2
$\begingroup$

Given the Langevin equation of a massless brownian particle:

$$ \gamma \dot{x}=\eta, $$

where $\gamma$ is the friction coefficent and $\eta$ the noise ($\langle\eta \rangle =0$ and $\langle\eta(t)\eta(s)\rangle=2k_BT\gamma\delta(t-s)$), I want to find the correlation $C(t,s)=\langle x(t)x(s)\rangle$ and the response function $G(t,s)=\langle \frac{\delta x(t)}{\delta \eta(s)} \rangle$.

I find that, since $x(t)=\int_0^tv(s)ds$ ($x_0=0$ for simplicity),

$$ C(t,s)=\frac{1}{\gamma^2}\int_0^tdu\int_0^sdv\langle\eta(u)\eta(v)\rangle=\frac{1}{\gamma}\int_0^s\int_0^sdudv2k_BT\delta(u-v)=\frac{2k_BT}{\gamma}s, $$

where I considered $t>s$.

I find also, from the Langevin equation,

$$ \frac{\delta}{\delta\eta(s)}\gamma\frac{\partial x(t)}{\partial t}=\frac{\delta\eta(t)}{\delta\eta(s)}=\delta(t-s), $$

then

$$ \int_{s-\epsilon}^{s+\epsilon}\frac{\partial}{\partial t}\frac{\delta x(t)}{\delta\eta(s)}dt=\int_{s-\epsilon}^{s+\epsilon}\frac{1}{\gamma}\delta(t-s)=\frac{1}{\gamma} $$

and

$$ \frac{\delta x(t)}{\delta \eta(s)}\Bigg|_{t\rightarrow s^+}=\frac{1}{\gamma}\;\;\;\;\;\text{and}\;\;\;\;\;\frac{\delta x(t)}{\delta \eta(s)}\Bigg|_{t\rightarrow s^-}=0\;\;\text{(by causality)}. $$

Hence, in the discontinuity I choose half of the value for positive $s$: $\frac{1}{2\gamma}$.

Then

$$ G(t,s)=\langle\frac{1}{2\gamma}\rangle=\frac{1}{2\gamma}. $$

Now I want to verify the fluctuation-dissipation theorem, which in this case I believe reads

$$ \frac{\partial}{\partial s} C(t,s)=k_BTG(t,s). $$

Obviously, the problem is that $\frac{\partial}{\partial s} C(t,s)=\frac{2k_BT}{\gamma}$, while $k_BTG(t,s)=\frac{k_BT}{2\gamma}$.

Am I doing something wrong? Isn't it weird that $C(t,s)$ doesn't depend on $t$? Shouldn't the fluctuation-dissipation theorem be satisfied?

$\endgroup$
1
$\begingroup$

Let us use the definition $\langle\eta(s)\eta(t)\rangle=\Gamma\gamma^2\delta(t-s)$. First of all, $C(s,t)$ depends on $t$ because $$C(s,t)=\Gamma\min(s,t).$$

It is clear, from causality, that $\frac{\delta x(t)}{\delta \eta(s)}=0$ if $t<s$. If $t>s$, compute the difference $\delta x(t)$ caused by two realisations of noise that differ only at time $s$ by an amount $\delta\eta$. This means that $\delta\eta(t)=\delta\eta\,\delta(t-s)$. Then you have $$\frac{\partial \delta x}{\partial t}=\gamma^{-1}\delta\eta \,\delta(t-s)$$ and therefore $\delta x(t)=\gamma^{-1}\delta\eta$ for $t>s$, which means that $$\frac{\delta x(t)}{\delta\eta(s)}=\frac1\gamma\,\Theta(t-s).$$

Now we compute $$\frac{\partial C}{\partial s}(s,t)=\Gamma\Theta(t-s).$$ So this agrees with the fluctuation-dissipation theorem if $$\Gamma=\frac1\gamma\qquad\text{or}\qquad\langle\eta(s)\eta(t)\rangle=\gamma\delta(t-s).$$

The value of $\Gamma$ that you proposed, $2k_{\text B}T\gamma$, has no meaning here, since the particle has no mass, it has no kinetic energy and it is therefore impossible for it to be thermalized.

$\endgroup$
  • $\begingroup$ Got it. I implicitly used $\frac{1}{2}m\langle v^2\rangle=\frac{1}{2}k_B T$. But wouldn't $\langle \eta(t) \eta(s)\rangle=\gamma \delta(t-s)$ be wrong, dimensionally speaking? $\endgroup$ – Charlie Jan 18 '16 at 14:32
  • $\begingroup$ Ok, it works with $\Gamma=\frac{k_BT}{2\gamma}$, using the half-maximum convention for the $\Theta$ function. $\endgroup$ – Charlie Jan 18 '16 at 21:35

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.