# Can we take $\sigma^2_N=\frac{\kappa_T}{\beta V}N^2$ to be an example of the fluctuation-dissipation theorem?

In statistical mechanics, the relation $\sigma^2_E=\langle E^2\rangle-\langle E\rangle^2=k_BT^2C_v$ is interpreted ad one of the examples of fluctuation-dissipation theorem. The fluctuation in energy is directly related to the ability of the system to absorb (or dissipate) energy.

In grand canonical ensemble, one finds that the number fluctuation is given by $$\sigma^2_N=\langle N^2\rangle-\langle N\rangle^2=\frac{\kappa_T}{\beta V}N^2.$$ Can this be regarded as an example of the fluctuation-dissipation theorem? If yes, which quantity on the RHS is the dissipative term? And why?

• What's $\kappa_T$? – Bzazz May 15 '17 at 14:36
• @Bzazz It represents the isothermal compressibility. – SRS May 15 '17 at 14:44
• I don't know then, I don't even know where that formula comes from :D – Bzazz May 15 '17 at 15:08
• @Bzazz Like you obtain $\sigma^2_E=k_BT^2(\frac{\partial U}{\partial T})_{V,N}=k_BT^2 C_V$. You can show, $\sigma_N^2=\langle N^2\rangle-\langle N\rangle\rangle^2=\frac{1}{\beta}(\frac{\partial N}{\partial \mu})_{V,T}=\frac{\kappa_T}{\beta V}N^2$ which is trivial to show. – SRS May 15 '17 at 15:16

## 1 Answer

I would say that neither one of these is a fluctuation-dissipation relation. The two formulas are fluctuation-susceptibility relations. The left hand side is a fluctuation observable, and the right hand side is a susceptibility (a derivative of of a thermodynamic potential wrt to a thermodynamic variable). Neither the LHS nor the RHS contains a dissipative coefficient (a conductivity, viscosity, etc).

The FD relation is a relation between the symmetrized and retarded Green functions $$G_s(\omega,k)=\coth\left(\frac{\omega}{2T}\right) \,{\rm Im}\, G_R(\omega,k).$$ Physically, the symmetrized correlator is a measure of fluctuations, and the retarded correlator is a measure of dissipation. For example, the retarded correlator of the stress tensor determines viscosity $$\eta(\omega) =\frac{1}{\omega}{\rm Im}\, G_R(\omega,0)\, .$$ There is some relation to what you wrote, because an observable like $\langle(\Delta N)^2\rangle$ is the spatial integral of $G_S^{nn}$ at equal time. The Kubo limit, which determines dissipation, is a different integral of the same function: $$\eta(0) = \lim_{T\to\infty} \int^T \!dt\, \int d^3x \, G_S(x,t)$$