Consider a two-dimensional power-potential Hamiltonian in polar coordinates:

$H=\frac{{p}_{r}^{2}}{2m}+\frac{{p}_{\phi}^{2}}{2m{r}^{2}}+\alpha {r}^{k}$

Hamilton-Jacobi theory determines the classical action from the first-order nonlinear differential equation

$H(r,\phi ;\frac{\partial S}{\partial r},\frac{\partial S}{\partial \phi})+\frac{\partial S}{\partial t}=0$

where linear momenta are replaced by partial derivatives of the action with respect to their conjugate coordinates. The equation for the system under consideration is

$\frac{1}{2m}(\frac{\partial S}{\partial r}{)}^{2}+\frac{1}{2m{r}^{2}}(\frac{\partial S}{\partial \phi}{)}^{2}+\alpha {r}^{k}+\frac{\partial S}{\partial t}=0$

Since the potential is a function of the radial variable only, the equation is separable if the action is a linear sum of the form

$S=L\phi -Et+f\left(r\right)$

where the coefficients of the first two terms are chosen for later physical significance. With this functional form the Hamilton-Jacobi equation becomes

$\frac{1}{2m}(\frac{\partial f}{\partial r}{)}^{2}+\frac{{L}^{2}}{2m{r}^{2}}+\alpha {r}^{k}=E$

Solving for the partial derivative, the action can be written

$S=L\phi -Et+\int dr\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}$

containing what is called the radial action integral.

The Hamilton-Jacobi equation is derived by treating the action as a generating function that induces a transformation of the Hamiltonian to generalized coordinates in which it is identically zero. If the Hamiltonian is identically zero,

${\stackrel{\xb7}{Q}}_{i}=\frac{\partial H}{\partial {P}_{i}}\equiv 0\phantom{\rule{5em}{0ex}}{\stackrel{\xb7}{P}}_{i}=-\frac{\partial H}{\partial {Q}_{i}}\equiv 0$

then the new generalized coordinates are all constant. The new momenta are traditionally designated with alphas and can be taken as

${P}_{1}={\alpha}_{1}=E\phantom{\rule{5em}{0ex}}{P}_{2}={\alpha}_{2}=L$

The new conjugate coordinates, traditionally designated with betas, are given by partial derivatives of the action:

${Q}_{i}={\beta}_{i}=\frac{\partial S}{\partial {\alpha}_{i}}$

For the system at hand, the first conjugate coordinate is

${\beta}_{1}=\frac{\partial S}{\partial E}=-t+m\int \frac{dr}{\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}$

This constant is just an offset of the temporal variable and can be taken as zero. This gives an integral relation between two variables:

$t=m\int \frac{dr}{\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}$

Taking a derivative provides the differential relation between the two variables:

$\frac{dr}{dt}=\frac{1}{m}\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}$

Since this relation can also be derived by setting ${p}_{r}=m\stackrel{\xb7}{r}$ and ${p}_{\phi}=L$ in the Hamiltonian, it can be designated the energy equation for this system. The temporal variable is thus determined by what can be called the energy integral.

The second conjugate coordinate is

${\beta}_{2}=\frac{\partial S}{\partial L}=\phi -L\int \frac{dr}{{r}^{2}\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}$

This constant offset can again be taken as zero, giving an integral relation between two different variables:

$\phi =L\int \frac{dr}{{r}^{2}\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}$

Taking a derivative provides the differential relation between these two variables:

$\frac{dr}{d\phi}=\frac{{r}^{2}}{L}\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}$

Applying the differential relation between radial and temporal variables to the previous integral and taking a derivative, one has

$\phi =L\int \frac{dt}{m{r}^{2}}\phantom{\rule{3em}{0ex}}\to \phantom{\rule{3em}{0ex}}m{r}^{2}\stackrel{\xb7}{\phi}=L$

which is simply the definition of angular momentum. The angular variable is thus determined by what can be called the angular momentum integral.

If the radial variable is expressed as a function of angular variable, then the area swept out by the orbit is

$A=\int r\phantom{\rule{.2em}{0ex}}drd\phi =\frac{1}{2}\int {r}^{2}\left(\phi \right)d\phi $

Knowing the differential relation to the two variables in the integral allows this to be written

$A=\frac{L}{2}\int \frac{dr}{\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}=\frac{Lt}{2m}$

via the energy integral. Kepler’s law concerning equal areas is thus a simple consequence of angular momentum conservation for central forces.

Comparing the forms of the energy and angular momentum integrals, it is simple to confirm that

$\frac{\partial}{\partial L}t\left(r\right)=-\frac{\partial}{\partial E}\phi \left(r\right)$

This is a statement of the behavior of orbits under a change in initial conditions. It clearly does not depend upon whether the integrals are restricted to integral periods of the orbits, but holds in general for equal upper endpoints.

A similar statement was introduced by Hénon in the context of orbit periods, but again can be interpreted more generally. It can be motivated by considering the Hamiltonian in Cartesian coordinates,

$H=\frac{{p}_{x}^{2}}{2m}+\frac{{p}_{y}^{2}}{2m}+\alpha {r}^{k}=E$

for which the equations of motion are

$\begin{array}{l}\stackrel{\xb7\xb7}{x}=\frac{{\stackrel{\xb7}{p}}_{x}}{m}=-\frac{1}{m}\frac{\partial H}{\partial x}=-\frac{k\alpha}{m}{r}^{k-2}x\\ \stackrel{\xb7\xb7}{y}=\frac{{\stackrel{\xb7}{p}}_{y}}{m}=-\frac{1}{m}\frac{\partial H}{\partial y}=-\frac{k\alpha}{m}{r}^{k-2}y\end{array}$

If spatial variables are scaled by some constant linear factor $\lambda $ , then the equations of motion remain unchanged if the temporal variable is simultaneously scaled by ${\lambda}^{1-k/2}$ . Under this transformation both energy and angular momentum acquire scaling factors as well:

$r\to \lambda r\phantom{\rule{3em}{0ex}}t\to {\lambda}^{1-k/2}t\phantom{\rule{3em}{0ex}}E\to {\lambda}^{k}E\phantom{\rule{3em}{0ex}}L\to {\lambda}^{1+k/2}L$

Applying this transformation to the integral determining the angular variable produces no change, which is to be expected for a dimensionless variable. Thus one can write

$\phi ({\lambda}^{k}E,{\lambda}^{1+k/2}L)=\phi (E,L)$

Now take a derivative of this equation with respect to the scaling parameter and then set the parameter equal to unity,

$kE\frac{\partial}{\partial E}\phi \left(r\right)+\frac{k+2}{2}L\frac{\partial}{\partial L}\phi \left(r\right)=0$

where the derivatives with respect to each slot on the left-hand side are effectively with respect to physical constants. Hénon then combines this statement with the previous one for the result

$\frac{\partial}{\partial L}t\left(r\right)=\frac{k+2}{2k}\frac{L}{E}\frac{\partial}{\partial L}\phi \left(r\right)$

If one assigns a particular energy to the system, then this last equation can be interpreted as a statement that the temporal and angular variable have simultaneously extrema as a function of angular momentum, which was the focus of Hénon. Again, the statement is more general than that, holding over the entire orbit.

It is perhaps more interesting to write an equation that does not include the temporal variable,

$L\frac{\partial}{\partial L}\phi \left(r\right)=-\frac{2k}{k+2}E\frac{\partial}{\partial E}\phi \left(r\right)$

since now the derivative combinations on each side indicate the dimensionality of the angular variable with respect to each of the physical constants. There is also an interesting factor on the right-hand side that appears in multiple contexts for power potential orbits.

A previous presentation explored the effect of the conformal mapping $R={r}^{(k+2)/2}$ on power potential orbits. The main result was to show that orbits in a potential proportional to ${r}^{k}$ become orbits in a potential proportional to ${r}^{-2k/(k+2)}$ under the mapping. The exponent of this second potential is precisely the factor on the right-hand side of the equation directly above.

Now consider transformations of the integrals in this presentation. Treating the conformal mapping as a change of real variable, the action integral becomes

$\int dr\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}=\frac{2}{k+2}\int dR\sqrt{2mE{R}^{-2k/(k+2)}-2m\alpha -\frac{{L}^{2}}{{R}^{2}}}$

and the exponent appears again, along with an interchange of energy and coupling constant. Applying the same change of variable to the integral determining the angular variable,

$\int \frac{dr}{{r}^{2}\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}=\frac{2}{k+2}\int \frac{dR}{{R}^{2}\sqrt{2mE{R}^{-2k/(k+2)}-2m\alpha -\frac{{L}^{2}}{{R}^{2}}}}$

and the same exponent and interchange appear again.

In both cases the change of variable leaves the basic form of the integral unchanged apart from the interchange noted. This will not be true of the energy integral and it is interesting to understand why. Leaving the basic form of the integral intact arises from a balance between the potential and angular momentum terms in the integrand. For an arbitrary power the transformation of these two terms is

$r={R}^{p}\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}2m\alpha {r}^{k}+\frac{{L}^{2}}{{r}^{2}}=2m\alpha {R}^{kp}+\frac{{L}^{2}}{{R}^{2}}{R}^{2-2p}$

For the integrand to behave as above, one must have

$kp=2-2p\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}p=\frac{2}{k+2}$

and the (inverse) power of the conformal mapping appears. Whether or not the integral is then unchanged depends on the power of the initial radial variable outside the square root. If this is some arbitrary power, then

${r}^{n}dr\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}\frac{2}{k+2}{R}^{(2n-k)/(k+2)}dR$

For integrals with the square root in the numerator, the general behavior under the change of variable is

$\begin{array}{l}\int {r}^{n}dr\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}\\ \phantom{\rule{5em}{0ex}}=\frac{2}{k+2}\int {R}^{2n/(k+2)}dR\sqrt{2mE{R}^{-2k/(k+2)}-2m\alpha -\frac{{L}^{2}}{{R}^{2}}}\end{array}$

so that only the action integral with $n=0$ remains unchanged in general form, apart from the interchange. For integrals with the square root in the denominator,

$\int \frac{{r}^{n}dr}{\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}=\frac{2}{k+2}\int \frac{{R}^{2(n-k)/(k+2)}dR}{\sqrt{2mE{R}^{-2k/(k+2)}-2m\alpha -\frac{{L}^{2}}{{R}^{2}}}}$

and now only the angular momentum integral with $n=-2$ remains unchanged. The energy integral also has $n=0$ , so that

$\int \frac{dr}{\sqrt{2mE-2m\alpha {r}^{k}-\frac{{L}^{2}}{{r}^{2}}}}=\frac{2}{k+2}\int \frac{{R}^{-2k/(k+2)}dR}{\sqrt{2mE{R}^{-2k/(k+2)}-2m\alpha -\frac{{L}^{2}}{{R}^{2}}}}$

and curiously the exponent of the transformed potential appears in the numerator.

*Uploaded 2022.10.20*
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