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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).

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  • $\begingroup$ 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. $\endgroup$
    – JalfredP
    Sep 18 at 8:26
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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).

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  • $\begingroup$ Hmm ... but $\langle x \dot{x} \rangle \ne 0$. $\endgroup$ Sep 17 at 20:26
  • $\begingroup$ Can you explain why it isn't 0? To me, by symmetry, it is. $\endgroup$
    – Alwin
    Sep 17 at 23:55
  • $\begingroup$ That is, the integral of an odd function is 0. $\endgroup$
    – Alwin
    Sep 18 at 0:12
  • $\begingroup$ 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 $\langle 2 x \dot{x} \rangle = 2D$. $\endgroup$ Sep 18 at 8:30
  • $\begingroup$ By the same reasoning you present in your edited question, $(d/dt)<xx> = <(d/dt)(xx)> + \int\int (xx)(df/dt)dxdv = <2x\dot{x}> + \int\int(xx)(df/dt)dxdv$. 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. $\endgroup$
    – Alwin
    Sep 18 at 12:07

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