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,

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


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


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

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


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

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?


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.

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  • $\begingroup$ 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

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