I've been working my way through Feynman's Lost Lecture (planetary motion). I'm not quite finished with it, but along the way I've picked up on a number of points that weren't immediately clear to me from the audio portion and the written companion.
The first portion of the lecture details the properties of an ellipse. One bit of unfinished business is to track down a copy of Principia, to at least familiarize myself with the obscure properties of conic sections that Feynman alludes to.
Early in the lecture, Feynman takes time to demonstrate that the reflective property of an ellipse is a consequence of the sum of the distances definition. A point here, which really ought to be obvious but wasn't, is that for any point p, there is always a line that makes the necessary angles to f and f' You can construct it by extending f' p, then bisecting the angle formed by this line and fp. What Feynman proves is that this line is necessarily tangent to the curve, as it intersects the curve at P and nowhere else.
What Feynman actually states is that the two definitions are equivalent - that you can from the reflection definition of an ellipse and prove the sum of distances property. I beat my head against this for a while, but thankfully Will Self (University of Montana) found an enlighting way of pointing out that I was being an idiot.
Assume the curve C has the property that at every point P, the tangent at P forms equal angles with f and f'. If we assume there is a point Q on C, such that d(fQ) + d(f'Q) > d(fP) + d(f'P), a contradiction arrises as follows: construct E, being a curve such that, for every point e on the curve, d(fe) + d(f'e) = 1/2 (d(fQ) + d(f'Q) > d(fP) + d(f'P)). For curve C to go from P to Q, it must cross E at some point X. However, the tangent of C at X must still have the reflection property, which means that it is the same line as the tangent of E at X (as demonstrated by Feynman's construction) if C is parallel to E at X, it certainly can't cross E at X. Rah.
Stumbling block number three was "why does equal areas imply that the change in velocity is toward the sun?". Consider an arbitrary change in direction. We can break that vector down into the sum of two vectors - one toward the sun, and the other perpendicular to the first. Now, the area of the triangle is base times amplitude. The base is independent of the change in velocity. The change in velocity toward the sun is parallel to the base, and thus makes no change in the area. However, the change in velocity perpendicular to the sun is in the direction of the amplitude. If there is any component here at all, the amplitude will change, which will modify the area of the triangle (the radial component, having no effect on base or amplitude, cannot possible compensate). Therefore, the change in velocity must be directed toward the sun.
May 25, 2003 1:24 PM
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