Now, years later it turns out that the lectures were taped, and that after some wrangling, a famous guy (let's call him Bill) bought the rights to this material. Recently, Bill made the video available online. There's just one small problem. I promised myself I would never again install any software made by Bill and his friends on my computer. But, in order to view the lectures, he requires me to install the Silverlight plug-in for my browser (it's undoubtedly a DRM thing). In typical Bill fashion, if you go to the link for the videos, the download of the plug-in starts automatically. Long story short---I held my nose and did it. So, thanks for the lectures (link).
In Feynman's second lecture (the Relation of Mathematics and Physics) he talks about Kepler's Second Law: "A line joining a planet and the sun sweeps out equal areas during equal intervals of time." Feynman uses an argument based on vectors to show this. I didn't understand what Feynman said (and the book is just a transcription of the talk), but I googled around and found a post here. Unfortunately, the post is truncated at the critical point. The author was kind enough to write me with a detailed explanation. In the spirit of this blog as the place I post my homework for a self-taught course in anything I'm interested in, here is my version of his explanation of Feynman's explanation of Kepler's law.
Feynman uses the cross-product of two vectors.
In the diagram, the vectors A and B originate at the same point and the angle between them is θ. The cross-product A x B is a vector with magnitude = |A| |B| sin θ and its direction is perpendicular to A in the plane formed by A and B. The area of the triangle formed by A and B is one-half the magnitude of the cross-product. (I wish I knew how to produce the arrows over the labels for the vectors using html but I don't, sorry).
In the context of our problem, A is the vector from the sun to our planet at time-zero, and B is the vector at time dt. The magnitude of B is not equal to that of A, in general, because the orbit is an ellipse. So we can replace the labels by A = r and B = r + dr. The area being swept out (or its tiny component dA in a very short time dt) is proportional to the cross-product (neglect the factor of one-half):
or, as Feynman wrote using the dot notation:
Now, what we are interested in is the proposition that the rate-of-change of the area is constant, that the rate-of-change of the rate-of-change of the area is zero.
So, we need to differentiate again with respect to time, and, apparently, we can use the product rule from ordinary differentiation on this cross-product:
(As Feynman says: it's just playing with dots). The second term has the cross-product of a vector with itself, θ is zero and sin θ is zero so the whole thing equals zero.
The first term has the cross-product of r with the acceleration, the rate-of-change of the rate-of-change of r with time. That is equal to F/m.
But the thing is that the force acts along the radius, so now we have the cross-product of a vector with another vector that is turned by 180° from it. This product is also zero, and therefore Ä = 0.