Greetings from The Amaz!ng Meeting!

This is the happiest I have yet been, here in the city I am learning to hate, the city with nothing to offer me and nothing to enjoy — except Joshua, Rebecca, PZ, Phil and nine hundred other friends. Yes, Gentle Reader, you know me well enough to guess that my most truly recreational experience here in Las Vegas has been sleeping late, eating a Toblerone in bed, tidying up a blag post from my drafts pile, and missing Michael Shermer’s talk. Joshua just “texted” me, as the kids say, with the following message: “Ok. Shermer needs his Powerpoint privileges revoked. How many text-dump slides is he going to use?”

Call me psychic, Gentle Reader, or at least give me a little credit for remembering what I read. . . .

Anyway, because I’m here to have fun, it’s time to talk physics. With equations.

When we studied the hydrogen atom, we found that an interaction potential which fell off inversely with distance was shape-invariant, implying all sorts of nice symmetry properties of the hydrogen atom’s state space. The classical analogue of this situation would be two objects interacting via an inverse-square force (remember that force is given by the derivative of the potential). Having recently taken a rather madcap tour of the history of classical mechanics, we can probe a little more deeply and investigate one item in more technical detail. Today’s subject will be defining and appreciating the Laplace-Runge-Lenz vector, which as we said earlier was not discovered by Laplace, Runge or Lenz. After finding out that this vector quantity is conserved, we’ll take a quick look at equations which define ellipses and then show that an inverse-square law of gravity can yield elliptical orbits. If any portion of this post is in error, please return the unused portion for a full refund.

DEFINING THE LRL VECTOR

We consider a central-force problem, in which Newton’s second law can be written as

$\displaystyle \dot{\vec{p}} = f(r) \hat{r} = f(r) \frac{\vec{r}}{r}.$

This equation means that the force depends only on the distance $r$ and is always directed along the line between the center of force and the moving object. Rotational symmetry tells us that the angular momentum $\vec{L}$ is conserved. (That’s the ghost of Emmy Noether whispering in our ears.) If we take the cross product of $\dot{\vec{p}}$ with $\vec{L}$, we find that

$\displaystyle \dot{\vec{p}} \times \vec{L} = \frac{mf(r)}{r} \left[\vec{r} \times (\vec{r} \times \dot{\vec{r}})\right],$

which the standard triple-cross-product identity lets us rewrite as

$\displaystyle \dot{\vec{p}} \times \vec{L} = \frac{mf(r)}{r} \left[\vec{r}(\vec{r}\cdot\dot{\vec{r}}) – r^2 \dot{\vec{r}}\right].$

Because the component of the velocity in the radial direction is just $\dot{r}$, we can write

$\displaystyle \vec{r}\cdot\dot{\vec{r}} = \frac{1}{2} \partial_t (\vec{r}\cdot\vec{r}) = r\dot{r},$

which lets us shuffle our expression for $\dot{\vec{p}}\times\vec{L}$ into the form

$\displaystyle \partial_t (\vec{p}\times\vec{L}) = -mf(r)r^2 \left(\frac{\dot{\vec{r}}}{r} – \frac{\vec{r}\dot{r}}{r^2}\right).$

One more turn of the algebra crank gives us

$\displaystyle \partial_t (\vec{p}\times\vec{L}) = -mf(r)r^2 \partial_t \hat{r}.$

If we know that $f(r)$ is inversely proportional to $r^2$, then

$\displaystyle \partial_t (\vec{p}\times\vec{L}) = \partial_t (mk \hat{r}),$

implying that for the inverse-square case, there exists a conserved vector quantity given by

$\displaystyle \boxed{\vec{A} = \vec{p} \times \vec{L} – mk\hat{r}.}$

This quantity is what we call the Laplace-Runge-Lenz (LRL) vector.

EQUATION FOR AN ELLIPSE

The equation for a circle in Cartesian coordinates is basically just Pythagoras:

$\displaystyle x^2 + y^2 = a^2,$

alternatively written

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{a^2} = 1.$

We get an ellipse by stretching this circle along one of the axes until the length from the middle to the edge is $b$, thus:

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

You can derive this equation from the definition of the ellipse we gave earlier, that the distance from one focus $F$ to a point $P$ on the curve and back to the other focus $F^\prime$ is constant. It’s a fun way to kill time on an airplane.

The distance from the center (the point where the major and minor axes cross) to either focus is some fraction of the semimajor axis $a$, call it $ea$. The parameter $e$ is called the eccentricity. A bit of clever thinking shows that the distance from one focus to the far end of the ellipse and back to the other focus is just $2a$, and since we know that this distance must be the same for any point chosen on the ellipse, we can employ the Pythagorean Theorem again to write

$\displaystyle b^2 + a^2e^2 = a^2.$

Our mission is to rewrite the ellipse’s equation in polar coordinates. We’re free to choose the origin of our coordinate system as we like it; because Kepler tells us that one focus of the ellipse has physical significance — the Sun is located there — we’ll pick the focus on the positive x-axis as our origin. This means that when $ r = 0$ in polar coordinates, $x = ea$ and $y = 0$. Generally speaking,

$\displaystyle x = ea + r\cos\theta,\ y = r\sin\theta.$

These will be our coordinate transformations from Cartesian to polar. Substituting them into our ellipse equation gives

$\displaystyle \frac{(ea + r\cos\theta)^2}{a^2} + \frac{r^2\sin^2\theta}{b^2} = 1.$

Using the Pythagorean identity

$\displaystyle \sin^2 \theta + \cos^2 \theta = 1$

and the relationship

$\displaystyle b = a\sqrt{1 – e^2},$

we can chug through some algebra it’s not really worth writing out in detail to get, in the end, the equation

$\displaystyle \boxed{\frac{1}{r} = \frac{1 + e\cos\theta}{a(1 – e^2)}.} $

This relates the radial distance from the focus of choice to the ellipse as a function of angle; the relationship is governed by the eccentricity $e$. If the eccentricity is zero, then we recover the polar equation for the circle,

$ \displaystyle r = a. $

ELLIPTICAL ORBITS

Because of the way cross products work, $\vec{p}\times\vec{L}$ is perpendicular to $\vec{L}$ and $\vec{r}\times\vec{p}$ is perpendicular to $\hat{r}$. Together, these imply that $\vec{A}$ is orthogonal to $\vec{L}$:

$\displaystyle \vec{A}\cdot\vec{L} = 0.$

In other words, $\vec{A}$ has to be a vector in the plane of the planet’s motion. We know from our work above that it is a fixed vector; because the radius vector $\vec{r}$ is always changing direction, the angle $\vec{r}$ makes with $\vec{A}$ must be changing. We can get at this angle by taking the dot product:

$\displaystyle \vec{A}\cdot\vec{r} = Ar\cos\theta = \vec{r}\cdot(\vec{p}\times\vec{L}) – mkr.$

The combination of a vector dotted with a cross product is a familiar one, which we can manipulate by cyclic permutations, a result which follows naturally when the cross product is written in terms of the Levi-Civita tensor. In this form, the i-th component of the cross product is

$\displaystyle (\vec{a}\times\vec{b})_i = \epsilon_{ijk} a_j b_k,$

so dotting this cross product with a third vector gives

$\displaystyle \vec{c}\cdot(\vec{a}\times\vec{b}) = c_i \epsilon_{ijk} a_j b_k.$

We can cyclically permute the subscripts on the Levi-Civita symbol:

$\displaystyle \epsilon_{ijk} c_i a_j b_k = \epsilon_{jki} c_i a_j b_k = \epsilon_{kij} c_i a_j b_k.$

Translating this back into our old language tells us that

$\vec{c}\cdot(\vec{a}\times\vec{b}) = \vec{b} \cdot(\vec{c}\times\vec{a}).$

The upshot is that we can say

$\displaystyle \vec{r}\cdot(\vec{p}\times\vec{L}) = \vec{L} \cdot (\vec{r}\times\vec{p}) = L^2,$

meaning that our dot product equation for the LRL vector becomes

$\displaystyle Ar\cos\theta = L^2 – mkr.$

A bit of algebra gives us

$\displaystyle \boxed{\frac{1}{r} = \frac{mk}{L^2} \left(1 + \frac{A}{mk}\cos\theta\right),}$

which is the equation for an ellipse of eccentricity $A/(mk)$. The LRL vector $\vec{A}$ points from the focus to the perihelion of the elliptical orbit.

READINGS

• Goldstein, Herbert, Charles Poole and John Safko (2002). Classical Mechanics: Third Edition. Chapter 3 covers central-force problems, and the LRL vector is in section 3.9.
• Wikipedia actually has a pretty good write-up on the LRL vector, at least at the moment.
• A general run-through on stuff elliptical can be found at PlanetMath.