Feynman's derivation of average square distance variation in Brownian motion

[l4teLearner]

TL;DR

I am puzzled by the calculations performed in https://www.feynmanlectures.caltech.edu/I_41.html#Ch41-S4 i.e. I cannot understand how the average distance square is calculated.

I am studying lesson 41 (volume I) of the Feynman lectures on physics.
You can find it at the following link
https://www.feynmanlectures.caltech.edu/I_41.html#Ch41-S4

What I don't understand, first, is this consideration:

"What is the rate of change of $x^2$
? It is $\frac{d x^2}{dt}=2x(\frac{dx}{dt})$
so what we have to find is the average of the position times the velocity. We shall show that this is a constant, and that therefore the mean square radius will increase proportionally to the time, and at what rate."

As $x$ is after all a stocastic variable, is it allowed to bring the derivative inside the average? Because the derivative of the average is non zero, of course (the particle's squared distance from start grows with time), instead the average of the velocity times the space, to me, could as well be the average of position times the average of velocity (they are uncorrelated) hence be zero.

After, the calculation, one of the terms found is indeed $m \frac{d[x(dx/dt)]}{dt}$. It is then said that this is zero: "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". This should have been proved but it is instead stated with an explanation that I canno understand. I find this a bit of a circular argument.

Any thoughts?

thanks

Manuel

 

[Herman Trivilino]

"What is the rate of change of $x^2$? It is $\frac{x^2}{dt}=2x(\frac{dx}{dt})$?

That should be $\frac{d}{dt}x^2=2x(\frac{dx}{dt})$.

As x is after all a stocastic variable, is it allowed to bring the derivative inside the average?

I don't understand this question. Feynman is simply taking the derivative of the average value of $x^2$. Yes, of course you are allowed to take the derivative of that quantity. He's not bringing the derivative inside the average, neither is he bringing the average inside the derivative, if that's what you meant. He's simply taking the derivative of the average value.

After, the calculation, one of the terms found is indeed md[x(dx/dt)]dt. It is then said that this is zero: "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". This should have been proved but it is instead stated with an explanation that I canno understand. I find this a bit of a circular argument.

It's certainly a hand-waving argument, but I don't see any circularity.

Any thoughts?

There's a whole lot of hand-waving going on in that entire passage. I find it hard to follow, too.

The Feynman Lectures on Physics was, according to Feynman, an experiment. He had agreed to teach the introductory course at Cal Tech shortly after arriving there, but he did it under the condition that it was to be a one off.

That was in the early 1960's. The series of books has certainly gained a lot of popularity and fame, but Feynman himself considered the entire experiment to be something of a failure.

I recall reading a journal article years ago, written by one of his colleagues there at Cal Tech at the time. He was rather critical of the lectures themselves as well as the series of books that followed. He reported that as the series of courses proceeded fewer and fewer students attended the lectures, but more and more physics professors attended, as observers. He considered it to be a grand view from above, rather than an approach that built knowledge.

I never found it to be a good source for the uninitiated wanting to learn the subject. But if you're an accomplished physicist you may find his well thought out approach to his own knowledge a wonder to behold. He had a well-developed understanding of physics, but that doesn't necessarily translate into good teaching.

Don't get me wrong. I'm a huge fan of Richard Feynman and find him to be someone who was a fascinating person. He understood physics in a way that is unequaled. His scientific foundation to a way of thinking about everything is something I absorbed from reading him, watching him, and reading about him. It has had a huge influence on the way I think, and the way I taught my children and grandchildren to think. Truly a legacy worth passing on.

 

[l4teLearner]

That should be $\frac{d}{dt}x^2=2x(\frac{dx}{dt})$.

I don't understand this question. Feynman is simply taking the derivative of the average value of $x^2$. Yes, of course you are allowed to take the derivative of that quantity. He's not bringing the derivative inside the average, neither is he bringing the average inside the derivative, if that's what you meant. He's simply taking the derivative of the average value

thanks for spotting the typo, I'll fix it.
as for the order of derivation and expectation, if you look at the beginning of the paragraph, the value to calculate is the derivative of the average of the distance squared, as we seek to find $\alpha$ in $<R^2>= \alpha$ and prove that it is constant.
later, when saying "What is the rate of change of #x^2#? It is $d(x2)/dt=2x(dx/dt)$, so what we have to find is the average of the position times the velocity." it seems to me he is averaging $2x(dx/dt)$ i.e. it he is averaging the derivative of $x^2$.
Indeed, when taking the average of the Langevin equation multiplied by $x$ it is implicitely calculating the average of the derivative of $x^2$.
finally, equation 41.20 instead, has the derivative of the average: $d(<x^2>)(t) = 2\frac{kT}{\mu}.$.
So it seems to me he is using d/dt<> and <d/dt> interchangeably.
Or am I missing something?

 

[Ibix]

So it seems to me he is using d/dt<> and <d/dt> interchangeably.

You can work through it from the definition of an average$$\begin{eqnarray*}
\frac{d}{dt}\left(\langle x\rangle\right)&=&\frac{d}{dt}\left(\frac1N\sum_{i=1}^N x_i\right)\\
&=&\frac 1N\sum_{i=1}^N\frac{dx_i}{dt}\\
&=&\langle \frac{dx}{dt}\rangle
\end{eqnarray*}
$$where I have assumed $\dfrac{dN}{dt}=0$, which is reasonable enough in this context. This works because an average is just a sum multiplied by a constant, and the derivative of a sum is the sum of the derivatives.