Confirming Einstein’s Theory of General Relativity With Calculus, Part 2d: Hyperbolas and Polar Coordinates

Calculus, Differential Equations, Physics, Precalculus 2 Minutes

In this series, I’m discussing how ideas from calculus and precalculus (with a touch of differential equations) can predict the precession in Mercury’s orbit and thus confirm Einstein’s theory of general relativity. The origins of this series came from a class project that I assigned to my Differential Equations students maybe 20 years ago.

In a previous post, we showed that the polar equation

r = \displaystyle \frac{a}{1 + e \cos \theta}

is equivalent to the rectangular equation

\displaystyle \frac{\left(x + \displaystyle \frac{\alpha e}{1-e^2} \right)^2}{\displaystyle \frac{\alpha^2}{(1-e^2)^2}} + \frac{y^2}{\displaystyle \frac{\alpha^2}{1-e^2}} = 1

as long as e \ne 0. Furthermore, if 0 < e < 1, then this represents an ellipse with eccentricity e whose major axis lies on the x-axis, with one focus located at the origin.

While not directly related to our discussion of precession, it turns out that this equation represents a hyperbola if e > 1. Under this assumption, 1-e^2 < 0 and e^2-1>0, so let me rewrite the previous equation in terms of e^2-1:

\displaystyle \frac{\left(x - \displaystyle \frac{\alpha e}{e^2-1} \right)^2}{\displaystyle \frac{\alpha^2}{(e^2-1)^2}} - \frac{y^2}{\displaystyle \frac{\alpha^2}{e^2-1}} = 1

This matches the form of a left-right hyperbola

\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,

where the center of the hyperbola is located at

(h,k) = \displaystyle \left( \frac{\alpha e}{e^2-1} , 0 \right)

Also, for a hyperbola, the distance c from the center to the foci satisfies

c^2 = a^2 + b^2,

so that

c^2 = \displaystyle \frac{\alpha^2}{(e^2-1)^2} + \displaystyle \frac{\alpha^2}{e^2-1}

c^2 = \displaystyle \frac{\alpha^2 + \alpha^2 (e^2 - 1)}{(e^2-1)^2}

c^2 = \displaystyle \frac{\alpha^2 e^2}{(e^2-1)^2}

c = \displaystyle \frac{\alpha e}{e^2-1}

The two foci are located a distance c to the left of the right of the center. Since it happened to happen that c = h, this means that the origin is, once again, one of the foci of the hyperbola.

Furthermore, the eccentricity c/a of the hyperbola is easily computed as

\displaystyle \frac{c}{a} = \frac{ \displaystyle \frac{\alpha e}{e^2-1} }{ \displaystyle \frac{\alpha}{e^2-1}} = e,

so that, once again, the well-chosen parameter e is the eccentricity.

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Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

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