Where is that located on the printer? Sep 8, 2019 at 19:11\n\nThis seems like a long-shot, but I've noticed at this time of year many 3D prints fail. We noticed 4 printers all went dead and had massive non-stick issues last year about this time. Turns out it was mostly around changes in temperature and humidity - the outside temperature changed inside AC settings\/wind-flow, etc.\n\nSo, you might think through some of the meta-causes of where the printer is, and if temp\/air\/humidity might be just enough chaos to not make the material stick. Right in September, I start putting a light layer of glue down on the glass bed under each print or increasing the use of rafts...\n\n\u2022 Ok, That could be it considering it is in the garage, Do you have any idea why most of the older sliced models work though? Sep 9, 2019 at 0:47\n\u2022 I\u2019ve found that when it\u2019s colder outside I slow down my print speed, especially on the first few layers. Maybe those older slicers had lower speeds? Sep 10, 2019 at 1:04\n\u2022 After looking at the gcode, nothing appears to have changed on the speed side, anything else? Sep 10, 2019 at 16:15","date":"2022-05-19 21:41:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.20163847506046295, \"perplexity\": 1592.8341286793502}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-21\/segments\/1652662530066.45\/warc\/CC-MAIN-20220519204127-20220519234127-00142.warc.gz\"}"}
| null | null |
Q: HttpWebRequest.Accept in java/Android I'm doing an application that performs a REST web service query.
I need to set the following headers:
application / vnd.sdmx.structure + xml; version = 2.0
in:
webRequest.Accept
but webRequest.Accept is a property of .NET how can I set this in Android / Java?
I tried doing this:
getRequest.setHeader("webRequest.Accept",
"application/vnd.sdmx.structure+xml;version=2.0");
but don't work.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,822
|
Some familiar faces are leaving Milwaukee TV news, taking with them more than 50 years of experience in local TV journalism.
Myra Sanchick, who has been a reporter and anchor at WITI-TV (Channel 6) for more than 30 years, said on social media this week that Friday will be her last day at the station.
The same day, WITI's Beverly Taylor, who joined the Fox affiliate as an investigative reporter in 1997, said on Twitter that Wednesday was her last day on the air at the station. Taylor, who came to Milwaukee from Alabama, recently earned a master's degree in dispute resolution and, she tweeted, "it's time to pursue that field and a few other ventures."
After 21 years at Fox6 tomorrow is my last day on air. Recently I got my Master's Degree in Dispute Resolution and it's time to pursue that field and a few other ventures. It will be hard to leave Fox6 and the wonderful people.Thank you to all who welcomed me into your homes.
Meanwhile, reporter TaTiana Cash has left WTMJ-TV (Channel 4). Hired as a traffic anchor on the station's morning newscast, Cash became a reporter, doing mainly feature stories during the station's daytime news shows.
And, on Wednesday, Jacob Kittilstad, a weekend anchor and reporter at WDJT-TV (Channel 58), said on Twitter that it also was his last day at the Milwaukee station. He even left his Twitter followers a greatest-hits video.
Local TV news always has a bit of a revolving-door feel to it, but the tally seems higher in Milwaukee in 2018.
Every Milwaukee station with its own newscast lost a news anchor this year, from George Mallet's exit from WTMJ in February and Jessob Reisbeck's departure from WITI in August, to Kathy Mykleby's retirement from WISN-TV (Channel 12) and Kate Chappell's leaving WDJT-TV (Channel 58) in November.
|
{
"redpajama_set_name": "RedPajamaC4"
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| 39
|
{"url":"https:\/\/yalmip.github.io\/_posts\/faq\/2016-09-17-maximize\/","text":"# How do I maximize a function instead of minimize?\n\nUpdated:\n\nYALMIP always assumes minimization. Hence, to maximize, you simply negate your objective function. Note that this will lead to a non-convex objective function if the objective is convex.","date":"2018-05-25 02:53:33","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9694317579269409, \"perplexity\": 321.55490729539326}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-22\/segments\/1526794866938.68\/warc\/CC-MAIN-20180525024404-20180525044404-00478.warc.gz\"}"}
| null | null |
Video Analysis: Ukrainian Anti-Tank Weapons Appear to be Decimating Russian Armored Vehicles
A former US Army Infantry Task Force Commander who led units which destroyed the Iraqi Republican Guard near the Baghdad Airport in 2003 provides analysis on the strength of Javelin Anti-Tank Missiles
Javelin Anti-Tank missiles may be decimating Russian armored vehicles attempting to close in on Kyiv, and they may be using an effective strategy to shut down or close off key "intersections" or "choke points" vital to any Russian advance. They may also been disabling or destroying key supply lines necessary for the Russian force to sustain any kind of advance.
On Feb 28th, CNN's Mathew Chance described video shots of a column of destroyed Russian armored vehicles in what looked like an attempt to cross a bridge. There were fallen soldiers and burnt and destroyed Russian vehicles in the video, in an area which looked like a bridge or narrow point of entry.
Javelin Anti-Tank Missiles
While the exact weapons used to destroy these vehicles may not be known for obvious security reasons, the destruction to the vehicles may have been caused by Javelins.
The Javelin Close Combat Missile System – Medium (CCMS-M) is a man-portable, medium-range tactical missile system that provides the U.S. Army and Marine Corps with precision direct-fire effects to defeat main battle tanks and other armored vehicles.
Experienced Iraqi war commanders do say that Javelins can destroy tanks and have been used in the past. At the same time, it could certainly be possible that that kind of damage to armored vehicles could also be caused by artillery or missiles fired from helicopters or aircraft.
Most artillery systems, however, are challenged to hit moving targets and often used as "area" weapons to destroy fixed areas and enable troops to maneuver with the cover of suppressive fire. However, precision-guided artillery does exist now, and some emerging artillery weapons have shown an ability in testing to hit moving targets with laser-guidance, Ukrainian forces may not have advanced precision-artillery systems.
Javelins, however, or air-fired missiles such as a HELLFIRE, are able to destroy moving vehicles trying to cross a bridge. Certainly small arms fire of any kind would not be able to destroy a small column of advancing armored vehicles.
The Indirect Fire Protection Capability increment 2-I Multi Mission Launcher fires a Longbow Hellfire missile during a test at White Sands Missile Range, N.M., in 2016. The proposed FY20 budget includes funds to field two "Iron Dome" batteries as int... (Photo Credit: U.S. Army)
It would make sense if Ukrainian forces were ambushing or closing off chokepoints or passageways crucial to a Russian attack given that it might be a way that a smaller number of well-armed soldiers with anti-tank missiles could potentially inflict a devastating combat impact without necessarily using tanks or heavy vehicles.
Interestingly, the concept of hitting "intersections and chokepoints" was described to Warrior by a former US Army Infantry Task Force Commander who led units which destroyed the Iraqi Republican Guard near the Baghdad Airport in 2003 during the opening invasion.
Retired Lt. Col. Scott Rutter told Warrior in a interview that, during his units attack on the Iraqi Republican Guard, enemy forces did have some marginal success trying to control or "shut down" intersections. At the beginning of the Russian invasion, Rutter told Warrior he anticipated that Ukrainian forces would use these kinds of tactics to stop a Russian advance. He may have been correct, at least when it comes to what the CNN video looks like.
"I think it's almost going to be like what the Republican Guard attempted to do to us as we were moving up, which was to try to find key terrain and key intersections to disrupt the formations. The Iraqis were a little bit effective. As we were moving up on key intersections, they tried to hold those intersections to destroy, disrupt and delay. That is probably what the Ukrainians are going to attempt to do," Rutter added.
A narrow area vehicles have to transit through, or urban areas between buildings, would seem to leave armored vehicles vulnerable to Javelin fire or destruction from other kinds of air-and-ground based anti-tank weapons.
T-90 Tank
This phenomenon may prove to be of even greater significance should Russian forces close in on urban areas, as buildings and narrow streets could offer Javelin armed dismounted units to attack armored vehicles from hidden or more "obscured" locations.
While there may be other weapons being used which are not being discussed for obvious security reasons, the Pentagon has been clear about its move to send Javelins. Javelins may be now proving to be a crucial factor in stopping Russian attacks and giving new life to the Ukrainian fighters.
Kris Osborn, Warrior Maven President, Center for Military Modernization
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 3,282
|
Q: How do I disable vim/gvim warnings about missing CR (W15) I routinely move files back and forth between Unix (Linux) and Windows 10 Home, and the files are always using the Unix line endings (fileformat=unix is set in my ~/.vimrc file on Windows). When I open those files on Windows, despite setting fileformat, I get warnings about a missing (W15).
A simple ESC or RETURN/ENTER (whichever you prefer to call it) clears those, but they are getting really annoying. So my question, which I hope has a simple answer is, how do I disable those W15 warnings so I just don't see them anymore?
I've read :help W15, and it seems to imply that having fileformat set should make it work (which I'm guessing means it shouldn't be whining), but that's not what I'm seeing...it's still whining every time.
Thanks,
--jim
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,595
|
Q: How would you deal with categorical data in a naive Bayesian classifier? I've built a little naive Bayesian classifier that works with Boolean and real values. Boolean distributions are dealt with via Bernoulli distributions, while real valued data are dealt with kernel mixture estimators. I'm currently in the process of adding count data in.
How would one deal with categorical data though, e.g. Monday, Tuesday, Wednesday, or Toyota, Honda, Ford?
My initial thoughts are to assign a number to each category, treat it as a normal real value and round to the nearest integer category on prediction. That seems very wrong to me though.
A: For an Naive Bayes classifier, categorical values are the easiest to deal with. All you are really after is P(Feature | Class). This should be easy for the days of the week. Compute P(Monday | Class=Yes) and so on.
A: The way to deal with categorical data is to create each category as a feature and with boolean values. Not only this way removes the limitation of categories for some of the libraries (no applicable here since you have written your own function) but also it's fast.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 2,199
|
BMW of San Antonio, Texas has an epic selection of 2019 BMW M240i is available for purchase!
Incredible performance. Brilliant technology. Head-turning design. The all-new 2019 BMW M240i has arrived on the scene and is making quite a splash amongst customers and critics alike. Jam-packed with exciting new updates for 2019, the all-new BMW M240i features undeniable style and wicked performance that will have you longing for more time on the road. Buckle up, get ready, and let's take a closer look at the all-new 2019 BMW M240i.
Performance: Built for ultimate speed and performance, the 2019 BMW M240i boasts a pretty mighty powertrain under its perfectly sculpted hood. Loaded up with a 3.0-liter M Performance TwinPower Turbo inline 6-cylinder, the newest BMW M240i generates a staggering 335 horsepower and 369 pound-feet of torque. That means that the 2019 BMW M240i can sprint from 0-60 miles per hour in a shocking 4.4 seconds when equipped with the automatic transmission, and 4.6 seconds when equipped with the manual! Got a need for speed? The all-new BMW M240i has got you covered.
Technology: Not only does the 2019 BMW M240i boast incredible driving performance, but it also features some of the latest and greatest automotive technology that can be found in today's market. BMW's iDrive infotainment system arrives standard in the all-new BMW M240i, granting effortless access to weather, music, messages and more directly through your new BMW. The 2019 BMW M240i also comes with SiriusXM® Satellite Radio with a complimentary 1-year subscription, so you can enjoy commercial-free tunes every time you drive.
Design: Featuring sweeping lines, a luxurious interior, and a ton of attitude, the 2019 BMW M240i has a irrevocable sense of style that just can't be beaten. Available in seven brilliant colors, the all-new BMW M240i was made to match your style. Choose from a Coupe, Convertible, Sedan, or Gran Coupe body style and the 2019 BMW M240i is ready for action, no matter what. And with a spectacular sunroof overhead, you'll be itching to roll the windows down and simply enjoy your time on the road in the all-new BMW M240i.
The largest BMW dealership in North America is BMW of San Antonio! Click here to visit us today.
|
{
"redpajama_set_name": "RedPajamaC4"
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| 6,920
|
Block placed at the back of the house.
Interior of older house – note the old window that was blocked in. Very impressive stone masonry around the windows.
The front corner – note the newer stone work – it is under a fireplace. Also, we see a window blocked in.
Old timber used to support first floor – this was also salvaged from even older buildings, it is rough -sawn.
Another view of the recycled timber. The workmanship here is very poor, unlike the workmanship on the original stone house. The builder simply stacked stone and set the timbers that he salvaged on top.
I have seen the older homes containing salvaged timber, that is rather common. Reusing timber from old houses or barns that were torn down would have made sense to the builder. The cost of materials vs. labor probably was more in favor of using the labor and trying to save on materials. Hopefully the investor that bought the house and the Architect will make use of the old house in the basement and make this a more distinctive house. This is one of the most interesting homes I've ever visited.
This entry was posted in Forensic/Foundation and Structural Repair Engineering, Interesting Houses and tagged in Forensic/Foundation and Structural Repair Engineeringc, historic structure, old house.
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{
"redpajama_set_name": "RedPajamaC4"
}
| 9,764
|
Alpine backpack fusing comfort and innovative functions, with 2-buckle lid with Opposite Opening™, storm collar, compression straps and front stretch pocket. Pole holder, Rain cover, Quick Axe™ ice-axe holder, Wave MBS™ back for ventilation and mobility. Shoulder straps with pocket and ergonomic ventilated hipbelt.
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{
"redpajama_set_name": "RedPajamaC4"
}
| 2,189
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Der Liechtensteiner Cup 1992/93 war die 48. Austragung des Fussballpokalwettbewerbs der Herren in Liechtenstein. Der FC Balzers gewann zum zehnten Mal den Titel.
Teilnehmende Mannschaften
Folgende 16 Mannschaften nahmen am Liechtensteiner Cup teil:
Achtelfinale
|}
Viertelfinale
|}
Halbfinale
|}
Finale
Das Finale fand am 20. Mai 1993 in Ruggell statt.
|}
Weblinks
Liechtensteiner Fussballverband
transfermarkt.de
rsssf.com
Cup 1992/93
Fußballsaison 1992/93
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,710
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Wiernsheim – miejscowość i gmina w Niemczech, w kraju związkowym Badenia-Wirtembergia, w rejencji Karlsruhe, w regionie Nordschwarzwald, w powiecie Enz, wchodzi w skład związku gmin Heckengäu. Leży w Heckengäu, ok. 12 km na wschód od Pforzheim.
Współpraca
Miejscowości partnerskie:
Ayancık, Turcja, od 1998
New Harmony, Stany Zjednoczone, od 1980
Pinasca, Włochy, od 1982
Powiat Enz
Gminy w Badenii-Wirtembergii
Miejscowości w Badenii-Wirtembergii
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,415
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The perfect combination of mobility and style, the Regal stroller offers the luxury of three wheels (traditionally reserved for larger, more expensive models) in an affordable, convenient stroller. The Regal stroller will hold a dog or cat up to 25 pounds and has numerous little touches for extra convenience. The front convertible wheel swivels on smooth surfaces or locks on rough surfaces for easy traveling, while the extra storage pouches are perfect for your water bottle and other accessories.
The Regal Plus pet stroller includes three of our exclusive Smart-Features. The Smart-Canopy™ folds up under the canopy hood (instead of into the pet compartment), giving your pet more space. The 38" Smart-Reach™ handle, compared to similarly-sized strollers, provides a greater amount of kick space for convenient walking at any pace. The Smart-Basket™ safely stores your belongings with a rear zipper for easy access.
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{
"redpajama_set_name": "RedPajamaC4"
}
| 6,779
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Aq Band (, also Romanized as Āq Band) is a village in Atrak Rural District, Dashli Borun District, Gonbad-e Qabus County, Golestan Province, Iran. At the 2006 census, its population was 1,294, in 246 families.
References
Populated places in Gonbad-e Kavus County
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,204
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Q: Using and Graphing The Results of a Crosstab Dataframe in Python I am trying to do something that should be pretty simple - create a crosstab from some data and then manipulate/graph the results.
Take the following code:
import pandas as pd
import numpy as np
df=pd.read_csv("https://raw.githubusercontent.com/wesm/pydata- book/master/ch08/tips.csv", sep=',')
df_out=df.pivot_table(index=["day"],values=["tip"], columns=["sex"],aggfunc=[np.sum])
Which gives me a pivot table of tips by day and looks like the following:
The problem is that I need a dataframe that looks like:
So I can interact with is and graph it
For example, I want to do
df[female]-df['male']
and I want to graph a seaborn factor plot of male and female by day
How do I get rid of the extraneous data here? I tried dropping columns, resetting the index, etc, but can't seem to figure it out
Thanks for your help - been fighting with this all day
A: I think I remember running into this with other aggregate functions. Will the following work?
new_df = df['sum']['tip']
new_df['delta'] = new_df['female'] - new_df['male']
A: Alternate method:
df_out = df_out['sum']['tip']
del df_out.columns.name
del df_out.index.name
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,081
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\section{Introduction and results}
\subsection{The general Boussinesq `abcd' model} In this work, we are concerned with the Boussinesq system
\begin{equation}
\label{b:1}
\left|
\begin{array}{l}
\eta_t+ u_x+(\eta u)_x+ a u_{xxx}-b \eta_{xxt}=0\\
u_t+\eta_x+u u_x+c \eta_{xxx}-d u_{xxt}=0.
\end{array}
\right.
\end{equation}
The first formal derivation for this system has appeared in the work of Bona-Chen-Saut, \cite{BCS}
to describe the \nolinebreak (essentially two dimensional) motion of small-amplitude long waves on the surface of an ideal fluid under the force of gravity. Here, $\eta$ represents the vertical deviation of the free surface from its rest position, while $u$ is the horizontal velocity at time $t$. In the case of zero surface tension $\tau=0$, the constants $a,b,c,d$ must satisfy in addition the consistency conditions $a+b=\f{1}{2}(\theta^2-1/3)$ and $c+d=\f{1}{2}(1-\theta^2)>0$. In the case of non-zero surface tension however, one only requires $a+b+c+d=\f{1}{3}-\tau$.
For this reason (as well as from the pure mathematical interest in the analysis of \eqref{b:1}), one may as well consider \eqref{b:1} for all values of the parameters.
Systems of the form \eqref{b:1} have been the subject of intensive investigation over the last decade. In particular, the role of the parameters $a,b,c,d$ in the actual fluid models has been explored in great detail in the original paper \cite{BCS} and later in \cite{BCS1}. It was argued that only models in the form \eqref{b:1}, for which one has linear and nonlinear well-posedness are physically relevant. We refer the reader to these two papers for further discussion and some precise conditions, under which one has such well-posedness theorems.
Regarding explicit traveling wave solutions, Chen, has considered various cases of interest in \cite{Ch1}, \cite{Ch2}. In fact, she has written down numerous traveling wave solutions (i.e. in the form $(\eta, u)=(\varphi(x-w t), \psi(x - w t))$, where in fact some of them are not necessarily homoclinic to zero at $\pm \infty$. In a subsequent paper, \cite{Ch3}, Chen has also found new and explicit
multi-pulsed traveling wave solutions.
In \cite{CCN}, Chen-Chen-Nguyen consider another relevant case, namely the BBM system, which ($a=c=0, b=d=\f{1}{6}$). They construct periodic traveling wave solutions for the BBM case, as well as in more general situations. In \cite{AABCW}, the authors explore the existence theory for the the BBM system as well as its
relations to the single BBM equation.
We wish to discuss another aspect of \eqref{b:1}, which is its Hamiltonian formulation. Since it is derived from the Euler equation by ignoring the effects of the dissipation, one generally expects such systems to exhibit a Hamiltonian structure. {\it This is however not generally the case, unless one imposes some further restrictions on the parameters}. Indeed, if $b=d$, one can easily check that
$$
H(\eta, u)=\int -c \eta_x^2-au_x^2+\eta^2+(1+\eta) u^2 dx
$$
Furthermore, $H(\eta, u)$ is positive definite only if $a, c<0$. From this point of view, it looks natural to consider the case $b=d$ and $a,c<0$. In order to focus our discussion, we shall concentrate then on this version
\begin{equation}
\label{1}
\left|
\begin{array}{l}
\eta_t+ u_x+(\eta u)_x+ a u_{xxx}-b \eta_{xxt}=0\\
u_t+\eta_x+u u_x+c \eta_{xxx}-b u_{xxt}=0.
\end{array}
\right.
\end{equation}
We will refer to \eqref{1} as the Boussinesq `abc' system.
It is a standard practice that stable coherent structures, such as traveling pulses etc. are produced as constrained minimizers of the corresponding (positive definite) Hamiltonians, with respect to a fixed conserved quantity.
In fact, this program has been mostly carried out, at least in the Hamiltonian cases, in a series of papers by Chen, Nguyen and Sun. More precisely, in \cite{CNS}, the authors have shown that traveling waves for \eqref{b:1} exist in the regime\footnote{which in particular requires that $a+b+c+d<0$, corresponding to a ``large'' surface tension $\tau>\f{1}{3}$} $b=d$,
$a, c<0, ac>b^2$. In addition, they have also shown stability of such waves in the sense of a `set stability' of the set of minimizers. In the companion paper \cite{CNS1}, the authors have considered the general case $b=d>0$, $a,c<0$, which in particular allows for small surface tension.
The existence of a traveling wave was proved for every speed
$|w| \in (0, \min(1,\f{\sqrt{ac}}{b}))$.This is the so-called subsonic regime. Finally, we point out to a recent work by Chen, Curtis, Deconinck, Lee and Nguyen, \cite{CCDLN} in which the authors study numerically various aspects of spectral stability/instability of some solitary waves of \eqref{b:1}, including the multipulsed solutions exhibited in \cite{Ch3}. In the same paper, the authors
also study (numerically) the transverse stability/instability of the same waves, viewed as solutions to the two dimensional problem.
The purpose of this paper is to study rigorously the spectral stability of some explicit traveling waves in the regime $b=d>0$, $a, c<0$. This would be achieved via the use of the instabilities indices counting formulas of Kapitula, Kevrekidis and Sandstede, \cite{KKS1}, \cite{KKS2} and the subsequent refinement by Kapitula, Stefanov \cite{KS}.
\subsection{The traveling wave solutions}
In this section, we follow almost verbatim the description of some explicit solutions of interest of \eqref{b:1}, given by Chen, \cite{Ch1}, see also the more detailed exposition of the same results in \cite{Ch2}. More precisely,
the solutions of interest are traveling waves, that is in the form
$$
\eta=\varphi(x-w t), \ \ u(x,t)=\psi(x-w t).
$$
A direct computation shows that if we require that the
pair $(\varphi, \psi)$ vanishes at $\pm \infty$, then it satisfies the system
\begin{equation}
\label{5}
\left|
\begin{array}{l}
(1+c\partial_x^2)\varphi-w(1-b\partial_x^2)\psi+\f{\psi^2}{2}=0\\
-w(1-b\partial_x^2)\varphi+(1+a\partial_x^2)\psi+\varphi \psi=0.
\end{array}
\right.
\end{equation}
The typical ansatz that one starts with, in order to simplify the system
\eqref{5} to a single equation is $\psi=B\varphi$. This has been worked out by Chen, \cite{Ch1}, \cite{Ch2}. The following result is contained in the said papers.
\newpage
\begin{theorem}(Chen, \cite{Ch1}, \cite{Ch2})
\label{mchen}
Let the parameters $a,b,c$ in the system satisfy one of the following
\begin{enumerate}
\item $a+b\neq 0$, $p=\f{c+b}{a+b}>0$, $(p-1/2)((b-a)p-b)>0$
\item $a=c=-b$, $b>0$
\end{enumerate}
Then, there are the following (pair of) exact traveling
wave solutions (i.e. solutions of \eqref{5}) $(\varphi(x- wt), \psi(x- wt))$, where
\begin{eqnarray*}
& & \varphi(x)=\eta_0 sech^2(\lambda x) \\
& & \psi(x)= B(\eta_0)\eta_0 sech^2(\lambda x)
\end{eqnarray*}
and
$$
w=w(\eta_0)= \pm \frac{3+2\eta_0}{\sqrt{3(3+\eta_0)}};\ \lambda= \f{1}{2} \sqrt{\f{2\eta_0}{3(a-b)+2b(\eta_0+3)}}; \ \ B(\eta_0)=\pm \sqrt{\f{3}{\eta_0+3}},
$$
and $\eta_0$ is a constant that satisfies
\begin{enumerate}
\item $\eta_0=\f{3(1-2p)}{2p}$ in Case $(1)$
\item $\eta_0>-3, \eta_0\neq 0$ in Case $(2)$.
\end{enumerate}
\end{theorem}
\subsection{Different notions of stability}
Before we state our results, we pause to discuss the various definitions of stability. First, one says that the solitary wave solution $(\varphi_w, \psi_w)$ is orbitally stable, if for every $\varepsilon>0$, there exists $\delta>0$, so that whenever
$\|(f,g)-(\varphi_w, \psi_w)\|_{X}<\delta$, one has that the corresponding solutions $(\eta, u): (f,g)=(\eta, u)|_{t=0}$
$$
\sup_{t>0} \inf_{x_0}\|(\eta(x-x_0,t), u(x -x_0,t))-(\varphi(x-w t), \psi(x - w t))\|_X<\varepsilon.
$$
Note that we have not quite specified a space $X$, since this usually
depends on the particular problem at hand (and mostly on the available conserved quantities), but suffices to say that $X$ is usually chosen to be a natural energy space for the problem.
This notion of (nonlinear) stability has been of course successfully used to treat a great deal of important problems, due to the versatility of the classical Benjamin and Grillakis-Shatah-Strauss approaches. However, it looks like these methods are not readily applicable (if at all) to the Boussinesq `abc' system. We encourage the interested reader to consult the discussion in \cite{CNS}, where a weaker, but related stability was established in the regime $ac>b^2$ and additional smallness assumption on the wave is required as well. This is why, one needs to develop an alternative approach to this important problem, which is one of the main goals of this work.
In this paper, we will concentrate on spectral stability. There is also (the closely related and almost equivalent) notion of linear stability, which we also mention below. In order to introduce the object of our study, as well as to motivate its relevance, let us perform a linearization of the nonlinear system \eqref{1}. Using the ansatz
$$
\left|
\begin{array}{l}
\eta=\varphi(x-w t)+v(t, x-w t)\\
u=\psi(x-w t)+z(t, x-w t),
\end{array}
\right.
$$
in \eqref{1} and ignoring all quadratic terms in the form $O(v^2), O(v z), O(z^2)$ leads to the following linearized problem
$$
(1-b\partial_x^2)\left(\begin{array}{c}
v \\ z
\end{array}\right)_t= -\partial_x \left(\begin{array}{c c}
0 & 1 \\ 1 & 0
\end{array}\right)\left(\begin{array}{cc}
1+c\partial_x^2 & b w \partial_x^2 +\psi-w \\
b w \partial_x^2 +\psi-w & 1+a \partial_x^2 +\varphi
\end{array}\right)
$$
Letting
\begin{equation}
\label{110}
L=\left(\begin{array}{cc}
1+c\partial_x^2 & b w \partial_x^2 +\psi-w \\
b w \partial_x^2 +\psi-w & 1+a \partial_x^2 +\varphi
\end{array}\right), J= -\partial_x (1-b\partial_x^2)^{-1}\left(\begin{array}{c c}
0 & 1 \\ 1 & 0
\end{array}\right)
\end{equation}
the linearized problem that we need to consider may be written in the form
\begin{equation}
\label{90}
u_t=J L u
\end{equation}
Note that in the whole line context, $L$ is a self-adjoint operator, when considered with the natural domain $D(L)=H^2(\mathbf R^1)\times H^2(\mathbf R^1) $. Letting $H:= J L$, we see that the problem \eqref{90} is in the form $u_t=H u$. The study of linear problems in this form is at the basis of
the deep theory of $C_0$ semigroups. Informally, if the Cauchy problem $u_t=H u$ has global solutions for all smooth and decaying data, we say that $H$ generates a $C_0$ semigroup $\{T(t)\}_{t>0}$ via the exponential map $T(t)=e^{t H}$. Furthermore, we say that we have {\it linear stability} for the linearized problem $u_t=H u$, whenever the growth rate of the semigroup is zero or equivalently $\lim_{t\to \infty} e^{-\delta t} \|T(t) f\|=0$ for all $\delta>0$ and for all sufficiently smooth and decaying functions
$f$. Finally, we say that the system is {\it spectrally stable}, if $\sigma(H)\subset \{z: \Re z\leq 0\}$. It is well-known that if $H$ generates a $C_0$ semigroup, then linear stability implies spectral stability, but not vice versa. Nevertheless, the two notions are very closely related and in many cases (including the ones under consideration), they are indeed equivalent. For the purposes of a formal definition, we proceed as follows
\begin{definition}
\label{defi:1}
We say that the problem \eqref{90} is unstable, if there is $\mathbf{f}\in H^2(\mathbf R^1)\times H^2(\mathbf R^1)$ and $\lambda: \Re \lambda>0$, so that
\begin{equation}
\label{100}
J L \mathbf{f}=\lambda \mathbf{f}.
\end{equation}
Otherwise, the problem \eqref{90} is stable. That is, stability is equivalent to the absence of solutions of \eqref{100} with $\lambda: \Re\lambda>0$.
\end{definition}
\subsection{Main results}
We are now ready to state our results. We chose to split them in two cases, just as in Theorem \ref{mchen}. For the case $a=c=-b, b>0$, we have
\begin{theorem}
\label{theo:1}
Let $a=c=-b, b>0$. Then, the traveling wave solutions of the `abc' system
\begin{equation}
\label{waves}
\left(\eta_0 sech^2\left(\f{x - w t}{2\sqrt{b}}\right), \pm \eta_0
\sqrt{\f{3}{\eta_0+3}} sech^2\left(\f{x-w t}{2\sqrt{b}}\right)\right)
\end{equation}
with speed $w=\pm \frac{3+2\eta_0}{\sqrt{3(3+\eta_0)}}$
are stable, for all $\eta_0:
\eta_0\in (-\f{9}{4},0).
$
Equivalently, all waves in \eqref{waves} are stable, for all speeds $|w|<1$.
\end{theorem}
\noindent Note that $|w|<1$ is equivalent to $\eta_0\in (-\f{9}{4},0)$, so we assume this henceforth.
In the remaining case, we assume only $a=c<0, b=d>0$, but observe that in this case, Theorem \ref{mchen} requires that $\eta_0=-3/2, w=0$, that is the traveling waves become standing waves.
\begin{theorem}
\label{theo:20}
Let $a=c<0, b=d>0$.
Then, the standing wave solutions of the Boussinesq system
$$
\varphi(x)=-\f{3}{2} sech^2\left(\f{x}{2\sqrt{-a}}\right), \psi(x)= \pm \f{3}{\sqrt{2}}
sech^2\left(\f{x}{2\sqrt{-a}}\right)
$$
are spectrally stable \underline{if and only if}
\begin{equation}
\label{1000}
\dpr{(a\partial_x^2+1-\varphi)^{-1}(\varphi- b \varphi'')}{(\varphi- b \varphi'')}\leq 8 \sqrt{-a}
\left( \f{9}{2}+\f{12}{5} \f{b}{|a|} -
\f{3}{10} \f{b^2}{a^2}\right).
\end{equation}
In particular, the condition \eqref{1000} holds ( and thus the waves are spectrally stable), whenever
$$
0\leq \f{b}{-a}< 8.00163,
$$
On the other hand, the condition \eqref{1000} fails ( and thus the waves are spectrally unstable), if
$$
\f{b}{-a}> 8.82864.
$$
\end{theorem}
{\bf Remark: }
Note that while, we cannot explicitly compute the value $(a\partial_x^2+1-\varphi)^{-1}(\varphi- b \varphi'')$ in \eqref{1000}, we obtain estimates, which imply some pretty good results for the stability/instability intervals. One can in fact push this further to narrow the gap between the stability and instability regions, predicted by \eqref{1000}. This can be done in principle with any degree of accuracy, but it increases the complexity the argument.
\section{Preliminaries}
In this section, we collect some preliminary results, which will be useful in the sequel.
\subsection{Some spectral properties of $L$}
We shall need some spectral information about the operator $L$. We collect the results in the following
\begin{proposition}
\label{prop:10}
Let $a,c<0$ and $w: 0\leq |w|<\min\left(1, \f{\sqrt{a c}}{|b|}\right)$. Then, the self-adjoint operator $L$ has the following spectral properties
\begin{itemize}
\item
Then the operator $L$ has an eigenvalue at zero, with an eigenvector
$\left(\begin{array}{c} \varphi' \\ \psi' \end{array} \right)$.
\item There is $\kappa>0$, so that the essential spectrum is in $\sigma_{ess}(L)\subset [\kappa, \infty)$.
\end{itemize}
\end{proposition}
\begin{proof}
The first property is easy to establish, this is the usual eigenvalue at zero generated by translational invariance. For the proof, all one needs to do is take a spatial derivative in the defining system \eqref{5}, whence $L\left(\begin{array}{c} \varphi' \\ \psi' \end{array} \right)= \left(\begin{array}{c} 0 \\ 0 \end{array} \right)$.
Regarding the essential spectrum, we reduce matters to the Weyl's theorem (using the vanishing of the waves at $\pm \infty$), which ensures that
$$
\sigma_{ess.}(L)=\sigma_{ess.}(L_0)=\sigma[\left(\begin{array}{cc}
1+c\partial_x^2 & b w \partial_x^2 -w \\
b w \partial_x^2 -w & 1+a \partial_x^2
\end{array}\right)]
$$
That is, it remains to check that the matrix differential operator $L_0>\kappa$.
By Fourier transforming $L_0$, it will suffice to check that the matrix
$$
L_0(\xi)=\left(\begin{array}{cc}
1-c\xi^2 & -w(b\xi^2 +1) \\
-w(b\xi^2 +1)& 1-a \xi^2
\end{array}\right)
$$
is positive definite for all $\xi\in \mathbf R^1$. Since $1-c\xi^2\geq 1$, it will suffice to check that the determinant has a positive minimum over $\xi\in \mathbf R^1$.
We have
$$
det(L_0(\xi))= \xi^4(ac-b^2 w^2)+\xi^2(-a-c-2b w^2)+(1-w^2)\geq (1-w^2)+
2\xi^2(\sqrt{a c}- |b| w^2),
$$
where in the last inequality, we have used $-a-c\geq 2 \sqrt{a c}$. The strict
positivity follows by observing that $\sqrt{ac}\geq |b| w \geq |b|w^2$, since $w<1$.
\end{proof}
\subsection{Instability index count} In this section, we introduce the instability indices counting formulas, which in many cases of interest can in fact be used to determine accurately both stability and instability regimes for the waves under consideration. As we have mentioned above, this theory has been under development for some time, see \cite{Mad}, \cite{Kap}, \cite{Pel}, but we use a recent formulation due to Kapitula-Kevrekidis and Sandstede (KKS), \cite{KKS1} (see also \cite{KKS2}). In fact, even the (KKS) index count formula is not directly applicable\footnote{due to a crucial assumption for invertibility of the skew-symmetric operator $J$, which is not satisfied for $\partial_x$ acting on $\mathbf R^1$} to the
problem of \eqref{90}, which is why Kapitula and Stefanov, \cite{KS} have found an approach, based on the KKS of the theory, which covers this situation.
In order to simplify the exposition, we will restrict to a corollary of the main result in \cite{KS}.
More precisely, a the stability problem in the form is considered in the form
\begin{equation}
\label{c:5}
\partial_x\mathcal L u= \lambda u,
\end{equation}
where $\mathcal L$ is a self-adjoint linear differential operator with domain
$D(\mathcal L)=H^s(\mathbf R^1)$ for some $s$. It is assumed that for the
operator $\mathcal L$,
\begin{enumerate}
\item there are $n(\mathcal L)=N<+\infty$ negative eigenvalues\footnote{We will henceforth denote by $n(M)$ the number of negative eigenvalues (counting multiplicities) of
a self-adjoint operator $M$} (counting multiplicity), so that each of the corresponding eigenvectors $\{f_j\}_{j=1}^N$ belong to $H^{1/2}(\mathbf R^1)$.
\item there is a $\kappa>0$ such that
$\sigma_{ess}(\mathcal L)\subset[\kappa^2,+\infty)$
\item $\dim[\ker(\mathcal L)]=1$, $\ker(\mathcal L)= span\{\psi_0\}$,
$\psi_0$ real-valued function, $\psi_0\in H^{\infty}(\mathbf R^1)\cap \dot{H}^{-1}(\mathbf R^1)$.
\end{enumerate}
Here, $\dot{H}^{-1}(\mathbf R^1)$ is the homogeneous Sobolev space, defined via the norm
$$
\|u\|_{\dot{H}^{-1}(\mathbf R^1)}:=\left( \int_{\mathbf R^1} \f{|\hat{u}(\xi)|^2}{|\xi|^2} d\xi\right)^{1/2}.
$$
or equivalently, $u=\partial_x z$ in sense of distributions, where $z\in L^2$ and
$\|u\|_{\dot{H}^{-1}(\mathbf R^1)}:=\|z\|_{L^2}$. In that case, we have
\begin{theorem}(Theorem 3.5, \cite{KS})
\label{t:index}
For the eigenvalue problem
\begin{equation}
\label{c:10}
\partial_x\mathcal L u=\lambda u,\quad u\in L^2(\mathbf R^1),
\end{equation}
where the self-adjoint operator $\mathcal L$ satisfies $D(\mathcal L)=H^s(\mathbf R^1)$ for some
$s>0$, assume that
\[
\langle\mathcal L^{-1}\partial_x^{-1}\psi_0,\partial_x^{-1}\psi_0\rangle\neq0.
\]
Then, the number of solutions of \eqref{c:5}, $n_{unstable}(\mathcal L)$,
with $\lambda:\Re \lambda>0$ satisfies\footnote{here $\partial_x^{-1} \psi_0$ is {\bf any} $L^2$ function $f$, so that $\psi_0=\partial_x f$ in distributional sense}
\begin{equation}
\label{120}
0\leq n_{unstable}(\partial_x \mathcal L)= n(\mathcal L)-n\left(\langle\mathcal L^{-1}\partial_x^{-1}\psi_0,\partial_x^{-1}\psi_0\rangle\right)\mod 2.
\end{equation}
\end{theorem}
Of course, our eigenvalue problem \eqref{100} does not immediately
fit the form of Theorem \ref{t:index}. First, Theorem \ref{t:index} applies for scalar-valued operators $\mathcal L$, while we need to deal with vector-valued operators. This is a minor issue and in fact, one sees easily that the arguments in \cite{KS} carry over easily in the case, where $\mathcal L$ is a vector-valued self-adjoint operator as well. A second, more substantive issue is that the form of \eqref{100} is not quite the one in \eqref{c:10}. Namely, we have that the operator $J$, while still skew-symmetric is not equal to $\partial_x$.
In order to fix that, we need to recast the eigenvalue problem \eqref{100} in a slightly different form. Indeed, letting $\mathbf{f}=(1-b\partial_x^2)^{-1/2}\mathbf{g}$ and taking $(1-b\partial_x^2)^{1/2}$ on both sides of \eqref{100}, we may rewrite it as follows
$$
-\partial_x \left(\begin{array}{c c}
0 & 1 \\ 1 & 0
\end{array}\right)(1-b\partial_x^2)^{-1/2} L (1-b\partial_x^2)^{-1/2} \mathbf{g}=\lambda \mathbf{g}.
$$
If we now introduce
$$
\tilde{J}:= -\partial_x \left(\begin{array}{c c}
0 & 1 \\ 1 & 0
\end{array}\right); \ \ \tilde{L}:=(1-b\partial_x^2)^{-1/2} L (1-b\partial_x^2)^{-1/2},
$$
we easily see that $\tilde{J}$ is still anti-symmetric, $\tilde{L}$ is
self-adjoint and we have managed to represent the eigenvalue problem in the form $\tilde{J} \tilde{L} \mathbf{g}=\lambda \mathbf{g}$. Note that the operator $\tilde{J}$ is very similar to $\partial_x$, except for the action of the invertible symmetric operator $\left(\begin{array}{c c}
0 & 1 \\ 1 & 0
\end{array}\right)$ on it. It is not hard to see that the result of Theorem \ref{t:index} applies to it (while it still fails the standard conditions of the KKS theory, due to the non-invertibility of $\tilde{J}$). Note that one needs to replace $\partial_x^{-1}$ by $\tilde{J}^{-1}$ in the formula \eqref{120}. Furthermore, the number of unstable modes for the two systems ($J L$ and $\tilde{J} \tilde{L}$) is clearly the same, due to the simple transformation $(1-b\partial_x^2)^{-1/2}$ connecting the corresponding eigenfunctions.
Thus, {\it if we can verify the conditions under which Theorem \ref{t:index} applies}, we get the stability index formula
\begin{equation}
\label{130}
n_{unstable}(J L)=n_{unstable}(\tilde{J} \tilde{L})=n(\tilde{L})-
n(\dpr{\tilde{L}^{-1} \tilde{J}^{-1}\psi_0}{\tilde{J}^{-1}\psi_0}) \mod 2.
\end{equation}
Since by Proposition \ref{prop:10}, $L\left(\begin{array}{c}\varphi'\\ \psi'\end{array}\right)=0$, we conclude that $\tilde{L}[(1-b \partial_x^2)^{1/2} \left(\begin{array}{c}\varphi'\\ \psi'\end{array}\right)]=0$. It follows that
$\psi_0=\partial_x (1-b \partial_x^2)^{1/2} \left(\begin{array}{c}\varphi \\ \psi\end{array}\right)$ and
\begin{eqnarray*}
\dpr{\tilde{L}^{-1} \tilde{J}^{-1}\psi_0}{\tilde{J}^{-1}\psi_0} &=&
\dpr{ L^{-1} [(1-b \partial_x^2) \left(\begin{array}{c c}
0 & 1 \\ 1 & 0
\end{array}\right) \left(\begin{array}{c}\varphi \\ \psi\end{array}\right)]}{ (1-b \partial_x^2) \left(\begin{array}{c c}
0 & 1 \\ 1 & 0
\end{array}\right) \left(\begin{array}{c}\varphi \\ \psi\end{array}\right)}=\\
&=& \dpr{L^{-1}[(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)]}{(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)}
\end{eqnarray*}
Thus, we conclude that we will have established spectral stability for \eqref{100},
if we can verify the conditions $(1), (2), (3)$ of Theorem \ref{t:index} for the operator $\tilde{L}$,
$n(\tilde{L})=1$ and
\begin{equation}
\label{150}
\dpr{L^{-1}[(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)]}{(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)}<0.
\end{equation}
and instability otherwise.
Concretely, we will verify the conditions on $\tilde{L}$ in Proposition \ref{prop:25} below, after which, we compute the quantity in \eqref{150} in Proposition
\ref{prop:30}.
\begin{proposition}
\label{prop:25}
The self-adjoint operator $\tilde{L}=(1-b \partial_x^2)^{-1/2}L (1-b \partial_x^2)^{-1/2}$ satisfies
\begin{enumerate}
\item $\sigma_{ess.}(\tilde{L})\subset [\kappa, \infty)$ for some positive $\kappa$.
\item $n(\tilde{L})=1$.
\item $Ker(\tilde{L})=span\{(1-b \partial_x^2)^{1/2}\left(\begin{array}{c} \varphi'\\ \psi'\end{array}\right)\}$.
\end{enumerate}
in the following cases
\nopagebreak
\begin{itemize}
\item $a=c=-b, b>0$, $B=\pm \sqrt{\f{3}{3+\eta_0}}$, $w=\pm \f{3+2\eta_0}{\sqrt{3(3+\eta_0)}},
\eta_0\in (-\f{9}{4}, 0)$.
\item $a=c<0$, $b>0$, $w=0, B=\pm \sqrt{2}$.
\end{itemize}
\end{proposition}
\begin{proposition}
\label{prop:30}
Regarding the instability index, we have
\begin{itemize}
\item For $a=c=-b, b>0$, $w=\pm \f{3+2\eta_0}{\sqrt{3(3+\eta_0)}}$, $B(\eta_0)=\pm \sqrt{\f{3}{3+\eta_0}}$, and for all $\eta_0\in (-\f{9}{4},0)$,
$$
\dpr{L^{-1}[(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)]}{(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)} < 0
$$
\item For $a=c<0$, $b>0$, $w=0$, $B=\pm \sqrt{2}$,
\begin{eqnarray*}
& & \dpr{L^{-1}[(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)]}{(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)} = \\
&=& \f{1}{3}\left(8 \sqrt{-a}
\left( -\f{9}{2}-\f{12}{5} \f{b}{|a|}+
\f{3}{10} \f{b^2}{a^2}\right)+ \dpr{(a\partial_x^2+1-\varphi)^{-1} f}{f}\right)
\end{eqnarray*}
In particular,
\begin{eqnarray*}
& & \dpr{L^{-1}[(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)]}{(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)}<0, \ \ 0<\f{b}{-a}<8.00163,\\
& & \dpr{L^{-1}[(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)]}{(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)}>0, \ \
\f{b}{-a}>8.82864.
\end{eqnarray*}
\end{itemize}
\end{proposition}
Theorem \ref{theo:1} follows by virtue of
Proposition \ref{prop:25} and Proposition \ref{prop:30}.
Thus, it remains to prove these two.
\section{Proof of Proposition \ref{prop:25}}
We start with the gap condition for $\sigma_{ess.}(\tilde{L})$ stated in
Proposition \ref{prop:25}.
\subsection{$\tilde{L}$ is strictly positive}
The idea is contained in Proposition \ref{prop:10}. Write
$$
\tilde{L}=(1-b \partial_x^2)^{-1/2}L (1-b \partial_x^2)^{-1/2}=(1-b \partial_x^2)^{-1/2}L_0 (1-b \partial_x^2)^{-1/2}+(1-b \partial_x^2)^{-1/2}(L-L_0)(1-b \partial_x^2)^{-1/2},
$$
where $L-L_0$ is a multiplication by smooth and decaying potential. It is also not hard to see that $(1-b\partial_x^2)^{-1/2}$ is given by a convolution kernel
$K: K(x)=\int_{-\infty}^\infty \f{e^{2\pi i x \xi}}{\sqrt{1+4\pi^2 b\xi^2}} d\xi$, which decays faster than polynomial at $\pm \infty$.
It follows that the operator $(1-b \partial_x^2)^{-1/2}(L-L_0)(1-b \partial_x^2)^{-1/2}$ is a compact operator on $L^2(\mathbf R^1)$ and hence By Weyl's theorem
$$
\sigma_{ess.}(\tilde{L})= \sigma_{ess.}((1-b \partial_x^2)^{-1/2}L_0 (1-b \partial_x^2)^{-1/2})=\sigma((1-b \partial_x^2)^{-1/2}L_0 (1-b \partial_x^2)^{-1/2})
$$
Thus, as we have explained in the proof of Proposition \ref{prop:10},
it will suffice to check that the matrix
$$
(1+4\pi^2 b \xi^2)^{-1/2} L_0(\xi) (1+4\pi^2 b \xi^2)^{-1/2}
$$
is positive definite. But since $L_0(\xi)$ is positive definite, the result follows. Note that this only shows that $\sigma_{ess.}(\tilde{L})\geq 0$. Since we need to show an actual gap between $\sigma_{ess.}(\tilde{L})$ and zero, it suffices to observe (by the arguments in Proposition \ref{prop:10}) that the eigenvalues of $L_0(\xi)$ have the rate of $O(\xi^2)$ for large $\xi$, which implies that the positive eigenvalues of $(1+4\pi^2 b \xi^2)^{-1/2} L_0(\xi) (1+4\pi^2 b \xi^2)^{-1/2}$ have the rate of $O(1)$.
\subsection{The negative eigenvalue and the zero eigenvalue are both simple}
We now pass to the harder task of establishing the existence and simplicity of a negative eigenvalue for $\tilde{L}$ as well as the simplicity of the zero eigenvalue. Note that as we have already observed $L\left(\begin{array}{c}\varphi' \\ \psi' \end{array}\right)=0$. It follows that
$$
\tilde{L}[(1-b\partial_x^2)^{1/2} \left(\begin{array}{c} \varphi'\\ \psi'\end{array}\right)]=
(1-b\partial_x^2)^{-1/2} [L \left(\begin{array}{c} \varphi'\\ \psi'\end{array}\right)]=0.
$$
Thus, we have already identified one element of $Ker(\tilde{L})$, but it still remains to prove that $dim(Ker(\tilde{L}))=1$, in addition to the existence and the simplicity of the negative eigenvalue of $\tilde{L}$.
Next, we find it convenient to introduce the following notation for the eigenvalues of a self-adjoint operator $\mathcal L$. Indeed, assume that $\mathcal L=\mathcal L^*$ is bounded from below, $\mathcal L\geq -c$, we order\footnote{We follow the standard convention that if an equality appears multiple times in the sequence of eigenvalues, that signifies that eigenvalue has the same multiplicity} the eigenvalues as follows
$$
\inf spec(\mathcal L)=\lambda_0(\mathcal L) \leq \lambda_1(\mathcal L)\leq \ldots.
$$
Recall also the following max min principle, due to Courant
$$
\lambda_0(\mathcal L)=\inf_{\|f\|=1} \dpr{\mathcal L f}{f}, \ \ \lambda_1(\mathcal L)=\sup_{g\neq 0}
\inf_{\|f\|=1, f\perp g} \dpr{\mathcal L f}{f}, \lambda_2(\mathcal L)=\sup_{g_1, g_2: g_1\neq a g_2}
\inf_{\|f\|=1, f\perp span[g_1, g_2]} \dpr{\mathcal L f}{f}.
$$
Clearly, our claims can be recast in the more compact form
\begin{equation}
\label{15}
\lambda_0(\tilde{L})<0=\lambda_1(\tilde{L})<\lambda_2(\tilde{L}).
\end{equation}
matters from $\tilde{L}$ to standard second order differential operators, like $L$.
\begin{lemma}
\label{le:100}
Let $a,c<0, b>0$ and $w: 0\leq |w|<\min\left(1, \f{\sqrt{a c}}{|b|}\right)$. Then
\begin{itemize}
\item all eigenvectors of $L$ from \eqref{110}, corresponding to non-positive eigenvalues, belong to \\
$H^\infty(\mathbf R^1)=\cap_{l=1}^\infty H^l(\mathbf R^1) $.
\item If $\mathcal L$ is any bounded from below
self-adjoint operator, for which \\ $\lambda_0(\mathcal L)<0=\lambda_1(\mathcal L)<\lambda_2(\mathcal L)$,
and $S$ is a bounded invertible operator, then
$$
\lambda_0(S^*\mathcal L S)<0=\lambda_1(S^*\mathcal L S)<\lambda_2(S^*\mathcal L S).
$$
\item If $L$ has the property $\lambda_0(L)<0=\lambda_1(L)<\lambda_2(L)$, then so does \\
$\tilde{L}=(1-b\partial_x^2)^{-1/2} L (1-b\partial_x^2)^{-1/2}$. That is, \eqref{15} holds.
\end{itemize}
\end{lemma}
\begin{proof}(Lemma \ref{le:100})
Take the eigenvector $\mathbf{f}$, corresponding to $-a^2, a\geq 0$, i.e. $L \mathbf{f}=-a^2 \mathbf{f}$. As observed in the proof of Proposition \ref{prop:10}, we can represent $L=L_0+\mathbf{V}$, where $\mathbf{V}$ is smooth and decaying matrix potential. In addition, recall $L_0\geq \kappa$, hence $L_0+a^2\geq \kappa Id$ and hence invertible. It follows that the eigenvalue problem at $-a^2$ can be rewritten in the equivalent form
$$
\mathbf{f}=-(L_0+a^2)^{-1}[\mathbf{V}\mathbf{f}]
$$
Clearly, $(L_0+a^2)^{-1}:L^2\to H^2$, whence we get immediately that $f\in H^2$, if $f\in L^2$. Bootstrapping this argument (recall $\mathbf{V}\in C^\infty$)
yields $f\in H^4, H^6$ etc. In the end, $\mathbf{f}\in H^\infty$.
Next, we have
$$
\lambda_0(S^*\mathcal L S)
=\inf_{f:\|f\|=1} \dpr{ S^*\mathcal L S f}{f}=\inf_{f\neq 0} \f{\dpr{\mathcal L S f}{S f}}{\|f\|^2}= \inf_{g\neq 0} \f{\dpr{\mathcal L g}{g}}{\|S^{-1} g\|^2} <0,
$$
since $\lambda_0(\mathcal L)=\inf_{g:\|g\|=1} \dpr{\mathcal L g}{g}<0$. Since $\lambda_1(\mathcal L)=0$, it follows that there is $h$, so that \\
$
\inf_{g\perp h} \dpr{\mathcal L g}{g} \geq 0.
$
Thus,
$$
\lambda_1(S^*\mathcal L S)\geq \inf_{f\perp S h} \f{\dpr{S^*\mathcal L S f}{f}}{\|f\|^2}=
\inf_{g\perp h} \f{\dpr{\mathcal L g}{g}}{\|S^{-1} g\|^2}\geq 0.
$$
Since $0$ is still an eigenvalue for $\mathcal L$ with say eigenvector $\chi$, it follows that $S^{-1} \chi$ is an eigenvector to $S^*\mathcal L S$, so $0$ is also an eigenvalue for
$S^*\mathcal L S$ and hence $\lambda_1(S^*\mathcal L S)=0$.
Regarding $\lambda_2(S^* \mathcal L S)$, we already know that
$\lambda_2(S^* \mathcal L S)>\lambda_1(S^* \mathcal L S)=0$. Assuming the contrary would mean that $\lambda_2(S^* \mathcal L S)=0$, that is $0$ is a double eigenvalue for $S^* \mathcal L S$, say with linearly independent eigenvectors $f_1, f_2$. From this and the invertibility of $S$, it follows that $S^{-1} f_1, S^{-1} f_2$ are two linearly independent vectors in $Ker(L)$, a contradiction with the assumption that $0$ is a simple eigenvalue for $L$.
The result regarding $(1-b\partial_x^2)^{-1/2} L (1-b\partial_x^2)^{-1/2}$ follows in a similar way, although clearly cannot go through the previous claim (since $(1-b\partial_x^2)^{-1/2}$ does not have a bounded inverse). To show that $\lambda_0(\tilde{L})<0$, take an eigenvector say $g_0:\|g_0\|=1$,
corresponding to the negative eigenvalue $-a^2$ for $L$. Note that by the first claim, such a $g_0$ is smooth, so in particular $(1-b\partial_x^2)^{1/2} g_0$ is well-defined, smooth and non-zero. We have
$$
\lambda_0(\tilde{L})\leq
\f{\dpr{\tilde{L} (1-b\partial_x^2)^{1/2} g_0}{(1-b\partial_x^2)^{1/2} g_0}}{\|(1-b\partial_x^2)^{1/2} g_0\|^2}=\f{\dpr{L g_0}{g_0}}{\|(1-b\partial_x^2)^{1/2} g_0\|^2}=-\f{a_0^2}{\|(1-b\partial_x^2)^{1/2} g_0\|^2}<0.
$$
Next, to show that $\lambda_1(\mathcal L)\geq 0$ (the fact that $0$ is an eigenvalue for $\tilde{L}$ was established already), recall that since $L$ has a simple negative eigenvalue, with eigenfunction $g_0$, we have
$$
\inf_{g: g\perp g_0} \dpr{L g}{g} = 0.
$$
It follows that
$$
\lambda_1(\tilde{L})\geq
\inf_{f\perp (1-b\partial_x^2)^{-1/2} g_0} \f{\dpr{\tilde{L} f}{f}}{\|f\|^2}=
\inf_{h\perp g_0} \f{\dpr{L h}{h}}{\|(1-b\partial_x^2)^{1/2} h\|^2}\geq 0.
$$
Regarding the proof of $\lambda_2(\tilde{L})>0$, we start with $\lambda_2(\tilde{L})\geq \lambda_1(\tilde{L})=0$ and we reach a contradiction as before (i.e. we generate two linearly independent vectors in $Ker(L)$), if
we assume that $\lambda_2(\tilde{L})=0$.
\end{proof}
Using Lemma \ref{le:100}, allows us to
reduce the proof of \eqref{15} to the proof of
\begin{equation}
\label{200}
\lambda_1(L)<0=\lambda_1(L)<\lambda_2(L),
\end{equation}
which we now concentrate on.
We have
\begin{eqnarray*}
L &=& \left(\begin{array}{cc}
1+a\partial_x^2 & b w \partial_x^2 +\psi-w \\
b w \partial_x^2 +\psi-w & 1+a \partial_x^2 +\varphi
\end{array}\right)= \\
&=& (1+a\partial_x^2 ) Id+(b w\partial_x^2-w) \left(\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}\right)+\left(\begin{array}{cc}
0 & \psi \\
\psi & \varphi
\end{array}\right)
\end{eqnarray*}
Introduce an orthogonal matrix $T=\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{array}
\right)$ and observe that
$$
\left(\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}\right)=T^{-1} \left(\begin{array}{cc}
1 & 0 \\
0 &-1
\end{array}\right)T.
$$
It follows that
$$
L=T^{-1}\left( (1+a\partial_x^2 ) Id +(b w\partial_x^2-w+\psi) \left(\begin{array}{cc}
1 & 0 \\
0 &-1
\end{array}\right) +\f{\varphi}{2} \left(\begin{array}{cc}
1 & 1 \\
1& 1
\end{array}\right) \right)T,
$$
whence, by unitary equivalence, it suffices to consider the operator inside the parentheses. That is, we consider
\begin{equation}
\label{700}
M=\left(\begin{array}{cc}
-\partial_x^2(-a-b w)+(1-w)+\psi+\f{\varphi}{2} & \f{\varphi}{2} \\
\f{\varphi}{2} & -\partial_x^2(-a+b w)+(1+w)-\psi+\f{\varphi}{2}
\end{array}
\right)
\end{equation}
We shall need the following
\begin{lemma}
\label{le:90}
Let $\alpha, \lambda>0$ and $Q\in \mathbf R^1$. Then, the Hill operator
$$
\mathcal L=-\partial_x^2 +\alpha^2-Q sech^2(\lambda x)\geq 0
$$
if and only if
\begin{equation}
\label{230}
\alpha^2+\alpha \lambda \geq Q.
\end{equation}
\end{lemma}
\begin{proof}
This is standard result, which follows from the ones found in the literature by a simple change of variables. First, if $Q\leq 0$, we see right away that $\mathcal L>0$ and also the inequality \eqref{230} is satisfied as well. So, assume $Q>0$. Consider $\mathcal L f = \sigma f$ and introduce $f(x)=g(\lambda x)$. We have (after dividing by $\lambda^2$ and assigning $y=\lambda x$)
$$
[- \partial_{yy}+\f{\alpha^2}{\lambda^2} - \f{Q}{\lambda^2} sech^2(y) ] g=\f{\sigma}{\lambda^2} g(y)
$$
Recall that the negative the operator $-\partial_{yy}-Z sech^2(y)$ are
$k_m=-\left[ \left(
Z+\frac{1}{4}\right)^{\frac{1}{2}}-m-\frac{1}{2}\right]^2$,
provided $\left(
Z+\frac{1}{4}\right)^{\frac{1}{2}}-m-\frac{1}{2}>0$, $m=0,1,2...$
[see \cite{Ab}]. Note that $k_0=\inf \sigma(-\partial_{yy}-Z sech^2(y))$ and hence, to avoid negative spectrum, we need to have
$$
0 \leq \f{\alpha^2}{\lambda^2} +k_0= \f{\alpha^2}{\lambda^2} - \left[ \left(
\f{Q}{\lambda^2}+\frac{1}{4}\right)^{\frac{1}{2}}-\frac{1}{2}\right]^2
$$
Solving this last inequality yields \eqref{230}.
\end{proof}
We are now ready to proceed with the count of $n(\tilde{L})$ in each particular case of consideration. \\
\\
{\bf Case I: $a=c=-b, b>0$} \\
\\
Going back to the operator $M$, we can rewrite it as
$$
M= S \left(\begin{array}{cc}
-\partial_x^2+\f{1}{b}+ \f{B+\f{1}{2}}{ b(1-w)}\varphi & \f{ \varphi }{2b\sqrt{1-w^2}} \\
\f{ \varphi }{2b\sqrt{1-w^2}} & -\partial_x^2+\f{1}{b}+ \f{-B+\f{1}{2}}{ b(1+w)}\varphi
\end{array}
\right) S
$$
where $S=\left(\begin{array}{cc}
\sqrt{b(1-w)} & 0 \\
0 & \sqrt{b(1+w)}
\end{array}
\right)$.
Thus, according to Lemma \ref{le:100}, we have reduced matters to
$$
M_1=(-\partial_x^2+\f{1}{b})Id+ \varphi \left(\begin{array}{cc}
\f{B+\f{1}{2}}{ b(1-w)} & \f{ 1 }{2b\sqrt{1-w^2}} \\
\f{ 1 }{2b\sqrt{1-w^2}} & \f{-B+\f{1}{2}}{ b(1+w)}
\end{array}
\right)
$$
Diagonalizing this last symmetric matrix yields the representation
$$
\left(\begin{array}{cc}
\f{B+\f{1}{2}}{ b(1-w)} & \f{ 1 }{2b\sqrt{1-w^2}} \\
\f{ 1 }{2b\sqrt{1-w^2}} & \varphi \f{-B+\f{1}{2}}{ b(1+w)}
\end{array}
\right)=U^* \left(\begin{array}{cc}
\frac{1+2 B w+\sqrt{4 B^2+4 B w+1} }{2 b \left(1-w^2\right)} & 0 \\
0 & \frac{1+2 B w - \sqrt{4 B^2+4 B w+1} }{2 b \left(1-w^2\right)}
\end{array} \right) U
$$
for some orthogonal matrix $U$. Factoring out $U^*, U$ again and using Lemma \ref{le:100} once more reduces us to the operator
$$
M_2= \left(\begin{array}{cc}
\mathcal L_1 & 0 \\
0 & \mathcal L_2
\end{array} \right)
$$
which contains the following Hill operators on the main diagonal
\begin{eqnarray*}
\mathcal L_1 &=& -\partial_x^2+\f{1}{b}+\eta_0 \frac{1+2 B w+\sqrt{4 B^2+4 B w+1} }{2 b \left(1-w^2\right)} sech^2\left(\f{x}{2\sqrt{b}}\right); \\
\mathcal L_2 &=& -\partial_x^2+\f{1}{b}+ \eta_0 \frac{1+2 B w - \sqrt{4 B^2+4 B w+1} }{2 b \left(1-w^2\right)} sech^2\left(\f{x}{2\sqrt{b}}\right)
\end{eqnarray*}
Note that $n(\tilde{L})=n(\mathcal L_1)+n(\mathcal L_2)$.
Using the formulas
$$
B(\eta_0)=\pm \sqrt{\f{3}{3+\eta_0}}, w(\eta_0)=\pm \f{3+2\eta_0}{\sqrt{3(3+\eta_0)}}
$$
yields
\begin{eqnarray*}
\mathcal L_1 &=& -\partial_x^2+\f{1}{b}-\f{3}{b} sech^2\left(\f{x}{2\sqrt{b}}\right) ; \\
\mathcal L_2 &=& -\partial_x^2+\f{1}{b}-\f{3\eta_0}{b(9+4\eta_0)}sech^2\left(\f{x}{2\sqrt{b}}\right)
\end{eqnarray*}
According to the formulas for the eigenvalues in Lemma \ref{le:90}
(with $\alpha=\f{1}{\sqrt{b}}, \lambda=\f{1}{2\sqrt{b}}$,$ Q=\f{3}{b}>0$) we have that
$$
\lambda_1(\mathcal L_1)= \f{\alpha^2}{\lambda^2}-\left(\sqrt{\f{Q}{\lambda^2}+\f{1}{4}}-\f{3}{2}\right)^2=2-(\sqrt{12.25}-1.5)^2=0,
$$
which indicates that $\mathcal L_1$ has one negative eigenvalue and the next one is zero, whence $n(\mathcal L_1)=1$ for all $\eta_0>-3$. Thus, $n(\tilde{L})=1+n(\mathcal L_2)$.
It is also immediately clear that for $\eta_0\in (-\f{9}{4},0)$, $\mathcal L_2>0$ and hence $n(\tilde{L})=1$.
\\
\\
{\bf Case II: $a=c<0, b=d>0,$ $a+b\neq 0$} \\
\\
In this case, we have $p=\f{c+b}{a+b}=1$, $\eta_0=\f{3(1-2p)}{2p}=-\f{3}{2}$ and thus $w(\eta_0)=w(-3/2)=0$, $\lambda=\f{1}{2\sqrt{-a}}, B(\eta_0)=\pm \sqrt{2}$.
This simplifies the computations quite a bit. In fact, starting from the operator $M$, defined in \eqref{700}, we see that it has the form
$$
M=(a\partial_x^2+1) Id +\left(\begin{array}{cc} B+\f{1}{2} & \f{1}{2} \\
\f{1}{2} & -B+\f{1}{2} \end{array}\right)\varphi
$$
Recall that here $B=\pm \sqrt{2}$. Consider first $B=\sqrt{2}$.
Diagonalizing the matrix vian an orthogonal matrix $S$ yields the representation
\begin{eqnarray*}
& &
\left(\begin{array}{cc} \sqrt{2}+\f{1}{2} & \f{1}{2} \\
\f{1}{2} & -\sqrt{2}+\f{1}{2} \end{array}\right)= S^{-1}
\left(\begin{array}{cc} 2 & 0 \\
0 & -1 \end{array}\right) S, \\
& & S=\f{1}{\sqrt{6}}\left(
\begin{array}{cc}
\sqrt{3+2 \sqrt{2}} &
\sqrt{3-2 \sqrt{2}} \\
-\sqrt{3-2 \sqrt{2}} &
\sqrt{3+2\sqrt{2}}
\end{array}
\right)
\end{eqnarray*}
Thus, in this case, we have represented the operator $L$ in the form
\begin{equation}
\label{710}
L= (S T)^* \left(\begin{array}{cc} -a\partial_x^2+1+2\varphi & 0 \\
0 & -a\partial_x^2+1-\varphi \end{array}\right)S T,
\end{equation}
where $S, T$ are explicit orthogonal matrices. It is now clear that since $\eta_0=-\f{3}{2}<0$, we have that $\varphi(x)<0$ and hence the operator
$a\partial_x^2+1-\varphi>0$. On the other hand,
$L_{KdV}= a\partial_x^2+1+2\varphi$ is well known to have a zero eigenvalue (with eigenfunction $\varphi'$) and an unique simple negative eigenvalue.
For the case $B=-\sqrt{2}$, we have \eqref{710}, with
$$
S=\f{1}{\sqrt{6}}\left(
\begin{array}{cc}
\sqrt{3-2 \sqrt{2}} &
\sqrt{3+2 \sqrt{2}} \\
\sqrt{3+2 \sqrt{2}} &
-\sqrt{3-2\sqrt{2}}
\end{array}
\right)
$$
\section{Proof of Proposition \ref{prop:30}}
\label{sec:4}
The purpose of this section is to compute the quantity appearing in \eqref{150}, whose negativity will be equivalent to the stability of the waves. Thus, we need to find
$$
L^{-1} [(1-b \partial_x^2)\left(\begin{array}{c} \psi \\
\varphi\end{array} \right)].
$$
Here, our considerations need to be split in two cases: $a=c=-b$, and $a=c<0, b>0$.
The case $a=c=-b$ is easier to manage, since in int we have a {\it a free parameter} $w=w(\eta_0)$ that we can differentiate with respect to in \eqref{5}. The remaining case is harder, because the parameter $\eta_0=-3/2$, whence $w=0$ and one cannot apply the same technique.
\subsection{The case $a=c=-b$, $b>0$}
Taking a derivative with respect to $w$ in \eqref{5}, we find
$$
L[\left(\begin{array}{c} \partial_w \varphi \\
\partial_w \psi\end{array} \right)= (1-b \partial_x^2)\left(\begin{array}{c} \psi \\
\varphi\end{array} \right),
$$
whence
$$
L^{-1} [(1-b \partial_x^2)\left(\begin{array}{c} \psi \\
\varphi\end{array} \right)]=\left(\begin{array}{c} \partial_w \varphi \\
\partial_w \psi\end{array} \right).
$$
We obtain
\begin{eqnarray*}
& & \dpr{L^{-1}[(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)]}{(1-b \partial_x^2) \left(\begin{array}{c}\psi \\ \varphi \end{array}\right)}= \dpr{(1-b \partial_x^2)\left(\begin{array}{c} \psi \\
\varphi\end{array} \right)}{\left(\begin{array}{c} \partial_w \varphi \\
\partial_w \psi\end{array} \right)}=\\
&=& \partial_w [\dpr{\varphi}{\psi}+ b \dpr{\varphi'}{\psi'}] =
\partial_w[B(\eta_0)\int \varphi(\xi)^2+b (\varphi'(\xi))^2 d\xi]= \\
&=&
B\partial_w [\int_{\mathbb{R}}{[\varphi^2(\xi)+b\varphi'^2(\xi)]}d\xi]+\partial_w B \int_{\mathbb{R}}{[\varphi^2(\xi)+b\varphi'^2(\xi)]}d\xi=\\
&=& \frac{16\sqrt{b}}{5}\left[ B\frac{d\eta_0^2}{dw}+\eta_0^2\frac{dB}{dw}\right]=\frac{16\sqrt{b}}{5}\left[ 2B+\eta_0\frac{dB}{d\eta_0}\right]\eta_0\frac{d\eta_0}{dw}=:d(w)
\end{eqnarray*}
We are now ready to compute this last expression in the cases of interest. \\
\subsubsection{$B(\eta_0)=-\sqrt{\f{3}{3+\eta_0}},
w=-\frac{3+2\eta_0}{\sqrt{3(3+\eta_0)}}$}
We have
$$
\frac{d\eta_0}{dw}=-\frac{2\sqrt{3}(3+\eta_0)^{\frac{3}{2}}}{2\eta_0+9},
\ \ \frac{dB}{d\eta_0}=\frac{\sqrt{3}}{2}\frac{1}{(3+\eta_0)^{\frac{3}{2}}}
$$
and
$$d(w)=-\frac{48\sqrt{3b}}{10(3+\eta_0)^{\frac{3}{2}}}(4+\eta_0)\eta_0\frac{d\eta_0}{dw}<0
$$
for $-\f{9}{4}<\eta_0<0$.
\subsubsection{$B(\eta_0)=\sqrt{\f{3}{3+\eta_0}}, w=\frac{3+2\eta_0}{\sqrt{3(3+\eta_0)}}$} We have
$$
\frac{d\eta_0}{dw}=\frac{2\sqrt{3}(3+\eta_0)^{\frac{3}{2}}}{2\eta_0+9},
\ \ \frac{dB}{d\eta_0}=-\frac{\sqrt{3}}{2}\frac{1}{(3+\eta_0)^{\frac{3}{2}}}
$$
hence
$$
d(w)=\frac{48\sqrt{3b}}{10(3+\eta_0)^{\frac{3}{2}}}(4+\eta_0)\eta_0\frac{d\eta_0}{dw}<0.
$$
for $-\f{9}{4}<\eta_0<0$.
\subsection{The case: $a=c<0, b>0$} As we have discussed above, we have explicit formulas for all the quantities involved. Namely,
we have $w=0, \lambda=\f{1}{2\sqrt{-a}}, B =\pm \sqrt{2}$. Thus,
$$
\varphi(x)=-\f{3}{2} sech^2\left(\f{x}{2\sqrt{-a}}\right).
$$
\subsubsection{Case $B=\sqrt{2}$}
We need to compute
$$
\dpr{L^{-1}\left( \begin{array}{c} (1-b\partial_x^2)\psi \\ (1-b\partial_x^2)\varphi \end{array} \right)}{\left( \begin{array}{c} (1-b\partial_x^2)\psi \\ (1-b\partial_x^2)\varphi \end{array} \right)}
$$
To that end, we use the representation \eqref{710}. We have
\begin{eqnarray*}
I &=& \dpr{L^{-1}\left( \begin{array}{c} (1-b\partial_x^2)\psi \\ (1-b\partial_x^2)\varphi \end{array} \right)}{\left( \begin{array}{c} (1-b\partial_x^2)\psi \\ (1-b\partial_x^2)\varphi \end{array} \right)} = \\
& & = \dpr{\left(\begin{array}{cc} a\partial_x^2+1+2\varphi & 0 \\
0 & a\partial_x^2+1-\varphi \end{array}\right)[ S T \left( \begin{array}{c} \sqrt{2} \\ 1 \end{array} \right)(1-b\partial_x^2)\varphi]}{S T \left( \begin{array}{c} \sqrt{2} \\ 1 \end{array} \right)(1-b\partial_x^2)\varphi }
\end{eqnarray*}
A direct computation shows that $S T \left( \begin{array}{c} \sqrt{2} \\ 1 \end{array} \right)=\left(\begin{array}{c} 2 \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{3}}\end{array}\right)$, whence our index $I$ can be computed as follows
$$
I=\f{8}{3}\dpr{(a\partial_x^2+1+2\varphi)^{-1}[(1-b\partial_x^2)\varphi]}{(1-b\partial_x^2)\varphi}+
\f{1}{3}\dpr{(a\partial_x^2+1-\varphi)^{-1}[(1-b\partial_x^2)\varphi]}{(1-b\partial_x^2)\varphi}
$$
Denote $f=(1-b\partial_x^2)\varphi$ and
\begin{eqnarray*}
L_{KdV} &=& a\partial_x^2+1+2\varphi \\
L_{Hill} &=& a\partial_x^2+1-\varphi
\end{eqnarray*}
Note that by Weyl's theorem
$ \sigma_{ess.}(L_{Hill})=[1, \infty)$. On the other hand, by the fact that $\varphi<0$, the potential $-\varphi>0$ and hence, by the results for
absence of embedded eigenvalues, $\sigma(L_{Hill})=\sigma_{ess.}(L_{Hill})=[1, \infty)$.
We now compute the index
$$
I=\f{1}{3}(8 \dpr{L_{KdV}^{-1} f}{f}+\dpr{L_{Hill}^{-1} f}{f}).
$$
To that end, we differentiate the equation
$$
a\varphi''+\varphi+\varphi^2=0
$$
with respect to $a$. We get\footnote{we use the notation $\varphi_a=\partial_a \varphi$ denotes the derivative with respect to $a$}
\begin{equation}
\label{720a}
L_{KdV}\varphi_a=-\varphi'',
\end{equation}
whence $L_{KdV}^{-1}[\varphi'']=-\varphi_a$.
Using that $L_{KdV}\varphi=\varphi^2=-a\varphi''-\varphi$ and the above relation, we obtain that
$$
-\varphi=aL_{KdV}^{-1}\varphi''+L_{KdV}^{-1}\varphi=-a\varphi_a+L_{KdV}^{-1}\varphi .
$$
It follows that
\begin{eqnarray}
\label{720b}
L_{KdV}^{-1}\varphi &=& a\varphi_a-\varphi, \\
\label{720c}
L_{KdV}^{-1}f &=& (a+b)\varphi_a-\varphi .
\end{eqnarray}
and
$$
\langle L_{KdV}^{-1}f,f\rangle=(a+b)\langle \varphi_a, \varphi
\rangle-b(a+b)\langle \varphi_a, \varphi'' \rangle-\langle \varphi, \varphi
\rangle+b\langle \varphi, \varphi'' \rangle.
$$
By direct computations
\begin{eqnarray*}
\langle \varphi_a, \varphi
\rangle &=& \frac{1}{2}\frac{d}{da}\int_{-\infty}^{+\infty}{\varphi^2}dx=
-\f{3}{2\sqrt{-a}}, \\
\langle \varphi, \varphi''
\rangle &=& -\int_{-\infty}^{+\infty}{\varphi'^2}dx=-\frac{6}{5\sqrt{-a}}, \\
\langle \varphi_a, \varphi''
\rangle &=& -\frac{1}{2}\frac{d}{da}\int_{-\infty}^{+\infty}{\varphi'^2}dx=-\frac{3}{10|a|\sqrt{-a}}, \\
\langle \varphi, \varphi
\rangle &=& \f{9}{2}\sqrt{-a}
\int_{-\infty}^{+\infty}{sech^4(y)}dy=6 \sqrt{-a}.
\end{eqnarray*}
As a consequence,
\begin{eqnarray*}
\langle
L_{KdV}^{-1}f,f\rangle &=& -\frac{3(a+b)}{2\sqrt{-a}}+\frac{3b(a+b)}{10|a|
\sqrt{-a}}-6\sqrt{-a}-\frac{6b}{5\sqrt{-a}}=\\
&=& -\f{9}{2}\sqrt{-a}-\f{12}{5}\f{b}{\sqrt{-a}} +
\f{3b^2}{10|a| \sqrt{-a}}= \sqrt{-a}
\left( -\f{9}{2}-\f{12}{5} \f{b}{|a|}+
\f{3}{10} \f{b^2}{a^2}\right).
\end{eqnarray*}
This yields the desired computation for the terms involving $L_{KdV}^{-1}$. We turn our attention to $L_{Hill}^{-1}$. The situation here is a bit trickier, since we cannot compute explicitly the quantities $L_{Hill}^{-1}[\varphi], L_{Hill}^{-1}[\varphi'']$, as required in the formula for $I$. Instead, we need to rely on estimates. To start with, observe that
$$
L_{Hill}[\varphi]=a\varphi''+\varphi-\varphi^2=-2\varphi^2=2a\varphi''+2\varphi,
$$
whence
\begin{equation}
\label{800}
L_{Hill}^{-1}[a\varphi''+\varphi]=\f{\varphi}{2}.
\end{equation}
Since we need to compute $ L_{Hill}^{-1}[f]=L_{Hill}^{-1}[\varphi-b\varphi'']$, we project the vector $f$ onto $a\varphi''+\varphi$ and its orthogonal subspace as follows
$$
f=\varphi-b\varphi''=\f{\dpr{\varphi-b \varphi''}{a\varphi''+\varphi}}{\|a\varphi''+\varphi\|^2}(a\varphi''+\varphi)+g
$$
Calculations then show that since
$$
\|\varphi''\|^2= \dpr{\varphi''}{\varphi''}=\f{6}{7 |a|\sqrt{-a}},
$$
we have that
$$
f=\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right) (a\varphi''+\varphi)+g
$$
whence
$$
L_{Hill}^{-1}[f]=\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right) \f{\varphi}{2}+L_{Hill}^{-1}[g].
$$
Thus, the quantity that needs to be computed is
\begin{eqnarray*}
\dpr{ L_{Hill}^{-1} f}{f} &=&
\f{1}{2}\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right)\dpr{\varphi-b\varphi''}{\varphi}
+ \dpr{ L_{Hill}^{-1} g}{f}=\\
&=& \f{1}{2}\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right)\dpr{\varphi-b\varphi''}{\varphi} +
\dpr{ L_{Hill}^{-1} g}{g}+\f{1}{2}\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right)\dpr{g}{\varphi}
\end{eqnarray*}
All of these can be computed explicitly, except for $ \dpr{ L_{Hill}^{-1} g}{g}$, which we estimate by \\ $ 0<\dpr{ L_{Hill}^{-1} g}{g}\leq \|g\|^2$, which holds since $\sigma(L_{Hill})\subset [1, \infty)$. Thus,
\begin{eqnarray*}
\dpr{ L_{Hill}^{-1} f}{f} &\leq & \f{1}{2}\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right)\dpr{\varphi-b\varphi''}{\varphi} + \|g\|^2 +\f{1}{2}\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right)\dpr{g}{\varphi}=\\
& = & \sqrt{-a}\left( \f{22}{45}\f{b^2}{a^2}+\f{2}{9} \f{b}{|a|}+\f{26}{9}\right)
\end{eqnarray*}
and on the other hand
$$
\dpr{ L_{Hill}^{-1} f}{f}>\f{1}{2}\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right)\dpr{\varphi-b\varphi''}{\varphi} + \f{1}{2}\left(\f{7}{9}+\f{2}{9}\f{b}{|a|}\right)\dpr{g}{\varphi}=
\sqrt{-a}\left( \f{4}{45} \f{b^2}{a^2} + \f{46}{45} \f{b}{|a| } +\f{ 112}{45} \right).
$$
Thus, we obtain the following {\it estimate} for the instability index $I$
\begin{eqnarray*}
3 I &=& 8 \dpr{L_{KdV}^{-1} f}{f}+\dpr{L_{Hill}^{-1} f}{f}\leq \sqrt{-a}\left( 8\left(-\f{9}{2}-\f{12 b}{5|a|}+
\f{3 b^2}{10 a^2}\right)
+ \left( \f{22}{45}\f{b^2}{a^2}+\f{2}{9} \f{b}{\sqrt{-a}}+\f{26}{9}\right)\right)=\\
&=& \f{2\sqrt{-a}}{45}\left( 65 \f{b^2}{a^2}-427 \f{b}{\sqrt{-a}}-745\right)
\end{eqnarray*}
On the other hand, we have the following estimate from below
\begin{eqnarray*}
3 I &=& 8 \dpr{L_{KdV}^{-1} f}{f}+\dpr{L_{Hill}^{-1} f}{f} > \sqrt{-a} \left( 8\left(-\f{9}{2}-\f{12 b}{5|a|}+
\f{3 b^2}{10 a^2}\right)+ \left( \f{4}{45} \f{b^2}{a^2} + \f{46}{45} \f{b}{|a| } +\f{ 112}{45} \right) \right)\\
&=& \f{2\sqrt{-a}}{45}\left(56 \f{b^2}{a^2}-409 \f{b}{|a| } -754 \right).
\end{eqnarray*}
The picture below shows the graphs of the two estimates of $3I/\sqrt{-a}$. If one solves the corresponding quadratic equations, we see that we have {\bf stability}, whenever
$$
0\leq \f{b}{-a}< \frac{1}{130} \left(427+3 \sqrt{41781}\right)\sim 8.00163.
$$
and {\bf instability}, when
$$
\f{b}{-a}> \frac{1}{112} \left(409+3 \sqrt{37353}\right)\sim 8.82864.
$$
\begin{figure}[h]
\centering
\includegraphics[width=15cm,height=7cm]{indexabcsystem}
\caption{The picture shows the graphs of the function
$\f{2}{45}\left( 65 z^2-427 z-745\right) $ is in blue, while
$ \f{2}{45} (56 z^2-409 z -754) $ in red. Note that the graphs do coincide for $z=1$, which is the case, since there $g=0$ and the computations becomes precise. }
\label{fig}
\end{figure}
\subsubsection{Case $B=-\sqrt{2}$}
In this case, the computation for the index is the same since
\begin{eqnarray*}
I &=& \dpr{L^{-1}\left( \begin{array}{c} (1-b\partial_x^2)\psi \\ (1-b\partial_x^2)\varphi \end{array} \right)}{\left( \begin{array}{c} (1-b\partial_x^2)\psi \\ (1-b\partial_x^2)\varphi \end{array} \right)} = \\
&=& \dpr{\left(\begin{array}{cc} a\partial_x^2+1+2\varphi & 0 \\
0 & a\partial_x^2+1-\varphi \end{array}\right)[ S T \left( \begin{array}{c} -\sqrt{2} \\ 1 \end{array} \right)(1-b\partial_x^2)\varphi]}{S T \left( \begin{array}{c} -\sqrt{2} \\ 1 \end{array} \right)(1-b\partial_x^2)\varphi }=\\
&=& \f{1}{3}(8 \dpr{L_{KdV}^{-1} f}{f}+\dpr{L_{Hill}^{-1} f}{f}),
\end{eqnarray*}
where in the last line, we have used that $S T \left( \begin{array}{c} \sqrt{2} \\ 1 \end{array} \right)=\left(\begin{array}{c} 2 \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{3}}\end{array}\right)$ as above. The rest of the argument proceeds in exactly the same way, since the exact same quantity is being computed.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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Бяргтангар () е най-западната точка на Исландия, както и на Европа (заедно с островната част). Бяргтангар се намира в Йойстюр-Бардащрандар - една от сислите на Исландия.
География на Исландия
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,649
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Die Landespolizeischule Niedersachsen (LPSN) war eine Polizeischule mit Hauptsitz in Hann. Münden, die von 1946 bis 1997 die Polizeiausbildung und -fortbildung der Landespolizei in Niedersachsen betrieben hat. 1997 wurde sie in das Bildungsinstitut der Polizei (BIP NI) umgewandelt, das 2007 in die Polizeiakademie Niedersachsen mit Sitz in Nienburg/Weser überging. Trotz der Umorganisierungen ist die Liegenschaft in Hann. Münden seit dem Zweiten Weltkrieg der zentrale Standort für die Aus- und Fortbildung der Polizei Niedersachsen, heute als stark ausgelastete Außenstelle der Polizeiakademie.
Liegenschaft
Die Polizeischule entstand 1946 in der ehemaligen Gneisenau-Kaserne der Wehrmacht in Hann. Münden. Die 1934 in der Zeit des Nationalsozialismus errichtete Militäranlage entstand auf dem Gimter Feld nahe dem Ortsteil Gimte. Wegen ihrer Nähe zur Weser gab es eine Nutzung durch Pioniereinheiten, die in Hann. Münden eine lange Tradition haben. Gegen Ende des Zweiten Weltkriegs war ein Teil des Oberkommandos des Heeres in der Kaserne einquartiert. Am 30. und 31. März 1945 war Hann. Münden Ziel von Luftangriffen durch US-amerikanische Bomberverbände, bei denen 32 Menschen getötet und 50 schwer verletzt wurden. Auch die Gebäude der Gneisenau-Kaserne waren durch Bombentreffer erheblich beschädigt worden.
Gründung auf der Weisung der Militärregierung
Unmittelbar nach Ende des Zweiten Weltkriegs entstand auf Weisung der britischen Militärregierung eine regionale Polizeischule in Hannover. Sie war für die damalige Region Hannover innerhalb der britischen Besatzungszone bestimmt, was etwa dem Gebiet des heutigen Niedersachsens entsprach. Die Schule begann im September 1945 als "Schule der Ordnungspolizei-Region Hannover" ihren Lehrbetrieb, der in einer Wehrmachts-Kaserne am Welfenplatz stattfand. Da die Briten beim Wiederaufbau des deutschen Polizei eine Kommunalisierung betrieben, gab es ab Ende 1945 weitere örtliche Polizeischulen bei den Regierungsbezirken (RB) im Gebiet des heutigen Niedersachsens. Das waren Unterrichtsorte in Aurich (RB Aurich), Querum (Stadt und RB Braunschweig), Eldagsen (RB Hannover), Bad Salzdetfurth (RB Hildesheim), Steinhorst (RB Lüneburg), Oldenburg (RB Oldenburg), Bramsche (RB Osnabrück), Neuhaus (Oste) (RB Stade).
Im Mai 1946 wurde auf Befehl der britischen Militärregierung die Polizeiausbildung im Gebiet des heutigen Niedersachsens zentralisiert und nach Hann. Münden verlegt. Um den Sitz als Polizeischule hatten sich 27 Städte in der britischen Besatzungszone beworben. Hauptkriterium für die Auswahl zugunsten von Hann. Münden waren die großzügigen räumlichen Bedingungen in der Gneisenau-Kaserne. Ein britisches Bataillon räumte sie, damit die Polizeischule aus Hannover im Mai 1946 einziehen konnte. Die anderen örtlichen Ausbildungsstellen bei den Regierungsbezirken wurden noch 1946 aufgelöst.
Lehrbetriebsanfänge
Am 11. Juni 1946 begann der erste Polizeianwärterlehrgang in Hann. Münden. Der Lehrbetrieb lief unter den schlechten äußeren Gegebenheiten der Nachkriegszeit an. Die Gebäude wiesen Schäden durch amerikanische Bombenangriffe und Plünderungen durch die Bevölkerung auf. Nachdem im November 1946 das Land Niedersachsen gegründet worden war, trug die Polizeischule ab dem 1. Januar 1947 die Bezeichnung Polizeischule des Landes Niedersachsen. In den ersten Jahren ihres Bestehens wurde der Lehrgangsbetrieb häufiger für längere Zeit unterbrochen. 1947 schloss die Schule für zwei Monate, weil kein Heizmaterial vorhanden war. Im selben Jahr ruhte der Betrieb nochmals, weil bis zu 500 Polizeianwärter bei der ersten Hannover Messe eingesetzt waren, die große Bedeutung für den wirtschaftlichen Aufschwung des neu gegründeten Landes hatte. 1948 gab es zeitweise keinen Unterricht, weil viele Beamte Dienst bei Verkehrskontrollen an der "Zonengrenze" im Zusammenhang mit der Berlin-Blockade verrichteten. Erst ab 1951 trug die Schule die langjährige Bezeichnung Landespolizeischule Niedersachsen.
Etablierung
Zu Beginn noch unter britischer Aufsicht und Kontrolle entwickelte sich die Polizeischule im Laufe der Zeit zu einer modernen Ausbildungsstätte der Polizei Niedersachsen. Die umfangreichen Aus- und Fortbildungsaufgaben umfassten unter anderem die:
Grundausbildung von Polizeianwärtern
Laufbahnlehrgänge
Aufstiegsausbildung
Fortbildungslehrgänge
Die Lehrgänge waren hauptsächlich für Angehörige der Schutz- und Kriminalpolizei ausgerichtet. Sie betrafen auch weitere Personenkreise, wie Übernahmelehrgänge für Bundeswehr- und Bundesgrenzschutzangehörige. Eine außergewöhnliche Beschulung war die Unterweisung von Lagerführern für die Lager von Displaced Persons kurz nach Ende des Zweiten Weltkriegs.
Die Baulichkeiten der früheren Kaserne wurden im Laufe der Zeit den polizeilichen Bedürfnissen angepasst. Es entstanden unter anderem neue Lehrsaalgebäude, weitere Wohnunterkünfte, eine Sporthalle, ein Schießkeller, Sportplätze, eine Bibliothek und ein großer Speisesaal. Auf einem angrenzenden Grundstück entstand 1996 das Logistik Zentrum Niedersachsen als Ausstatter der Polizei und anderer niedersächsischer Behörden.
In den 1970er Jahren hatte die Landespolizeischule ihre höchste Auslastungszahl in der Aus- und Fortbildung von Polizeibeamten. Verpflegungsstärken bis zu 1400 Personen pro Tag waren keine Seltenheit. Die Gesamtzahl der ausgebildeten Personen an der LPSN ist wegen der Vielzahl der angebotenen Lehrgänge nicht bekannt, sie dürfte aber im Bereich von mehreren Zehntausend Personen liegen. Ihre Grundausbildung absolvierten zwischen 1950 und 1976 14.000 Polizeibeamte in einem Grundlehrgang, den 12.500 bestanden.
Weitere Standorte
Aufgrund der hohen Auslastung der Landespolizeischule in Hann. Münden ab den 1970er Jahren wurden in Niedersachsen Zweigstellen eingerichtet um die große Anzahl der Polizeibeamten beschulen zu können. Es gab zeitweise weitere Standorte in:
Uelzen 1964–1979
Sögel 1970–1974
Bad Iburg 1973–2004 (Iburger Schloss)
Wolfenbüttel 1974–1983
Huntlosen 1975–1983
Hohegeiß 1979–1996
Liebenau 1980–1990
Auflösung unter Standortbeibehaltung
Die Landespolizeischule Niedersachsen wurde am 30. April 1997 nach 51 Jahren ihres Bestehens aufgelöst und in das Bildungsinstitut der Polizei Niedersachsen (BIP NI) umgewandelt. Der Hauptsitz dieser neuen Polizeieinrichtung blieb weiterhin in Hann. Münden. Sie übernahm die gesamte Fortbildung der etwa 22.500 Bediensteten der Polizei in Niedersachsen. Die Neuausbildung ging als an die "Fakultät Polizei" der Niedersächsischen Fachhochschule für Verwaltung und Rechtspflege in Hildesheim. Hann. Münden fungierte dabei als ihre Außenstelle.
Bei der Reform der polizeilichen Aus- und Fortbildung in Niedersachsen kam es am 1. Oktober 2007 zu einer weiteren Umorganisation. Die Fachhochschule und das Bildungsinstitut wurden zur Polizeiakademie Niedersachsen mit Hauptsitz in Nienburg/Weser zusammengeführt. Der seitdem weiterhin gut ausgelastete Studienort Hann. Münden ist neben Oldenburg eine von zwei Zweigstellen der Akademie.
Literatur
13. Juni 1946 – 30 Jahre Landespolizeischule in Hann. Münden, Hrsg.: Mündener Gemeinschaft für polizeiliches Fachschriftentum an der Landespolizeischule, 1976.
Hans-Joachim Bartholdt: Die "Landespolizeischule Niedersachsen" – eine moderne Aus- und Fortbildungsstätte in: Niedersachsen und seine Polizei: Herausgegeben vom Niedersächsischen Ministerium des Innern. Polizei-Technik-Verkehr-Verlagsgesellschaft, Wiesbaden 1979, S. 192–196.
40 Jahre Landespolizeischule Niedersachsen Hann. Münden 1946–1986, Hrsg.: Landespolizeischule Niedersachsen.
50 Jahre Landespolizeischule Niedersachsen in Hann. Münden 1946–1996, Hrsg.: Landespolizeischule Niedersachsen.
Heinz-Joachim Brauleke: LPSN Hannoversch Münden: Wetterleuchten einer demokratischen Polizeiinstitution – Dokumentarische Einblicke in die Entstehungsgeschichte der heutigen Landespolizeischule Niedersachsen, in: Archiv für Polizeigeschichte 1991.
Förderkreis der Polizeigeschichtlichen Sammlung Niedersachsen e.V. (Hg.): Polizeigeschichte miterlebt. Zeitzeugnisse aus den Anfängen der niedersächsischen Polizei, Braunschweig 2013. ISBN 978-3-00-039992-3
Einzelnachweise
Polizeiausbildung in Deutschland
Schule im Landkreis Göttingen
Ehemalige Schule in Niedersachsen
Organisation der Landespolizei (Deutschland)
Polizei (Niedersachsen)
Organisation (Hann. Münden)
Polizeigeschichte (Deutschland)
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
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The orchestra meets as a class during the school day, and has 3 to 4 mandatory performances during the year and is directed by Norm Vagn. There are numerous voluntary performance opportunities for you as well if music is your thing. The RHS Orchestra participates in FOA events. We perform for Solo and Ensemble and Concert Musical Performance Assessments. Student also have the opportunity to audition for the FMEA All State Orchestra. If making music is something you enjoy, we have the group for you! If you enjoy playing a variety of music from different composers and time periods, we have the group for you! If having fun while striving to achieve musical goals sounds good to you, then you sound good to us! Players of any and all experience levels wanted!
Norm was born in Chicago and grew up in Arlington Heights, Illinois. He received his Bachelors Degree in Music Education from Northern Illinois University in Dekalb, Illinois. He played bass trombone in the NIU Jazz Ensemble under the direction of Ron Modell. He received his Masters Degree in Wind Conducting from the University of Kansas. Norm taught for 10 years in McHenry County, which is an hour west of Chicago. Norm currently teaches at Riverview High School in Sarasota. He teaches orchestra, jazz ensemble, AP and Regular Music Theory and IB Music. Norm also works with the Kiltie Band. Norm Currently sits on the board of directors for the Jazz Club of Sarasota. Norm has been involved in church music for a long time and is currently the Director of Worship Music at Beautiful Savior Lutheran Church.
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{
"redpajama_set_name": "RedPajamaC4"
}
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\section{Introduction}
\label{sec:intro}
During the last decade, robots with multimodal locomotion capability have undergone intensive development, and demonstrated the versatile maneuvering in multiple domains (i.e., terrestrial, aerial, and aquatic) \cite{multimodal-locomotion:MUQA-IROS2013, multimodal-locomotion:qaudrotor-HyTAQ-IROS2014, multimodal-locomotion:Snake-ACM-R5-ISR2005,multimodal-locomotion:Salamander-Science2007, multimodal-locomotion:LoonCopter-JFR2018} which can benefit the performance in various situations, such as disaster response and industrial surveillance.
Among various forms for multimodal locomotion, the legged type has the advantage in unstructured terrains, and several bipedal models have been developed to demonstrate the promising walking that is interlaced with flying motion \cite{flying-bipedal-robot:LEONARDO-SciRo2021,flying-bipedal-robot:KXR-Humanoids2020}. The legged robots are desired to provide not only the advanced locomotion capability, but also the manipulation ability by the limb end-effector. Then, multilegged (multilimbed) is considered effective from the aspect of both the stability in the terrestrial locomotion and the freedom in manipulation.
In spite of the achievement of a simple arm motion in midair by the flying humanoid robot in \cite{flying-bipedal-robot:KXR-Humanoids2020}, the centralized rotor arrangement in most of the existing robot platforms can hardly handle the large change of Center-of-Gravity (CoG) caused by the joint motion in midair. Besides, the external force at the limb end during the aerial manipulation is also difficult to compensate by the centralized rotors due to the large moment arm.
Hence, a distributed rotor arrangement is necessary for aerial manipulation.
Then, we propose a novel quadruped robot platform in this work, which can both walk and fly by using the spherically vectorable rotors distributed in each link unit as illustrated in \figref{figure:intro}.
\begin{figure}[t]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/intro.ps}
\vspace{-6mm}
\caption{{\bf Air-ground amphibious quadruped robot SPIDAR}: {\bf SP}her{\bf I}cally vectorable and {\bf D}istributed rotors assisted {\bf A}ir-ground amphibious quadruped {\bf R}obot. {\bf (A)} walk on the ground like an usual quadruped robot, but with the assistance of thrust force; {\bf (B)} fly and transform in midair.}
\label{figure:intro}
\end{center}
\end{figure}
One of the efficient type of air-ground hybrid robot platforms equips an ordinary multirotor for flying, and then deploys rolling cage or wheels to perform terrestrial locomotion
\cite{multimodal-locomotion:qaudrotor-HERALD-IROS2014, multimodal-locomotion:qaudrotor-HyTAQ-IROS2014, multimodal-locomotion:qaudrotor-FCSTAR-RAL2021}. Although the rolling mechanism can achieve the stable terrestrial locomotion without any complex control, it is relatively difficult for this type to handle the unstructured terrains (e.g., very few foothold). Then, legged model is proposed in several researches \cite{flying-bipedal-robot:LEONARDO-SciRo2021,flying-bipedal-robot:KXR-Humanoids2020, flying-bipedal-robot:Niiyama-ROBIO2018, flying-bipedal-robot:iCub-RAL2018, flying-bipedal-robot:Jet-HR2-RAL2022} to solve this issue by the walking motion. Most of these robots are bipedal model which attaches the multirotor unit in their torso for aerial maneuvering. Compared with the bipedal model, it is relatively easier to achieve the walking stability by the multilegged model because of the larger support polygon. Besides, the legs can be considered as the limbs for manipulation in midair, and thus more limbs can provide more end-effectors for complex manipulation task.
Therefore, we choose the quadruped model that can offer both the stable terrestrial locomotion and the potential of aerial manipulation.
For most of the quadruped robot, the form is bio-inspired and specialized for terrestrial locomotion \cite{quadruped-robot:LittleDog-IJRR2011, quadruped-robot:ANYmal-IROS2016, quadruped-robot:MiniCheetah-ICRA2019}. However, in our work, the robot is also desired to perform manipulation task in multiple domains. Several ape-like quadruped robots show the manipulation ability by the hands attached on the leg ends \cite{quadruped-robot:JPL-RoboSimian-JFR2015, quadruped-robot:WAREC-1-SSRR2017}. The point symmetry is the crucial difference of these robots from the skeleton design for common quadruped robots. In our work, we also adopt the point symmetry for our hybrid quadruped robot to enable the omni-directional maneuvering and manipulation in both terrestrial and aerial domains.
Regarding the rotor arrangement, it is difficult for rotors centralized in the robot torso like \cite{flying-bipedal-robot:LEONARDO-SciRo2021} to handle the change of CoG caused by the joint motion. Besides, the external force acted at the end-effectors can also induce a large rotational load for the centralized rotors due to the large moment arm. To ensure a sufficient control margin for the stable joint motion in midair, \cite{flying-bipedal-robot:iCub-RAL2018} proposes a distributed rotor arrangement that deploys the thrust units at each limb end. However, this rotor arrangement deprives the robot of manipulation ability. Therefore, a fully-distributed rotor design proposed by \cite{aerial-robot:DRAGON-RAL2018} is applied in this work. In this design, spherically vectorable rotor apparatus is embedded in each link and thus can generate individual three-dimensional thrust force for the promising maneuvering and manipulation in midair as presented in \cite{aerial-robot:DRAGON-IJRR2022}.
For the stable flying motion, the whole platform should be lightweight, which leads a relatively compact and weak joint actuator for legs. Hence, the thrust force is also required to assist the walking motion by reducing the load from gravity. For model with centralized rotor arrangement, both the model-based \cite{flying-bipedal-robot:LEONARDO-SciRo2021} and the policy-based \cite{flying-bipedal-robot:KXR-IROS2022} methods are developed to obtain the assistive thrust. However, these methods are not suitable for the model with rotors distributed in all links. Then, \cite{flying-bipedal-robot:Jet-HR1-IROS2020} proposes an assistive thrust control method for a bipedal robot with the vectorable rotor attached at each foot, whereas \cite{aerial-robot:DRAGON-IJRR2022} presents a control method for a fully-distributed rotor model in the aerial domains.
Based on these methods, a comprehensive investigation of the modeling and control for the multilegged model is performed in this work to handle multiple types of torques and forces (i.e., the joint torque, the contact force on each limb end, the gravity of each link, and the thrust force from each vectorable rotor).
The main contributions of this work can be summarized as follows:
\begin{enumerate}
\item We propose a unique mechanical design for air-ground quadruped robot with the spherically vectorable rotors distributed in all links.
\item We present a modeling and control methods for this multilegged platform for hybrid locomotion in both terrestrial and aerial domains.
\item We achieve the seamless and stable motion that involves walking, flying and joint motion in midair by the prototype of quadruped robot.
\end{enumerate}
The remainder of this paper is organized as follows. The mechanical design for this unique quadruped robot is introduced in \secref{design}. The modeling of our robot is presented in \secref{model}, followed by the integrated control method for hybrid locomotion in \secref{control}. We then show the experimental results in \secref{experiment} before concluding in \secref{conclusion}.
\section{Design}
\label{sec:design}
In this section, we present the mechanical design for the quadruped robot that is capable of terrestrial/aerial hybrid locomotion. The key of the whole structure is the spherically vectorable rotor embedded in each link unit, along with the unique quadruped shape that differs from the common bio-inspired type.
\begin{figure}[t]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/design.ps}
\caption{{\bf Mechanical design for the air-ground hybrid quadruped robot}. {\bf (A)} skeleton model for common bio-inspired mammal-type quadruped robot. {\bf (B)} proposed point-symmetric skeleton model for hybrid quadruped model. {\bf (C1)/(C2)} two-DoF joint module for each limb, where the yaw axis $q_{i\_\mathrm{yaw}}$ comes first followed by the pitch axis $q_{i\_\mathrm{pitch}}$. {\bf (D)} spherically vectorable rotor apparatus with two vectoring angles ($\phi, \theta$), and a combined thrust force $\lambda_i$ from the counter rotating dual rotors. There is a small offset between two vectoring axes because two rods cannot intersect with each other.}
\label{figure:design}
\end{center}
\end{figure}
\subsection{Skeleton Model}
\label{subsec:skeleton_model}
As shown in \figref{figure:design}(A), the common skeleton for quadruped robot is mammal-type, which puts a priority on the forward motion. Thus, the model is plane symmetric, and each leg has three DoF (two in the hip, and one in knee). However, our robot is desired to enable not only the terrestrial/aerial hybrid locomotion, but also the manipulation in midair. Therefore, the omni-directional movement is a critical feature for the skeleton design. Then, a point symmetric structure is introduced as depicted in \figref{figure:design}(B). This is similar to the sprawling-type quadruped design proposed by \cite{quadruped-robot:TITAN-XIII-IROS2013}, which can provide s wider supporting polygon and also a lower CoG than the mammal-type. According to this design concept, each limb consists of two links that have the same length, and is connected to the center torso with a joint module that has two Degree-of-Freedom (DoF). For this joint module, the yaw axis $q_{i\_\mathrm{yaw}}$ comes first, which is followed by the pitch axis $q_{i\_\mathrm{pitch}}$ to allow a larger swinging range for walking as shown in \figref{figure:design}(C1)/(C2). The crucial difference from the ordinary sprawling-type is that we also introduce an identical two-DoF joint module to connect neighboring links, which can provide a four-DoF manipulation capability by each limb end without the help of the torso motion. Eventually, this robot is composed of 8 links with 16 joints for walking and flying. Given that the lightweight design is significantly important for the flight performance, we deploy a compact servo motor to individually actuate each joint at the expense of the torque power. Nevertheless, the shortage of the joint torque can be compensated by the rotor thrust in our robot.
\subsection{Spherically Vectorable Rotor}
\label{subsec:vectorable_rotor}
Rotors embedded in links are used to achieve flight with arbitrary joint motion in midair. Besides, it is also necessary to use the rotor thrust to assist leg lifts for walking, because the joint actuator is weak due to the lightweight design as mentioned in \subsecref{skeleton_model}. Therefore, the rotor is required to point arbitrary direction to handle the change in link orientation. In other words, it is required to generate a three-dimensional thrust force by each rotor module to interact with not only the gravity and also the external force (i.e., the contact force on each foot). Then a spherically vectorable apparatus proposed by \cite{aerial-robot:DRAGON-RAL2018} is equipped in each link as depicted in \figref{figure:design}(D).
To achieve the spherical vectoring around the link unit, two rotation axes is necessary. We first introduce a roll axis $\phi_i$ around the link rod. Then, we need the second orthogonal vectoring axis. If we use a single rotor, the collision between the propeller and the link rod will be inevitable while performing the second vectoring. Therefore, we apply the counter-rotating dual-rotor module to avoid the collision as shown in \figref{figure:design}(D). In addition, this dual-rotor can also counteract the drag moment and gyroscopic moment. Then, we define the pitch axis across dual rotors as the second vectoring axis $\theta_i$. Each vectoring axis is also actuated by an individual compact servo motor. Regarding the thrust force, since we assume that a pair of rotors rotate with the same speed, it is possible to introduced a combined thrust $\lambda_i$ for each spherically vectorable rotor module. Eventually, there are three control input (two vectoring angles $\phi_i, \theta_i$, and one thrust $\lambda_i$) for each vectorable rotor module, and totally 8 sets are used to control the whole quadruped model.
\section{Modeling}
\label{sec:model}
In this section, we describe the modeling for this robot which can be divided into two parts: the thrust model and the whole dynamics model.
\begin{figure}[b]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/model.ps}
\caption{{\bf Dynamics model of the proposed quadruped robot}. The entire dynamics involves the joint torque ${\bm \tau}_j$, the contact force at each limb end ${\bm f}_{c_i}$, the gravity of each link $m_i\bm{g}$, and the thrust force from each vectorable rotor ${\bm f}_{i}$. $\{L_i\}$ and $\{F_i\}$ denote the frame for the $i$-th link and rotor respectively, whereas $\{CoG\}$ is the CoG frame for the whole model. For the free leg during walking, the contact force ${\bm f}_{c_i}$ at the limb end disappears.}
\label{figure:model}
\end{center}
\end{figure}
\subsection{Spherically Vectorable Thrust}
\label{subsec:allocation}
Based on the kinematic model depicted in \figref{figure:model}, the three-dimensional force $\bm{f}_i$ generated by the $i$-th rotor module can be written as:
\begin{align}
\label{eq:thrust_model}
\bm{f}_{i} &= \lambda_i \bm{u}_{i}, \\
\label{eq:u_vector}
\bm{u}_{i} &= {}^{CoG}R_{F_i}({\bm q}, \phi_i, \theta_i) \begin{bmatrix}0&0&1\end{bmatrix}^{\mathsf{T}},
\end{align}
where ${}^{CoG}R_{F_i}$ denotes a rotation matrix of the rotor frame $\{F_i\}$ w.r.t. the frame $\{CoG\}$. For this robot, we define the frame $\{CoG\}$ to have an origin at the CoG point as depicted in \figref{figure:model}, and a xyz coordinate that is identical to the baselink at the center torso.
$\bm{u}_{i}$ denotes the unit vector for the spherically vectorable mechanism that is effected by two vectoring angles $\phi_i$ and $\theta_i$. Besides, this vector also depends on the joint angles ${\bm q} \in {\mathcal R}^{N_J}$, $N_J$ is the number of joints.
Then the total wrench in the frame $\{CoG\}$ can be given by
\begin{align}
\label{eq:total_wrench}
\begin{bmatrix} \bm{f}_{\lambda} \\ \bm{\tau}_{\lambda} \\ \end{bmatrix}
=
\begin{bmatrix}
{\displaystyle \sum_{i = 1}^{N_r}} \bm{f}_{i} \\
{\displaystyle \sum_{i = 1}^{N_r}} \bm{p}_{i} \times \bm{f}_{i} \\
\end{bmatrix}
= Q \bm{\lambda},
\end{align}
\begin{align}
\label{eq:q_matrix}
Q &=
\begin{bmatrix}
\bm{u}_1 & \bm{u}_2 & \cdots & \bm{u}_{N_{\mathrm{r}}} \\
\bm{p}_{1} \times \bm{u}_1 & \bm{p}_{2} \times \bm{u}_2 & \cdots & \bm{p}_{N_{\mathrm{r}}} \times \bm{u}_{N_{\mathrm{r}}} \\
\end{bmatrix}
, \\
\bm{\lambda} &=
\begin{bmatrix}
\lambda_1 & \lambda_2 & \cdots & \lambda_{N_{\mathrm{r}}} \\
\end{bmatrix}^{\mathsf{T}}, \nonumber
\end{align}
where $\bm{p}_{i}$ is the position of the frame $\{F_i\}$ origin from the frame $\{CoG\}$ that is influenced by the joint angles $\bm{q}$ and the first vectoring angle $\phi_i$ for the $i$-th rotor. $N_{\mathrm{r}}$ is the number of rotors.
\subsection{Dynamics of Multilinked Model}
\label{subsec:dynamics}
The whole dynamic model can be written as follows:
\begin{align}
\label{eq:translational_dynamics}
\dot{P}_{\Sigma} =& R_c \bm{f}_{\lambda} - m_{\Sigma} {\bm g} + {\displaystyle \sum_{i = 1}^{N_{\mathrm{c}}}} \bm{f}_{\mathrm{c}_i}, \\
\label{eq:rotational_dynamics}
\dot{{\mathcal {\bm L}}}_{\Sigma} =& \bm{\tau}_{\lambda} + {\displaystyle \sum_{i = 1}^{N_{\mathrm{c}}}} \bm{p}_{\mathrm{c}_i} \times R_c^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}, \\
\label{eq:joint_dynamics}
M_J(\bm{q})\ddot{\bm{q}} + c(\bm{q},\dot{\bm{q}}) =& \displaystyle \bm{\tau}_{q}
+ \sum^{N_{\mathrm{c}}}_{i=1} J_{\mathrm{c}_i}^{\mathsf{T}} \bm{f}_{\mathrm{c}_i} \nonumber \\
& + \sum^{N_r}_{i=1} J_{\mathrm{r}_i}^{\mathsf{T}} \bm{f}_i
+ \sum^{N_s}_{i=1} J_{\mathrm{s}_i}^{\mathsf{T}} m_{\mathrm{s}_i}\bm{g}.
\end{align}
\equref{eq:translational_dynamics} and \equref{eq:rotational_dynamics} denote the centroidal dynamics for the whole multibody model.
$P_{\Sigma}$ and ${\mathcal {\bm L}}_{\Sigma}$ are the entire linear and rotational momentum described in the inertial frame $\{W\}$ and the frame $\{CoG\}$, respectively.
These momentum are both affected by the joint angles, vectoring angles, and their velocities (i.e., $\bm{q}, \dot{\bm{q}}, \bm{\phi}, \dot{\bm{\phi}}, \bm{\theta}, \dot{\bm{\theta}}$).
$R_c$ is the orientatin of the frame $\{CoG\}$ w.r.t. the frame $\{W\}$, and is identical to $R_b$ that is the orientatin of the baselink.
${\bm f}_{\lambda}$ and ${\bm \tau}_{\lambda}$ corresponds to the total wrench described in \equref{eq:total_wrench}.
$\bm{f}_{\mathrm{c}_i}$ is the contact force at the $i$-th limb end (foot) w.r.t. the frame $\{W\}$, whereas $\bm{p}_{\mathrm{c}_i}$ is the position of this contact point from the frame $\{CoG\}$ which is also influenced by the joint angles $\bm{q}$. $N_{\mathrm{c}}$ is the number of contact points (i.e., standing legs).
$m_{\Sigma}$ is the total mass, and $\bm{g}$ is a three-dimensional vector expressing gravity.
\equref{eq:joint_dynamics} corresponds to the joint motion. $M_J(\bm{q})$ denotes the inertial matrix, whereas $c(\bm{q},\dot{\bm{q}})$ is the term related to the centrifugal and Coriolis forces in joint motion.
$J_{{\mathrm{*}}_i}\in{\mathcal R}^{3 \times N_J}$ is the Jacobian matrix for the frame of the $i$-th contact point ($* \rightarrow \mathrm{c}$), the $i$-th rotor ($* \rightarrow \mathrm{r}$), and the $i$-th segment's CoG ($* \rightarrow \mathrm{s}$), respectively. $\bm{\tau}_q \in \mathcal{R}^{N_J}$ is the vector of joint torque and $\bm{f}_i$ is the vectoring thrust force corresponding to \equref{eq:thrust_model}.
\eqref{eq:translational_dynamics} $\sim$ \equref{eq:joint_dynamics} are highly complex due to the joint motion. Then the realtime feedback control based on these nonlinear equations is significantly difficult for an onboard computational resource. Therefore, for the joint motion, a crucial assumption is introduced in our work to simplify the dynamics, i.e., all the joints are actuated slowly by individual servo motors. This is also called the quasi-static assumption that allows $\dot{\bm{q}} \approx \bm{0}; \ddot{\bm{q}} \approx \bm{0}$ during the joint motion.
Under this assumption, the original dynamic model can be approximated as follows:
\begin{align}
\label{eq:approx_translational_dynamics}
&m_{\Sigma} \ddot{\bm r}_c(\bm{q}) = R_c \bm{f}_{\lambda}
- m_{\Sigma} {\bm g} + {\displaystyle \sum_{i = 1}^{N_{\mathrm{c}}}} \bm{f}_{\mathrm{c}_i}, \\
\label{eq:approx_rotational_dynamics}
&I_{\Sigma}(\bm{q})\dot{\bm \omega} + {\bm \omega} \times I_{\Sigma}(\bm{q}) {\bm \omega} = \bm{\tau}_{\lambda} + {\displaystyle \sum_{i = 1}^{N_{\mathrm{c}}}} \bm{p}_{\mathrm{c}_i} \times \bm{f}_{\mathrm{c}_i}, \\
\label{eq:approx_joint_dynamics}
& \bm{0} = \displaystyle \bm{\tau}_{q}
+ \sum^{N_{\mathrm{c}}}_{i=1} J_{\mathrm{c}_i}^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}
+ \sum^{N_r}_{i=1} J_{\mathrm{r}_i}^{\mathsf{T}} \bm{f}_i
+ \sum^{N_s}_{i=1} J_{\mathrm{s}_i}^{\mathsf{T}} m_{\mathrm{s}_i}\bm{g},
\end{align}
where $\bm{r}_c$ is the position of the frame $\{CoG\}$ w.r.t the frame $\{W\}$, which can be calculated using the forward-kinematics from the baselink states with the joint angles $\bm{q}$. ${\bm \omega}_c$ is the angular velocity of the frame $\{CoG\}$ w.r.t the frame of $\{CoG\}$, and is identical to ${\bm \omega}_b$ that is the angular velocity of the baselink. $I_{\Sigma} (\bm{q})$ is the total inertia tensor that is also influenced by the joint angles $\bm{q}$.
\equref{eq:approx_translational_dynamics} and \eqref{eq:approx_rotational_dynamics} show the property of a single rigid body, whereas \equref{eq:approx_joint_dynamics} indicates the equilibrium between the forces and torques for the joint motion. Given that we apply the quasi-static assumption for joint motion, only the slow terrestrial motion, such as the static walk gait, is allowed
\section{Control}
\label{sec:control}
In this section, we first describe a unified control framework as depicted in \figref{figure:control}, and then present the modification for the aerial and terrestrial locomotion, respectively.
\begin{figure}[b]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/control.ps}
\caption{{\bf Unified control framework for terrestrial/aerial locomotion}. ``Model approximation'' presented in \subsecref{dynamics} is followed by the vectorable rotor control based on the centroidal motion. The joint control is performed independently.}
\label{figure:control}
\end{center}
\end{figure}
\subsection{Centroidal Motion Control}
For the approximated dynamics of \eqref{eq:approx_translational_dynamics} and \eqref{eq:approx_rotational_dynamics}, the position feedback control based on an ordinary PID control is given by
\begin{align}
\label{eq:pid_pos}
{\bm f}_{\lambda}^d &= m_{\Sigma} R_c^{\mathsf{T}} (K_{f, p} \bm{e}_{\bm{r}} + K_{f, i} \int \bm{e}_{\bm{r}} + K_{f, d} \dot{\bm{e}}_{\bm{r}}) \nonumber \\
& \hspace{5mm} + R_c^{\mathsf{T}} (m_{\Sigma} {\bm g} - {\displaystyle \sum_{i = 1}^{N_{\mathrm{c}}}} \bm{f}_{\mathrm{c}_i}),
\end{align}
where $\bm{e}_{\bm{r}} = \bm{r}_c^d - \bm{r}_c$, and $K_{f, \ast}$ are the PID gain diagonal matrices.
The attitude control follows the SO(3) control method proposed by \cite{aerial-robot:SE3-Control-CDC2010}:
\begin{align}
\label{eq:pid_rot}
{\bm \tau}^d_{\lambda} &= I_{\Sigma} (K_{\tau, p} \bm{e}_{R} + K_{\tau, i} \int \bm{e}_{R} + K_{\tau, d} \bm{e}_{\bm{\omega}}) \nonumber \\
& \hspace{3mm}+ {\bm \omega}_c \times I_{\Sigma} {\bm \omega}_c -{\displaystyle \sum_{i = 1}^{N_{\mathrm{c}}}} \bm{p}_{\mathrm{c}_i} \times R_c^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}, \\
\label{eq:rotation_error}
\bm{e}_{R} &= \frac{1}{2}\left[R_c^{\mathsf{T}}R_c^d - R_c^{d\mathsf{T}}R_c\right]^{\vee}, \\
\label{eq:omega_error}
\bm{e}_{\bm{\omega}} &= R_c^{\mathsf{T}}R_c^d\bm{\omega}_c^d - \bm{\omega}_c,
\end{align}
where $\left[\star\right]^{\vee}$ is the inverse of a skew map.
Then, the desired wrench w.r.t the frame $\{CoG\}$ can be summarized as follows:
\begin{align}
\label{eq:desired_wrench}
\bm{\mathrm{w}} ^d = \begin{bmatrix} \bm{f}_{\lambda} ^d && \bm{\tau}_{\lambda} ^d \end{bmatrix}^{\mathsf{T}}.
\end{align}
\subsection{Control Allocation}
The goal of vectorable rotor control is to obtain the control input (the desired thrust $\bm{\lambda}^d$ and the desired vectoring angles $\bm{\phi}^d, \bm{\theta}^d$) from the desired wrench ${\bm{\mathrm{w}}}^d$. Meanwhile, it is also important to suppress the rotor output and the joint load from the aspect of the energy consumption. Then, an optimization problem should be design to obtain the desired control input.
Since the vectoring angles $\bm{\phi}$ and $\bm{\theta}$ demand the trigonometric function, nonlinear constraints would appear in the optimization problem and thus lead a complex computation. To decrease the computational load during the realtime control loop, we introduce an alternative three-dimensional forces $\bm{f}^{'}_{i}$ that is the vectorable thrust w.r.t, the related link frame $\{L_i\}$: $\bm{f}^{'}_{i} = {}^{L_i}R_{F_i}(\phi_i, \theta_i) \begin{bmatrix}0&0&\lambda_i\end{bmatrix}^{\mathsf{T}}$. The definition of the frames of $\{L_i\}$ and $\{F_i\}$ can be found in \figref{figure:model}. Then the above optimization problem can be modified as follows:
\begin{align}
\label{eq:rough_allocation_cost}
& \displaystyle \min_{\bm{f}^{'}_{i}, \bm{\tau}_q, \bm{f}_{c_i}} \hspace{3mm} w_1 {\displaystyle \sum_{i = 1}^{N_{\mathrm{r}}}} \| \bm{f}^{'}_{i} \|^2 + w_2 \|\bm{\tau}_q\|^2,\\
\label{eq:wrench_allocation_constraint}
& s.t. \hspace{3mm}
\bm{\mathrm{w}} ^d = {\displaystyle \sum_{i = 1}^{N_{\mathrm{r}}}} {Q}_{i} \bm{f}^{'}_{i},
\hspace{3mm}
{Q}_{i} =
\begin{bmatrix}
E_{3\times 3} \\
\left[\bm{p}_i \times \right]
\end{bmatrix} {}^{CoG}R_{L_i},\\
\label{eq:quasi_static_constraint}
& \hspace{8mm} \displaystyle \bm{\tau}_{q} =
- \sum^{N_{\mathrm{c}}}_{i=1} J_{\mathrm{c}_i}^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}
- \sum^{N_{\mathrm{r}}}_{i=1} J_{\mathrm{r}_i}^{\mathsf{T}} \bm{f}_i
- \sum^{N_s}_{i=1} J_{\mathrm{s}_i}^{\mathsf{T}} m_{\mathrm{s}_i}\bm{g}, \\
\label{eq:thrust_constraint}
& \hspace{8mm} 0 < \lambda_i < \bar{\lambda}, \\
\label{eq:joint_constraint}
& \hspace{8mm} - \bar{\tau}_q < \tau_{q_i} < \bar{\tau}_q, \\
\label{eq:contact_force_constraint}
& \hspace{8mm} 0 < f_{c_i}(2),
\end{align}
where $w_1$ and $w_2$ in \equref{eq:rough_allocation_cost} are the weights for the cost of rotor thrust and joint torque, respectively.
\equref{eq:wrench_allocation_constraint} is the modified form of wrench allocation from \equref{eq:total_wrench} by using the alternative variable $\bm{f}^{'}_{i}$. $\bm{p}_i$ is defined in \equref{eq:total_wrench}, whereas ${}^{CoG}R_{L_i}$ is the orientation of the frame $\{L_i\}$ w.r.t. the frame $\{CoG\}$. $E_{3\times 3}$ is a 3 $\times$ 3 identity matrix and $\left[\cdot \times \right]$ denotes the skew symmetric matrix of a three dimensional vector.
\equref{eq:quasi_static_constraint} denotes the equilibrium between the joint torque $\bm{\tau}_q$, the contact force $\bm{f}_{c_i}$, the thrust force $\bm{f}_i$, and the segment gravity $m_{\mathrm{s}_i}\bm{g}$ to satisfy the joint quasi-static assumption. \equref{eq:thrust_constraint} and \equref{eq:joint_constraint} denote the bounds for the rotor thrust and joint torque, respectively. The contact force $\bm{f}_{c_i}$ is also considered as the searching variable, and the $z$ element $f_{c_i}(2)$ should be always non-negative as shown in \equref{eq:contact_force_constraint}.
\if 0
The goal of vectorable rotor control is to obtain the control input (the desired thrust $\bm{\lambda}^d$ and the desired vectoring angles $\bm{\phi}^d, \bm{\theta}^d$) from the desired wrench ${\bm{\mathrm{w}}}^d$. Meanwhile, it is also important to suppress the rotor output and the joint load from the aspect of the energy consumption. Then, following optimal problem can be given to obtain the desired control input.
\begin{align}
\label{eq:allocation_cost}
& \displaystyle \min_{\bm{\lambda}, \bm{\theta}, \bm{\phi}, \bm{\tau}_q, \bm{f}_{c_i}} \hspace{3mm} w_1 \|\bm{\lambda}\|^2 + w_2 \|\bm{\tau}_q\|^2, \\
\label{eq:wrench_allocation_constraint}
& s.t. \hspace{3mm} \bm{\mathrm{w}} ^d = Q(\bm{\theta}, \bm{\phi}) \bm{\lambda}, \\
\label{eq:quasi_static_constraint}
& \hspace{8mm} \displaystyle \bm{\tau}_{q} =
- \sum^{N_{\mathrm{c}}}_{i=1} J_{\mathrm{c}_i}^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}
- \sum^{N_{\mathrm{r}}}_{i=1} J_{\mathrm{r}_i}^{\mathsf{T}} \bm{f}_i
- \sum^{N_s}_{i=1} J_{\mathrm{s}_i}^{\mathsf{T}} m_{\mathrm{s}_i}\bm{g}, \\
\label{eq:thrust_constraint}
& \hspace{8mm} 0 < \lambda_i < \bar{\lambda}, \\
\label{eq:joint_constraint}
& \hspace{8mm} - \bar{\tau}_q < \tau_{q_i} < \bar{\tau}_q, \\
\label{eq:contact_force_constraint}
& \hspace{8mm} 0 \leq f_{c_i}(2),
\end{align}
where $w_1$ and $w_2$ are the weights for the cost of rotor thrust and joint torque, respectively.
\equref{eq:wrench_allocation_constraint} corresponds to the wrench allocation based on \eqref{eq:q_matrix}.
\equref{eq:quasi_static_constraint} denotes the equilibrium between the joint torque $\bm{\tau}_q$, the contact force $\bm{f}_{c_i}$, the thrust force $\bm{f}_i$, and the segment gravity $m_{\mathrm{s}_i}\bm{g}$ to satisfy the joint quasi-static assumption. \equref{eq:thrust_constraint} and \equref{eq:joint_constraint} denote the bounds for the rotor thrust and joint torque, respectively. The contact force $\bm{f}_{c_i}$ is also considered as the searching variable, and the $z$ element $f_{c_i}(2)$ should be always non-negative as shown in \equref{eq:contact_force_constraint}.
The cost function \equref{eq:allocation_cost} is quadratic form; however \equref{eq:wrench_allocation_constraint} and \equref{eq:quasi_static_constraint} are nonlinear due to the trigonometric function with the vectoring angles $\bm{\phi}$ and $\bm{\theta}$. Although nonlinear optimization algorithm such as SQP can be applied, to further decrease the computational load during the realtime control loop, we introduce an alternative three-dimensional forces $\bm{f}^{'}_{i}$ that is the vectorable thrust w.r.t, the related link frame $\{L_i\}$: $\bm{f}^{'}_{i} = ^{L_i}R_{F_i}(\phi_i, \theta_i) \begin{bmatrix}0&0&\lambda_i\end{bmatrix}^{\mathsf{T}}$. The definition of the frames of $\{L_i\}$ and $\{F_i\}$ can be found in \figref{figure:model}. Then the above optimization problem can be modified as follows:
\begin{align}
\label{eq:rough_allocation_cost2}
& \displaystyle \min_{\bm{f}^{'}_{i}, \bm{\tau}_q, \bm{f}_{c_i}} \hspace{3mm} w_1 {\displaystyle \sum_{i = 1}^{N_{\mathrm{r}}}} \| \bm{f}^{'}_{i} \|^2 + w_2 \|\bm{\tau}_q\|^2,\\
\label{eq:wrench_allocation_constraint2}
& s.t. \hspace{3mm}
\bm{\mathrm{w}} ^d = {\displaystyle \sum_{i = 1}^{N_{\mathrm{r}}}} {Q}_{i} \bm{f}^{'}_{i},
\hspace{3mm}
{Q}_{i} =
\begin{bmatrix}
E_{3\times 3} \\
\left[\bm{p}_i \times \right]
\end{bmatrix} {}^{CoG}R_{L_i},\\
& \hspace{8mm} \displaystyle \bm{\tau}_{q} =
- \sum^{N_{\mathrm{c}}}_{i=1} J_{\mathrm{c}_i}^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}
- \sum^{N_{\mathrm{r}}}_{i=1} J_{\mathrm{r}_i}^{\mathsf{T}} \bm{f}_i
- \sum^{N_s}_{i=1} J_{\mathrm{s}_i}^{\mathsf{T}} m_{\mathrm{s}_i}\bm{g}, \tag{\ref{eq:quasi_static_constraint}} \\
& \hspace{8mm} 0 < \lambda_i < \bar{\lambda}, \tag{\ref{eq:thrust_constraint}} \\
& \hspace{8mm} - \bar{\tau}_q < \tau_{q_i} < \bar{\tau}_q, \tag{\ref{eq:joint_constraint}} \\
& \hspace{8mm} 0 < f_{c_i}(2), \tag{\ref{eq:contact_force_constraint}}
\end{align}
where \equref{eq:wrench_allocation_constraint2} shows the modified form of wrench allocation by using the alternative variable $\bm{f}^{'}_{i}$. $\bm{p}_i$ is defined in \equref{eq:total_wrench}, whereas $^{CoG}R_{L_i}$ is the orientation of the frame $\{L_i\}$ w.r.t. the frame $\{CoG\}$. $E_{3\times 3}$ is a 3 $\times$ 3 identity matrix and $\left[\cdot \times \right]$ denotes the skew symmetric matrix of a three dimensional vector.
\fi
Given that all constraints \equref{eq:wrench_allocation_constraint} $\sim$ \equref{eq:contact_force_constraint} are linear, an ordinary algorithm for quadratic problem can be applied. Once the optimized thrust force $\tilde{\bm{f}}^{'}_{i}$ is calculated, the true control input for the spherically vector rotor apparatus can be obtained as follows:
\begin{align}
\label{eq:pseudo_inverse_desired_thrust_force}
&\lambda_i = \| {\bm{f}}^{'}_{i} \| ,\\
\label{eq:pseudo_inverse_desired_vectoring_phy}
&\phi_i = tan^{-1}(\frac{-f^{'}_{i}(1)}{f^{'}_{i}(2)}),\\
\label{eq:pseudo_inverse_desired_vectoring_theta}
&\theta_{i} = tan^{-1}(\frac{f^{'}_{i}(0)}{-f^{'}_{i}(1) sin(\phi_{i}) + f^{'}_{i}(2) cos(\phi_{i})}),
\end{align}
where $f^{'}_{i}(0)$, $f^{'}_{i}(1)$, and $f^{'}_{i}(2)$ are the $x,y$, and $z$ element of the vector.
As a unique mechanical feature of the spherically vectorable apparatus depicted in \figref{figure:design}(D), the result of vectoring angles $\bm{\phi}$ and $\bm{\theta}$ from \eqref{eq:pseudo_inverse_desired_vectoring_phy} and \eqref{eq:pseudo_inverse_desired_vectoring_theta} will deviate the position $\bm{p}_i$ in \eqref{eq:wrench_allocation_constraint} because of the small offset between two vectoring axes as depicted in \figref{figure:design}(D). Then, the results of \eqref{eq:pseudo_inverse_desired_thrust_force} $\sim$ \eqref{eq:pseudo_inverse_desired_vectoring_theta} will no longer satisfy the constraint \eqref{eq:wrench_allocation_constraint} because ${Q_i}$ has changed. To solve this problem, we apply the iteration process that is based on the gradient of a residual term $\bm{\epsilon} := \bm{\mathrm{\mathrm{w}}} ^d - Q(\bm{\theta}, \bm{\phi}) \bm{\lambda}$, and finally we can obtain the convergent values of $\bm{\phi}^d$, $\bm{\theta}^d$ and $\bm{\lambda}^d$. The detail can be found in \cite{aerial-robot:DRAGON-IJRR2022}.
\subsection{Joint Control}
The proposed optimization problem of \equref{eq:rough_allocation_cost} can also provides the joint torque that however only satisfies the quasi-static assumption for joint motion. In addition, the measurement bias and noisy from the joint encoders along with the slight deformation of the link and joint structure can also induce the model error. To handle this model error, it is necessary to apply a feed-back control to track the desired position for joints. Therefore, a simple PD control for joint position is introduced for each joint:
\begin{align}
\label{eq:joint_pd_control}
\tau_i^d = k_{\mathrm{j},p}(q_i^d - q_i) - k_{\mathrm{j},d} \dot{q}_i,
\end{align}
where $q_i^d$ is the desired joint angle from the walking gait or the aerial transformation planning. $k_{\mathrm{j},p}$ and $k_{\mathrm{j},d}$ are the P and D control gains.
It is also notable that we also used the same PD control for the rotor vectoring angles $\phi_i$ and $\theta_i$.
\subsection{Aerial Locomotion}
\label{subsec:flight_control}
The control mode for aerial locomotion follows the flow shown in \figref{figure:control} but without the contact force $\bm{f}_{c_i}$. Then the constraint of \equref{eq:contact_force_constraint} and the first term $\sum^{N_{\mathrm{c}}}_{i=1} J_{\mathrm{c}_i}^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}$ at the right side of \equref{eq:quasi_static_constraint} can be omitted. The joint control is executed independently to follow the trajectory given by other task planing.
\subsection{Terrestrial Locomotion}
\label{subsec:walk_control}
\subsubsection{Torso altitude control}
The terrestrial locomotion is totally based on the quasi-static joint motion. Therefore the centroidal motion should be also assumed to be static, which results in a desired wrench only handling gravity ($\bm{\mathrm{w}} ^d = \begin{bmatrix} 0 & 0 & -m_{\Sigma}\bm{g} & 0 & 0 & 0 \end{bmatrix}$) for \equref{eq:wrench_allocation_constraint}.
Despite of the joint position control proposed in \equref{eq:joint_pd_control}, a small error regarding the torso (i.e., baselink) pose, particularly along the altitude direction, would still remain mainly due to the influence of gravity. Therefore, we apply a feedback control using the rotor thrust for the torso altitude. Instead of the PID position control for the centroidal motion as proposed in \equref{eq:pid_pos}, a truncated feedback control for the torso altitude is introduced as follows:
\begin{align}
\label{eq:pid_torso_altitude}
f^d_z = k_{b} (z_b^d - z_b),
\end{align}
where $k_{b}$ is the P gain, and $z_b$ is the torso altitude.
We assume this altitude control is for the ``floating'' baselink even in the terrestrial locomotion mode. Therefore, instead of considering $f^d_z$ in \equref{eq:wrench_allocation_constraint} and \equref{eq:quasi_static_constraint}, we introduce another independent control allocation to obtain the additional thrust force as follows:
\begin{align}
\label{eq:torso_altitude_allocation}
\Delta \bm{\mathrm{w}} ^d = {\displaystyle \sum_{i = 1}^{\frac{N_{\mathrm{r}}}{2}}} {Q}_{2i} \Delta \bm{f}^{'}_{2i},
\end{align}
where $\Delta \bm{\mathrm{w}} ^d = \begin{bmatrix} 0 & 0 & f^d_z & 0 & 0 & 0 \end{bmatrix}^{\mathsf{T}}$. It is notable that we only choose the rotors in the inner link of each leg to suppress the influence on the joint quasi-static motion as presented in \equref{eq:quasi_static_constraint}. Then $\Delta \bm{f}^{'}_{2i}$ can be given by
\begin{align}
\label{eq:torso_altitude_allocation_inv}
\Delta \bm{f}^{'} &= \tilde{Q}^{\#} \Delta \bm{\mathrm{w}} ^d, \\
\tilde{Q} &= \begin{bmatrix} {Q}_{0} & {Q}_{2} & \cdots & {Q}_{N_{\mathrm{r}}} \end{bmatrix}, \nonumber \\
\Delta \bm{f}^{'} &= \begin{bmatrix} \Delta \bm{f}^{'}_{0} & \Delta \bm{f}^{'}_{2} & \cdots & \Delta \bm{f}^{'}_{N_{\mathrm{r}}} \end{bmatrix}^{\mathsf{T}}, \nonumber
\end{align}
where $\tilde{Q}^{\#}$ is the psuedo-inverse matrix of $\tilde{Q}$. Finally, $\bm{f}^{'}_{2i} \rightarrow \bm{f}^{'}_{2i} + \Delta \bm{f}^{'}_{2i}$ is performed before substituting it into \equref{eq:pseudo_inverse_desired_thrust_force}$\sim$\equref{eq:pseudo_inverse_desired_vectoring_theta}.
\if 0
To consider the additional force $f^d_z$ for both the centroidal and joint motion, we modify the constraints of \equref{eq:wrench_allocation_constraint} and \equref{eq:quasi_static_constraint} as follows for the optimization problem of \equref{eq:rough_allocation_cost}:
\begin{align}
\label{eq:wrench_allocation_constraint_torso}
& \bm{\mathrm{w}} ^d + \Delta \bm{\mathrm{w}} ^d = {\displaystyle \sum_{i = 1}^{N_{\mathrm{r}}}} \tilde{Q}_{i} \bm{f}^{'}_{i}, \\
\label{eq:quasi_static_constraint_torso}
& \displaystyle \bm{\tau}_{q} =
- \sum^{N_{\mathrm{c}}}_{i=1} J_{\mathrm{c}_i}^{\mathsf{T}} \bm{f}_{\mathrm{c}_i}
- \sum^{N_{\mathrm{r}}}_{i=1} J_{\mathrm{r}_i}^{\mathsf{T}} \bm{f}_i
- \sum^{N_s}_{i=1} J_{\mathrm{s}_i}^{\mathsf{T}} m_{\mathrm{s}_i}\bm{g} + J_{\mathrm{b}}^{\mathsf{T}} f^d_z,
\end{align}
where $J_{\mathrm{b}}$ is the Jacobian matrix regarding the baselink, and $\Delta \bm{\mathrm{w}} ^d = \begin{bmatrix} 0 & 0 & f^d_z & 0 & 0 & 0 \end{bmatrix}$.
\fi
\subsubsection{Static walking gait}
In this work, we only focus on the static walking gait. Hence only one leg is allowed to lift during walking. As the update of the foot step for the lifting leg, we analytically solve the inverse-kinematics for the related three joint angles: $q_{i\_\mathrm{yaw}}, q_{i\_\mathrm{pitch}}$, and $q_{i+1\_\mathrm{pitch}}$ as depicted in \figref{figure:design}, which can be uniquely determined. Regarding the gait for linear movement, we design a creeping gait that lifts the front-left, front-right, rear-right, and rear-left legs in order for one gait cycle, and also solely moves the torso in standing mode just after the two front legs have moved to the new position. To enable the repetition of the gait cycle, the stride length of all feet is set equal to the moving distance of torso.
We further assume the robot only walks on a flat floor, and thus the height of feet should be always zero. Then, we first set an intermediate target position right above the new foot step with a small height offset. Thus $q_{i\_\mathrm{yaw}}$ and $q_{i+1\_\mathrm{pitch}}$ are identical to the final target, whereas $q_{i\_\mathrm{pitch}}$ is smaller than the final value. Once the lifting leg moves to this intermediate pose, the robot starts lowering the leg to reach the new foot step only by changing $q_{i\_\mathrm{pitch}}$. Given that there is no tactile sensor on the foot, we introduce a threshold $\Delta q_{c}$ for the joint angle error of $q_{i\_\mathrm{pitch}}$ to detect touchdown. That is, if $q^d_{i\_\mathrm{pitch}} - q_{i\_\mathrm{pitch}} < \Delta q_{c}$, then switch the lifting leg to the standing mode, and thus the number of the contact force $\bm{f}_{c_i}$ changes from three to four.
\section{Experiment}
\label{sec:experiment}
\subsection{Robot Platform}
In this work, we developed a prototype of SIDAR as shown in \figref{figure:platfrom}, and the basic specification is summarized in \tabref{table:specification}. Given the lightweight design, we employed CFRP material for link rod where cables can pass through. For the joint module, we used the Aluminum sheet to connect links, whereas the joint servo was Dynamixel XH430-V350R of which the torque was enhanced by pulley made from PLA. The range of joint angle was $\left[-90^{\circ} \ 90^{\circ}\right]$.
For the vectorable rotor module, a pair of counter-rotating plastic propellers were enclosed by ducts with the aim of safety and increase of thrust, whereas Dynamixel XL430-W250T was used for the rotor vectoring.
Batteries are distributed in each link unit in parallel as shown in \figref{figure:platfrom}(G) which can provide a flight duration up to \SI{9}{min} and a longer walk duration up to \SI{20}{min}. A hemisphere foot with anti-slip tape was equipped to ensure the stable point contact during the terrestrial locomotion.
\begin{figure}[b]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/platform.ps}
\vspace{-5mm}
\caption{{\bf Prototype of SPIDAR}: {\bf (A)} center torso that employed an original red MCU called {\it Spinal} and a high level processor (Nvidia Jetson TX2); {\bf (B)} spherically vectorable dual-rotor module; {\bf (C1)(C2)} two-DoF joint module for the ``hip'' and the ``knee'', respectively; {\bf (D)} single leg (limb) that had the maximum length of \SI{1.1}{m}; {\bf (E)} small relay board called ``Neuron'' for each link unit that was connected with ``Spinal'' via CAN; {\bf (F)} hemisphere foot with anti-slip tape; {\bf (G)} distributed battery attached at each link unit.}
\label{figure:platfrom}
\end{center}
\end{figure}
On the center torso as shown in \figref{figure:platfrom}(A), NVIDIA Jetson TX2 and an original MCU board called {\it Spinal} were deployed to perform the realtime control framework as presented in \figref{figure:control}. For each link unit, there was a distributed MCU board called {\it Neuron} that served as relay node between {\it Spinal} and each actuator. {\it Neuron}s and {\it Spinal} were connected by CAN cable. The detail of the onboard communication can be found in \cite{aerial-robot:DRAGON-RAL2018}. Besides, an external motion capture system was applied in our experiment to obtain the state of the baselink (i.e., $\bm{r}_b$, $\dot{\bm r}_b$, $R_b$, and ${\bm \omega}_b$), which were used to calculate the state of centroidal motion based on forward-kinematics.
\begin{table}[t]
\begin{center}
\caption{Prototype Specifications}
\begin{tabular}{c|ccc|c}
\multicolumn{2}{c}{1. Main Feature} & & \multicolumn{2}{c}{3. Vectorable Rotor} \\
Attribute & Value && Attribute & Value \\ \cline{1-2} \cline{4-5}
total mass & \SI{15.2}{kg} && rotor KV & 1550 \\ \cline{1-2} \cline{4-5}
max size (dia.) & \SI{2.7}{m} && propeller diameter & \SI{5}{inch} \\ \cline{1-2} \cline{4-5}
max flight time & \SI{9}{min} && max thrust ($\bar{\lambda}$) & \SI{42}{N} \\ \cline{1-2} \cline{4-5}
max walk time & \SI{20}{min} && pulley ratio & 1:1.5 \\ \cline{4-5}
\multicolumn{2}{c}{} && max vectoring torque & \SI{1.5}{Nm} \\ \cline{4-5}
\multicolumn{2}{c}{2. Link and Joint} && max vectoring speed & \SI{4.2}{rad/s} \\ \cline{4-5}
Attribute & Value && \multicolumn{2}{c}{} \\ \cline{1-2} \cline{1-2}
link length & \SI{0.54}{m} && \multicolumn{2}{c}{4. Lipo Battery} \\ \cline{1-2}
joint pulley ratio & 1:2 && Attribute & Value \\ \cline{1-2} \cline{4-5}
max joint speed & \SI{0.34}{rad/s} && capacity & 6S 3Ah \\ \cline{1-2} \cline{4-5}
max torque ($\bar{\tau}_q$) & \SI{6.5}{Nm} && amount & 8 \\ \cline{1-2} \cline{4-5}
\end{tabular}
\label{table:specification}
\end{center}
\end{table}
\if 0
\begin{table}[h]
\begin{center}
\caption{Vectorable Rotor}
\begin{tabular}{|c|c|c|}
\hline
Attribute & Value \\ \hline \hline
rotor KV & 1550 \\ \hline
propeller diameter & \SI{5}{inch} \\ \hline
max rotor thrust & \SI{42}{N} \\ \hline
max vectoring speed & \SI{3}{rad/s} \\ \hline
max vectoring torque & \SI{1.5}{Nm} \\ \hline
pulley ratio & 1:1.5 \\ \hline
\end{tabular}
\label{table:specification}
\end{center}
\end{table}
\fi
\subsection{Basic Experimental Evaluation}
\subsubsection{Aerial transformation}
\begin{figure}[b]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/flight_test.ps}
\vspace{-5mm}
\caption{{\bf Stable joint motion in midair}: {\bf (A)} extended pose that has diameter of \SI{2.6}{m}; {\bf (B)} standing pose, implying the feasibility to takeoff directly from the terrestrial mode.}
\label{figure:flight}
\end{center}
\vspace{-5mm}
\end{figure}
\begin{figure}[!b]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/flight_test_plot.ps}
\vspace{-3mm}
\caption{{\bf Plots related to \figref{figure:flight} }: {\bf (A)} positional errors of $\{CoG\}$; {\bf (B)} rotational errors of $\{CoG\}$ described in XYZ Euler angles; {\bf (C)} joint trajectories for leg1 ($q_{1\_{\mathrm{yaw}}}$, $q_{1\_{\mathrm{pitch}}}$, $q_{2\_{\mathrm{yaw}}}$, and $q_{2\_{\mathrm{pitch}}}$), other legs followed the same joint trajectories; {\bf (D)} torques for those joints.}
\label{figure:flight_plot}
\end{center}
\end{figure}
A unique feature of SPIDAR is the aerial maneuvering with joint motion (i.e., aerial transformation). To validate the stability during flight, simple transformation as shown in \figref{figure:flight} was performed. All limbs changed their joints with the same trajectories as plotted in \figref{figure:flight_plot}(C).
For the control gains in \equref{eq:pid_pos} and \equref{eq:pid_rot}, we set $K_{f, p}$, $K_{f, i}$, $K_{f, d}$, $K_{\tau, p}$, $K_{\tau, i}$, and $K_{\tau, d}$ as $D(3.6, 3.6, 2.8)$, $D(0.03, 0.03, 1.2)$, $D(4, 4, 2.8)$, $D(15, 15, 10$), $D(0.3, 0.3, 0.1)$, and $D(5,5,5)$, where $D(*,*,*) \in \mathcal{R}^{3\times3}$ is a diagonal matrix. For the optimization problem of \equref{eq:rough_allocation_cost}, we omitted the second term (i.e., $w_2 = 0$) to put a priority on the minimization of the thrust force.
\figref{figure:flight_plot}(A) and (B) plotted the positional and rotational errors during the flight and transformation, and the RMS of those errors were [0.014, 0.023, 0.038] \si{m} and [0.81, 0.69, 0.92]$^{\circ}$. The altitude error $e_{r_z}$ indicated a relatively large deviation during the joint motion, which was caused by the violation of the quasi-static assumption. Nevertheless, this deviation rapidly decreased once the joint motion finished. \figref{figure:flight_plot}(C) and (D) showed that all joint were well controlled by the PD control as presented in \equref{eq:joint_pd_control}. Eventually, these results demonstrated the stability of both the baselink pose and the joint motion in aerial locomotion.
\subsubsection{Leg lifting}
The key to achieve walking by legged robot is the stability while lifting the leg.
Then, we evaluated the proposed control method by performing a long-term single leg lifting as shown in \figref{figure:raise_test}. The cost weights in \equref{eq:rough_allocation_cost} were set as $w_1 = 1, w_2 = 1$, and the bound for joint torque $\bar{\tau}_{q}$ was decreased to $\SI{1.5}{Nm}$ to ensure sufficient margin for joint control. Besides, the gain $k_{b}$ in \equref{eq:pid_torso_altitude} was set as 25.
Leg1 was lifted by changing $q_{1\_{\mathrm{pitch}}}$ from \SI{-16}{^{\circ}} to \SI{-28}{^{\circ}}, and the lifting motion lasted around \SI{30}{s} as shown in \figref{figure:raise_plot}(A). Other joints were kept constant in the whole motion as shown in \figref{figure:raise_plot}(A) and (C), and their torques were within the bounds as depicted in \figref{figure:raise_plot}(B) and (D). These results demonstrated the stability of joint motion against the influence of thrust force. Besides the stability of the baselink pose can be confirmed in \figref{figure:raise_plot}(E) and (F), where both the positional and rotational errors converged to the sufficiently small value (i.e., \SI{0.01}{m} and \SI{0.5}{^{\circ}}). \figref{figure:raise_plot}(G) showed the large increase of the thrust forces in the lifting leg, whereas \figref{figure:raise_plot}(H) showed small changes in other standing legs. In addition, these plots also confirmed the stable transition between standing mode and leg lifting mode. In particular, the shift back to the standing model around \SI{40}{s} demonstrated the smooth touchdown, which indicates the promising terrestrial locomotion.
\begin{figure}[t]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/raise_test.ps}
\vspace{-5mm}
\caption{{\bf lifting a single leg from standing mode}: {\bf (A)} standing mode where all feet have contact with the ground; {\bf (B)} lifting single leg and keeping the raised pose with the assistance of rotor thrust.}
\label{figure:raise_test}
\end{center}
\end{figure}
\begin{figure}[t]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/raise_test_plot.ps}
\vspace{-5mm}
\caption{{\bf Plots related to \figref{figure:raise_test}}: {\bf (A)} trajectories for hip pitch joints. $q_{7\_{\mathrm{pitch}}}$ was omitted due to the symmetric pose of leg4 related to leg2; {\bf (B)} torques of joints in (A); {\bf (C)} trajectories for knee pitch joints; {\bf (D)} torques of joints in (C); {\bf (E)} positional errors of baselink ; {\bf (F)} rotational errors of baselink; {\bf (G)} thrust forces in leg1; {\bf (H)} thrust forces in other legs.}
\label{figure:raise_plot}
\end{center}
\vspace{-5mm}
\end{figure}
\subsection{Seamless Terrestrial/Aerial Hybrid Locomotion}
\label{subsec:seamless_motion}
\begin{figure}[!b]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/seamless_motion.ps}
\vspace{-5mm}
\caption{{\bf Seamless Terrestrial/Aerial Hybrid Locomotion}: \textcircled{\scriptsize 1} $\sim$ \textcircled{\scriptsize 3} shows the representative phases (moving the front-left leg, the torso, and rear-left leg) in one creeping gait cycle. After five gait cycles, robot switched to the aerial locomotion directly from the terrestrial pose as shown in \textcircled{\scriptsize 4}.}
\label{figure:seamless_motion}
\end{center}
\end{figure}
We further evaluated the feasibility of seamless locomotion transition as shown in \figref{figure:seamless_motion}. \figref{figure:seamless_motion_plot}(A) and (C) demonstrated the baselink pose trajectory during walking with five gait cycles. We observed that the translational drift along the walking direction ($x$ axis) and the orthogonal direction ($y$ axis) finally grew to \SI{0.18}{m} and \SI{0.10}{m}, whereas the rotational drift along the yaw axis also increased to \SI{9}{^{\circ}}. These drifts can be attributed to the feed-froward gait planing where the target baselink pose was updated based on the last target values but not the actual values. Nevertheless, these drifts can be considered relatively small compared to the total displacement, and are possible to be suppressed by adding a feed-back loop in planning as a future work. Furthermore, the deviations regarding the $z$, roll, and pitch axes were sufficiently small, which demonstrated the efficiency of the proposed control method presented in \secref{control}.
As shown in \figref{figure:seamless_motion_plot}(B) and (D), the transition to the aerial locomotion was smooth and stable, and the stability in midair was also confirmed, Thus, these results demonstrated the feasibility of the mechanical design, modeling and control methods for the terrestrial/aerial hybrid quadruped platform.
\begin{figure}[t]
\begin{center}
\includegraphics[width=1.0\columnwidth]{figs/seamless_motion_plot.ps}
\vspace{-5mm}
\caption{{\bf Plots related \figref{figure:seamless_motion}}.{\bf (A)/(B)} trajectories of baselink position during the terrestrial locomotion and the aerial locomotion, respectively; {\bf (C)/(D)} trajectories of baselink orientation during the terrestrial locomotion and the aerial locomotion, respectively.}
\label{figure:seamless_motion_plot}
\end{center}
\end{figure}
\section{Conclusion}
\label{sec:conclusion}
In this paper, we presented the achievement of the terrestrial/aerial hybrid locomotion by the quadruped robot SPIDAR that were equipped with the vectorable rotors distributed in all links. We first proposed the mechanical design for this unique quadruped platform, and then developed the modeling and control methods to enable static walking and transformable flight. The feasibility of the above methods were verified by the experiment of seamless terrestrial/aerial hybrid locomotion with the prototype of SPIDAR.
A crucial issue remained in this work is the oscillation and deviation of the baselink pose and joint angles during walking. To improve the stability, the rotor thrust should be directly used in the joint position control to replace the current simple PD control.
Furthermore, the gait planning should be also robust against the drift by adding a feed-back loop as discussed in \subsecref{seamless_motion}.
Last but not least, the dynamic walking and the aerial manipulation will be investigated to enhance the versatility of this robot in both maneuvering and manipulation.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,587
|
title: Script Tag
---
## Script Tag
The HTML Script tag is used to either contain client side JavaScript or reference an external JavaScript file using the script "src" attribute.
The `<script>` tag/element is used to incorporate client-side Javascript into your HTML file which can be used to add interactivity and logic to your website
```
<script>
//JavaScript code is written here
</script>
<script src="js/app.js">
```
The tag can be used to encompass actual Javascript code right in the HTML itself like this
```
<script>
alert('hello this is my Javascript doing things!');
</script>
```
Or you can use it as a way to reference an external javascript file like this
```
<script src="main.js" />
```
Here the `src` attribute of the element takes in a path to a Javascript file
Script tags are loaded into your HTML in-order and syncronously so it is usually best practice to add your scripts right before the ending of your `<body>` tag in your HTML like so
```
<script src="main.js" />
<script>
alert('hello this is my Javascript doing things!');
</script>
</body>
```
You can see the official documentation for the script element on the [MDN Docs](https://developer.mozilla.org/en-US/docs/Web/HTML/Element/script)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,851
|
\section{Appendix A: The third group cohomology $\cH^3[G,U(1)]$ for symmetry $G$}
\label{Gcoh}
In this section, we will briefly describe the group cohomology theory. As we are focusing on 2D SPT phases, we will be interested in the third cohomology group.
For a group $G$, let $M$ be a G-module, which is an abelian group (with
multiplication operation) on which $G$ acts compatibly with the multiplication
operation (\ie the abelian group structure) on $M$:
\begin{align}
\label{gm}
g\cdot (ab)=(g\cdot a)(g\cdot b),\ \ \ \ g\in G,\ \ \ \ a,b\in M.
\end{align}
For the cases studied in this paper, $M$ is simply the $U(1)$ group and $a$ an
$U(1)$ phase. The multiplication operation $ab$ is the usual multiplication of
the $U(1)$ phases. The group action is trivial: $g\cdot a=a$, $g\in G$, $a=\in
U(1)$.
Let $\om_n(g_1,...,g_n)$ be a function of $n$ group
elements whose value is in the G-module $M$. In other words, $\om_n:
G^n\to M$. Let $\cC^n[G,M]=\{\om_n \}$ be the space of all such
functions.
Note that $\cC^n[G,M]$ is an Abelian group
under the function multiplication
$ \om''_n(g_1,...,g_n)= \om_n(g_1,...,g_n) \om'_n(g_1,...,g_n) $.
We define a map $d_n$ from $\cC^n[G,U(1)]$ to $\cC^{n+1}[G,U(1)]$:
\begin{align}
&\ \ \ \
(d_n \om_n) (g_1,...,g_{n+1})=
\nonumber\\
&
g_1\cdot \om_n (g_2,...,g_{n+1})
\om_n^{(-1)^{n+1}} (g_1,...,g_{n}) \times
\nonumber\\
&\ \ \ \ \
\prod_{i=1}^n
\om_n^{(-1)^i} (g_1,...,g_{i-1},g_ig_{i+1},g_{i+2},...g_{n+1})
\end{align}
Let
\begin{align}
\cB^n[G,M]=\{ \om_n| \om_n=d_{n-1} \om_{n-1}| \om_{n-1} \in \cC^{n-1}[G,M] \}
\end{align}
and
\begin{align}
\cZ^n[G,M]=\{ \om_{n}|d_n \om_n=1, \om_{n} \in \cC^{n}[G,M] \}
\end{align}
$\cB^n[G,M]$ and $\cZ^n[G,M]$ are also Abelian groups
which satisfy $\cB^n[G,M] \subset \cZ^n[G,M]$ where
$\cB^1[G,M]\equiv \{ 1\}$. $\cZ^n[G,M]$ is the group of $n$-cocycles and
$\cB^n[G,M]$ is the group of $n$-coboundaries.
The $n$th cohomology group of $G$ is defined as
\begin{align}
\cH^n[G,M]= \cZ^n[G,M] /\cB^n[G,M]
\end{align}
In particular, when $n=3$, from
\begin{align}
&\ \ \ \ (d_3 \om_3)(g_1,g_2,g_3,g_4)
\nonumber\\
&= \frac{ \om_3(g_2,g_3,g_4) \om_3(g_1,g_2g_3,g_4)\om_3(g_1,g_2,g_3) }
{\om_3(g_1g_2,g_3,g_4)\om_3(g_1,g_2,g_3g_4)}
\end{align}
we see that
\begin{align}
& \cZ^3[G,U(1)]=\{ \om_3|
\\
&\ \ \ \frac{ \om_3(g_2,g_3,g_4) \om_3(g_1,g_2g_3,g_4)\om_3(g_1,g_2,g_3) }
{\om_3(g_1g_2,g_3,g_4)\om_3(g_1,g_2,g_3g_4)}
=1
\} .
\nonumber
\end{align}
and
\begin{align}
& \cB^3[G,U(1)]=\{ \om_3| \om_3(g_1,g_2,g_3)=\frac{
\om_2(g_2,g_3) \om_2(g_1,g_2g_3)}{\om_2(g_1g_2,g_3)\om_2(g_1,g_2)}
\},
\label{3coboundary}
\end{align}
which give us the third cohomology group
$\cH^3[G,U(1)]=\cZ^3[G,U(1)]/\cB^3[G,U(1)]$.
\section{Appendix B: Matrix Product Operator Representation of Symmetry}
\label{MPUO}
In \Ref{CLW1141} the symmetry operators on the edge of bosonic SPT phases were represented in the matrix product operator formalism from which their connection to group cohomology is revealed and the non-existence of gapped symmetric states was proved. In this section, we review the matrix product representation of the unitary symmetry operators and how the corresponding cocycle can be calculated from the tensors in the representation.
A matrix product operator acting on a 1D system is given by,\cite{MCP1012}
\be
O=\sum_{\{i_k\},\{i_k'\}}Tr(T^{i_1,i'_1}T^{i_2,i'_2}...T^{i_N,i'_N})|i'_1i'_2...i'_N\>\<i_1i_2...i_N|
\ee
where for fixed $i$ and $i'$, $T^{i,i'}$ is a matrix with index $\alpha$ and $\beta$. Here we are interested in symmetry transformations, therefore we restrict $O$ to be a unitary operator $U$. Using matrix product representation, $U$ does not have to be an onsite symmetry. $U$ is represented by a rank-four tensor $T^{i,i'}_{\alpha,\beta}$ on each site, where $i$ and $i'$ are input and output physical indices and $\alpha$, $\beta$ are inner indices.
If $U(g)$'s form a representation of group $G$, then they satisfy $U(g_1)U(g_2)=U(g_1g_2)$. Correspondingly, the tensors $T(g_1)$ and $T(g_2)$ should have a combined action equivalent to $T(g_1g_2)$. However, the tensor $T(g_1,g_2)$ obtained by contracting the output physical index of $T(g_2)$ with the input physical index of $T(g_1)$, see Fig. \ref{P12}, is usually more redundant than $T(g_1g_2)$ and can only be reduced to $T(g_1g_2)$ if certain projection $P_{g_1,g_2}$ is applied to the inner indices (see Fig. \ref{P12}).
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.5]{P12}
\end{center}
\caption{Reduce combination of $T(g_2)$ and $T(g_1)$ into $T(g_1g_2)$.
}
\label{P12}
\end{figure}
$P_{g_1,g_2}$ is only defined up to an arbitrary phase factor $e^{i\mu(g_1,g_2)}$. If the projection operator on the right side $P_{g_1,g_2}$ is changed by the phase factor $e^{i\mu(g_1,g_2)}$, the projection operator $P^{\dagger}_{g_1,g_2}$ on the left side is changed by phase factor $e^{-i\mu(g_1,g_2)}$. Therefore the total action of $P_{g_1,g_2}$ and $P^{\dagger}_{g_1,g_2}$ on $T(g_1,g_2)$ does not change and the reduction procedure illustrated in Fig.\ref{P12} still works. In the following discussion, we will assume that a particular choice of phase factors have been made for each $P_{g_1,g_2}$.
Nontrivial phase factors appear when we consider the combination of three symmetry tensors $T(g_1)$, $T(g_2)$ and $T(g_3)$. See Fig. \ref{P123}.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.5]{P123}
\end{center}
\caption{Different ways to reduce combination of $T(g_3)$, $T(g_2)$ and $T(g_1)$ into $T(g_1g_2g_3)$. Only the right projection operators are shown. Their combined actions differ by a phase factor $\phi(g_1,g_2,g_3)$.
}
\label{P123}
\end{figure}
There are two different ways to reduce the tensors. We can either first reduce the combination of $T(g_1)$, $T(g_2)$ and then combine $T(g_3)$ or first reduce the combination of $T(g_2)$,$T(g_3)$ and then combine $T(g_1)$. The two different ways should be equivalent. More specifically, they should be the same up to phase on the unique block of $T(g_1,g_2,g_3)$ which contributes to matrix contraction along the chain. Denote the projection onto the unique block of $T(g_1,g_2,g_3)$ as $Q_{g_1,g_2,g_3}$. We find that
\be
\begin{array}{l}
Q_{g_1,g_2,g_3}(I_3\otimes P_{g_1,g_2})P_{g_1g_2,g_3}= \\
\phi(g_1,g_2,g_3) Q_{g_1,g_2,g_3}(P_{g_2,g_3}\otimes I_1)P_{g_1,g_2g_3}
\end{array}
\ee
>From this we see that the reduction procedure is associative up to a phase factor $\phi(g_1,g_2,g_3)$. If we then consider the combination of four symmetry tensors in different orders, we can see that $\phi(g_1,g_2,g_3)$ forms a 3-cocycle of group $G$. That is, $\phi(g_1,g_2,g_3)$ satisfies
\be
\frac{ \phi(g_2,g_3,g_4) \phi(g_1,g_2g_3,g_4)\phi(g_1,g_2,g_3) }
{\phi(g_1g_2,g_3,g_4)\phi(g_1,g_2,g_3g_4)}
=1
\ee
The arbitrary phase factor of $P_{g_1,g_2}$ contributes a coboundary term to $\phi(g_1,g_2,g_3)$. That is, if we change the phase factor of $P_{g_1,g_2}$ by $\mu(g_1,g_2)$, then $\phi(g_1,g_2,g_3)$ is changed to
\be
\t \phi(g_1,g_2,g_3) = \phi(g_1,g_2,g_3)\frac{\mu(g_2,g_3)\mu(g_1,g_2g_3)}{\mu(g_1,g_2)\mu(g_1g_2,g_3)}
\ee
$\t \phi(g_1,g_2,g_3)$ still satisfies the cocycle condition and belongs to the same cohomology class as $\phi(g_1,g_2,g_3)$.
\section{Appendix C: Cohomology class of symmetry operator $U^{(M)}_N$ in Eqn.(\ref{UMN})}
\label{UMN_coh}
In this section, we discuss the property of the symmetry operator $U^{(M)}_N$ given in Eqn.(\ref{UMN}). First we show that $U^{(M)}_N$ indeed generates a $\mathbb{Z}_N$ symmetry. Next from its matrix product unitary operator representation we find that the transformation among the tensors are indeed related to the $M$th element in the cohomology group $\cH^3[\mathbb{Z}_N,U(1)]$. The calculation of cohomology class goes as described in the previous section. We repeat the definition of $U^{(M)}_N$ here
\be
U^{(M)}_N = \prod_{i} CP^{(M)}_{i,i+1} \prod_i e^{i2\pi L_i/N}
\label{UMN_a}
\ee
where $CP^{(M)}_{i,i+1}$ acts on two neighboring rotors and depends on $M$ as
\begin{equation*}
CP^{(M)}_{i,i+1} = \int d\varphi_i d\varphi_{i+1} e^{iM(\varphi_{i+1}-\varphi_i)_r/N} \ket{\varphi_i\varphi_{i+1}}\bra{\varphi_i\varphi_{i+1}}
\end{equation*}
Note that $(\varphi_{i+1}-\varphi_i)_r$ represents $\varphi_{i+1}-\varphi_i$ to be confined within $(-\pi,\pi]$.
As $\prod_i e^{i2\pi L_i/N}$ rotates all the $\varphi_i$'s by the same angle and $\prod_{i} CP^{(M)}_{i,i+1}$ only depends on the difference between neighboring $\varphi$'s, the two parts in the symmetry operator commutes. Therefore
\begin{equation}
\left(U^{(M)}_N\right)^N = \prod_{i} \left(CP^{(M)}_{i,i+1}\right)^N \prod_i \left(e^{i2\pi L_i/N}\right)^N
\end{equation}
As $\prod_{i} \left(CP^{(M)}_{i,i+1}\right)^N = I$ and $\prod_i \left(e^{i2\pi L_i/N}\right)^N = e^{i2\pi L}=I$, $U^{(M)}_N$ indeed generators a $\mathbb{Z}_N$ symmetry on the 1D rotor system.
The matrix product representation of $U^{(M)}_N$ is given by
\begin{equation}
\begin{array}{l}
(T^{\varphi_0,\varphi_1})^{(M)}_N(1) = \delta(\varphi_1-(\varphi_0+\frac{2\pi}{N}))\times \\
\int d\varphi_{\alpha} d\varphi_{\beta} \ket{\varphi_{\beta}}\bra{\varphi_{\alpha}} \delta(\varphi_{\beta}-\varphi_0)e^{iM(\varphi_{\alpha}-\varphi_0)_r/N}
\end{array}
\end{equation}
And the tensors representing $\left(U^{(M)}_N\right)^k$, $k\in \mathbb{Z}_N$ are given by
\begin{equation}
\begin{array}{l}
(T^{\varphi_0,\varphi_1})^{(M)}_N(k) = \delta(\varphi_1-(\varphi_0+\frac{2k\pi}{N}))\times \\
\int d\varphi_{\alpha} d\varphi_{\beta} \ket{\varphi_{\beta}}\bra{\varphi_{\alpha}} \delta(\varphi_{\beta}-\varphi_0)e^{ikM(\varphi_{\alpha}-\varphi_0)_r/N}
\end{array}
\end{equation}Following the calculation described in the previous section, we find that the projection operation when combining $T^{(M)}_N(m_1)$ and $T^{(M)}_N(m_2)$ into $T^{(M)}_N((m_1+m_2)_N)$ is
\begin{equation}
\begin{array}{r}
P^{(M)}_N(m_1,m_2)=\int d\varphi_0\ket{m_2\frac{2\pi}{N}+\varphi_0}\ket{\varphi_0}\bra{\varphi_0} \times \\
e^{-iM\varphi_0(m_1+m_2-(m_1+m_2)_N)/N}
\end{array}
\end{equation}
where $(m_1+m_2)_N$ means addition modulo $N$.
When combining $T^{(M)}_N(m_1)$, $T^{(M)}_N(m_2)$ and $T^{(M)}_N(m_3)$, the phase angle in combining $m_1$ with $m_2$ first and then combining $(m_1+m_2)_N$ with $m_3$ is
\begin{equation}
\begin{array}{ll}
&M\varphi_0(-m_1-m_2+(m_1+m_2)_N -(m_1+m_2)_N- \\
&m_3+((m_1+m_2)_N+m_3)_N)/N\\
=&M\varphi_0(-(m_1+m_2+m_3)+(m_1+m_2+m_3)_N)/N
\end{array}
\end{equation}
the phase angle in combining $m_2$ with $m_3$ first and then combining $m_1$ with $(m_2+m_3)_N$ is
\begin{equation}
\begin{array}{ll}
&M\varphi_0(-m_2-m_3+(m_2+m_3)_N-m_1-\\
&(m_2+m_3)_N+ (m_1+(m_1+m_2)_N)_N)/N+ \\
&Mm_1\frac{2\pi}{N}(-m_2-m_3+(m_2+m_3)_N)/N \\
=&M\varphi_0(-(m_1+m_2+m_3)+(m_1+m_2+m_3)_N)/N + \\
&Mm_1\frac{2\pi}{N}(-m_2-m_3+(m_2+m_3)_N)/N
\end{array}
\end{equation}
Therefore, the phase difference is
\begin{equation}
\begin{array}{l}
\phi^{(M)}_N(m_1,m_2,m_3) = \\
e^{iMm_1\frac{2\pi}{N}(-m_2-m_3+(m_2+m_3)_N)/N}
\end{array}
\end{equation}
We can check explicitly that $\phi^{(M)}_N(m_1,m_2,m_3)$ satisfies the cocycle condition
\begin{equation}
\begin{array}{l}
\frac{\phi^{(M)}_N(m_2,m_3,m_4)\phi^{(M)}_N(m_1,(m_2+m_3)_N,m_4)\phi^{(M)}_N(m_1,m_2,m_3)}{\phi^{(M)}_N((m_1+m_2)_N,m_3,m_4)\phi^{(M)}_N(m_1,m_2,(m_3+m_4)_N)}\\
=1
\end{array}
\end{equation}
Also we see that $\{\phi^{(M)}_N\}$, $M=0,...,N-1$, form a $\mathbb{Z}_N$ group generated by $\phi^{(1)}_N$. Therefore, the tensor $T^{(M)}_N$ corresponds to the $M$th element in the cohomology group $\cH^3[\mathbb{Z}_N,U(1)]$.
Similar calculation holds for the $U(1)$ symmetry generated by $e^{i\alpha(Kl+K'm)}$, $K,K' \in \mathbb{Z}$. The cohomology class is labeled by $M=KK'$.
\section{Appendix D: Interpretation in terms of fermionization}
The free boson theory given in Eqn.(\ref{Lboson}) can be fermionized and the
low energy effective action of the symmetry discussed here can be reinterpreted
in terms of a free Dirac fermion. In particular, the fermionized theory has
Lagrangian density
\be
\mathcal{L}_f = \sum_{i=1,2} \psi^L_i(\partial_t + \partial_x)\psi^L_i + \psi^R_i(\partial_t - \partial_x)\psi^R_i
\label{Lfermion}
\ee
where $\psi_1$ and $\psi_2$ are two real fermions, out of which a complex
fermion can be defined $\Psi=\psi_1+i\psi_2$. Note that in order to have a
state to state correspondence between the boson and fermion theory, the fermion
theory contains both the periodic and anti-periodic sectors.
Since the $\mathbb{Z}_2$ symmetry in the nontrivial $\mathbb{Z}_2$ SPT phase only act on, say, the right moving sector,
one may naively guess that only $\psi_1^R$ change sign, while $\psi_2^R$,
$\psi_1^L$, and $\psi_2^L$ do not change under the $\mathbb{Z}_2$
transformation: $ (\psi_1^R, \psi_2^R, \psi_1^L, \psi_2^L) \to (-\psi_1^R,
\psi_2^R, \psi_1^L, \psi_2^L) $. In this case, the fermion mass term, such as
$(\psi_2^R)^\dagger \psi_2^L$, will be allowed by the $\mathbb{Z}_2$ symmetry.
Such a mass term will reduce the $c=1$ edge state to a $c=\frac 12$ edge state
without breaking the $\mathbb{Z}_2$ symmetry. In the following, we will show
that the $\mathbb{Z}_2$ symmetry is actually realized in a different way. The
$c=1$ edge state is stable if the $\mathbb{Z}_2$ symmetry is not broken. So
the $c=1$ edge state represents the minimal edge state for the $\mathbb{Z}_2$
(as well as the $\mathbb{Z}_N$ and $U(1)$) SPT phases.
The situation is best illustrated with explicit Jordan-Wigner transformation of the $XY$ model in Eqn.(\ref{H_Z2_XY}). Consider a system of size $N=4n$, $n \in \mathbb{Z}_+$. After the Jordan Wigner transformation
\begin{align}
\Psi_i=e^{i\pi\sum^{i-1}_{j=1}Z_j}(X_i+iY_i) \\ \nonumber
\Psi^{\dagger}_i=e^{i\pi\sum^{i-1}_{j=1}Z_j}(X_i-iY_i)
\end{align}
The Hamiltonian becomes
\begin{equation}
\begin{array}{l}
H=H_a+H_b \\
H_a=\sum^N_{i=1} (\Psi^{\dagger}_{i+1}\Psi_i+\Psi_i^{\dagger}\Psi_{i+1}) \\
H_b=-(P+1)(\Psi^{\dagger}_{1}\Psi_N+\Psi_N^{\dagger}\Psi_1)
\end{array}
\end{equation}
where $P=e^{i\pi\sum_{i=1}^N \Psi_j^{\dagger}\Psi_j}$ is the total fermion parity in the chain and $H_b$ is the boundary term which depends on $P$. Therefore, the fermion theory contains two sectors, one with an even number of fermions and therefore anti-periodic boundary condition and one with an odd number of fermions and periodic boundary condition. Without terms mixing the two sectors, we can solve the free fermion Hamiltonian in each sector separately. After Fourier transform, the Hamiltonian becomes
\be
H=\sum_k \cos\left(\frac{2\pi k}{N}\right) \Psi_k^{\dagger}\Psi_k
\ee
where $k$ takes value $0$, $1$, ..., $N-1$ in the periodic sector and $\frac{1}{2}$, $\frac{3}{2}$, ... $\frac{2N-1}{2}$ in the anti-periodic sector. The ground state in each sector has all the modes with energy $\leq 0$ filled. Note that with this filling the parity constraint in each sector is automatically satisfied. The ground state energy in the periodic sector is higher than in the anti-periodic sector and the difference is inverse proportional to system size $N$.
Now let's consider the effect of various perturbations on the system.
The $(l,m)=(1,0)$ operator or the $(-1,0)$ operator in the boson theory (as
shown in Fig. \ref{fig:XY_Z2X}) corresponds to changing the boundary condition
of the Dirac fermion from periodic to anti-periodic.
Such operators would totally gap out the edge states. However, from Eqn.
(\ref{UMN}) and Eqn. (\ref{UMN1}), we see that both operators carry nontrivial
quantum number in all
$\mathbb{Z}_N$ (and $U(1)$) SPT phases, therefore it is forbidden by the
symmetry.
The $(l,m) = (2,0)$ operator in the boson theory corresponds
to the pair creation operator $\Psi^{\dagger}_L\Psi^{\dagger}_R$ in the fermion
theory. Its combination with the $(-2, 0)$ operator ($\Psi_R\Psi_L$
in the fermion theory) would gap out the system, but due
to the existence of the two sectors the ground state would be
two fold degenerate. To see this more explicitly, consider the $XY$ model again where
the combination of $(l,m)=(2,0)$ and $(-2,0)$ operators can be realized with an anisotropy term
\be
H^{XY}_{(2,0)} = \gamma \sum_i X_{i-1}X_i - Z_{i-1}Z_i
\ee
Under Jordan Wigner transformation, it is mapped to the p-wave pairing term
\begin{equation}
\begin{array}{l}
H_{(2,0)}=H_{a,(2,0)}+H_{b,(2,0)} \\
H_{a,(2,0)}=\gamma \sum^N_{i=1} (\Psi^{\dagger}_{i+1}\Psi^{\dagger}_i+\Psi_i\Psi_{i+1}) \\
H_{b,(2,0)}=-\gamma(P+1)(\Psi^{\dagger}_{1}\Psi_N+\Psi_N^{\dagger}\Psi_1)
\end{array}
\end{equation}
Again, we have period boundary condition for $P=-1$ and anti-periodic boundary condition for $P=1$. After Fourier transform, the Hamiltonian at each pair of $k$ and $N-k$ is
\begin{equation}
\begin{array}{lll}
H_{k,N-k} &=& \cos\left(\frac{2\pi k}{N}\right)(\Psi_k^{\dagger}\Psi_k+\Psi_{N-k}^{\dagger}\Psi_{N-k}) + \\
& &i\gamma\sin\left(\frac{2\pi k}{N}\right) (-\Psi_k^{\dagger}\Psi^{\dagger}_{N-k}+\Psi_{N-k}\Psi_{k})
\end{array}
\end{equation}
The Bogoliubov modes changes smoothly with $\gamma$ and the ground state parity
remains invariant. The ground state energy is $\frac{1}{2}\sum_k
\left(1-(1-\gamma^2)\sin^2\left(\frac{2\pi k}{N}\right)\right)^{1/2}$ and
explicit calculation shows that the energy difference of the two sectors (with
$k = $ int. and $k=$ int. $+\frac12$) becomes exponentially small with nonzero
$\gamma$. Therefore, upon adding the $(l,m)=(2,0)$ and $(-2,0)$ terms, the
ground state becomes two fold degenerate. Such an operator does carry trivial
quantum number in the nontrivial $\mathbb{Z}_2$ SPT phase and renders the
gapless edge unstable. However, a two fold degeneracy would always be left over
in the ground states, indicating a spontaneous $\mathbb{Z}_2$ symmetry
breaking at the edge.
The $(0,1)$ operator in the boson theory corresponds to a scattering term
between the left and right moving fermions $\Psi^{\dagger}_L\Psi_R$. Its
combination with the $(0,-1)$ operator ($\Psi^{\dagger}_R\Psi_L$ in the fermion
theory) would gap out the system. Unlike the $(2,0)$ operator, there is no
degeneracy left in the ground state. In the $XY$ model, this corresponds to a staggered coupling constant
\be
H^{XY}_{(0,1)} = \gamma \sum_i(-1)^i \left(X_{i-1}X_i + Z_{i-1}Z_i\right)
\ee
Mapped to fermions, the Hamiltonian at $k$ and $k+\frac{N}{2}$ becomes
\begin{equation}
\begin{array}{lll}
H_{k,k+\frac{N}{2}} &=& \cos\left(\frac{2\pi k}{N}\right)(\Psi_k^{\dagger}\Psi_k-\Psi_{k+\frac{N}{2}}^{\dagger}\Psi_{k+\frac{N}{2}}) + \\
& &i\gamma\sin\left(\frac{2\pi k}{N}\right) (-\Psi_k^{\dagger}\Psi_{k+\frac{N}{2}}+\Psi^{\dagger}_{k+\frac{N}{2}}\Psi_{k})
\end{array}
\end{equation}
For each pair of $k$ and ${k+\frac{N}{2}}$,
there is one positive energy mode and one negative energy mode and we want to
fill the negative energy mode with a fermion to obtain to ground state.
For the anti-periodic sector, such a construction works
since there is a $N/2$ = even number of negative energy modes, and
the anti-periodic sector contains an even number of fermions.
However, for the periodic sector, such a construction fails
since there is a $N/2$ = even number of negative energy modes, and
the periodic sector must contain an odd number of fermions.
So we have to add an fermion to a positive energy mode
(or have a hole in a negative energy mode), to have an odd number of
fermions.
Therefore, the ground state in the periodic sector has a finite energy gap
above the anti-periodic one and the ground state of the whole system is
nondegenerate. However, because this term carries nontrivial quantum number in
any nontrivial $\mathbb{Z}_N$ (and $U(1)$) SPT phases, it is forbidden by the
symmetry. For the trivial $\mathbb{Z}_2$ SPT phase, the $(0,\pm 1)$ operators
are $\mathbb{Z}_2$ symmetric operators, and can be added to the edge effective
Hamiltonian. The presence of the $(0,\pm 1)$ operators will gap the edge state
and remove the ground state degeneracy.
\end{document}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,026
|
\section{Introduction}
NGC1977 is an \hbox{H\,{\sc ii}}\ region located at the northern edge of the Orion molecular
cloud at a distance of about 0.5 kpc from the Sun. The region has
an emission measure of 5 $\times$ 10$^3$ pc cm$^{-6}$ and is believed to be
an example of an interface between the \hbox{H\,{\sc ii}}\ region and the adjacent molecular
cloud (Shaver \& Goss 1970). A 6-cm VLA image of the \hbox{H\,{\sc ii}}\ region made by
Kutner et al.~ (1985) shows a strong peak that coincides with the brightest
part of the optical bright rim. They investigated whether the
radio peak is due to a compact \hbox{H\,{\sc ii}}\ region or dense clump of gas ionized by
the central star in NGC1977. Both these possibilities were ruled out based
on the expected radio and IR emission from such ionized gas.
Subrahmanyan, Goss, \& Malin (2001) imaged NGC~1977 with the VLA at 330,
1420, 4860 and 8440 MHz. They detected the bright radio peak as an
unresolved object designated as J0535$-$0452. The estimated flux density of the compact object
is 320, 50, 11.4 and 5.2 mJy at 330, 1420, 4860 and 8440 MHz respectively.
The spectrum of the object is straight, with no turn over down to 330 MHz.
The spectral index is $-$1.3 over this range of frequencies.
The 8440 MHz observations are used to obtain an upper-limit
on the source size of $\sim$ 100 mas. The implied brightness temperature is
$7 \times 10^8$~K. Subrahmanyan et al.~\ (2001) have concluded that the object is a
non-thermal source.
Kutner et al.~\ (1985) argued that the non-thermal source could be an
extragalactic background source located behind NGC1977. Compact steep spectrum
extragalactic sources with spectral index close to $-$1.3 are typically
high redshift objects ($\stackrel >{_\sim}$ 2; \nocite{md08}Miley \& De Breuck 2008).
For these redshifts, the size of J0535$-$0452 is $\stackrel <{_\sim}$ 780 pc, assuming standard
cosmological parameters. Compact steep spectrum sources of this size typically show
spectral turn over near a few GHz, which is not the case for J0535$-$0452.
Thus if J0535$-$0452 is an extragalactic source, then it should be one
of the rare steep-spectrum, high redshift radio galaxies (for example,
see \nocite{ketal00}Kaplan et al.~\ 2000).
Another possibility is that the compact source
in NGC1977 is a pulsar. The spectral indices of both normal and
millisecond pulsar are in the range $-$1 to $-$2.5 with a mean
value of $-$1.65 (\nocite{ketal98}Karmer et al. 1998).
Thus the compactness and spectral index of J0535$-$0452 may suggest that
it is a pulsar. An earlier attempt to detect pulsed emission near 330 MHz was not successful
(Ramachandran, R. 1995; unpublished). However,
J0535$-$0452 would have eluded detection in earlier pulsar
searches due to extreme scattering. We show in Section~\ref{onobs}
that the best frequencies to search for pulsed emission from this pulsar
are $\ge$ 5 GHz depending on the pulsar period. We used the GBT
to search for pulsed emission at frequencies near 4.8 and 14.8 GHz,
but did not detect pulsed emission. The
observations and results are given in Section~\ref{gbtobs} and \ref{result},
respectively.
\section{On observing pulsed emission from J0535$-$0452}
\label{onobs}
\begin{figure*}
\centering
\includegraphics[width=\textwidth]{fig1.eps}
\caption{(a)Schematic showing the scattering geometry. The \hbox{H\,{\sc ii}}\ region--molecular cloud
interface (scattering screen) is located at $\sim$ 450 pc from the observer and the
distance to the compact source is $\ge$ 2.5 kpc.
(b) The expected pulse broadening for three frequencies (330 MHz, 5 and 14 GHz)
is shown in the pulsar period--angular broadening parameter space. The curves are
the locus of points for pulse broadening equal to the pulsar period.
For each frequency the parameter space above the curve can be probed by the pulsar search
observation. The pulse broadening estimation includes contribution from an assumed intrinsic
duty cycle of 10\% of the pulsar period and broadening due to DM.} For these calculation,
the smallest pulsar period is taken as 1 msec. The range of angular broadening
is constrained by observations between 10 and 100 mas. For all frequencies $\stackrel <{_\sim}$ 14 GHz,
the pulse broadening is dominated by scattering. At 14 GHz, for angular broadening $\stackrel <{_\sim}$ 60 mas,
the pulse broadening is limited by the assumed intrinsic duty cycle.
\label{fig1}
\end{figure*}
An attempt was made in 2003 to image J0535$-$0452 using the VLBA at 8.4 GHz
(Roshi et al.~\ 2003 unpublished). The source was not detected in
this observation. A possible interpretation of the non-detection in the VLBA
observations is that the source would have angular broadened due
to scattering in the \hbox{H\,{\sc ii}}\ region -- molecular cloud interface. Assuming
that the source is scatter broadened, a lower limit can be obtained for angular
broadening which turns out to be $\sim$ 10 mas.
If J0535$-$0452 is a pulsar, then the pulsed emission will be broadened due
to scattering in the \hbox{H\,{\sc ii}}\ region -- molecular cloud interface.
We estimate the pulse broadening
using the measured limits on angular broadening of J0535$-$0452 along with
the source-scattering screen geometry shown in Fig~\ref{fig1}a. The distance to the
compact object, $D_{pul}$, as measured from \hbox{H\,{\sc i}}~\ absorption studies is
$\ge$ 2.5 kpc (Subrahmanyan et al.~\ 2001) and the distance
to NGC1977, $D_{hii}$, is $\sim$ 450 pc (Genzel et al.~\ 1981).
The $e^{-1}$ pulse broadening time in sec due to scattering is obtained using the equation
(Cordes \& Lazio 1997)
\begin{equation}
t_s = \frac{D_{hii}}{D_{pul} - D_{hii}} \times
\frac{D_{pul}\; \theta_{s,8.4GHz}^2}{8 \; \mbox{ln}(2)\; c} \times \left(\frac{f_{GHz}}{8.44}\right)^{-4},
\end{equation}
where $c$ is the velocity of light in cm s$^{-1}$, the units of distances are in cm,
$f_{GHz}$ is the observing frequency in GHz and $\theta_{s,8.4GHz}$ is the angular
broadening at 8.4 GHz in rad. The pulse broadening time is considered
to scale with the 4$^{th}$ power of
frequency, appropriate for strong scattering. $\theta_{s,8.4GHz}$
range between 10 and 100 mas, which are obtained from the VLA (Subrahmanyan et al.~\ 2001)
and VLBA observations near 8.4 GHz (Roshi et al.~\ 2003 unpublished).
In Fig~\ref{fig1}b we plot the estimated pulse broadening time
for three observing frequencies (330 MHz, 5 and 14 GHz) in the pulse period--
angular broadening parameter space. The pulse broadening time in this plot
also includes an assumed intrinsic duty cycle of 10\% of the pulsar period
(minimum pulsar period is taken as 1 msec). The spectral resolution
used for the calculation is 0.78 MHz. For a given dispersion measure (DM),
this finite spectral resolution produces temporal broadening, which is also
added to the plotted pulse broadening. Any temporal broadening due to DM griding
in the pulsar search algorithm is neglected.
Continuum observations toward J0535$-$0452 can be used to estimate limits on the DM.
The measured spectrum of the compact source does not show any turnover at 330 MHz
due to the continuum optical depth of the \hbox{H\,{\sc ii}}-region--molecular cloud interface.
This fact is used to obtain an upper limit on the emission measure (EM) of $\sim 10^4$ cm$^{-6}$ pc
for an assumed ionized gas temperature of 8000 K and $\tau_c \sim$ 0.05 at 330 MHz.
The 3 sigma uncertainty of the 330 MHz map of NGC1977 region (Subrahmanyan et al.~\ 2001)
is used to infer an upper limit on $\tau_c$ of $\sim$ 0.05. A DM
corresponding to a given EM can be obtained as $DM = \sqrt{EM\, \phi\, L}$,
where $\phi$ is the filling factor of the ionized gas in the intervening \hbox{H\,{\sc ii}}\ region
and $L$ is the line of sight extent of the \hbox{H\,{\sc ii}}\ region. Taking $\phi \sim 1$ and
$L \sim 10$ pc, a good fraction of the size of the Orion molecular cloud, the DM
we get is $\sim$ 300 pc cm$^{-3}$. Contribution to DM due to the distributed ionized gas
has to be added to the above estimate. NE2001 model (Cordes \& Lazio 2002) provides
a total DM of $\sim$ 50 pc cm$^{-3}$ in the direction of NGC1977. Thus the upper
limit obtained on DM is $\sim$ 350 pc cm$^{-3}$. For estimating pulse broadening we used
DM values up to 650 pc cm$^{-3}$.
The result of the pulse broadening calculation shows that for all frequencies $\stackrel <{_\sim}$ 14 GHz
the pulse broadening is dominated by scattering (see Fig~\ref{fig1}b). At 14 GHz, for $\theta_{s,8.4GHz} < 60$ mas,
pulse broadening plotted in Fig~\ref{fig1}b is limited by the assumed intrinsic
duty cycle. For each observing frequency indicated
in Fig~\ref{fig1}b pulsed emission can be detected in the parameter space
above the corresponding curve.
The estimated continuum flux density at 14.0 GHz is 2.5 mJy.
If the observed continuum emission is the mean flux density of the pulsar, then the
pulsar can be detected at frequencies 5 and 14 GHz (depending on the pulsar period)
with signal-to-noise ratio $>$ 100 in 1 hr of observing time with the GBT.
Thus, as seen in Fig.~\ref{fig1}b,
such an observation will be sensitive to both MSP and ordinary pulsars.
Note that earlier searches would not have detected pulsars for the measured
angular broadening limits (see Fig~\ref{fig1}b).
\section{The GBT Observations and Data Analysis}
\label{gbtobs}
The GBT observations on J0535$-$0452 were made at 4.8 and 14.8 GHz
on 15 May 2011. The GUPPI backend with 800 MHz bandwidth, 1.6 MHz
spectral resolution and
41 $\mu$sec time resolution was used for the observations.
Flux density calibration was done using the source 3C161 for both
frequencies. The flux densities of 3C161 were taken to be 6.7 and
2 Jy at 4.8 and 14.8 GHz (Ott et al.~\ 1994). The measured telescope
gains were 2 and 1.9 K/Jy and system temperatures were 18.5 and 26 K
at 4.8 and 14.8 GHz respectively.
Reference pointing was done on B0540-0415 at 14.8 GHz
and on 3C161 at 4.8 GHz. The total on-source observing time at 14.8 GHz
is 27 minutes and that at 4.8 GHz is 29 minutes.
The data processing was performed using PRESTO\footnote{
\url{http://www.nrao.edu/~sransom/presto}} (Ransom, Eikenberry \& Middleditch 2002).
A DM range of 0 to 720 pc cm$^{-3}$ was searched with full time resolution.
The value of 720 pc cm$^{-3}$ is about two times larger than the upper limit on
DM estimated in Section~\ref{onobs}. In addition we searched for pulsar
acceleration in the range 0 to 2.1 $\times 10^4 \times$ $P$
m s$^{-2}$, where $P$ is the period of the pulsar in sec.
\section{Results and Conclusion}
\label{result}
We did not detect pulsed emission from J0535$-$0452. The upper limit
obtained for pulsed emission at 4.8 GHz is 55 $\mu$Jy for a pulsar period
of 10 msec and pulse broadening of 7 msec corresponding to an angular
broadening of 60 mas (see Fig~\ref{fig1}b). At 14.8 GHz, the upper limit
obtained is 30 $\mu$Jy for a pulsar period of 3 msec, typical for MSP,
assuming 10\% intrinsic duty cycle.
Acceleration searches are most sensitive to binary pulsars with orbital period greater
than 10 times the observation time (\nocite{rce03}Ransom, Cordes \& Eikenberry 2003).
This means that our data analysis will rule out compact sources in binary system
with orbital period $\stackrel >{_\sim}$ 5 hrs (assuming circular orbit).
We have searched for accelerations up to 2.1 $\times 10^4 \times$ $P$ m s$^{-2}$.
For orbital period of $\sim$ 5 hrs, this upper limit rules out companion object
of mass $\stackrel <{_\sim}$ 3 ~M$_{\odot}$~ for an assumed pulsar period of 3 msec. If the companion
objects is of higher mass (ie $>$ 3 ~M$_{\odot}$~) then for orbital periods $\stackrel >{_\sim}$
5 hrs the object cannot be an ordinary star. This is because the orbital radius
becomes comparable to the size of the star. The companion object could be
a stellar mass black hole. However, we are not aware of any other observational
evidence for the presence of a stellar mass black hole in the direction of NGC1977.
Thus the only possibility that remains is that J0535$-$0452 is a binary MSP with
short orbital period (ie $\stackrel <{_\sim}$ 5 hrs).
\begin{acknowledgements}
We thank the anonymous referee for the critical comments on the manuscript.
\end{acknowledgements}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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