Timeline for Feynman Lectures Vol I 41-4: Find the number of collisions received by a water molecule per second
Current License: CC BY-SA 3.0
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| Mar 30, 2018 at 10:42 | comment | added | Michael A. Gottlieb | 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." | |
| Mar 29, 2018 at 16:10 | comment | added | Steven Thomas Hatton | 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. | |
| Mar 29, 2018 at 1:23 | vote | accept | Steven Thomas Hatton | ||
| Mar 29, 2018 at 1:23 | comment | added | Steven Thomas Hatton | 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 $3.75\times 10^{13}$. | |
| Mar 28, 2018 at 18:26 | history | edited | Michael A. Gottlieb | CC BY-SA 3.0 |
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| Mar 28, 2018 at 18:06 | history | edited | Michael A. Gottlieb | CC BY-SA 3.0 |
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| Mar 28, 2018 at 15:50 | history | answered | Michael A. Gottlieb | CC BY-SA 3.0 |