The Runge-Lenz vector

The closed Kepler orbit and the Runge-Lenz vector

This page works the conserved quantity and does not derive the symmetry behind it. The second half is flagged below.

The puzzle

A planet moving under an inverse-square attraction traces a closed ellipse that does not precess. Under any other central force law it precesses: the orbit is still bounded, but the long axis slowly turns.

Energy and angular momentum are conserved for every central force. So they cannot be what closes the orbit. Something extra is conserved in the inverse-square case only, and it was noticed long before anyone could say why.

The quantity, and the check

With $V=-k/r$, momentum $p=m\dot{r}$, and angular momentum $L=r\times p$, define

$A=p\times L-mk\hat{r}\mathrm{(}\mathrm{1}\mathrm{)}$

Differentiate along the motion. Since $L$ is conserved,

$\frac{dA}{dt}=\dot{p}\times L-mk\frac{d\hat{r}}{dt}\mathrm{(}\mathrm{2}\mathrm{)}$

The force is $\dot{p}=-k\hat{r}/r^2$. Using the vector identity for a triple product,

$\dot{p}\times L=-\frac{k}{r^2}\hat{r}\times (r\times p)=-\frac{k}{r^2}[r(\hat{r}\cdot p)-pr]=\frac{mk}{r^2}[r\dot{r}-r(\hat{r}\cdot \dot{r})]\mathrm{(}\mathrm{3}\mathrm{)}$

And differentiating the unit vector,

$\frac{d\hat{r}}{dt}=\frac{\dot{r}}{r}-\frac{r(\hat{r}\cdot \dot{r})}{r^2}=\frac{1}{r^2}[r\dot{r}-r(\hat{r}\cdot \dot{r})]\mathrm{(}\mathrm{4}\mathrm{)}$

The bracket in (3) and the bracket in (4) are the same. Substituting both into (2), the two terms cancel exactly and

$\frac{dA}{dt}=0\mathrm{(}\mathrm{5}\mathrm{)}$

Where the inverse square enters. At (3), and nowhere else. Any other power leaves a residue the second term cannot cancel, and the orbit precesses.

The vector points along the long axis of the ellipse, which is why its constancy is the statement that the axis does not turn.

The symmetry, not derived

UNVERIFIED. The symmetry responsible is stated in the literature to be a rotational symmetry of a four-dimensional space, larger than the ordinary rotations of the three-dimensional problem, and acting on a combination of position and momentum rather than on position alone. Exhibiting it and running it through (N12) is not done here.

Two things are worth saying without it.

It is not visible in the Lagrangian. The transformation mixes the coordinates with their derivatives. That is exactly the case Noether takes trouble to allow — see prolongation — and the case most textbook statements of the theorem exclude by assumption.

This is the reverse direction earning its keep. A conserved quantity turned up with no known symmetry. The converse says one exists. It did.

Provenance: (1) to (5) are this volume's own derivation. The symmetry is flagged UNVERIFIED and no source was consulted. See Sources.