This short presentation will consider the relationship between the generalized Runge vector for spherically symmetric potentials and the tensor that describes angular momentum. The general solution found is

$\mathbf{R}=\{\phantom{\rule{.5em}{0ex}}\begin{array}{l}\frac{R}{L}\phantom{\rule{.4em}{0ex}}[\phantom{\rule{.4em}{0ex}}pcos\alpha \phantom{\rule{.5em}{0ex}}\mathbf{r}-rcos\phi \phantom{\rule{.5em}{0ex}}\mathbf{p}\phantom{\rule{.3em}{0ex}}]\\ \frac{R}{L}\phantom{\rule{.4em}{0ex}}[-psin\alpha \phantom{\rule{.5em}{0ex}}\mathbf{r}+rsin\phi \phantom{\rule{.5em}{0ex}}\mathbf{p}\phantom{\rule{.3em}{0ex}}]\end{array}$

where φ is the angle between the fixed direction of the Runge vector and the radius vector **r** and α is the angle between the same fixed direction and the linear momentum vector **p**.

In a three-dimensional space angular momentum can be described by a vector, but this does not hold true for spaces of higher dimension. The explicit definition for three dimensions is

$\begin{array}{l}\mathbf{L}=\mathbf{r}\times \mathbf{p}=\left|\phantom{\rule{.5em}{0ex}}\begin{array}{lll}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ {r}_{1}& {r}_{2}& {r}_{3}\\ {p}_{1}& {p}_{2}& {p}_{3}\end{array}\phantom{\rule{.5em}{0ex}}\right|\\ \phantom{\mathbf{L}}=[\phantom{\rule{.3em}{0ex}}{r}_{2}{p}_{3}-{r}_{3}{p}_{2}\phantom{\rule{.3em}{0ex}},\phantom{\rule{.3em}{0ex}}{r}_{3}{p}_{1}-{r}_{1}{p}_{3}\phantom{\rule{.3em}{0ex}},\phantom{\rule{.3em}{0ex}}{r}_{1}{p}_{2}-{r}_{2}{p}_{1}\phantom{\rule{.3em}{0ex}}]\end{array}$

and it is clear that the three vector components of angular momentum are cyclic assignments of the components of a tensor with two indices,

${L}_{i}={\epsilon}_{ijk}{L}_{jk}={\epsilon}_{ijk}({r}_{j}{p}_{k}-{r}_{k}{p}_{j})\phantom{\rule{4em}{0ex}}i,j,k=1,2,3$

where there is no summation over repeated indices. In spaces of higher dimension, it is more convenient to work with the tensor

${L}_{ij}={r}_{i}{p}_{j}-{r}_{j}{p}_{i}$

but since this tensor has

$\frac{d{r}_{i}}{dt}=\frac{{p}_{i}}{m}\phantom{\rule{5em}{0ex}}\frac{d{p}_{i}}{dt}=-\frac{{r}_{i}}{r}\frac{dV}{dr}$

The tensor is even applicable to a two-dimensional space, where it has a single component

${L}_{12}={r}_{1}{p}_{2}-{r}_{2}{p}_{1}$

Now consider the explicit form of the generalized Runge vector in two dimensions. The angles appearing in the vector represent projections of dynamic vectors on a fixed direction. If the fixed direction is taken to be the unit vector

$\mathbf{n}=[{n}_{1},{n}_{2}]\phantom{\rule{5em}{0ex}}{n}_{1}^{2}+{n}_{2}^{2}=1$

the first choice for the Runge vector gives

$\begin{array}{l}\frac{R}{L}\phantom{\rule{.3em}{0ex}}[\phantom{\rule{.2em}{0ex}}({p}_{1}{n}_{1}+{p}_{2}{n}_{2})[{r}_{1},{r}_{2}]-({r}_{1}{n}_{1}+{r}_{2}{n}_{2})[{p}_{1},{p}_{2}]\phantom{\rule{.3em}{0ex}}]\\ \phantom{\rule{4em}{0ex}}=\frac{R}{L}[\phantom{\rule{.3em}{0ex}}({r}_{1}{p}_{2}-{r}_{2}{p}_{1}){n}_{2}\phantom{\rule{.3em}{0ex}},\phantom{\rule{.3em}{0ex}}({r}_{2}{p}_{1}-{r}_{1}{p}_{2}){n}_{1}\phantom{\rule{.3em}{0ex}}]\end{array}$

By appropriate choice of the fixed direction the Runge vector can be made to lie along any direction in the plane. What is curious, however, is that the constancy of the vector depends upon the appearance of the single component of the angular momentum tensor, and this behavior extends to higher dimensions. In three dimensions the unit vector is

$\mathbf{n}=[{n}_{1},{n}_{2},{n}_{3}]\phantom{\rule{5em}{0ex}}{n}_{1}^{2}+{n}_{2}^{2}+{n}_{3}^{2}=1$

and the Runge vector can be written

$\begin{array}{l}\frac{R}{L}\phantom{\rule{.3em}{0ex}}[\phantom{\rule{.2em}{0ex}}({p}_{1}{n}_{1}+{p}_{2}{n}_{2}+{p}_{3}{n}_{3})[{r}_{1},{r}_{2},{r}_{3}]-({r}_{1}{n}_{1}+{r}_{2}{n}_{2}+{r}_{3}{n}_{3})[{p}_{1},{p}_{2},{p}_{3}]\phantom{\rule{.3em}{0ex}}]\\ \phantom{\rule{3em}{0ex}}=\frac{R}{L}[\phantom{\rule{.3em}{0ex}}({r}_{1}{p}_{2}-{r}_{2}{p}_{1}){n}_{2}+({r}_{1}{p}_{3}-{r}_{3}{p}_{1}){n}_{3}\phantom{\rule{.3em}{0ex}},\\ \phantom{\rule{6.5em}{0ex}}({r}_{2}{p}_{1}-{r}_{1}{p}_{2}){n}_{1}+({r}_{2}{p}_{3}-{r}_{3}{p}_{2}){n}_{3}\phantom{\rule{.3em}{0ex}},\\ \phantom{\rule{8em}{0ex}}({r}_{3}{p}_{1}-{r}_{1}{p}_{3}){n}_{1}+({r}_{3}{p}_{2}-{r}_{2}{p}_{3}){n}_{2}\phantom{\rule{.3em}{0ex}}]\end{array}$

where all three components of the angular momentum tensor appear. In four dimensions the unit vector is

$\mathbf{n}=[{n}_{1},{n}_{2},{n}_{3},{n}_{4}]\phantom{\rule{5em}{0ex}}{n}_{1}^{2}+{n}_{2}^{2}+{n}_{3}^{2}+{n}_{4}^{2}=1$

and the Runge vector can now be written

$\begin{array}{l}\frac{R}{L}\phantom{\rule{.3em}{0ex}}[\phantom{\rule{.2em}{0ex}}({p}_{1}{n}_{1}+{p}_{2}{n}_{2}+{p}_{3}{n}_{3}+{p}_{4}{n}_{4})[{r}_{1},{r}_{2},{r}_{3},{r}_{4}]\\ \phantom{\rule{5em}{0ex}}-({r}_{1}{n}_{1}+{r}_{2}{n}_{2}+{r}_{3}{n}_{3}+{r}_{4}{n}_{3})[{p}_{1},{p}_{2},{p}_{3},{p}_{4}]\phantom{\rule{.3em}{0ex}}]\\ \phantom{\rule{3em}{0ex}}=\frac{R}{L}[\phantom{\rule{.3em}{0ex}}({r}_{1}{p}_{2}-{r}_{2}{p}_{1}){n}_{2}+({r}_{1}{p}_{3}-{r}_{3}{p}_{1}){n}_{3}+({r}_{1}{p}_{4}-{r}_{4}{p}_{1}){n}_{4}\phantom{\rule{.3em}{0ex}},\\ \phantom{\rule{6.5em}{0ex}}({r}_{2}{p}_{1}-{r}_{1}{p}_{2}){n}_{1}+({r}_{2}{p}_{3}-{r}_{3}{p}_{2}){n}_{3}+({r}_{2}{p}_{4}-{r}_{4}{p}_{2}){n}_{4}\phantom{\rule{.3em}{0ex}},\\ \phantom{\rule{6.5em}{0ex}}({r}_{3}{p}_{1}-{r}_{1}{p}_{3}){n}_{1}+({r}_{3}{p}_{2}-{r}_{2}{p}_{3}){n}_{2}+({r}_{3}{p}_{4}-{r}_{4}{p}_{3}){n}_{4}\phantom{\rule{.3em}{0ex}}]\phantom{\rule{.3em}{0ex}},\\ \phantom{\rule{8em}{0ex}}({r}_{4}{p}_{1}-{r}_{1}{p}_{4}){n}_{1}+({r}_{4}{p}_{2}-{r}_{2}{p}_{4}){n}_{2}+({r}_{4}{p}_{3}-{r}_{3}{p}_{4}){n}_{3}\end{array}$

where all six components of the angular momentum tensor appear. Seen this way, the Runge vector is essentially a rearrangement of the angular momentum tensor to give a vector extensible to any number of dimensions. This helps to explain why the angular momentum tensor plus the Runge vector always have the group structure *SO*(*n*+1)

Lest one think that this feature is an artifact of the particular form of the general solution, consider rewriting the standard solution for the Kepler problem. The Runge vector as given on page 103 of both the second and third editions of Goldstein’s *Classical Mechanics* is

$\mathbf{R}=\mathbf{p}\times \mathbf{L}-\frac{mk}{r}\mathbf{r}$

With the definition **L** = **r** × **p**

$\mathbf{R}=({p}^{2}-\frac{mk}{r})\mathbf{r}-(\mathbf{r}\xb7\mathbf{p})\mathbf{p}$

The corresponding Hamiltonian for the Kepler problem is

$H=\frac{{p}^{2}}{2m}-\frac{k}{r}=E$

and the squares of the two composite vectors are

$\begin{array}{l}{L}^{2}={r}^{2}{p}^{2}-(\mathbf{r}\xb7\mathbf{p}{)}^{2}\\ {R}^{2}=2mE{L}^{2}+{m}^{2}{k}^{2}\end{array}$

where the first comes from a vector quadruple product. If the Runge vector is taken to define the *x*-axis, then

$\mathbf{p}\xb7\mathbf{R}={p}_{x}R=-\frac{mk}{r}(\mathbf{r}\xb7\mathbf{p})$

from which follows

${p}_{y}^{2}={p}^{2}-{p}_{x}^{2}=\frac{{L}^{2}}{{R}^{2}}(2mE{p}^{2}+\frac{{m}^{2}{k}^{2}}{{r}^{2}})=\frac{{L}^{2}}{{R}^{2}}({p}^{2}-\frac{mk}{r}{)}^{2}$

Similarly for the spatial coordinates

$\mathbf{r}\xb7\mathbf{R}=xR={L}^{2}-mkr$

from which follows

${y}^{2}={r}^{2}-{x}^{2}=\frac{{L}^{2}}{{R}^{2}}(2mE{r}^{2}+2mkr-{L}^{2})=\frac{{L}^{2}}{{R}^{2}}(\mathbf{r}\xb7\mathbf{p}{)}^{2}$

Taking positive square roots, the Runge vector can be written

$\mathbf{R}=\frac{R}{L}({p}_{y}\mathbf{r}-y\mathbf{p})$

which is the negative of the second choice for the Runge vector given at the beginning of this presentation. Funny that this simple form does not appear in Goldstein. Presumably this is because of the insistence on using only system vectors, *i.e.* vectors composed only of system variables, in evaluating Poisson brackets: see pages 417-418 of the second edition or pages 409-410 of the third. Introducing a fixed direction along the *x*-axis for the Runge vector breaks the Poisson bracket relations, even though the Runge vector is essentially a rearrangement of the angular momentum tensor.

*Uploaded 2013.12.06 — Updated 2014.08.07*
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