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[...] There we discovered that the mean square of the distance from one end to the other of the chain of random steps, which was the intensity of the light, is the sum of the intensities of the separate pieces. And so, by the same kind of mathematics, we can prove immediately that if $\mathbf{R}_N$ is the vector distance from the origin after N steps, the mean square of the distance from the origin is proportional to the number N of steps. That is, $\langle R^2_N\rangle=NL^2$, where $L$ is the length of each step. Since the number of steps is proportional to the time in our present problem, the mean square distance is proportional to the time:

$$\langle R^2\rangle = \alpha t \tag{41.17} $$

This does not mean that the mean distance is proportional to the time. If the mean distance were proportional to the time it would mean that the drifting is at a nice uniform velocity. The sailor is making some relatively sensible headway, but only such that his mean square distance is proportional to time. That is the characteristic of a random walk.

We may show very easily that in each successive step the square of the distance increases, on the average, by $L^2$. For if we write $\mathbf{R}_N=\mathbf{R}_{N−1}+\mathbf{L}$, we find that $\bf{R}^2_N$ is $$\mathbf{R}_N\cdot\mathbf{R}_N=\mathbf{R}^2_N=R^2_{N−1}+2\mathbf{R}_{N−1}\cdot\mathbf{L}+L^2,$$ and averaging over many trials, we have $\langle R^2_N\rangle=\langle R^2_{N−1}\rangle+L^2$, since $\langle\mathbf{R}_{N−1}\cdot\mathbf{L}\rangle=0$. Thus, by induction, $$ R^2_N=NL^2.\tag{41.18}$$

This is taken from Richard Feynman's Lectures on Physics, chapter 41, part 4.

There are two things I don't understand here.

  1. Why is $\langle \vec R_{N-1} \cdot \vec L \rangle = 0$
    Can someone show the detailed calculation or reason?
    My guess is that he is considering them perpendicular. In case that is true, how come? In brownian motion, $\vec L$ has arbitrary direction, why should it be perpendicular to the (N-1)-th position vector?

  2. Shouldn't the left-hand term of equation (41.18) be enclosed by brackets ?
    (That is $\langle R^2_{N} \rangle = NL^2$ instead of the given expression?) What allows to remove them?

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There is no correlation between $\vec{R}_{N-1}$ and $\vec{L}$. Their angle is random, and the average of $\vec{R}_{N-1} . \vec{L}$ is the average of a cosine, which is 0.

I would think that (41.18) would indeed be better written with brackets, as you said.

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$\langle\mathbf{R}_{N-1}\cdot\mathbf{L}\rangle$ is zero because $\bf L$ is equally likely to be in any direction, and so is as likely to be anti-parallel to $\mathbf{R}_{N-1}$ (and thus negative) as is to be parallel. It may be helpful to rewrite this in terms of angles:

$$ \langle\mathbf{R}_{N-1}\cdot\mathbf{L}\rangle=\langle R_{N-1}L\cos\theta\rangle=\langle R_{N-1}\rangle \langle L\rangle\langle\cos\theta\rangle=0$$

where the second equality holds and the expectation values of $L$, $\cos\theta$, and $R_{N-1}$ are uncorrelated, and the last holds because $\langle\cos\theta\rangle$ is zero.

For your second question, that looks like a typo.

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  • $\begingroup$ This solves my doubt, but I am not sure if L is constant, it's not stated in this section. In chapter 6, part 3 were he first introduces random walks, Feymann starts off with a random walk of constant step and then in part 4 generalizes it to one of variable length with average one $\endgroup$
    – J.C.VegaO
    Jun 26 '19 at 13:47
  • $\begingroup$ It doesn't change the result if it's not constant, since its value is still uncorrelated with $R $ and $\theta $. $\endgroup$
    – Chris
    Jun 26 '19 at 13:54
  • $\begingroup$ I know, just for sake of correctness, L cannot come out of the spectation operator $\endgroup$
    – J.C.VegaO
    Jun 26 '19 at 14:36
  • $\begingroup$ @juancarlosvegaoliver The original text is clearly treating $L$ as a constant, but I put some $\langle\rangle$ around $L$ anyway. $\endgroup$
    – Chris
    Jun 26 '19 at 14:50

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