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# 8.13: Appendix IV- Correlations in the Langevin formalism

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As shown above, integrating the Langevin equation $$\Dp+\gamma p = F + \eta(t)$$ yields $p(t)=p(0)\,e^{-\gamma t} +{F\over\gamma}\>\big(1-e^{-\gamma t}\big) + \int\limits_0^t\!\!ds\>\eta(s)\,e^{\gamma (s-t)}\ .$. Thus, the momentum autocorrelator is $\begin{split} \blangle p(t)\,p(t')\brangle -\blangle p(t)\brangle \blangle p(t')\brangle &= \int\limits_0^t\!\!ds\!\int\limits_0^{t'}\!\!ds'\> e^{\gamma(s-t)}\,e^{\gamma(s'-t')}\,\blangle\eta(s)\,\eta(s')\brangle\\ &=\Gamma\,e^{-\gamma(t+t')}\!\! \int\limits_0^{t\ns_{min}}\!\!\!ds\>e^{2\gamma s} =M\kT\,\Big( e^{-\gamma |t-t'|} - e^{-\gamma(t+t')}\Big)\ , \end{split}$ where $t\ns_{min}={min}(t,t')= \begin{cases} t&{if}\ t<t' \\ t'&{if}\ t'<t \end{cases}$ is the lesser of $$t$$ and $$t'$$. Here we have used the result $\begin{split} \int\limits_0^t\!\!ds\!\int\limits_0^{t'}\!\!ds'\> e^{\gamma(s+s')}\,\delta(s-s')&= \int\limits_0^{t\ns_{min}}\!\!\!ds\!\int\limits_0^{t\ns_{min}}\!\!\!ds'\>e^{\gamma(s+s')}\,\delta(s-s')\\ &=\int\limits_0^{t\ns_{min}}\!\!\!ds\>e^{2\gamma s}={1\over 2\gamma}\Big(e^{2\gamma t\ns_{min}}-1\Big)\ . \end{split}$ One way to intuitively understand this result is as follows. The double integral over $$s$$ and $$s'$$ is over a rectangle of dimensions $$t\times t'$$. Since the $$\delta$$-function can only be satisfied when $$s=s'$$, there can be no contribution to the integral from regions where $$s>t'$$ or $$s'>t$$. Thus, the only contributions can arise from integration over the square of dimensions $$t\ns_{min}\times t\ns_{min}$$. Note also $t+t'-2\,{min}(t,t')=|t-t'|\ .$

Let’s now compute the position $$x(t)$$. We have $\begin{split} x(t)&=x(0)+{1\over M}\!\int\limits_0^t\!\!ds\>p(s)\\ &=x(0) + \int\limits_0^t\!\!ds\, \Bigg[\bigg(v(0)-{F\over\gamma M}\bigg)\,e^{-\gamma s} + {F\over \gamma M}\Bigg] +{1\over M}\!\int\limits_0^t\!\!ds\!\int\limits_0^{s}\!\!ds\ns_1\>\eta(s\ns_1)\,e^{\gamma(s\ns_1-s)}\\ &=\blangle x(t)\brangle + {1\over M}\!\int\limits_0^t\!\!ds\!\int\limits_0^{s}\!\!ds\ns_1\>\eta(s\ns_1)\,e^{\gamma(s\ns_1-s)}\ , \end{split}$ with $$v=p/M$$. Since $$\big\langle\eta(t)\big\rangle=0$$, we have $\begin{split} \blangle x(t)\brangle&=x(0)+\int\limits_0^t\!\!ds\, \Bigg[\bigg(v(0)-{F\over\gamma M}\bigg)\,e^{-\gamma s} + {F\over \gamma M}\Bigg]\\ &=x(0)+{Ft\over \gamma M} + {1\over\gamma}\bigg(v(0)-{F\over\gamma M}\bigg)\,\big(1-e^{-\gamma t}\big)\ . \end{split}$ Note that for $$\gamma t\ll 1$$ we have $$\big\langle x(t)\big\rangle = x(0)+v(0)\, t + \half M^{-1} F t^2 + \CO(t^3)$$, as is appropriate for ballistic particles moving under the influence of a constant force. This long time limit of course agrees with our earlier evaluation for the terminal velocity, $$v\ns_\infty=\big\langle p(\infty)\big\rangle/M = F/\gamma M$$.

We next compute the position autocorrelation: \begin{aligned} \blangle x(t)\,x(t')\brangle - \blangle x(t)\brangle \blangle x(t')\brangle & = {1\over M^2}\!\int\limits_0^t\!\!ds\! \int\limits_0^{t'}\!\!ds'\>e^{-\gamma(s+s')}\!\int\limits_0^s\!\!ds\ns_1\!\int\limits_0^{s'}\!\!ds'_1\> e^{\gamma(s\ns_1+s\ns_2)}\,\blangle\eta(s\ns_1)\,\eta(s\ns_2)\brangle\nonumber\\ &={\Gamma\over 2\gamma M^2}\int\limits_0^t\!\!ds\!\int\limits_0^{t'}\!\!ds'\> \Big(e^{-\gamma|s-s'|}-e^{-\gamma(s+s')}\Big)\end{aligned} We have to be careful in computing the double integral of the first term in brackets on the RHS. We can assume, without loss of generality, that $$t\ge t'$$. Then $\begin{split} \int\limits_0^t\!\!ds\!\int\limits_0^{t'}\!\!ds'\>e^{-\gamma|s-s'|}&=\int\limits_0^{t'}\!\!ds'\,e^{\gamma s'}\!\int\limits_{s'}^t\!\!ds\> \>e^{-\gamma s}+ \int\limits_0^{t'}\!\!ds'\,e^{-\gamma s'}\!\int\limits_0^{s'}\!\!ds\>e^{\gamma s}\\ &=2\gamma^{-1}t' + \gamma^{-2}\big(e^{-\gamma t} + e^{-\gamma t'} - 1 - e^{-\gamma (t-t')} \big)\ . \end{split}$ We then find, for $$t>t'$$, $\blangle x(t)\,x(t')\brangle - \blangle x(t)\brangle \blangle x(t')\brangle = {2\kT\over \gamma M}\> t' + {\kT\over \gamma^2 M}\,\big(2e^{-\gamma t} + 2 e^{-\gamma t'} -2 -e^{-\gamma(t-t')} - e^{-\gamma(t+t')} \big)\ .$ In particular, the equal time autocorrelator is $\blangle x^2(t)\brangle - \blangle x(t)\brangle^{\!2} = {2\kT\over \gamma M}\> t + {\kT\over \gamma^2 M}\,\big(4 e^{-\gamma t} - 3 - e^{-2\gamma t}\big)\ .$ We see that for long times $\blangle x^2(t)\brangle - \blangle x(t)\brangle^{\!2} \sim 2Dt\ ,$ where $$D=\kT/\gamma M$$ is the diffusion constant.