0
$\begingroup$

What is the Martin-Siggia-Rose-DeDominicis-Janssen action corresponding to the overdamped Langevin stochastic equation $$\frac{d\mathbf{x}}{dt} = -\mathbf{\nabla}V + \mathbf{\eta}$$ where V is the external potential and $\mathbf{\eta}$ a Gaussian noise corresponding to a thermal bath. We assume that $\mathbf{\eta}$ has zero mean and constant variance : $$\left<\eta_\alpha(t)\eta_\beta(t')\right> = 2T\delta(t-t')\delta_{\alpha\beta}$$

$\endgroup$
  • $\begingroup$ @Kostas it's not in MSR, but it is in Janssen's paper (and why his name is associated to the formalism). $\endgroup$ – fqq Nov 8 at 19:43
  • $\begingroup$ Thank you for your answers. Searching a little more, I found answers in that blog post : inordinatum.wordpress.com/2012/09/27/… $$ $$ I think I understand most of the derivation, except the identity : $$\delta (\partial_tx(t) - F(x(t), t) - \xi(x(t), t)) = \int\mathcal{D}[\tilde{x}]\exp\left({-\int_t i\tilde{x}(t)(\partial_tx(t) - F(x(t), t) - \xi(x(t), t)})\right)$$ How can the minus sign and the $t$ integral come from the formula : $$\delta(q) \propto \int_\tilde{x}e^{i\tilde{x}q}$$ $\endgroup$ – Emmy Nov 8 at 20:40
  • $\begingroup$ @Emmy I think the Langevin equation should have $dv/dt$ on the left, not $dx/dt$, after all if $\eta=0$ this should be the Newtons Law. $\endgroup$ – Kostas Nov 8 at 20:55
  • $\begingroup$ I'm working with the overdamped Langevin equation, in other words Newton's equation : $$m\frac{d^2\mathbf x}{dt^2} + \gamma\frac{d\mathbf x}{dt} = -\mathbf\nabla V + \mathbf\eta$$ for which inertia has been neglected. Please excuse me for the confusion $\endgroup$ – Emmy Nov 8 at 21:51
  • $\begingroup$ @Emmy You should edit your question to reflect the change. My posted answer would change very little, $\nabla V -> \nabla V + \gamma/m p$ $\endgroup$ – Kostas Nov 10 at 17:42
0
$\begingroup$

This so-called Langevin equation is just the Newtons second Law with an extra random force: $$\frac{d\mathbf{p}(t)}{dt} = -\mathbf{\nabla}V(x) + \mathbf{\eta}(t)$$ together with $\mathbf{p}=\frac{d\mathbf{x}(t)}{dt}$ obviously.

It is better behaved mathematically if you write this in the integral form: $$ \mathbf{p}(t) = \int dt ~ \big[-\mathbf{\nabla}V(x) + \mathbf{\eta}(t)\big]$$, which is the starting point for Ito calculus.

The action from which stuff can be computed by perturbation theory is: $$L = T \frac {\tilde{\mathbf{x}}^2} 2 - \tilde{\mathbf{x}} \big[\frac{d\mathbf{p}(t)}{dt}+\mathbf{\nabla}V(x) \big]$$ Here $\tilde{\mathbf{x}}$ is conjugate variable to $\mathbf{p}$.

I am not sure about the coefficient of the first term, it may be $T$ or $iT$ depending on conventions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.