Fluctuation–Dissipation Relations

In thermal equilibrium, connected stress fluctuations and the absorptive part of stress response are not independent inputs. The Kubo–Martin–Schwinger (KMS) condition fixes their ratio at every nonzero frequency. The result is exact for a stationary KMS state under the hypotheses below; it is not a generic identity for an evaporating black hole, a squeezed state, or any other nonequilibrium geometry.

Required background. Partial Differential Equations and Green Functions supplies causal response kernels, Fourier transforms, and distributional boundary values.

Helpful background. Quantum Fields in Curved Spacetime supplies KMS states and stress observables, Einstein Equations supplies their gravitational coupling, Black-Hole Geometry and Mechanics supplies stationary horizons, and Semiclassical Gravity supplies the response problem to which the relation is applied.

Thermal spectral relation

Let $A(t)$ be a Hermitian, renormalized stress-tensor observable smeared with a real compactly supported tensor profile and translated by a stationary time flow. Stationarity makes its one-point function constant. Set

$$\delta A(t)=A(t)-\langle A\rangle$$

and define the connected Wightman functions

$$G^\gt (t)=\langle \delta A(t)\delta A(0)\rangle, \qquad G^\lt (t)=\langle \delta A(0)\delta A(t)\rangle,$$

with Fourier convention $F(\omega)=\int dt\,e^{i\omega t}F(t)$, consistent with mode dependence $e^{-i\omega t}$. The noise spectrum and commutator spectrum are

$$\widetilde N(\omega) =\frac12\left[G^\gt (\omega)+G^\lt (\omega)\right], \qquad \rho(\omega)=G^\gt (\omega)-G^\lt (\omega).$$

Centering is essential: without it, $\widetilde N$ contains the disconnected term $2\pi\langle A\rangle^2\delta(\omega)$, while $\rho$ does not. In stochastic gravity this is exactly why the noise kernel is defined with $\widehat t_{ab}=\widehat T_{ab}-\langle\widehat T_{ab}\rangle$ in Hu and Verdaguer 2008, § 3.2, Eqs. (3.11)–(3.12), printed p. 10.

Theorem. Suppose the background and state are stationary; the state is KMS at inverse temperature $\beta\gt 0$ for the same time flow; $A$ is Hermitian and sufficiently smeared that the connected two-point distributions and their Fourier transforms exist; the perturbation couples linearly to $A$; and local contact terms have been separated from the nonlocal response. Then, for $\omega\ne0$,

$$\widetilde N(\omega) =\frac12\coth\!\left(\frac{\beta\omega}{2}\right)\rho(\omega).$$

To fix the susceptibility sign, choose a perturbing Hamiltonian $H_{\mathrm{int}}(t)=-j(t)A(t)$. Then $\delta\langle A(t)\rangle=\int dt'\,\chi^{\mathrm{ret}}(t-t')j(t')$, with

$$\chi^{\mathrm{ret}}(t)=i\theta(t)\langle[A(t),A(0)]\rangle, \qquad \operatorname{Im}\chi^{\mathrm{ret}}(\omega)=\frac{\rho(\omega)}{2}.$$

Thus $\widetilde N=\coth(\beta\omega/2)\operatorname{Im}\chi^{\mathrm{ret}}$ away from zero frequency in this source convention. Reversing the sign of the coupling reverses the susceptibility sign, not the noise. The metric response has the additional factor $1/2$ from $\delta S_{\mathrm m}=\frac12\int T^{ab}h_{ab}$. The influence-functional identification of noise and dissipation is reviewed in Hu and Verdaguer 2008, §§ 4.1–4.3 and § 8.2, while the general linear-response relation originates in Kubo 1957, pp. 570–586.

Proof from the KMS condition

The proof has four short steps.

1. Analytic continuation. The KMS hypothesis says that the two boundary values are related by translation through the thermal strip: $G^\gt (t)=G^\lt (t+i\beta)$, in the distributional sense appropriate to the smeared observable.

2. Detailed balance. Fourier transforming and shifting the contour within that strip gives

   $$G^\gt (\omega)=e^{\beta\omega}G^\lt (\omega).$$

   This is where equilibrium enters. Stationarity alone permits a Fourier transform but does not supply the exponential factor.

3. Eliminate the Wightman spectra. From detailed balance,

   $$\rho(\omega) =\left(e^{\beta\omega}-1\right)G^\lt (\omega),$$

   whereas

   $$2\widetilde N(\omega) =\left(e^{\beta\omega}+1\right)G^\lt (\omega).$$

   Their ratio is $(e^{\beta\omega}+1)/(e^{\beta\omega}-1)=\coth(\beta\omega/2)$ for $\omega\ne0$. This division does not constrain a spectral atom supported at $\omega=0$.

4. Connect to causal response. Linear response expresses the retarded susceptibility through a step function times the commutator. Its discontinuity across the real frequency axis is therefore $\rho$. Thus the same spectral density that controls absorption also fixes the equilibrium noise.

The zero-frequency expression extends by a limit only in a regular dissipative channel with no zero-frequency atom. There $\rho(\omega)$ is odd and begins linearly in $\omega$, canceling the $1/\omega$ behavior of the hyperbolic cotangent. A conserved component of $A$ can instead produce a static connected fluctuation proportional to $\delta(\omega)$; KMS detailed balance alone does not reconstruct it from the commutator. At zero temperature and away from such a zero mode, $\coth(\beta\omega/2)$ tends to $\operatorname{sgn}\omega$, leaving quantum vacuum fluctuations rather than zero noise.

Tensor form in semiclassical gravity

For stress tensors, $A$ may be any smeared polarization $A_f=\int dV\,f^{ab}T_{ab}$. Applying the theorem to every pair of test tensors produces a bitensor distributional identity between $N_{ab,c'd'}$ and the commutator spectral kernel. Ward identities restrict both sides to conserved channels. The Einstein–Langevin solution then weights the equilibrium spectrum by two retarded gravitational propagators.

In this tensor setting, $N_{ab,c'd'}$ is precisely the noise kernel, while the commutator supplies the nonlocal dissipative response. Local real counterterms can shift the dispersive part of the response but cannot be inferred from the noise kernel. They must still be fixed by the renormalized gravitational couplings.

A horizon temperature supplies a KMS relation only when the state is genuinely thermal with respect to the stationary generator in the region where the observables are defined. The Hartle–Hawking setting is the model case. The Unruh state relevant to evaporation has outgoing flux and is not a global equilibrium bath, so imposing a global factor $\coth(\beta\omega/2)$ would add an assumption rather than derive a result.

Counterexample when KMS is dropped

Consider two independent stationary oscillators of frequencies $\Omega_1$ and $\Omega_2$, each in a density matrix diagonal in number states, with mean occupations $n_1$ and $n_2$. Stationarity holds. At the positive-frequency line of oscillator $j$,

$$\frac{\widetilde N_j}{\rho_j}=n_j+\frac12.$$

Choose $n_1$ and $n_2$ so that no single $\beta$ satisfies $n_j=(e^{\beta\Omega_j}-1)^{-1}$ for both modes. There is then no common thermal factor $\frac12\coth(\beta\omega/2)$, even though each response is causal and each noise spectrum is positive. This explicit stationary nonequilibrium state shows why time-translation invariance cannot replace the KMS hypothesis.

Other failure modes are distinct. In a nonstationary state, the kernels depend on both times rather than only their difference, so a single-frequency theorem is unavailable. In a nonlinear response problem, higher-point functions enter and the two-point relation does not close the dynamics. Neither failure invalidates the definitions of noise or retarded response separately.

Independent checks and limiting cases

Three limits check the proof without repeating it. First, detailed balance implies $G^\gt (-\omega)=G^\lt (\omega)$ for a Hermitian stationary observable, so $\rho$ is odd and $\widetilde N$ is even; the product of the odd functions $\coth(\beta\omega/2)$ and $\rho$ is therefore even. Second, for $\beta\lvert\omega\rvert\ll1$, the relation becomes $\widetilde N(\omega)\simeq\rho(\omega)/(\beta\omega)$, the classical thermal limit. Third, for $\beta\lvert\omega\rvert\gg1$, thermal occupation is suppressed but the sign-function limit retains vacuum noise.

Positivity provides another independent diagnostic. For positive $\omega$, a passive KMS state has $\rho(\omega)\ge0$ in every Hermitian smeared channel, while $\coth(\beta\omega/2)\gt 0$. The predicted noise spectrum is therefore nonnegative. A numerical implementation that produces negative power in such a channel has a sign, Fourier, smearing, or state-normalization error before it has a gravitational interpretation.

Exercise

For one harmonic oscillator with $A=x$, frequency $\Omega$, mass $m$, and thermal occupation $n=(e^{\beta\Omega}-1)^{-1}$, calculate the positive-frequency weights of $G^\gt$, $G^\lt$, $\rho$, and $\widetilde N$. Verify the theorem at $\omega=\Omega$.

Solution

Using $x=(a+a^\dagger)/\sqrt{2m\Omega}$ and $a(t)=ae^{-i\Omega t}$,

$$G^\gt (\omega)\supset\frac{2\pi}{2m\Omega}(n+1)\delta(\omega-\Omega), \qquad G^\lt (\omega)\supset\frac{2\pi}{2m\Omega}n\delta(\omega-\Omega).$$

Thus the positive-frequency weights of $\rho$ and $\widetilde N$ are respectively $2\pi/(2m\Omega)$ and $[2\pi/(2m\Omega)](2n+1)/2$. Their ratio is

$$n+\frac12 =\frac12\frac{e^{\beta\Omega}+1}{e^{\beta\Omega}-1} =\frac12\coth\!\left(\frac{\beta\Omega}{2}\right),$$

which is precisely the stated relation.

The calculation also performs the limiting checks. At high temperature, $n\simeq1/(\beta\Omega)$ and the ratio is dominated by the classical term $1/(\beta\Omega)$. At zero temperature, $n\to0$ but the ratio approaches $1/2$, showing explicitly that the ground-state noise does not disappear. The commutator weight is temperature independent, whereas the symmetric covariance contains both thermal and vacuum parts.

References

• Hu, B. L., and Verdaguer, E. 2008. “Stochastic Gravity: Theory and Applications.” Living Reviews in Relativity 11, 3. DOI.
• Kubo, R. 1957. “Statistical-Mechanical Theory of Irreversible Processes. I. General Theory and Simple Applications to Magnetic and Conduction Problems.” Journal of the Physical Society of Japan 12, 570–586. DOI.