Friday, August 28, 2009

Feynman and Kepler

When I was in college I bought a book containing lectures that Richard Feynman gave at Cornell in 1964 (the Messenger series). The book is called The Character of Physical Law. Quite simply, it was and is wonderful.

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):

dA = r x (r + dr)

or, as Feynman wrote using the dot notation:

.       .
A = r x r

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.

A = d2A/dt2 = 0

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:

..            .
A = d/dt (r x r)

.. . .
= r x r + r x r

(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.