Given the Langevin equation $$\text{d}v=-\gamma v\text{d}t+\text{d}W,$$ where $v$ is the speed, $\text{d}W$ is a Wiener process with $\langle \text{d}W\rangle=0$ and $\langle (\text{d}W)^2\rangle=\text{d}t$, and $\gamma$ is a constant, we have the solution $$v = v_0e^{-\gamma t}+\int_0^te^{-\gamma (t-s)}\text{d}W(s),$$ which can be used to determine the position $$x(t)=x_0+\int_0^tv(s)\text{d}s.$$ Now, I wish to calculate the variance of the position, which I know from my lecture notes to be $$\langle(\delta x)^2\rangle=\langle (x-\langle x\rangle)^2\rangle=\frac{1}{\gamma^2}\left(t+\frac{1}{\gamma}(e^{-\gamma t}-1)-\frac{1}{2\gamma}(1-e^{-\gamma t})^2\right).$$ I have determined that $$x-\langle x\rangle=\int_0^t\int_0^se^{-\gamma (s-s')}\text{d}W(s')\text{d}s,$$ and using $$\left\langle\text{d}W(s')\text{d}W(s''')\right\rangle=\delta(s'-s''')\text{d}s',$$ I have come this far \begin{align*} \langle(\delta x)^2\rangle&=\left\langle\int_0^t\int_0^se^{-\gamma (s-s')}\text{d}W(s')\text{d}s\int_0^t\int_0^{s''}e^{-\gamma (s''-s''')}\text{d}W(s''')\text{d}s''\right\rangle\\ &=\int_{s=0}^t\int_{s''=0}^t\int_{s'=0}^s\int_{s'''=0}^{s''}e^{-\gamma (s-s'+s''-s''')}\underbrace{\left\langle\text{d}W(s')\text{d}W(s''')\right\rangle}_{=\delta(s'-s''')\text{d}s'}\text{d}s\text{d}s''\\ &=\int_{s=0}^t\int_{s''=0}^t\int_{s'=0}^s e^{-\gamma (s-2s'+s'')}\text{d}s'\text{d}s\text{d}s'' =\left\{\begin{array}{l} u=s-2s'+s''\\ \text{d}u=-2\text{d}s'\end{array}\right\}\\ &=-\frac{1}{2}\int_{s=0}^t\int_{s''=0}^{t} \int_{u=s+s''}^{s''-s}e^{-\gamma u}\text{d}u\text{d}s\text{d}s''=\frac{1}{2\gamma}\int_{s=0}^t\int_{s''=0}^{t}(e^{\gamma(s-s'')}-e^{-\gamma(s+s'')})\text{d}s\text{d}s'', \end{align*} at which point it's clear that integrating further can not produce a term proportional to $t$, so something must have gone wrong. I'm also not quite sure if my integrals are ordered correctly or if I've even squared $x-\langle x\rangle$ correctly; any help is appreciated.
-
1$\begingroup$ In a less abstract version of this theory, with which I am familiar, the derivatives $F(s) = dW(s)/ds$ are delta-correlated. And in the expression for $\delta x = x - \langle x\rangle$ the double integral is rewritten as a single integral. Then Einstein's formula is easily recovered. $\endgroup$Gec– Gec2025-02-09 07:50:59 +00:00Commented Feb 9 at 7:50
-
$\begingroup$ @Gec what do you mean by Einstein's formula? This has nothing to do with Einstein as far as I'm aware. $\endgroup$MrPillow– MrPillow2025-02-09 14:18:41 +00:00Commented Feb 9 at 14:18
-
$\begingroup$ In a course on nonequilibrium statistical mechanics I was told that the formula $\langle (\delta x)^2 \rangle \approx 2D t $ for $t \gg 1/\Gamma$ is named after Einstein. I did not check whether the author of the course had any grounds for this. Nevertheless, I call this formula Einstein's formula by inertia. You are not obliged to do the same. $\endgroup$Gec– Gec2025-02-09 14:28:34 +00:00Commented Feb 9 at 14:28
-
$\begingroup$ All these stochastic processes described by the Langevin equations arose from the theory of Brownian motion. Einstein was the author of one of the theories of Brownian motion. Look at the beginning of the article on wiki link $\endgroup$Gec– Gec2025-02-09 14:49:36 +00:00Commented Feb 9 at 14:49
-
$\begingroup$ Now that you said it, I recall the connection between Brown and Einstein. $\endgroup$MrPillow– MrPillow2025-02-09 15:00:48 +00:00Commented Feb 9 at 15:00
2 Answers
You can simplify your calculations by integrating $s$: $$ x-\langle x\rangle=\frac1\gamma\int_0^t(1-e^{-\gamma(t-s)})dW(s) $$ To compute the variance: $$ \begin{align} \langle(x-\langle x\rangle)^2\rangle &= \frac1{\gamma^2}\int_0^t\int_0^t(1-e^{-\gamma(t-s)})(1-e^{-\gamma(t-s')})\langle dW(s)dW(s')\rangle \\ &= \frac1{\gamma^2}\int_0^t(1-e^{-\gamma(t-s)})^2ds \\ &= \frac1{\gamma^3}\left[\gamma t-2(1-e^{-\gamma t})+\frac{1-e^{-2\gamma t}}2\right]\\ &\sim \frac t{\gamma^2} \end{align} $$ You therefore do get a diffusive scaling for large times.
Hope this helps.
-
$\begingroup$ This is certainly a good and simpler way to it, thanks. Though, my question was more about handling the four-fold integral, so I've accepted the other answer. $\endgroup$MrPillow– MrPillow2025-02-09 17:18:20 +00:00Commented Feb 9 at 17:18
It would be more correct to write $$ \langle dW(s')dW(s''')\rangle = \delta(s'-s''')ds'ds''', $$ but this is not the reason for the error in your calculations. In a fourfold integral, the domain of integration is $s'\leq s\leq t$, $s'''\leq s''\leq t$. Using the delta function leads to taking the integral over $s'''$, which reduces to substituting $s''' = s'$ into the integrand, but also to the condition $s' \leq s''$. Therefore, the correct form of the triple integral is the following $$ \langle (\delta x)^2\rangle = \int\limits_0^t ds\int\limits_0^s ds'\int\limits_{s'}^t ds''\ e^{-\gamma(s+s''-2s')}, $$ which after integration gives the desired result $$ \langle (\delta x)^2\rangle = \frac1{\gamma^3}\left(\gamma t-\frac32 + 2e^{-\gamma t} -\frac12e^{-2\gamma t}\right). $$
-
$\begingroup$ Okay. Is there a more obvious way to see this? Seems like an easy subtelty to miss, which I did. I've also noticed that you ordered the integrals as $\int ds\int ds'\int ds''\int ds'''$. Is that better than what I did$\int ds\int ds''\int ds'\int ds'''$? $\endgroup$MrPillow– MrPillow2025-02-09 15:20:02 +00:00Commented Feb 9 at 15:20
-
$\begingroup$ I'm not sure there's a more obvious way to avoid missing this subtlety. For non-trivial integration domains, you just need to keep track of whether it is possible to satisfy the equality $s' = s'''$ depending on the values of the other variables. There is no subtlety if we go from the double integral for $\delta x$ to a single integral, as I mentioned in the first comment and LPZ wrote in his answer. As for ordering the integrals, I think you just have to keep track of which ones are inner and which ones are outer, depending on the domain of integration. $\endgroup$Gec– Gec2025-02-09 15:33:33 +00:00Commented Feb 9 at 15:33