8.13: Appendix IV- Correlations in the Langevin formalism
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As shown above, integrating the Langevin equation ˙p+γp=F+η(t) yields p(t)=p(0)e−γt+Fγ(1−e−γt)+t∫0dsη(s)eγ(s−t) .. Thus, the momentum autocorrelator is ⟨p(t)p(t′)⟩−⟨p(t)⟩⟨p(t′)⟩=t∫0dst′∫0ds′eγ(s−t)eγ(s′−t′)⟨η(s)η(s′)⟩=Γe−γ(t+t′)t∗min∫0dse2γs=MkBT(e−γ|t−t′|−e−γ(t+t′)) , where t∗min=min(t,t′)={tif t<t′t′if t′<t is the lesser of t and t′. Here we have used the result t∫0dst′∫0ds′eγ(s+s′)δ(s−s′)=t∗min∫0dst∗min∫0ds′eγ(s+s′)δ(s−s′)=t∗min∫0dse2γs=12γ(e2γt∗min−1) . One way to intuitively understand this result is as follows. The double integral over s and s′ is over a rectangle of dimensions t×t′. Since the δ-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∗min×t∗min. Note also t+t′−2min(t,t′)=|t−t′| .
Let’s now compute the position x(t). We have x(t)=x(0)+1Mt∫0dsp(s)=x(0)+t∫0ds[(v(0)−FγM)e−γs+FγM]+1Mt∫0dss∫0ds∗1η(s∗1)eγ(s∗1−s)=⟨x(t)⟩+1Mt∫0dss∫0ds∗1η(s∗1)eγ(s∗1−s) , with v=p/M. Since ⟨η(t)⟩=0, we have ⟨x(t)⟩=x(0)+t∫0ds[(v(0)−FγM)e−γs+FγM]=x(0)+FtγM+1γ(v(0)−FγM)(1−e−γt) . Note that for γt≪1 we have ⟨x(t)⟩=x(0)+v(0)t+12M−1Ft2+O(t3), 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∗∞=⟨p(∞)⟩/M=F/γM.
We next compute the position autocorrelation: ⟨x(t)x(t′)⟩−⟨x(t)⟩⟨x(t′)⟩=1M2t∫0dst′∫0ds′e−γ(s+s′)s∫0ds∗1s′∫0ds′1eγ(s∗1+s∗2)⟨η(s∗1)η(s∗2)⟩=Γ2γM2t∫0dst′∫0ds′(e−γ|s−s′|−e−γ(s+s′)) 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≥t′. Then t∫0dst′∫0ds′e−γ|s−s′|=t′∫0ds′eγs′t∫s′dse−γs+t′∫0ds′e−γs′s′∫0dseγs=2γ−1t′+γ−2(e−γt+e−γt′−1−e−γ(t−t′)) . We then find, for t>t′, ⟨x(t)x(t′)⟩−⟨x(t)⟩⟨x(t′)⟩=2kBTγMt′+kBTγ2M(2e−γt+2e−γt′−2−e−γ(t−t′)−e−γ(t+t′)) . In particular, the equal time autocorrelator is ⟨x2(t)⟩−⟨x(t)⟩2=2kBTγMt+kBTγ2M(4e−γt−3−e−2γt) . We see that for long times ⟨x2(t)⟩−⟨x(t)⟩2∼2Dt , where D=kBT/γM is the diffusion constant.