Confirming Einstein’s Theory of General Relativity With Calculus, Part 5b: Deriving Orbits under Newtonian Mechanics 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.

We previously 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 Newtonian mechanics (i.e., without general relativity) the motion of the planet follows the differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ ,

where $u = 1/r$ and $\alpha$ is a certain constant. We will also impose the initial condition that the planet is at perihelion (i.e., is closest to the sun), at a distance of $P$ , when $\theta = 0$ . This means that $u$ obtains its maximum value of $1/P$ when $\theta = 0$ . This leads to the two initial conditions

$u(0) = \displaystyle \frac{1}{P} \qquad \hbox{and} \qquad u'(0) = 0$ ;

the second equation arises since $u$ has a local extremum at $\theta = 0$ .

In the previous post, we confirmed that

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

solved this initial-value problem. However, the solution was unsatisfying because it gave no indication of where this guess might have come from. In this post, I suggest a series of questions that good calculus students could be asked that would hopefully lead them quite naturally to this solution.

Step 1. Let’s make the differential equation simpler, for now, by replacing the right-hand side with 0:

$u''(\theta) + u(\theta) = 0$ ,

or

$u''(\theta) = -u(\theta)$ .

Can you think of a function or two that, when you differentiate twice, you get the original function back, except with a minus sign in front?

Answer to Step 1. With a little thought, hopefully students can come up with the standard answers of $u(\theta) = \cos \theta$ and $u(\theta) = \sin \theta$ .

Step 2. Using these two answers, can you think of a third function that works?

Answer to Step 2. This is usually the step that students struggle with the most, as they usually try to think of something completely different that works. This won’t work, but that’s OK… we all learn from our failures. If they can’t figure out, I’ll give a big hint: “Try multiplying one of these two answers by something.” In time, they’ll see that answers like $u(\theta) = 2\cos \theta$ and $u(\theta) = 3\sin \theta$ work. Once that conceptual barrier is broken, they’ll usually produce the solutions $u(\theta) = a \cos \theta$ and $u(\theta) = b \sin \theta$ .

Step 3. Using these two answers, can you think of anything else that works?

Answer to Step 3. Again, students might struggle as they imagine something else that works. If this goes on for too long, I’ll give a big hint: “Try combining them.” Eventually, we hopefully get to the point that they’ll see that the linear combination $u(\theta) = a \cos \theta + b \sin \theta$ also solves the associated homogeneous differential equation.

Step 4. Let’s now switch back to the original differential equation $u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ . Let’s start simple: $u''(\theta) + u(\theta) = 5$ . Can you think of an easy function that’s a solution?

Answer to Step 4. This might take some experimentation, and students will probably try unnecessarily complicated guesses first. If this goes on for too long, I’ll give a big hint: “Try a constant.” Eventually, they hopefully determine that if $u(\theta) = 5$ is a constant function, then clearly $u'(\theta) = 0$ and $u''(\theta) = 0$ , so that $u''(\theta) + u(\theta) = 5$ .

Step 5. Let’s return to $u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ . Any guesses on an answer to this one?

Answer to Step 5. Hopefully, students quickly realize that the constant function $u(\theta) = \displaystyle \frac{1}{\alpha}$ works.

Step 6. Let’s review. We’ve shown that anything of the form $u(\theta) = a\cos \theta + b \sin \theta$ is a solution of $u''(\theta) + u(\theta) = 0$ . We’ve also shown that $u(\theta) = \displaystyle\frac{1}{\alpha}$ is a solution of $u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ . Can you think use these two answers to find something else that works?

Answer to Step 6. Hopefully, with the experience learned from Step 3, students will guess that $u(\theta) = a\cos \theta + b\sin \theta + \displaystyle \frac{1}{\alpha}$ will work.

Step 7. OK, that solves the differential equation. Any thoughts on how to find the values of $a$ and $b$ so that $u(0) = \displaystyle \frac{1}{P}$ and $u'(0) = 0$ ?