I'm interested in the existence of a Lagrangian field theory description of Bronwnian motion, does such a thing exist? Given a particle of some spin $\sigma$, which has a Lagrangian associated with it $\mathcal{L}_{\sigma}$ (which, using the Euler-Lagrange equations, produces Klein-Gordon for $\sigma$ = 0 etc) is there a way I can allow for Brownian type freedoms in this description? Hopefully such a stochastic freedom is allowed in the Lagrangian description.

  • 1
    $\begingroup$ This is not an answer to your question! You might want to take a look at the Feynman-Kac formula. $\endgroup$ – suresh Jul 19 '14 at 10:12
  • 1
    $\begingroup$ From what has already been said by others, take the partition function $Z = \int \mathcal{D}x e^S$ (in QM a "$i/\hbar$" also enters) where the action $S = \int_0^T dt \mathcal{L}$ and $\mathcal{L}$ is the "Lagrangian", often of the form $\dot{x}^2$. That it is a Lagrangian is true only for the semiclassical approximation, i.e. for the trajectory that doesn't have fluctuations. You might also be interested in Malliavin calculus, although I've never seen it used in physics. As it is supposed to be variational calculus with stochastic stuff, it might be related? $\endgroup$ – alarge Jul 19 '14 at 15:45
  • $\begingroup$ Thank you, @amlrg. I have encountered the Malliavin calculus before in the context of variational approaches for PDE with, say, $L^{1}$ measure data right hand sides. I hadn't thought of it in this context though $\endgroup$ – Arthur Suvorov Jul 19 '14 at 22:54
  • $\begingroup$ Look up the Caldeira-Leggett model. $\endgroup$ – DanielSank Jul 13 '19 at 20:51

The Brownian motion $x(t)$ is non-differentiable, so a particular trajectory $x(t)$ can't extremize an action $S$ which would be a functional of $x(t)$ and its derivative, $\dot x(t)$, because the derivative isn't even well-defined and any expression of the type $\int [\dot x(t)]^2 dt$, the usual kinetic term in the action, diverges. (See e.g. middle of page 2 of this paper to see the statement that there is no Lagrangian, too. The paper does its best to construct something that is "as close as possible" to the normal Lagrangian formulation.)

However, when you mention field theory, it's interesting to point out that the typical trajectories $x(t)$ that contribute to Feynman's path integral computation of ordinary quantum mechanics do resemble the Brownian trajectories very closely. But the amount of zigzag motion is determined by the uncertainty principle and Planck's constant, not by adjustable collisions with the molecules of a liquid etc. There are many other differences in the physical interpretation, too.

  • $\begingroup$ Thank you @Lubos, I was hoping there might be a Wiener–Khinchin type theorem (such as Page-Lampard for non-stationary) that could be applied to bound $\int \dot{x}(t)^2 dt$, but perhaps not! I was thinking one could think of the quantum vacuum as a kind of fluid (by performing a kind of decomposition of, say, the Euler-Heisenberg Lagrangian into Maxwell + Magfluid) wherein we introduce a field whose excitations give rise to a kind of Brownian motion. $\endgroup$ – Arthur Suvorov Jul 19 '14 at 22:59
  • $\begingroup$ These are interesting research projects, I think, not questions about established science. ;-) $\endgroup$ – Luboš Motl Jul 20 '14 at 4:54
  • $\begingroup$ I realize this is over a year later, but I would like to add a comment. Brownian motion is not classically differentiable, thus the kinetic energy term diverges. However, it has been shown in various available papers that Browninan motion is fractional differentiable, and fractional derivatives are related to the fractal dimension of a non-rectifiable curve. It may be possible to calculate a finite kinetic energy using fractional calculus, which would allow for a Lagrangian for Brownian motion. $\endgroup$ – soultrane Dec 28 '15 at 15:46
  • $\begingroup$ That's a very interesting proposal. $\endgroup$ – Luboš Motl Dec 29 '15 at 5:17

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.