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Simplifying plasma balls and black holes with nonlinear diffusion Behan, Connor Classen 2014

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Simplifying plasma balls and black holes with nonlineardiffusionbyConnor Classen BehanB. Science, Queen’s University, 2011A THESIS SUBMITTED IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFMaster of ScienceinTHE FACULTY OF GRADUATE AND POSTDOCTORALSTUDIES(Physics)The University of British Columbia(Vancouver)July 2014c© Connor Classen Behan, 2014AbstractThe AdS / CFT dictionary, while still incomplete, hints at deep connections be-tween thermal field theories and the dynamics of black holes. Without specifying aLagrangian, we develop a non-standard approximation for field theories dominatedby thermal noise in order to show that many black hole features are universal. Ourmodel is a nonlinear partial differential equation which may be derived, as it waslast year, by examining random equilibration of energy on a collection of sites. Anextension pairing energy with other conserved quantities is also proposed. For typ-ical holographic gauge theories, the linear versions of our models show that Hage-dorn densities of states are associated with long lived lumps of deconfined plasma.With the help of numerical and mathematical results, we show that the nonlineardiffusion properties are more subtle and discuss the implications for using thesemodels to study unsolved problems in holography.iiPrefaceThis thesis is based on a model that appeared in [1], a 2013 paper by CB, KlausLarjo, Nima Lashkari, Brian Swingle and Mark Van Raamsdonk. In describing thebasic model, Chapter 3 paraphrases the relevant sections of this paper. Chapters 4,5 and 7 have some overlap with the paper and are intended to follow up on its looseends.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiAcknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Aspects of holography . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.1 In free field theory . . . . . . . . . . . . . . . . . . . . . 62.1.2 In string theory . . . . . . . . . . . . . . . . . . . . . . . 92.2 Black holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2.1 Useful metrics . . . . . . . . . . . . . . . . . . . . . . . 142.2.2 Hawking radiation . . . . . . . . . . . . . . . . . . . . . 182.3 Strong coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.3.1 Gauge theory phases . . . . . . . . . . . . . . . . . . . . 272.3.2 Plasma balls . . . . . . . . . . . . . . . . . . . . . . . . 343 Treating energy stochastically . . . . . . . . . . . . . . . . . . . . . . 423.1 Main equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.1.1 Physical assumptions . . . . . . . . . . . . . . . . . . . . 43iv3.1.2 The continuum limit . . . . . . . . . . . . . . . . . . . . 463.2 Interesting features . . . . . . . . . . . . . . . . . . . . . . . . . 493.2.1 Static situations . . . . . . . . . . . . . . . . . . . . . . . 493.2.2 Mean-field variances . . . . . . . . . . . . . . . . . . . . 514 Nonlinear diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.1 Basic properties on a bounded domain . . . . . . . . . . . . . . . 554.1.1 Conservation of energy . . . . . . . . . . . . . . . . . . . 564.1.2 The maximum principle . . . . . . . . . . . . . . . . . . 564.1.3 Existence of a steady-state . . . . . . . . . . . . . . . . . 594.2 Time scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.2.1 The concentration comparison theorem . . . . . . . . . . 624.2.2 Estimates in one dimension . . . . . . . . . . . . . . . . 644.2.3 Higher dimensional generalization . . . . . . . . . . . . . 684.2.4 Upper and lower bounds . . . . . . . . . . . . . . . . . . 734.2.5 Remaining problems . . . . . . . . . . . . . . . . . . . . 764.3 Comments on unbounded domains . . . . . . . . . . . . . . . . . 784.3.1 Barriers to uniqueness . . . . . . . . . . . . . . . . . . . 784.3.2 Barenblatt profiles . . . . . . . . . . . . . . . . . . . . . 835 Numerical analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865.1 Implementation details . . . . . . . . . . . . . . . . . . . . . . . 875.1.1 The Crank-Nicolson method . . . . . . . . . . . . . . . . 895.1.2 Convergence tests . . . . . . . . . . . . . . . . . . . . . . 935.2 Simulation results . . . . . . . . . . . . . . . . . . . . . . . . . . 945.2.1 Short time dynamics . . . . . . . . . . . . . . . . . . . . 945.2.2 Long time dynamics . . . . . . . . . . . . . . . . . . . . 975.2.3 Higher dimensions . . . . . . . . . . . . . . . . . . . . . 996 Room for improvement . . . . . . . . . . . . . . . . . . . . . . . . . 1016.1 A common approximation . . . . . . . . . . . . . . . . . . . . . 1016.1.1 Basic hydrodynamics . . . . . . . . . . . . . . . . . . . . 1026.1.2 The fluid / gravity correspondence . . . . . . . . . . . . . 1086.2 Restricting the density of states . . . . . . . . . . . . . . . . . . . 110v6.2.1 Some conformal field theory . . . . . . . . . . . . . . . . 1106.2.2 Convoluted functions . . . . . . . . . . . . . . . . . . . . 1167 Entropic dynamics of momentum . . . . . . . . . . . . . . . . . . . 1197.1 Proceeding by analogy . . . . . . . . . . . . . . . . . . . . . . . 1207.1.1 Allowed transitions and rates . . . . . . . . . . . . . . . . 1217.1.2 A tensor identity . . . . . . . . . . . . . . . . . . . . . . 1227.1.3 Setting up the equations . . . . . . . . . . . . . . . . . . 1257.2 The continuum limit . . . . . . . . . . . . . . . . . . . . . . . . 1277.2.1 To first non-vanishing order . . . . . . . . . . . . . . . . 1277.2.2 To first consistent order . . . . . . . . . . . . . . . . . . . 1307.2.3 Static configurations . . . . . . . . . . . . . . . . . . . . 1357.3 Consistency check . . . . . . . . . . . . . . . . . . . . . . . . . 1367.3.1 Linearized equations . . . . . . . . . . . . . . . . . . . . 1377.3.2 Comparison with hydrodynamics . . . . . . . . . . . . . 1397.3.3 One more simulation . . . . . . . . . . . . . . . . . . . . 1418 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147A Cumbersome derivatives . . . . . . . . . . . . . . . . . . . . . . . . 156B Discontinuous PDE solutions . . . . . . . . . . . . . . . . . . . . . . 158C Crank-Nicolson code . . . . . . . . . . . . . . . . . . . . . . . . . . . 161viList of FiguresFigure 1.1 By focusing on two adjacent sites in this energy distribution,one may check whether the system favours a homogeneous oran inhomogeneous state. The properties of this entropic evolu-tion are determined by the local density of states. . . . . . . . 1Figure 2.1 Photon A leaves a light source and starts heading toward r = 0.After a certain time interval, photon B does the same. If thespacetime is Minkowski, the distance between the two photonswill not change. Conversely, if a black hole at r = 0 forms atjust the right time, photon A will escape but photon B will staytrapped at the event horizon forever. . . . . . . . . . . . . . . 21Figure 2.2 For the function with +iε , it is not difficult to show that thesemi-circular arc has no contribution to the integral. Thereforethe integral between −∞ and ∞ is zero. . . . . . . . . . . . . 24Figure 2.3 For an inverse temperature like the one shown, the gravity sidemust choose the geometry that minimizes the free energy. . . 34viiFigure 2.4 The green sheet is the boundary at z = 0. While Poincare´ AdSlooks like empty space ending on this boundary, the black holeand soliton solutions are more interesting. Instead of enclosinga region of finite volume, the horizon is a sheet at z = z0 likethat of a black brane. To remind us that there is a Scherk-Schwarz circle at each point, we have drawn the horizon as aset of black circles inside cylinders that extend infinitely far tothe left. These cylinders become cigar shaped regions in thesoliton solution that has the infrared wall. On the right, weshow an interpolation between these behaviours. For y = ∞,the IR wall does not exist. For finite positive y it is behind thehorizon and for negative y it is in front. . . . . . . . . . . . . 38Figure 2.5 On the field theory side, the domain wall shows up as a sud-den jump in the energy density. The minimum is close to zerowhile the maximum is the expected energy at the deconfine-ment temperature. The existence of a plasma ball is the ad-ditional assumption that we can make this domain wall “wraparound”. The decay of the resulting object might look like dif-fusion governed by the heat equation. . . . . . . . . . . . . . 39Figure 2.6 This localized lump of energy is certainly something like thet = 0 slice of a plasma ball. We are claiming that if it forms ona sphere, its evolution is too dynamic to be considered that ofa plasma ball. Its motion cannot be undone by a boost becausethis turns it back into the uniform profile of (2.59). . . . . . . 40Figure 4.1 To emphasize the four phases expected in holographic gaugetheories, we have drawn a piecewise β (E) function. It is morerealistic to expect an approximation to this function that is dif-ferentiable whenever E > 0. The function plotted in the mid-dle figure is still difficult to work with because there is a smallrange of energies for which it is increasing. Assuming that thisphase is negligible is the best way to predict the behaviour ofour nonlinear PDE. . . . . . . . . . . . . . . . . . . . . . . . 54viiiFigure 4.2 Resricting the size of the domain allows us to avoid the van-ishingly small energies for which our PDE no longer applies. . 55Figure 4.3 The two energy distributions shown have equal minima, equalmaxima and equal masses. They also have the central peak astheir only extreme value. A solution to (4.3) that alternatedbetween these functions indefinitely would satisfy all of theproperties that we have proven so far but not have a steady-state. 60Figure 4.4 This shows how we must change the function that appears in-side the Laplacian in order to satisfy the hypotheses of the the-orem. Bounding the time scales arising from β is no moredifficult than bounding the time scales arising from Φ. . . . . 62Figure 4.5 We were able to show that our filtration equation had a welldefined steady-state by comparing it to a heat equation. Withthis linear function as our bounding function, an estimate forthe decay time can also be obtained. However, this would leadto a crude bound because the linear function above is not verysimilar to the true Φ. . . . . . . . . . . . . . . . . . . . . . . 63Figure 4.6 Except for the piece on top that will quickly diffuse, these en-ergy distributions are qualitatively similar to the step functionwe are considering. . . . . . . . . . . . . . . . . . . . . . . . 65Figure 4.7 The union of the red and purple curves is what we are lookingfor. It is (4.6), the solution to the (4.5) filtration equation forthe (4.4) Cauchy data. The solution to the auxillary problemthat we use to find it is the union of the blue and purple curves.This is F , the solution to the heat equation for the HeavisideCauchy data. . . . . . . . . . . . . . . . . . . . . . . . . . . 67Figure 4.8 The curve where the blue sheet intersects the evolving surfaceis x∗(t). The area of the blue sheet between the surface and theco-ordinate planes is a2−V ∗(t). We are instead calculating thearea V ∗(t) which has an equal and opposite rate of change. . . 70ixFigure 4.9 The piecewise-linear bounding functions must relate to our fil-tration function in this way. On the left, Φ1 could cross over toHagedorn behaviour at some energy E1 ≥ EH and on the right,Φ2 could cross over to Hagedorn behaviour at some energyE2 ≤ EH. The tightest possible time scale bounds are achievedwhen E1 = EH = E2. . . . . . . . . . . . . . . . . . . . . . . 74Figure 4.10 The upper bound for the decay time, given by Φ˜2 depends onwhich of the above cases is realized. The figure on the leftwould lead to a situation much like the one for Φ2 that wesaw before. However, on the right, the Hagedorn energy islarge enough to make Φ˜ change concavity before reaching theHagedorn phase. . . . . . . . . . . . . . . . . . . . . . . . . 75Figure 5.1 One the left is the derivative of (5.1) showing the three phases.The simplest choice for removing the clustering phase leads usto differentiate (5.2), the plot on the right. . . . . . . . . . . . 87Figure 5.2 For this function, the flat section begins at E = 1.0. The region0.5<E < 1.0 is used to interpolate between the flat section andthe E−110 power law. The flat section ends at E = 0.9EF and theE−14 power law begins at E = EF. In the region 0.9EF < E <EF, a parabola again interpolates. The plot is for EF = 10.0 butthis general rule has been followed for all EF. . . . . . . . . . 90Figure 5.3 A standard demonstration that the forward Euler method is nu-merically unstable for the heat equation. . . . . . . . . . . . . 91Figure 5.4 Typically we will work with an initial condition like that shownin red. The width of the domain is chosen so that all of thefunction’s mass can fit below the Hagedorn energy in blue. Theplot on the left shows that E0(x) is very flat for most of the xvalues. An adaptive dx makes sense and will be chosen so thatthe minimum dx is much smaller than the distance between theinflection points. The plot on the right shows the graph of thesame function but more conveniently. . . . . . . . . . . . . . 92xFigure 5.5 The factor Q(t) plotted for simulations that lasted t = 90.0.The grid was uniform and five step sizes were used. . . . . . . 94Figure 5.6 These initial conditions with the same total mass both evolvetoward a state that has a flat line in the high energy phase nearEF = 10.0. . . . . . . . . . . . . . . . . . . . . . . . . . . . 95Figure 5.7 Two plots of the thermalization time for field theory energyEF = 10.0 and different values of k in the initial condition. . . 95Figure 5.8 A plot of how the peak energy moves down with time. Thiscorresponds to the k = 9 decay in Figure 5.7. . . . . . . . . . 96Figure 5.9 This shows one of our initial conditions diffusing all the waythrough the Hagedorn regime. It takes about a hundred timeslonger to do this than it does to reach the thermalization time. 96Figure 5.10 A plot of the decay time for EF = 10.0, EH = 1.0 and fivevalues of k. Even though they specify different shapes for theinitial condition, the times are all within 4% of eachother. . . . 97Figure 5.11 A plot of the decay time for k = 9, EH = 1.0 and five valuesof EF. Even though the profiles begin to flatten out at differentheights, their final decay times are all within 0.4% of eachother. 98Figure 5.12 On the left is a plot of the decay time in two dimensions fork = 9, EH = 1.0 and five values of EF. These are much smallerthan the one-dimensional decay times in Figure 5.11. Thezoomed out version on the right shows that even though thesedots do not vary linearly, they are sandwiched between twobounds which do vary linearly. For the sake of the plot, theblue lines have been understated. In reality the lower bound ismuch closer to being horizontal and the upper bound is muchcloser to being vertical. . . . . . . . . . . . . . . . . . . . . . 99Figure 6.1 These two theories live on a circle of volume V but their tem-peratures are different. They are related through multiplicationby Vβ which is a conformal transformation. . . . . . . . . . . . 111xiFigure 6.2 The theory we have discussed so far resembles a harmonic po-tential where arbitrarily many non-interacting bosons can bepiled into each level. . . . . . . . . . . . . . . . . . . . . . . 115Figure 7.1 If one uses a density of states for two spatial dimensions, bothof these situations are compatible with the lattice being a line.One has an overall dimensionality of 2 + 1, the other 3 + 1.The task of introducing momentum to our model forces us toaddress this ambiguity. . . . . . . . . . . . . . . . . . . . . . 120Figure 7.2 Energy and momentum profiles for the simulation that only hasan x-axis. They are shown up until the time when momentumvalues of ±0.9 form. . . . . . . . . . . . . . . . . . . . . . . 141Figure 7.3 On the left are slices of the E distributions along the x-axis.These are spherically symmetric. On the right are Px distri-butions along the x-axis. These smoothly approach a y-axisvalue of zero as one rotates the direction along which they areplotted. The Py distributions behave in an equal and oppositeway. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142xiiAcknowledgmentsFirstly, I would like to thank my supervisor, Mark Van Raamsdonk, for involvingme in his research and lending his strong intuition whenever it was needed. Apartfrom proofreading, he works hard to ensure that students have projects matchingtheir interests. Gordon Semenoff agreed to proofread this thesis as well. I thankhim, not only for that, but for delivering the lectures that first taught me stringtheory. I wish to thank my collaborators Klaus Larjo, Nima Lashkari and BrianSwingle who discussed many of the problems that came up in our work, and pos-sessed the skill and motivation to solve them. I am grateful to many fellow students,especially Michael McDermott, Fernando Nogueira and Jared Stang, for sharingtheir progress and taking an interest in mine. In the first year of this work, I wassupported by the Nation Science and Engineering Research Council of Canada.Lastly, it is a pleasure to thank my parents for all of their love and support. Thisthesis is dedicated to my cousin Greg, whose wedding I missed while pursuing thisdegree.xiiiChapter 1IntroductionThis thesis deals with a recently proposed toy model for the dynamics of energydistributions in thermal field theories. These include the conformal field theoriesand deformations of them that have gravity duals according to the AdS / CFT corre-spondence [2]. As argued in [1], our model suggests that certain important aspectsof gravitational physics emerge for thermodynamic reasons. From this perspective,it is related to the ideas of entropic gravity in [3–6].Figure 1.1: By focusing on two adjacent sites in this energy distribution, onemay check whether the system favours a homogeneous or an inhomo-geneous state. The properties of this entropic evolution are determinedby the local density of states.Before deriving the equations of our model, it is helpful to consider Figure 1.11— energy quanta that randomly hop between sites in a line. To each site, we ascribea density of states ρ(n) counting the number of ways for it to have n units of energy.The growth of this function lets us determine which scenario is more likely: site 1giving a quantum to site 2 or site 2 giving a quantum to site 1. Asking this questionis equivalent to comparing the sizes of ρ(n1−1)ρ(n2+1) and ρ(n1+1)ρ(n2−1).Positing a form logρ(n) ∝ nα , we see that site 1 is most likely to give up energywhen α < 1 and site 2 is most likely to give up energy when α > 1. Thus, we seethat this random walk leads to diffusion when the density of states is log-concaveand clustering when the density of states is log-convex. In the diffusion case e.g., auniform energy distribution is the inevitable final state, even when the microscopicphysics are completely reversible. Special attention is paid to the Hagedorn phaseα = 1 which is almost completely static.Even though the essence of our model is this simple statement, it takes the formof a nonlinear partial differential equation that accepts a ρ(E) function as input. Aubiquitous density of states, which we derive using the AdS / CFT correspondence,consists of four phases. One of the narrow phases is omitted throughout this thesisfor simplicity. The three that are left consist of a diffusive phase at high energies,a Hagedorn phase at intermediate energies and another diffusive phase at low en-ergies. Roughly speaking, these respectively correspond to a black hole forming,living for a long time and ultimately evaporating away. Less ambitiously, we maysay that they correspond to balls of plasma in a purely field theoretic setting [7].We derive rigorous bounds on the decay times for these objects in our model andcompare them to the hadronization times in [7]. We find that our times are longerin one dimension and much shorter in higher dimensions.To address these problems, a second model is proposed that treats momentumas another quantity that moves stochastically through a lattice. Since the secondmodel is much more complicated, the discussion of its properties remains at a spec-ulative level. Even though evolution equations for energy and momentum soundsimilar to the spirit of hydrodynamics, we compare our equations to the hydroequations and only find agreement in the crudest approximation. Despite takingthe form of classical PDEs, we hasten to emphasize that our models include quan-tum effects when functions like the density of states are chosen appropriately.This thesis begins with theoretical background in Chapter 2. This chapter fo-2cuses on the tools needed to derive thermodynamic quantities via the AdS / CFTcorrespondence and contains some lengthy derivations. The main model is derivedafterward in Chapter 3. In Chapter 4, various results from the mathematical liter-ature on nonlinear diffusion equations are applied to our PDE and used to derivethe time scales for black hole evaporation. The suspicious features of our resultsare discussed in this chapter as well. Chapter 5 introduces numerical methods thatare suitable for our PDE and uses them to check most of our results. The methodchosen for most problems is the implicit Crank-Nicolson approach. Chapters 6 and7 contain the newer results that were derived after [1] appeared. Their focus is theextension of our model that includes momentum. Just as our first model dependson a density of states, our second model depends on a momentum restricted densityof states. An expression for this quantity is derived that allows a small amount ofnumerical work to be done. Code forming the basis for all of our simulations ispresented in the appendix.3Chapter 2Aspects of holographyOf all the conjectures that have been made about quantum gravity, the one thathas had the largest impact so far is the AdS / CFT correspondence proposed byJuan Maldacena [2]. Known by various other names like holography or gauge-gravity duality, it states that string theory in anti-de Sitter space is equivalent to aconformally invariant quantum field theory living on the boundary of that space.Questions about string theory can therefore be recast in the language of quantumfield theory without gravity. Deriving the evidence for the AdS / CFT correspon-dence would exceed the scope of this thesis [8]. Instead, we will explore certaindynamical processes that can be best understood with the correspondence. The ef-fect that will demand most of our attention is black hole evaporation. Hawking’sderivation of black hole evaporation is one of the most successful uses of quantumfield theory in curved spacetime and any eventual theory of quantum gravity is ex-pected to account for it. Many studies of Hawking radiation have been done usingstring theory and the AdS / CFT correspondence in particular [9–12].Naturally, the first such studies focused on the original version of the cor-respondence in which the background is AdS5 × S5 [2]. If one writes the six-dimensional Euclidean Dirac matrices asΓi =[0 C¯iCi 0]i ∈ {1, . . . ,6} ,4the conformal field theory is specified by the Lagrangian [13]L = −12g2YM∫R312Tr(FµνFµν +2DµφiDµφ i− [φi,φ j][φ i,φ j])+Tr(λ¯σµDµλ −Ciλ [φ i,λ ]−C¯iλ¯ [φ i, λ¯ ])dx . (2.1)Typically the gauge group is U(N) or SU(N) meaning that the scalars, spinors andvectors that show up are really N×N matrices consisting of those types of fields.This is called theN = 4 Super Yang-Mills theory or sometimes the field theory ofD3-branes. A less than encouraging fact about string theory is that AdS5×S5 is farfrom the only background we need to consider. There is really a whole landscape ofvacua whose boundary field theories may look very different. Indeeed CFT dualshave been proposed for AdS4×CP3 [14], AdS3×S3×T4 [15], AdS3×S3×S3×S1[16] and many others.Calculations involving these theories are difficult. Even showing that (2.1) hasconformal symmetry is not trivial. Something that allows us to explore Hawking’sprocess from the holographic viewpoint without choosing a specific Lagrangian isthe intimate connection between black holes and thermodynamics.2.1 ThermodynamicsA number of different field theories have the same thermodynamic potentials. Auseful example of this appears in a conformal field theory. Neglecting the Casimireffect, energy and entropy are both extensive so they must be proportional to thevolume. A conformal theory has no intrinsic scale so the only dimensionful quan-tity that can multiply this volume is the temperature. This leads to the expressionsE ∝V T d+1 and S ∝V T d . Substituting them into eachother yieldsS ∝V1d+1 Edd+1 . (2.2)The density of states will turn out to play a fundamental role in our model so wewill sometimes exponentiate this expression.In the calculations that follow we will see some situations in which this formulafor the entropy does not hold. In general, the rule is that (2.2) becomes true for5non-conformal theories if the energy is much larger than any other scale. Differentlow energy behaviours can be introduced if one compactifies a CFT like (2.1) on asphere.2.1.1 In free field theoryAn exercise done in [17] is finding the partition function of a free field theory. Start-ing with the fact that E = p in a massless theory, Z∗(p) = 1+e−β p is the contribu-tion of a single fermionic mode and Z(p) = 1+e−β p+e−2β p+ · · ·=(1− e−β p)−1is the contribution of a single bosonic mode. Using s∗ and s for the number ofinternal states, the partition function is given byZ =∏pZ(p)sZ∗(p)s∗.If we take the log, the product turns into a sum and if we take the momentum spec-trum to be continuous, the sum turns into an integral. Remembering the integrationmeasure for momentum space, we havelogZ ≈∫Rds∗ log(1+ e−β p)− s log(1− e−β p) V dp(2pi)d=dωdV(2pi)d∫ ∞0s∗ log(1+ e−β p)− s log(1− e−β p)pd−1dp=d!ωdV(2piβ )d [sζ (d +1)+ s∗ζ ∗(d +1)]≡AVβ d . (2.3)Here, ζ (σ) = ∑∞n=1 1nσ is the Riemann zeta function, ζ∗(σ) = ∑∞n=1(−1)n−1nσ is thealternating zeta function and ωd is the volume of a unit ball in Rd . We may nowuse S = ddT (T logZ) and E = T2 ddT (logZ) to show that (2.2) holds with a propor-tionality constant of[(d+1)d+1dd A] 1d+1.If one is interested in the density of states, the exponential of this entropy iscertainly the first term in ρ(E). However, there are an infinite number of otherterms that come from the differences between the canonical and microcanonicalensembles. The second term is a standard result that comes from treating Z(β ) as6the Laplace transform of ρ(E). Performing a saddle point approximation,ρ(E) = 12pi∫ ∞−∞Z(iβ )eiβEdβ≡12pi∫ ∞−∞e f (iβ )dβ∼12pi∫ ∞−∞e f (iβ0)−β22 f′′(iβ0)dβ=1√2pi f ′′(iβ0)e f (iβ0) .The higher asymptotic terms cannot be found in the same way because the integralof e f (iβ0)−β22 f′′(iβ0)−i β36 f′′′(iβ0) has no closed form solution. Instead, powers of βafter the first two must be Taylor expanded again so that the above becomesρ(E) ∼ 12pi∫ ∞−∞e f (iβ0)−β22 f′′(iβ0)(1− iβ 36f ′′′(iβ0)−β 672f ′′′(iβ0)2 + . . .)(1+β 424f ′′′′(iβ0)+β 81152f ′′′′(iβ0)2 + . . .). . .dβ .These calculations require us to consider an ever-growing number of ways in whicha power of β can be made. Nevertheless, this method is still practical for findingthe third term in ρ(E) and the resulting expression isρ(E) ∼ 1√2pi(d(d +1)d+1AV E−d−2) 12(d+1)exp((d +1)d+1ddAV Ed) 1d+1−(d +2)(2d +1)24(d +1)(dAV Ed)− 1d+1 .(2.4)In [17], (2.4) is found in a different way. The inverse Laplace transform ofZ(β ) is found exactly via a Hankel contour but as a Taylor series, not an asymptoticseries. From this seriesρ(E)∼∞∑j=1(AV ) jEd j−1j!(d j−1)!, (2.5)the first three asymptotic terms are picked off. An advantage of this is that (2.5)7can be compared to a recent expression for the d = 1 density of states due to Loran,Sheikh-Jabbari and Vincon [18]:ρ(E)∼ picV3I1(√2picV E/3)√2picV E/3. (2.6)Neither is a generalization of the other because d is arbitrary in (2.5) and the inter-actions are arbitrary in (2.6).In the partition function we have constructed, the β variable is conjugate tothe energy. There are also conjugate variables associated with each momentumdirection. Something special that we can do in 1+1 dimensions is combine theseinto a complex number. Let p be a positive momentum. If there are N excitationsof p and N˜ excitations of−p, this state has an energy of p(N+N˜) and a momentumof p(N− N˜). Therefore, generalized partition functions we can write down are:Z∗(p) =1∑N=01∑N˜=0e−pVτ2(N+N˜)+pViτ1(N−N˜) =(1+ e−pV (τ2−iτ1))(1+ e−pV (τ2+iτ1))Z(p) =∞∑N=0∞∑˜N=0e−pVτ2(N+N˜)+pViτ1(N−N˜) =(1− e−pV (τ2−iτ1))−1(1− e−pV (τ2+iτ1))−1.Taking the product of Z(p)sZ∗(p)s∗over all positive momenta, we havelogZ ≈∫ ∞0s∗[log(1+ e−pV (τ2−iτ1))+ log(1+ e−pV (τ2+iτ1))]−s[log(1− e−pV (τ2−iτ1))+ log(1− e−pV (τ2+iτ1))]V dp2pi=12pi(1τ2− iτ1+1τ2 + iτ1)[sζ (2)+ s∗ζ ∗(2)]= −ℑ(1τ)pi [sζ (2)+ s∗ζ ∗(2)]= −piℑ(1τ)6(s+s∗2). (2.7)The dimensionless number τ = τ1+ iτ2 is called the modular parameter. If τ ≡ iβV ,(2.7) becomes the regular partition function (2.3). The quantity c≡ s+ s∗2 appearingin (2.7) is central charge that we would use in (2.6) if we wanted to apply it to a8free theory.2.1.2 In string theoryThe worldsheet theory of a string can be regarded as a conformal field theory in1+ 1 dimensions. However, S ∝√E would not be correct for a macroscopic ob-server who has different notions of energy and dimensionality. The worldsheetLagrangian for a supersymmetric string theory in flat space isL = −14piα ′∫ 2pi0∂aXµ∂ aXµ − iΨ¯µγa∂aΨµdσ1= −12piα ′∫ 2pi02∂Xµ ∂˜Xµ − iψµ ∂˜ψµ − iψ˜µ∂ψ˜µdσ1 . (2.8)In the second form we have split each Dirac spinor field into two Majorana spinorfields. We have also written derivatives with respect to σ± = σ0±σ1 as ∂ and ∂˜ .What makes this different from a usual quantum field theory is that the D scalarfields Xµ can be interpreted as positions in a D-dimensional target space. Theworldsheet energy comes from σa 7→ σa + δσa but the energy we should use forcounting states is the conserved quantity associated with Xµ 7→ Xµ + δXµ . Theworldsheet has SO(1,1) Lorentz symmetry regardless of how many fields thereare, but the SO(D− 1,1) Lorentz symmetry of the target space is more sensitive.To survive quantization it requires that D = 10 [13]. If we had left the fermions outof (2.8) to construct bosonic string theory, the same calculation would tell us thatD = 26.To calculate the free energy of a gas of strings, we will begin in the same wayas before.F =1β∫RD−1∑mlog(1− e−β√p2+m2)−∑m∗log(1+ e−β√p2+m∗2) V dp(2pi)D−1= −1β∞∑n=1∫RD−1∑m1ne−βn√p2+m2−∑m∗(−1)nne−βn√p2+m∗ V dp(2pi)D−1This expression has a sum over the masses of bosons and a sum over the massesof fermions. To arrive at (2.3), we set these masses to zero and replaced the sumsby degeneracy factors. This was valid because the masses became negligible in the9high temperature limit. The high temperature limit of a string theory is differentbecause it supports arbitrarily large masses. To continue, we will use the trick∫ ∞0e−a2s2 −b22sds√s=√2piae−abto rewrite the free energy density.FV= −∞∑n=1∫ ∞0∫RD−1∑me−β2n2s2 −p2+m22s − (−1)n∑m∗e−β2n2s2 −p2+m∗22sdp(2pi)D−1ds√2pis= −∞∑n=1∫ ∞0e− β2n24piα ′τ2[∑me−piτ2α′m2− (−1)n∑m∗e−piτ2α′m∗2](4pi2α ′τ2)−D2dτ2τ2(2.9)Above, we have made the substitution τ2 = 12piα ′s . To proceed further, we need toknow the mass spectrum of our theory.For concreteness we will work in Type II which is a theory of closed strings.This is natural because evidence of the AdS / CFT correspondence was first dis-covered with Type IIB string theory [2]. Very little would change if we used TypeI or heterotic strings. The mode expansions for the scalar fields are identical to theones that describe the closed bosonic string:Xµ(σ0,σ1) = xµ +α ′pµ σ++σ−2+ i√α ′2 ∑n6=01n(α˜µn e−inσ+ +αµn e−inσ−).For the closed superstring, the left and right movers (ψ and ψ˜) are independent andhave the mode expansionsψµ(σ0,σ1) =√α ′ ∑r∈Z+vbµr eirσ+ψ˜µ(σ0,σ1) =√α ′ ∑r∈Z+vb˜µr e−irσ− .Since fermions can have two different types of boundary conditions, the parameterv ∈{0, 12}denotes which one we are using. For Ramond fermions, which areperiodic, v = 0. For Neveu-Schwarz fermions which are antiperiodic, v = 12 . The10creation and anhilation operators above obey the relations [19][αµm ,ανn ] = mηµνδm+n,0 = [α˜µm , α˜νn ]{bµr ,bνs }= ηµνδr+s,0 = {b˜µr , b˜νs } .To build up the spectrum from this, we need to consider gauge symmetries.The action (2.8) came from a more general action in which the worldsheet metricwas dynamical. Choosing gab = ηab restricts the physical Hilbert space to onlythose states which are anhilated by the Virasoro generators:Lm =12 ∑n∈Z: αm−n ·αn : +14 ∑r∈Z+v(2r−m) : bm−r ·br :−aδm,0Gr = ∑n∈Zαn ·br−n .Analogous expressions hold for L˜m and G˜r. Like D, the normal ordering constanta = v is an anomaly that can be fixed by demanding Lorentz invariance [19]. Wemay use the relativistic dispersion relation, the mode expansions and the Virasorogenerators to write down a formula for the mass operator.m2 = −pµ pµ= −2α ′α0 ·α0= −2α ′[L0−∑n>0α−n ·αn−∑r>vrb−r ·br +a]=2α ′[∑n>0α−n ·αn +∑r>vrb−r ·br−a](2.10)We must have m2 = m˜2. This translates into a condition known as level matchingrequiring every state to have the same number of left and right moving excitations.Despite accounting for a gauge symmetry in this way, the action (2.8) still hassome gauge symmetry left. A common technique for dealing with this redundancyis fixing the lightcone gauge. This essentially means that any Lorentz index µrunning from 0 to D−1 becomes a regular index i running from 1 to D−2 [19].We now have everything we need to derive the massless spectrum of Type II11string theory. In typical examples of a Fock space, the ground state is unique. It isa singlet with respect to any symmetry group of interest and denoted most often by|0〉. This is not the case for the superstring. For Ramond fermions, the operatorsbi0 commute with m2 meaning that many states have zero mass. This degenerateground state in fact transforms as a spinor in ten dimensions. Moreover it can besplit into two chiralities |+〉 and |−〉. This is different from four dimensions whichwould make the split into Weyl spinors inconsistent with the split into Majoranaspinors that we have already performed [13]. For Neveu-Schwarz fermions, thelowest lying state has negative m2. However, one of the advantages of the super-string is that it allows us to avoid this tachyon and start at the massless states. Theseare also degenerate and are denoted by bi− 12|0〉. Even though b is an anticommut-ing operator for the worldsheet, the i index here makes this a vector particle inthe target space. We have shown that massless R states are spacetime fermionswhile massless NS states are spacetime bosons. The choice between R and NS canbe made for the left and right movers separately. This means that Type II stringtheories have four sectors [19].(v˜,v) Type IIA Type IIBR-R (0,0) |−〉⊗ |+〉 |+〉⊗ |+〉NS-NS(12 ,12)b˜i− 12|0〉⊗bi− 12|0〉 b˜i− 12|0〉⊗bi− 12|0〉NS-R(12 ,0)b˜i− 12|0〉⊗ |+〉 b˜i− 12|0〉⊗ |+〉R-NS(0, 12)|−〉⊗bi− 12|0〉 |−〉⊗bi− 12|0〉(2.11)Each sector is 64-fold degenerate.Our expression for the free energy density has terms like e−piτ2α′m2 summedover masses. These sums look like familiar partition functions if we substitute(2.10) in for m2. Going back to (2.9), it is almost correct to replace the term insquare brackets with 64Z0,0(τ2)+64Z 12 ,12(τ2)−128(−1)nZ0, 12 (τ2). In the notation12being usedZv˜,v = e2(v+v˜)piτ2 ∑{Nin,N˜in}exp(−2piτ28∑i=1∞∑n=1n(Nin + N˜in))[∑{Mir}exp(−2piτ28∑i=1∞∑r=v+1rMir)] ∑{M˜ir}exp(−2piτ28∑i=1∞∑r=v˜+1rM˜ir) ,(2.12)Nin and N˜in are bosonic occupation numbers while Mir and M˜ir are fermionic occu-pation numbers. Accounting for level matching is the one correction that needs tobe made. This can be done by inserting a Kronecker deltaδL,R =∫ 12− 12e2piiτ1(L−R)dτ1where L = ∑i,n nN˜in +∑i,r rM˜ir and R = ∑i,n nNin +∑i,r rMir are the left and rightexcitations respectively. Multiplying this by (2.12), we see that the quantity beingintegrated is nothing but the generalized partition function for 8 fermions and 8bosons. Rewriting (2.9),FV= −∞∑n=1∫ ∞0∫ 12− 12e− β2n24piα ′τ2 64ZB(τ)8[ZR(τ)8 +ZNS(τ)8−2(−1)n√ZR(τ)ZNS(τ)8](4pi2α ′τ2)−5dτ1dτ2τ2. (2.13)In terms of our old notation, ZB(τ) = Z(τ) while ZR(τ) and ZNS(τ) approach Z∗(τ)in the small τ limit.Our goal is to investigate the high temperature limit of (2.13). Since this cor-responds to β → 0, the integral is dominated by the n = 1 term of the sum andthe small τ limits of the worldsheet partition functions. A curious fact about stringtheories is that at a high enough temperature, called the Hagedorn temperature TH,the free energy density diverges. We will solve for βH. This can be done by lookingat any one of the four terms in the integrand of (2.13). Substituting the generalized13partition function (2.7), the function we are integrating is64e− β24piα ′τ2 e−2piℑ(1τ ) (4pi2α ′τ2)−5τ2.The value of βH is reached when the overall exponent is zero. This meansβ 2H = 8pi2α ′τ22τ21 + τ22≈ 8pi2α ′ . (2.14)The partition function for a system first diverges when the density of states becomesexponential and the decay of the Boltzmann factor can no longer overpower suchgrowth. This is equivalent to saying that S ∝ E. The proportionality constant canbe read off from (2.14) because β = dSdE must give βH. The result of this, in contrastto (2.2) is:S = 2pi√2α ′E . (2.15)2.2 Black holesIt is clear how entropy arises in the field theories we have discussed. If we onlyknow the energy E of a field, the corresponding ensemble of particles can be in anyone of ρ(E) microstates contributing to our lack of knowledge about the system.Microstates of this form do not appear to be present for black holes. Classically,one can learn everyting about a black hole from just three numbers: mass, chargeand angular momentum. The discovery that black holes have entropy as well, hasled to some of the deepest results in theoretical physics [20].2.2.1 Useful metricsBlack hole metrics in arbitrary dimension have seen increasing interest since thediscovery of the AdS / CFT correspondence [21]. Most authors take the Einsteinequations to be fundamental so that they readRµν −12Rgµν +Λgµν = 8piGTµν (2.16)14regardless of how many dimensions there are. The same cannot be said of the New-tonian limit. If (2.16) describes the full theory of gravity in d +1 dimensions, onecan show that dimension dependent prefactors necessarily appear in the Poissonequation [21]:∆Φ = 8piGd−2d−1ρ= 8piGd−2d−1Mδ (0) . (2.17)The second form above specializes to a point mass of M.The neutral, irrotational black hole in arbitrary dimension is called the Schwarzschild-Tangherlini solution.ds2 =−(1−µrd−2)dt2 +(1−µrd−2)−1dr2 + r2dΩ2d−1 (2.18)We will take this opportunity to review some of the basic properties that black holemetrics should have.1. It can be checked that (2.18) solves Einstein’s equation with no cosmologicalconstant and no stress-energy tensor. In fact, it is the unique sphericallysymmetric and time independent solution. The requirement that it be timeindependent is redundant if d = 3.2. It is clear that (2.18) is asymptotically flat. Therefore, the notion of “escap-ing” from a potential well in this metric is well defined.3. It is also clear that beyond a certain radius, one can no longer escape. Faraway from the origin, t is timelike and r is spacelike but this reverses whenr falls below µ1d−2 . On the inside of this event horizon, the flow of time in(2.18) is such that an object is inexorably drawn toward the centre.4. This implies that the mass generating the event horizon has been compressedto a point. This, along with the fact that metrics for equal point masses shouldbe indistinguishable, tells us that (2.18) represents the most efficient packingof said mass into a sphere of radius µ1d−2 .155. The mass may be computed by taking the Newtonian limit. For those unfa-miliar with the ADM procedure [22], we will consider a test particle far awayfrom the origin, moving radially outward. If the motion is non-relativistic,the timelike geodesic condition becomes gtt t˙2 ≈ −1 or t˙2 ≈(1− µrd−2)−1.Substituting this into the geodesic equation,r¨ ≈ −Γrtt t˙2≈12∂rgttgrr1− µrd−2= −∇( µ2rd−2).The function inside the gradient should be a solution to (2.17). Recallingthe Green’s function for the d-dimensional Laplacian, this is only true ifµ = 16piGMd(d−1)ωd .A further generalization of interest to us is the metric for a black hole that isasymptotically AdS. Anti-de Sitter space is the maximally symmetric solution toEinstein’s equations when they have a negative cosmological constant. A cosmo-logical constant defines a length scale for the spacetime withΛ≡−d(d−1)2L2(2.19)by convention. Because of this, AdS posesses a conformal boundary. Its radialco-ordinate is infinite but an observer is able to reach it in a finite amount of propertime. The solution isds2 =−(1+r2L2)dt2 +(1+r2L2)−1dr2 + r2dΩ2d−1 (2.20)and a black hole metric asymptotic to this isds2 =−(1+r2L2−µrd−2)dt2 +(1+r2L2−µrd−2)−1dr2 + r2dΩ2d−1 . (2.21)Uniqueness of (2.21) in the same sense as (2.18) is suspected but not known. Thehorizon radius is a solution to µrd−20= 1+ r20L2 with the outermost one being the point16of no return. The mass is again given by µ = 16piGMd(d−1)ωd . To see this, a non-relativisticradial trajectory satisfiesr¨ ≈ −Γrtt t˙2≈12∂rgtt= −∇( µ2rd−2)−rL2.While this lacks the sophistocation of methods like [23], we obtain the right answerif we simply subtract the acceleration that a particle would have in pure AdS.It will be convenient to rewrite this metric in Eddington-Finkelstein co-ordinates.This can be done using either the retarded time or advanced time, which take theformu = t− r∗v = t + r∗respectively. Differentiating these to arrive atu˙ = t˙−dr∗drr˙v˙ = t˙ +dr∗drr˙ ,we see that r∗ should be chosen so that its radial derivative is the factor relating t˙and r˙. For a null geodesic,t˙ =±(1+r2L2−µrd−2)−1r˙where the positive sign corresponds to an outgoing particle and the negative signcorresponds to an ingoing particle. We now havedr∗dr=(1+r2L2−µrd−2)−1.In certain cases, this can be integrated to give r∗ explicitly. However, this is not17needed for replacing dt. Performing the change of variables,ds2 = −(1+r2L2−µrd−2)du2−2dudr+ r2dΩ2d−1ds2 = −(1+r2L2−µrd−2)dv2 +2dvdr+ r2dΩ2d−1 (2.22)are the desired metrics.2.2.2 Hawking radiationTo derive the relation between entropy and the area of a black hole, we will followHawking’s original paper [24] as well as the clarifications in [25, 26]. What wewill see is that two observers — one observing spacetime before a black hole hasformed, the other after — will have different definitions of the quantum vacuum.The Klein-Gordon equation for a field in curved spacetime is∇µ∇µφ −m2φ = 01√−g∂µ(√−ggµν∂νφ)−m2φ = 0 . (2.23)For any two solutions φ1 and φ2, the Klein-Gordon inner product〈φ1,φ2〉=∫S[φ ∗1∇µφ2−φ2∇µφ ∗1 ]dΣµ (2.24)will be conserved. The notation above suggests a scalar field, but this does not haveto be the case. Fields with multiple components like vectors and spinors satisfy theKlein-Gordon equation componentwise. To indicate that Hawking radiation is amixture of all types of particles, we will write creation and anhilation operators asaI † and aI where I is an index set. One simplification we will make, however,is that the fields are massless.We may write a basis of solutions to (2.23) as fi and choose them to be or-thonormal with respect to (2.24). If we do this, the field operator takes the formφI =∑ifiaIi + f∗i aI †i .18In other words, positive frequency modes multiply anhilation operators while neg-ative frequency modes multiply creation operators. We will let these fi representany particles that can be seen before a black hole forms. Since there are no suchparticles, the past observer will see the vacuum state |0〉a defined as the state thatis anhilated by all aIi . The future observer sees a different metric and in particu-lar a different time component of the metric. This means he will have a differentdefinition of positive and negative frequency. WritingφI =∑igibIi +g∗i bI †i ,each gi representing a particle in the black hole spacetime should be expressibleas a linear combination of the fi. A positive frequency gi may therefore includea contribution from a negative frequency fi and vice versa. If so, the bIi willnot anhilate the aIi vacuum and the aIi will not anhilate the bIi vacuum. Thisdiscrepancy between |0〉a and |0〉b means that the future observer will see radiationprecisely because the past observer did not.It is not correct to say that the only modes of φI are fi waves that the pastobserver can see and gi waves that the future observer can see. There are also hiwaves in the future that cannot be seen because they are behind the event horizonof the black hole. We would have to consider these if we wanted to write the pastmodes as linear combinations of the future modes. As it happens, we will only needto write the future modes as linear combinations of the past modes. Convertinggi = ∑jαi j f j +βi j f ∗jhi = ∑jσi j f j + τi j f ∗jinto a set of relations between operators, we arrive at the so-called Bogoliubovtransformation:bIi = ∑jα∗i jaIj −β ∗i jaI †jcIi = ∑jσ∗i jaIj − τ∗i jaI †j .19This tells us that the number of particles detected as belonging to gi in the future isgiven byni = a 〈0|bI †i bIi |0〉a =∑j∣∣βi j∣∣2 . (2.25)Assuming that we are dealing with bosons, we also have1 = a 〈0|[bIi ,bI †i]|0〉a =∑j∣∣αi j∣∣2−∣∣βi j∣∣2 . (2.26)We will now be more explicit about what the modes are so that we may plug theminto the inner product and find the αi j and βi j coefficients.Spherical waves are convenient choices, but it is important not to use the ex-pressions for flat space spherical waves when we are really in a curved space. Byconstruction, outgoing null geodesics are lines of constant u while ingoing nullgeodesics are lines of constant v. Therefore, the advanced and retarded timesshould be used in place of t± r giving usfω,l1,...,ld−1(v,θ1, . . . ,θd−1) =eiωv√(ωr)d−1dωdYl1,...,ld−1(θ1, . . . ,θd−1)gω,l1,...,ld−1(u,θ1, . . . ,θd−1) =eiωu√(ωr)d−1dωdYl1,...,ld−1(θ1, . . . ,θd−1) (2.27)as approximate solutions for large r. We could similarly consider fω,l1,...,ld−1(u,θ1, . . . ,θd−1)and gω,l1,...,ld−1(v,θ1, . . . ,θd−1) but these would affect the result very little. The in-teresting effects come from waves that switch from ingoing to outgoing while theblack hole is forming. By this, we mean that waves of constant v travel toward thecollapsing mass at r = 0. As long as an event horizon has not formed yet, suchwaves may emerge from the other side and start moving away with constant u. Anatural question to ask is which constant u? That is, what will u be in terms of the vthat the wave used to have? This is the key question that must be answered beforewe can take an inner product and derive Hawking’s result. The difficulty in relatingthese is explained in Figure 2.1. A better way to compare objects A and B is toimagine that A throws a ball C backwards until it is caught by B. The proper timefor C to travel should be equal along all stages of the journey. Instead of propertime, we will use a difference of affine parameters which is appropriate for signals20AB(a) BeforeAB(b) AfterFigure 2.1: Photon A leaves a light source and starts heading toward r = 0.After a certain time interval, photon B does the same. If the spacetimeis Minkowski, the distance between the two photons will not change.Conversely, if a black hole at r = 0 forms at just the right time, photonA will escape but photon B will stay trapped at the event horizon forever.travelling at the speed of light.Waves hoping to escape the black hole must start off with a v smaller than theone posessed by photon B in Figure 2.1. We will call this largest advanced timev0. If the spacetime is Minkowski, long before the black hole has formed, v = t + rand affine parametrizations aret(λ ) = t(0)+ λ2rA(λ ) = rA(0)−λ2rB(λ ) = rB(0)−λ2rC(λ ) = rA(0)+λ2.By the time C makes it to B, it will have the same position and time co-ordinateas B so it must have the same v as B. Therefore subtracting the advanced times21corresponding to A and C, we havev− v0 = rA(λ )− rC(λ ) =−λ .We may therefore call v0− v an affine parameter that vanishes for the wave thatstays at the event horizon. This means that after the horizon at r0 has formed, awave’s radial co-ordinate must look like r = r0− λ . We will substitute this intothe retarded time for C noting that C is not a wave of constant u because it travelsbackwards from A back to B.u˙ = t˙−dr∗drr˙= 2(1+r2L2−µrd−2)−1= 2(1+r2L2−(r0r)d−2(1+r20L2))−1= 2(1+(r0−λ )2L2−(r0r0−λ)d−2(1+r20L2))−1(2.28)This does not have a closed form integral, but the interesting effects come fromwaves that are close to the horizon. Keeping only the lowest order in λ ,u˙ ≈ −2r0λL2(d−2)L2 +dr20u ≈ −2r0L2(d−2)L2 +dr20log(λC)= −2r0L2(d−2)L2 +dr20log(v0− vC).The equation relating u to v has now been found, so we may substitute (2.27)into (2.24) for a surface whose normal derivative is ∂ r. We will abbreviate the22l1, . . . , ld−1 dependence as l and the θ1, . . . ,θd−1 dependence as θ .αω′,l′ω,l =〈gω,l, fω ′,l′〉= i∫Sd−1∫ v0−∞f ∗ω ′,l′(v,θ)∂ rgω,l(v,θ)−gω,l(v,θ)∂ r f ∗ω ′,l′(v,θ)dvdΩ= i∫Sd−1∫ v0−∞f ∗ω ′,l′(v,θ)[∂v +(1+r2L2−µrd−2)∂r]gω,l(v,θ)−gω,l(v,θ)[∂v +(1+r2L2−µrd−2)∂r]f ∗ω ′,l′(v,θ)dvdΩ=idωd−1√ωω ′d−1∫Sd−1Yl(θ)Y ∗l′ (θ)dΩeiω′v∂ve2iωr0 L2(d−2)L2+dr20log( v0−vC )− e2iωr0 L2(d−2)L2+dr20log( v0−vC )∂veiω′vdv=δ l′1l1δ l′2l2. . .δ l′d−1ld−1dωd−1√ωω ′d−1∫ v0−∞(2ωr0v0− vL2(d−2)L2 +dr20+ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC )eiω′vdvThe other Bogoliubov coefficient is found similarly. The only difference is thatwhen integrating two spherical harmonics without a complex conjugate, we haveto use the identity Yl1,l2,...,ld−1 = (−1)l1Y ∗−l1,l2,...,ld−1 .βω′,l′ω,l =〈gω,l, f∗ω ′,l′〉=(−1)l1δ l′1−l1δ l′2l2. . .δ l′d−1ld−1dωd−1√ωω ′d−1∫ v0−∞(2ωr0v0− vL2(d−2)L2 +dr20−ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC )e−iω′vdvThe integrands above have a branch cut on the real axis because of the log( v0−vC).In order to manipulate them with complex analysis, it is convenient to displacethem with ±iε . The α and β integrals become∫ v0−∞(2ωr0v0− vL2(d−2)L2 +dr20+ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC +iε)eiω′vdv (2.29)∫ v0−∞(2ωr0v0− vL2(d−2)L2 +dr20−ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC −iε)e−iω′vdv (2.30)respectively. The signs for ±iε above are dictated by our requirement that e±iω ′vvanish at infinity.23v0 ∞-∞Figure 2.2: For the function with +iε , it is not difficult to show that the semi-circular arc has no contribution to the integral. Therefore the integralbetween −∞ and ∞ is zero.The integral in (2.29) vanishes if the domain is the contour in Figure 2.2. Wemay therefore split it up as follows.∫ v0−∞(2ωr0v0− vL2(d−2)L2 +dr20+ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC +iε)eiω′vdv= −∫ ∞v0(2ωr0v0− vL2(d−2)L2 +dr20+ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC +iε)eiω′vdv= −eiω′v0∫ 0−∞(2ωr0v′L2(d−2)L2 +dr20+ω ′)e2iωr0 L2(d−2)L2+dr20log(−(−v′C −iε))eiω′v′dv′In the second step, we made the substitution v′ = v0− v. This integral is writ-ten with the understanding that it should be evaluated with a contour in the lowerhalf plane. We must therefore write log(−A) = log(A)− ipi instead of log(−A) =log(A)+ ipi to avoid crossing the branch cut. We will perform this step and thenmake another substitution v′ = v− v0.∫ v0−∞(2ωr0v0− vL2(d−2)L2 +dr20+ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC +iε)eiω′vdv= −eiω ′v0+2piωr0 L2(d−2)L2+dr20∫ 0−∞(2ωr0v′L2(d−2)L2 +dr20+ω ′)e2iωr0 L2(d−2)L2+dr20log(−v′C −iε)eiω′v′dv′= e2iω ′v0+2piωr0 L2(d−2)L2+dr20∫ v0−∞(2ωr0v0− vL2(d−2)L2 +dr20−ω ′)e2iωr0 L2(d−2)L2+dr20log( v0−vC −iε)eiω′vdvBy manipulating (2.29), we have turned it into a multiple of (2.30). This implies24the relation:∣∣∣αω′,l′1,...,l′d−1ω,l1...,ld−1∣∣∣= e2piωr0 L2(d−2)L2+dr20∣∣∣βω′,l′1,...,l′d−1ω,l1...,ld−1∣∣∣ . (2.31)Going back to (2.25) and (2.26), the integral∑l′1,...,l′d−1∫ ∞0∣∣∣αω′,l′1,...,l′d−1ω,l1...,ld−1∣∣∣2−∣∣∣βω′,l′1,...,l′d−1ω,l1...,ld−1∣∣∣2dω ′describes what the black hole will absorb. The integral∑l′1,...,l′d−1∫ ∞0∣∣∣βω′,l′1,...,l′d−1ω,l1...,ld−1∣∣∣2dω ′describes what the black hole will emit. Without worrying about normalization,(2.31) tells us that the ratio between a mode’s absorbtion and emission cross sec-tions isnω =(e4piωr0 L2(d−2)L2+dr20 −1)−1. (2.32)This is precisely the Bose-Einstein thermal factor for a blackbody at temperatureTBH =14pir0(d−2+dr20L2)(2.33)Had we used an anticommutator in (2.26), we would have seen the Fermi-Diracfactor for the same temperature. The famous Bekenstein-Hawking entropy, S = A4G ,clearly follows from this if L is large. For a general L, we will use the fact thatµ = rd−20 +rd0L2 to writedM =d(d−1)ωd16piG dµ =d(d−1)ωd16piG[(d−2)rd−30 +dL2rd−10]dr0 .Then integrating,SBH =∫dETBH=∫dMTBH=d(d−1)ωd4G∫rd−20 dr0=dωd4Grd−10=A4G(2.34)and we see that the entropy formula is exactly the same in AdSd+1.25Hints that the area of a black hole somehow describes an entropy were alreadyknown in 1973 when Bekenstein proposed the proportionality with a coefficient“close to” log28pi [27]. Apart from improving the coefficient to14 , Hawking’s 1975paper established that (2.34) is the genuine entropy of a thermal spectrum [24].Modern techniques can derive (2.34) much more quickly but at the cost of onceagain obscuring the nature of this entropy [28, 29]. In 1981, Bekenstein noticedthat (2.34) is more than just the entropy of a black hole. It is an upper boundon the entropy that any system occupying the same volume can have [30]. Theargument, which was strong motivation for the AdS / CFT correspondence [8], isremarkably simple. Suppose that a non-black hole system fills a ball of radius r0and has more entropy than A4G . Its mass must be less than that of a black hole withhorizon radius r0 and therefore, it can be turned into said black hole through theaddition of mass. Such a procedure would give the system an entropy of A4G lateron, violating the second law of thermodynamics. Incidentally, two major openproblems in physics are related to the evaporation of black holes. A featurelessobject described uniquely by mass, charge and angular momentum should not haveentropy and yet we have seen that it contains more entropy than anything else.While their exact nature remains unknown, some methods for elucidating blackhole microstates are provided by string theory [31]. A more serious problem isthe black hole information paradox. This is concerned with the fact that a bathof radiation cannot contain information about the formation of a black hole. If athermal state is all that a black hole leaves behind after it evaporates, one effectivelyhas a pure state evolving into a mixed state which is a violation of unitarity. Stringtheoretic resolutions to this have been proposed as well but are, at the time ofwriting, much more speculative [32, 33].2.3 Strong couplingThe AdS / CFT correspondence delivers on a 1974 promise to make strongly cou-pled U(N) and SU(N) gauge theories more tractable when N is large [34]. Moreprecisely, the quantum gravity theory in the bulk that is dual to a CFT becomesincreasingly classical as we take N → ∞ with λ = g2YMN fixed. This is known asthe large N, planar or ‘tHooft limit. The ‘tHooft coupling λ which only needs to be26fixed, is the parameter that would have to be small for the usual Feynman diagramexpansion to be valid. When a series of Feynman diagrams is written down usingpowers of 1N instead of gYM, the expansion looks very similar to that of a closedstring theory with coupling gs. For this reason, the identificationgs =λ4piN (2.35)appears in the duality [8]. This small string coupling allows a perturbative calcula-tion to be done in AdS when the field theory on the boundary is strongly coupled.Some of the coupling strengths not covered by this limit (e.g. large gYM and N)can be explored with the help of string dualities. For instance, a weak-strong sym-metry known as S-duality is often associated with Type IIB string theory [35]. IfIIB in AdS is equivalent to SYM on the boundary, this statement implies that (2.1)is invariant under gYM 7→ 1gYM . Although such a result could have been discoveredthrough holography, it was discovered earlier using some of the same evidence thatled to the correspondence [36, 37]. In addition, we should note that even if onebelieves AdS / CFT, the idea of S-duality holding for Type IIB string theory is alsoa conjecture [35].Using classical gravity to approximate field theories in the strong couplingregime has become the most widely explored aspect of the AdS / CFT corre-spondence [38, 39]. For this application, questions about whether string theoryis realized in nature, are irrelevant. We will go through an example of this duality,whereby the interacting spectrum of (2.1) can be understood through our seeminglyunrelated calculations regarding free field theories and black holes.2.3.1 Gauge theory phasesWhen compactified on a sphere, there are at least four interesting phases posessedby Super Yang-Mills. The transitions in and out of these phases are gradual, asthey must be for a theory with finitely many fields. However, the transitions maybecome sharp in the strict N→ ∞ limit. Following our pattern above, we will givean expression for the entropy of each phase in order of increasing energy.The first thing we need to know is that in the original version of the corre-spondence, the bulk geometry is AdS5×S5 where the length scale of AdS5 and the27radius of S5 are equal [2]. This radius, which we will call L is given by the dualityprescription asL4 = 4pigsα ′2N . (2.36)There is also a radius for the S3 of the the field theory, which we will call R. Itis natural to compare the dimensionless energies of the string theory EL to “somemultiple” of the dimensionless energies of the field theory ER. By studying theKlein-Gordon equation (2.23) for a graviton propagating in AdS5× S5, one mayshow that the second-lowest energy it can have isE1 =1L. (2.37)While this would not be the case for a general gauge theory, a highly supersymmet-ric theory like SYM has some excited state energies that can be computed withoutthe correspondence. This is a harder calculation but an analysis of chiral primaryoperators [8, 40] tells us that E1 = 1R on the field theory side. The multiple inquestion is therefore 1 and we will be able to replace EL with ER in what follows.The low energy behaviour of the bulk is described by a free gas of strings intheir worldsheet ground states. Refering to (2.11), this is a gas of 128 bosons and128 fermions — essentially gravitons and their superpartners. The free field theoryresult (2.3) includes a volume V , which is only well defined if there is a clearseparation between space and time, i.e.ds2 =−dt2 +gi jdxidx jwhere gi j is a Riemanian metric. Since the line elementds2 =−(1+r2L2)dt2 +(1+r2L2)−1dr2 + r2dΩ23 +L2dΩ25 (2.38)is not in this form, the brute force calculation of the partition function would have tostart with the Klein-Gordon equation. Solving the relevant Klein-Gordon equationis certainly a useful exercise. In addition to energy eigenvalues like (2.37), it wouldallow us to derive a bound on the mass that any particle in AdS must satisfy [41, 42].However, there is another method that can tell us the appropriate V more quickly.28Performing a conformal transformation on (2.38), we may turn it intods2 =−dt2 +(1+r2L2)−2dr2 +(1+r2L2)−1(r2dΩ23 +L2dΩ25). (2.39)The massless version of (2.23) (called the minimally coupled Klein-Gordon equa-tion) is not invariant under such a rescaling, but the conformally coupled Klein-Gordon equation is. This equation is∇µ∇µφ + d−14dRφ = 0 (2.40)where R is the Ricci scalar. The volume associated with the (2.39) metric wouldtherefore appear in the partition function for this graviton gas in the conformallycoupled case.V =∫ ∞0∫∫S3×S5(1+r2L2)−5L5r3dΩ3dΩ5dr= 15ω3ω5L5∫ ∞0(1+r2L2)−5r3dr=1524ω3ω5L9Because the difference between minimal and conformal coupling only appears be-yond the leading order thermodynamics [8, 17], it is sufficient to substitute d = 9,s = s∗ = 128 and the V above. This yieldsS1 =(10109915249!ω3ω5ω9(2pi)9 128(ζ (10)+ζ∗(10))E9L9) 110= 10(9!ω3ω5ω9152410234ζ (10)) 110(EL18pi) 910= 10(27289355pi8) 110(ER9) 910(2.41)as the entropy of the lowest energy phase.As energy increases, the infinite tower of worldsheet vibrations becomes im-portant and the entropy enters the Hagedorn regime. Starting with (2.15), we just29need to use the (2.35) and (2.36) identifications to get the entropy for the secondphase in terms of gauge theory parameters:S2 = 2pi√2α ′E= 2pi(L4pigsN) 14E= 2pi(4λ) 14ER . (2.42)As with any proper string theory, the AdS5× S5 background is not static. Itreceives a backreaction from stringy states that becomes more significant as theenergy increases. The highest energy phases of Super-Yang Mills will thereforeinvolve Newton’s constant G. The formulaG5L5 = 8pi3g2sα ′4 (2.43)is the last piece of the correspondence that we need [43]. We mentioned previouslythat Einstein’s equations (and Newton’s constant in particular) should be the samein all dimensions. Thus, it may seem strange to refer to a five-dimensional gravi-tational constant G5. The explanation is that G5 is not the gravitational constant atall, but rather an illusion created by the presence of compact dimensions. The trueG appears as a prefactor in the Einstein-Hilbert actionS =116piG∫AdS5×S5(R−2Λ)√−gdx .If the radius of the sphere is small enough, a macroscopic observer only sees AdS5.Integrating out the S5 and looking at the prefactor once again will tell us the re-lation between G5 and G. Using the fact that Ricci scalars add for direct product30manifoldsS =116piG∫AdS5×S5(RAdS5−2Λ)√−gdx+116piG∫AdS5×S5RS5√−gdxSeff =116piG∫S5√gS5dx∫AdS5(RAdS5−2Λ)√−gAdS5dx+Sshift=6ω6L516piG∫AdS5(RAdS5−2Λ)√−gAdS5dx+Sshift .We have turned the action into an effective action by evaluating part of it. Thismakes it clear that G5 = G6ω6L5 . The fact that these dimensionally reduced Newtonconstants are generally much smaller than G has led to the hypothesis that theapparent strength of gravity increases when the distance is very small. Indeed,proponents of extra-dimension phenomenology have discussed the possibility offorming black holes at the LHC [44–46].With these constants in hand, we need to calculate the entropy associated withthe geometry that develops in the third phase. Since entropy increases with energy,it is only logical that our spacetime should eventually achieve the geometry thathas a monopoly on entropy — that of a black hole. The mysterious microstatesof this black hole can be put in a one-to-one correspondence with the well definedmicrostates of the CFT. When the event horizon r0 first forms, it is smaller than theradius L. It is therefore a good approximation to describe it with the Schwarzschildsolution (2.18) involving all ten spacetime dimensions. Using the entropy formula31(2.34),S3 =9ω94Gr80=9ω94Gµ87=94(Gω9) 17(2piE9) 87=94(6L5G5ω6ω9) 17(2piE9) 87=94(3piω6N2ω9) 17(2piEL9) 87=94(1890N2) 17(piER9) 87. (2.44)As the black hole grows to a radius r0  L, the five small dimensions becomenegligible allowing us to use the asymptotically AdS black hole (2.21). This alsomakes it a good approximation to say µ = r20(1+ r20L2)≈ r40L2 . Inserting this into(2.34),S4 =ω4G5r30=ω4G5µ34=(ω4G5) 14(4piEL23) 34=(2N2ω4piL3) 14(4piEL23) 34= pi√N(43ER) 34. (2.45)This matches the behaviour that a conformal theory must have at high energies(2.2).We have yet to give estimates for the energy ranges where these phases arevalid. Prefactors for these energies would be suspicious due to the gradual nature of32the phase transitions. We will therefore only keep factors that may be comparableto N2. To determine when the Hagedorn phase becomes important, we should set Eto the mass of an excited string. From (2.10), we see that this is of order 1√α ′ ∝λ14R .Strings have a characteristic length and a black hole with this length as its hori-zon radius has a characteristic energy. When the energy of a string gas exceedsthis, it is expected to collapse to the small black hole that we discussed before.Of course there are some non-black hole geometries having energies of this mag-nitude (e.g. a giant graviton [47]) but these are “rare”. This is consistent withthe “heat death” proposal in which a black hole is the inevitable final state of asystem that evolves via thermal fluctuations. An equivalent statement on the CFTside is that as the dimensions of gauge invariant operators increase, the fraction ofthem that describe black holes approaches unity [48]. The transition for this blackhole “probably forming” can be found by checking when the Hagedorn entropy be-comes comparable to the small black hole entropy. Setting (2.42) equal to (2.44),this energy is of order N2λ74 R.Finally, the midpoint between the small black hole and the large black holeoccurs when r0 = R. Expressing the event horizon radius in terms of the mass,E ∝ R7G ∝R2G5∝ N2R . Putting this together we see that the entropy for strongly coupledSYM is given byS(E) =10(27289355pi8) 110(ER9) 910 ER λ 142pi(4λ) 14 ER λ 14  ER λ− 74 N294(1890N2) 17(piER9) 87 λ− 74 N2 ER N2pi√N(43 ER) 34 N2 ER. (2.46)Notice that if we were to find the entropy of free Super-Yang Mills by substitutingd = 3 and s = s∗ = 8N2 in (2.3), the result would be 43pi√N (ER)34 . The entropiesdiffer by a factor of(43) 14 or equivalently, the free energies differ by a factor of 43 .WritingF =−16h(λ )pi2N2V T 4 (2.47)with h(0) = 1 and limλ→∞ h(λ ) = 34 , various authors have studied how h interpo-lates between these limits using curvature corrections on the string theory side [49]33and loop diagrams on the field theory side [50, 51]. It was later found that interpo-lating between weakly coupled and strongly coupled free energy is not as simpleas multiplying by h. Corrections to (2.47) involving T 2 need to be multiplied bydifferent functions of the ‘tHooft coupling [43].2.3.2 Plasma ballsThe microcanonical entropy of strongly coupled SYM on S3 (2.46) is a formulathat we will use repeatedly. Part of its derivation relied on the fact that the theory’sdual description involved black holes radiating a thermal spectrum. The goal ofthis thesis is to argue for the converse: an arbitrary field theory with an entropysufficiently similar to (2.46) exhibits dynamics that are indicative of black holeformation and evaporation. There is a large class of field theory solutions, calledplasma balls, that have been shown to be of this type [7]. Most studies of them arenumerical [7, 52, 53] but at least one has been constructed analytically [54].Consider the canonical phases of Super Yang-Mills found by fixing the temper-ature instead of the energy. We may differentiate the entropy in (2.46) to plot β asa function of E. At low temperatures, the system must be in the graviton gas phase.EβPerturbed AdSHagedornSmall black holeLarge black holeFigure 2.3: For an inverse temperature like the one shown, the gravity sidemust choose the geometry that minimizes the free energy.34One could raise the temperature (lower the dotted line) all the way to the Hagedorntemperature at which the canonical ensemble ceases to exist but the most interest-ing situation occurs for an intermediate value where there is competition betweenthree phases. From (2.46), a straightforward calculation of the graviton gas freeenergy yieldsF =−27289355pi8R9T 10 (2.48)and we have already written the large black hole free energy (2.47). If we were tocalculate the small black hole free energy in the same way, we would find that itis positive, so (2.48) and (2.47) are the only ones we need. Setting them equal, wefind a first order phase transition atTD =1R(9355N216368pi5) 16. (2.49)Were it not for the complication of the internal manifold S5, this would be theHawking-Page transition [28] showing that a sufficiently large black hole in AdScan come to equilibrium with the radiation it emits. Since the energy, entropy andtemperature of a black hole are all known in terms of its event horizon radius r0,F = E−T S=dωd16piGrd−20(1−r20L2).When r0 < L, this is minimized for an r0 that rolls to zero. When r0 > L, this isminimized for as large an r0 as possible. Substituting r0 = L into (2.33), we findTHP =d−12piL . (2.50)This phenomenon on the field theory side has the interpretation of a deconfinementphase transition related to the scale R. As R→ ∞, the temperature (2.49) vanishesand there is no confinement as expected for a CFT in Minkowski space. Sincethe confining theory of greatest physical interest (quantum chromodynamics) livesin infinite volume, it has little in common with Super Yang-Mills on S3. A holo-graphic study of QCD requires one to introduce a scale to SYM in a more drastic35way.This can be done by compactifying some but not all of the directions in aMinkowski CFT. Witten’s model [29] e.g. compactifies SYM on a Scherk-Schwarzcircle — S1 with antiperiodic fermions. Since the other directions are extended, it ishelpful to rewrite (2.20) so that they manifestly appear as Minkowski space. GlobalAdS is large enough for this to be done several times yielding disjoint patches sep-arated by co-ordinate singularities. A given patch has the following metric knownas Poincare´ AdSds2 =L2z2[−dτ2 +dz2 +dxidxi](2.51)where the boundary is located at z = 0. The transformationt = arctan(2LτL2 + z2 + x2− τ2)r =Lz√z2 + τ2 + 14L2(L2 + z2 + x2− τ2)2 (2.52)sinθi . . .sinθi−1 cosθi =xi√z2 + τ2 + 14L2 (L2 + z2 + x2− τ2)2(2.53)converts between the global and Poincare´ metrics [8]. Some sources assume r Lbefore deriving (2.51) in order to write the simpler transformation z = L2r [55].This gives the false impression that (2.51) is only approximately equal to a patchof AdS. Applying (2.52) to (2.21) gives another form of the AdS black hole:ds2 =L2z2[−(1−zdzd0)dτ2 +(1−zdzd0)−1dz2 +dxidxi]. (2.54)If we compactify a spatial direction and let j take on fewer values than i, the re-sulting metric is:ds2 =L2z2[−(1−zdzd0)dτ2 +(1−zdzd0)−1dz2 +dx jdxj +dθ 2]. (2.55)An interesting procedure, that would not have worked for any of the previous met-36rics, is available to be used on (2.55). We may Euclideanize, exchange the θ circlewith the τ circle and switch back to a Minkowskian signature. This yields a space-time known as the AdS soliton without us having to solve Einstein’s equationsagain.ds2 =L2z2[−dτ2 +(1−zdzd0)−1dz2 +dx jdxj +(1−zdzd0)dθ 2](2.56)The deconfinement we saw earlier was a transition between two spacetimes thatshared the same boundary: empty AdS and the AdS black hole. We now have(2.55) and (2.56) competing for the same boundary. However, the horizon positionz0 has a very different interpretation in (2.56) because it causes the θ circle toshrink to zero size. A horizon that observers can safely cross changes the signatureaccording to (−,+,+, . . . ,+,+) 7→ (+,−,+, . . . ,+,+). On the other hand, if weallowed z > z0, we would see (−,+,+, . . . ,+,+) 7→ (−,−,+, . . . ,+,−). Becausethis is a Lorentzian theory, z0 is simply a point where the spacetime ends. Thisenduring scale, called the infrared wall, is what leads to a mass gap [56].The Witten model with these two backgrounds is the typical arena for seeingplasma balls. These were conjectured [7] based on the observation that stable do-main walls should exist between solutions like (2.55) and (2.56). Roughly, sucha domain wall is constructed by choosing a special direction y and making z0 afunction of y. Choosing this function appropriately, the bulk metric can be made tolook like the black hole at y = ∞ and the soliton at y = −∞. This solution, whichcannot be found analytically, may look like the one in Figure 2.4. In order for itto be stable, the pressure of the deconfined phase must be small enough to balancethe domain wall tension at some temperature. Intuitive arguments for this are givenin [7] with the final confirmation being numerical. This process can be repeated tofurther localize the domain wall. Instead of going from the confining vacuum aty = −∞ to the deconfined plasma at y = ∞, one may change the solution so thatit goes from confined to deconfined and back [56]. This can also be done usingdirections other than y to make the area of the black hole horizon finite. The blackhole made in this way decays in a process that looks like some combination ofshrinking in y and hitting the infrared wall in z. The field theory state dual to this37yzx yzx(a) Separate solutionsyz(b) InterpolationFigure 2.4: The green sheet is the boundary at z = 0. While Poincare´ AdSlooks like empty space ending on this boundary, the black hole and soli-ton solutions are more interesting. Instead of enclosing a region of finitevolume, the horizon is a sheet at z = z0 like that of a black brane. Toremind us that there is a Scherk-Schwarz circle at each point, we havedrawn the horizon as a set of black circles inside cylinders that extendinfinitely far to the left. These cylinders become cigar shaped regions inthe soliton solution that has the infrared wall. On the right, we show aninterpolation between these behaviours. For y =∞, the IR wall does notexist. For finite positive y it is behind the horizon and for negative y it isin hole near the IR is called a plasma ball. The dual decay process consists ofhadrons leaving the ball and travelling outwards. Because they are travelling intoa confining vacuum, they must be color singlets, leading to a lifetime proportionalto N2 [7].When deriving (2.46), the black holes we discussed were dual to energy eigen-states of SYM on S3. In analogy with free theories (whose momentum eigenstatesare completely delocalized), these states have uniform energy density of order N2R4on the whole sphere. The situation is very different for plasma balls. They havenon-uniform energy densities like those in Figure 2.5 because their dual black holescome from an interpolation of gravity saddle points. When we construct plasmaballs (in a completely different manner), we should keep in mind that small com-pact directions and infrared walls are likely to appear in the corresponding ge-ometries. As a check, it is interesting to see what goes wrong when trying toconstruct a plasma ball for Super-Yang Mills on S3. In principle, one could pre-pare a state in the CFT that has Figure 2.5’s energy density at t = 0. Rather than38xyE(a) Domain wallxyE(b) Plasma ballFigure 2.5: On the field theory side, the domain wall shows up as a suddenjump in the energy density. The minimum is close to zero while themaximum is the expected energy at the deconfinement temperature. Theexistence of a plasma ball is the additional assumption that we can makethis domain wall “wrap around”. The decay of the resulting object mightlook like diffusion governed by the heat equation.a thermalization process dual to Hawking radiation, this state’s future is governedby the phenomenon of collective oscillations [57]. When quantized on a sphere,the generators of conformal transformations obey the same commutation relationsas traditional raising and lowering operators. Rewriting them as Li+ and Li−, [57]constructed undamped oscillating states by applying a function of them to a densitymatrix: ρ 7→N g(L1±, . . . ,Ld±)ρg†(L1±, . . . ,Ld±). In analogy with coherent statesof the harmonic oscillator, explicit functions were given such as the simplest one:g(Li+,Li−)= eαLi++βLi− . (2.57)39When α = −β ∗, (2.57) is unitary and N = 1, but [57] gave normalization con-stants for other α and β as well. Crucially, the AdS isometry dual to (2.57) is nomore complicated than a boost. This allows its effect on strongly coupled statesto be found with the AdS / CFT correspondence. The example considered in [57]starts with a three-dimensional spacetime known as the BTZ black hole [58]:ds2 =−(r2R2−r20R2)dt2 +(r2R2−r20R2)−1dr2 +r2R2dx2 . (2.58)As before, the boundary theory dual to this has a uniform stress-energy tensor whencompactified on the xR ∼xR +2pi sphere:Ttt = Txx =r2016piGR3Ttx = Txt = 0 . (2.59)We could use AdS3 / CFT2 analogues of (2.36) and (2.43) to replace r0 and G withxEFigure 2.6: This localized lump of energy is certainly something like the t = 0slice of a plasma ball. We are claiming that if it forms on a sphere, itsevolution is too dynamic to be considered that of a plasma ball. Itsmotion cannot be undone by a boost because this turns it back into theuniform profile of (2.59).40gauge theory expressions above. Because AdS has a boundary, boosting (2.58) toa velocity of v yields a black hole that oscillates about the origin indefinitely. TheCFT state dual to this bouncing black hole has the stress-energy tensor:Ttt = Txx =132piGR3[(1− v2)(R2 + r20)(vcos(t−xR)−1)2 +(1− v2)(R2 + r20)(vcos(t+xR)−1)2 −2R2]Ttx = Txt =v(1− v2)(R2 + r20)8piGR3[sin(tR)sin(xR)(vcos(tR)cos(xR)−1)(vcos(t−xR)−1)2 (vcos(t+xR)−1)2].Thus we see that a valid CFT solution having Figure 2.6 as its t = 0 energy densityis not a meta-stable ball at a fixed position, but a stable flow with an oscillatingposition [57].41Chapter 3Treating energy stochasticallyOur goal is to model certain features of an interacting field theory, without recourseto what the specific interaction is. One way to accomplish this is to construct amodel that is based on the system’s density of states. A thermodynamic quantitylike this is easier to understand than the Hamiltonian because it is only an indica-tion of the spectrum of the Hamiltonian. The result of our derivation will be anevolution equation for the energy density at each point in space.3.1 Main equationsTo begin our analysis, we consider a cubic lattice of identical sites in d dimensionsand keep track of the number of units of energy that can be found on each site.Following [1], we will write {nr} to mean n1 units on site 1, n2 units on site 2, n3units on site 3, etc. Each configuration has a certain probability of being realized.This probability is P({Nr}(t) = {nr}). Uppercase letters have been used for ran-dom variables with the lowercase versions denoting specific values. However, wewill often shorten this to P({nr}). In a stochastic process with continuous time,the probabilities as a function of time obey the master equation [59]:∂P({nr})∂ t = ∑{n′r}[P({n′r})W{n′r}→{nr}−P({nr})W{nr}→{n′r}]. (3.1)42The W quantities which determine the process are called the transition rates andare defined by:W{nr}→{n′r} = limτ→0P({Nr}(t + τ) = {n′r}|{Nr}(t) = {nr})τ .To convert (3.1) into something more concrete, we will make three physical as-sumptions: local energy conservation, detailed balance and entropic dominance.3.1.1 Physical assumptionsInline with our first assumption, we declare that any transition which is nonlocal ordoes not conserve energy has a W value of zero. In the transition rates that are left,energy is transferred between two sites and those sites must be nearest neighbours.Instead of listing all {n′r} configurations that can be reached from {nr}, we maysimply choose a pair of sites 〈a,b〉 and a number k to transfer between them. Themaster equation therefore becomes∂P({nr})∂ t = ∑〈a,b〉∑k 6=0[P(. . . ,na + k,nb− k, . . .)W(na+k,nb−k)→(na,nb)−P(. . . ,na,nb, . . .)W(na,nb)→(na+k,nb−k)]. (3.2)We will not work directly with probabilities, but rather the expectation of a partic-ular site’s energy:n¯c ≡ 〈Nc〉= ∑{nr}P({nr})nc . (3.3)The next step is to differentiate (3.3) and substitute (3.2):∂ n¯c∂ t = ∑{nr}nc∂P({nr})∂ t= ∑{nr}nc ∑〈a,b〉∑k 6=0[P(. . . ,na + k,nb− k, . . .)W(na+k,nb−k)→(na,nb)−P(. . . ,na,nb, . . .)W(na,nb)→(na+k,nb−k)].Every P(. . . ,na+k,nb−k, . . .)W(na+k,nb−k)→(na,nb) in the sum is a P(. . . ,na,nb, . . .)W(na,nb)→(na+k,nb−k)for some other {nr} and the negative k value. As long as nc does not appear, these43cancel with the same coefficient. We may therefore let a = c and reindex.∂ n¯c∂ t = ∑〈b,c〉∑k 6=0∑{nr}[ncP(. . . ,nb + k,nc− k, . . .)W(nb+k,nc−k)→(nb,nc)−ncP(. . . ,nb,nc, . . .)W(nb,nc)→(nb−k,nc+k)]= ∑〈b,c〉∑k 6=0∑{nr}[(nc + k)P(. . . ,nb,nc, . . .)W(nb,nc)→(nb−k,nc+k)−ncP(. . . ,nb,nc, . . .)W(nb,nc)→(nb−k,nc+k)]= ∑〈b,c〉∑k 6=0∑{nr}kP(. . . ,nb,nc, . . .)W(nb,nc)→(nb−k,nc+k)=〈∑〈b,c〉∑k 6=0kW(nb,nc)→(nb−k,nc+k)〉≈ ∑〈b,c〉∑k 6=0kW(n¯b,n¯c)→(n¯b−k,n¯c+k) (3.4)We have yet to show that it is safe to replace random variables by their expectationsin the last step. Since (3.4) is still quite general, further work is required to narrowdown our choices for W .The configurations {nr} represent collections of several microstates. Introduc-ing the function ρ(nr) giving the number of ways for site r to have energy nr, itis easy to count the number of ways in which our configurations can be realized.There are ∏r ρ(nr) microstates with the distribution {nr}. The most familiar situ-ation in statistical mechanics is that of thermal equilibrium. State µ in equilibriumis achieved with probability P(µ) = 1Z e−βE . Since this is just one microstate, weshould add up a sufficient number of them to getP({nr}) =1Ze−βE∏rρ(nr) . (3.5)These are equilibrium probabilities so the master equation should vanish when theyare inserted. This condition provides a constraint on the possible transition ratesbut we will make use of a stronger condition; the principle of detailed balance.Detailed balance, which holds for reversible Markov chains, states that all terms in44(3.2) should separately vanish in equilibrium instead of just the entire sum:P(. . . ,na+k,nb−k, . . .)W(na+k,nb−k)→(na,nb) =P(. . . ,na,nb, . . .)W(na,nb)→(na+k,nb−k) .(3.6)This principle was famously used by Einstein to predict spontaneous emission ratesbefore quantum field theory had been developed [60]. Substituting (3.5) into (3.6),our condition becomesρ(na + k)ρ(nb− k)W(na+k,nb−k)→(na,nb) = ρ(na)ρ(nb)W(na,nb)→(na+k,nb−k) . (3.7)There are many solutions to this system of equations but some make more sensethan others.One solution to (3.7) has W(na+k,nb−k)→(na,nb) proportional to the number of finalstates ρ(na)ρ(nb). This type of transition rate is the one most compatible with theergodic principle. When fluctuations are completely thermal, a higher number offinal states should be the only thing favouring one transition over another. We writeW(na,nb)→(na+k,nb−k) ∝C(na,nb)ρ(na + k)ρ(nb− k)where there can be some additional dependence on k. Substituting this into (3.7),we findC(na,nb) =C(na + k,nb− k) .Using this relation repeatedly, we may set k equal to na or nb, telling us that C isonly a function of the total energy. For this to be valid, any configuration must bereachable from the configuration obtained by having one site shift all of its energyto a neighbour. This is the same as saying that there are no superselection sectors.We now haveW(na,nb)→(na+k,nb−k) ∝C(na +nb2)ρ(na + k)ρ(nb− k) (3.8)where we have included the factor of 12 for later convenience. We should note thatW(na+k,nb−k)→(na,nb) is not proportional to the number of configurations that have naon site a and nb on site b. This would be ρ(na)ρ(nb)∏r/∈{a,b}ρ(nr). A transition45rate involving all of these factors would be inconsistent with the nearest neighbourlogic we have been using. The only transition we have allowed is one in which siteb sends k units of energy to site a. If a transition rate compatible with the ergodicprinciple also depends linearly on ρ(nc) for some other site c, this is not a transitionrate for (na,nb)→ (na + k,nb− k) but rather (na,nb,nc)→ (na + k,nb− k,nc). Inother words, site c has itself undergone a transition to some other internal state thatkeeps the same energy nc. This amounts to two transition happening in the sametimestep. Moreover, there is no way to tell that this is indeed a nearest neighbourtransition paired with a transition between internal states far away. It could havebeen site b sending k units of energy to site c followed immediately by c sending kunits of energy to site a, thus violating locality.3.1.2 The continuum limitThe equation we wish to build on is∂nc∂ t = ∑〈b,c〉∑k 6=0kW(nc,nb)→(nc+k,nb−k) (3.9)where the transition rates are given by (3.8). So far, we have been assuming thatthe energy and the spatial co-ordinate both vary by discrete amounts. One way towrite this is to have site c labelled by x which means that site b is x±ae for someunit vector e. To consider a continuous version of (3.9), the lattice constant a mustapproach zero. Additionally, the sum over k 6= 0 must become a sum over ±ε withε also going to zero. Our formula (3.8) becomesW(E(x),E(x+ae))→(E(x)+ε,E(x+ae)−ε) =C(E(x)+E(x+ae)2)ρ(E(x)+ε)ρ(E(x+ae)−ε)(3.10)where we use E instead of n to make it clear that we are talking about energydensities that are being incremented continuously. One should keep in mind that Chas units of inverse time in order for W to be a rate. Since our differential equationfor the energy density is now a function of the small parameters ε and a, a usefulapproximation to it can be derived with a Taylor expansion.46Using (3.10) in the continuous version of (3.9), we have∂E(x)∂ t = ε ∑e∈{±e1,...,±ed}[W(E(x),E(x+ae))→(E(x)+ε,E(x+ae)−ε)−W(E(x),E(x+ae))→(E(x)−ε,E(x+ae)+ε)]= ε ∑e∈{±e1,...,±ed}C(E(x)+E(x+ae)2)[ρ(E(x)+ ε)ρ(E(x+ae)− ε)−ρ(E(x)− ε)ρ(E(x+ae)+ ε)] . (3.11)We will define∂E∂ t = X(ε,a) = εX˜(ε,a)in which case the relevant Taylor series becomes∂E∂ t =∞∑m=0∞∑n=0εmanm!n!∂m+nX∂εm∂an∣∣∣∣a=0ε=0. (3.12)We can see from (3.11) that X˜(0,a) = 0 = X˜(ε,0), so any term in (3.12) that sur-vives, must involve at least three derivatives of X : two with respect to ε and onewith respect to a. In fact, the number of derivatives we need to take is even higher.Differentiating something like ρ(E(x+ ae)) with respect to a would contribute a∂ρ∂E ∂iEei term inside the sum. If we add up the ith components of e where e runsover all positive and negative standard basis vectors, the result is zero. This meanswe need at least one more derivative with respect to a and the approximation weseek is:∂E∂ t =14ε2a2 ∂4X∂ε2∂a2∣∣∣∣a=0ε=0. (3.13)We will use the abbreviated notation ρ+ = ρ(E(x+ae)) and C+ =C(E(x)+E(x+ae)2)which satisfy∂ρ+∂a = e j∂ jρ+∂C+∂a =12e j∂ jC+ . (3.14)Also, the derivatives with respect to ε are not calculated here but in the appendix.47Picking up from where the appendix leaves off,∂ 4X∂ε2∂a2∣∣∣∣a=0ε=0= 2∂ 3X˜∂ε∂a2∣∣∣∣a=0ε=0= 4 ∑e∈{±e1,...,±ed}∂ 2∂a2[C+(ρ+dρdE−ρ dρ+dE)]∣∣∣∣a=0= 4 ∑e∈{±e1,...,±ed}ei∂∂a[12∂iC+(ρ+dρdE−ρ dρ+dE)+C+(∂iρ+dρdE−ρ∂idρ+dE)]∣∣∣∣a=0= 4[C(∂i∂ jρdρdE−ρ∂i∂ jdρdE)−∂iC(ρ∂ jdρdE−∂ jρdρdE)]∑e∈{±e1,...,±ed}eie j= 8[C(∂i∂iρdρdE−ρ∂i∂idρdE)−∂iC(ρ∂idρdE−∂iρdρdE)]. (3.15)It is not immediately obvious but (3.15) simplifies to a more compact expressioninvoling a logarithm. If we expand−Cρ2∂idlogρdE= −Cρ2∂idρdEρ= Cρ2∂iρ dρdE −ρ∂idρdEρ2= C(∂iρdρdE−ρ∂idρdE),we get something whose ∂i derivative is (3.15). This shows that∂E∂ t =−2ε2a2∂i(C(E)ρ2(E)∂idlogρ(E)dE). (3.16)Checking the dimensions of (3.16), the left hand side is an energy density over atime. On the right hand side, we have an energy density in the form of ε becausethe other ε cancels with the dE. We also have an inverse time because the functionC had inverse time units. The a2 cancels with the two spatial derivatives. Fromnow on, we will drop unknown dimensionful parameters by absorbing them into48the time. The main differential equation of our model is∂E∂ t =−∂i(Cρ2∂idlogρdE)(3.17)in which C is assumed to be a dimensionless function. Common choices for it willbe 1 and ρ−2.3.2 Interesting featuresOur continuum limit equation has particularly nice things to say about a systemwith microcanonical phases like (2.46). At least two of these phases only appearat energies that are large compared to the spatial volume. After considering someinsights in [1] concerned with the case ∂E∂ t = 0, we will see that large energies arerequired to even trust the model at a basic level.3.2.1 Static situationsA special role is played by the density of states whose logarithm is linear in E.This is the Hagedorn density of states that we saw appearing in string theory andSuper Yang-Mills: ρ(E) ∝ eβHE . In this case dlogρ(E)dE = βH, a constant. The ∂iacting on this constant will set the left hand side of (3.17) to zero. Under Hagedornbehaviour, the energy distribution E(x, t) does not change with time.The gauge theories with holographic duals have a Hagedorn regime as well asother phases. As stated before, we expect logρ(E) ∝ Eα where we could haveα < 1, α > 1 or α = 1 depending on the energy. Since the dynamics are frozenwith a purely Hagedorn density of states, we expect changes in the energy to takeplace very slowly if α = 1 is the widest phase. The α = 1 phase can equivalentlybe described as the energy range for which the inverse temperature β (E) is flat.Using the fact that logρ(E) is the microcanonical entropy, we can rewrite ourmain equation in terms of β (E) as well:d logρ(E)dE=dS(E)dE= β (E)49and our equation becomes∂E∂ t =−∂i(Cρ2∂iβ). (3.18)The phases can be characterized by whether β is decreasing (α < 1), increasing(α > 1) or neither (α = 1).With a Hagedorn density of states, ∂E∂ t vanishes for any energy distribution.Conversely with a uniform energy distribution, ∂E∂ t vanishes for any density ofstates. This equilibrium distribution may be stable or unstable depending on thephase we are in. We will decompose the energy asE(x, t) = E0 + E˜(x, t)where E˜ is small, allowing us to keep only one power of it in the PDE (3.18). First,∂iβ (E) ≈ ∂i(β (E0)+ E˜dβ (E0)dE)=dβ (E0)dE∂iE˜ .This expression with one power of E˜ is multiplied by C(E)ρ2(E). A first orderexpansion of C(E)ρ2(E) would give an overall result that is second order in E˜ sowe only expand it to zeroth order:∂ E˜∂ t ≈−C(E0)ρ2(E0)dβ (E0)dE∂i∂iE˜ . (3.19)This is either the heat equation or the reverse heat equation depending on whetherthe overall coefficient is negative or positive. The sign of dβ0dE is what matters be-cause C and ρ are positive functions. Agreeing with our earlier intuition about theentropic dynamics of energy, we have the following three cases:• dβdE < 0 is a decreasing inverse temperature, a concave entropy and a log-concave density of states. It leads to diffusion or inhomogeneities that de-crease with time due to the heat equation.• dβdE > 0 is an increasing inverse temperature, a convex entropy and a log-50convex density of states. It leads to clustering or inhomogeneities that in-crease with time due to the reverse heat equation.• dβdE = 0 is a constant inverse temperature, a linear entropy and a Hagedorndensity of states. It leads to static behaviour.Understanding the detailed properties of the diffusion and to a lesser extent theclustering caused by this PDE will be the focus of the next chapter.3.2.2 Mean-field variancesA loose end in this chapter has been the assumption that we may work only withexpected values in (3.4). In general, mean-field approximations may be used onquantities that have a small variance. An energy with a small variance is also one ofthe desired features of our model. After all, the model is an attempt at connectingthe excitations of field theory degrees of freedom to Einstein gravity, somethingthat is completely deterministic.If variances are initially small, we want to make sure that they grow slowly sothat our model stays valid for a long time. Just as we derived an expression for ∂ n¯c∂ tfrom the master equation, we can repeat the calculation for ∂ n¯2c∂ t .∂ n¯2c∂ t = ∑{nr}n2c∂P({nr})∂ t= ∑〈b,c〉∑k 6=0∑{nr}[n2cP(. . . ,nb + k,nc− k, . . .)W(nb+k,nc−k)→(nb,nc)−n2cP(. . . ,nb,nc, . . .)W(nb,nc)→(nb−k,nc+k)]= ∑〈b,c〉∑k 6=0∑{nr}[((nc + k)2−n2c)P(. . . ,nb,nc, . . .)W(nb,nc)→(nb−k,nc+k)]= ∑〈b,c〉∑k 6=0∑{nr}(2knc + k2)P(. . . ,nb,nc, . . .)W(nb,nc)→(nb−k,nc+k)≈ ∑〈b,c〉∑k 6=0(2knc + k2)W(n¯b,n¯c)→(n¯b−k,n¯c+k) (3.20)51If we combine (3.20) with (3.4),σ2c = n¯2c− n¯2c∂σ2c∂ t =∂ n¯2c∂ t −2n¯c∂ n¯c∂ t= ∑〈b,c〉∑k 6=0k2W(n¯b,n¯c)→(n¯b−k,n¯c+k)= ∑〈b,c〉∑k 6=0k2C(n¯b + n¯c2)ρ(n¯b− k)ρ(n¯c + k)where we have substituted the W solution (3.8). This equation for the variance canbe examined in the continuum limit and the key is that we do not need as manyorders as ε2a2 in the subsequent Taylor expansion. The continuum limit is∂σ2(x)∂ t = ε2 ∑e∈{±e1,...,±ed}[W(E(x),E(x+ae))→(E(x)+ε,E(x+ae)−ε)+W(E(x),E(x+ae))→(E(x)−ε,E(x+ae)+ε)]= ε2 ∑e∈{±e1,...,±ed}C(E(x)+E(x+ae)2)[ρ(E(x)+ ε)ρ(E(x+ae)− ε)+ρ(E(x)− ε)ρ(E(x+ae)+ ε)] (3.21)where the positive sign is due to the fact that we have k2 instead of k. Again, define∂σ2∂ t = X(ε,a) = ε2X˜(ε,a) .When Taylor expanding X , we need at least two derivatives with respect to ε be-cause of the ε2 prefactor. This is all we need because X˜(0,0) is nonzero.∂σ2∂ t =12ε2 ∂2X∂ε2∣∣∣∣a=0ε=0= ε2X˜(0,0)= 2ε2 ∑e∈{±e1,...,±ed}Cρ2= 4dε2Cρ2 (3.22)Unlike (3.16), it is not the same unknown function that appears on the left and right52hand sides of (3.22). One must first solve for E(x, t) in order to solve for σ(x, t).Since the expression for ∂E∂ t has a small prefactor of a2 while that for ∂σ2∂ t doesnot, it would seem that the rate of change of the variance is parametrically larger,something we wished to avoid. However, it does not make sense to compare thesequantities directly. The energy has the same units as its standard deviation so weshould compare ∂σ∂ t to∂E∂ t or more conveniently,∂σ2∂ t to∂E2∂ t .∂σ2∂ t∂E2∂ t=∂σ2∂ t2E ∂E∂ t(3.23)Factors of ε2 in the numerator and denominator cancel leaving Ea2 in the denomi-nator. This tells us that such a ratio of derivatives can indeed be small if the energyis large enough. In other words, a PDE like (3.17) can be trusted to model high en-ergy phenomena. Thinking about gravity, this includes the extreme environmentsof black holes but not the everyday motion of test particles around them. On amore practical level, it would be difficult to even write down the equation (3.17)if we were concerned with it holding for low energies. For the field theories weare interested in, only asymptotic expressions are known for the density of states.Even for situations in which the exact number of states is known for all energies,this ρ(E) is not continuous.53Chapter 4Nonlinear diffusionWe now cover some of the properties of equations like (3.17) that are known analyt-ically. The main assumption we will use throughout this chapter is that β is weaklydecreasing, i.e. dβdE ≤ 0. This is a slight departure from the strongly coupled gaugetheory result as shown in Figure 4.1. If C(E) = ρ−2(E),Eβ(a) Sharp phasesEβ(b) Smooth phasesEβ(c) Purely diffusiveFigure 4.1: To emphasize the four phases expected in holographic gauge the-ories, we have drawn a piecewise β (E) function. It is more realistic toexpect an approximation to this function that is differentiable wheneverE > 0. The function plotted in the middle figure is still difficult to workwith because there is a small range of energies for which it is increas-ing. Assuming that this phase is negligible is the best way to predict thebehaviour of our nonlinear PDE.∂E∂ t =−∂i(C(E)ρ2(E)∂iβ (E))(4.1)54becomes∂E∂ t =−∂i∂iβ˜ (E) (4.2)where β˜ (E) = β (E). We will in fact consider (4.2) regardless of C. This is because(4.1) is always (4.2) for some other β˜ . Simply define β˜ ′ = Cρ2β ′. Because C andρ are positive functions, β˜ is decreasing if and only if β is. Therefore we will dropthe tilde and write∂E∂ t =−∆β (E) (4.3)from now on. Of course the β (E) in (4.3) no longer has to be of the form plottedin Figure 4.1 but this will be unimportant for most of the results that follow.4.1 Basic properties on a bounded domainWe will start by assuming that x ∈ Ω where Ω is an open, bounded domain in Rd .This allows the initial energy density E(x,0) to be integrable without decaying tozero. As shown in Figure 4.2, a potential problem with (4.3) is thus avoided be-cause the low energies for which β diverges are not realized. The initial conditionxEEmaxEmin(a) Energy distributionEβEminEmax(b) Realized range of energiesFigure 4.2: Resricting the size of the domain allows us to avoid the vanish-ingly small energies for which our PDE no longer applies.will be denoted E0(x) reflecting the fact that we choose an energy distribution toevolve forward in time, i.e. E0 is an input to the Cauchy problem.554.1.1 Conservation of energyEnergy conservation was one of the properties we demanded from the start. Aswith any Cauchy problem, whether or not energy is conserved depends on theboundary conditions. As with the heat equation, Neumann boundary conditionsare the appropriate ones to consider. When discussing these mathematical results,we will use “mass” to refer to the integral of E(x, t) over space rather than a gap inthe spectrum of a field theory.Theorem 1. If E solves∂E∂ t (x, t) =−∆β (E(x, t)) (x, t) ∈Ω× (0,∞)E(x,0) = E0(x) x ∈Ω∇E(x, t) ·n = 0 (x, t) ∈ ∂Ω× (0,∞)then M(t) =∫Ω E(x, t)dx is constant.Proof. We can show that the derivative of M is zero using Green’s first identity.dMdt=∫Ω∂E∂ t dx=∫Ω∆β (E)dx=∫Ω∇ ·(β ′(E)∇E)dx=∫Ωβ ′(E)∆E +∇(β ′(E))·∇Edx=∫∂Ωβ ′(E)(∇E ·n)dSx= 0 .Dirichlet boundary conditions would not lead to conserved energy unless we finelytuned β ′ to be zero on the boundary.4.1.2 The maximum principlePerhaps the most ubiquitous tool in the study of elliptic and parabolic equations isthe maximum principle. Although it is sometimes introduced as a tool for studying56linear equations, many nonlinear versions of it have appeared over the years [61].Our proof of a suitable maximum principle will be very similar to the one in [62].Theorem 2. Suppose that ∂u∂ t ≥ −∆β (u) and∂v∂ t ≤ −∆β (v) for a monotonicallydecreasing β . If u(x,0) > v(x,0) for all x ∈ Ω, there does not exist a spacetimepoint (x0, t0) ∈Ω× (0,∞) for which u(x0, t0)< v(x0, t0).Proof. First suppose that w ∈ C2(Ω× (0,∞)) is initially non-negative but not al-ways non-negative. Then there must exist some point (x0, t0) ∈ Ω× (0,∞) suchthat w(x0, t0)< 0. We can let x0 be the position of the minimum of w(·, t0) so that∇w(x0, t0) = 0 and ∆w(x0, t0) ≥ 0 are also satisfied. If t0 is the first time such apoint occurs, ∂w∂ t (x0, t0)≤ 0.Now let w(x, t) = e−At [β (v(x, t))−β (u(x, t))] where A is some positive con-stant. Saying that w is non-negative is the same as saying that u ≥ v because βis monotone. Therefore if u ≥ v initially but not always, the above says that theremust be a point (x0, t0) ∈Ω× (0,∞) such that:• β (u(x0, t0))> β (v(x0, t0))• β ′(u(x0, t0))∇u(x0, t0) = β ′(v(x0, t0))∇v(x0, t0)• −∆β (u(x0, t0))≥−∆β (v(x0, t0))• β ′(u(x0, t0)) ∂u∂ t (x0, t0)−β′(v(x0, t0)) ∂v∂ t (x0, t0)≥A [β (u(x0, t0))−β (v(x0, t0))]Since u is a supersolution and v is a subsolution, ∂u∂ t ≥−∆β (u) and∂v∂ t ≤−∆β (v)can be combined to give ∂∂ t (u− v)+∆ [β (u)−β (v)] ≥ 0. On the other hand, ourfour conditions above can be combined into an inequality that contradicts thistelling us that u cannot drop below v if u(x,0)≥ v(x,0).First we use the fact that ∆β (u(x0, t0)) ≤ ∆β (v(x0, t0)) to write ∂∂ t (u(x0, t0)−v(x0, t0))+∆ [β (u(x0, t0))−β (v(x0, t0))] ≤ ∂∂ t (u(x0, t0)− v(x0, t0)). We must nowshow that the difference of the time derivatives is less than zero. To do this, wenote that at least one of the following must be true:β ′(u(x0, t0))∂u∂ t (x0, t0)−β′(v(x0, t0))∂v∂ t (x0, t0)≤ β′(u(x0, t0))∂∂ t (u(x0, t0)− v(x0, t0))β ′(u(x0, t0))∂u∂ t (x0, t0)−β′(v(x0, t0))∂v∂ t (x0, t0)≤ β′(v(x0, t0))∂∂ t (u(x0, t0)− v(x0, t0)) .57The first one is true if ∂v∂ t (x0, t0) and β′(v(x0, t0))− β ′(u(x0, t0)) have the samesign. The second one is true if they have different signs. Use β ′0 to denote whateverprefactor appears in the correct statement, either β ′(u(x0, t0)) or β ′(v(x0, t0)). Thekey is that this is a negative constant. The last inequality in our list of four can turnthis intoβ ′0∂∂ t (u(x0, t0)− v(x0, t0)) ≥ β′(u(x0, t0))∂u∂ t (x0, t0)−β′(v(x0, t0))∂v∂ t (x0, t0)≥ A [β (u(x0, t0))−β (v(x0, t0))] .Dividing through by β ′0, we get∂∂ t (u(x0, t0)− v(x0, t0))≤Aβ ′0[β (u(x0, t0))−β (v(x0, t0))] .Using the first inequality in the list of four, we see that this is less than zero, com-pleting the proof.Even though this could still be generalized further [63], it is already more gen-eral than the maximum principle. To get the maximum principle, let v be a solution(a special case of a subsolution) and u be max{v(x,0) : x ∈ Ω¯}. Since u is a con-stant, it is also a solution and therefore a supersolution. The theorem above nowtells us that u must continue to upper bound v at all later times which means themaximum of v decreases with time. As always, this is equivalent to the minimumprinciple which states that the minimum increases with time. We have already seenin (3.19) that our PDE turns into the heat equation when the energy distribution isclose to uniform. The maximum principle tells us that our PDE is similar to theheat equation in some respects for general energy distributions as well.In addition to talking about the value of the maximum of E, we can gain someinformation about its position. If our initial condition E0 is spherically symmetric,58our equation gives∂E∂ t = −r1−d ∂∂ r(rd−1∂β (E)∂ r)∂∂ t∂E∂ r = −∂∂ r[r1−d∂∂ r(rd−1∂β (E)∂ r)]= −∂∂ r[d−1rβ ′(E)∂E∂ r +β′′(E)(∂E∂ r)2+β ′(E)∂2E∂ r2]=d−1r2β ′(E)∂E∂ r −d−1rβ ′′(E)(∂E∂ r)2−d−1rβ ′(E)∂2E∂ r2 −β′′′(E)(∂E∂ r)3−3β ′′(E)∂E∂ r∂ 2E∂ r2 −β′(E)∂ 3E∂ r3 .Now suppose that E0 achieves its maximum at the origin and has no other localminima or maxima. In other words ∂E0∂ r < 0 for r 6= 0. If∂E∂ r were to becomepositive away from the origin at some later time, it would have to first vanish. Also,since an extreme point has not formed yet, (r0, t0) is an inflection point satisfying∂ 2E∂ r2 (r0, t0) = 0 and∂ 3E∂ r3 (r0, t0) < 0. Plugging these into our expression above, wesee that ∂∂ t∂E∂ r (r0, t0) = −β′(E(r0, t0)) ∂3E∂ r3 (r0, t0) < 0. If the radial derivative ofE ever gets to zero, it starts decreasing again and never passes zero to becomepositive. Therefore the origin is the only local extremum at all times.4.1.3 Existence of a steady-stateWe now have everything we need to discuss the behaviour of solutions in the limitof infinite time. By setting ∂E∂ t equal to zero, we can see that the limiting energydistribution satisfies {∆β (E(x)) = 0 x ∈Ω∇E(x) ·n = 0 x ∈ ∂Ω.The only solutions to the Neumann problem for Laplace’s equation are constants.Therefore β (E) must be constant in space if ∂E∂ t = 0. There are two ways forthis to happen. One is for E to solely occupy the Hagedorn regime. If not, β isinvertible in a neighbourhood of at least one energy in the steady-state and E must59be constant. By energy conservation, the value of this constant is of courseE¯0 =1|Ω|∫ΩE0(x)dx .In other words, we know what limt→∞ E(·, t) must be if it exists but it is not yetobvious that it exists. Even for an energy distribution whose extrema stay in thesame place and smooth out over time, it is possible for the intermediate regions toconstantly oscillate without ever converging to any function. Our proof that thisdoes not happen for the (4.3) will compare it to the heat equation.xEFigure 4.3: The two energy distributions shown have equal minima, equalmaxima and equal masses. They also have the central peak as theironly extreme value. A solution to (4.3) that alternated between thesefunctions indefinitely would satisfy all of the properties that we haveproven so far but not have a steady-state.Theorem 3. Suppose that E solves∂E∂ t (x, t) =−∆β (E(x, t)) (x, t) ∈Ω× (0,∞)E(x,0) = E0(x) x ∈Ω∇E(x, t) ·n = 0 (x, t) ∈ ∂Ω× (0,∞)60for a spherically symmetric E0 with a central maximum as its only local extremepoint. If β ′(E) ≤ 0 becomes strict whenever Emin < E < E¯0, then E converges toits average.Proof. Letting BdR denote a ball of radius R, the mass satisfiesMR =∫BdREdxdMRdt=∫BdR∂E∂ t dx= −∫BdR∆β (E)dx= −∫Sd−1Rβ ′(E)∇E ·ndSx≤ 0 .The inequality came from the fact that the integrand is non-negative. The fluxis non-positive because the value of E must always decrease as we move awayfrom the origin and β ′(E) is non-positive by assumption. Let R(t,ε) be the largestradius at which E(R, t) = E¯0− ε . If R(t) ever reaches the radius of Ω, the solutionwill have converged to E¯0 by energy conservation and the maximum principle.Therefore assume the opposite and let R=max{R(t,ε) : t > 0}. Since E(R, t)≤ E¯0,we can make the integral larger by replacing β ′(E) with β ′(E¯0).dMRdt≤−β ′(E¯0)∫Sd−1R∇E ·ndSx < 0We now see that MR for our equation (4.3) shrinks more quickly than MR for a heatequation with diffusion constant −β ′(E¯0). The standard Fourier series methodshows that the solution to said heat equation with initial data E0 will converge toE¯0. Thus Sd−1R , the sphere where E¯0 is first achieved, eventually encloses a constantfunction. By energy conservation and the maximum principle, E must be equal tothe same constant outside this sphere as well. Therefore for any ε > 0, the energyat the edge always comes within ε of E¯0 which means that “the outside of thissphere” must have not existed all along.614.2 Time scalesSo far we have shown that the nonlinear diffusion we are interested in shares a num-ber of intuitive properties with linear diffusion: energy is conserved, peaks smoothout over time and reasonable configurations of energy will decay to a distributionthat is completely uniform. However, we still have relatively little information onhow quickly these energy distributions decay. Defining and bounding time scalesfor the diffusion will allow us to compare our model to what is already knownabout gravitational bound states.4.2.1 The concentration comparison theoremEβΦFigure 4.4: This shows how we must change the function that appears insidethe Laplacian in order to satisfy the hypotheses of the theorem. Bound-ing the time scales arising from β is no more difficult than bounding thetime scales arising from Φ.If we continued applying the basic properties, we would be able to derive somepowerful results that are found in the literature. One of these is the concentra-tion comparison theorem [64] which applies to equations of the form (4.3) calledfiltration-equations. The only difference is that filtration equations are typicallywritten ∂E∂ t = ∆Φ(E) where Φ is weakly increasing. A fundamental property of62EΦFigure 4.5: We were able to show that our filtration equation had a well de-fined steady-state by comparing it to a heat equation. With this linearfunction as our bounding function, an estimate for the decay time canalso be obtained. However, this would lead to a crude bound becausethe linear function above is not very similar to the true Φ.diffusion is that the mass contained within a fixed ball at the origin MR(t) shoulddecrease. The total mass M(t) which we found to be constant in time is M∞(t).The concentration comparison theorem, which we state below, is used to convertan inequality involving Φ to an inequality involving MR.Theorem 4. Let Φ1,Φ2 ∈C1(R) be increasing functions sending 0 to 0 such thatΦ′1 ≥ Φ′2. Suppose that u1 having mass M1,R(t) and u2 having mass M2,R(t) aresolutions to ∂u∂ t =∆Φ1(u) and∂u∂ t =∆Φ2(u) respectively with spherically symmetricdata. If M1,R(0)≤M2,R(0) for all R then M1,R(t)≤M2,R(t) for all R and t.To use this theorem, we must convert our function β (E) into something satis-fying Φ(0) = 0. This is easily done with Φ(E) = β (Emin)−β (E +Emin). Figure4.4 shows what kind of function Φ is if β is the function in Figure 4.1.∂∂ t (E +Emin) =−∆β (E +Emin)63is now the same as saying∂E∂ t = ∆Φ(E)where E0 is now some energy distribution equal to zero on ∂Ω. If we can findan exact solution to another filtration equation whose mass shrinks more quicklythan that of E, we will have found a “decay time” that is shorter than the one weare looking for. Similarly, a mass that shrinks more slowly would be associatedwith a longer “decay time”. Time scales can therefore be determined if we choosebounding functions that are “steeper” than Φ or “flatter” than Φ. The crudest thingwe could do is find a lower bound on the decay time by drawing a linear functionabove Φ. The next logical step is to find proper estimates for the decay time bycomparing our filtration equation to something more non-trivial.4.2.2 Estimates in one dimensionAs stated before, we are considering a density of states that has two microcanonicalphase transitions. We will call the energy of the first one EH for Hagedorn and theenergy of the second one EF for field theory. A useful definition of decay time forus will be the time required for the maximum of an energy distribution to descendfrom EF to EH. Specifically, we will look at the initial conditionE0(x) =EF |x1|, . . . , |xd |< a0 otherwise(4.4)and see how long it takes until E(x, t) < EH for all values of x. The steep verticaljump through the Hagedorn regime is a feature that (4.4) has in common withsharply peaked initial energy distributions. The sharply peaked functions do notneed to be constant at EF but this will happen anyway once they are allowed toevolve. The distributions in Figure 4.6 will flatten out in a relatively short timebecause diffusion dominates above EF but not below. This suggests that the decaytime for a general peak is dominated by the decay time of (4.4). Further evidencethat (4.4) is representative of more general initial conditions will appear in the nextchapter on numerics. Finding the decay time of (4.4) with respect to a generalfiltration function is still too hard, so we will pick a simple filtration function with64xEEFEHFigure 4.6: Except for the piece on top that will quickly diffuse, these en-ergy distributions are qualitatively similar to the step function we areconsidering.the goal of using the concentration comparison theorem. The one to pick isΦ(E) =E E < EHEH E ≥ EH. (4.5)If we take an energy profile solving ∂E∂ t = ∆Φ(E) and look at it at some instant oftime, the parts with energy above EH should be static and the parts with energybelow EH should be satisfying the heat equation. Energy crosses EH at a particulardistance x∗(t). Using the method of [1], we will construct a one-dimensional solu-tion where x∗ starts off at a and eventually shrinks to zero indicating that the decaytime T has been reached.E(x, t) =EF |x|< x∗(t)F(x, t) |x|> x∗(t)(4.6)where F solves the regular heat equation. Clearly F cannot be just any solutionto the heat equation. To obey the initial condition (4.4), we must have F(x,0) = 065for |x| > a. Also, we must impose conservation of energy. The mass containedbetween 0 and x∗(t) is x∗(t)EF. The mass everywhere else is∫ ∞x∗(t) F(x, t)dx. Wewant their rates of change to be equal and opposite soEFdx∗(t)dt= −ddt∫ ∞x∗(t)F(x, t)dx= F(x∗(t), t)dx∗(t)dt−∫ ∞x∗(t)∂F∂ t (x, t)dx= EHdx∗(t)dt−∫ ∞x∗(t)∆F(x, t)dx= EHdx∗(t)dt+∂F∂x (x∗(t), t)where we have differentiated under the integral sign. Therefore(EF−EH)dx∗(t)dt=∂F∂x (x∗(t), t) (4.7)is a necessary condition for the ansatz (4.6) to work. It is also sufficient as we showin the appendix. The initial condition for F that will give us these necessary andsufficient conditions is F(x,0) = AΘ(x−a) where A is a yet undetermined constantthat will turn out to be between EH and 2EH. Figure 4.7 shows the basic setup. Tosolve for F , we first note that the heat kernel isΦH(x, t) =1(4pit) d2e−x24t .The heat equation’s Cauchy problem is solved by taking the convolution of theinitial condition with the heat kernel. ThereforeF(x, t) =A√4pit∫ ∞−∞θ(y−a)e−(x−y)24t dy=A√4pit∫ a−∞e−(x−y)24t dy=A2(1+ erf(a− x2√t)). (4.8)66xEEHAEF(a) t = 0xEEHAEF(b) Some later timeFigure 4.7: The union of the red and purple curves is what we are lookingfor. It is (4.6), the solution to the (4.5) filtration equation for the (4.4)Cauchy data. The solution to the auxillary problem that we use to findit is the union of the blue and purple curves. This is F , the solution tothe heat equation for the Heaviside Cauchy data.Differentiating F is now straightforward and the position of the interface can befound by setting F(x, t) = EH. From this we obtain∂F∂x (x, t) = −A√4pite−(x−a)24t (4.9)x∗(t) = a−2√terf−1(2EHA−1)(4.10)and the ratio between (4.9) at the interface and the derivative of (4.10) is constant.∂F∂x (x∗(t), t)dx∗(t)dt=A√4pie−erf−1(2EHA −1)2erf−1(2EHA −1)We must set this equal to EF−EH. The A we obtain from doing so is given byA =2EH1+ erf(I)∈ [EH,2EH] (4.11)√piIeI2(1+ erf(I)) = EHEF−EH. (4.12)67The transcendental equation defining I can be solved because the range of the lefthand side includes all positive real numbers. The last thing we need to do is finda more explicit value of A for the case when EF  EH. If we were to consider√piIeI2(1+ erf(I)) = 0, the solution would simply be I = 0. Since EH is close tozero but not quite, it is appropriate to linearize the left hand side of (4.12) givingus√piI = EHEF−EH . Plugging this into (4.11) givesA≈2EH1+ erf(EH/√piEF−EH) .Finally, the time scale can be found by setting x∗(T ) = 0 or F(0,T ) = EH.T =14(aerf−1(2EHA −1))2≈14(a(EF−EH)EH/√pi)2≈pi4[aEFEH]2(4.13)4.2.3 Higher dimensional generalizationThe procedure above works in an arbitrary number of dimensions if the initial con-dition has energy EF inside a hypercube of side length 2a. We will take F(x,0) =Aθ(x1− a) . . .θ(xd − a), solve the heat equation for F with this Cauchy data andwriteE(x, t) =EF x1 < x∗1(t), . . . ,xd < x∗d(t)F(x, t) otherwise. (4.14)68Performing the convolution with the heat kernel is straightforward because the so-lution has a product form.F(x, t) =A(4pit) d2∫ ∞−∞. . .∫ ∞−∞θ(y1−a) . . .θ(yd−a)e−(x−y)24t dy1 . . .dyd=A(4pit) d2∫ a−∞. . .∫ a−∞e−(x1−y1)2+···+(xd−yd )24t dy1 . . .dyd=A2d(1+ erf(a− x12√t)). . .(1+ erf(a− xd2√t))(4.15)From this we see that the surface x∗(t) where F takes on the value EH is the locusof pointsx∗(t) =a−2√terf−1 (z1−1)...a−2√terf−1 (zd−1) (4.16)where ∏i zi = 2d EHA . Differentiating F and evaluating it on this surface, we get∂F∂xi(x∗(t), t) =2EHzi1√4pie−erf−1(zi−1)2 . (4.17)The analogue of (4.7) to the higher dimensional case is(EF−EH)dV ∗(t)dt=∫x∗(t)∇F(x, t) ·ndSx (4.18)where V ∗(t) is ad minus the volume enclosed by the surface (4.16) in the positiveorthant. This can be visualized in Figure 4.8. Clearly V ∗(0) = 0. If we want tocalculate the volume V ∗(t), we are looking for xi values that go from the x∗i (t)69curve to a instead of from 0 to the x∗i (t) curve. This is the same as saying thatz1 ∈[1,2dEHA]z2 ∈[1,2dz1EHA]...zd ∈[1,2dz1 . . .zd−1EHA].Each xi only depends on one zi so it is trivial to find the Jacobian determinant.xyE(a) Early timexyE(b) Later timeFigure 4.8: The curve where the blue sheet intersects the evolving surface isx∗(t). The area of the blue sheet between the surface and the co-ordinateplanes is a2−V ∗(t). We are instead calculating the area V ∗(t) which hasan equal and opposite rate of change.dxidzi= −√piteerf−1(zi−1)2∣∣∣∣dx1dz1. . .dxddzd∣∣∣∣ = (pit)d2d∏i=1eerf−1(zi−1)2Also, by multiplying all but one of these eigenvalues together, we can find the areaelement required to calculate the surface integral in (4.18). An expression for the70volume isV ∗(t) =∫ 2dEHA1. . .∫ 2dz1 ...zd−1EHA1∣∣∣∣dx1dz1. . .dxddzd∣∣∣∣dzd . . .dz1= (pit)d2∫ 2dEHA1. . .∫ 2dz1 ...zd−1EHA1d∏i=1eerf−1(zi−1)2dzd . . .dz1 . (4.19)We only need to know the length of ∇F to compute the surface integral. Workingthis out,∫x∗(t)∇F(x, t) ·ndSx =∫ 2dEHA1. . .∫ 2dz1 ...zd−2EHA1|∇F(x∗(t), t)|∣∣∣∣dx1dz1. . .dxd−1dzd−1∣∣∣∣dzd−1 . . .dz1=1√4pit(pit)d−12∫ 2dEHA1. . .∫ 2dz1 ...zd−1EHA12EH√d∑i=11z2ie−2erf−1(zi−1)2d−1∏i=1eerf−1(zi−1)2dzd−1 . . .dz1 (4.20)where we have arbitrarily chosen zd as the variable to be determined from z1, . . . ,zd−1.The fact that (4.19) is proportional to one more factor of t than (4.20) is what allows(4.18) to be obeyed. There is a problem with these expressions however that onlyrears its head when d > 1; not all points on the surface x∗(t) reach the co-ordinateplanes at the same time. In two dimensions for instance, the endpoint[a−2√terf−1(4EHA −1)a]will reach (0,a) before the midpointa−2√terf−1(√4EHA −1)a−2√terf−1(√4EHA −1)reaches (0,0). In other words, for late times, the volume (4.19) includes someregions outside the positive orthant. It is not correct to integrate xi values from ato the x∗(t) curve. We must integrate from a to max(x∗(t),0). The decay time71obtained by setting (4.15) to EH isT =14aerf−1(2(EHA) 1d −1)2.The expression (4.19) is most accurate when T is large, i.e. when A≈ 2dEH. If thisis true, we can obtain a reasonable approximation to this complicated integral bysetting V ∗(T )≈ ad . Doing this, we see that∫ 2dEHA1. . .∫ 2dz1 ...zd−1EHA1d∏i=1eerf−1(zi−1)2dzd . . .dz1≈(2√pi)derf−1(2(EHA) 1d−1)d.With this, EF−EH becomes much easier to express.EF−EH =∫ 2dEHA1 . . .∫ 2dz1 ...zd−2EHA1 2EH√∑di=11z2ie−2erf−1(zi−1)2 ∏d−1i=1 eerf−1(zi−1)2dzd−1 . . .dz1pid∫ 2dEHA1 . . .∫ 2dz1 ...zd−1EHA1 ∏di=1 eerf−1(zi−1)2dzd . . .dz1≈2EH√d(2√pi)d−1erf−1(2(EHA) 1d−1 −1)d−1pid(2√pi)2erf−1(2(EHA) 1d −1)d≈EH√piderf−1(2(EHA) 1d −1) lim2d EHA →1erf−1(2(EHA) 1d−1 −1)erf−1(2(EHA) 1d −1)d−1=EH√piderf−1(2(EHA) 1d −1)(dd−1)d−1In the first step, we have used the fact that√∑di=11z2ie−2erf−1(zi−1)2 ≈√d becausez1, . . . ,zd are very close to 1 while in the last step, we have used l’Hoˆpital’s rule.Linearizing the inverse error function in the denominator and solving for A, we see72that2dEHA=[1+EHEF−EH1√pid(dd−1)d−1]≈ 1 ,which is precisely the condition we needed in the first place to be able to say thatT was large and use this approximation. Therefore when EF EH, the time scaleT in d dimensions becomesT =pid4[aEF−EHEH(d−1d)d−1]2≈pid4[aEFEH(d−1d)d−1]2. (4.21)4.2.4 Upper and lower boundsIt is time to leverage these results to improve the decay time estimates for thefiltration function in Figure 4.5. To achieve the tightest possible bounds, we willmake sure that our more-steep and less-steep functions flatten out at EH. Figure4.9 shows what this looks like. The function on the left Φ1 has a slope Φ′1(0) equalto Φ′(0). The function on the left Φ2 has a slope Φ′2(0) equal to limE→E−H Φ′(E).When these slopes are determined, the concentration comparison theorem tells usthat our time scale will satisfypid4Φ′1(0)[aEFEH(d−1d)d−1]2≤ T ≤pid4Φ′2(0)[aEFEH(d−1d)d−1]2. (4.22)This is the main part of the text that will use a specific formula for β (E), so firstconsider the one that is valid when C(E) = ρ−2(E). Recall that this is β (E) =cEα−1 = β (Emin)E1−αmin Eα−1 in the lowest energy regime. The slopes that we mustinsert are simple:Φ′1(0) = −β ′ (Emin) = (1−α)β (Emin)Emin(4.23)Φ′2(0) = −β ′ (EH) = (1−α)β (Emin)Eα−2HEα−1min. (4.24)73EΦ1EHEF(a) Comparing Φ to Φ1EΦ2EHEF(b) Comparing Φ to Φ2Figure 4.9: The piecewise-linear bounding functions must relate to our filtra-tion function in this way. On the left, Φ1 could cross over to Hagedornbehaviour at some energy E1 ≥ EH and on the right, Φ2 could cross overto Hagedorn behaviour at some energy E2 ≤ EH. The tightest possibletime scale bounds are achieved when E1 = EH = E2.Plugging (4.23) and (4.24) into (4.22), this time scale is now solved:pid4(1−α)β (Emin)EminE2H[aEF(d−1d)d−1]2≤T ≤pid4(1−α)β (Emin)Eα−1minEαH[aEF(d−1d)d−1]2.(4.25)This dependence on (aEF)2 is a general feature of long Hagedorn regions as wewill see in our next time scale. Recall that when we had C(E) = 1, the effectiveinverse temperature function to consider β˜ was defined byβ˜ ′(E) = ρ2(E)β ′(E) .We will construct a filtration function from this in the same way as before. Φ˜(E) =β˜ (Emin)− β˜ (E +Emin). In the low energy regime, we haveΦ˜′(E) = −ρ2 (E +Emin)β ′ (E +Emin) = ρ2 (E +Emin)Φ′(E)= (1−α)β (Emin)e2αβ(Emin)(E+Emin)αEα−1min(E +Emin)α−2Eα−1min.74EΦEH(a) EH < E∗EΦEHE*(b) EH > E∗Figure 4.10: The upper bound for the decay time, given by Φ˜2 depends onwhich of the above cases is realized. The figure on the left would leadto a situation much like the one for Φ2 that we saw before. However,on the right, the Hagedorn energy is large enough to make Φ˜ changeconcavity before reaching the Hagedorn phase.The first thing to notice is that EH is the same for Φ and Φ˜. One derivative vanishesif and only if the other does. Also, at an energy of Emin, the number of states is veryclose to 1 so Φ˜′(0) =Φ′(0). Finding the slopes of appropriate bounding functionsΦ˜1 and Φ˜2 comes down to extremizing Φ˜′.Φ˜′′(E∗) = 02β (Emin)Eα−1min(1−α)Eα−1min β (Emin)(E∗+Emin)2α−3 = (2−α)(1−α)Eα−1min β (Emin)(E∗+Emin)α−3E∗ =[2−α2β (Emin)E1−αmin] 1α−EminTherefore Φ˜ starts off with a steep derivative at 0 and becomes less steep untilreaching the inflection point E∗. This means that there are two important cases toconsider: EH < E∗ and EH > E∗. The flattest point occurs at EH in the former case75and at E∗ in the latter.Φ˜′1(0) = (1−α)β (Emin)Emin(4.26)Φ˜′2(0) =(1−α)β (Emin)Eα−2HEα−1mine2αβ(Emin)EαHEα−1min EH < E∗(1−α)(2−α2)α− 2α β (Emin)2α E2α−2min e2α−1 EH > E∗(4.27)Plugging (4.26) and (4.27) into (4.22), the next time scale is solved as well:pid4(1−α)β (Emin)EminE2H[aEF(d−1d)d−1]2≤T ≤CT (Emin,EH)[aEF(d−1d)d−1]2(4.28)whereCT (Emin,EH)=pid4(1−α)β (Emin)Eα−1minEαHe− 2αβ(Emin)EαHEα−1min EH <[2−α2β (Emin)E1−αmin] 1αpid4(1−α)(2−α2) 2α−α β (Emin)−2α E2− 2αminE2He1−2α EH >[2−α2β (Emin)E1−αmin] 1α.There is a third possibility; EH being so large that Φ′ (EH) surpasses Φ′(0). How-ever, Emin can always be made small enough to stop this from happening.4.2.5 Remaining problemsA useful question to ask is how our decay time differs from a decay time that wouldarise from linear diffusion. Solivng the heat equation with the (4.4) initial data,E(x, t) =EF(4pit) d2∫ a−a. . .∫ a−ae−(x−y)24t dy1 . . .dyd=EF2dd∏i=1(erf(xi +a2√t)− erf(xi−a2√t)).The time needed for the central peak to come down to EH isT =a24erf−1(EHEF)− 2d≈a24(2EF√piEH) 2d.76For d > 1, this decay time is parametrically less than the O((aEF)2)result that wefound for diffusion with a Hagedorn regime. However, linear diffusion in d = 1 ismuch slower. To get a sense of why the d = 1 times are similiar, we will look at theamount of energy outside B1a after a short amount of time [1]. For diffusion with aHagedorn regime, we findaEF−Ma(t) = A∫ ∞a1+ erf(a− x2√t)dx= 2A√tpi≈ 4EH√tpi .For the heat equation with the same slope, the result isaEF−Ma(t) = EF∫ ∞aerf(x+a2√t)− erf(x−a2√t)dx= 2EF√tpi(1− e−a2t)+2EFa(1− erf(a√t))≈ 2EF√tpiwhere in the last step we have assumed t  a2. Therefore, diffusion without aHagedorn regime is faster at the beginning, but not over longer time scales. Wecan explain this effect by noticing that a lump of energy diffusing linearly becomesremarkably flat towards the “end” of its diffusion. Small spatial derivatives leadto small time derivatives and a long time constant. Diffusion with a Hagedornregime does not have this problem. The discontinuous solutions we saw had steepprofiles during all stages of what we called the diffusion. A different effect, thelarge static phase, is what lengthened the time constant. Even though diffusiontimes are similar with and without a Hagedorn phase, the numerics will show thatif the energy distribution begins with a peak well above EF, its diffusion will slowdown significantly upon reaching EF. This does not contradict our analysis becausethe diffusion in the high energy phase comes with a much larger constant than thediffusion that crosses over to Hagedorn behaviour in the other two phases.77Regardless of how many dimensions there are, we should expect that our modeloverestimates the decay time of a plasma ball. The energy in our model can onlydiffuse away to infinity so quickly once it leaves a ball of radius a. In a realsticplasma ball, energy escaping the central region is quickly ejected because it istravelling through a vacuum.4.3 Comments on unbounded domainsFor unbounded domains such as the whole space, the only sensible energy distri-butions are the ones that decay to zero and filtration equations like (4.3) take onmany new properties. Even though the diffusion of energy governed by a masterequation should certainly be well defined for a space that extends infinitely, wefound in the last chapter that our PDE is only a useful model for such behaviourwhen the energy is high. Therefore, it is not necessarily true that a result derivedfor Ω= Rd applies to the entropic limit of a field theory.4.3.1 Barriers to uniquenessThe first thing we note is that∂E∂ t (x, t) =−∆β (E(x, t)) (x, t) ∈ Rd× (0,∞)E(x,0) = E0(x) x ∈ Rdwould have many solutions if we simply looked for them in the set of all differen-tiable functions. This property is well known for the heat equation. In fact, for theheat equation [65] and similar equations [66] in the whole space, nonzero solutionshave been found that vanish at t = 0. These authors have found a restriction thatmust be made in order to recover uniqueness; there must exist a paraboloid in Rdsuch that logE stays below it at all times. Because our equation is different fromthe heat equation, it is not yet clear that the same condition, or a stronger one, willyield unique solutions to (4.3). Nevertheless, we will impose the stronger conditionthat E be uniformly bounded in spacetime.Requiring uniform boundedness is natural because it forces solutions to stillobey the maximum principle. The maximum principle phrased in terms of subso-78lutions and supersolutions made no assumption about the domain being bounded,but in applying it, we set one solution equal to the maximum of the other. Ourassumption of uniform boundedness is the only reason that we can still writeu = max{v(x,0) : x ∈ Rd} because the maximum of a continuous function is onlyguaranteed to exist on a compact domain. Having to make assumptions about howsolutions grow is a common theme when extending maximum principles to the un-bounded case [61]. We now turn to the question of uniqueness which is intimatelyrelated to energy conservation.Consider the function β (E) = − 1m Em which is monotonically decreasing forall m ∈ R. The equation (4.3) for this choice∂E∂ t = ∇ ·(Em−1∇E)(4.29)has been studied extensively [67–70]. It is called the porous medium equation form ∈ (1,∞), the fast diffusion equation for m ∈ (0,1] and the very fast diffusionequation for m ∈ (−1,0]. We are mainly interested in very fast diffusion becausewe had 0 < α ≤ 1 and m is essentially α − 1. We are only writing m instead ofα−1 here for consistency with the literature.Trying to naively derive conservation of energy for (4.29) does not work. Thisis related to the fact that ∂E∂ t does not need to be bounded at a given time, even ifE is uniformly bounded. To illustrate this, let us work with the very fast diffusionequation in one dimension. For the initial condition E0, it is perfectly valid tochoose a function that decays like |x|−k where k > 1. For large x, this leads to atime derivative given by:∂E∂ t (x,0) =∂∂x[Em−1(x,0)∂E∂x (x,0)]=∂∂x[Em−10 (x)∂E0∂x (x)]∼∂∂x(x(1−m)kx−k−1)∼ x−mk−2 .The exponent tells us that it is possible for E0 to decay “too quickly”. If k > 2−m79then ∂E∂ t is not bounded at t = 0. This affects our ability to computedMdt=ddt∫ ∞−∞Edx .In order to take the derivative inside the integral, there must exist an ε > 0 suchthat ∂E∂ t is bounded on (−∞,∞)× [0,ε]. We cannot do this with all k values, buteven if we could, we would still not be out of the woods.Pick a value of k satisfying 1−m < k <2−m . Since k <2−m∂E∂ t as derived abovemust be finite at t = 0. Computing the time derivative of the total mass,dMdt(0) =∫ ∞−∞∂E∂ t (x,0)dx=∫ ∞−∞∂∂x[Em−1(x,0)∂E∂x (x,0)]dx= Em−10 (x)∂E0∂x (x)∣∣∣∣∞−∞= 2 lim|x|→∞Em−10 (|x|)∂E0∂x (|x|)∼ lim|x|→∞|x|−mk−1 .Our choice k > 1−m tells us that the time derivative of M is initially infinite. There-fore M cannot possibly be a constant function on [0,∞). We have demonstratedthat there are infinitely many functions E0 ∈ L1(R) such that there is no energyconserving soluton to the following:∂E∂ t (x, t) =∂∂x[Em−1(x, t)∂∂xE(x, t)](x, t) ∈ R× (0,∞)E(x,0) = E0(x) x ∈ R. (4.30)As it turns out, however, for any E0 ∈ L1(R), there is [67, 69] an energy conservingsolution to the similar problem:∂E∂ t (x, t) =∂∂x[Em−1(x, t)∂∂xE(x, t)](x, t) ∈ R× (0,∞)limt→0E(x, t) = E0(x) x ∈ R. (4.31)80Differential equations can only hold in open sets so the notation E(x,0) = E0(x) re-ally means that any derivative of E(·, t) will converge to the corresponding deriva-tive of E0 as t → 0. The subtle distinction between (4.30) and (4.31) is that theonly thing that needs to converge is the function itself. The E(·, t) will converge inL1(R) to E0 but the ∂∂ t E(·, t) need not converge to∂∂x(Em−10∂E0∂x). In other words,dM(0)dt appeared to diverge because we were still calculating it incorrectly.Esteban, Rodriguez and Va´zquez [67] showed that an energy conserving solu-tion to the very fast diffusion Cauchy problem exists in one dimension. However,there are also solutions that vanish in finite time such asE(x, t) = (T − t)1−(1− e−√2x1+ e−√2x)2 .This solves (4.31) for m = 0 and does not conserve energy. The theorem of [67]still applies to initial conditions likeE0(x) = T1−(1− e−√2x1+ e−√2x)2 ,it just yields a different solution. In fact, for all −1 < m ≤ 0 and all initial condi-tions, there are short lived solutions to (4.31) in addition to the one that conservesenergy. One of the following equivalent conditions is needed to make sure we arechoosing the right solution.1.∫ ∞−∞ E(x, t)dx =∫ ∞−∞ E0(x)dx for t ∈ (0,∞).2. limx→±∞ Em−1(x, t) ∂E∂x (x, t) = 0 for t ∈ (0,∞).3. E(x, t)> 0 for (x, t) ∈ R× (0,∞).In fact, [68] generalized this to arbitrary flux functions.Theorem 5. Suppose E0 ∈ L1(R) and f ,g ∈ L∞((0,∞)) are non-negative. Then81there exists a unique solution to∂E∂ t (x, t) =∂∂x(Em−1(x, t)∂∂xE(x, t))(x, t) ∈ R× (0,∞)limt→0E(x, t) = E0(x) x ∈ Rlimx→∞Em−1(x, t)∂∂xE(x, t) =− f (t) t ∈ (0,∞)limx→−∞Em−1(x, t)∂∂xE(x, t) = g(t) t ∈ (0,∞)whose total mass satisfies∫ ∞−∞E(x, t)dx =∫ ∞−∞E0(x)dx−∫ t0f (s)+g(s)ds .The energy conserving solution that interests us is precisely the case f = g = 0.It is important to remember that even though the vanishing flux of E is posed asa constraint, we are still free to consider an initial profile E0 that does not havea vanishing flux. This is an example of a problem with inconsistent initial andboundary conditions. Some numerical methods have been written specifically toaddress this [71]. Infinite propogation speed ensures that the conditions are onlyinconsistent on a set of measure zero. Depending on what we specify for the flux,the distribution might “jump” to a function that decays quickly or to a functionthat decays slowly. When m > 0, energy conservation in (4.31) is known to holdwithout the complication of non-uniqueness [69].We may summarize this discussion by saying that the maximum principle onlyholds for a general (4.3) if we specify uniform boundedness and that uniquenessonly holds for a general (4.3) if we specify energy conservation. The very fastdiffusion equation in one dimension provides an example of uniqueness not holdingfor (4.3) if we only specify uniform boundedness. Even when we have a uniqueenergy conserving solution, the time derivative of the total mass will not makesense at t = 0 for the perfectly good initial conditions that are inconsistent with theboundary conditions. For those, the energy flux that vanishes for the solution at allpositive times will not vanish at t = 0.824.3.2 Barenblatt profilesSolutions to (4.3) on a bounded domain converge to their average values whichmeans that solutions to the same equation on the whole space should convergeto zero. This will be an L∞ convergence and not an L1 convergence because ofenergy conservation. One should check whether the associated decay times on Rdhave anything to do with our main time scales (4.25) and (4.28). Some knownresults about the very fast diffusion equation will help us do this. There is animportant family of exact solutions to the porous medium, fast diffusion and veryfast diffusion equations given by:U(x, t) =[ (41−m −2d)t|x|2 +Bt22−d(1−m)] 11−m(4.32)where B is a positive constant that determines the mass [70]. These self-similarfunctions called Barenblatt solutions have a Dirac delta as their initial conditions.Solving for the mass,M(t) = dωd∫ ∞0U(r, t)rd−1dr= dωd[(41−m−2d)t] 11−m∫ ∞0dd−1(r2 +Bt22−d(1−m)) 11−mdr= dωd[(41−m −2d)tBt22−d(1−m)] 11−m ∫ ∞0Bd2 td2−d(1−m) sd−1(s2 +1)11−mds= dωd(41−m−2d) 11−mBd2−11−m∫ ∞0sd−1(s2 +1)11−mds=dωd2(41−m−2d) 11−mBd2−11−mΓ(d2)Γ(11−m −d2)Γ(11−m) (4.33)83where we have recognized a beta function. The time that it takes for a Barenblattprofile’s peak to reach EH is given by:T =EH(B41−m −2d) 11−m(1−m)− 2d. (4.34)If we wanted to call this a Barenblatt profile of mass aEF, we could use the massrelation (4.33) to replace B:T =(41−m−2d) 2m−d(1−m)d(1−m)Ed(1−m)−2dH2aEFdωdΓ( d2 )Γ( 11−m− d2 )Γ( 11−m)4d −2(1−m)2−d(1−m). (4.35)Even though this decay time is concerned with a Dirac delta of mass aEF, it upperbounds the decay time for a box of mass aEF by the concentration comparisontheorem. Things are even nicer than this; the Barenblatt profiles attract all solutionsto the Cauchy problems for these nonlinear diffusion equations. As time goes on,general solutions E(·, t) will converge to U(·, t) and the rate of convergence hasbeen found [70]:limt→∞td2−d(1−m) ||E(·, t)−U(·, t)||L∞ = 0 .This means that (4.35) still depends on the mass in the same way as the true decaytime. Since O((aEF)4d −2(1−m)2−d(1−m))and O((aEF)2)are very different, we should nothave used results from an unbounded domain to derive (4.25) and (4.28). Implicitin all of this is a certain relation between d and m. Consider this list of conditions:1. The constant 41−m − 2d must be positive for the Barenblatt solutions to bewell defined.2. The overall power of t in (4.32) must be negative in order for the peak valueto decay with time.3. In order for the mass (4.33) to be finite, sd−1(s2+1)11−mmust approach zero fasterthan 1s .844. An O((aEF)4d −2(1−m)2−d(1−m))decay time should decrease with the mass.Any one of these four requirements will tell us that m > d−2d . Since diffusion inour model happens when 0 < α < 1, our primary interest is −1 < m < 0. This isonly above the critical exponent in one dimension providing another reason why itwas necessary to demand a bounded domain. We know from [67, 68] that whend = 1, the very fast diffusion equation yields infinitely many solutions and oneenergy conserving solution. The latter is indeed asymptotic to the Barenblatt profilehaving the same mass. However, in d = 2 we already see that there is no energyconserving solution to the very fast diffusion equation. It would be interestingto find some analogue of (4.32) for bounded domains that acts as a fundamentalsolution and stays well defined for all m and d.85Chapter 5Numerical analysisThe expressions for our time scales rely on the fact that EF  EH in a domainthat has a large but finite size. To see how sensitively the inequalities depend onthese factors, we have undertaken a numerical test of our time scales and (4.25) inparticular. The code used for most simulations can be found in the appendix. Onething that is immediately visible in the code is our decision to use initial conditionsthat decay according to a power law:E0(x) = Emax(11+ x2) k2.For the very fast diffusion equation on an unbounded domain, we saw that it waspossible for the initial condition to decay too quickly (k > 1−m ). Since the nicestfunctions to simulate are the ones with a finite dMdt (0), this rules out functions E0that decay exponentially or have compact support. For low energies, the relevantexponent is m+ 1 = α = 910 . This confines us to using k values smaller than 10and not so much smaller that the decay takes forever. The numerics are done on abounded domain so dMdt (0) is technically finite for any k. However, if we were topick k > 10, the flux at the edges would experience unbounded growth as we madethe domain larger casting doubt on our ability to trust those results.865.1 Implementation detailsIn (4.3), we have been considering a function β = dSdE which has diffusive behaviourfor low and high energies with a Hagedorn phase and possibly a clustering phase inbetween. The chosen entropy should be of the Super Yang-Mills type (2.46) whereS(E)∼E910 E < EHE EH < E < E ′FE87 E ′F < E < EFE34 E > EF.A smooth approximation to this can be accomplished with the function:S(E) =(EEH) 910(1+EEH) 110(1+EE ′F) 17(1+EEF)− 1128. (5.1)However, the clustering phase where S is convex is difficult to simulate numeri-Eβ(a) E ′F 6= EFEβ(b) E ′F = EFFigure 5.1: One the left is the derivative of (5.1) showing the three phases.The simplest choice for removing the clustering phase leads us to dif-ferentiate (5.2), the plot on the right.cally and our results including (4.25) have neglected it. Assuming that E ′F = EF so87that this phase does not exist, we arrive at:S(E) =(EEH) 910(1+EEH) 110(1+EEF)− 14. (5.2)Plotting the β (E) derived from these two functions, we see a slight problem. Afterthe flat region of (5.1), β (E) increases reaching a local maximum. After this itbegins to decrease more and more quickly until it reaches a point of inflectionand asymptotically approaches 0. The plot for (5.2), on the other hand, does notlook very flat in any region and seems to always be asymptotically approaching 0.There is no point of inflection where some other behaviour “crosses over” to E−14behaviour. We will now show by brute force that there is no inflection point, rulingout the possibility that this only happens for certain EH and EF.Claim 1. For any N > 0, β : R>0→ R defined byβ (x)= 910x−110 (1+x)110(1+xN)− 14+110x910 (1+x)−910(1+xN)− 14−14Nx910 (1+x)110(1+xN)− 54has no point of inflection.Proof. First we will writeβ ′(x) = − 9100(1+xN)− 14[x−1110 (1+ x)110 −2x−110 (1+ x)−910 + x910 (1+ x)−1910]−120N(1+xN)− 54[9x−110 (1+ x)110 + x910 (1+ x)−910]+516N2x910 (1+ x)110(1+xN)− 94.88Now the equation β ′′(x) = 0 becomes:91000(1+xN)− 14[11x−2110 (1+ x)110 −3x−1110 (1+ x)−910 −27x−110 (1+ x)−1910 +19x910 (1+ x)−2910]+27400N(1+xN)− 54[x−1110 (1+ x)110 −2x−110 (1+ x)−910 + x910 (1+ x)−1910]+332N2(1+xN)− 94[9x−110 (1+ x)110 + x910 (1+ x)−910]−4564N3x910 (1+ x)110(1+xN)− 134= 03125(1+xN)3 [11(1+ x)3−3x(1+ x)2−27x2(1+ x)+19x3]+950N(1+xN)2 [x(1+ x)3−2x2(1+ x)2 + x3(1+ x)]+14N2(1+xN)[9x2(1+ x)3 + x3(1+ x)2]−158N3x3(1+ x)3 = 0625x6 +(2500N +1625)x5 +(7250N +2275)x4 +(9250N +819)x3+(2340N2 +3402N)x2 +(720N3 +972N2)x+264N3 = 0 .Even though we cannot factor this sixth degree polynomial, we may conclude thatit has no positive real roots via Descartes’ rule of signs.What we should take away from this is that removing the clustering phaseby hand is too crude to give us a good filtration function to use in the numerics.Instead we will use a function like the one plotted in Figure 5.2 where quadraticinterpolation has been used to guarantee that it is differentiable and decreasing withan almost-flat section and two inflection points.5.1.1 The Crank-Nicolson methodNow that the form of (4.3) has been decided, we need a method capable of findingapproximate solutions to it. Our PDE involves time and space, so it is natural todiscretize space and reduce it to the following set of coupled ODEs:ddtEi(t) = fi(E(t), t)fi(E(t), t) = −β (Ei+1(t))−2β (Ei(t))+β (Ei−1(t))(dx)2.890 5 10 15E00.511.52βFigure 5.2: For this function, the flat section begins at E = 1.0. The region0.5 < E < 1.0 is used to interpolate between the flat section and theE−110 power law. The flat section ends at E = 0.9EF and the E−14 powerlaw begins at E = EF. In the region 0.9EF < E < EF, a parabola againinterpolates. The plot is for EF = 10.0 but this general rule has beenfollowed for all EF.The simplest method for integrating this type of ODE system is the forward Eulermethod:Ei(t +dt) = Ei(t)+ fi(E(t), t)dt . (5.3)One might expect a method like (5.3) to be accurate as long as dx and dt aresmall. However, one can use the heat equation to show that things are not sosimple. Figure 5.3 shows that making the discretization smaller is not always animprovement. To avoid numerical instabilities, the restriction that must be obeyedis dt < 12(dx)2. This is the precise restriction for the heat equation but a nonlinearequation would suffer from a similar restriction. This is especially harsh for in-vestigating the long decay times associated with a Hagedorn phase. Although theyare more difficult to apply, there are other methods known to be stable for largetimesteps [72]. One of these is the backward Euler method:Ei(t +dt) = Ei(t)+ fi(E(t +dt), t +dt)dt . (5.4)90-2 -1 0 1 2x0. = 0.0t = 0.025(a) dt = 0.005, dx = 0.1-2 -1 0 1 2x0. = 0.0t = 0.025(b) dt = 0.005, dx = 0.05Figure 5.3: A standard demonstration that the forward Euler method is nu-merically unstable for the heat equation.In (5.3), we evolve E forward in time by adding the derivative that it has now. In(5.4) which looks similar, we add the derivative that it will have after we add it.Since some inversion is clearly necessary, we refer to this type of method as animplicit method. In this work, we will use the average of the forward and backwardEuler methods, known as the Crank-Nicolson method. In order to isolate Ei(t +dt) in (5.4) when treating the heat equation, we must solve a system of linearequations - one for each site in the lattice. For a nonlinear equation like (4.3), thesebecome nonlinear algebraic equations, and hence often require numerical methodsthemselves. Something that might look familiar is:xn+1 = xn−f (xn)f ′(xn).This is Newton’s method where the sequence converges to a zero of f unless theinitial guess x0 is sufficiently far away. The derivative becomes the Jacobian whenwe have more than one variable.g′(En(t +dt))i j(En+1j (t +dt)−Enj (t +dt)) =−gi(En(t +dt))gi(E(t +dt)) = Ei(t)+ fi(E(t +dt))dt−Ei(t +dt) (5.5)91We should solve this (tridiagonal) linear system for an initial guess of E0i (t +dt) =Ei(t) since dt is small enough to not change the energy much with each step. TheCrank-Nicolson method for us is thenEi(t +dt) =12[Ei(t)+ fi(E(t))dt +E∞i (t +dt)]fi(E(t)) = −β (Ei+1(t))−2β (Ei(t))+β (Ei−1(t))(dx)2, (5.6)with E∞i (t +dt) generated in the (5.5) way. One aspect of this which is still unde--200 -100 0 100 200x50100150E(a) Linear scale-6 -4 -2 0 2 4 6sgn(x) * log(1 + x)12345log(1 + E)(b) Modified logarithmic scaleFigure 5.4: Typically we will work with an initial condition like that shown inred. The width of the domain is chosen so that all of the function’s masscan fit below the Hagedorn energy in blue. The plot on the left showsthat E0(x) is very flat for most of the x values. An adaptive dx makessense and will be chosen so that the minimum dx is much smaller thanthe distance between the inflection points. The plot on the right showsthe graph of the same function but more conveniently.sirable is having a constant dx. Given the energy profiles we wish to evolve, it willbe more convenient to have dx depend on i so that the lattice can be finely grainedin regions where the energy profile changes the most. Figure 5.4 shows the needfor a variable spatial step and also establishes the conventions we will use to havelogarithmic axes.925.1.2 Convergence testsWe wish to test this algorithm, but do not have any exact solutions at our disposal.To verify that our finite difference scheme converges to a solution as O(h2) ≡O((dx)2), there is a useful test suggested by [73] based on the Richardson ex-pansion:Eh(x, t) = E(x, t)+ e2(x, t)h2 + e4(x, t)h4 + . . . . (5.7)Here Eh is the approximate solution computed by the Crank-Nicolson method,which differs from the exact solution according to functions e2, e4, etc. If wechoose different discretizations, dx ∈ {h, 2h, 4h} for example, (5.7) states that fora given (x, t),Q(t) = limh→0E4h(x, t)−E2h(x, t)E2h(x, t)−Eh(x, t)= limh→016h2−4h24h2−h2= 4 .As a better check, we will compare the 2-norms of the numerical solutions insteadof choosing a single point x.Q(t) = limh→0∣∣∣∣E4h(·, t)−E2h(·, t)∣∣∣∣`2||E2h(·, t)−Eh(·, t)||`2= 4 (5.8)Plotting this for five different values of h, it looks like Q(t) = 2. This would beexpected for a non-centred finite difference method whose Richardson expansionlooks like:Eh(x, t) = E(x, t)+ e1(x, t)h+ e2(x, t)h2 + . . . .Our departure from the expected error O(h2) is likely a result of the nonlinear alge-braic equations. As soon as we use Newton’s method, Crank-Nicolson is not beingfollowed exactly and an O(h) error can be introduced. Linear combinations suchas 2Eh−E2h can be used to cancel this error term in a technique called Richardsonextrapolation. This has been successfully used for other diffusion equations withthe Crank-Nicolson method [74].93Figure 5.5: The factor Q(t) plotted for simulations that lasted t = 90.0. Thegrid was uniform and five step sizes were used.5.2 Simulation resultsGoing ahead with the Crank-Nicolson method, one can print out slices of the en-ergy every so often to see how it is diffusing. The numerics make it clear that thereare two time scales of interest. What we have predicted in (4.25) is the long decaytime needed for a distribution of energy to descend below EH. Before this, the dis-tribution will reach EF or some number close to it, at what we call the termalizationtime. Figure 5.6 shows that for a variety of initial conditions, the energy profilesare qualitatively similar to step functions when this happens.5.2.1 Short time dynamicsThe left plot in Figure 5.6 uses the E0(x) = 200(11+x2) k2form. It is fairly clearfrom the figure that the profile is flattening out somewhere above log(1+EF) =log11. After interpolation, the effective Hagedorn and field theory energies be-come slightly shifted with respect to the parameters EH and EF. Therefore to mea-sure thermalization times as a function of k, we have used 32 EF as a rule of thumb94-6 -4 -2 0 2 4 6sgn(x) * log(1 + x)12345log(1 + E)t = 0t = 550t = 1100t = 1650(a) Wider function-6 -4 -2 0 2 4 6sgn(x) * log(1 + x)12345log(1 + E)t = 0t = 550t = 1100t = 1650(b) Narrower functionFigure 5.6: These initial conditions with the same total mass both evolve to-ward a state that has a flat line in the high energy phase near EF = 10.0.(a) Same peak (b) Same massFigure 5.7: Two plots of the thermalization time for field theory energy EF =10.0 and different values of k in the initial condition.for where the peak energy should be. After making a plot for five values of k, wesee that the thermalization time is much more sensitive to variations in LEF/M thanEF/Emax.Another trend we may investigate is how E(0, t) behaves as a function of t.Plotting this for k = 9 will give us Figure 5.8. After a slow start at early times, theplot becomes steeper before levelling off again. This basic shape holds for other k95Figure 5.8: A plot of how the peak energy moves down with time. This cor-responds to the k = 9 decay in Figure 5.7.-4 -2 0 2 4sgn(x) * log(1 + x)12345log(1 + E)t = 0t = 1650t = 9950t = 45800Figure 5.9: This shows one of our initial conditions diffusing all the waythrough the Hagedorn regime. It takes about a hundred times longerto do this than it does to reach the thermalization time.96Figure 5.10: A plot of the decay time for EF = 10.0, EH = 1.0 and five valuesof k. Even though they specify different shapes for the initial condition,the times are all within 4% of eachother.values as well. It is different from the decay of a Barenblatt profile which wouldalways have E(0, t) as a power law. The time t∗ when E(0, t) is changing mostquickly appears somewhere in Figure 5.8. If we work with the functions havingsimilar thermalization times (the ones normalized to have the same mass), we findthat t∗ ≈ 0.6T ∗ in all five cases. The ratio between ddt E(0, t∗) and the average rateof change in the interval (0,T ∗) is about 1.8. We have not found a way to predictthese numbers analytically.5.2.2 Long time dynamicsThe next step is to wait until E(0, t)< EH. An example of what this diffusion lookslike is in Figure 5.9. This time, T  T ∗ is predicted in (4.25) to depend on thesquared mass. After all, a cylinder of radius a and height EF has a volume of 2aEFif the base is one-dimensional. When investigating the short time dynamics, initial97conditions of five different shapes were normalized to have the same mass. SinceT ∗ was found to be almost the same for them, it is no surprise that T shares thisproperty. Figure 5.10 shows a plot of these times. As long as the mass is keptconstant, early and late time dynamics are largely insensitive to k so we will usethe initial conditionE0(x) = 200(11+ x2) 92(5.9)from now on. The major quantity we have not changed yet is the EF/EH ratio. Wehave always had EF = 10.0 and EH = 1.0 so far. As we know in one dimension,T is proportional to M2 and not a power of EF. Whether or not we can change EFand leave T invariant will be the real test of (4.25). The condition 1.0 = EH EF Emax = 200.0 needs to be satisfied so we will choose a few 10.0≤ EF ≤ 20.0values. The decay times for these choices, plotted in Figure 5.11, are remarkablyFigure 5.11: A plot of the decay time for k = 9, EH = 1.0 and five values ofEF. Even though the profiles begin to flatten out at different heights,their final decay times are all within 0.4% of eachother.98close. The exact prefactors in (4.25) still need to be checked but it turns out thatthey are not very constraining. We know that α = 910 and taking the length L fromthe code, we can plug it into (5.9) to find Emin = E0(L). Because Emin is so small,the prefactor multiplying M2 in the upper bound from (4.25) is very different fromthe prefactor multiplying M2 in the lower bound from (4.25). In this simulation,they differ by a factor ofEα−2min =(8.69 ·10−19) 910−2 = 7.36 ·1019 ,which is much larger than any of the numbers put into the simulation by hand.5.2.3 Higher dimensions(a) Zoomed in (b) Zoomed outFigure 5.12: On the left is a plot of the decay time in two dimensions fork = 9, EH = 1.0 and five values of EF. These are much smaller than theone-dimensional decay times in Figure 5.11. The zoomed out versionon the right shows that even though these dots do not vary linearly,they are sandwiched between two bounds which do vary linearly. Forthe sake of the plot, the blue lines have been understated. In reality thelower bound is much closer to being horizontal and the upper bound ismuch closer to being vertical.Good agreement between our prediction and the numerical results has been99demonstrated in one dimension. In our case, extending the numerics to higherdimensions is simple because we are focusing on spherically symmetric data. Ourequation becomes∂E∂ t = −r1−d ∂∂ r(rd−1∂β (E)∂ r)= −∂ 2β (E)∂ r2 −d−1r∂β (E)∂ r .This adds a term with a single spatial derivative to our one-dimensional equationfrom before. A plot like Figure 5.11 should no longer be expected because T in(4.25) is no longer proportional to M2. In two dimensions, (aEF)2 =(a2EF)EF ∝MEF. This means we are looking for a decay time that varies linearly with the fieldtheory energy. After modifying the Crank-Nicolson code in the appendix, two-dimensional simulations testing this have been performed. The results, plotted inFigure 5.12, do not appear to be proportional to EF. The EF = 20.0 decay time ismuch less than twice the EF = 10.0 decay time. Even though the times scale dif-ferently from the bounds in (4.25), they do not actually violate the bounds becauseof the extreme prefactors that differ by a factor of 1019.The typical values for T differ greatly between dimensions one and two. Moregenerally, the decay times when d ≥ 2 are much smaller than when d = 1. Onedimension is special because it is only in this case that the α < 1 Barenblatt profileexists. Figure 5.12 shows decay times that are only nonzero because of the boundeddomain - a phenomenon known as instantaneous extinction. This states that T → 0as Emin→ 0, or in other words, the process we are simulating is not well defined ininfinite volume.100Chapter 6Room for improvementThe dynamics of our model have shown many similarities to plasma balls but thereare three main problems. To start, our solutions show that energy diffuses to in-finity rather slowly once it escapes the central region. The phenomenon of decayvia hadron ejection is not realized. A more serious problem is the instantaneousextinction we have seen for a lattice with two or more dimensions. Finally, wehave seen that some punishing prefactors allow the admissible decay times to spanseveral orders of magnitude.Though one can imagine several ad hoc changes to our model that might elim-inate these problems, an extension that refers to conserved quantities other thanenergy is physically well motivated. The remainder of this thesis will focus ondeveloping an explicit PDE description for the momentum in a Poincare´ invarianttheory.6.1 A common approximationThere is already a widely applicable PDE framework called hydrodynamics thatputs energy and momentum on equal footing. Hydrodynamics is concerned withsystems that are close to equilibrium. Because of this, a system may deviate from atranslation invariant state only slightly leading to slowly varrying charges E(x, t),P(x, t) and Q(x, t). These are associated with long distance variations in theirsources T (x, t), v(x, t) and µ(x, t) respectively.101A theory with a symmetric stress-energy tensor and a U(1) current has d+1+(d+1)(d+2)2 independent components. It is therefore not possible to describe everysystem using the d+2 hydrodynamic variables above. Nevertheless, hydrodynam-ics provides a good description of field theories with high occupation numbers andhas demonstrated a particular aptitude for problems relating to plasma balls andblack holes.6.1.1 Basic hydrodynamicsThe hydrodynamics equations are nothing but the local conservation laws∂µT µν = 0∂µJµ = 0 . (6.1)The conserved currents are built from T , µ and uµ making this a system of d + 2equations for d +2 unknown functions (the velocity satisfies u2 = −1). The rulesfor writing (6.1) in terms of the hydrodynamic variables are called the constitutiverelations. There is no limit to how complicated the constitutive relations might beso it is helpful to use the assumption that the functions are slowly varrying. Thisallows one to consider different versions of hydrodynamics based on how manyderivatives are kept. In zeroth order hydrodynamics, the temperature, velocity andchemical potential are not differentiated in the expressions for T µν and Jµ . In firstorder hydrodynamics, they are differentiated at most once. These orders are oftenreferred to as “ideal” and “dissipative” hydrodynamics. It is important to note thatthis is different from the linearization of hydrodynamics. Nonlinear equations (e.g.the Navier-Stokes equations) can easily arise from zeroth or first order hydrody-namics, so it is common to introduce a further order-by-order expansion that dropsterms with too many variables multiplied together.The first step in developing constitutive relations is decomposing the currentsinto components that are transverse and longitudinal with respect to uµ . This leadstoT µν = E uµuν +P∆µν +(qµuν +qνuµ)+ tµνJµ = N uµ + jµ (6.2)102where we have defined the projector ∆µν = ηµν +uµuν . Transverse quantities likejµ , qµ and tµν cannot be built out of T , µ and uµ without derivatives. Thereforeideal hydrodynamics takes the form:T µν = εuµuν +P∆µνJµ = nuµ . (6.3)The pressure P, while not a charge or a source, is usually given in terms of sourcesby an equation of state P(T,µ). Going to the fluid’s rest frame makes it clear thatε is the energy density and n is the charge density.Things become more complicated when we include one order of dissipation[75]. Looking at (6.2), many quantities such as uµ , qµ and jµ depend on positionfor a general non-equilibrium configuration. There is nothing that prevents us fromchanging uµ to a different function of spacetime as long as qµ and jµ change aswell to keep T µν and Jµ invariant. This redundancy, similar to a gauge freedom,is called frame invariance in hydrodynamics [76]. Consider a shift u′µ(x, t) =uµ(x, t)+δuµ(x, t). To preserve the normalization, δuµ is transverse to uµ . Usingthe inverse of (6.2), corresponding changes in the coefficients can be calculated tofirst order.E = uµuνTµνδE = 2uµδuνT µν = 2uµδuνtµν ≈ 0P =1d∆µνTµνδP = 2duµδuνT µν ≈ 0N = −uµJµδN = −δuµJµ =−δuµ jµ ≈ 0 .Note that purely dissipative quantities like jµ and tµν become second order whenmultiplied by δuµ . These are all zero which explains why we could write down(6.3) without worrying about frame invariance. The same analysis would give103δ tµν = 0 as well. Conversely,qµ = −∆µαuβTαβδqµ = −δuβTµβ −2uµδuαuβTαβ −δuµuαuβTαβ= −Pδuµ −qβuµδuβ −δuβ tµβ +2uµδuαqα −E δuµ≈ −(E +P)δuµjµ = ∆µνJνδ jµ = uµδuνJν +uνδuµJν= uµδuν jν −N δuµ≈ −N δuµ .We see that a suitable definition of local velocity reduces the number of terms in(6.2). The one that makes jµ = 0 is called the Eckart frame while the one thatmakes qµ = 0 is called the Landau frame. Naturally, T ′(x, t) = T (x, t)+ δT (x, t)and µ ′(x, t) = µ(x, t) + δµ(x, t) are allowed shifts of the other hydro variables.Under these redefinitions, the coefficients from (6.3) becomeε(T ′,µ ′) = ε(T,µ)+ ∂ε∂T δT +∂ε∂µ δµP(T ′,µ ′) = P(T,µ)+ ∂P∂T δT +∂P∂µ δµn(T ′,µ ′) = n(T,µ)+ ∂n∂T δT +∂n∂µ δµ .This means we can define temperature and chemical potential such that E = ε andN = n [75].Choosing the Landau frame, constitutive relations are expressions for the scalarP , the transverse vector jµ and the transverse traceless symmetric tensor tµν .These may contain any combination of T , µ and uµ with one derivative. However,the equations of zeroth order hydrodynamics (6.3) give relations between manyof these terms up to higher order corrections. The Landau frame expressions that104follow from this are:T µν = εuµuν +(P−ζ∂λuλ)∆µν −η∆µα∆νβ(∂αuβ +∂βuα −2dηαβ∂λuλ)Jµ = nuµ −σT∆µν∂ν(µT)+χT∆µν∂νT . (6.4)The frame invariant parameters ζ , η and σ are functions of T and µ that must bedetermined from experiment or the microscopic theory. They are called the bulkviscosity, shear viscosity and conductivity respectively. It turns out that χT mustbe zero for time-reversal invariance to be satisfied. The linearization of (6.4) willbe important in what follows.We linearize around the equilibrium solution which has constant hydro vari-ables and zero velocity. This allows us to write uµ = (1,vi). Explicitly, what weseek are evolution equations forJ0(x, t) = n(T (x, t),µ(x, t))≈ n(T0,µ0)+(∂n∂T)0δT (x, t)+(∂n∂µ)0δµ(x, t)≡ n0 + n˜(x, t)T 00(x, t) = ε(T (x, t),µ(x, t))≈ ε(T0,µ0)+(∂ε∂T)0δT (x, t)+(∂ε∂µ)0δµ(x, t)≡ ε0 + ε˜(x, t)T 0i(x, t) = (ε(T (x, t),µ(x, t))+P(T (x, t),µ(x, t)))vi(x, t)≈ (ε0 +P0)vi(x, t)≡ P˜i(x, t) .It is important to remember that most coefficients in (6.3) and (6.4) are functionsof T and µ . Their derivatives evaluated at T0 and µ0 should therefore appear in thelinearization. These derivatives, known as susceptibilities, do not have a functionaldependence on T and µ anymore; they are simply numbers. For this reason, weshould not be worried if the linearized hydrodynamics equations appear to havemore coefficients than the six we have seen so far. One equation of hydrodynamics105is always the continuity equation ∂ε∂ t = ∂iT0i. It is easy to see that this becomes∂ ε˜∂ t = ∂iP˜i (6.5)because it is already linear. The equation involving stresses linearizes to∂ P˜ j∂ t = ∂i[(P−ζ∂kvk)δ i j−η(∂ iv j +∂ jvi− 2dδ i j∂kvk)]= ∂i[(P−ζε0 +P0∂kP˜k)δ i j− ηε0 +P0(∂ iP˜ j +∂ jP˜i− 2dδ i j∂kP˜k)]= ∂i[δ i j(∂P∂ε)0ε˜+δ i j(∂P∂n)0n˜+δ i j2dη0−ζ0ε0 +P0∂kP˜k−η0ε0 +P0(∂ iP˜ j +∂ jP˜i)]=(∂P∂ε)0∂ jε˜+(∂P∂n)0∂ jn˜+2−dd η0−ζ0ε0 +P0∂ j∂kP˜k−η0ε0 +P0∂k∂ kP˜ j . (6.6)Finally, to handle the U(1) current∂ n˜∂ t = ∂i[nvi−σTδ i j∂ j(µT)]= ∂i[n0ε0 +P0P˜i−σTδ i j∂ j(µT)]= ∂i[n0ε0 +P0P˜i−σ0Tδ i j(∂µ/T∂ε)0∂ jε˜−σ0Tδ i j(∂µ/T∂n)0∂ jn˜]=n0ε0 +P0∂iP˜i−σ0T(∂µ/T∂ε)0∂i∂ iε˜−σ0T(∂µ/T∂n)0∂i∂ in˜ . (6.7)It is interesting to note that the continuity equation (6.5) effectively splits into twoequations in non-relativistic hydrodynamics. In this limit, one must insert the speedof light back into (6.4) and collect powers of c. Taking c→ ∞, inverse powersvanish leaving only O(c) and O(1) terms in the hydro equations. Since these orderscan be considered separate, they yield conservation of mass and conservation ofkinetic energy [77].Matching the predictions of hydrodynamics with those of the interaction pic-ture leads to the Kubo formulas. A simple Kubo formula can be derived for thediffusion constant in the heat equation ∂n∂ t −D∂ 2n∂x2 = 0. Performing a two-sided106Fourier transform in space and a one-sided Fourier transform in time, the equationcan be solved asn(k,z) =n0(k)Dk2− iz=χµ0(k)Dk2− iz. (6.8)We assume that the initial buildup of charge is due to a local chemical potentialthat existed before t = 0. If this chemical potential increased adiabatically startingat t =−∞, the corresponding perturbation to the Hamiltonian is given byH 7→ H−∫ ∞−∞µ(x, t)n(x, t)dx= H−∫ ∞−∞eεtµ0(x)θ(−t)n(x, t)dx≡ H−δH .Heisenberg’s equation now givesddt〈n(x, t)〉 = −i〈[n(x, t),δH(x, t)]〉〈n(x, t)〉 = −i∫ t−∞∫ ∞−∞eεt′µ0(x)θ(−t ′)〈[n(x, t),n(x′, t ′)]〉dx′dt ′= −i∫ 0−∞∫ ∞−∞eεt′µ0(x)θ(t− t ′)〈[n(x, t),n(x′, t ′)]〉dx′dt ′= −∫ o−∞∫ ∞−∞eεt′µ0(x)GRnn(t− t ′,x− x′)dx′dt ′ .Here GRnn(t− t′,x− x′) is the retarded Green’s function of n with itself. The onefact we need is that the zero mode of GRnn is the susceptibility −χ . It is a matter ofcomplex analysis to show that〈n(k,z)〉 = −µ0(k)∫ ∞−∞GRnn(ω,k)1(ε+ iω)(ε+ i(ω− z))dω2pi=µ0(k)iz[GRnn(0,k)−GRnn(z,k)]=iµ0(k)z[χ+GRnn(z,k)]. (6.9)Comparing (6.8) to (6.9), we see that a quantum theory well described by linear dif-fusion should be one whose momentum space retarded Green’s function is χDk2iz−Dk2 .107A calculation that is longer but equally straightforward has the linear hydrodynam-ics equations in place of the heat equation. This is done in [75]. Additionally a fieldtheory is developed that allows one to compute corrections to the Green’s functionsarising from small nonlinearities in the hydrodynamics equations.6.1.2 The fluid / gravity correspondenceCorrelation functions in certain CFTs are related to supergravity amplitudes in cer-tain AdS backgrounds. The link between Green’s functions and linearized hydro-dynamics opens up the possibility of describing fluid phenomena using gravity.This was demonstrated in [78] which found that the shear viscosity in Super Yang-Mills theory is given by η = pi8 N2T 3. By now it is known that in arbitrary dimen-sion, the holographic shear viscosity differs from the entropy density by a factor of14pi [79]. A major result of 2007 is that all transport coefficients in the fluid stress-energy tensor can be algorithmically found from the gravity side as well [80]. Thisexpansion, which is valid to all orders, is called the fluid / gravity correspondence.In the examples discussed previously, we related the energy of a state on theboundary of AdS to the mass of a black hole in the bulk. More generally, there is asystematic way to find the boundary stress-energy tensor corresponding to a givenbulk metric [81, 82]. Because every metric yields a conserved stress-energy tensor,it is not surprising that the fluid / gravity correspondence exists. However, the limitin which temperature and velocity suffice to describe T µν could have correspondedto an intractable gµν . Showing otherwise, [80] started with the metricds2 = L2[−ρ2(1−1zd0ρd)dv2 +2dvdρ+ρ2dxidxi]. (6.10)This is nothing but the black brane (2.54) written in ingoing Eddington-Finkelsteinco-ordinates with ρ = 1z . Boosting the brane to a particular velocity, this becomesds2 = L2[−ρ2(1−1zd0ρd)uµuνdxµdxν −2uµdxµdρ+ρ2∆µνdxµdxν].(6.11)Greek indices have been used for all co-ordinates except ρ which is not in the fieldtheory. The key step is to promote z0 (which determines the temperature) and uµ108to slowly varrying functions of spacetime. One could repeat the calculation for acharged black hole if she wanted chemical potential to vary as well. Clearly, thesefunctions are heavily constrained for (6.11) to still solve Einstein’s equations. Theconstraints turn out to be those of hydrodynamics with infinitely many orders ofdissipation. Put another way, we may say that temperature and velocity functionssatisfying SYM hydrodynamics at a given order, cause (6.11) to only violate Ein-stein’s equations at a higher order.Instead of making z0 and uµ functions of x, [80] makes them functions of εxto keep track of derivatives before setting ε = 1. Using g(0) to denote the metric of(6.11), g(0) with unconstrained z0 and uµ has a stress-energy tensor like (6.3) andviolates the Einstein equations at order ε . To correct this and make the violationorder ε2, one must do two things. The first is to write z0 = z(0)0 + εz(1)0 , uµ =u(0)µ + εu(1)µ so that g(0) picks up terms of order ε . When these corrections areexplicitly calculated, z(1)0 e.g. will be some multiple of a derivative of z(0)0 . Thesecond is to add a new piece of the metric g = g(0) + εg(1). If we plug this newmetric into Einstein’s equation, a vanishing ε term will dictate that z(0)0 and u(0)µsatisfy ideal hydrodynamics (6.3). It will also allow us to compute the correctedmetric and show that it corresponds to the stress-energy tensor (6.4). The result ofthis is [56]ds2 = L2[−ρ2(1−1z(0)d0 ρd)u(0)µ u(0)ν dxµdxν −2u(0)µ dxµdρ+ρ2∆(0)µνdxµdxν+2ρ2z(0)0 F(z(0)0 ρ)(∆(0)µα∆(0)µβ(∂αu(0)β +∂ βu(0)α)−1d−1∆(0)µν∂λu(0)λ)dxµdxν+2d−1ρu(0)µ u(0)ν ∂λu(0)λdxµdxν −ρu(0)λ∂λ(u(0)µ u(0)ν)dxµdxν](6.12)whereF(x) =∫ ∞xyd−1−1y(yd−1)dy .Of course (6.12) is still in the form of (6.11). The choice to not collect all thedxµdxν terms makes it easy to see the non-derivative part whose componentssatisfy ideal hydrodynamics. Continuing the procedure with z(2)0 and u(2)µ andg(2), these same components are constrained to satisfy dissipative hydrodynam-109ics. Rather than (6.12) associated with the dissipative stress-energy tensor (6.4),Einstein’s equations then yield a longer metric associated with the stress-energytensor for a conformal fluid at two orders of dissipation. This logic continues in-ductively showing many non-trivial constitutive relations. Beyond just conformalfluids, similar constitutive relations may be derived for fluids in spaces with com-pactified directions. These relations have been used to study plasma balls and otherobjects that appear in the Witten model [83–85]. These studies assume that a blackhole is already present because the process of black hole formation is outside theregime of hydrodynamics [56].Simply replacing the stochastic model of this thesis with hydrodynamics is notwhat we plan to do. For one thing, it is hoped that the initial conditions of Figure5.4 correspond to black holes that have not formed yet. For hydrodynamics toapply, we must wait for the inhomogeneities to be smoothed out. Therefore, weshould only expect agreement between hydrodynamics and a suitable extension ofour model if we linearize both of them. Another reason to extend our model comesfrom the importance of the density of states. When we only had energy beingtransferred, this function allowed us to read off diffusing or clustering behaviour.It is plausible that something similar can be done when we include momentum.6.2 Restricting the density of statesIf momentum is viewed as a quantity exchanged between nearest neighbour sites,the rate for each transition naturally depends on a momentum restricted density ofstates ρ(E,P). While expressions for ρ(E,P) appear in some models of electronstructure [86], the restricted density of states for a field theory is a barely stud-ied quantity. Here, we attempt to rectify this by deriving some properties of therestricted density of states for simple field theories. A strongly coupled SYM ex-pression analogous to (2.46) is most likely beyond our reach.6.2.1 Some conformal field theorySome of the most interesting techniques for counting states are applicable to two-dimensional CFTs. Consider a Minkowskian theory on S1×R where the circlehas radius 1 by convention. At thermal equilibrium, this can be thought of as a110Euclidean theory on S1 × S1 since temperature and time are related by a Wickrotation. The radius of the second circle must be β . If we rescale each S1 by thesame factor, one such rescaling gives us another Euclidean theory on a circle ofradius 1 as shown in Figure 6.1. Conformal invariance then tells us that inverseβVVV2 / βFigure 6.1: These two theories live on a circle of volume V but their temper-atures are different. They are related through multiplication by Vβ whichis a conformal transformation.temperature β and inverse temperature V 2β are completely equivalent [87]. In termsof the modular parameter we defined for (2.7), τ = i β2pi becomes i4pi2/β2pi = −1τ .For a τ that is not imaginary, the factor we trace over to compute the generalizedpartition function is e−2piτ2H+2piiτ1P. Clearly, this does not change if we increase thereal part of τ by 1. Symmetries of the generalized partition function can thereforebe written asτ 7→ τ+1τ 7→ −1τ . (6.13)These two transformations generate the modular group which is the discrete butinfinite group SL(2,Z). The utility of modular invariance for studying thermo-dynamics was first noticed by Cardy in 1986 [88]. Our previous expressions forthe partition function of a free theory do not posess modular invariance. This isbecause they are only asymptotic expressions for the high temperature limit of Z.Low temperature information is lost when we convert sums to integrals. It is per-haps for this reason that most sources prefer to write infinite products or specialfunctions when simple expressions like (2.3) and (2.7) are all that are needed.111For an infinite collection of oscillators with frequencies 1,2,3, . . . , the energyof an arbitrary state |N1,N2,N3, . . .〉 is N1 + 2N2 + 3N3 + . . . . The degeneracy ofenergy n is simply the number of ways to make n by picking a certain number ofones, a certain number of twos, a certain number of threes and so on. This well-studied object p(n) is usually called “the partition function” in number theory. Wewill call this microcanonical quantity “the partition sum” to distinguish it from thecanonical partition function we have been using so far. The first asymptotic p(n)∼14√3nepi√23 n due to Hardy and Ramanujan was based on transformation propertiesof the generating functionF(q) =∞∑n=1p(n)qn=[1+q+q2 + . . .][1+q2 +q4 + . . .]. . .=∞∏n=1(1−qn)−1 . (6.14)If we take q = e2piiτ (and τ = i β2pi as before), (6.14) becomes a sum of Boltzmannfactors. It is not quite the partition function of our oscillator family because wehave not yet included the Casimir effect. Accounting for zero point energies, eachterm in the product (6.14) picks up a factor of qn2 giving usZ1(τ) =∞∏n=1qn2 (1−qn)−1= q−124∞∏n=1(1−qn)−1= η−1(τ) . (6.15)We have analytically continued and written this partition function in terms of theDedekind eta functionη(τ) = epiiτ12∞∏n=1(1− e2piinτ). (6.16)112Referring to the eta function, Rademacher gave a convergent series for p(n)p(n) =1√2pi∞∑k=1Ak(n)√kddnsinh(pik√23(n− 124))√n− 124 (6.17)Ak(n) = ∑m∈(Z/kZ)∗epii(s(m,k)−2nmk )where the coefficients are written in terms of the Dedekind sums(m,k) =k−1∑l=1lk(mlk−⌊mlk⌋−12).Just as we treated ρ(E) as the inverse Laplace transform of Z(β ), the partition sumis an integral transform of its generating function.p(n) =12pii∫γF(q)qn+1dqInstead of a vertical line, the contour γ that leads to (6.17) is a beautiful shape builtup from fractals in the complex plane. A very readable thesis on this derivation is[89].The main fact about F(q) used in the proof is that it is an imaginary exponentialtimes a function which plays nicely with the modular group — as it would be evenif our theory were not free. Under the modular generators (6.13), the Dedekind etafunction transforms in the following way:η(τ+1) = epii12 η(τ)η(−1τ)=√−iτη(τ) . (6.18)The first of these equalities is trivial. The second is not but there is a clever proofcontaining no more than a page of algebra [90]. A function with transformationproperties similar to (6.18) is called a modular form. More precisely, a holomor-113phic function f in the upper half plane is a modular form of weight w if∣∣∣∣ f(−1τ)∣∣∣∣= τw | f (τ)| .According to this definition, η is a modular form of weight 12 . The Rademacherformula can be rederived for (multiples of) other modular forms. The result is[91, 92]pw(n) = 2pi ∑m− c24<0(n− c24∣∣m− c24∣∣)w−12pw(m)∞∑k=11kKl(n−c24,m−c24;k)I1−w(4pik√∣∣∣m−c24∣∣∣(n−c24))(6.19)where the Kloosterman sum is defined byKl(n,m;k) = ∑d∈(Z/kZ)∗e2piik (dn+d−1m) .For our purposes, we will choose a weight w and use k = m = 0. Uniqueness of thevacuum gives pw(0) = 1 and the central charge is 2 when we have left movers andright movers. This formula can be understood as the origin of the Bessel functionin (2.6).In order to use the partition sum and its asymptotics to study ρ(E,P), there aresome changes that must be made to our theory of positively indexed oscillators. Tostart, we should introduce negatively indexed oscillators to make it more like a fieldtheory in momentum space. An excitation of any oscillator increases the energy.Whether it increases or decreases momentum depends on whether the frequency ispositive or negative. This means that in the contribution of the newly introducedoscillators, τ2 has the same sign as in Z1(τ) and τ1 has the opposite. Including asum of e−2piinτ∗Boltzmann factors along with our sum of e2piinτ Boltzmann factorssuggests that the partition function for the doubly infinite family of oscillators is|η(τ)|−2. This is almost correct, but we need to remember that the Casimir contri-bution doubles instead of vanishes when we add more oscillators. Fixing this, the114Figure 6.2: The theory we have discussed so far resembles a harmonic po-tential where arbitrarily many non-interacting bosons can be piled intoeach level.appropriate partition function is given byZ2(τ) = η−1(τ)η−1(−τ∗)∗ . (6.20)This is still not modular invariant because we have not yet included the oscillatorwith zero frequency. The zero mode has a continuous spectrum which is quadraticin the free variable ξ [87]. This can be seen e.g. in the scalar field mode expan-sion that was discussed in the context of the string action (2.8). The most exactexpression for the partition function of scalar field theory is thereforeZ3(τ) = Z2(τ)∫ ∞−∞e−piℑτξ2 dξ2pi=12pi√ℑτη−1(τ)η−1(−τ∗)∗ . (6.21)To summarize these results, the partition function Z1 is a modular form of weight−12 . By (6.19), its density of states is asymptotic to the partition sum; p− 12(n) ∼14√3nepi√23 n. Doubling the number of oscillators, we made the partition functionZ2 which is a modular form of weight −1. Its density of states has the behaviour115p−1(n) ∼ 31412n54e2pi√13 n. The partition function Z3, since it describes a CFT, is amodular form of weight 0. If we take the density of states p0(n) ∼ 314√12n34e2pi√13 nand substitute n = V E2pi for the conformal weight, our expression agrees with (2.4)which gave ρ(E) for a free CFT from an entirely different setup.6.2.2 Convoluted functionsEvidently, the Casimir term and the zero mode do not affect log pw(n) to leadingorder. It is therefore valid to use F(q)2 instead of Z3(τ) as the generating function.With this approximation, the convolution formula gives a very clear picture of howthe density of states is comprised.p0(n) ≈1n!dndqnF(q)2=n∑k=0p(k)p(n− k)The first term p(0)p(n) has n right movers and 0 left movers and corresponds toenergy n and momentum n. The last term p(n)p(0) has 0 right movers and n leftmovers and corresponds to energy n and momentum −n. The middle term p(n2)2has a left mover for every right mover and corresponds to energy n and momentum0. This tells us that in a 1+1-dimensional possibly interacting CFT,ρ(E,P) = 12p(V4pi (E−P))p(V4pi (E +P))(6.22)is a good approximation to the momentum restricted density of states. This onlybecomes difficult to evaluate when |P| 6= E are of the same order.Essentially we derived (6.22) by starting with a function whose asymptoticbehaviour is log p(n) ∼ pi√23 n. The convolution of two of them yields a functionthat follows log p0(n) ∼ pi√43 n. In other words, convolution doubles the centralcharge. At least for free theories, this phenomenon can be readily seen in higher116dimensions. The crudest form of (2.4) islogρ(E)∼((d +1)d+1ddAV Ed) 1d+1.It will be helpful to write this density of states explicitly as ρ(E;A) from now on.For a bosonic theory, we had A proportional to s and for a fermionic theory, we hadA proportional to s∗. Writing the partition function ∏p Z(p)sZ∗(p)s∗, it becomesclear that ρ(·;A) is ρ(·; A2)convolved with itself. Performing substitutions in theresulting integral,ρ(E;A) =∫ E0ρ(ξ ; A2)ρ(E−ξ ; A2)dξ=12∫ E−Eρ(E−P2;A2)ρ(E +P2;A2)dP=∫ E0ρ(E−P2;A2)ρ(E +P2;A2)dP=∫ ∞0ρ(E−P2;A2)ρ(E +P2;A2)dP=1dωd∫Rdρ(E−|P|2;A2)ρ(E + |P|2;A2)dP . (6.23)In the second last step we have used the fact that densities of states are zero fornegative arguments. Part of the definition of the restricted density of states isρ(E;A) =∫Rdρ(E,P;A)dP . (6.24)Comparing (6.23) to (6.24), a plausible formula for the restricted density of statesin a free CFT islogρ(E,P)∼((d +1)d+12(2d)dAV (E−|P|)d) 1d+1+((d +1)d+12(2d)dAV (E + |P|)d) 1d+1.(6.25)In order to use the convolution formula, we factored the partition function in away that is not always valid. This means we again have results for a free CFT inarbitrary dimension and an arbitrary CFT in two dimensions. It is possible that fur-117ther results could be obtained using an entropy proposed by Erik Verlinde in 2000[93, 94]. This Cardy-Verlinde formula suggests that higher dimensional CFTs arethermodynamically more similar to their two-dimensional cousins than previouslythought.118Chapter 7Entropic dynamics of momentumRetracing our steps, there are nonlinear PDEs associated with changes in the mo-mentum distribution on a lattice. Each lattice direction, which we will call a largedirection, needs to have one evolution equation associated with it. Additionally,each site is allowed to have compact or small directions which do not show upin the PDEs. Therefore the ρ(E,P) we should use comes from a more restrictedfunction ρ(E,P,P∗) for the whole field theory and has P∗ for the small directionsintegrated out. Most of the simulations for our previous model had one large direc-tion. Because we used the density of states (2.46) for SYM on S3, it is tempting tosay that these simulations had three small directions. This is misleading as Figure7.1 shows. If each site is viewed as a miniature field theory whose spatial direc-tions are all orthogonal to the lattice, there is nothing left after integrating out P∗.The result is simply an unrestricted density of states ρ(E) precluding our attemptsto extend the model. What we really need are sites that include the directions ofthe lattice in addition to their compact directions. Therefore what we consideredpreviously was not S3 with radius r but S2× [0,r]. This is a background with onlytwo small directions because the non-periodic [0,r] matches up with the lattice. Byarguments in [7], the density of states for such a theory should be similar to (2.46).Including time, this theory with one large direction and two small ones lives onR2×S2. The Witten model, which is known to host plasma balls, has the back-ground R4×S1 suggesting that its lattice sites look like S1× [0,r]3. The equationsthat follow will have d referring to the number of large spatial dimensions only.119(a) Sharing no directions(b) Sharing one directionFigure 7.1: If one uses a density of states for two spatial dimensions, both ofthese situations are compatible with the lattice being a line. One has anoverall dimensionality of 2+1, the other 3+1. The task of introducingmomentum to our model forces us to address this ambiguity.7.1 Proceeding by analogyIn order to derive our PDE for the energy∂E∂ t = ∂i(C(E)ρ2(E)∂idlogρ(E)dE), (7.1)we took the continuum limit of the equation∂nc∂ t = ∑〈b,c〉∑k 6=0kW(nc,nb)→(nc+k,nb−k) . (7.2)The main steps involved were assuming that k was some small amount of energy±ε and writing nc as E(x) where site b was “one lattice vector” away from site c.Then (7.2) became∂E(x)∂ t = ε ∑e∈{±e1,...,±ed}[W(E(x),E(x+ae))→(E(x)+ε,E(x+ae)−ε)−W(E(x),E(x+ae))→(E(x)−ε,E(x+ae)+ε)](7.3)and we took ε and a to zero. The sum runs over positive and negative versions ofthe standard basis vectors so that in a cubic lattice, all 2d nearest neighbours areaccessed.1207.1.1 Allowed transitions and ratesIn (7.3) above, ∂E(x)∂ t was proportional to ε , the smallest amount of energy allowedto move in one timestep. Therefore, we should presumably add an equation formomentum where ∂P(x)∂ t is proportional to some q, the smallest momentum vectorthat moves in one timestep. This can be written as qe, the length times some unitvector, where q→ 0 just like ε and a.However, q and ε are not independent. A CFT in d = 1 for instance has energyand momentum operators that can be written in terms of left movers and rightmovers:E = NR +NL−c12P = NR−NL .The only way to add to E is to excite either NR or NL with ε units of energy andthis necessarily adds ε to or subtracts ε from P. In this case we must have q = ε .We will make the assumption that the other theories we wish to model also havethe energy transferred equal to the magnitude of the momentum transferred.Now that we know E increases or decreases by ε whenever P increases ordecreases by εe, an allowed transition takes the form(E(x),P(x),E(x+ae),P(x+ae))→ (E(x)±ε,P(x)−εe′,E(x+ae)∓ε,P(x+ae)+εe′)where e is a unit vector indicating the two sites involved and e′ is a priori someother unit vector. However, it does not make sense for momentum in the verticaldirection to move between sites that are horizontally displaced and it does notmake sense for momentum pointing in the up direction to move down. Thereforewe require that e′ = e. This is where we see that the vector of large momentumcomponents P lives in the same number of dimensions as the lattice vector x.When focusing on energy distributions, the form of the transition rates in (7.3)followed from thermodynamic arguments asW(E(x),E(x+ae))→(E(x)+ε,E(x+ae)−ε) =C(E(x)+E(x+ae)2)ρ(E(x)+ε)ρ(E(x+ae)−ε)(7.4)121with the number of states having energy E given by ρ(E). The analogous transitionrate involving momentum should depend on ρ(E,P) in the same way. We are ledto:W(E(x),P(x),E(x+ae),P(x+ae))→(E(x)+ε,P(x)−εe′,E(x+ae)−ε,P(x+ae)+εe′) =δe,e′C(E(x)+E(x+ae)2,P(x)+P(x+ae)2)ρ(E(x)+ ε,P(x)− εe′)ρ(E(x+ae)− ε,P(x+ae)+ εe′) . (7.5)7.1.2 A tensor identityConsider what happens when we differentiate E(x+ae) twice with respect to a.∂∂aE(x+ae) = ei∂iE(x+ae)∂ 2∂a2 E(x+ae) = eie j∂i∂ jE(x+ae) (7.6)Note that ei and e j without the boldface are not the ith and jth standard basis vectors,they are the ith and jth components of the general unit vector e. All of the terms thatwe must differentiate with respect to a occur inside a sum over e so we must knowhow to deal with sums of components of unit vectors. When we sum eie j, there isonly one standard basis vector for which ei is nonzero and only one standard basisvector for which e j is nonzero. They have to be the same one for the resulting sumto be nonzero, so it is clear that:∑e∈{±e1,...,±ed}eie j = 2δi j .We used this fact when deriving the PDE for our first model. For reasons that arenot initially clear, we will show that one can get the same answer up to a constantby replacing the sum over lattice vectors with a surface integral over a sphere.Claim 2. ∫Sd−1eie jdSe = ωdδ i jwhere ωd is the volume of the unit ball in Rd .122Proof. If i = j, we simply apply the divergence theorem to the identity vector field.Let v(e) = e, defined on the unit ball Bd . Then,∫Bd∇ ·vde =∫∂Bdv ·ndSefor this vector field readsd∫Bdde =∫Sd−1(e1)2 + · · ·+(ed)2dSebecause if the unit vector e extends from the origin to a point on Sd−1, the unitnormal to the surface at this point is just e. By symmetry, the right hand side is dcopies of what we are trying to compute, so this cancels the d on the left hand sideyielding:ωd =∫Sd−1(ei)2dSe .To see that this integral vanishes when i and j are different, parametrize the unitsphere asx1 = cosθ1x2 = sinθ1 cosθ2. . .xd−1 = sinθ1 . . .sinθd−2 cosθd−1xd = sinθ1 . . .sinθd−2 sinθd−1and without loss of generality, choose ei = xd and e j = xd−1. The desired integralis then∫ 2pi0∫ pi0. . .∫ pi0[sin2 θ1 . . .sin2 θd−2 sinθd−1 cosθd−1]sind−2 θ1 sind−3 θ2 . . .sinθd−2dθ1 . . .dθd−2dθd−1where the part in brackets is eie j. Since the area element only goes up to θd−2,the part involving θd−1 is just a sine and a cosine which we know integrates tozero.Unlike in our first model, equations involving momenta will have several places123in which there are four components of e. This will make it important that weintegrate rather than sum. A sum of four components will give zero unless theindices are all the same.∑e∈{±e1,...,±ed}eie jekel =2 i = j = k = l0 otherwiseThe continuous version, on the other hand, is a much nicer object.Claim 3. ∫Sd−1eie jekeldSe =ωdd +2(δi jδkl +δikδ jl +δilδ jk)where ωd is the volume of the unit ball in Rd .Proof. We will first prove the special case when i = j = k = l. Define a vector fieldv : Bd → Rd which cubes every component of its argument:vi(e) = (ei)3 .Using the divergence theorem on this, we get:∫Bd∇ ·vde =∫∂Bdv ·ndSe3∫Bd(e1)2 + . . .(ed)2de =∫Sd−1(e1)4 + · · ·+(ed)4dSe3dωd∫ 10rd+1dr = d∫Sd−1(ei)4dSe3ωdd +2=∫Sd−1(ei)4dSe .124We will now apply this to the case when i and k are different.∑k 6=i∫Sd−1(ei)2(ek)2dSe =d∑k=1∫Sd−1(ei)2(ek)2dSe−∫Sd−1(ei)4dSe=∫Sd−1(ei)2dSe−∫Sd−1(ei)4dSe= ωd−3ωdd +2= ωdd−1d +2(7.7)We summed k over all but one of the d choices, so by symmetry, this should bed−1 times the actual value. This shows that:∫Sd−1eie j(ek)2dSe =ωdd +2δi j .We have computed the integral when k and l are the same. Similarly, if k and iwere the same, the result would be proportional to δ jl and if k and j were the same,the result would be proportional to δil . This covers the three ways to have pairs ofequal indices. It remains to be seen that the expression vanishes when i, j, k and lare all different. As before, we can show this using explicit co-ordinates.ei = sinθ1 . . .sinθd−2 sinθd−1e j = sinθ1 . . .sinθd−2 cosθd−1ek = sinθ1 . . .sinθd−3 cosθd−2el = sinθ1 . . .sinθd−4 cosθd−3Again, ei and e j contribute an odd function of θd−1 while the surface measure doesnot depend on θd−1. The other components cannot change this because ek and elare both distinct from ei and e j.7.1.3 Setting up the equationsIn (7.3), ∂E(x)∂ t was schematically given by “W for a transition that adds ε to sitex” minus “W for a transition that subtracts ε from site x”, summed over e and125multiplied by ε . We will write out this type of expression except we will integrateinstead of sum.∂E(x)∂ t = ε∫Sd−1W(E(x),P(x),E(x+ae),P(x+ae))→(E(x)+ε,P(x)−εe,E(x+ae)−ε,P(x+ae)+εe)− W(E(x),P(x),E(x+ae),P(x+ae))→(E(x)−ε,P(x)−εe,E(x+ae)+ε,P(x+ae)+εe)dSe(7.8)Note that P(x) always receives a −εe contribution because we are adopting a con-vention where x+ae is the “other site”. If our convention had x−ae as the othersite, P(x) would receive a +εe contribution. The fact that there are two ways toraise the energy E(x) 7→ E(x) + ε (adding a left mover from the right site andadding a right mover from the left site) is accounted for by the integral that causese to change direction. Writing down the momentum equation and being carefulabout the same type of thing, we get:∂P(x)∂ t = −ε∫Sd−1e[W(E(x),P(x),E(x+ae),P(x+ae))→(E(x)+ε,P(x)−εe,E(x+ae)−ε,P(x+ae)+εe)+ W(E(x),P(x),E(x+ae),P(x+ae))→(E(x)−ε,P(x)−εe,E(x+ae)+ε,P(x+ae)+εe)]dSe . (7.9)Substituting our transition rates (7.5), these become:∂E(x)∂ t = ε∫Sd−1C(E(x)+E(x+ae)2,P(x)+P(x+ae)2)[ρ(E(x)+ ε,P(x)− εe)ρ(E(x+ae)− ε,P(x+ae)+ εe)−ρ(E(x)− ε,P(x)− εe)ρ(E(x+ae)+ ε,P(x+ae)+ εe)]dSe∂P(x)∂ t = −ε∫Sd−1eC(E(x)+E(x+ae)2,P(x)+P(x+ae)2)[ρ(E(x)+ ε,P(x)− εe)ρ(E(x+ae)− ε,P(x+ae)+ εe)+ρ(E(x)− ε,P(x)− εe)ρ(E(x+ae)+ ε,P(x+ae)+ εe)]dSe .(7.10)Our task in the following section will be to differentiate the right hand sides withrespect to ε and a and plug in ε = 0 and a = 0. Recall that in our model for theenergy, the only non-vanishing derivative of order≤ 4 was ∂4∂ε2∂a2 . The momentum126equation in (7.10) has other low order derivatives that do not vanish. For example,set C = 1 and compute ∂4∂ε∂a3 . Differentiating with respect to ε once and settingε = 0 simply removes the factor of ε in front. Therefore this contribution to ∂Pi∂ t is:−2∫Sd−1eiρ(E(x),P(x))∂ 3∂a3 ρ(E(x+ae),P(x+ae))|a=0 dSe= −2ρ(E(x),P(x))∂ j∂k∂lρ(E(x),P(x))∫Sd−1eie jekeldSe=−6ωddρ(E(x),P(x))∂i∂ j∂ jρ(E(x),P(x)) . (7.11)This produces a rotationally covariant equation precisely because of the integral.From the discrete sum of eie jekel that we computed previously, we can see that thiswould have given us −4ρ∂i∂i∂iρ which does not transform as a vector.7.2 The continuum limitRewriting (7.10) for brevity,∂E∂ t = X(ε,a) = εX˜(ε,a)∂Pi∂ t = Yi(ε,a) = εY˜i(ε,a) .We must Taylor expand X and Yi around small arguments.7.2.1 To first non-vanishing orderIt is clear that if we do not differentiate at all with respect to ε , X and Yi will vanishat ε = 0 no matter how many times were differentiate with respect to a. It is alsotrue that X˜ and Y˜i will vanish identically if we do not differentiate them with respectto a. The integrand in X˜ becomes the zero function of ε once we plug in a = 0.The integrand in Y˜i does not but it becomes proportional to ei which is odd.X(0,a) = 0 = X(ε,0) and Yi(0,a) = 0 = Yi(ε,0) so the lowest order deriva-tive that could possibly survive is ∂2∂ε∂a . Looking at X(ε,a) once more,∂X∂ε∣∣∣ε=0=X˜(a,0) which is the zero function of a. Therefore, ∂2X∂ε∂a∣∣∣a=0ε=0= 0. We will now127compute this same derivative for Yi.∂ 2Yi∂ε∂a∣∣∣∣a=0ε=0=∂Y˜i∂a∣∣∣∣a=0ε=0= −2∫Sd−1eiρ(E(x),P(x))∂∂aC(E(x)+E(x+ae)2,P(x)+P(x+ae)2)ρ(E(x+ae),P(x+ae))|a=0 dSe= −2ρ(12ρ∂ jC+C∂ jρ)∫Sd−1eie jdSe= −ωd∂i(Cρ2)What this shows is that to second order,∂E∂ t = 0∂Pi∂ t = −εaωd∂i(Cρ2) . (7.12)This is consistent with the intuition about which way a distribution of momen-tum should move. In (7.12), Cρ2 is a function of the energy (which is static) andthe momentum magnitude |P|. Consider the simplest case where we let d = 1 andTaylor expand Cρ2 as some constant plus |P|:∂P(x, t)∂ t ∝−∂ |P(x, t)|∂x .A solution to the above equation is P(x, t) = f (x− t) where f is a non-negativefunction. Because of the absolute value, another solution is P(x, t) = − f (x+ t).This confirms that the direction of motion for a disturbance is equal to the sign ofthe disturbance; f (x− t) is a lump of positive (right) momentum that moves to theright, while− f (x+ t) is a lump of negative (left) momentum that moves to the left.This behaviour appears to rectify a shortcoming of our last model; an ejected pieceof a plasma ball propagating through empty space.For our purposes, it is not enough to stop at the second order expansion wherethe energy is constant in time. As we did with our first model, we will compute asmany derivatives of X as we need to see the energy dynamics. We argued above128that we must differentiate X at least once with respect to a and at least twice withrespect to ε . In fact, we must differentiate even more than this; ∂ 3X∂ε2∂a still vanishes.What we find by looking at the fourth order derivatives is:∂E∂ t =14ε2a2 ∂4X∂ε2∂a2∣∣∣∣a=0ε=0. (7.13)When evaluating this and similar expressions, we will surpress the steps in calcu-lating the ε derivatives, saving them for the appendix. Additionally, we will adoptthe notation ρ+ = ρ(E(x+ae),P(x+ae)) and C+ =C(E(x)+E(x+ae)2 ,P(x)+P(x+ae)2)and recall that:∂ρ+∂a = e j∂ jρ+∂C+∂a =12e j∂ jC+ . (7.14)If we apply this to (7.13), our steps are:∂ 4X∂ε2∂a2∣∣∣∣a=0ε=0= 2∂ 3X˜∂ε∂a2∣∣∣∣a=0ε=0= 4∫Sd−1∂ 2∂a2[C+(ρ+∂ρ∂E −ρ∂ρ+∂E)]∣∣∣∣a=0dSe= 4∫Sd−1ei∂∂a[12∂iC+(ρ+∂ρ∂E −ρ∂ρ+∂E)+C+(∂iρ+∂ρ∂E −ρ∂i∂ρ+∂E)]∣∣∣∣a=0dSe= 4∫Sd−1eie j[C(∂i∂ jρ∂ρ∂E −ρ∂i∂ j∂ρ∂E)−∂iC(ρ∂ j∂ρ∂E −∂ jρ∂ρ∂E)]dSe= 4ωd[C(∂i∂iρ∂ρ∂E −ρ∂i∂i∂ρ∂E)−∂iC(ρ∂i∂ρ∂E −∂iρ∂ρ∂E)]= −4ωd∂i(Cρ2∂idlogρdE).This shows that to first non-vanishing order,∂E∂ t =−ε2a2ωd∂i(Cρ2∂i∂ logρ∂E). (7.15)Notice that when we only considered the energy, we dropped factors of ε and a129from (7.1). We must keep factors of ε and a in the PDEs for energy and momentumbecause they appear with different exponents. The dominant contribution to ∂P∂ tincludes εa, while in the dominant contribution to ∂E∂ t , it is (εa)2. The ratio betweenthese factors is a physical quantity because it tells us how long the diffusion timescale in (7.15) is relative to the transport time scale in (7.12).7.2.2 To first consistent orderWe now have non-trivial differential equations for both energy and momentum.However, the former is correct to fourth order while the latter is only correct tosecond order. To be consistent, we will find further terms in the momentum PDE.The terms are∂Pi∂ t = εa∂ 2Yi∂ε∂a∣∣∣∣a=0ε=0+16ε3a ∂4Yi∂ε3∂a∣∣∣∣a=0ε=0+14ε2a2 ∂4Yi∂ε2∂a2∣∣∣∣a=0ε=0+16εa3 ∂4Yi∂ε∂a3∣∣∣∣a=0ε=0,(7.16)130which we will calculate one by one.∂ 4Yi∂ε3∂a∣∣∣∣a=0ε=0= 3∂ 3Y˜i∂ε2∂a∣∣∣∣a=0ε=0= 6∫Sd−1ei∂∂a[C+(∂ρ∂E − e j∂ρ∂Pj)(∂ρ+∂E − ek∂ρ+∂Pk)+C+(∂ρ∂E + e j∂ρ∂Pj)(∂ρ+∂E + ek∂ρ+∂Pk)−C+ρ+(∂ 2ρ∂E2 + e jek∂ 2ρ∂Pj∂Pk)−C+ρ(∂ 2ρ+∂E2 + e jek∂ 2ρ+∂Pj∂Pk)]∣∣∣∣a=0dSe= 6∫Sd−1eiel[2∂ρ∂E(12∂lC∂ρ∂E +C∂l∂ρ∂E)+2∂ρ∂Pje jek(12∂lC∂ρ∂Pk+C∂l∂ρ∂Pk)−(12∂lCρ+C∂lρ)(∂ 2ρ∂E2 + e jek∂ 2ρ∂Pj∂Pk)−ρ(12∂lC∂ 2ρ∂E2 +12∂Ce jek∂ 2ρ∂Pj∂Pk+C∂l∂ 2ρ∂E2 +Ce jek∂l∂ 2ρ∂Pj∂Pk)]dSe=6ωdd +2[∂lC(∂ρ∂E)2δ jk +∂lC∂ρ∂Pj∂ρ∂Pk+C∂l(∂ρ∂E)2δ jk+C∂l(∂ρ∂Pj∂ρ∂Pk)−ρ∂lC∂ 2ρ∂E2 δ jk−ρ∂lC∂ 2ρ∂Pj∂Pk−C∂lρ∂ 2ρ∂E2 δ jk−C∂lρ∂ 2ρ∂Pj∂Pk−Cρ∂l∂ 2ρ∂E2 δ jk−Cρ∂l∂ 2ρ∂Pj∂Pk](δi jδkl +δikδ jl +δilδ jk)= −6ωdd +2∂l[Cρ2(∂ 2 logρ∂E2 δ jk +∂ 2 logρ∂Pj∂Pk)](δi jδkl +δikδ jl +δilδ jk)131In the next one, the steps are very similar.∂ 4Yi∂ε2∂a2∣∣∣∣a=0ε=0= 2∂ 3Y˜i∂ε∂a2∣∣∣∣a=0ε=0= 4∫Sd−1eie j∂ 2∂a2[C+ρ+∂ρ∂Pj−C+ρ∂ρ+∂Pj]∣∣∣∣a=0dSe= 4∫Sd−1eie jek∂∂a[(12∂kC+ρ++C+∂kρ+)∂ρ∂Pj−(12∂kC+∂ρ+∂Pj+C+∂k∂ρ+∂Pj)ρ]∣∣∣∣a=0dSe= 4∫Sd−1eie jekel[(∂kC∂lρ+C∂k∂lρ)∂ρ∂Pj−(∂kC∂l∂ρ∂Pj+C∂k∂l∂ρ∂Pj)ρ]dSe=4ωdd +2(∂kC∂lρ∂ρ∂Pj+C∂k∂lρ∂ρ∂Pj−∂kC∂l∂ρ∂Pjρ−C∂k∂l∂ρ∂Pjρ)(δi jδkl +δikδ jl +δilδ jk)= −4ωdd +2∂k(Cρ2∂l∂ logρ∂Pj)(δi jδkl +δikδ jl +δilδ jk)The final expression for ∂4Yi∂ε3∂a∣∣∣a=0ε=0involves no more than one derivative of C be-caues there is only one derivative with respect to a. Since ∂4Yi∂ε2∂a2∣∣∣a=0ε=0has two aderivatives, there is a chance that the expression will involve two derivatives of C.However, we know from above that the double derivatives of C cancel out. Thiswill not happen when we compute ∂4Yi∂ε∂a3∣∣∣a=0ε=0. In this case all three derivatives of Cwill survive and we will have to extend (7.14) so that we know how to deal with upto three derivatives of C+. A simple C(E(x)) can be used to illustrate the problem.∂C∂x =∂C∂E∂E∂x∂ 2C∂x2 =∂ 2C∂E2(∂E∂x)2+∂C∂E∂ 2E∂x2∂ 3C∂x3 =∂ 3C∂E3(∂E∂x)3+3∂ 2C∂E2∂E∂x∂ 2E∂x2 +∂C∂E∂ 3E∂x3132These expansions are straightforward. However, when we consider C+, we do notget the benefit of being able to write multiple derivatives in a compact form.∂C+∂a∣∣∣∣a=0=12∂C∂E∂E∂x=12∂C∂x∂ 2C+∂a2∣∣∣∣a=0=∂∂a(12∂C+∂E∂E+∂x)∣∣∣∣a=0=14∂ 2C∂E2(∂E∂x)2+12∂C∂E∂ 2E∂x26=14∂ 2C∂x2∂ 3C+∂a3∣∣∣∣a=0=∂∂a[14∂ 2C+∂E2(∂E+∂x)2+12∂C+∂E∂ 2E+∂x2]∣∣∣∣∣a=0=18∂ 3C∂E3(∂E∂x)3+34∂ 2C∂E2∂E∂x∂ 2E∂x2 +12∂C∂E∂ 3E∂x36=18∂ 3C∂x3 (7.17)133We must be careful to use these relations in the last part of the calculation we arecarrying out.∂ 4Yi∂ε∂a3∣∣∣∣a=0ε=0=∂ 3Y˜i∂a3∣∣∣∣a=0= −2∫Sd−1eiρ∂∂a3 (C+ρ+)∣∣∣∣a=0dSe= −2ωdd +2[C∂ j∂k∂lρ+32∂ jC∂k∂lρ+3∂ jρ(14∂ 2C∂E2 ∂kE∂lE +12∂ 2C∂E∂Pm∂kPm∂lE+14∂ 2C∂Pm∂Pn∂kPm∂lPn +12∂C∂E ∂k∂lE +12∂C∂Pm∂k∂lPm)+ρ(18∂ 3C∂E3 ∂ jE∂kE∂lE+38∂ 3C∂E2∂Pm∂ jPm∂kE∂lE +38∂ 3C∂E∂Pm∂Pn∂ jPm∂kPn∂lE +18∂ 3C∂Pm∂Pn∂Po∂ jPm∂kPn∂lPo+34∂ 2C∂E2 ∂ j∂kE∂lE +34∂ 2C∂Pm∂E∂ jPm∂k∂lE +34∂ 2C∂E∂Pm∂ j∂kPm∂lE+34∂ 2C∂Pm∂Pn∂ j∂kPm∂lPn +12∂C∂E ∂ j∂k∂lE +12∂C∂Pm∂ j∂k∂lPm)](δi jδkl +δikδ jl +δilδ jk)We saw in (7.11) that the fourth order term in ∂Pi∂ t involving εa3 is quite simplewhen C = 1. The above shows that it is considerably more messy for a general C.134Putting together all of the derivatives that have been computed, the end result is:∂E∂ t = −ε2a2ωd∂i(Cρ2∂i∂ logρ∂E)∂Pi∂ t = −εaωd∂i(Cρ2)− ε3a ωdd +2∂l[Cρ2(∂ 2 logρ∂E2 δ jk +∂ 2 logρ∂Pj∂Pk)](δi jδkl +δikδ jl +δilδ jk)−ε2a2 ωdd +2∂k(Cρ2∂l∂ logρ∂Pj)(δi jδkl +δikδ jl +δilδ jk)−εa3 ωd3(d +2)[C∂ j∂k∂lρ+32∂ jC∂k∂lρ+3∂ jρ(14∂ 2C∂E2 ∂kE∂lE +12∂ 2C∂E∂Pm∂kPm∂lE+14∂ 2C∂Pm∂Pn∂kPm∂lPn +12∂C∂E ∂k∂lE +12∂C∂Pm∂k∂lPm)+ρ(18∂ 3C∂E3 ∂ jE∂kE∂lE+38∂ 3C∂E2∂Pm∂ jPm∂kE∂lE +38∂ 3C∂E∂Pm∂Pn∂ jPm∂kPn∂lE +18∂ 3C∂Pm∂Pn∂Po∂ jPm∂kPn∂lPo+34∂ 2C∂E2 ∂ j∂kE∂lE +34∂ 2C∂Pm∂E∂ jPm∂k∂lE +34∂ 2C∂E∂Pm∂ j∂kPm∂lE +34∂ 2C∂Pm∂Pn∂ j∂kPm∂lPn+12∂C∂E ∂ j∂k∂lE +12∂C∂Pm∂ j∂k∂lPm)](δi jδkl +δikδ jl +δilδ jk). (7.18)These equations take up a lot of space and that is all because of the last term.7.2.3 Static configurationsFor a general density of states ρ(E,P) and function C(E,P), (7.18) is only staticwhen energy and momentum are both uniform. If only one of them is, derivativesof ρ and C will not vanish and the equations will introduce non-uniformities. If themomentum is initially zero everywhere, we can compare the energy equation justderived with the one from our first model:∂E∂ t = −ε2a2ωd∂i(C(E,0)ρ2(E,0)∂i∂ logρ(E,0)∂E)∂E∂ t = −ε2a2ωd∂i(C(E)ρ2(E)∂idlogρ(E)dE).The unrestricted density of states ρ(E) should be qualitatively similar to the zeromomentum density of states ρ(E,0). Therefore these models predict similar diffu-sive behaviour at early times. However, this diffusion generates momentum which135can cause very different behaviour to occur at late times once the non-uniformitiesbecome significant. For example, a Hagedorn density of states with P(x,0) = 0requires ∂E∂ t to be zero (indicating slow diffusion) at t = 0 but not necessarily latertimes. This supports the idea that our first model overestimates the amount of timetaken for a cluster of energy to diffuse.Before, we saw that the dynamics were frozen for any energy distribution whenρ(E) was a Hagedorn density of states. If such a density of states were to exist forthis system of PDEs, logρ(E,P) would have to be linear in both E and P. Ofthe five terms in (7.18), this form causes the three involving logarithms to vanish.What about the term in momentum proportional to εa? This will not vanish unlesswe choose a special C function as well. The one to choose is C ∝ ρ−2. It isnot immediately obvious, but this choice causes the one remaining term (the oneproportional to εa3) to vanish as well. This is because∂ 4Yi∂ε∂a3∣∣∣∣a=0ε=0= −2∫Sd−1ei∂∂a3 (C+ρ+ρ)∣∣∣∣a=0dSe= −2∫Sd−1ei∂∂a3(e−2(aE+b jPj2 +βHE++βHv jPj+2)eβHE++βHv jPj+eβHE+βHv jPj)∣∣∣∣∣a=0= −2∫Sd−1ei∂∂a3 1∣∣∣∣a=0dSe= 0 .Therefore C ∝ ρ−2 and C ∝ 1 appear to be natural choices once again. We will seethat the part of our equation that looks complicated for a more general C simplifiesconsiderably if we take the linearization.7.3 Consistency checkHydrodynamics is concerned with fluctuations of a system around some equilib-rium state. This system is often a field theory like the one being modelled by (7.18).An encouraging result we have seen already is the solution at order εa in terms ofleft and right moving waves. These are exact solutions to hydrodynamics for a CFTin two spacetime dimensions. It will not be possible to compare solutions in theother cases of interest so we will focus on matching coefficients in the linearized136equations.7.3.1 Linearized equationsWe will use (E0,P0) as our equilibrium state and insertE(x, t) = E0 + E˜(x, t)P(x, t) = P0 + P˜(x, t)into our equations of motion leaving only one power of the perturbations (E˜, P˜).Working with the term inside the derivatives in our energy equation (7.18), we maywrite∂ logρ(E,P)∂E ≈∂ logρ(E0,P0)∂E + E˜∂ 2 logρ(E0,P0)∂E2 + P˜m∂ 2 logρ(E0,P0)∂E∂Pm=∂ logρ0∂E + E˜∂ 2 logρ0∂E2 + P˜m∂ 2 logρ0∂E∂Pmwhere we have used the subscript 0 to denote evaluation at (E0,P0). Taking onederivative kills the constant so we have:Cρ2∂i∂ logρ∂E ≈ Cρ2∂i[∂ logρ0∂E + E˜∂ 2 logρ0∂E2 + P˜m∂ 2 logρ0∂E∂Pm]= Cρ2(∂ 2 logρ0∂E2 ∂iE˜ +∂ 2 logρ0∂E∂Pm∂iP˜m)≈ C0ρ20(∂ 2 logρ0∂E2 ∂iE˜ +∂ 2 logρ0∂E∂Pm∂iP˜m).When we use this type of logic on the momentum equation in (7.18), a nice thinghappens. Only a few terms in the long εa3 contribution do not have derivativesof E˜ and P˜ multiplied together: the one with ∂ j∂k∂lρ , the one with ∂ j∂k∂lE andthe one with ∂ j∂k∂lPm. Carrying out the straightforward linearization, we see that137(7.18) becomes:∂ E˜∂ t = −ε2a2ωdC0ρ20(∂ 2 logρ0∂E2 ∂i∂iE˜ +∂ 2 logρ0∂E∂Pm∂i∂iP˜m)∂ P˜i∂ t = −εaωdρ20[(∂C0∂E +2C0∂ logρ0∂E)∂iE˜ +(∂C0∂Pm+2C0∂ logρ0∂Pm)∂iP˜m]−ε3a ωdd +2[[C0ρ20(∂ 3 logρ0∂E3 δ jk +∂ 3 logρ0∂E∂Pj∂Pk)+ρ20(∂ 2 logρ0∂E2 δ jk +∂ 2 logρ0∂Pj∂Pk)(∂C0∂E +2C0∂ logρ0∂E)]∂lE˜+[C0ρ20(∂ 3 logρ0∂E2∂Pmδ jk +∂ 3 logρ0∂Pj∂Pk∂Pm)+ρ20(∂ 2 logρ0∂E2 δ jk +∂ 2 logρ0∂Pj∂Pk)(∂C0∂Pm+2C0∂ logρ0∂Pm)]∂lP˜m](δi jδkl +δikδ jl +δilδ jk)−ε2a2 ωdd +2C0ρ20[∂ 2 logρ0∂Pj∂E∂k∂lE˜ +∂ 2 logρ0∂Pj∂Pm∂k∂lP˜m](δi jδkl +δikδ jl +δilδ jk)(7.19)−εa3 ωd2(d +2)ρ20[(∂C0∂E +2C0∂ logρ0∂E)∂i∂ j∂ jE˜ +(∂C0∂Pm+2C0∂ logρ0∂Pm)∂i∂ j∂ jP˜m].These equations are still rather long. The εa3 term is now only one line but thelinearization has made the ε3a term expand. The situation can be improved if weassume that the momentum P0 around which we linearize is not only a constantfunction but a constant close to zero. This is reminiscent of a common assumptionin hydrodynamics where the fluid velocity must be much less than the speed ofsound to yield propagating hydrodynamic modes [75]. We also used this assump-tion previously when we linearized hydrodynamics. This allows us to remove allcoefficients above that involve an odd number of derivatives with respect to com-ponents of P. After all, functions like C and logρ are spherically symmetric in P,so their odd order derivatives vanish at the origin. Moreover, second derivatives138become proportional to δ jk:∂ 2 logρ∂Pj∂Pk=∂∂Pj(∂ logρ∂ |P|Pk|P|)=∂ 2 logρ∂ |P|2PjPk|P|2+∂ logρ∂ |P|∂∂Pj(Pk|P|)=∂ 2 logρ∂ |P|2PjPk|P|2+∂ logρ∂ |P||P|δ jk−PjPk|P||P|2=∂ logρ∂ |P|δ jk|P|+PjPk|P|2(∂ 2 logρ∂ |P|2 −1|P|∂ logρ∂ |P|)≈∂ logρ∂ |P|δ jk|P|.If we substitute this into (7.19), we arrive at:∂ E˜∂ t = −ε2a2ωdC0ρ20∂ 2 logρ0∂E2 ∂i∂iE˜∂ P˜i∂ t = −εaωdρ20(∂C0∂E +2C0∂ logρ0∂E)∂iE˜−ε3aωd[C0ρ20(∂ 3 logρ0∂E3 +1|P0|∂ 2 logρ∂E∂ |P|)+ρ20(∂C0∂E +2C0∂ logρ0∂E)]∂iE˜−ε2a2 ωdd +2C0ρ201|P0|∂ logρ0∂ |P|(∂ j∂ jP˜i +2∂i∂ jP˜j)−εa3 ωd2(d +2)ρ20(∂C0∂E +2C0∂ logρ0∂E)∂i∂ j∂ jE˜ . (7.20)7.3.2 Comparison with hydrodynamicsWe are now in a position to compare (7.20) to hydrodynamics. Since our equationshave second and third derivatives, we should not attempt to relate them to idealhydrodynamics. In fact, ideal hydrodynamics has a conserved entropy and oneof the fundamental assumptions in our model was that the entropy was driven toincrease [95]. We will have to use a stress-energy tensor that includes derivatives— a situation we referred to as dissipative hydrodynamics.In the Landau frame, dissipative hydrodynamics came from (6.4). For our pur-139poses, it is enough that these equations have two derivatives even though (7.20) hasthree. For a proper comparison, we need the linearization of the hydro equations.An alternative expression for the system (6.5) and (6.6) with no charge is∂ ε˜∂ t + ikP˜‖ = 0∂ P˜‖∂ t + ik∂P0∂ε ε˜+ γsk2P˜‖ = 0∂ P˜⊥∂ t + γηk2P˜⊥ = 0 .In this form, we have implicitly Fourier transformed the fields and decomposed themomentum into a part parallel to k and a part perpendicular to k. The coefficientsγs =d−2d η0 +ζ0ε0 +P0γη =η0ε0 +P0have been defined as in [75]. The most obvious difference we see is that the energydecouples in (7.20) instead of satisfying a continuity equation. If a model withonly energy is governed by the heat equation, one hopes that this is an effectivedescription of a continuity equation with Fick’s law: Pi ∝ ∂iE. The linearization of(7.18) has shown that this is not the case; the heat equation for energy is still anexplicit part of our model.One way around this is to consider an incompressible fluid: ∂iP˜i = 0. In hy-drodynamics this is equivalent to the approximation that the energy is constant intime. For this to be true in the model (7.20), we need an energy that is alreadyuniform so that only momentum is flowing. The simplest comparison we can makeis between the incompressible hydro equations and a pure momentum version ofour model. Since this is the ε2a2 term in (7.20), the relationsγη = ε2a2ωdd +2C0ρ201|P0|∂ logρ0∂ |P|γs = 2γη (7.21)are produced. This implies ζ = 4−d2d η , which is much different from the results140of [78]. For a conformal fluid in any number of dimensions, tracelessness of thestress-energy tensor demands ζ = 0. This also differs from the results of [96]which considered the hydrodynamics of a non-conformal theory of holographicQCD. One would have to contend with this problem even if she found a way ofgoing beyond incompressible hydrodynamics (e.g. introducing auxiliary conservedcurrents to cancel the problematic terms in (7.20)).These problems (no continuity equation and the coefficients of ∂ j∂ jPi and∂i∂ jPj not being independent) suggest that our entropic model is fundamentallyincompatible with the long distance effective description that is hydrodynamics.Some evidence for this can be seen in our expression for the mean-field variance(3.23). To have energy variances grow more slowly than squared energies, our Ein the denominator had to be large. Similarly, requiring |P| to be large is the mostobvious way of ensuring that the growth of the momentum covariance matrix issmall. While not necessarily incompatible with hydrodynamics, this is certainlyincompatible with the linearization of it.7.3.3 One more simulation-2 -1 0 1 2x0. = 0.1t = 0.3t = 0.5t = 0.7(a) Energy-1 1x-0.50.5Pxt = 0.1t = 0.3t = 0.5t = 0.7(b) MomentumFigure 7.2: Energy and momentum profiles for the simulation that only hasan x-axis. They are shown up until the time when momentum values of±0.9 form.The most fundamental problem with our first model is that it only had welldefined decay times in one dimension. The only way to make this model sensible141is to remove the low energy phase from the density of states, allowing (3.23) to bestrictly obeyed. This has the effect of removing decay times altogether leading toinfinitely long lived black holes. A pressing question is whether the decay timesof (7.18) remain comparable when moving from one dimension to the next. Since-2 -1 0 1 2x0. = 0t = 0.2t = 0.4(a) Energy-1 1x-0.50.5Pxt = 0t = 0.2t = 0.4(b) MomentumFigure 7.3: On the left are slices of the E distributions along the x-axis. Theseare spherically symmetric. On the right are Px distributions along the x-axis. These smoothly approach a y-axis value of zero as one rotates thedirection along which they are plotted. The Py distributions behave inan equal and opposite way.instantaneous extinction is a low energy effect, it suffices to use much easier initialconditions than the ones in Figure 5.4. Its presence is also agnostic to whetherthe field theory is strongly coupled or weakly coupled. Therefore, we may use therestricted density of states expressions (6.25) that are valid for free fields. Unfortu-nately, a numerical investigation of (7.18) shows that momenta grow very quicklyeven if they start from zero. This leads to a time scale for satisfying ||P||∞ ≈ ||E||∞.After this time, momenta at various points are comparable to the energies at thosepoints and (6.25) is no longer valid. This time scale is shorter than the black holedecay time and probably also shorter than the black hole thermalization time thatwe defined earlier. Nevertheless, we present results showing that this “momentumgeneration time” is similar in d = 1 and d = 2. There may be arguments involvingthe other time scales that follow from this.Figure 7.2 shows the results for d = 1. In this case, the sites only have direc-tions that are aligned with the lattice so the d = 1 restricted density of states is used.142Choices made for (7.18) are ε = a = 0.1 and C = 1. Since we are only interested inshort time dynamics, the simplest forward difference Euler method has been used.It is clear that these nonlinear equations do not satisfy the maximum principle.The centre of the distribution splits into left and right moving waves instead. Thed = 2 simulation has also been done with no extra small dimensions leading to thed = 2 formula (6.25). Even though the momentum generation time is shorter inthis case, Figure 7.3 shows that it still has the same order of magnitude.If more accurate expressions for the restricted density of states were to befound, it would be interesting to return to the study of decay times for (7.18) using abetter numerical method. In addition to being more trustworthy than the estimates(4.25) and (4.28), these decay times are also likely to be shorter and therefore easierto find numerically.143Chapter 8ConclusionMotivated by the universality of black hole physics in a variety of holographicmodels, we studied general properties of field theories where excitations movethroughout a lattice purely due to statistical noise. This led to two interesting mod-els: one in which only energy is dynamical and one in which momentum is as well.Many shortcomings of the first model appeared to be resolved by the second whenthe most basic calculations were carried out. A more quantitative comparison be-tween them is likely to be difficult because both models require one to deal withnonlinear partial differential equations.To elabourate on our first model (3.17), a so-called filtration equation, we foundthat it made sense for high energies and predicted frozen dynamics for systems thatare governed by a Hagedorn density of states [1]. As input to the PDE, we chosea filtration function based on the thermodynamics of N = 4 Super Yang-Mills,which have been famously explored using the AdS / CFT correspondence [2, 8].Additional references to the correspondence have been made throughout the thesis,since we likened the solutions of (3.17) to the plasma balls in confining gaugetheories that were discovered numerically in 2005 [7]. To simplify the analysis ofour PDE, we removed the “small black hole” phase of SYM which is responsiblefor a first order phase transition in the original background for holography. Becausewe still saw solutions that were similar to plasma balls, our results suggest one oftwo things. Either the assumptions about the phase transitions in [7] can be relaxedor it is true that first order phase transitions in large field theories emerge whenever144small field theories with a Hagedorn phase are assembled on a lattice.A detailed analysis of the model (3.17) revealed several problems. One is theaforementioned difficulty of simulating a phase that has a convex microcanonicalentropy. The resulting equation is unstable for the same reason as the reverse heatequation. Even though the reverse heat equation cannot be simulated starting withonly Cauchy data, the situation can be improved if further constraints on the solu-tions are imposed. A very physical one is positivity [97]. Reverse heat equationmethods developed recently [98, 99] have the potential to fill in the missing phasein our numerics but this effort might not be justified in view of the other prob-lems. To see this, the decay times (4.25) and (4.28) derived for our model haveupper and lower bounds proportional to E2F where EF is the critical energy of theplasma ball. In a large N gauge theory, such an energy is proportional to N2 leadingto a predicted decay time of O(N4). This conflicts with the O(N2) prediction of[7]. Another problem with these time scales is that they contain extreme prefac-tors: Emin which is very small and Eα−1min which is very large. It is not possible tofix these values due to a vicious cycle that plagues any realistic simulation. Therequirement of a small Hagedorn energy EH  EF forces the domain kept in thesimulation to be very large — large enough to accommodate the entire mass of theinitial condition below EH. The large domain forces any distribution with powerlaw tails to have a very small Emin. In one dimension, our inability to find morerestrictive prefactors can be viewed as a purely mathematical problem.In dimension two and higher, this is not the case. Our numerics and the theoryof Barenblatt profiles agreed on the behaviour of the plasma ball decay time. Ithugs the lower bound in (4.25) and (4.28) leading to a decay time of zero in theinfinite volume limit. The source of this pathology (a diverging diffusion constant)is clear but again not easy to fix. The high energy parts of energy distributions,for which our model is valid, have their dynamics contaminated by the low energyparts which necessarily appear in the same distributions. This serious problem wasthe main motivation for our second model (7.18).The second model consisted of PDEs for energy and momentum that resultedin faster nonlinear diffusion, possibly consistent with the O(N2) plasma ball pre-diction. This system also appeared not to suffer from instantaneous extinction butthe only check that was possible with out current understanding was limited to very145short times. Exploring this nonlinear model further requires one to derive a mo-mentum restricted density of states that is valid for small differences between E and|P|. In the linear regime, we attempted to strengthen the model’s connection withholography by comparing the limit (7.20) to the equations for hydrodynamics. Theequations were found to have only superficial similarities and not offer meaningulpredictions for transport coefficients. As it makes no explicit reference to a La-grangian, our model seems to apply equally well to strongly coupled and weaklycoupled field theories. Conversely, the procedure for calculating transport coeffi-cients at weak coupling [100–102] is more difficult to carry out than the methodrevealed by the fluid / gravity correspondence [80]. Any future method claiming tofix this aspect of our model would be worthwhile to pursue, but it may be that theassumptions of (7.18) are fundamentally incompatible with those of hydrodynam-ics.Using two widely applicable stochastic models and some basic input about theSuper Yang-Mills theory, were were able to show that long lifetimes of black holesare associated with the Hagedorn density of states of a string worldsheet. Despiteour calculation being insufficient in practice, we showed that it is possible to studynonlinear diffusion, both analytically and numerically as a means of predicting thetime required for these black holes to evaporate through Hawking radiation. Oursecond model in particular showed promise as the number of large spatial dimen-sions was not restricted to one. 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Physical Review D, 79(5), 2009. arXiv:0811.0729. → pages146155Appendix ACumbersome derivativesIn a few parts of the thesis we have dealt with discrete amounts of energy andmomentum denoted by ε and εe respectively. In order to convert energy and mo-mentum to continuous quantities we have had to differentiate expressions whereε appears inside the argument of the density of states ρ . These calculations arecollected here for reference.In the calculation of (7.13), we had the quantityX˜(a,ε) = ρ(E(x)+ ε,P(x)− εe)ρ(E(x+ae)− ε,P(x+ae)+ εe)−ρ(E(x)− ε,P(x)− εe)ρ(E(x+ae)+ ε,P(x+ae)+ εe)(A.1)(actually there was a C multiplying this expression and a surface integral aroundthe whole thing) which we had to differentiate once. The steps involved are:∂ X˜∂ε∣∣∣∣ε=0= ρ+(∂ρ∂E − e j∂ρ∂Pj)+ρ(−∂ρ+∂E + e j∂ρ+∂Pj)−ρ+(−∂ρ∂E − e j∂ρ∂Pj)−ρ(∂ρ+∂E + e j∂ρ+∂Pj)= 2ρ+∂ρ∂E −2ρ∂ρ+∂E .The quantity that we had to differentiate in the calculation of (3.13) was the sameas (A.1) except without the P dependence. Because the terms with an explicit Pcancelled above, the answer for this case is the same.156A similar quantity to this one came up in the calculation of (7.16):Y˜ (a,ε) = −ρ(E(x)+ ε,P(x)− εe)ρ(E(x+ae)− ε,P(x+ae)+ εe)−ρ(E(x)− ε,P(x)− εe)ρ(E(x+ae)+ ε,P(x+ae)+ εe) .(A.2)Since the middle sign is all that distinguishes (A.2) from (A.1), it is easy to see thatthe energy pieces will cancel this time instead of the momentum pieces:∂Y˜∂ε∣∣∣∣ε=0= −ρ+(∂ρ∂E − e j∂ρ∂Pj)−ρ(−∂ρ+∂E + e j∂ρ+∂Pj)−ρ+(−∂ρ∂E − e j∂ρ∂Pj)−ρ(∂ρ+∂E + e j∂ρ+∂Pj)= 2e jρ+∂ρ∂Pj−2e jρ∂ρ+∂Pj.However, there is one more calculation we have to do: differentiating (A.2) twicebefore plugging in ε = 0.∂ 2Y˜∂ε2∣∣∣∣ε=0= −∂∂ε[(∂ρ∂E (E + ε,P− εe)− e j∂ρ∂Pj(E + ε,P− εe))ρ(E+− ε,P++ εe)+ρ(E + ε,P− εe)(−∂ρ∂E (E+− ε,P++ εe)+ e j∂ρ∂Pj(E+− ε,P++ εe))+(−∂ρ∂E (E− ε,P− εe)− e j∂ρ∂Pj(E− ε,P− εe))ρ(E++ ε,P++ εe)+ρ(E− ε,P− εe)(∂ρ∂E (E++ ε,P++ εe)+ e j∂ρ∂Pj(E++ ε,P++ εe))]∣∣∣∣ε=0= −ρ+(∂ 2ρ∂E2 −2e j∂ 2ρ∂E∂Pj+ e jek∂ 2ρ∂Pj∂Pk)−ρ(∂ 2ρ+∂E2 −2e j∂ 2ρ+∂E∂Pj+ e jek∂ 2ρ+∂Pj∂Pk)−2(∂ρ∂E − e j∂ρ∂Pj)(−∂ρ+∂E + ek∂ρ+∂Pk)−ρ+(∂ 2ρ∂E2 +2e j∂ 2ρ∂E∂Pj+ e jek∂ 2ρ∂Pj∂Pk)−ρ(∂ 2ρ+∂E2 +2e j∂ 2ρ+∂E∂Pj+ e jek∂ 2ρ+∂Pj∂Pk)−2(−∂ρ∂E − e j∂ρ∂Pj)(∂ρ+∂E + ek∂ρ+∂Pk)= 2(∂ρ∂E − e j∂ρ∂Pj)(∂ρ+∂E − ek∂ρ+∂Pk)+2(∂ρ∂E + e j∂ρ∂Pj)(∂ρ+∂E + ek∂ρ+∂Pk)−2ρ+(∂ 2ρ∂E2 + e jek∂ 2ρ∂Pj∂Pk)−2ρ(∂ 2ρ+∂E2 + e jek∂ 2ρ+∂Pj∂Pk)157Appendix BDiscontinuous PDE solutionsWhen deriving our main bounds on black hole decay times, a key step was findinga solution to ∂E∂ t =∂∂x(θ(E−EH) ∂E∂x). We argued that such a solution is given byE(x, t) =EF |x|< x∗(t)F(x, t) |x|> x∗(t)where F solves the heat equation and satisfies the mass conservation condition(EF−EH)dx∗(t)dt=∂F∂x (x∗(t), t) .Here, x∗(t) is defined byF(x∗(t), t) = EHA function constructed this way retains the intuitive properties that we expect asolution to have but in some sense it is not a solution; it is discontinuous andderivatives acting on a discontinuous function have no meaning. We need to showthat it is a solution in a precise generalized sense. Consider the solution to the heatequation F . Since∂F∂ t −∂ 2F∂x2 = 0 , (B.1)158it is clearly true that∫ T0∫ ∞−∞[∂F∂ t −∂ 2F∂x2]ϕdxdt = 0 (B.2)where ϕ ∈C∞0 (R× (0,T )) is a test function with compact support. The importantpart is that (B.1) and (B.2) are not equivalent. If F were not differentiable, (B.2)would still make sense because ϕ is smooth and the derivatives can be shiftedonto ϕ through integration by parts. A function solving a differential equation butonly inside an integral with a test function like this is called a weak solution. Thefollowing two expressionsddt∫ x∗−x∗ϕdx = dx∗dtϕ∣∣∣∣x∗−x∗+∫ x∗−x∗∂φ∂ t dxddt∫ x∗−x∗Fϕdx = F dx∗dtϕ∣∣∣∣x∗−x∗+∫ x∗−x∗∂∂ t (Fϕ)dx= EHdx∗dtϕ∣∣∣∣x∗−x∗+∫ x∗−x∗F∂ϕ∂ t +ϕ∂F∂ t dxwhich use Feynman’s trick of differentiating under the integral sign, will be usefulin showing that E is a weak solution.159∫ T0∫ ∞−∞[∂E∂ t −∂ 2Φ(E)∂x2]ϕdxdt = −∫ T0∫ ∞−∞E∂ϕ∂ t dxdt +∫ T0∫ ∞−∞∂Φ(E)∂x∂ϕ∂x dxdt= −∫ T0∫R\(−x∗,x∗)E∂ϕ∂ t dxdt−∫ T0∫ x∗−x∗E∂ϕ∂ t dxdt+∫ T0∫R\(−x∗,x∗)∂Φ(E)∂x∂ϕ∂x dxdt +∫ T0∫ x∗−x∗∂Φ(E)∂x∂ϕ∂x dxdt= −∫ T0∫R\(−x∗,x∗)F∂ϕ∂ t dxdt−EF∫ T0∫ x∗−x∗∂ϕ∂ t dxdt+∫ T0∫R\(−x∗,x∗)∂F∂x∂ϕ∂x dxdt=∫ T0∫ x∗−x∗F∂ϕ∂ t dxdt−∫ T0∫ ∞−∞F∂ϕ∂ t dxdt−EF∫ T0∫ x∗−x∗∂ϕ∂ t dxdt+∫ T0∫ ∞−∞∂F∂x∂ϕ∂x dxdt−∫ T0∫ x∗−x∗∂F∂x∂ϕ∂x dxdt=∫ T0∫ x∗−x∗(F−EF)∂ϕ∂ t −∂F∂x∂ϕ∂x dxdt=∫ T0∫ x∗−x∗(F−EF)∂ϕ∂ t dx−∂F∂x ϕ∣∣∣∣x∗−x∗+∫ x∗−x∗∂ 2F∂x2 ϕdxdt=∫ T0∫ x∗−x∗(F−EF)∂ϕ∂ t dx− (EF−EH)dx∗dtϕ∣∣∣∣x∗−x∗+∫ x∗−x∗∂ 2F∂x2 ϕdxdt=∫ T0∫ x∗−x∗F∂ϕ∂ t +∂ 2F∂x2 ϕdxdt +EF∫ T0dx∗dtϕ∣∣∣∣x∗−x∗dt−EF∫ T0dx∗dtϕ∣∣∣∣x∗−x∗+∫ x∗−x∗∂ϕ∂ t dxdt=∫ T0∫ x∗−x∗F∂ϕ∂ t +∂ 2F∂x2 ϕdxdt +EF∫ T0dx∗dtϕ∣∣∣∣x∗−x∗dt−EF∫ T0ddt∫ x∗−x∗ϕdxdt=∫ T0∫ x∗−x∗∂ 2F∂x2 ϕdxdt +∫ T0∫ x∗−x∗F∂ϕ∂ t dx+EFdx∗dtϕ∣∣∣∣x∗−x∗dt=∫ T0∫ x∗−x∗∂ 2F∂x2 ϕdxdt +∫ T0ddt∫ x∗−x∗Fϕdx−∫ x∗−x∗ϕ ∂F∂ t dxdt=∫ T0∫ x∗−x∗∂ 2F∂x2 ϕdxdt−∫ T0∫ x∗−x∗ϕ ∂F∂ t dxdt = 0160Appendix CCrank-Nicolson code/* Changing float to double might help to avoid crashes. */#include <stdio.h>#include <stdlib.h>#include <stdbool.h>#include <math.h>#define SMALL_ERROR 5e-7#define TOO_MANY_LOOPS 30/* Parameters for the initial condition. */#define PMAX 200.0#define EXP 1.3333/* Ten magic constants handle the interpolation. */#define EF 10.0float c0, c1, c2, c3, c4, c5, c6, c7, c8, c9;/* The maximum distance, the maximum time and the* individual steps. */float length, dx_min, dx_max, dt, max_time;161/* The number of sites in our discretized space. */int n_s;/* Arrays of this size needed for the solution. */float *energy;float *new_energy;float *jac_sub_diag;float *jac_diag;float *jac_sup_diag;float *constant_part;float *diff;float *xs;float beta(float x) {if (x < 0.5) {return pow(x, -0.1);} else if ((x > 0.5) && (x < 1.0)) {return c0 * x * x + c1 * x + c2;} else if ((x > 1.0) && (x < 0.9 * EF)) {return c3 * x + c4;} else if ((x > 0.9 * EF) && (x < EF)) {return c5 * x * x + c6 * x + c7;} else {return c9 * pow(x - c8, -0.25);}}float beta_pr(float x) {if (x < 0.5) {return -0.1 * pow(x, -1.1);} else if ((x > 0.5) && (x < 1.0)) {return 2.0 * c0 * x + c1;} else if ((x > 1.0) && (x < 0.9 * EF)) {162return c3;} else if ((x > 0.9 * EF) && (x < EF)) {return 2.0 * c5 * x + c6;} else {return -0.25 * c9 * pow(x - c8, -1.25);}}/* The tridiagonal matrix algorithm which fills result* with the Ax = b solution from Wikipedia.*/void tridiag(int num, float *sub_diag, float *diag,float *sup_diag, float *res) {int i;sup_diag[0] /= diag[0];res[0] /= diag[0];for (i = 1; i < num; i++) {float factor = 1.0 / (diag[i] - sub_diag[i - 1] * sup_diag[i - 1]);sup_diag[i] *= factor;res[i] = (res[i] - sub_diag[i - 1] * res[i - 1]) * factor;}for (i = num - 2; i >= 0; i--) {res[i] -= sup_diag[i] * res[i + 1];}}void step_crank_nicolson() {float *temp;float error;163bool converged = false;int i, j = 0;/* Initial guess for Newton’s method. */for (i = 0; i < n_s; i++) {new_energy[i] = energy[i];}/* Do not recalculate this in every Newton iteration. */constant_part[0] = energy[0] - 0.5 *(dt / ((xs[1] - xs[0]) * (xs[1] - xs[0]))) *(beta(energy[1]) - beta(energy[0]));for (i = 1; i < (n_s - 1); i++) {constant_part[i] = energy[i] -(dt / (xs[i + 1] - xs[i - 1])) *(((beta(energy[i + 1]) - beta(energy[i])) / (xs[i + 1] - xs[i])) -((beta(energy[i]) - beta(energy[i - 1])) / (xs[i] - xs[i - 1])));}constant_part[n_s - 1] = energy[n_s - 1] - 0.5 *(dt / ((xs[n_s - 1] - xs[n_s - 2]) * (xs[n_s - 1] - xs[n_s - 2]))) *(beta(energy[n_s - 2]) - beta(energy[n_s - 1]));while (!converged) {diff[0] = constant_part[0] - 0.5 *(dt / ((xs[1] - xs[0]) * (xs[1] - xs[0]))) *(beta(new_energy[1]) - beta(new_energy[0])) - new_energy[0];jac_diag[0] = 1.0 - 0.5 * dt * beta_pr(new_energy[0]) /((xs[1] - xs[0]) * (xs[1] - xs[0]));jac_sup_diag[0] = 0.5 * dt * beta_pr(new_energy[1]) /((xs[1] - xs[0]) * (xs[1] - xs[0]));164for (i = 1; i < (n_s - 1); i++) {diff[i] = constant_part[i] -(dt / (xs[i + 1] - xs[i - 1])) *(((beta(new_energy[i + 1]) - beta(new_energy[i])) /(xs[i + 1] - xs[i])) -((beta(new_energy[i]) - beta(new_energy[i - 1])) /(xs[i] - xs[i - 1]))) -new_energy[i];jac_sub_diag[i - 1] = -1.0 * dt * beta_pr(new_energy[i - 1]) *(-1.0 / ((xs[i + 1] - xs[i - 1]) * (xs[i] - xs[i - 1])));jac_diag[i] = 1.0 -(dt * beta_pr(new_energy[i]) / (xs[i + 1] - xs[i - 1])) *((1.0 / (xs[i + 1] - xs[i])) + (1.0 / (xs[i] - xs[i - 1])));jac_sup_diag[i] = -1.0 * dt * beta_pr(new_energy[i + 1]) *(-1.0 / ((xs[i + 1] - xs[i - 1]) * (xs[i + 1] - xs[i])));}diff[n_s - 1] = constant_part[n_s - 1] - 0.5 *(dt / ((xs[n_s - 1] - xs[n_s - 2]) *(xs[n_s - 1] - xs[n_s - 2]))) *(beta(new_energy[n_s - 2]) - beta(new_energy[n_s - 1])) -new_energy[n_s - 1];jac_sub_diag[n_s - 2] = 0.5 * dt * beta_pr(new_energy[n_s - 2]) /((xs[n_s - 1] - xs[n_s - 2]) * (xs[n_s - 1] - xs[n_s - 2]));jac_diag[n_s - 1] = 1.0 - 0.5 * dt * beta_pr(new_energy[n_s - 1]) /((xs[n_s - 1] - xs[n_s - 2]) * (xs[n_s - 1] - xs[n_s - 2]));tridiag(n_s, jac_sub_diag, jac_diag, jac_sup_diag, diff);error = 0.0;j++;165for (i = 0; i < n_s; i++) {error += diff[i] * diff[i];new_energy[i] += diff[i];}if (((error / n_s) < SMALL_ERROR) || (j == TOO_MANY_LOOPS))converged = true;}/* Enforce Neumann boundary conditions. */new_energy[0] = new_energy[1];new_energy[n_s - 1] = new_energy[n_s - 2];temp = new_energy;new_energy = energy;energy = temp;}void simulate_crank_nicolson() {char *filename = malloc(10 * sizeof(char));FILE *fp;bool exit_loop = false;int i, j = 0;float t = 0.0;float time_since_write = max_time;/* If a certain simulation is proving very difficult,* we might want to perform heuristics like increasing* the timestep as time goes on and falling back to a* backup copy of the energy if we accidentally make it166* too big.*/while (!exit_loop) {step_crank_nicolson();if (time_since_write > 10000.0 * dt) {sprintf(filename, "f%d.dat", j);fp = fopen(filename, "w");fprintf(fp, "# t = %f\n", t);for (i = 0; i < n_s; i++) {fprintf(fp, "%f\t%f\n",log(1.0 + xs[i]), log(1.0 + energy[i]));}j++;time_since_write = 0.0;fclose(fp);}/* We quit if time is up or the peak has reached* the Hagedorn energy. */if ((energy[0] < 1.0) || (t > max_time)) exit_loop = true;t += dt;time_since_write += dt;}printf("Final time: %f\n", t);free(filename);}int main(int argc, char **argv) {167int i;float x = 0.0;float read_number;bool reading = false;FILE *fp;/* Normally, the initial condition is hard coded. However,* if a file is specified on the command line, the file will* be read in order to determine the initial condition. This* is ideal if the program runs for awhile and then crashes.* The last successfully generated file can be passed to make* the program pick up where it left off.*/if (argc > 1) {reading = true;fp = fopen(argv[1], "r");fscanf(fp, "# t = %f", &read_number);}/* Choose a, b, c such that axˆ2 + bx + c agrees with the* value and slope of xˆ(-1/10) when x = 0.5. It should also* have a slope of -0.01 at x = 1.0.*/c0 = -0.01 + 0.1 * pow(2.0, -1.1);c1 = -0.01 - 2.0 * c0;c2 = pow(2.0, 0.1) - 0.5 * c1 - 0.25 * c0;/* Now choose a, b such that ax + b agrees with the value and* slope of the above at x = 1.0.*/c3 = -0.01;c4 = c0 + c1 + c2 - c3;/* Choose a, b, c such that axˆ2 + bx + c agrees with the value168* and slope of the above at x = 0.9 * EF. It should also have* half that value at x = EF.*/c5 = (-0.5 * (c3 * (0.9 * EF) + c4) - c3 * 0.1 * EF) /(0.1 * EF * 0.1 * EF);c6 = c3 - 2.0 * c5 * (0.9 * EF);c7 = 0.5 * (c3 * (0.9 * EF) + c4) - c6 * EF - c5 * EF * EF;/* Finally choose a, b such that a(x - b)ˆ(-1/4) agrees with* the value and slope of the above at x = EF.*/c8 = EF + 0.25 * (c5 * EF * EF + c6 * EF + c7) /(2.0 * c5 * EF + c6);c9 = (c5 * EF * EF + c6 * EF + c7) * pow(EF - c8, 0.25);/* Makes the total mass enough to fit twice below the* Hagedorn interface. */length = PMAX * exp(gamma(0.5) + gamma(EXP - 0.5) - gamma (EXP));/* Two things we pick are a "dt" that is "reasonably small" and* a "dx" that fits 10 times between the inflection points. If we* want the "dx" to vary with position, the above is really the* minimum "dx" value. This example code is for performing a* convergence test after a short amount of time so we want the* grid spacing to be uniform. The code will be rerun with "dx"* multiplied by (1/2), (1/4), (1/8), etc. When we run this code* to generate data, we will make the grid non-uniform and simulate* for MUCH longer.*/dx_min = 0.1 * sqrt(1.0 / (1.0 + 2.0 * EXP));dx_max = dx_min;//dx_max = 0.01 * length;dt = 0.001;max_time = 100000.0 * dt;169//max_time = 999999.9;n_s = 0;/* The space step should vary linearly. */while (x < length) {n_s++;x += dx_min + ((dx_max - dx_min) / length) * x;}/* Now that we know the number of sites, we can allocate the* arrays that will be needed in the numerical solution. The* numerical solution at a given time is "energy" and we have* a copy to advance forward.*/energy = malloc(n_s * sizeof(float));new_energy = malloc(n_s * sizeof(float));/* The numbers in the Jacobian matrix used by Newton’s method. */jac_sub_diag = malloc((n_s - 1) * sizeof(float));jac_diag = malloc(n_s * sizeof(float));jac_sup_diag = malloc((n_s - 1) * sizeof(float));/* Other things that should be stored for Newton’s method. */constant_part = malloc(n_s * sizeof(float));diff = malloc(n_s * sizeof(float));/* This stores where our grid points are. */xs = malloc(n_s * sizeof(float));x = 0.0;/* This sets up the initial condition and assigns the grid points. */170for (i = 0; i < n_s; i++) {if (reading) {fscanf(fp, "%f", &read_number);fscanf(fp, "%f", &read_number);energy[i] = exp(read_number) - 1.0;} else {energy[i] = PMAX * pow(1.0 / (1.0 + x * x), EXP);}x += dx_min + ((dx_max - dx_min) / length) * x;xs[i] = x;}if (reading) fclose(fp);/* Everything is set up so we can call the main loop. */printf("L: %f, dx: %f-%f, dt: %f, sites: %d\n",length, dx_min, dx_max, dt, n_s);simulate_crank_nicolson();free(energy);free(new_energy);free(jac_sup_diag);free(jac_diag);free(jac_sub_diag);free(constant_part);free(diff);free(xs);}171


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