# Fluctuation-dissipation theorem and Kramers-Kronig relations

Is there any connection between fluctuation dissipation theorem and Kramers-Kronig relations? They are often described together under linear response theory but I do not see any exact connection (like one being special case of another).

• Kubo uses Kramers-Kronig on p. 22 of his paper The Fluctuation-Dissipation Theorem, but only once. Mar 21, 2016 at 10:43
• Fluctuations and dissipations are the real and imaginary parts of the response function, and are related to each other via Kramers-Kronig relations. The general mathematical statement behind everything is the Hilbert transform. See these key-words (Kramers-Kronig / fluctuation-dissipation / Hilbert transform) on Wikipedia, and answer your question yourself :-) Mar 22, 2016 at 21:28

Kramers-Kronig relations can be understood as requirement that $$\operatorname{Re} \chi(\omega)$$ is an even function, while $$\operatorname{Im} \chi(\omega)$$ is an odd function - see this answer and the linked Wikipedia therein.

The general Fluctuation-Dissipation theorem usually written as proportionality between fluctuation energy spectrum and imaginary part of fourier of responce function: $$S_x(\omega) = \frac{2kT}{\omega}\operatorname{Im} \chi(\omega)$$ While the lhs of the above expression $$S_x(\omega)=\langle x(\omega)x^*(\omega)\rangle$$ is easily interpreted as "fluctuation" part. It is not at all obvious that $$\operatorname{Im} \chi(\omega)$$ is related to "dissipation". The standard derivation below uses the the Kramers-Kronig symmetry requirements and links $$\operatorname{Im} \chi(\omega)$$ to the dissipation of the system.

Assuming that we have the observable $$x(t)$$, the stochastic external force $$f(t)$$ and the response function $$\chi(t)$$ are linked via: $$x(t) = \int_{-\infty}^{\infty}d\tau\; \chi(t-\tau)f(\tau)$$

We want to compute dissipated power $$f(t)\dot x(t)$$ under periodic force $$f(t) = A \cos\Omega t$$. To do that we first simplify the expression for $$\dot{x}(t)$$. Taking the derivative gives us factor $$-i\omega$$ and integration over $$\tau$$ gives a double-$$\delta$$ fourier transform of the cosine:

$$\dot{x}(t) = \int \frac{d\omega}{2\pi}(-i\omega) e^{-i\omega t}\chi(\omega) \pi A[\delta(\omega - \Omega) + \delta(\omega+\Omega)]$$

Integrating out the $$\delta$$s we'll get: $$\dot{x}(t) =-iA\frac\Omega2\left[\chi(\Omega)e^{-i\Omega t} - \chi(-\Omega)e^{i\Omega t} \right]$$

Now we multiply by $$f(t)$$ and average over one period:

$$\langle W\rangle = \frac{\Omega}{2\pi}\int_0^{2\pi/\Omega}dt\;f(t)\dot{x}(t) = -\frac{i A^2\Omega}{4} \left[\chi(\Omega) - \chi(-\Omega)\right] = \frac12 A^2\Omega\operatorname{Im}\chi(\Omega)$$ In the last equality we've used that real part of $$\chi(\omega)$$ is even, while imaginary is odd.
This links $$\operatorname{Im} \chi(\omega)$$ to dissipation using Kramers-Kronig relations.

• What I understand from your answer is that there is no deeper connection between FD theorem and KK relations apart from the fact that $\chi(\omega)$ appears in both the relations and it represents dissipation. Apr 24, 2021 at 19:10
• @mithusengupta123 I would say that KK relations are just mathematical equalities for a Fourier transform of certain class of functions. So KK don't really say anything about what $Im \chi$ represents. You can show that $Im \chi$ represents dissipation using KK relations (as done in the answer). Apr 24, 2021 at 19:53
• I liked your answer. Particularly the second part. But I think the oddness of the imaginary part of $\chi(\omega)$ follows directly from the defintion of $\chi(\omega)$ which is the Fourier transform of the response function, and reality of theresponse function. One need not use KK relations to show that. I would like to know your comment on this. Apr 24, 2021 at 20:02