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research35 min studyLesson 798 in Physics

Holographic transport and Kubo calculations

Connect retarded response to horizon conditions while limiting the scope of the Einstein viscosity result.

Before you begin
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The question for today

What you’ll understand

  • Fix the Fourier and retarded-function signs.
  • Explain why incoming horizon conditions select retarded response.
  • Compute the shear ratio under its isotropic two-derivative assumptions.

A thermal model's equilibrium equation of state does not determine how it relaxes after a small perturbation. Transport coefficients require real-time response functions. In holographic calculations the relevant bulk fluctuation must satisfy a causal horizon condition as well as its boundary source condition. That prescription gives conditional predictions for specified strongly coupled models. It does not turn every observed dissipative fluid into a gravity dual, and it does not make one Einstein-gravity viscosity ratio a universal theorem for all quantum matter.

Fix a retarded convention and the dissipative limit

Use time dependence exponential minus i omega t and define the retarded stress correlator with minus i times the step function and commutator. Its Fourier transform uses exponential plus i omega t. In this convention the low-positive-frequency imaginary part of a passive shear response is negative, and viscosity is minus its slope. Mixing a Fourier sign from one paper with a Green-function sign from another can incorrectly produce negative dissipation.

The shear Kubo limit uses zero spatial momentum and small frequency in a suitable hydrodynamic regime. The operator normalization must match how a boundary metric perturbation couples to stress. A finite-frequency peak is not itself the zero-frequency transport coefficient, and contact terms must be treated in the proper response convention. Conductivity is a different correlator; momentum conservation can yield a delta function in electrical response even when the shear viscosity is finite.

Gxy,xyR(ω,0)=−i∫0∞dt∫dd−1x eiωt⟨[Txy(t,x),Txy(0)]⟩,η=−lim⁡ω→0+Im⁡Gxy,xyR(ω,0)ωG^R_{xy,xy}(\omega,\mathbf0)=-i\int_0^\infty dt\int d^{d-1}x\,e^{i\omega t}\langle[T_{xy}(t,\mathbf x),T_{xy}(0)]\rangle,\qquad \eta=-\lim_{\omega\to0^+}\frac{\operatorname{Im}G^R_{xy,xy}(\omega,\mathbf0)}{\omega}
Natural units are used. The source perturbation has e^(−iωt) time dependence. The definition and minus sign together fix the positive-viscosity convention.↔ Scroll wide equations horizontally.

Select the causal bulk fluctuation

A boundary shear source excites an appropriate metric fluctuation in an isotropic equilibrium black-brane geometry. Near a regular future horizon, two local wave branches are available. The incoming branch carries disturbances into the future horizon and selects retarded response. The outgoing choice gives a different causal prescription; arbitrary real mixtures need not describe the desired thermal Green function. Euclidean regularity alone cannot be replaced by a naive Lorentzian boundary derivative without specifying this condition.

In the two-derivative isotropic Einstein setting, the shear mode reduces to a minimally coupled massless fluctuation under the appropriate symmetry assumptions. A conserved radial flux relates its boundary imaginary response to horizon absorption. This is why a leading coefficient can depend simply on horizon area. The derivation relies on the action and decoupling structure; higher-derivative couplings, anisotropy or broken translations can change the fluctuation equation and invalidate the same shortcut.

hxy(t,z)∝e−iωt(1−z/zh)−iω/(4πT)near the future horizon, incoming branchh_{xy}(t,z)\propto e^{-i\omega t}(1-z/z_h)^{-i\omega/(4\pi T)}\qquad\text{near the future horizon, incoming branch}
This expression fixes the nonextremal incoming power for the stated z-coordinate orientation and Fourier convention. A full solution must also have the prescribed boundary source and appropriate normalization.↔ Scroll wide equations horizontally.

Read the viscosity ratio with its assumptions attached

For the neutral planar Einstein model, the leading shear coefficient equals the horizon area density divided by sixteen pi Newton constant. Entropy density uses the same area divided by four Newton constant, so their ratio is one over four pi in natural units. Restoring constants gives reduced Planck constant divided by four pi Boltzmann constant. The result is robust across a particular broad model class, but the calculation's hypotheses are part of the result.

One should not promote that ratio into an unconditional lower bound for every fluid or every consistent gravitational effective theory. Finite coupling, quantum corrections, anisotropic response and other dynamics require separate analysis. Nor does matching one measured ratio establish a specific duality. The damped-oscillator lab can illustrate a deterministic dissipation timescale across several cycles; it has no stress commutator, horizon absorption, hydrodynamic limit or quantum Kubo calculation.

η=Ld−116πGd+1zhd−1,s=Ld−14Gd+1zhd−1,ηs=14π(ℏ4πkB with constants restored)\eta=\frac{L^{d-1}}{16\pi G_{d+1}z_h^{d-1}},\qquad s=\frac{L^{d-1}}{4G_{d+1}z_h^{d-1}},\qquad \frac{\eta}{s}=\frac1{4\pi}\quad\left(\frac{\hbar}{4\pi k_B}\ \text{with constants restored}\right)
The formulas assume classical two-derivative isotropic Einstein dynamics with the specified regular planar horizon and a decoupled shear mode. In natural units η and s both have length^(−d+1) units.↔ Scroll wide equations horizontally.

Work it through

The same horizon normalization in two coefficients

For the d = 4 model let κ = L³/(16πG5) = 1 and zh = 2 in the same length unit. Compute η, use the previously derived entropy density and find η/s. What low-frequency imaginary slope follows in the stated Kubo convention?

  1. The viscosity is η=κ/zh3=1/8\eta=\kappa/z_h^3=1/8 in inverse-volume units. The entropy density is s=4πκ/zh3=π/2s=4\pi\kappa/z_h^3=\pi/2.

  2. Their ratio is (1/8)/(π/2)=1/(4π)≈0.07958(1/8)/(\pi/2)=1/(4\pi)\approx0.07958 in natural units. The ratio cancels the shared Newton and area factors.

  3. The Kubo definition then requires Im⁡GR=−ω/8+o(ω)\operatorname{Im}G^R=-\omega/8+o(\omega) at positive small frequency. A positive imaginary slope with the same definitions would instead give negative viscosity and would need a sign or physical-assumption check.

The result

η = 1/8, s = π/2 and η/s = 1/(4π); the retarded imaginary slope is −1/8.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

A dissipation envelope only

Reset and set BOTH Damping ratio ζ = 0.10 and Natural frequency ω₀ = 1. Reset and set ζ = 0.50 and ω₀ = 1 for a fresh comparison at equal times; predict faster decay of the oscillation envelope across several cycles. This prepared deterministic release is not a driven stress response, horizon fluctuation, thermal spectral function or Kubo solver.

A dissipation envelope onlyUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. 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 relaxation analogy does not determine a transport coefficient without the correct response function.

Pause & reason

Check your understanding

Which horizon condition selects the retarded shear response in the stated nonextremal black-brane calculation?

Your turn

Try it yourself

A computation claims a passive stable shear channel but reports Im⁡GR(ω,0)=+0.20 ω+o(ω)\operatorname{Im}G^R(\omega,\mathbf0)=+0.20\,\omega+o(\omega) for small positive ω, using exactly the definitions above. Find the inferred η and identify the necessary diagnostic before interpreting it as a physical violation of the Einstein ratio.

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

Give me a hint

Apply the specified minus sign first; a convention mismatch can imitate a negative transport coefficient.

Compare with the solution

The inferred viscosity is −0.20-0.20 in the chosen viscosity units, contrary to passive nonnegative dissipation under the stated assumptions. Check the Fourier convention, retarded definition, boundary-source normalization and incoming versus outgoing horizon branch. A failed sign check is not evidence that the physical Einstein ratio has been violated. If all conventions are consistent, the assumed stable passive setup itself must be reconsidered before assigning the result a transport interpretation.

Keep the idea

Take it with you

  • Retarded signs and horizon conditions jointly fix dissipation.
  • The shear coefficient requires a specified hydrodynamic limit.
  • The Einstein ratio is conditional rather than a universal statement about all matter.

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