A previous presentation showed that the addition of a generalized Runge vector to the symmetry group *SO*(*n*)*SO*(*n*+1)

$\begin{array}{c}\frac{R}{L}[\phantom{\rule{1em}{0ex}}pcos[\int \frac{dp}{p\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{r}-rcos[\int \frac{dr}{r\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{p}\phantom{\rule{1em}{0ex}}]\\ \frac{R}{L}[\phantom{\rule{1em}{0ex}}psin[\int \frac{dp}{p\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{r}+rsin[\int \frac{dr}{r\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{p}\phantom{\rule{1em}{0ex}}]\end{array}$

are invariants of a spherically symmetric system, either of which can be taken as the Runge vector.

These explicit forms provide motivation to return to the question of the enlargement of the dynamic symmetry group by the Runge vector and to investigate by direct evaluation how this enlargement occurs. Much of this presentation will duplicate the former, apart from changes in the letter denoting the spatial variable, in order to provide a self-contained argument with evaluations more efficiently organized. The value of the explicit forms for the Runge vector is surprisingly low with respect to the algebraic structure of the symmetry group.

In the mathematical equations below, repeated indices represent sums over those indices, as is commonly done in coordinate-based presentations of classical general relativity.

Begin with the representation of the algebra for *SO*(*n*)*n*-dimensional space, the angular momentum tensor is

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

and has

$[A,B]=\frac{\partial A}{\partial {r}_{k}}\frac{\partial B}{\partial {p}_{k}}-\frac{\partial A}{\partial {p}_{k}}\frac{\partial B}{\partial {r}_{k}}$

The Poisson bracket of components of the angular momentum tensor is

$[{L}_{ij},{L}_{kl}]=-{\delta}_{jk}{L}_{il}-{\delta}_{il}{L}_{jk}+{\delta}_{ik}{L}_{jl}+{\delta}_{jl}{L}_{ik}$

and while this looks complicated, what is important is that the bracket of two components of the angular momentum tensor always leads to another component. Coordinates and momenta are vectors in this *n*-dimensional space, and their brackets with components of the angular momentum tensor will describe how vectors are transformed by rotation:

$\begin{array}{l}[{r}_{i},{L}_{jk}]=-{\delta}_{ij}{r}_{k}+{\delta}_{ik}{r}_{j}\\ [{p}_{i},{L}_{jk}]=-{\delta}_{ij}{p}_{k}+{\delta}_{ik}{p}_{j}\end{array}$

The angular momentum tensor corresponds to the symmetry of rotational invariance because the bracket of any component with a square of a vector or a dot product of two vectors is always zero,

$[{r}_{k}{r}_{k},{L}_{ij}]=[{p}_{k}{p}_{k},{L}_{ij}]=[{r}_{k}{p}_{k},{L}_{ij}]=0$

so that squares and dot products of dynamic vectors are unchanged by rotation of the system. This is not true of dot products that include nondynamic constant vectors, *i.e.* vectors whose constancy does not have its source in the dynamic system. This can be seen by evaluating the brackets

$\begin{array}{l}[{c}_{k}{r}_{k},{L}_{ij}]=-{c}_{i}{r}_{j}+{c}_{j}{r}_{i}\\ [{c}_{k}{p}_{k},{L}_{ij}]=-{c}_{i}{p}_{j}+{c}_{j}{p}_{i}\end{array}$

where the vectors *c _{k}* consist of arbitrary constants. Dot products of dynamic with nondynamic vectors behave like vectors when evaluating Poisson brackets.

A scalar is by definition rotationally invariant, having an identically zero bracket with components of the angular momentum tensor. This means that scalar functions appearing in rotationally symmetric systems can only depend on squares or dot products of dynamic vectors. The bracket of a general scalar with a component of the angular momentum tensor is

$[f[{r}^{2},{p}^{2},\left(rp\right)],{L}_{ij}]=({r}_{i}\frac{\partial}{\partial {r}_{j}}-{r}_{j}\frac{\partial}{\partial {r}_{i}}+{p}_{i}\frac{\partial}{\partial {p}_{j}}-{p}_{j}\frac{\partial}{\partial {p}_{i}})f[{r}^{2},{p}^{2},\left(rp\right)]$

with the notation

${r}^{2}={r}_{k}{r}_{k}\phantom{\rule{3em}{0ex}}{p}^{2}={p}_{k}{p}_{k}\phantom{\rule{3em}{0ex}}\left(rp\right)={r}_{k}{p}_{k}$

for the quantities that can appear in a rotationally symmetric function. When the individual derivatives are applied to the rotationally symmetric function, their effective action can be written

$\begin{array}{l}\frac{\partial}{\partial {r}_{i}}=2{r}_{i}\frac{\partial}{\partial {r}^{2}}+{p}_{i}\frac{\partial}{\partial \left(rp\right)}\\ \frac{\partial}{\partial {p}_{i}}=2{p}_{i}\frac{\partial}{\partial {p}^{2}}+{r}_{i}\frac{\partial}{\partial \left(rp\right)}\end{array}$

Substitution of these forms into the bracket above verifies that it is indeed identically zero. Use of these effective derivatives is unambiguous as long as all dynamic quantities are written explicitly in terms of coordinates and momenta.

Now consider the two choices of generalized Runge vector given above, each of which can be written schematically as

${R}_{i}=A[{r}^{2},{p}^{2},\left(rp\right)]\phantom{\rule{.3em}{0ex}}{r}_{i}-B[{r}^{2},{p}^{2},\left(rp\right)]\phantom{\rule{.3em}{0ex}}{p}_{i}$

Since brackets of scalar quantities with components of the angular momentum tensor are identically zero, the bracket with the Runge vector corresponds to those of the coordinate and momentum vectors,

$[{R}_{i},{L}_{jk}]=A[{r}^{2},{p}^{2},\left(rp\right)]\phantom{\rule{.3em}{0ex}}[{r}_{i},{L}_{jk}]-B[{r}^{2},{p}^{2},\left(rp\right)]\phantom{\rule{.3em}{0ex}}[{p}_{i},{L}_{jk}]=-{\delta}_{ij}{R}_{k}+{\delta}_{ik}{R}_{j}$

and this bracket forms part of the algebra for *SO*(*n*+1)*SO*(*n*+1)*i* ≠ *j* when evaluating the bracket:

$\begin{array}{l}[{R}_{i},{R}_{j}]={L}_{ij}[B,A]+A({r}_{j}\frac{\partial}{\partial {p}_{i}}-{r}_{i}\frac{\partial}{\partial {p}_{j}})A+B({p}_{i}\frac{\partial}{\partial {r}_{j}}-{p}_{j}\frac{\partial}{\partial {r}_{i}})B\\ \phantom{\rule{5em}{0ex}}+A({p}_{i}\frac{\partial}{\partial {p}_{j}}-{p}_{j}\frac{\partial}{\partial {p}_{i}})B+B({r}_{j}\frac{\partial}{\partial {r}_{i}}-{r}_{i}\frac{\partial}{\partial {r}_{j}})A\end{array}$

Replacing the derivatives in parentheses with their effective action upon rotationally symmetric functions, the bracket can be written compactly as

$\begin{array}{c}[{R}_{i},{R}_{j}]=F[{r}^{2},{p}^{2},\left(rp\right)]\phantom{\rule{.3em}{0ex}}{L}_{ij}\\ \\ F=[B,A]-\frac{\partial {A}^{2}}{\partial {p}^{2}}-\frac{\partial {B}^{2}}{\partial {r}^{2}}-\frac{\partial AB}{\partial \left(rp\right)}\end{array}$

where the Poisson bracket that has not been expressed in terms of effective action represents

$\begin{array}{l}[B,A]=4\left(rp\right)[\frac{\partial B}{\partial {r}^{2}}\frac{\partial A}{\partial {p}^{2}}-\frac{\partial B}{\partial {p}^{2}}\frac{\partial A}{\partial {r}^{2}}]+2{r}^{2}[\frac{\partial B}{\partial {r}^{2}}\frac{\partial A}{\partial \left(rp\right)}-\frac{\partial B}{\partial \left(rp\right)}\frac{\partial A}{\partial {r}^{2}}]\\ \phantom{\rule{10em}{0ex}}-2{p}^{2}[\frac{\partial B}{\partial {p}^{2}}\frac{\partial A}{\partial \left(rp\right)}-\frac{\partial B}{\partial \left(rp\right)}\frac{\partial A}{\partial {p}^{2}}]\end{array}$

This Poisson bracket can be written more compactly as

$[B,A]=2\left|\begin{array}{ccc}\frac{\partial A}{\partial {r}^{2}}& \frac{\partial B}{\partial {r}^{2}}& {p}^{2}\\ \frac{\partial A}{\partial {p}^{2}}& \frac{\partial B}{\partial {p}^{2}}& {r}^{2}\\ \frac{\partial A}{\partial \left(rp\right)}& \frac{\partial B}{\partial \left(rp\right)}& -2\left(rp\right)\end{array}\right|$

Since angular momentum is related to the dynamic variables by

${L}^{2}={r}^{2}{p}^{2}-(rp{)}^{2}$

the determinant is equivalent to

$[B,A]=2\left|\begin{array}{ccc}\frac{\partial A}{\partial {r}^{2}}& \frac{\partial B}{\partial {r}^{2}}& \frac{\partial {L}^{2}}{\partial {r}^{2}}\\ \frac{\partial A}{\partial {p}^{2}}& \frac{\partial B}{\partial {p}^{2}}& \frac{\partial {L}^{2}}{\partial {p}^{2}}\\ \frac{\partial A}{\partial \left(rp\right)}& \frac{\partial B}{\partial \left(rp\right)}& \frac{\partial {L}^{2}}{\partial \left(rp\right)}\end{array}\right|$

Although the coefficients *A* and *B* are functionally related via the square of the Runge vector, this determinant is not identically zero.

There now arises the question of how to interpret the action of these derivatives upon the coefficient functions *A* and *B*. The integrals in these functions are meant to be evaluated in terms of one variable, either *r* or *p*, with total angular momentum *L* treated as constant with respect to the integration. For evaluating Poisson brackets, however, the resulting integrated functions must then be written purely in terms of dynamic variables, with the constants composed of dynamic variables (energy, angular momentum and the magnitude of the Runge vector) replaced by the appropriate combinations of coordinates and momenta. Since the integrals can only be explicitly evaluated for a handful of cases, it is clearly not possible to complete the process for an arbitrary spherically symmetric potential.

One can at this point write a relationship between the coefficients *A* and *B* by squaring the Runge vector in schematic form,

${r}^{2}{A}^{2}-2\left(rp\right)AB+{p}^{2}{B}^{2}={R}^{2}$

and solving for either coefficient function:

$\begin{array}{l}A=\frac{1}{{r}^{2}}[\left(rp\right)B\pm \sqrt{{r}^{2}{R}^{2}-{L}^{2}{B}^{2}}]\\ B=\frac{1}{{p}^{2}}[\left(rp\right)A\pm \sqrt{{p}^{2}{R}^{2}-{L}^{2}{A}^{2}}]\end{array}$

Having the determinant above then simplifies an evaluation in terms of either coefficient function to a certain extent for the following reason. Considering *A* resp. *B* as a function of *B* resp. *A*, any action of derivatives on instances of *B* resp. *A* appearing in *A* resp. *B* will produce a column in the determinant that is proportional to the second resp. first column. Since a determinant with proportional columns or rows is zero, these derivatives will not contribute to the result, and the only contributions are from derivatives of *A* resp. *B* acting upon the explicit effective variables. The resulting algebra is nowhere near the “fair amount of tedious manipulation” described on page 421 of the second edition and page 414 of the third edition of Goldstein’s *Classical Mechanics*, but rather quite manageable as will be demonstrated below.

Even the manageable amount of algebra can be avoided, however, by appealing to the definition of the Runge vector as a constant of the motion. With a Hamiltonian of the form

$H=\frac{{p}^{2}}{2m}+V\left(r\right)$

the bracket of the Runge vector and the Hamiltonian is

$[{R}_{i},H]=[A{r}_{i}-B{p}_{i},H]=(\frac{B}{r}\frac{dV}{dr}+[A,H]){r}_{i}+(\frac{A}{m}-[B,H]){p}_{i}$

where the two intermediate brackets are

$\begin{array}{l}[A,H]=2\left|\begin{array}{ccc}\frac{1}{2r}\frac{dV}{dr}& \frac{\partial A}{\partial {r}^{2}}& {p}^{2}\\ \frac{1}{2m}& \frac{\partial A}{\partial {p}^{2}}& {r}^{2}\\ 0& \frac{\partial A}{\partial \left(rp\right)}& -2\left(rp\right)\end{array}\right|\\ \phantom{[A,H]}=\frac{2\left(rp\right)}{m}\frac{\partial A}{\partial {r}^{2}}-\frac{2\left(rp\right)}{r}\frac{dV}{dr}\frac{\partial A}{\partial {p}^{2}}+(\frac{{p}^{2}}{m}-r\frac{dV}{dr})\frac{\partial A}{\partial \left(rp\right)}\end{array}$

$\begin{array}{l}[B,H]=2\left|\begin{array}{ccc}\frac{1}{2r}\frac{dV}{dr}& \frac{\partial B}{\partial {r}^{2}}& {p}^{2}\\ \frac{1}{2m}& \frac{\partial B}{\partial {p}^{2}}& {r}^{2}\\ 0& \frac{\partial B}{\partial \left(rp\right)}& -2\left(rp\right)\end{array}\right|\\ \phantom{[B,H]}=\frac{2\left(rp\right)}{m}\frac{\partial B}{\partial {r}^{2}}-\frac{2\left(rp\right)}{r}\frac{dV}{dr}\frac{\partial B}{\partial {p}^{2}}+(\frac{{p}^{2}}{m}-r\frac{dV}{dr})\frac{\partial B}{\partial \left(rp\right)}\end{array}$

As a dynamic constant, the bracket of the Runge vector with the Hamiltonian is identically zero. Since the two dynamic variables *r _{i}* and

$\frac{m}{r}\frac{dV}{dr}=\frac{2\left(rp\right)\frac{\partial A}{\partial {r}^{2}}+{p}^{2}\frac{\partial A}{\partial \left(rp\right)}}{2\left(rp\right)\frac{\partial A}{\partial {p}^{2}}+{r}^{2}\frac{\partial A}{\partial \left(rp\right)}-B}=\frac{2\left(rp\right)\frac{\partial B}{\partial {r}^{2}}+{p}^{2}\frac{\partial B}{\partial \left(rp\right)}-A}{2\left(rp\right)\frac{\partial B}{\partial {p}^{2}}+{r}^{2}\frac{\partial B}{\partial \left(rp\right)}}$

Rearranging the second equality gives

$[B,A]=\frac{\partial {A}^{2}}{\partial {p}^{2}}+\frac{\partial {B}^{2}}{\partial {r}^{2}}+{r}^{2}\frac{\partial {A}^{2}}{\partial (rp{)}^{2}}+{p}^{2}\frac{\partial {B}^{2}}{\partial (rp{)}^{2}}-\frac{AB}{\left(rp\right)}$

so that the coefficient function that appears in the bracket of two Runge vector components is

$\begin{array}{l}F=[B,A]-\frac{\partial {A}^{2}}{\partial {p}^{2}}-\frac{\partial {B}^{2}}{\partial {r}^{2}}-\frac{\partial AB}{\partial \left(rp\right)}\\ \phantom{F}=\frac{\partial}{\partial (rp{)}^{2}}[{r}^{2}{A}^{2}-2\left(rp\right)AB+{p}^{2}{B}^{2}]\\ F=\frac{\partial {R}^{2}}{\partial (rp{)}^{2}}\end{array}$

This is a rather remarkable result, not depending explicitly on the actual form of the Runge vector as given an the outset. The structural steps to arrive at this result are

1) For a spherically symmetric system, assume a Hamiltonian additively separable in squares of the two dynamic vectors **r** and **p**;

2) Assume that there is a dynamically constant vector for this Hamiltonian with terms proportional to the two dynamic vectors; then

3) The bracket of components of the constant vector is proportional to the derivative of the square of the constant vector with respect to the square of the dot product of the dynamic vectors.

In order to realize the algebra for *SO*(*n*+1)

${R}^{2}=(pr{)}^{2}-{r}^{2}{p}^{2}+f(\frac{{p}^{2}}{2m}+V\left(r\right))=f\left(E\right)-{L}^{2}$

and then the bracket of components of the Runge vector will satisfy

$[{R}_{i},{R}_{j}]={L}_{ij}$

The inclusion of the arbitrary function of energy is important for keeping the square of the Runge vector positive as required for *SO*(*n*+1)*SO*(*n*,1)

The result for the choice of coefficient to scale the bracket is consistent with Equation (4.32) of a seminal paper by Fradkin, even though his result was developed for relativistic systems. Clearly the distinction between classical and relativistic systems is irrelevant to the algebra. The relativistic Hamiltonian used by Fradkin has the property of additive separation of squares of the dynamic variables, which is itself sufficient to ensure the existence of the generalized Runge vector.

Since angular momentum is linear in the square of the dot products of the dynamic vectors, one can write

$[{R}_{i},{R}_{j}]=-\frac{\partial {R}^{2}}{\partial {L}^{2}}{L}_{ij}$

as Fradkin does in his Equation (4.29), as well as Ben-Ya’acov in his Equation (5) of this preprint. Both of these approaches depend upon knowing that the square of the Runge vector is a function of energy and angular momentum, whether or not this can be shown explicitly for a general potential. Working instead exclusively with effective variables rather than dynamic constants, the same effective answer appears relatively easily. Since there is no coordinate transformation available between effective variables and dynamic constants, and it appears preferable to work with the former rather than the latter.

There is a particularly interesting choice for dynamically constant vectors that arises from the bracket of components of the Runge vector. If the squares of these vectors are independent of the dot product of the two dynamic variables, then the bracket of components will be identically zero. If in particular the square is set equal to the nondynamic constant of unity,

${r}^{2}{A}^{2}-2\left(rp\right)AB+{p}^{2}{B}^{2}=1$

then the resulting dynamic constant vectors

$\mathbf{U}=\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}\frac{1}{L}[pcos{I}_{p}\mathbf{r}-rcos{I}_{r}\mathbf{p}]\\ \frac{1}{L}[psin{I}_{p}\mathbf{r}+rsin{I}_{r}\mathbf{p}]\end{array}$

with the notation

${I}_{r}=\int \frac{dr}{r\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}\phantom{\rule{5em}{0ex}}{I}_{p}=\int \frac{dp}{p\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}$

are unit Runge vectors with zero bracket between components. To see that these are unit vectors, first evaluate the sum of the two integrals. Angular momentum is to be treated as a constant with respect to integration, so that with the substitution $u=\frac{rp}{L}$ one has

${I}_{r}+{I}_{p}=\int \frac{du}{u\phantom{\rule{.2em}{0ex}}\sqrt{{u}^{2}-1}}=-\int \frac{d\left(\frac{1}{u}\right)}{\sqrt{1-\frac{1}{{u}^{2}}}}=-{sin}^{-1}\left(\frac{1}{u}\right)=-{sin}^{-1}\left(\frac{L}{rp}\right)$

Using addition identities for circular functions one can write

$\begin{array}{l}sin{I}_{r}=-\frac{\left(rp\right)}{rp}sin{I}_{p}-\frac{L}{rp}cos{I}_{p}\\ cos{I}_{r}=\frac{\left(rp\right)}{rp}cos{I}_{p}-\frac{L}{rp}sin{I}_{p}\\ sin{I}_{p}=-\frac{\left(rp\right)}{rp}sin{I}_{r}-\frac{L}{rp}cos{I}_{r}\\ cos{I}_{p}=\frac{\left(rp\right)}{rp}cos{I}_{r}-\frac{L}{rp}sin{I}_{r}\end{array}$

which can be rearranged to

$\begin{array}{r}rpsin{I}_{r}+\left(rp\right)sin{I}_{p}=-Lcos{I}_{p}\\ rpcos{I}_{r}-\left(rp\right)cos{I}_{p}=-Lsin{I}_{p}\\ rpsin{I}_{p}+\left(rp\right)sin{I}_{r}=-Lcos{I}_{r}\\ rpcos{I}_{p}-\left(rp\right)cos{I}_{r}=-Lsin{I}_{r}\end{array}$

The basis for these identities is that the two integrals are relative angles between the Runge vector and the dynamic vectors, as pointed out at the end of the presentation of the generalized Runge vector. It is now simple to show that

$\begin{array}{r}[pcos{I}_{p}\mathbf{r}-rcos{I}_{r}\mathbf{p}{]}^{2}={L}^{2}\\ [psin{I}_{p}\mathbf{r}+rsin{I}_{r}\mathbf{p}{]}^{2}={L}^{2}\end{array}$

which means that ${\mathbf{U}}^{2}=1$ as stated. It is as simple to show that

$[pcos{I}_{p}\mathbf{r}-rcos{I}_{r}\mathbf{p}]\xb7[psin{I}_{p}\mathbf{r}+rsin{I}_{r}\mathbf{p}]=0$

so that the unit Runge vectors are orthogonal. The zero bracket of components of either vector follows from the general result, but can be evaluated directly fairly painlessly. Solve the expression for the square of the Runge vector for one of the coefficient functions:

$A=\frac{\left(rp\right)}{{r}^{2}}B\pm \frac{1}{{r}^{2}}\sqrt{{r}^{2}-{L}^{2}{B}^{2}}$

When this function is differentiated in the determinant part of the bracket, the action of derivatives on instances of B appearing in A will produce a column that is proportional to the second column. These derivatives will not contribute to the result, and the only contributions are from derivatives of A acting upon the explicit effective variables. Remembering that *L* is a function of the effective variables, the intermediate parts of the bracket evaluation are

$\begin{array}{l}[B,A]=2\left|\begin{array}{ccc}-\frac{1}{{r}^{4}}[\left(rp\right)B\pm \sqrt{{r}^{2}-{L}^{2}{B}^{2}}\phantom{\rule{.2em}{0ex}}]\pm \frac{1-{p}^{2}{B}^{2}}{2{r}^{2}\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}& \frac{\partial B}{\partial {r}^{2}}& {p}^{2}\\ \mp \frac{{B}^{2}}{2\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}& \frac{\partial B}{\partial {p}^{2}}& {r}^{2}\\ \frac{B}{{r}^{2}}\pm \frac{\left(rp\right){B}^{2}}{{r}^{2}\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}& \frac{\partial B}{\partial \left(rp\right)}& -2\left(rp\right)\end{array}\right|\\ \phantom{[B,A]}=2B\frac{\partial B}{\partial {r}^{2}}+\frac{2}{{r}^{4}}[[(rp{)}^{2}-{L}^{2}]B\pm \frac{\left(rp\right)}{\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}({r}^{2}-2{L}^{2}{B}^{2})]\frac{\partial B}{\partial {p}^{2}}\\ \phantom{\rule{5em}{0ex}}+\frac{2}{{r}^{2}}[\left(rp\right)B\pm \frac{1}{2\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}({r}^{2}-2{L}^{2}{B}^{2})]\frac{\partial B}{\partial \left(rp\right)}\end{array}$

$\begin{array}{l}\frac{\partial {A}^{2}}{\partial {p}^{2}}=\frac{2}{{r}^{4}}[[(rp{)}^{2}-{L}^{2}]B\pm \frac{\left(rp\right)}{\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}({r}^{2}-2{L}^{2}{B}^{2})]\frac{\partial B}{\partial {p}^{2}}\\ \phantom{\rule{5em}{0ex}}-\frac{{B}^{2}}{{r}^{2}}\mp \frac{\left(rp\right){B}^{3}}{{r}^{2}\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}\end{array}$

$\begin{array}{l}\frac{\partial AB}{\partial \left(rp\right)}=\frac{2}{{r}^{2}}[\left(rp\right)B\pm \frac{1}{2\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}({r}^{2}-2{L}^{2}{B}^{2})]\frac{\partial B}{\partial \left(rp\right)}\\ \phantom{\rule{5em}{0ex}}+\frac{{B}^{2}}{{r}^{2}}\pm \frac{\left(rp\right){B}^{3}}{{r}^{2}\sqrt{{r}^{2}-{L}^{2}{B}^{2}}}\end{array}$

and when combined the bracket of two components of either unit Runge vector is

$[{U}_{i},{U}_{j}]=[[B,A]-\frac{\partial {A}^{2}}{\partial {p}^{2}}-\frac{\partial {B}^{2}}{\partial {r}^{2}}-\frac{\partial AB}{\partial \left(rp\right)}]\phantom{\rule{.3em}{0ex}}{L}_{ij}\equiv 0$

It is worth noting that Fradkin endured “much tedious algebra” in deriving this zero bracket for components of unit Runge vectors. That seems an excessive statement in comparison with the evaluation as organized here.

To recap, the angular momentum tensor

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

and either choice for the scaled Runge vector

$\mathbf{R}=\{\phantom{\rule{.5em}{0ex}}\begin{array}{}\frac{\sqrt{f\left(E\right)-{L}^{2}}}{L}[\phantom{\rule{1em}{0ex}}pcos[\int \frac{dp}{p\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{r}-rcos[\int \frac{dr}{r\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{p}\phantom{\rule{1em}{0ex}}]\\ \frac{\sqrt{f\left(E\right)-{L}^{2}}}{L}[\phantom{\rule{1em}{0ex}}psin[\int \frac{dp}{p\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{r}+rsin[\int \frac{dr}{r\phantom{\rule{.2em}{0ex}}\sqrt{(\frac{rp}{L}{)}^{2}-1}}]\phantom{\rule{.3em}{0ex}}\mathbf{p}\phantom{\rule{1em}{0ex}}]\end{array}$

with an arbitrary function $f\left(E\right)$ and the constants of motion

${L}^{2}={r}^{2}{p}^{2}-(rp{)}^{2}\phantom{\rule{4em}{0ex}}E=\frac{{p}^{2}}{2m}+V\left(r\right)$

satisfy the algebra of *SO*(*n*+1)

*Uploaded 2014.12.30 — Updated 2015.01.13*
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