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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?

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    $\begingroup$ Is he not using $\frac{d}{d x} f(x+y) = \frac{d}{d y} f(x+y)$ to show inconsistency? $\endgroup$
    – Sunyam
    Aug 12, 2020 at 21:22
  • $\begingroup$ @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 $\langle v(t_1)v(t_2)\rangle$. $\endgroup$
    – SRS
    Aug 12, 2020 at 21:41
  • $\begingroup$ In your equation (4), shouldn't the condition be $t' \geq 0$? $\endgroup$
    – Daniel
    Aug 18, 2020 at 3:05
  • $\begingroup$ Yes, you are right, corrected! @Daniel $\endgroup$
    – SRS
    Jun 7, 2021 at 2:09

4 Answers 4

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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.

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  • $\begingroup$ books.google.co.in/… $\endgroup$
    – SRS
    Aug 22, 2020 at 17:35
  • $\begingroup$ 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 $\endgroup$
    – SRS
    Aug 22, 2020 at 17:35
  • $\begingroup$ 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? $\endgroup$
    – Daniel
    Aug 22, 2020 at 18:07
  • $\begingroup$ 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 $\endgroup$
    – SRS
    Aug 22, 2020 at 19:28
  • $\begingroup$ Yes, definitely. $\endgroup$
    – Daniel
    Aug 22, 2020 at 20:19
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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 $$\frac{d}{dt'} v(t+t') = \dot v (t+t') \cdot \frac{d}{dt'}(t+t') = \dot v(t+t')$$


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, $\gamma(t-t')=\gamma_0 \delta(t-t')$.


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

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  • $\begingroup$ "I'm not sure what you mean by "there cannot be an inconsistency"..." All I was saying is that Eq.(3) was derived from Eq.(1) which makes use of stationarity only! Now, the same stationarity property is obeyed by Eq.(4) which is clear from the explicit expression derived from the Langevin model i.e. $\sim e^{-\gamma |t'|}$ is independent of $t$. Had he differentiated (4) w.r.t $t$, he would have obtained zero. Then by setting $t'=0$, he would get back Eq.(3) in place of Eq. (5). Thus no contradiction would have arisen! @J.Murray $\endgroup$
    – SRS
    Jun 7, 2021 at 20:19
  • $\begingroup$ In short, Eq.(5) is not the same as Eq.(3), i.e., there is an "apparent" inconsistency because he chose to differentiate w.r.t $t'$. @J.Murray $\endgroup$
    – SRS
    Jun 7, 2021 at 20:19
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    $\begingroup$ @SRS Forgive me, but it seems to me that you are just repeatedly emphasizing his point. The correlation function $C(t,t')=\langle v(t),v(t+t')\rangle$ obeys $\lim_{t'\rightarrow 0^+}\frac{1}{2}\frac{\partial C}{\partial t} = \lim_{t'\rightarrow 0^+} \frac{\partial C}{\partial t'}$ by explicit construction, and yet the decaying exponential does not have this property. The conclusion is that the exponential is simply not the correct form for the correlation function for very short timescales. $\endgroup$
    – J. Murray
    Jun 7, 2021 at 21:48
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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.

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  • $\begingroup$ So your answer is that the apparent contradiction is only because $dv/dt$ and $dv/dt'$ are not distinguished from each other (and both are denoted by $\dot{v}$ as I suspected in my question). Is that right? @NAMcMahon $\endgroup$
    – SRS
    Jun 7, 2021 at 2:08
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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.

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  • $\begingroup$ 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 $\endgroup$
    – SRS
    Aug 12, 2020 at 21:37
  • $\begingroup$ 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. $\endgroup$
    – hyportnex
    Aug 13, 2020 at 0:26
  • $\begingroup$ 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 $\endgroup$
    – SRS
    Aug 13, 2020 at 4:25
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    $\begingroup$ 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. $\endgroup$
    – hyportnex
    Aug 13, 2020 at 8:48
  • $\begingroup$ 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. $\endgroup$
    – Daniel
    Aug 22, 2020 at 18:10

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