research35 min studyLesson 797 in Physics

Black-brane thermodynamics and field-theory temperature

Read an uncharged planar Einstein saddle in a fixed ensemble and verify its thermodynamic identities.

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What you’ll understand

  • Compute temperature and entropy density with fixed normalizations.
  • State the uncharged canonical ensemble and pressure reference.
  • Check conformal scaling without generalizing to every black hole.

In a thermal holographic model, a planar AdS black-brane saddle can encode an equilibrium boundary plasma. Its Hawking temperature matches the boundary temperature once the boundary time normalization is fixed. Horizon area gives a leading entropy, while the renormalized action determines thermodynamic potentials. These statements are conditional calculations in the model and its approximation regime. A planar horizon, a spherical AdS black hole and a charged solution can have different stability and ensemble structure, so their formulas should not be interchanged.

Fix the neutral planar geometry and its temperature

Use d plus one bulk dimensions, with d boundary spacetime dimensions, the Lorentzian signature already stated and natural units throughout. The neutral planar Einstein solution modifies the Poincaré metric by a blackening function that vanishes at a radial horizon. The boundary time is normalized so the boundary representative is Minkowski space. Arbitrarily rescaling that time without translating temperature would change the apparent answer while leaving the underlying state description inconsistent.

Continue time to Euclidean signature and demand regularity at the horizon. The near-horizon radial-time plane must have no conical singularity, fixing the inverse-temperature period. This gives a temperature proportional to the inverse horizon coordinate. The derivation assumes a nonextremal regular horizon in the stated metric. Charged or higher-derivative solutions need their own near-horizon analysis rather than the neutral expression with an extra parameter inserted by analogy.

ds2=L2z2[−f(z)dt2+dz2f(z)+dx2],f(z)=1−(z/zh)d,T=d4πzhds^2=\frac{L^2}{z^2}\left[-f(z)dt^2+\frac{dz^2}{f(z)}+d\mathbf x^2\right],\qquad f(z)=1-(z/z_h)^d,\qquad T=\frac{d}{4\pi z_h}
zh and L are lengths, T has inverse-length units and x has d − 1 spatial components. The formula uses Einstein gravity with negative cosmological constant and no conserved charge density.↔ Scroll wide equations horizontally.

Separate densities and the thermodynamic potential

A planar horizon has infinite area when the parallel spatial volume is infinite. Divide by that same boundary coordinate volume to define entropy density. The induced horizon metric then gives an area density proportional to the radius-to-horizon-coordinate ratio raised to d minus one. The Einstein area law supplies the factor inverse four Newton constant. In natural units entropy is dimensionless; restoring Boltzmann constant changes the entropy units, not the geometric scaling.

For the neutral planar saddle, choose the standard flat-boundary zero-temperature pressure reference. The pressure and energy density obey the conformal equation of state, and the Helmholtz free-energy density is minus pressure at fixed volume with no charge. If a chemical potential is introduced, one must distinguish fixed-charge Helmholtz and fixed-potential grand ensembles and their boundary terms. Even a familiar saddle must be compared with other allowed saddles before declaring it globally dominant.

κ=Ld−116πGd+1,p=κzhd,ε=(d−1)p,s=4πκzhd−1,f=−p\kappa=\frac{L^{d-1}}{16\pi G_{d+1}},\qquad p=\frac{\kappa}{z_h^d},\quad \varepsilon=(d-1)p,\quad s=\frac{4\pi\kappa}{z_h^{d-1}},\quad \mathfrak f=-p
κ is dimensionless with the declared Newton units. p, ε and free-energy density f have length^(−d) units; s has length^(−d+1) units. These expressions refer to the stipulated neutral planar Einstein saddle.↔ Scroll wide equations horizontally.

Check identities and the range of the comparison

The temperature and entropy density give an enthalpy density equal to energy plus pressure. Differentiating with respect to the horizon parameter also verifies the neutral first law at fixed volume. Eliminating that parameter gives conformal temperature powers for the thermal densities. These are useful normalization checks: an omitted factor of four pi or a mismatch of boundary volumes will spoil identities even if each expression appears dimensionally plausible.

The planar neutral saddle has positive thermal heat capacity in this regime. This does not establish that every black hole has positive heat capacity or that every holographic system has the same phase diagram. Spherical boundaries and charged ensembles permit other structures, and finite-N or string corrections change leading coefficients. The function lab can compare a declared quadratic reference power but has no horizon, Euclidean period, ensemble or gravitational free energy and cannot determine a black-brane phase transition.

Ts=ε+p=dp,dε=T dsat fixed boundary volume and zero charge,s∝Td−1Ts=\varepsilon+p=dp,\qquad d\varepsilon=T\,ds\quad\text{at fixed boundary volume and zero charge},\qquad s\propto T^{d-1}
The differential identity concerns the family of neutral equilibrium saddles, with fixed couplings and volume. It is not a statement about arbitrary nonequilibrium changes of horizon position.↔ Scroll wide equations horizontally.

A complete density check in d = 4

In the declared d = 4 neutral model set κ = 1 and zh = 2 in a fixed length unit. Find T, s, p, ε and the Helmholtz density. Check the enthalpy identity without changing the boundary-volume convention.

  1. Temperature is 4/(4π 2)=1/(2π)4/(4\pi\,2)=1/(2\pi) in inverse-length units. Entropy density is 4π/23=π/24\pi/2^3=\pi/2 in inverse-volume units.

  2. Pressure is 1/24=1/161/2^4=1/16 and energy density is 3/163/16. The Helmholtz density is −1/16-1/16, using the specified neutral pressure reference.

  3. The product Ts=[1/(2π)](π/2)=1/4Ts=[1/(2\pi)](\pi/2)=1/4. The enthalpy density is ε+p=3/16+1/16=1/4\varepsilon+p=3/16+1/16=1/4, so the normalization check agrees. These are model thermal densities, not absolute entropy of an infinite horizon.

The result

T = 1/(2π), s = π/2, p = 1/16, ε = 3/16 and the Helmholtz density is −1/16 in the declared length units.

Explore the model

Interactive lab

Interactive lab · function

A reference power rather than a thermal saddle

Reset and set BOTH Coefficient a = 1 and Position x = 0.50; predict y = 0.25. Reset and set a = 1 and x = 1; predict y = 1. This quadratic reference shows how a power law changes under doubling its argument; it is not the d = 4 plasma equation of state, a horizon metric or a thermodynamic ensemble.

A reference power rather than a thermal saddleThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA temperature exponent must come from the stated theory and dimension, not a generic plotted power.

Check your understanding

Which potential is identified as minus pressure times volume for the stated neutral fixed-temperature, fixed-volume saddle?

Try it yourself

At fixed d = 4 and κ, halve zh in the neutral planar family. Find the ratios of the new T, s, p and ε to their previous values. What does the temperature-energy relation imply for its fixed-volume heat capacity?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the separate powers of zh in each density rather than assuming they all scale alike.

Compare with the solution

Temperature doubles, entropy density grows by 23=82^3=8, and pressure and energy density each grow by 24=162^4=16. Equivalently, the energy density is proportional to T4T^4 with a positive coefficient, so its derivative with respect to temperature is positive at fixed volume. This is a property of the stipulated uncharged planar Einstein saddle and its leading approximation, not a general theorem about the heat capacity of spherical, charged or asymptotically flat black holes.

Take it with you

  • Boundary time and volume normalization fix thermal densities.
  • Ensembles and subtraction references determine which free energy is computed.
  • Neutral planar scaling cannot be transferred unchanged to every black hole.

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