Rotationally symmetric systems in classical mechanics always have the symmetry group *SO*(3)*SO*(4)

The purpose of this discussion is to show that the generalized Runge vector described in this presentation always leads to an enlarged symmetry group for rotationally symmetric systems, no matter what the functional dependence of the potential on the radial variable. This result is independent of the number of spatial dimensions, so that the addition of the generalized Runge vector to the symmetry group *SO*(*n*)*SO*(*n*+1)

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

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

and has

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

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:

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

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,

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

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

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

where the vectors *c _{k}* consist of arbitrary constants. Dot products of dynamic with nondynamic vectors behave like vectors when evaluating Poisson brackets, a point that will lead to an interesting puzzle at the end of this presentation.

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

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

with the notation

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

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

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

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 introduce the generalized Runge vector from this presentation. For simplicity, take all of the β_{i} of that presentation equal, so that the vector is

${R}_{i}=m(\beta {\stackrel{\xb7}{q}}_{i}-\stackrel{\xb7}{\beta}{q}_{i})=\beta {p}_{i}-m\stackrel{\xb7}{\beta}{q}_{i}$

where the scalar function β is determined through the dynamical quantities by

$\stackrel{\xb7\xb7}{\beta}=-\frac{1}{mq}\frac{dV}{dq}\beta =-\frac{2}{m}\frac{dV}{d{q}^{2}}\beta $

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,

$[{R}_{i},{L}_{jk}]=\beta [{p}_{i},{L}_{jk}]-m\stackrel{\xb7}{\beta}[{q}_{i},{L}_{jk}]=-{\delta}_{ij}{R}_{k}+{\delta}_{ik}{R}_{j}$

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

$\begin{array}{l}[{R}_{i},{R}_{j}]={L}_{ij}[\beta ,m\stackrel{\xb7}{\beta}]+\beta ({p}_{i}\frac{\partial}{\partial {q}_{j}}-{p}_{j}\frac{\partial}{\partial {q}_{i}})\beta +m\stackrel{\xb7}{\beta}({q}_{j}\frac{\partial}{\partial {p}_{i}}-{q}_{i}\frac{\partial}{\partial {p}_{j}})m\stackrel{\xb7}{\beta}\\ \phantom{\rule{5em}{0ex}}+\beta ({q}_{j}\frac{\partial}{\partial {q}_{i}}-{q}_{i}\frac{\partial}{\partial {q}_{j}})m\stackrel{\xb7}{\beta}+m\stackrel{\xb7}{\beta}({p}_{i}\frac{\partial}{\partial {p}_{j}}-{p}_{j}\frac{\partial}{\partial {p}_{i}})\beta \end{array}$

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

$\begin{array}{c}[{R}_{i},{R}_{j}]=F[{q}^{2},{p}^{2},\left(qp\right)]\phantom{\rule{.3em}{0ex}}{L}_{ij}\\ \\ F=[\beta ,m\stackrel{\xb7}{\beta}]-\frac{\partial {\beta}^{2}}{\partial {q}^{2}}-\frac{\partial (m\stackrel{\xb7}{\beta}{)}^{2}}{\partial {p}^{2}}-\frac{\partial (m\stackrel{\xb7}{\beta}\beta )}{\partial \left(qp\right)}\end{array}$

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

$\begin{array}{l}[\beta ,m\stackrel{\xb7}{\beta}]=4\left(qp\right)[\frac{\partial \beta}{\partial {q}^{2}}\frac{\partial m\stackrel{\xb7}{\beta}}{\partial {p}^{2}}-\frac{\partial \beta}{\partial {p}^{2}}\frac{\partial m\stackrel{\xb7}{\beta}}{\partial {q}^{2}}]+2{q}^{2}[\frac{\partial \beta}{\partial {q}^{2}}\frac{\partial m\stackrel{\xb7}{\beta}}{\partial \left(qp\right)}-\frac{\partial \beta}{\partial \left(qp\right)}\frac{\partial m\stackrel{\xb7}{\beta}}{\partial {q}^{2}}]\\ \phantom{\rule{10em}{0ex}}-2{p}^{2}[\frac{\partial \beta}{\partial {p}^{2}}\frac{\partial m\stackrel{\xb7}{\beta}}{\partial \left(qp\right)}-\frac{\partial \beta}{\partial \left(qp\right)}\frac{\partial m\stackrel{\xb7}{\beta}}{\partial {p}^{2}}]\end{array}$

The coefficient *F* is a complicated function of coordinates and momenta, but consists solely of combinations of constants of the motion. To understand this, remember that the Poisson brackets satisfy the Jacobi identity

$[A,[B,C\left]\right]+[B,[C,A\left]\right]+[C,[A,B\left]\right]=0$

which can be verified by writing out the twenty-four index summed terms in this statement and canceling them in pairs. When this identity is applied to the constants of the motion *R _{i}*,

$[{R}_{i},[{R}_{j},H\left]\right]+[{R}_{j},[H,{R}_{i}\left]\right]+[H,[{R}_{i},{R}_{j}\left]\right]=0$

the inner brackets of the first two terms are zero by definition of a constant of the motion, so that the last term must also be zero and its inner bracket a constant in its own right. This is a general result: the bracket of two constants of the motion is always another constant. Since this inner bracket is the function *F* multiplying a known constant of the motion *L _{ij}*,

$[H,[{R}_{i},{R}_{j}\left]\right]=0=[H,{L}_{ij}F]=[H,F]{L}_{ij}$

the function *F* must also have a zero bracket with the Hamiltonian, and can only consist of constants of the motion. The constants of the motion that can be part of a scalar function are the Hamiltonian, the squared contraction of the angular momentum tensor, and the square of the Runge vector:

$\begin{array}{l}\phantom{\rule{.3em}{0ex}}H=\frac{{p}_{k}{p}_{k}}{2m}+V\left({q}_{k}{q}_{k}\right)=\frac{{p}^{2}}{2m}+V\left(q\right)\\ {L}^{2}=\frac{1}{2}{L}_{kl}{L}_{kl}={q}^{2}{p}^{2}-(qp{)}^{2}\\ {R}^{2}={R}_{k}{R}_{k}={p}^{2}{\beta}^{2}-2\left(qp\right)m\stackrel{\xb7}{\beta}\beta +{q}^{2}(m\stackrel{\xb7}{\beta}{)}^{2}\end{array}$

These correspond with the three quantities *q*^{2}, *p*^{2} and *q**p*)

In order to demonstrate fully the algebra of *SO*(*n*+1)*F* that appears in

$\begin{array}{l}[{L}_{ij},{L}^{2}]={L}_{kl}[{L}_{ij},{L}_{kl}]=0\\ [{L}_{ij},{R}^{2}]={R}_{k}[{L}_{ij},{R}_{k}]=0\end{array}$

which means that any scalar function of the squared constants of the motion can be brought inside a bracket with any component of the angular momentum tensor. A complication arises for brackets containing individual Runge vector components, for which

$\begin{array}{l}[{R}_{i},{L}^{2}]={L}_{kl}[{R}_{i},{L}_{kl}]=2{R}_{k}{L}_{ki}\\ [{R}_{i},{R}^{2}]=2{R}_{k}[{R}_{i},{R}_{k}]=-2F{R}_{k}{L}_{ki}\end{array}$

Scalar functions cannot simply be brought inside such brackets, which means that one cannot blithely multiply Runge-vector components by an arbitrary function of squared constants of the motion. As an aside, the right-hand sides of these equations are generalizations of the three-dimensional cross product to an arbitrary number of dimensions, and are in general nonzero.

Fortunately for the goal of this presentation, the components of the Runge vector replicate a structure seen earlier. Consider first the explicit effect on the bracket of rescaling the Runge vector components by multiplying with an arbitrary power of a scalar function to be determined,

$[{G}^{u}{R}_{i},{G}^{u}{R}_{j}]={G}^{2u}[{R}_{i},{R}_{j}]+u{G}^{2u-1}[{R}_{j}[{R}_{i},G]-{R}_{i}[{R}_{j},G]]$

where the power *u* can be chosen for convenience. The trick is what happens when the difference of the two brackets is expanded,

$\begin{array}{l}{R}_{j}[{R}_{i},G]-{R}_{i}[{R}_{j},G]={L}_{ij}[m\stackrel{\xb7}{\beta}[\beta ,G]-\beta [m\stackrel{\xb7}{\beta},G]]\\ \phantom{\rule{5em}{0ex}}+{\beta}^{2}({p}_{i}\frac{\partial}{\partial {q}_{j}}-{p}_{j}\frac{\partial}{\partial {q}_{i}})G+(m\stackrel{\xb7}{\beta}{)}^{2}({q}_{j}\frac{\partial}{\partial {p}_{i}}-{q}_{i}\frac{\partial}{\partial {p}_{j}})G\\ \phantom{\rule{5em}{0ex}}+m\stackrel{\xb7}{\beta}\beta ({q}_{j}\frac{\partial}{\partial {q}_{i}}-{q}_{i}\frac{\partial}{\partial {q}_{j}})G+m\stackrel{\xb7}{\beta}\beta ({p}_{i}\frac{\partial}{\partial {p}_{j}}-{p}_{j}\frac{\partial}{\partial {p}_{i}})G\end{array}$

which in terms of effective action becomes

$\begin{array}{c}{R}_{j}[{R}_{i},G]-{R}_{i}[{R}_{j},G]=D\left(G\right)\phantom{\rule{.3em}{0ex}}{L}_{ij}\\ \\ D\left(G\right)=m\stackrel{\xb7}{\beta}[\beta ,G]-\beta [m\stackrel{\xb7}{\beta},G]-2{\beta}^{2}\frac{\partial G}{\partial {q}^{2}}-2(m\stackrel{\xb7}{\beta}{)}^{2}\frac{\partial G}{\partial {p}^{2}}-2m\stackrel{\xb7}{\beta}\beta \frac{\partial G}{\partial \left(qp\right)}\end{array}$

where this last quantity is a linear differential operator applied to the function to be determined. The bracket of the rescaled components is now

$[{G}^{u}{R}_{i},{G}^{u}{R}_{j}]=[{G}^{2u}F+u{G}^{2u-1}D\left(G\right)]\phantom{\rule{.3em}{0ex}}{L}_{ij}$

and can be made to conform to the algebra of *SO*(*n*+1) by setting the coefficient equal to unity and solving a differential equation. For the choice
$u=\frac{1}{2}$
this is the linear first-order inhomogeneous differential equation

$\frac{1}{2}D\left(G\right)+F[{q}^{2},{p}^{2},\left(qp\right)]G=1$

When a solution to this equation has been found, the constants of the motion
${L}_{ij}$
and
$\sqrt{G}{R}_{i}$
will satisfy the algebra of *SO*(*n*+1)

While this indicates that the scaling function *G* always exists in principle, determining it is not simple given the complexity of the differential equations. The Kepler potential

$\beta =\left(qp\right)\phantom{\rule{5em}{0ex}}m\stackrel{\xb7}{\beta}={p}^{2}-\frac{mk}{q}$

from which follows

$F=-{p}^{2}+\frac{2mk}{q}=2m(-E)=2m(-H)$

which is a positive quantity for bound states. Since the Hamiltonian has a bracket of zero with all constants of the motion, one can immediately conclude that

It is worth noting that the differential equation to be solved is equivalent to

$[\beta \sqrt{G},m\stackrel{\xb7}{\beta}\sqrt{G}]-\frac{\partial \left[{\beta}^{2}G\right]}{\partial {q}^{2}}-\frac{\partial [(m\stackrel{\xb7}{\beta}{)}^{2}G]}{\partial {p}^{2}}-\frac{\partial \left[m\stackrel{\xb7}{\beta}\beta G\right]}{\partial \left(qp\right)}=1$

which could have been found by simply including factors of
$\sqrt{G}$
when evaluating the coefficient *F* before setting it equal to unity. The bit of extra work for an arbitrary power of the scaling function confirms that the square root is the most convenient choice in that it leads to a linear differential equation.

And now for the puzzle mentioned above. For a Lagrangian

${\stackrel{\xb7\xb7}{q}}_{i}=-\frac{1}{mq}\frac{dV}{dq}{q}_{i}$

which is exactly the same as the first form of the equation determining β. Since these are linear differential equations, the coordinates span the function space used to represent β. By standard existence theorems, the function β must be a linear superposition of the coordinates and its temporal derivative the same linear superposition of momenta:

$\beta ={c}_{k}{q}_{k}\phantom{\rule{5em}{0ex}}\stackrel{\xb7}{\beta}={c}_{k}{\stackrel{\xb7}{q}}_{k}=\frac{1}{m}{c}_{k}{p}_{k}$

It was pointed out above that a dot product with nondynamic constant vectors does not behave like a scalar with respect to angular momentum, so this solution will not produce the correct algebra under Poisson brackets. In fact, writing explicitly

${R}_{i}={c}_{k}{q}_{k}{p}_{i}-{c}_{k}{p}_{k}{q}_{i}={c}_{k}{L}_{ki}$

the Runge vector is in a definite sense a linear superposition of components of the angular momentum tensor. While this is true numerically, it will of course not lead to an extended algebra in this form but apparently only when the function β is expressed in terms of the dynamic variables *q*^{2}, *p*^{2} and *qp*)

*Uploaded 2012.03.25 — Updated 2014.12.30*
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