Confirming Einstein’s Theory of General Relativity With Calculus, Part 5a: Confirming 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 next few posts, we’ll discuss the solution of this initial-value problem. Today’s post would be appropriate for calculus students, which is confirming that

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

solves this initial-value problem, where $\epsilon = \displaystyle \frac{\alpha-P}{P}$ . Since $r$ is the reciprocal of $u$ , we infer that

$r = \displaystyle \frac{\alpha}{1 + \epsilon \cos \theta}$ .

As we’ve already seen in this series, this means that the orbit of the planet is a conic section — either a circle, ellipse, parabola, or hyperbola. Since the orbit of a planet is stable and $\epsilon = 0$ is extremely unlikely, this means that the planet orbits the Sun in an ellipse, with the Sun at one focus of the ellipse.

So, for a calculus student to verify that planets move in ellipses, one must check that

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

is a solution of the initial-value problem

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

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

$u'(0) = 0$ .

The second line is easy to check:

$u(0) = \displaystyle \frac{1 + \epsilon \cos 0}{\alpha}$

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

$= \displaystyle \frac{1 + \displaystyle \frac{\alpha-P}{P}}{\alpha}$

$= \displaystyle \frac{1}{\alpha} \frac{P + \alpha - P}{P}$

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

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

The third line is also easy to check:

$u'(\theta) = \displaystyle \frac{-\epsilon \sin \theta}{\alpha}$

$u'(0) = \displaystyle \frac{-\epsilon \sin 0}{\alpha} = 0$ .

To check the first line, we first find $u''(\theta)$ :

$u''(\theta) = \displaystyle \frac{-\epsilon \cos \theta}{\alpha}$ ,

so that

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

thus confirming that $u(\theta) = \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha}$ solves the initial-value problem.

While the above calculations are well within the grasp of a good Calculus I student, I’ll be the first to admit that this solution is less than satisfying. We just mysteriously proposed a solution, seemingly out of thin air, and confirmed that it worked. In the next post, I’ll proposed a way that calculus students can be led to guess this solution. Then, we talk about finding the solution of this nonhomogeneous initial-value problem using standard techniques from differential equations.

Definition of e

Source: https://xkcd.com/2768

Confirming Einstein’s Theory of General Relativity With Calculus, Part 4c: Newton’s Second Law and Newton’s Law of Gravitation

In this post, following from the previous two posts, we will show 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. Deriving this governing differential equation will require some principles from physics. If you’d rather skip the physics and get to the mathematics, we’ll get to solving this differential equations in the next post.

From Newton’s second law, the gravitational force on the planet as it orbits the Sun satisfies

${\bf F} = m{\bf a}$ ,

where the force ${\bf F}$ and the acceleration ${\bf a}$ are vectors. When written in polar coordinates, this becomes

${\bf F} = m \displaystyle \left[ \frac{d^2r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 \right] {\bf u}_r + m \left(r \frac{d^2 \theta}{d t^2} + 2 \frac{dr}{dt} \frac{d\theta}{dt} \right) {\bf u}_\theta$ ,

where ${\bf u}_r$ is a unit vector pointing away from the origin and ${\bf u}_\theta$ is a unit vector perpendicular to ${\bf u}_r$ that points in the direction of increasing $\theta$ .

Furthermore, from Newton’s Law of Gravitation, if the Sun is located at the origin, then the gravitational force on the planet is

${\bf F} = \displaystyle -\frac{GMm}{r^2} {\bf u}_r$ ,

where $M$ is the mass of the sun, $m$ is the mass of the planet, and $G$ is the gravitational constant of the universe (which is a constant, no matter what Q from Star Trek: The Next Generation says).

Since these are the same force, the ${\bf u}_r$ components must be the same. (Also, the ${\bf u}_\theta$ component must be zero, but we won’t need to use that fact.) Therefore,

$m \displaystyle \left[ \frac{d^2r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 \right] = \displaystyle -\frac{GMm}{r^2}$ ,

$\displaystyle \frac{d^2r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 = \displaystyle -\frac{GM}{r^2}$ .

In a previous post, we showed that

$\displaystyle \frac{d\theta}{dt} = \frac{\ell}{mr^2}$ ,

where $\ell$ is a constant, and

$\displaystyle \frac{d^2r}{dt^2} = - \frac{\ell^2}{m^2 r^2} \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right)$ .

Substituting, we find

$\displaystyle - \frac{\ell^2}{m^2 r^2} \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right) - r \left( \frac{\ell}{mr^2} \right)^2 = \displaystyle -\frac{GM}{r^2}$

$\displaystyle \frac{\ell^2}{m^2 r^2} \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right) + r \left( \frac{\ell^2}{m^2 r^4} \right) = \displaystyle \frac{GM}{r^2}$

$\displaystyle \frac{\ell^2}{m^2 r^2} \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right) +\frac{\ell^2}{m^2 r^3} = \displaystyle \frac{GM}{r^2}$

$\displaystyle \frac{\ell^2}{m^2 r^2} \left[ \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right) + \frac{1}{r} \right] = \displaystyle \frac{GM}{r^2}$

$\displaystyle \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right) + \frac{1}{r} = \displaystyle \frac{GM}{r^2} \cdot \frac{m^2 r^2}{\ell^2}$

$\displaystyle \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right) + \frac{1}{r} = \displaystyle \frac{GMm^2}{\ell^2}$ .

So, substituting $u = 1/r$ and $\alpha = \displaystyle \frac{\ell^2}{GMm^2}$ , we finally obtain the governing equation

$\displaystyle \frac{d^2 u}{d\theta^2} + u = \displaystyle \frac{1}{\alpha}$ .

This is the governing differential equation of planetary motion under Newtonian mechanics. For now, it’s not obvious why we chose $\displaystyle \frac{1}{\alpha}$ as the constant on the right-hand side instead of just $\alpha$ , but the reason for this choice will become apparent in future posts.

In the next few posts, we use differential equations (or, if you’d prefer, just calculus) to show that Newtonian mechanics predicts that planets orbit the Sun in ellipses.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 4b: Acceleration in Polar Coordinates

In this part of the series, we will show 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

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

Part of the derivation of this governing differential equation will involve Newton’s Second Law

${\bf F} = m {\bf a}$ ,

where $m$ is the mass of the planet and the force ${\bf F}$ and the acceleration $a$ are vectors. In usual rectangular coordinates, the acceleration vector would be expressed as

${\bf a} = x''(t) {\bf i} + y''(t) {\bf j}$ ,

where the components of the acceleration in the $x-$ and $y-$ directors are $x''(t)$ and $y''(t)$ , and the unit vectors ${\bf i}$ and ${\bf j}$ are perpendicular, pointing in the positive $x$ and positive $y$ directions.

Unfortunately, our problem involves polar coordinates, and rewriting the acceleration vector in polar coordinates, instead of rectangular coordinates, is going to take some work.

Suppose that the position of the planet is $(r,\theta)$ in polar coordinates, so that the position in rectangular coordinates is ${\bf r} = (r\cos \theta, r \sin \theta)$ . This may be rewritten as

${\bf r} = r \cos \theta {\bf i} + r \sin \theta {\bf j} = r ( \cos \theta {\bf i} + \sin \theta {\bf j}) = r {\bf u}_r$ ,

where

${\bf u}_r = \cos \theta {\bf i} + \sin \theta {\bf j}$

is a unit vector that points away from the origin. We see that this is a unit vector since

$\parallel {\bf u}_r \parallel = {\bf u}_r \cdot {\bf u}_r = \cos^2 \theta + \sin^2 \theta =1$ .

We also define

${\bf u}_\theta = -\sin \theta {\bf i} + \cos \theta {\bf j}$

to be a unit vector that is perpendicular to ${\bf u}_r$ ; it turns out that ${\bf u}_\theta$ points in the direction of increasing $\theta$ . To see that ${\bf u}_r$ and ${\bf u}_\theta$ are perpendicular, we observe

${\bf u}_r \cdot {\bf u}_\theta = -\sin \theta \cos \theta + \sin \theta \cos \theta = 0$ .

Computing the velocity and acceleration vectors in polar coordinates will have a twist that’s not experienced with rectangular coordinates since both ${\bf u}_r$ and ${\bf u}_\theta$ are functions of $\theta$ . Indeed, we have

$\displaystyle \frac{d{\bf u}_r}{d\theta} = \frac{d \cos \theta}{d\theta} {\bf i} + \frac{d\sin \theta}{d\theta} {\bf j} = -\sin \theta {\bf i} + \cos \theta {\bf j} = {\bf u}_\theta$ .

Furthermore,

$\displaystyle \frac{d{\bf u}_\theta}{d\theta} = -\frac{d \sin \theta}{d\theta} {\bf i} + \frac{d\cos \theta}{d\theta} {\bf j} = -\cos \theta {\bf i} - \sin \theta {\bf j} = -{\bf u}_r$ .

These two equations will be needed in the derivation below.

We are now in position to express the velocity and acceleration of the orbiting planet in polar coordinates. Clearly, the position of the planet is $r {\bf u}_r$ , or a distance $r$ from the origin in the direction of ${\bf u}_r$ . Therefore, by the Product Rule, the velocity of the planet is

${\bf v} = \displaystyle \frac{d}{dt} (r {\bf u}_r) = \displaystyle \frac{dr}{dt} {\bf u}_r + r \frac{d {\bf u}_r}{dt}$

We now apply the Chain Rule to the second term:

${\bf v} = \displaystyle \frac{dr}{dt} {\bf u}_r + r \frac{d {\bf u}_r}{d\theta} \frac{d\theta}{dt}$

$= \displaystyle \frac{dr}{dt} {\bf u}_r + r \frac{d\theta}{dt} {\bf u}_\theta$ .

Differentiating a second time with respect to time, and again using the Chain Rule, we find

${\bf a} = \displaystyle \frac{d {\bf v}}{dt} = \displaystyle \frac{d^2r}{dt^2} {\bf u}_r + \frac{dr}{dt} \frac{d{\bf u}_r}{dt} + \frac{dr}{dt} \frac{d\theta}{dt} {\bf u}_\theta + r \frac{d^2\theta}{dt^2} {\bf u}_\theta + r \frac{d\theta}{dt} \frac{d{\bf u}_\theta}{dt}$

$= \displaystyle \frac{d^2r}{dt^2} {\bf u}_r + \frac{dr}{dt} \frac{d{\bf u}_r}{d\theta} \frac{d\theta}{dt} + \frac{dr}{dt} \frac{d\theta}{dt} {\bf u}_\theta + r \frac{d^2\theta}{dt^2} {\bf u}_\theta + r \frac{d\theta}{dt} \frac{d{\bf u}_\theta}{d\theta} \frac{d\theta}{dt}$

$= \displaystyle \frac{d^2r}{dt^2} {\bf u}_r + \frac{dr}{dt} \frac{d\theta}{dt} {\bf u}_\theta + \frac{dr}{dt} \frac{d\theta}{dt} {\bf u}_\theta + r \frac{d^2\theta}{dt^2} {\bf u}_\theta - r \left(\frac{d\theta}{dt} \right)^2 {\bf u}_r$

$= \displaystyle \left[ \frac{d^2r}{dt^2} - r \left(\frac{d\theta}{dt} \right)^2 \right] {\bf u}_r + \left[ 2\frac{dr}{dt} \frac{d\theta}{dt} + r \frac{d^2\theta}{dt^2} \right] {\bf u}_\theta$ .

This will be needed in the next post, when we use both Newton’s Second Law and Newton’s Law of Gravitation, expressed in polar coordinates.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 4a: Angular Momentum

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

One principle from physics that we’ll need is the Law of Conservation of Angular Momentum. Mathematically, this is expressed by

$mr^2 \displaystyle \frac{d\theta}{dt} = \ell$ ,

where $\ell$ is a constant. Of course, this can be written as

$\displaystyle \frac{d\theta}{dt} = \displaystyle \frac{\ell}{mr^2}$ ;

this will be used a couple times in the derivation below.

As we’ll soon see, we will need to express the second derivative $\displaystyle \frac{d^2 r}{d t^2}$ in a form that depends only on $\theta$ . To do this, we use the Chain Rule to obtain

$r' = \displaystyle \frac{dr}{dt}$

$= \displaystyle \frac{dr}{d\theta} \cdot \frac{d\theta}{dt}$

$= \displaystyle \frac{\ell}{mr^2} \frac{dr}{d\theta}$

$= \displaystyle - \frac{\ell}{m} \frac{d}{d\theta} \left( \frac{1}{r} \right)$ .

This last step used the Chain Rule in reverse:

$\displaystyle \frac{d}{d\theta} \left( \frac{1}{r} \right) = \frac{d}{dr} \left( \frac{1}{r} \right) \cdot \frac{dr}{dt} = -\frac{1}{r^2} \cdot \frac{dr}{dt}$ .

To examine the second derivative $\displaystyle \frac{d^2 r}{d t^2}$ , we again use the Chain Rule:

$\displaystyle \frac{d^2 r}{d t^2} = \displaystyle \frac{dr'}{dt}$

$= \displaystyle \frac{dr'}{d\theta} \cdot \frac{d\theta}{dt}$

$= \displaystyle \frac{\ell}{mr^2} \frac{dr'}{d\theta}$

$= \displaystyle \frac{\ell}{mr^2} \frac{d}{d\theta} \left[ \frac{dr}{dt} \right]$

$= \displaystyle \frac{\ell}{mr^2} \frac{d}{d\theta} \left[ - \frac{\ell}{m} \frac{d}{d\theta} \left( \frac{1}{r} \right) \right]$

$= \displaystyle - \frac{\ell^2}{m^2r^2} \frac{d}{d\theta} \left[ \frac{d}{d\theta} \left( \frac{1}{r} \right) \right]$

$= \displaystyle - \frac{\ell^2}{m^2r^2} \frac{d^2}{d\theta^2} \left( \frac{1}{r} \right)$ .

While far from obvious now, this will be needed when we rewrite Newton’s Second Law in polar coordinates.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 3: Method of Successive Approximations

One technique that will be necessary for this confirmation is the method of successive approximations. This will be needed in the context of a differential equation; however, we can illustrate the concept by finding the roots of a polynomial. Consider the quadratic equation

$x^2 - x - 1 = 0$ .

(Naturally, we can solve for $x$ using the quadratic formula; more on that later.) To apply the method of successive approximation, we will rewrite this so that $x$ appears on the left side and some function of $x$ appears on the right side. I will choose

$x^2 = x + 1$ , or

$x = 1 + \displaystyle \frac{1}{x}$ .

Here’s the idea of the method of successive approximations to obtain a recursively defined sequence that (hopefully) convergence to a solution of this equation:

Start with an initial guess $x_0$ .
Plug $x_0$ into the right-hand side to get a new guess, $x_1$ .
Plug $x_1$ into the right-hand side to get a new guess, $x_2$ .
And repeat.

For example, suppose that we choose $x_0 = 1$ . Then

$x_1 = 1 + \displaystyle \frac{1}{x_0} = 1 + \displaystyle \frac{1}{1} = 2$

$x_2 = 1 + \displaystyle \frac{1}{x_1} = 1 + \displaystyle \frac{1}{2} = \displaystyle \frac{3}{2} = 1.5$

$x_3 = 1 + \displaystyle \frac{1}{x_2} = 1 + \displaystyle \frac{1}{3/2} = \displaystyle \frac{5}{3} \approx 1.667$

$x_4 = 1 + \displaystyle \frac{1}{x_3} = 1 + \displaystyle \frac{1}{5/3} = \displaystyle \frac{8}{5} = 1.6$

$x_5 = 1 + \displaystyle \frac{1}{x_4} = 1 + \displaystyle \frac{1}{8/5} = \displaystyle \frac{13}{8} = 1.625$

$x_6 = 1 + \displaystyle \frac{1}{x_5} = 1 + \displaystyle \frac{1}{13/8} = \displaystyle \frac{21}{13} \approx 1.615$

$x_7 = 1 + \displaystyle \frac{1}{x_6} = 1 + \displaystyle \frac{1}{21/13} = \displaystyle \frac{34}{21} \approx 1.619$

This sequence can be computed by entering $1$ into a calculator, then entering $1 + 1 \div \hbox{Ans}$ , and then repeatedly hitting the $=$ button.

We see that the sequence appears to be converging to something, and that something is a root of the equation $x^2 - x - 1 = 0$ , which we now find via the quadratic formula:

$x = \displaystyle \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$ .

So it looks like the above sequence is converging to the positive root $(1 + \sqrt{5})/2 \approx 1.618$ .

(Parenthetically, you might notice that the Fibonacci sequence appears in the numerators and denominators of this sequence. As you might guess, that’s not a coincidence.)

Like most numerical techniques, this method doesn’t always work like we think it would. Another solution is the negative root $(1 - \sqrt{5})/2 \approx -0.618$ . Unfortunately, if we start with a guess near this root, like $x_0 = -0.62$ , the sequence unexpectedly diverges from $-0.618\dots$ but eventually converges to the positive root $1.618\dots$ :

$x_1 = 1 + \displaystyle \frac{1}{x_0} = 1 + \displaystyle \frac{1}{-0.62} = -0.6129\dots$

$x_2 = 1 + \displaystyle \frac{1}{x_1} = 1 + \displaystyle \frac{1}{-0.6129\dots} = -0.6315\dots$

$x_3 = 1 + \displaystyle \frac{1}{x_2} = 1 + \displaystyle \frac{1}{-0.6315\dots} = -0.5833\dots$

$x_4 = 1 + \displaystyle \frac{1}{x_3} = 1 + \displaystyle \frac{1}{-0.5833\dots} = -0.7142\dots$

$x_5 = 1 + \displaystyle \frac{1}{x_4} = 1 + \displaystyle \frac{1}{-0.5833\dots} = -0.4$

$x_6 = 1 + \displaystyle \frac{1}{x_5} = 1 + \displaystyle \frac{1}{-0.4\dots} = -1.5$

$x_7 = 1 + \displaystyle \frac{1}{x_6} = 1 + \displaystyle \frac{1}{-1.5\dots} = 0.3333\dots$

$x_8 = 1 + \displaystyle \frac{1}{x_7} = 1 + \displaystyle \frac{1}{0.3333\dots} = 4$

$x_9 = 1 + \displaystyle \frac{1}{x_8} = 1 + \displaystyle \frac{1}{4} = 1.25$

$x_{10} = 1 + \displaystyle \frac{1}{x_9} = 1 + \displaystyle \frac{1}{1.25} = 1.8$

$x_{11} = 1 + \displaystyle \frac{1}{x_{10}} = 1 + \displaystyle \frac{1}{1.8} = 1.555\dots$

$x_{12} = 1 + \displaystyle \frac{1}{x_{11}} = 1 + \displaystyle \frac{1}{1.555\dots} = 1.6428\dots$

$x_{13} = 1 + \displaystyle \frac{1}{x_{12}} = 1 + \displaystyle \frac{1}{1.6428\dots} = 1.6086\dots$

$x_{14} = 1 + \displaystyle \frac{1}{x_{13}} = 1 + \displaystyle \frac{1}{1.6086\dots} = 1.6216\dots$

$x_{15} = 1 + \displaystyle \frac{1}{x_{14}} = 1 + \displaystyle \frac{1}{1.6216\dots} = 1.6166\dots$

$x_{16} = 1 + \displaystyle \frac{1}{x_{15}} = 1 + \displaystyle \frac{1}{1.6216\dots} = 1.6185\dots$

I should note that the method of successive approximations generally converges at a slower pace than Newton’s method. However, this method will be good enough when we use it to predict the precession in Mercury’s orbit.

A Postcard from Spokane

A brief aside from the current series on general relativity — and the mysterious 43 seconds of arc per century in Mercury’s orbit — that turned into further discussion about angle measurement.

A few months ago, I received this clever postcard from someone visiting Spokane, Washington. The sender clearly knew the recipient (me) well: rather than sending me a postcard showing the jaw-dropping beauty of the Spokane area, I was impressed with the mathematical precision given for Spokane’s location.

I started wondering about exactly how precisely the postcard was measuring the location of Spokane — was it the location of City Hall or some other important landmark? — and I went to Google Maps to find out. (For what it’s worth, xkcd had a comic about this some time ago.)

And then it finally hit me, after far longer than it should have taken, that the postcard is utterly nonsensical.

We would never say that someone’s height is 4 feet, 20 inches. There are 12 inches in a foot, and so we would instead say that the height is 5 feet, 8 inches.

Likewise, when specifying an angle with minutes and seconds, there are (just like with ordinary time) 60 seconds in a minute and 60 minutes in a degree (so that there are 3600 seconds in a degree). Therefore, specifying an angle with 67′ or 66″, as in the postcard, makes absolutely no sense.

Furthermore, if converted into standard notation, we obtain a location of $48^{\circ} 7' 36''$ north, $117^{\circ} 42' 6''$ west, which is about 40 miles NNW of Spokane. (Images made by https://www.gps-coordinates.net/). Note on the conversion into decimal:

$47 + \displaystyle \frac{67}{60} + \displaystyle \frac{36}{3600} = 48 + \displaystyle \frac{7}{60} + \displaystyle \frac{36}{3600} = 48.12666\dots$

and

$117 + \displaystyle \frac{41}{60} + \displaystyle \frac{66}{3600} = 117 + \displaystyle \frac{42}{60} + \displaystyle \frac{6}{3600} = 117.701666\dots$

It’s a shame that the designer of the postcard made this error, as I genuinely thought this was a clever and aesthetically pleasing design idea for a postcard.

While I’m not sure how this mistake happened, my best guess is that the designer used the location of $47.6736^\circ$ north, $117.4166^\circ$ west — which is indeed in Spokane — and then misconverted from decimal notation to minutes and seconds.

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

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.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 2c: Circles, Parabolas, and Polar Coordinates

In the previous post, we showed that the polar equation

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

converts to

$x^2 (1-e^2) + 2 \alpha e x + y^2 = \alpha^2$

in rectangular coordinates. Furthermore, if $0 < e < 1$ , then this represents an ellipse with eccentricity $e$ whose semi-major axis lies along the $x-$ axis with one focus at the origin.

It turns out that, for different non-negative values of $e$ , the same polar equation represents different conic sections. These are not particularly relevant for our study of precession, but I’m including this anyway in this series as a small tangential discussion.

Let’s take a look at the easy case of $e = 0$ . With this substitution, the equation in rectangular coordinates simplifies to

$x^2 + y^2 = \alpha^2$ .

Of course, this is the equation of a circle that is centered at the origin with radius $\alpha$ .

The other easy case is $e = 1$ , so that $1-e^2 = 0$ . Then the equation in rectangular coordinates simplifies to

$2 \alpha x + y^2 = \alpha^2$

$y^2 = -2\alpha x + \alpha^2$

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

$y^2 = -4 \cdot \displaystyle \frac{\alpha}{2} \left( x - \frac{\alpha}{2} \right)$

This matches the form of a parabola that opens to the left with a horizontal axis of symmetry:

$(y-k)^2 = -4 p (x-h)$ .

In this case, the vertex of the parabola is located at

$(h,k) = \displaystyle \left( \frac{\alpha}{2} , 0 \right)$ ,

while the focus of the parabola is located a distance $p = \displaystyle \frac{\alpha}{2}$ to the left of the vertex. In other words, the origin is the focus of the parabola. (For what it’s worth, the directrix of the parabola would be the vertical line $y = \alpha$ , located $p$ to the right of the vertex.)

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

As part of our derivation, we’ll need to use the fact that, in polar coordinates, the graph of

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

turns out to be an ellipse if $0 < e < 1$ , with the origin at one focus.

We now prove this. Clearing the denominator, we obtain

$r + re \cos \theta = \alpha$ .

Switching to rectangular coordinates, this becomes

$\sqrt{x^2 + y^2} + e x = \alpha$

$\sqrt{x^2 + y^2} = \alpha - ex$

$x^2 + y^2 = (\alpha - ex)^2$

$x^2 + y^2 = \alpha^2 - 2\alpha ex + e^2 x^2$

$x^2 (1-e^2) + 2 \alpha e x + y^2 = \alpha^2$

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

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

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

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

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

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

Since we assumed that $0 < e < 1$ , we have $0 < 1 - e^2 < 1$ so that

$\displaystyle \frac{\alpha^2}{(1-e^2)^2} > \displaystyle \frac{\alpha^2}{1-e^2}$ .

Therefore, this matches the usual form of an ellipse in rectangular coordinates

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

where the center of the ellipse is located at

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

the semi-major axis is horizontal with length

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

and the semi-minor axis is vertical with length

$b = \displaystyle \frac{\alpha}{\sqrt{1-e^2}}$ .

Furthermore, the distance $c$ of the foci from the center of the ellipse satisfies the equation

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

so that

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

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

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

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

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

From this, we derive two nice properties of the ellipse. First, looking back on previous work, we see that $c = -h$ . Therefore, since the foci of the ellipse are distance $c$ away from the center along the major axis, we conclude that one focus of the ellipse is located at $(-h+c,0)$ , or $(0,0)$ . That is, the origin is one focus of the ellipse. (For the little it’s worth, the other focus is located at $(-2h,0)$ .

Second, the eccentricity of the ellipse is defined to be the ratio $c/a$ . This is now easily computed:

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

In other words, the letter $e$ was well-chosen to represent the eccentricity of the ellipse.

For what it’s worth, here’s an alternate derivation of the formulas for $a$ and $b$ . For this ellipse, the planet’s closest approach to the Sun occurs at $\theta = 0$ :