# The Laplace-Runge-Lenz Vector

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.

• 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.

## 15 thoughts on “The Laplace-Runge-Lenz Vector”

1. Joshua says:

Funny you’d link to something that also complains about Shermer’s Randist dogma, as well as his ineptitude at Powerpoint. What I didn’t get to tell you about earlier was Shermer’s “Government of the Gaps” theory on liberalism.

No, seriously. He said that liberalism (didn’t use the word, but you know what he meant) is the political equivalent of ID. He cited a straw version that went something like, “X is a problem. I can’t imagine a private solution for X. Therefore, X must be solved publicly.” Which is all well and good, but completely ignores the very real possibility that there may not be a “private” solution for the problem. In fact, in many cases, “private” enterprise and the market create the problem in the first place.

And then there’s the totally arbitrary concession that “some” things are appropriate for the government to do. Suspiciously, the line always seems to be drawn in a location that seems terribly convenient for the speaker…

2. efrique says:

Blake, I have a mathematics major (way back), so I am not scared of symbols. However, I don’t have physics past high school (decades ago) so I don’t know what all the notation is referring to. Where’s the notation at the start defined? What quantities do the letters stand for?

3. Blake Stacey says:

Yeah, I could probably have set that up a bit better, but let’s see: we’re talking about a particle in a potential well, with $$\vec{p}$$ denoting its momentum and $$\vec{r}$$ its position. The funny-lookin $$\hat{r}$$ is the unit vector in the direction of $$\vec{r}$$, the angular momentum is denoted by $$\vec{L}$$ and following common practice a constant of proportionality is $$k$$.

4. efrique says:

Thanks. What’s f(.)?

The arrow over the top presumably denotes a vector. Is the dot over the arrow a derivative w.r.t. time?

5. Greg Egan says:

That’s another very enjoyable post, Blake! (Efrique, f(r) is the force on the orbiting body as a function of its distance r from the centre of attraction.)

There’s a nice easy way to show that an inverse square force law is necessary in order for all bound obits to form closed curves with one perihelion per orbit: you can look at the period of an infinitesimal radial perturbation of a body in a circular orbit, and show that this will only equal the orbital period for f(r) = -k r-2.

If you analyse the force law f(r) = -k r(m/n)^2-3, where m and n are integers, that gives you m cycles of radial perturbation for every n orbits, so the orbits infinitesimally close to circular ones will form closed curves. For example, for m=2, n=1, that’s a 2-dimensional harmonic oscillator: a weight “orbiting” on the end of a spring. Since m=2, you get peanut-shaped orbits, with two “perihelia” and two “aphelia” per orbit!

But I’m not sure whether anything other than the inverse-square law gives closed curves once the perturbation away from a circular orbit is finite.

6. Blake Stacey says:

Yes, the dot is a derivative with respect to time (by convention, dots are time derivatives and primes are space derivatives, when both dependencies exist). $$f(r)$$ is a scalar function of the radial distance $$r$$, meaning that the strength of the interaction force depends on the distance, not on the momentum or anything else kooky. A specific form might be $$f(r) \propto 1/r^2$$, for example.

7. Greg Egan says:

A comment I just made seems to have disappeared, but in case it materialises here’s a preemptive correction: I suggested that the orbits of a 2-dimensional harmonic oscillator were “peanut shaped” (!); of course they’re not, they’re ellipses with the centre of attraction at the centre of the ellipse (which gives two points of equal closest approach and two points of equal furthest distance, per orbit). And of course they are all exactly closed orbits.

8. Greg Egan says:

My apologies for some half-baked comments (and some superscript formatting that didn’t stick around). I’m pretty sure now that the only central force laws for which all the bound orbits are closed curves are the inverse square law and the harmonic oscillator.

The law f(r)=-k r^((m/n)^2-3) will give closed curves for orbits infinitesimally close to circular, for all integers m and n, but only m=1, n=1 (inverse square) and m=2, n=1 (harmonic oscillator) are closed for finite perturbations.

9. efrique says:

Okay, thanks. Now that it’s clear what you’re writing about there, I need to go back and study my old calculus and analytic geometry texts so I can figure out how you’re getting from one line to the next there.

Interesting stuff.

10. Greg Egan says:

It turns out that a certain Monsieur Beltrand proved back in 1873 that only the inverse square and harmonic oscillator force laws produced closed orbits. I think I might have known this a couple of decades ago, then forgotten …

11. Greg Egan says:

Bertrand, sorry, not Beltrand. Proof here.

12. Blake Stacey says:

Heck, I knew that when I was reading the Goldstein textbook last week, but my travels this past weekend left me too scatterbrained to look it up again. Thanks for finding the guy’s name!

For future reference, you can do $$\LaTeX$$ markup in the comments by enclosing expressions in [ tex ] and [ /tex ] tags (remove the spaces inside the square brackets).

13. Laelaps says:

“DEFINING THE LRL VECTOR”

I first read that as “Defining the LOL Vector.” I can has central-force problem?

14. robert says:

The RL vector becomes even more interesting in the quantum domain. The corresponding quantum operator commutes with the Hamiltonian (as you might expect for a conserved quantity); the associated symmetry (SO(4))accounts for the degeneracy of hydrogenic orbitals with the same principal quantum number, but different angular momentum quantum numbers – e.g 2s and 2p orbitals. Pauli exploited the conservation of the RL vector in his matrix solution of the hydrogen atom problem. A related observation: the Schroedinger equation for the hydrogen atom separates in more than one co-ordinate system – spherical polars – as in all the text books – and parabolic coordinates – as when you are looking at the stark effect for example. The parallels with the 3-dimensional harmonic oscillator are quite marked – the SE in this case separates in spherical polars and Cartesian coordinates and ‘extra’ degenracies are found in the energy elvels. One puzzle you might like to consider in this context is that the Feynman path integral solution to the quantum harmonic oscillator problem is almost trivial – doing the same thing for the hydrogen atom is really hard (and was actually beyond the wit of Feynman himself).

15. Blake Stacey says:

I was actually planning on going into the quantum analogue next (well, once I’ve cleared a few other items off my queue).