Confirming Einstein’s Theory of General Relativity With Calculus, Part 6b: Checking Solution of New Differential Equation with Calculus

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 the last post, we showed that if the motion of a planet around the Sun is expressed in polar coordinates $(r,\theta)$ , with the Sun at the origin, then under general relativity the motion of the planet follows the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

where $u = \displaystyle \frac{1}{r}$ , $\displaystyle \frac{1}{\alpha} = \frac{GMm^2}{\ell^2}$ , $\delta = \displaystyle \frac{3GM}{c^2}$ , $G$ is the gravitational constant of the universe, $m$ is the mass of the planet, $M$ is the mass of the Sun, $\ell$ is the constant angular momentum of the planet, $c$ is the speed of light, and $P$ is the smallest distance of the planet from the Sun during its orbit (i.e., at perihelion).

I won’t sugar-coat it; the solution is a big mess:

$u(\theta) = \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\epsilon \delta}{\alpha^2} \theta \sin \theta - \frac{\epsilon^2 \delta}{6\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta$ .

That said, it is an elementary, if complicated, exercise in calculus to confirm that this satisfies all three equations above. We’ll start with the second one:

$u(0) = \displaystyle \frac{1 + \epsilon \cos 0}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\epsilon \delta}{\alpha^2} \cdot 0 \sin 0 - \frac{\epsilon^2 \delta}{6\alpha^2} \cos 0 - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos 0$

$= \displaystyle \frac{1 + \epsilon}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} - \frac{\epsilon^2 \delta}{6\alpha^2} - \frac{\delta(3+\epsilon^2)}{3\alpha^2}$

$= \displaystyle \frac{ 6\alpha(1+\epsilon) + 6\delta + 3 \delta \epsilon^2 - \delta \epsilon^2 - 2\delta (3+\epsilon^2)}{6\alpha^2}$

$= \displaystyle \frac{ 6\alpha(1+\epsilon) + 6\delta + 3 \delta \epsilon^2 - \delta \epsilon^2 - 6\delta - 2\delta \epsilon^2}{6\alpha^2}$

$= \displaystyle \frac{ 6\alpha(1+\epsilon)}{6\alpha^2}$

$= \displaystyle \frac{ 1+\epsilon}{\alpha}$

$= \displaystyle \frac{1}{P}$ ,

where in the last step we used the equation $P = \displaystyle \frac{\alpha}{1 + \epsilon}$ that was obtained earlier in this series.

Next, to check the initial condition $u'(0) = 0$ , we differentiate:

$u'(\theta) = \displaystyle -\frac{\epsilon \sin \theta}{\alpha} + \frac{\epsilon \delta}{\alpha^2} (\sin \theta + \theta \cos \theta) + \frac{\epsilon^2 \delta}{3\alpha^2} \sin 2\theta + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \sin\theta$

$u'(0) = \displaystyle -\frac{\epsilon \sin 0}{\alpha} + \frac{\epsilon \delta}{\alpha^2} (\sin 0 + 0 \cdot \cos 0) + \frac{\epsilon^2 \delta}{3\alpha^2} \sin 0 + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \sin 0 = 0$ .

Finally, to check the differential equation itself, we compute the second derivative:

$u''(\theta) = \displaystyle -\frac{\epsilon \cos \theta}{\alpha} + \frac{\epsilon \delta}{\alpha^2} (2 \cos \theta - \theta \sin \theta) + \frac{2\epsilon^2 \delta}{3\alpha^2} \cos 2\theta + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos\theta$ .

Adding $u''(\theta)$ and $u(\theta)$ , we find

$u''(\theta) + u(\theta) = \displaystyle \frac{1 + \epsilon \cos \theta - \epsilon \cos \theta}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\epsilon \delta}{\alpha^2} (\theta \sin \theta + 2 \cos \theta - \theta \sin \theta)$

$\displaystyle - \frac{\epsilon^2 \delta}{6\alpha^2} \cos 2\theta + \frac{2\epsilon^2 \delta}{3\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos\theta$ ,

which simplifies considerably:

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{2 \epsilon \delta}{\alpha^2} \cos \theta + \frac{\epsilon^2 \delta}{2\alpha^2} \cos 2\theta$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} \left( 1 + \frac{\epsilon^2}{2} + 2 \epsilon \cos \theta + \frac{\epsilon^2}{2} \cos 2\theta \right)$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} \left( 1 + 2 \epsilon \cos \theta + \epsilon^2 \frac{1+\cos 2\theta}{2} \right)$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} \left( 1 + 2 \epsilon \cos \theta + \epsilon^2 \cos^2 \theta \right)$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} (1 + \epsilon \cos \theta)^2$

$= \displaystyle \frac{1}{\alpha} + \delta \left( \frac{1+\epsilon \cos \theta}{\alpha} \right)^2$ ,

where we used the power-reduction trigonometric identity

$\cos^2 \theta = \displaystyle \frac{1 + \cos 2\theta}{2}$

on the second-to-last step.

While we have verified the proposed solution of the initial-value problem, and the steps for doing so lie completely within the grasp of a good calculus student, I’ll be the first to say that this solution is somewhat unsatisfying: the solution appeared seemingly out of thin air, and we just checked to see if this mysterious solution actually works. In the next few posts, I’ll discuss how this solution can be derived using standard techniques from first-semester differential equations.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6b: Checking Solution of New Differential Equation with Calculus

Published by John Quintanilla

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