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In The Feynman Lectures on Physics Vol. I Ch. 41: The Brownian Movement, $\S4$ The random walk we are told:

The reader may easily verify that the number of collisions a single molecule of water receives in a second is about $10^{14}$.

This may be discussed in the exercises, but there has been very little in the Lectures, to this point, which would suggest an approach to finding this number. Indeed, I'm not completely sure what the statement means. For example, in the following chapter the "volume occupied" by one molecule of liquid is the inverse of the number of molecules per volume. That suggests that the molecules are always in mutual contact. What, then, is meant by a liquid molecule receiving a collision?

Even if I had some notion of what a collision really meant, I'm not sure how the reader is expected to determine the number per second. I might be able to come up with a way of solving the problem, but I would like to know what approach Feynman intended in this context.

Any suggestions?

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What Feynman actually said at this point in his lecture was, "The number of collisions --Figure this one out; see if you don't get the same answer -- The number of collisions that a single molecule of water receives in a second is 10^14..." Feynman was challenging his students to figure this out on their own, without their having been told how to do it. This is typical of the kinds of problems that are assigned to physics students at Caltech: Caltech students are expected to learn how to think creatively.

The average number of collisions a single molecule of water receives in a second equals 1/tau, where tau is the mean time between collisions. The way to calculate that is discussed in The Feynman Lectures on Physics, Chapter 43, Diffusion.

Michael A. Gottlieb

Editor, The Feynman Lectures on Physics

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  • $\begingroup$ Indeed they assign difficult problems at Caltech. For example, the exercises for the Feynman Lectures ask for a proof of the Kepler conjecture in the first chapter. Not explaining that it had gone unproven for centuries. My approach to estimating the number of collisions was to treat the molecule like an oscillator confined to a box and find the Planck frequency associated with the average energy per molecule. I reasoned that each complete oscillation would constitute two collisions. The result was $3.75\times 10^{13}$. $\endgroup$
    – Steven Thomas Hatton
    Commented Mar 29, 2018 at 1:23
  • $\begingroup$ Mike, I was "playing dumb". I was aware of the subsequent material which provides an approach. I asked from the perspective of a student without the benefit of that material. Often a derivation left to the reader draws from ideas introduced in a recent discussion. The lectures were given in the context of a scientific community with lots of ideas flying about; multiple professors (Feynman was the "front man" for a team); teaching assistants; recitation sessions; etc. I'm trying to get a feel for how Feynman expected his student to approach such a challenge. $\endgroup$
    – Steven Thomas Hatton
    Commented Mar 29, 2018 at 16:10
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    $\begingroup$ Feynman tossed this problem out as an aside in his lecture. It's possible it was discussed later in a recitation section, but nothing of a similar nature (that I can find) appeared on tests, quizzes or homework until after the students heard the lecture on Diffusion. While I can't be certain without searching through Feynman's notes (which are not well organized), I suspect he solved the problem from first principles when he was preparing for the lecture, and that is why he said "see if you don't get the same answer." Recall that Feynman's motto was "What I cannot create I do not understand." $\endgroup$
    – Michael A. Gottlieb
    Commented Mar 30, 2018 at 10:42

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