Recently I tried to get in touch with some statistical mechanics. I am completely new to the field and hence many things are unclear to me. Currently, I have a question regarding the dissipation-fluctuation relation for Langevin equations induced by Hamiltonian systems for particles.

Let $q$ be a vector of particle positions and $p$ be a velocity vector. I was told that for a separable Hamiltonian $$H(q,p) = V(q) + \frac{1}{2} p^T M^{-1} p$$ where $V$ is a potential and $M$ is a mass matrix, the corresponding Langevin equation is given by $$ dq(t) = M^{-1}p(t) \, dt, \quad dp(t) = -\nabla V(q(t)) \, dt - \gamma(q(t)) M^{-1} p(t) dt + \sigma(q(t)) \, dW_t$$ where $W_t$ is Brownian motion and $\gamma, \sigma$ are matrices. Furthermore, I was told that the fluctuation-dissipation relation holds if $\sigma \sigma^T = 2\gamma k_B T$ is true.

Now my question: What happens if I change the kinetic term in the Hamiltonian? I.e., if I replace the term $\frac{1}{2} p^T M^{-1} p$ by a more general kinetic energy $E_{kin}(p)$? Can I still state the fluctuation-dissipation relation as $\sigma \sigma^T = 2\gamma k_B T$ or does it change?

Many thanks for your help!


1 Answer 1


Welcome to Physics StackExchange! But I notice that you have plenty of experience on Math StackExchange, where you might get more detailed and rigorous responses to this question, certainly in comparison to the fairly basic answer I'm about to give.

The short answer is that the fluctuation-dissipation relation will remain unchanged, but the equations themselves are altered. Changing the kinetic energy expression has two major consequences, which must be linked together. Firstly the equilibrium distribution at temperature $T$ in the canonical ensemble, $$ \rho_{eq}(q,p) = \exp[-V(q)/k_BT] \, \exp[-E_{kin}(p)/k_BT] $$ will no longer be a multivariate Gaussian function of the momenta. That statement basically arises without considering the dynamics, but of course, the dynamics have to be consistent with it! Secondly, the classical (Hamiltonian) equations of motion, to which the Langevin equations should reduce in the absence of frictional and random force terms, will take their more general form $$ dq = \nabla_p E_{kin}(p)\, dt , \qquad dp = -\nabla_q V(q) \, dt $$ The Langevin equations then become $$ dq = \nabla_p E_{kin}(p)\, dt , \qquad dp = -\nabla_q V(q) \, dt -\gamma \nabla_p E_{kin}(p)\, dt + \sigma dW $$ where $\sigma\sigma^T = 2\gamma k_BT$. So the right-hand side of the position evolution equation, and the friction term in the momentum evolution equation, both have their simple linear dependence on momentum replaced by the appropriate expression derived from $E_{kin}$. The fluctuation-dissipation expression is unchanged. I have assumed that $\gamma$ and $\sigma$ are constants, but I suspect that the same equations can be written for position-dependent functions.

The essential point is that it should be possible to convert the Langevin equation into a Fokker-Planck equation for the distribution function $\rho(q,p,t)$, and then show that $\rho_{eq}(q,p)$ is a stationary solution of that equation. I believe that the Hamiltonian form of the deterministic parts of the above equations helps to establish this, although I can't claim to have worked through it myself. I suppose that the Hamiltonian flow (in phase space) together with the flow in momentum space due to the combined dissipative and stochastic terms, may be shown to leave the canonical distribution invariant, provided that $\sigma$ is given by the FD expression. The derivation should be similar to the one that applies for a quadratic kinetic energy.

I'm sure that textbooks on stochastic differential equations will give more information, but I saw the above equations in this paper by Stoltz and Trstanova, which was eventually published in Multiscale Model Simul, 16, 777 (2018), as well as Trstanova's PhD thesis which can be found on this page. The emphasis of those publications is on the numerical solution of the Langevin equations, but the background material is quite enlightening.

  • $\begingroup$ Thank you very much for this wonderful answer, which even goes more into detail than I had hoped for. I would like to emphasize how especially the hint towards the literature on Langevin dynamics with general kinetics is probably worth a try for everyone who would like to learn more. $\endgroup$
    – Murp
    Commented Sep 11, 2018 at 22:38

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