Nail-Tinted Glasses

Equal Areas in Equal Times

The first demonstration that we need is that Newton's Laws of Motion are equivalent to Kepler's observation that a planet sweeps equal areas in equal times. To be more precise, what we are demonstrating is that a body constrained by Newton's laws would necessarily sweep equal areas in equal times.

We begin by examining the simple case of an object that travels with a uniform velocity.

As the velocity is uniform, the object is covering the same distance in any given amount of time. This tells us that the base of the triangular area being swept is the same. Furthermore, because the velocity is constant, the base of each triangle is on the same line, which means that the height of these triangles is the same; the height is the altitude from the sun to the line, which is a distance equal to the closest approach of the line to the sun. So for any fixed interval Δt, the area swept is 1/2 h V Δt.

Now we complicate the problem slightly, by assuming that there can be a change in velocity, but that change in velocity is toward the sun. To demonstrate that the planet still sweeps equal areas in equal times, we compare the swept area to the triangle produced by uniform motion.

Any amount of velocity in the direction R, added to the original velocity V, for a time Δt, is going to result in a position on the line R', which is parallel to R and passes through the position that would be achieved by an undisturbed orbit. So all of these triangles have the same base R, and the same altitude h, and therefore the same area 1/2 Rh.

If you listen to the post lecture discussion, you'll hear one of the students ask about this point - and Feynman explains that he can't be done, because he hasn't used the inverse square law yet. What he means here is that we've just demonstrated that any velocity change in the correct direction will produce equal areas in equal times, so we still need to understand the physics that determine the magnitude of the change....

TOC: Feynman's Lost Lecture - Motion of Planets Around the Sun

I've been studying Feynman's Lost Lecture on and off for years now, trying to come to terms with the nagging feeling that something was missing.

So as I work thought the demonstration yet again, I'm trying to fill the gaps.

My first problem was actually understanding what he was claiming to demonstrate - from the language he uses, it sounds as though it is, in total, much more grand and complete than what he delivers, and the delivery is sufficiently compelling that upon reaching the end, you may not notice the lot.

The overall pattern of the demonstration works like this.

First, Feynman repeats Newton's demonstration that equal areas in equal times is equivalent to Newton's first two laws - undisturbed motion is constant, changes in motion are in the direction of the disturbance.

Next, Feynman does a quick hand wave of the equivalence of Kepler's observation of the period of an orbit with Newton's law of gravitation. Feynman is a bit quick through here, so you may miss the scope of the observation - he moves from the period of circles to the inverse square law, and then stops. We never return to demonstrate that the relationship holds in the general case (ellipses). It does, of course, but it's not relevant to Feynman's demonstration.

Now the true demonstration begins, as "cooked up" by Feynman. First comes the transformation of equal areas in equal times to equal changes in velocity in equal orbital angles. As the text points out, he's a bit loose with his triangles - but it's all right; presenting the rigor doesn't make things more right.

Equal changes in equal angles does produce a polygon, and Archimedes demonstrated long ago that regular regular polygon becomes a circle as we make the angles infinitesmal. Since the changes in velocity are always directed toward the sun, it must be represented by the center of this circle, but the velocities themselves radiate from an off-center point.

Feynman now rotates the velocity diagram to get everything to line up, and we have almost everything that we need - the velocity is given by the length of the arrow on our diagram, the direction is perpendicular to the arrow (because we turned the diagram around), the angle in the orbit must match the angle from the sun to the head of the arrow, and the distance must be....

Whoops - Feynman says "take the perpendicular bisector" without ever justifying the choice of that particular perpendicular. As it happens, you can demonstrate that it is necessary from equal areas in equal times.

At this point, where are we? We have three laws from Newton. From these we can demonstrate easily enough that Kepler's observation of equal areas in equal times, and as Feynman shows we can produce from these three laws a demonstration that closed orbits solutions must be ellipses. It remains to be shown that these solutions have the right period, and in particular that the period is a function of the length of the major axis, but not a function of the eccentricity.

Before then moving on to Rutherford scattering, Feynman tosses off two additional observations: that you can use the same kind of velocity diagram to produce parabolic orbits and hyperbolic orbits. Most unsatisfactory - if you place the second focus on the velocity circle, then the perpendicular bisectors we had been using now all go through the sun. That Can't Be Right [tm]. After all, the magic of a parabolic mirror is to but the light source at the focus - not to embed it within the mirror itself.

Use Google, Tintin!

If they decide to relocate Secret of the Unicorn to the modern era, it'll be a five minute short.