l4teLearner
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- TL;DR Summary
- 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
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
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