Trying to understand the Langevin Equation. In particular, this passage from a Wikipedia article has me confused (section: "Thermal Noise in an Electrical Resistor"):

$\frac{dU}{dt} =-\frac{U}{RC}+\eta \left( t\right),\;\;$ $\left\langle\eta \left( t\right) \eta \left( t^{\prime }\right)\right\rangle = \frac{2k_{B}T}{RC^{2}}\delta \left(t-t^{\prime }\right).$

This equation may be used to determine the correlation function

$\left\langle U\left(t\right) U\left(t^{\prime }\right) \right\rangle =\left( k_{B}T/C\right) \exp \left( -\left\vert t-t^{\prime }\right\vert /RC\right) \approx 2Rk_{B}T\delta \left( t-t^{\prime}\right),$

Here's my derivation of a rather different result:

$\displaystyle \frac{dU(t)}{dt} =-\frac{U(t)}{RC}+\eta \left( t\right)$

$\displaystyle U(t+t^\prime)\frac{dU(t)}{dt} =-\frac{U(t+t^\prime)U(t)} {RC}+U(t+t^\prime)\eta(t)$

Now concentrate on LHS:

$\displaystyle U(t+t^\prime)\frac{dU(t)}{dt} = \frac{d}{dt}(U(t+t^\prime)U(t))-U(t)\frac{d}{dt}U(t+t^\prime)= \frac{d}{dt}(U(t+t^\prime)U(t))-U(t)\frac{d}{dt^\prime}U(t+t^\prime)$ $\displaystyle=\frac{d}{dt}(U(t+t^\prime)U(t))-\frac{d}{dt^\prime}(U(t)U(t+t^\prime))$

now do ensemble averaging and assuming system is statistically stationary. Then the first term on the new LHS vanishes as does the second term on the RHS, leaving:

$\displaystyle -\frac{d}{dt^\prime}\langle U(t+t^\prime)U(t)\rangle = -\frac{\langle U(t+t^\prime)U(t)\rangle} {RC}$

which gives the obviously unphysical

$\displaystyle \langle U(0)U(t^\prime)\rangle=\langle U(0)U(0) \rangle \exp(+t^\prime / RC)$

Can anyone tell me where I went wrong and supply the derivation of the Wikipedia result?


1 Answer 1


You start by solving the differential equation. It is a first order, linear differential equation with constant coefficients. So the solution of the homogenous system is quite simple: $U(t) = c\cdot e^{-t/RC}$.

Now we solve the particular system with variation of the constant $c$, which means we try the Ansatz $U(t) = c(t)\cdot e^{-t/RC}$. This give the differential equation $c'(t) = \eta(t)e^{t/RC}$, which can simply integrate to $c(t) = \int\limits_{0}^{t} \eta(\hat{t})e^{\hat{t}/RC}d\hat{t} + U_0$.

So the solutions of the langevin equation is

$U(t) = e^{-t/RC}\left(\int\limits_{0}^{\hat{t}} \eta(t)e^{\hat{t}/RC}d\hat{t} + U_0\right)$.

Now we can calculate all moments. The first moment is pretty strait forward

$\langle U(t)\rangle = \left\langle e^{-t/RC}\left(\int\limits_{0}^{t} \eta(\hat{t})e^{\hat{t}/RC}d\hat{t} + U_0\right) \right\rangle = U_0e^{-t/RC}$.

The second moment needs a little more effort.

$\langle U(t)U(t')\rangle = \left\langle e^{-(t+t')/RC}\left(\int\limits_{0}^{t}\int\limits_{0}^{t'} \eta(\hat{t})\eta(\hat{t'})e^{(\hat{t}+\hat{t'})/RC}d\hat{t}d\hat{t'} + 2U_0\int\limits_{0}^{t} \eta(\hat{t})e^{\hat{t}/RC}d\hat{t} + U_0^2\right)\right\rangle % =e^{-(t+t')/RC}\left(\int\limits_{0}^{t}\int\limits_{0}^{t'} \frac{2k_BT}{RC^2}\delta(\hat{t}-\hat{t}') e^{(\hat{t}+\hat{t'})/RC}d\hat{t}d\hat{t'} + U_0^2\right)$

To be correct, we have to treate two different cases, one for $t<t'$ and one for $t>t'$. But let's focus on $t>t'$ at the moment. Then the next step would be

$\langle U(t)U(t')\rangle =e^{-(t+t')/RC}\left(\frac{2k_BT}{RC^2}\int\limits_{0}^{t}e^{2\hat{t}/RC}d\hat{t} + U_0^2\right) % =e^{-(t+t')/RC}\left(\frac{k_BT}{C}e^{2t/RC} + U_0^2\right) % =\frac{k_BT}{C}e^{(t-t')/RC} + U_0^2e^{-(t+t')/RC}$.

For $t<t'$ you'll get

$\langle U(t)U(t')\rangle =\frac{k_BT}{C}e^{(t'-t)/RC} + U_0^2e^{-(t+t')/RC}$.

So we can combine both results to

$\langle U(t)U(t')\rangle =\frac{k_BT}{C}e^{-|t-t'|/RC} + U_0^2e^{-(t+t')/RC}$.

If you choose the initial voltage $U_0 = 0$, this gives you the exakt result from wikipedia.

  • $\begingroup$ Much obliged. But I don't think I am really going to understand things until I know why my derivation failed. Any ideas about that? $\endgroup$ Sep 17, 2015 at 21:21
  • $\begingroup$ You set $\langle U (t)\eta (t)\rangle =0$ in your third step. Now, if you look at the solution of the langevin equation, this is clearly not the case, since $U (t) $ itself contains a term proportional to $\eta (t) $. $\endgroup$
    – manthano
    Sep 18, 2015 at 6:02
  • $\begingroup$ Dang, if it had been a snake, it'd a bit me! Thanks! $\endgroup$ Sep 18, 2015 at 13:18
  • $\begingroup$ Suggestion: You can use $\$\$\mbox{[equation]}\$\$$ on your main equations. So they will be centralized in the answer, and in nicer form. $\endgroup$ Oct 17, 2015 at 21:03

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