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

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?

• Is he not using $\frac{d}{d x} f(x+y) = \frac{d}{d y} f(x+y)$ to show inconsistency? – Sunyam Aug 12 '20 at 21:22
• @Sunyam I don't think so! The autocorrelation is not a function of $t+t'$. If you assume stationarity then it is a function of $t'$ only, and if you assume nothing then it is separately a function of the time arguments $t_1$ and $t_2$ for $\langle v(t_1)v(t_2)\rangle$. – SRS Aug 12 '20 at 21:41
• In your equation (4), shouldn't the condition be $t' \geq 0$? – Daniel Aug 18 '20 at 3:05

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.

• You are right about Eq.$(4)$ but he writes for $t\geq 0$. But how does it answer my question? You show that $\frac{d}{d\tau}K(\tau)=E[\dot{X}(t+\tau)X(t)]=0$ but note that he says it is nonzero. @hyportnex – SRS Aug 12 '20 at 21:37
• if you differentiate eq 4 you should get 0 not eq 5, as I proved it; and the reason for the error is the missing abs || in the exponent. – hyportnex Aug 13 '20 at 0:26
• Please note that your notation is inconsistent. You replaced $d/d\tau$ by $d/d\eta$ i.e. started differenting w.r.t $\tau$ and ended up writing a derivative w.r.t $\eta$. Also, you are not showing $\langle \dot{X}(t)X(t)\rangle=0$ but $\langle \dot{X}(t+\tau)X(t)\rangle=0$ which does not follow from stationarity. It's not clear to me why should the derivative of an even function be zero. @hypnortex – SRS Aug 13 '20 at 4:25
• all even functions $f(x)=f(-x)$ have derivative $f'(x) =0$ at $x=0$ if exists; just draw it! the derivative of |x| is |x|/x, which of course does not exist at x=0, but you can approach it by smoothing |x| around x=0. And $\eta$ is not the variable, it is an infinitesimal around the variable $\tau$, t is not a variable of the correlation because it depends on the difference of two instance that is $\tau$, because the process is stationary. – hyportnex Aug 13 '20 at 8:48
• Note that if you allow asymmetric smoothing functions, you can take limits in a way that makes $f'(0)$ whatever you want (within $[-\gamma,\gamma]$). The assumption of symmetry is doing all of the work here. – Daniel Aug 22 '20 at 18:10

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 (with probability 1) nowhere differentiable, so neither (3) nor (5) is correct. $$\langle v\dot{v}\rangle$$ is undefined.

• books.google.co.in/… – SRS Aug 22 '20 at 17:35
• Read the paragraph below 15.73. Nonexistence of derivative at the origin is not the issue here. He is interested in the right derivative. The problem, in my opinion, is that he does not distinguish between derivative w.r.t $t$ and $t'$ as I explained in my post. @Daniel – SRS Aug 22 '20 at 17:35
• I don't have access to the book, unfortunately. But the use of $t$ and $t'$ looks right to me, since the author ends up calculating $\langle v(u) \frac{d}{du}v(u)\rangle$ both ways. Do you agree that $\langle f(t) \rangle$ is well-defined iff $f(t)$ is well-defined almost everywhere? – Daniel Aug 22 '20 at 18:07
• Assuming the derivative exists, if you took a derivative of $(1)$ w.r.t $t'$, you would not get zero because the RHS is not independent of $t'$. However, if you took a derivative of $(1)$ w.r.t $t$, you would indeed get zero because the RHS is independent on $t$ by stationarity. Do you agree on this part? @Daniel – SRS Aug 22 '20 at 19:28
• Yes, definitely. – Daniel Aug 22 '20 at 20:19

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.