Consider an additively separable spherically symmetric *n*-dimensional Hamiltonian

$H(r,p)=T\left(p\right)+V\left(r\right)$

that is a function of the squared dynamic variables

$r=\sqrt{{r}_{k}{r}_{k}}\phantom{\rule{8em}{0ex}}p=\sqrt{{p}_{k}{p}_{k}}$

with repeated indices summed over the arbitrary number of dimensions here and in the sequel. This form describes relativistic and nonrelativistic classical motion apart from electrodynamics. The equations of motion for the dynamic variables are

${\stackrel{\xb7}{r}}_{i}=\frac{\partial H}{\partial {p}_{i}}=\frac{dT}{dp}\frac{{p}_{i}}{p}\phantom{\rule{4em}{0ex}}{\stackrel{\xb7}{p}}_{i}=-\frac{\partial H}{\partial {r}_{i}}=-\frac{dV}{dr}\frac{{r}_{i}}{r}$

With these equations, each component of the *n*-dimensional angular momentum tensor

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

is easily seen to be a dynamic constant due to its skew structure:

$\frac{d}{dt}{L}_{ij}=\frac{dT}{dp}\frac{{p}_{i}{p}_{j}-{p}_{j}{p}_{i}}{p}-\frac{dV}{dr}\frac{{r}_{i}{r}_{j}-{r}_{j}{r}_{i}}{r}=0$

One can also construct a general dynamically constant vector of the form

${R}_{i}=A(r,p){r}_{i}-B(r,p){p}_{i}$

The inclusion of both dynamic vectors is necessary to allow balancing of the coefficients of these independent variables under temporal differentiation,

$\frac{d}{dt}{R}_{i}=[\frac{B}{r}\frac{dV}{dr}+\frac{dA}{dt}]{r}_{i}+[\frac{A}{p}\frac{dT}{dp}-\frac{dB}{dt}]{p}_{i}\equiv 0$

which implies the coupled differential equations

$\frac{dA}{dt}=-\frac{B}{r}\frac{dV}{dr}\phantom{\rule{4em}{0ex}}\frac{dB}{dt}=\frac{A}{p}\frac{dT}{dp}$

The existence of energy as a constant of the motion implies that for the purposes of integrating equations of motion either variable can be considered a function of the other,

$r={V}^{-1}[E-T\left(p\right)]\phantom{\rule{4em}{0ex}}p={T}^{-1}[E-V\left(r\right)]$

since both variables are implicit functions of time. This allows choosing the intermediate variable that allows cancellation of the explicit derivatives of the kinetic and potential energies. With the temporal derivatives

$\begin{array}{l}\frac{dr}{dt}=\frac{{r}_{k}}{r}{\stackrel{\xb7}{r}}_{k}=\frac{\left(rp\right)}{rp}\frac{dT}{dp}\\ \frac{dp}{dt}=\frac{{p}_{k}}{p}{\stackrel{\xb7}{p}}_{k}=-\frac{\left(rp\right)}{rp}\frac{dV}{dr}\end{array}\phantom{\rule{4em}{0ex}}\left(rp\right)\equiv {r}_{k}{p}_{k}$

the coupled differential equations for the coefficient functions *A* and *B* can be written

$\frac{dA}{dp}=\frac{p}{\left(rp\right)}B\phantom{\rule{5em}{0ex}}\frac{dB}{dr}=\frac{r}{\left(rp\right)}A$

where the dot product of the two dynamic variables is related to their squares through the square of the angular momentum tensor:

${L}^{2}=\frac{1}{2}{L}_{kl}{L}_{kl}={r}^{2}{p}^{2}-(rp{)}^{2}$

Forming the square of the dynamically constant vector

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

one can solve for either coefficient function,

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

which then allows decoupling of the differential equations:

$\frac{dA}{dp}=\frac{1}{p\left(rp\right)}[\left(rp\right)A\pm \sqrt{{p}^{2}{R}^{2}-{L}^{2}{A}^{2}}]\phantom{\rule{3em}{0ex}}\frac{dB}{dr}=\frac{1}{r\left(rp\right)}[\left(rp\right)B\pm \sqrt{{r}^{2}{R}^{2}-{L}^{2}{B}^{2}}]$

with the substitutions $A=p\mathcal{A}$ and $B=r\mathcal{B}$ these simplify to

$\frac{d\mathcal{A}}{dp}=\pm \frac{1}{p\left(rp\right)}\sqrt{{R}^{2}-{L}^{2}{\mathcal{A}}^{2}}\phantom{\rule{5em}{0ex}}\frac{d\mathcal{B}}{dr}=\pm \frac{1}{r\left(rp\right)}\sqrt{{R}^{2}-{L}^{2}{\mathcal{B}}^{2}}$

These equations are easily integrated to

$\mathcal{A}=\{\begin{array}{c}\frac{R}{L}sin\int \frac{L\phantom{\rule{.2em}{0ex}}dp}{p\left(rp\right)}\\ \frac{R}{L}cos\int \frac{L\phantom{\rule{.2em}{0ex}}dp}{p\left(rp\right)}\end{array}\phantom{\rule{4em}{0ex}}\mathcal{B}=\{\begin{array}{c}\frac{R}{L}sin\int \frac{L\phantom{\rule{.2em}{0ex}}dr}{r\left(rp\right)}\\ \frac{R}{L}cos\int \frac{L\phantom{\rule{.2em}{0ex}}dr}{r\left(rp\right)}\end{array}$

Since the sum of an inverse sine and cosine is identically equal to $\frac{\pi}{2}$, their derivatives have the opposite signs appearing here before integration.

Correlation of circular functions in the final dynamically constant vectors is determined by the coupled differential equations along with identities between the integrals appearing in the arguments of the circular functions. Designate these integrals as

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

where the dot product of the dynamic vectors has been replaced using the square of angular momentum. Since these integrals arise from equations of motion, angular momentum is to be treated as a constant with respect to integration. With the substitution $u=\frac{rp}{L}$ one has

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

Addition identities for circular functions give

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

Now write the coupled differential equation for the coefficient function *A* and insert the two options for coefficient function *B*:

$\begin{array}{l}A=\frac{\left(rp\right)}{r}\frac{dB}{dr}\\ \phantom{A}=\frac{R}{L}\frac{\left(rp\right)}{r}\frac{d}{dr}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}rsin{I}_{r}\\ rcos{I}_{r}\end{array}\\ \phantom{A}=\frac{R}{L}\frac{\left(rp\right)}{r}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}sin{I}_{r}+\frac{L}{\left(rp\right)}cos{I}_{r}\\ cos{I}_{r}-\frac{L}{\left(rp\right)}sin{I}_{r}\end{array}\\ A=\frac{R}{L}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}-psin{I}_{p}\\ pcos{I}_{p}\end{array}\end{array}$

Repeat with the coupled differential equation for the coefficient function *B* and the two options for coefficient function *A*:

$\begin{array}{l}B=\frac{\left(rp\right)}{p}\frac{dA}{dp}\\ \phantom{B}=\frac{R}{L}\frac{\left(rp\right)}{p}\frac{d}{dp}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}psin{I}_{p}\\ pcos{I}_{p}\end{array}\\ \phantom{B}=\frac{R}{L}\frac{\left(rp\right)}{p}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}sin{I}_{p}+\frac{L}{\left(rp\right)}cos{I}_{p}\\ cos{I}_{p}-\frac{L}{\left(rp\right)}sin{I}_{p}\end{array}\\ B=\frac{R}{L}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}-rsin{I}_{r}\\ rcos{I}_{r}\end{array}\end{array}$

In both cases, cosines correlate with the same sign and sines with a change in sign. With the inclusion of the sign in the original form, the two choices for the dynamically constant vector are

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

These two vectors correspond in three dimensions to the Runge-Lenz vector and its cross product with the angular momentum vector. In an arbitrary number of dimensions, the quantities in braces are still dynamically constant orthogonal vectors in an invariant hyperplane. To demonstrate orthogonality first rearrange the identities between integrals:

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

It is now simple to show that

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

It is as 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 the following are dynamically constant unit vectors in the invariant plane:

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

The general dynamically constant vector, or generalized Runge vector, enlarges the group symmetry *SO*(*n*)*SO*(*n*+1)*R* and *L* have been introduced as constants of integration in the solution of equations of motion, but when evaluating Poisson brackets

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

they must be treated as functions of the dynamic variables *r*^{2}, *p*^{2} and (*rp*). The Poisson bracket of components of the angular momentum tensor is

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

Coordinates and momenta are vectors in this *n*-dimensional space, and their brackets with components of the angular momentum tensor 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 bracket of any component of the angular momentum tensor with a square of a dynamic vector or a dot product of two dynamic vectors is always zero:

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

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,

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

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

The identically zero bracket of a general scalar with a component of the angular momentum tensor is evaluated as

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

When the individual derivatives are applied to the spherically 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 identically zero. Use of these effective derivatives is unambiguous as long as all dynamic quantities are written explicitly in terms of coordinates and momenta. These effective variables are a bookkeeping method for evaluating Poisson brackets.

Writing the general dynamically constant vector as a function of effective dynamic variables,

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

its bracket with components of angular momentum corresponds to those of the coordinate and momentum vectors,

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

so that it behaves as a vector as needed for the algebra of the group *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 on spherically 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

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

Knowing the relation of the square of angular momentum to the dynamics variables, 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 general dynamically constant vector, this determinant is not identically zero.

The determinant can be simplified appreciably by appealing to the definition of a dynamic constant as a combination of dynamic variables whose bracket with the Hamiltonian is identically zero. That bracket is

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

where the two intermediate brackets are

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

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

Since the two dynamic variables *r _{i}* and

$\frac{\frac{1}{r}\frac{dV}{dr}}{\frac{1}{p}\frac{dT}{dp}}=\frac{2\left(rp\right)\frac{\partial A}{\partial {r}^{2}}+{p}^{2}\frac{\partial A}{\partial \left(rp\right)}}{2\left(rp\right)\frac{\partial A}{\partial {p}^{2}}+{r}^{2}\frac{\partial A}{\partial \left(rp\right)}-B}=\frac{2\left(rp\right)\frac{\partial B}{\partial {r}^{2}}+{p}^{2}\frac{\partial B}{\partial \left(rp\right)}-A}{2\left(rp\right)\frac{\partial B}{\partial {p}^{2}}+{r}^{2}\frac{\partial B}{\partial \left(rp\right)}}$

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 components of the general dynamically constant vector is

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

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

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

and then the bracket of vector components 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 basis for the identities above between the integrals *I _{r}* and

$\frac{\mathbf{r}\xb7(-\mathbf{R})}{rR}=-\frac{1}{L}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}rpcos{I}_{p}-\left(rp\right)cos{I}_{r}\\ rpsin{I}_{p}+\left(rp\right)sin{I}_{r}\end{array}=\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}sin{I}_{r}\\ cos{I}_{r}\end{array}$

and the corresponding projection of the momentum vector on the same is

$\frac{\mathbf{p}\xb7(-\mathbf{R})}{pR}=-\frac{1}{L}\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}\left(rp\right)cos{I}_{p}-rpcos{I}_{r}\\ \left(rp\right)sin{I}_{p}+rpsin{I}_{r}\end{array}=\{\phantom{\rule{.5em}{0ex}}\begin{array}{c}-sin{I}_{p}\\ cos{I}_{p}\end{array}$

The negative sign in the projection of the momentum vector ensures that the general dynamic vector remains antisymmetric when expressed in terms of projected quantities. Up to an overall sign, the two forms of the vector can now be written in the unified form

$\mathbf{R}\sim \frac{1}{L}\left[\right(\mathbf{p}\xb7\mathbf{R})\mathbf{r}-(\mathbf{r}\xb7\mathbf{R}\left)\mathbf{p}\right]$

In three dimensions the right-hand side is a vector triple product of **r**, **p** and **R**, which is to be expected for a vector in the invariant plane. The origin of the identities between integrals is apparent from this expression, apart from signs.

From the unified form it is clear that the vector is numerically equal to a linear superposition of angular momentum tensor components. Numerically equal is not the same, however, as dynamically equal from the point of view of Poisson brackets. If the vector is provisionally written

${R}_{i}\stackrel{?}{=}{c}_{k}{L}_{ki}\phantom{\rule{5em}{0ex}}{c}^{2}={c}_{k}{c}_{k}=1$

with normalized nondynamic constants as superposition coefficients, the bracket of two components is

$[{R}_{i},{R}_{j}]\stackrel{?}{=}{c}_{k}{c}_{l}[{L}_{ki},{L}_{lj}]\stackrel{?}{=}{L}_{ij}+{c}_{j}{R}_{i}-{c}_{i}{R}_{j}+{\delta}_{ij}{c}_{k}{R}_{k}$

which is clearly not the desired result. The superposition coefficients must themselves be functions of the dynamic variables: this is achieved with the integrals.

In summary, the angular momentum tensor

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

and either choice for the scaled general dynamically constant 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

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

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

This presentation is a summary, generalization and extension of the following presentations appearing on this website:

A Generalized Runge Vector for Spherically Symmetric Potentials

*SO*(*n*+1) and the Runge Vector

*SO*(*n*+1) and the Runge Vector Redux

The Runge Vector and Angular Momentum

As a final note, the motiviation for starting with an additively separable Hamiltonian, beyond the fact that that is what is used in practice, is to ensure that the energy equation can be solved for either dynamic variable. If solvability is assumed then the Hamiltonian need not even be separable for the entire discussion to remain valid: it must then merely be a function of the squared dynamic variables but not their dot product.

*Uploaded 2015.01.19*
analyticphysics.com