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I found a nice discussion of Brownian motion in the Feynman lectures, reproduced online here: https://www.feynmanlectures.caltech.edu/I_41.html

Feynman considers a particle undergoing a Brownian motion, treated in one dimension in the first instance, such that the equation of motion (Newton's second law) is $$ m \ddot{x} = F - \alpha \dot{x}, $$ where $F$ is a rapidly fluctuating force and $\alpha$ is a constant viscous drag coefficient. (He uses the letter $\mu$ but I prefer to use $\alpha$ because $\mu$ is widely used for mobility.) (This is also called the Langevin equation). At one point in the argument we want to know the value of $$ \left\langle \frac{d}{dt}(x \dot{x}) \right\rangle, \tag{1} $$ where I believe the average is over all paths setting out from a given start point and continuing for a given time $t$. Feynman states that this quantity will be zero, but the reason he gives is not quite convincing to me. He asserts

"Now $x$ times the velocity has a mean that does not change with time, because when it gets to some position it has no remembrance of where it was before, so things are no longer changing with time. So this quantity, on the average, is zero."

The trouble is that the phrase "$x$ times the velocity has a mean that does not change with time" is the statement $$ \frac{d}{dt} \langle x \dot{x} \rangle = 0. \tag{2} $$ This quantity is indeed zero, but it is not self-evident that it is equal to (1).

Given (2), one way to obtain (1) would be to show that $$ \frac{d}{dt} \langle x \dot{x} \rangle = \left\langle \frac{d}{dt} (x \dot{x}) \right\rangle . \tag{3} $$

My question is: is (3) easy to prove (without assuming (1)!)? (If so please provide proof). Or is there some better way to prove that the quantity in (1) is equal to zero?

To forestall answers merely claiming "you can reverse order of integration and differentiation" here is why that alone is not the answer: $$ \frac{d}{dt} \langle x v \rangle = \frac{d}{dt} \iint x v f(x,v,t) dx dv $$ where $f(x,v,t)$ is the probability density function. This gives $$ \frac{d}{dt} \langle x v \rangle = \iint \frac{d}{dt} \left( x v f \right ) dx dv \\ = \iint f \frac{d}{dt}(xv) + (xv) \frac{df}{dt} dx dv = \langle \frac{d}{dt}(xv) \rangle + \iint x v \frac{df}{dt} dx dv $$ so either I have muddled something about the distribution function, or one has to show that the extra term is zero (keeping in mind that $\langle x v \rangle$ is not zero).

Andrew Steane
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    From your added part, just assume your are in steady state i.e. df/dt=0 as at equilibrium mean values should not change. Or, if that is not satisfactory, try to see what happens if f(x, v, t)=f(x, t)f(v, t) i.e. speed does not depend on position and vice versa. I expect you get $\langle x \rangle$ or $\langle v \rangle $ terms that are zero. – JalfredP Sep 18 '21 at 08:26
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    Feynman says that he's interested in the time average of the values, not averages over space or velocity or noise. So, isn't it just following standard calculus rules here that taking the derivative of an integral is equal to integrating the derivative? – Kyle Kanos Aug 27 '22 at 01:50

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Whether you take the mean to be the result of a sum over particles or over realizations, the question is whether the derivative of the mean is the same as the mean of the derivative.

This is the same as swapping the order of integration/summation and differentiation, since taking the mean is taking an integral or a sum.

So the sketch of the proof is that this is given by Leibniz.

Conceptually, I believe a symmetry argument can show that for every situation with increasing $x\dot{x}$, there is an equal and opposite situation with decreasing $x\dot{x}$ to cancel it out. For any collision/kick at given x, there will exist a statistical particle with equal and opposite change in the velocity. For any constant velocity drift particle at x with velocity $\dot{x}$, there will be a statistical particle at -x with velocity $\dot{x}$ to cancel it (both have the same velocity but one gets closer to x=0 and the other gets further).

Alwin
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  • Hmm ... but $\langle x \dot{x} \rangle \ne 0$. – Andrew Steane Sep 17 '21 at 20:26
  • Can you explain why it isn't 0? To me, by symmetry, it is. – Alwin Sep 17 '21 at 23:55
  • That is, the integral of an odd function is 0. – Alwin Sep 18 '21 at 00:12
  • We have $\langle x^2 \rangle = 2 D t$ so $(d/dt)\langle x^2 \rangle = 2D$. If we think that $(d/dt)\langle x^2 \rangle = \langle (d/dt)x^2 \rangle$ then it follows that $\langle 2 x \dot{x} \rangle = 2D$. – Andrew Steane Sep 18 '21 at 08:30
  • By the same reasoning you present in your edited question, $(d/dt) = <(d/dt)(xx)> + \int\int (xx)(df/dt)dxdv = <2x\dot{x}> + \int\int(xx)(df/dt)dxdv$. This time, the extra term is an integral of an even function which does not evaluate to 0. But $\int\int (xv)(df/dt)dxdv$ is an integral of an odd function which evaluates to 0. Anyway, I agree that swapping order is not alone the answer, it's swapping order + symmetry. – Alwin Sep 18 '21 at 12:07
  • Thanks for your further thoughts. So now it appears we might argue that $\langle x \dot{x} \rangle =0$ while $(d/dt) \langle x^2 \rangle \ne 0$. If that is right then it contradicts the argument offered by Feynman, and I would be surprised if he got it wrong. But it may be that he is talking about some other average and I have misunderstood that. – Andrew Steane Sep 18 '21 at 14:45
  • I agree with you that $$ must be non-zero, but I fail to see how $\int xv f$ and $\int xv df/dt$ can have different parity. Why would f be asymmetric whereas $df/dt$ be symmetric? I still wonder if it's possible that $\int xv df/dt = 0$ by the even/odd argument, which I must be misunderstanding, because my picture of the symmetry would argue both integrals to be 0. – Alwin Sep 18 '21 at 16:19
  • @AndrewSteane May be this is useful https://physics.stackexchange.com/questions/667633/correlation-of-position-and-velocity-in-brownian-motion?noredirect=1&lq=1 – user1420303 May 31 '22 at 11:21
  • @user1420303 err ... that was a follow-up question from me! – Andrew Steane May 31 '22 at 12:14