Is there really an inconsistency with the original Langevin equation (as claimed in the book Nonequilibrium Statistical Mechanics - V. Balakrishnan)?

Question

I am reading the book Nonequilibrium Statistical Mechanics by V. Balakrishnan. In chapter $17$ (page $244$) he argues that the original Langevin equation has inconsistencies and should, therefore, be replaced by generalized Langevin equation. He shows the inconsistency as follows (which I am not fully convinced about).

If the velocity process to $v(t)$ be a stationary random process with zero mean, the velocity autocorrelation function $\langle v(t)v(t+t')\rangle_{\rm eq}$ in equilibrium depends only on the difference of the time arguments i.e. $$\langle v(t)v(t+t')\rangle_{\rm eq}=\text{independent of }t.\tag{1}$$ Then he takes a derivative w.r.t $t$ to obtain, $$\langle \dot{v}(t)v(t+t')\rangle_{\rm eq}+\langle v(t)\dot{v}(t+t')\rangle_{\rm eq}=0,\tag{2}$$ where $\dot{v}=\frac{dv}{dt}$. Then by setting, $t'\to 0$, one immediately finds that $$\langle \dot{v}(t)v(t)\rangle_{\rm eq}=0\tag{3}$$ if we assume $v(t)$ to be classical commuting variable. Eq.$(3)$ simply means that for the stationary random velocity process, the instantaneous velocity and acceleration must be uncorrelated. Note that the derivation upto Eq.$(3)$ does not assume any functional form of the correlation function, but only stationarity!

In the original Langevin model, one can calculate this autocorrelation function and its explicit form turns out that $$\langle v(t)v(t+t')\rangle_{\rm eq}=\frac{k_BT}{m} e^{-\gamma t'}, t'\geq 0.\tag{4}$$ Note that it satisfies the criterion $(1)$. For this, if we go through the steps $(1)$-$(3)$, we indeed see that $\langle \dot{v}(t)v(t)\rangle_{\rm eq}=0,$ as expected. But in the book, he first took derivative w.r.t $t'$ (instead of $t$), continued to call $\dot{v}=\frac{dv}{dt'}$ (instead of $\dot{v}=\frac{dv}{dt}$), and then set $t'\to 0$, to derive that $$\langle \dot{v}(t)v(t)\rangle_{\rm eq}=-\frac{\gamma k_BT}{m}\neq 0\tag{5}$$ to argue that there is an inconsistentcy between $(3)$ and $(5)$!

Question But in my opinion, there cannot be an inconsistency because the first derivation is independent of the functional form of $\langle v(t)v(t+t')\rangle_{\rm eq}$, and hence, conclusion $(3)$ should be true irrespective of the form of $\langle v(t)v(t+t')\rangle_{\rm eq}$. Indeed if we follow steps $(1)$-$(3)$, we do not get any contradiction! The first part (i.e., up to $(3)$) not even uses Langevin's equation, and valid for the autocorrelation of any stationary random variable.

Can someone comment whether I am correct and the book is wrong or vice-versa?

Answer 1

The author's argument essentially reduces to a statement that $$\frac{d}{dt}\bigg|_{t' = 0} e^{-\gamma|t'|} = -\gamma $$ It should be clear to you that this is not true.

The way he argues this is by saying that $e^{-\gamma|t'|} = e^{-\gamma t'} $ for $t' > 0$, then taking the derivative and taking the limit as $t' \rightarrow 0^+$. Because the function is not smooth, this argument doesn't work. The derivative is actually undefined at $t' = 0$.

Of course, this still looks like a contradiction. Is $\langle v\dot{v}\rangle$ zero, or is it undefined? Equations (2) and (3) assume that $v(t)$ is differentiable almost everywhere. In the Langevin model, $v(t)$ is almost-nowhere differentiable, so neither (3) nor (5) is correct. $\langle v\dot{v}\rangle$ is undefined.

Answer 2

The first thing that needs to be addressed is the issue of differentiation with respect to $t$ vs $t'$. If you have a function $f$ of one variable, then we can call its derivative $f'$. You don't need to specify its argument; $f'$ is the derivative of $f$, full stop.

If you compose $f$ with another function $g$, then the chain rule says that $(f\circ g)' = (f'\circ g) \cdot g'$. This is more commonly written $\frac{d}{dt}f\big(g(t)\big) = f'\big(g(t)\big) \cdot g'(t)$.

Bearing that in mind (and switching to a dot rather than a prime to denote a derivative), $$\frac{d}{dt}v(t+t')= \dot v (t+t') \cdot \frac{d}{dt}(t+t') = \dot v(t+t')$$ and $$\frac{d}{dt'} v(t+t') = \dot v (t+t') \cdot \frac{d}{dt'}(t+t') = \dot v(t+t')$$

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But in my opinion, there cannot be an inconsistency because the first derivation is independent of the functional form of $\langle v(t)v(t+t′)\rangle_{eq}$, and hence, conclusion (3) should be true irrespective of the form of $\langle v(t)v(t+t′)\rangle_{eq}$.

I'm not sure what you mean by "there cannot be an inconsistency" - what you point out is precisely the inconsistency that Balakrishnan is talking about. If we assume that $v$ is a stationary random process in thermal equilibrium, then it follows immediately that $\langle v(t) \dot v(t)\rangle = 0$ by the argument you laid out.

On the other hand, multiplying the Langevin equation by $v$ and taking the ensemble average yields $$\langle v(t) \dot v(t)\rangle = -\frac{\gamma}{m} \langle v(t)^2\rangle + \frac{1}{m}\langle v(t)\eta(t)\rangle = -\frac{\gamma k_BT}{m} + 0 \neq 0$$ where we've used the fact that, because the particle has inertia, $\eta(t)$ is correlated with $\dot v(t)$ but not $v(t)$.

In words, the Langevin equation predicts a correlation between the velocity and acceleration at any given moment of time, but the requirement that the velocity be a stationary process implies otherwise. The reason for this apparent inconsistency - which I inadvertently alluded to in a comment to my answer to your related question - is that the Langevin equation includes a term $-\gamma v(t)$ which assumes that the frictional response of the fluid is instantaneous.

However, under close inspection this is physically untenable. Imagine that the particle is moving with some speed through the fluid before receiving a little kick which brings it (momentarily) to rest. The term $-\gamma v(t)$ suggests that the fluid friction would also drop to zero at that moment. However, the motion of the particle causes the fluid itself to flow, and it will take some (very small) time for the fluid to adjust after the particle has come to rest.

As long as we're not concerned with such short time scales, the Langevin equation should be adequate. However, the inconsistency noted above arises as we take the limit as $t'\rightarrow 0$, at which point we are taking into consideration the behavior of the particle at arbitrarily short time scales. To resolve the issue, we generalize to a non-instantaneous fluid response by including memory effects via the memory kernel

$$-\gamma v(t) \mapsto -\int_{-\infty}^t \gamma(t-t')v(t') \mathrm dt'$$

which reduces to the standard Langevin equation if we let the fluid response be instantaneous, $\gamma(t-t')=\gamma_0 \delta(t-t')$.

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Further reading: Non-Markovian Langevin Equations, p. 19 of Nonequilibrium Statistical Mechanics by Zwanzig.

Answer 3

There is no contradiction here, instead its a mis-interpretation of the equations. To see this it is useful to think about the physical meaning of the two derivatives. The derivative with respect to $t$ is saying that the correlations are the same if we shifted our time origin point (i.e. the derivative is zero). The derivative with respect to $t'$ instead is asking how do the correlations change as the two times get closer or further apart. If you think about the physical meaning the results both make sense. In fact the two derivatives that are taken are in orthogonal directions so the 'contradiction' is the same as claiming that $$\frac{df(x,y)}{dx} \neq \frac{df(x,y)}{dy}$$ is a contradiction.

To see this is might make sense to think about equation (4) like this: $$f(t_1, t_2) = Ke^{-\gamma |t_1 - t_2|}.$$ Just looking at this we expect the derivatives to be non-zero if we take the derivative with respect to $t_1$ or $t_2$ then we should get a non-zero number. However changing the coordinates to $\tau = \frac{1}{2}(t_1 - t_2)$ and $t = \frac{1}{2} (t_1 + t_2)$ then we get: $$\tilde{f}(t,\tau) = Ke^{-\gamma |\tau|}$$ And so since $\tilde{f}(t,\tau) = \langle v(t-\tau)v(t+\tau)\rangle_{eq}$ we can see that the two derivatives used to get equations (3) and (5) are in orthogonal directions of the $t_1$, $t_2$ plane.

Answer 4

I think $(4)$ should be $$\langle v(t)v(t+t')\rangle_{\rm eq}=\frac{k_BT}{m} e^{-\gamma |t'|}$$ for any $t$.

In general, for a stationary process define the autocorrelation $$K(t_1-t_2) = K(t_2-t_1)=\mathbf{E}[X(t_1)X(t_2)]$$ therefore $K(\tau)=K(-\tau)$.

Furthermore, $$\frac{d}{d\tau}K(\tau)=\frac{d}{d\tau}\mathbf{E}[X(t+\tau)X(t)]= lim_{\eta \to 0} \mathbf{E}\big[\frac{X(t+\tau+\eta)-X(t+\tau)}{\eta}X(t)]\\ =\mathbf{E}[\dot X(t+\tau)X(t)\big]$$ but this $0$ because $K(\tau)$ is even symmetrical.