A previous presentation described a relationship in the context of the Schrödinger equation between the two power potentials

$r^k\leftrightarrow r^{\kappa }=r^{-2k/(k+2)}$

which essentially interchanges energy and coupling constant in transforming between the two potentials. It was pointed out at the end of that presentation that the same relationship holds for classical radial action.

This presentation extends that relationship to the classical evolution of the systems, demonstrating the usefulness of complex variables in a real-valued context. The main techniques involved are described concisely in the first appendix of a book by Arnol’d entitled Huygens and Barlow, Newton and Hooke.

For future reference, given a power-potential Hamiltonian of the form

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

the equations of motion in Cartesian coordinates are

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

where all variables here are real. The two equations can be combined in terms of a single complex radial variable as

$\stackrel{··}{r}=\stackrel{··}{x}+i\stackrel{··}{y}=-\frac{k\alpha }{m}|r{|}^{k-2}r$

Note that for potentials with negative exponents, the coupling constant $\alpha$ must be taken as negative to produce oscillatory motion. This means the combination $k\alpha$ is always positive.

Also note that using the polar form of any complex variable,

$\left|z^p\right|=\left|\right(r{e}^{i\phi }{)}^p|=|r^p{e}^{ip\phi }|=r^p=|z{|}^p$

that is, the absolute value of a power is the power of the absolute value.

Consider first the simplest and best-known example of this relationship: the connection between a two-dimensional simple harmonic oscillator and the Kepler problem. In terms of frequency, the equations of motion for the oscillator are

$\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x\frac{d^{2}y}{d{t}^2}=-{\omega }^{2}y$

For appropriate initial conditions, the solution to these two equations can be written compactly as

$r\left(t\right)=x+iy=acos\omega t+ibsin\omega t$

which is an ellipse in the complex plane with semimajor axis a and semiminor axis b. The center of the ellipse is located at the origin, and its foci and eccentricity are

$c=\pm \sqrt{a^2-b^2}e=\sqrt{1-\frac{b^2}{a^2}}$

The motion around the orbit, with the given temporal parametrization and a frequency of one, looks like this:

Now take a square of this solution, not an absolute value, converting squared functions to double-angle functions:

$\begin{array}{l}{r}^2=a^2{cos}^2\omega t-b^2{sin}^2\omega t+2iabcos\omega tsin\omega t\\ \phantom{r^2}=\frac{a^2}{2}({cos}^2\omega t-{sin}^2\omega t+1)-\frac{b^2}{2}({sin}^2\omega t-{cos}^2\omega t+1)\\ +iabsin2\omega t\\ r^2=\frac{a^2+b^2}{2}cos2\omega t+iabsin2\omega t+\frac{a^2-b^2}{2}\equiv R\end{array}$

This is again an ellipse in the complex plane, but with different semimajor and semiminor axes and shifted along the x-axis. To understand the significance of the latter, evaluate the positive focus:

$C=\sqrt{A^2-B^2}=\sqrt{(\frac{a^2+b^2}{2}{)}^2-a^2{b}^2}=\frac{a^2-b^2}{2}$

The ellipse is shifted to place the negative focus at the origin. Again with a frequency of one but retaining the doubled temporal behavior, here is how the new radial variable R looks in action:

The center of attraction is now clearly at a focus, which is precisely the case for Keplerian orbits. Squaring the ellipse of the simple harmonic oscillator gives the ellipse of a gravitational system!

A word of caution is in order concerning temporal parametrization at this stage of the presentation: it may not always reflect physical reality. The body moving on the second ellipse revolves at a constant rate, which is not how Kepler orbits behave. Each of the two parametrizations needs to be checked.

For the simple harmonic oscillator, one expects the temporal behavior of the angle of the radius to be

$tan\phi =\frac{y}{x}=\frac{bsin\omega t}{acos\omega t}=\frac{b}{a}tan\omega t$

That this is accurate can be verified by first evaluating the angular momentum of the oscillator:

$\begin{array}{l}\mathbf{r}=[acos\omega t,bsin\omega t]\\ \mathbf{p}=m\stackrel{·}{\mathbf{r}}=m\omega [-asin\omega t,bcos\omega t]\\ l=(\mathbf{r}\times \mathbf{p}{)}_z=r_x{p}_y-r_y{p}_x=m\omega ab\end{array}$

The equation defining angular momentum in terms of basic variables is

$l=m|r{|}^2\stackrel{·}{\phi }=m(a^2{cos}^2\omega t+b^2{sin}^2\omega t)\stackrel{·}{\phi }=m\omega ab$

where an absolute square of the radial variable is required this time. The solution to this equation is then

$\begin{array}{l}\phi =\omega ab\int \frac{dt}{a^2{cos}^2\omega t+b^2{sin}^2\omega t}\\ \phantom{\phi }=\omega ab\int \frac{dt}{a^2{cos}^2\omega t(1+\frac{b^2}{a^2}{tan}^2\omega t)}\\ \phi =\int \frac{d(\frac{b}{a}tan\omega t)}{1+\frac{b^2}{a^2}{tan}^2\omega t}={tan}^{-1}(\frac{b}{a}tan\omega t)\end{array}$

and the temporal behavior is confirmed as physically accurate.

Now repeat the process for the second ellipse. Working in squared circular functions for convenience, the angular momentum this time is

$\begin{array}{l}\mathbf{R}=[a^2{cos}^2\omega t-b^2{sin}^2\omega t,2abcos\omega tsin\omega t]\\ \mathbf{P}=m\stackrel{·}{\mathbf{R}}=2m\omega [-(a^2+b^2)cos\omega tsin\omega t,ab({cos}^2\omega t-{sin}^2\omega t\left)\right]\\ L=(\mathbf{R}\times \mathbf{P}{)}_z=R_x{P}_y-R_y{P}_x=2m\omega ab(a^2{cos}^2\omega t+b^2{sin}^2\omega t)\end{array}$

This is already a problem, since the angular momentum is not constant with the given temporal parametrization. Attempting to use this nonconstant value in the equation defining angular momentum gives

$\begin{array}{c}L=m|R{|}^2\stackrel{·}{\Phi }=m(a^2{cos}^2\omega t+b^2{sin}^2\omega t{)}^2\stackrel{·}{\Phi }\\ \stackrel{·}{\Phi }=\frac{2\omega ab}{a^2{cos}^2\omega t+b^2{sin}^2\omega t}\\ \Phi =2{tan}^{-1}(\frac{b}{a}tan\omega t)\end{array}$

which somewhat unexpectedly is the same as the first solution with an extra factor of two. Using a double-angle identity for the tangent then gives

$tan\Phi =\frac{2\frac{b}{a}tan\omega t}{1-\frac{b^2}{a^2}{tan}^2\omega t}=\frac{2abcos\omega tsin\omega t}{a^2{cos}^2\omega t-b^2{sin}^2\omega t}\equiv \frac{Y}{X}$

which oddly enough leads to the expected relationship to the Cartesian coordinates. The lack of constancy for the angular momentum, however, renders this relationship unphysical.

The requirement to keep angular momentum constant can be used to introduce a second temporal parametrization that is physically accurate. Write each statement of constancy as

$m|r{|}^2\frac{d\phi }{dt}=c_{r}2m|R{|}^2\frac{d\phi }{dT}=c_R$

where the second ellipse covers twice the angle as the first. Combine these two statements to eliminate the angular variable, while allowing an overall free constant for later use,

$\frac{dT}{dt}=c\frac{|R{|}^2}{|r{|}^2}=c|r{|}^2$

where $\left|r^2\right|=|r{|}^2=r\overline{r}$ . Now evaluate the second derivative of the squared variable with respect to this new temporal variable:

$\begin{array}{l}\frac{d^{2}R}{d{T}^2}=\frac{1}{c|r{|}^2}\frac{d}{dt}\frac{1}{c|r{|}^2}\frac{d}{dt}{r}^2\\ \phantom{\frac{d^{2}R}{d{T}^2}}=\frac{2}{c^2|r{|}^2}\frac{d}{dt}\frac{\stackrel{·}{r}}{\overline{r}}=\frac{2}{c^2|r{|}^2}(\frac{\stackrel{··}{r}}{\overline{r}}-\frac{\stackrel{·}{r}\stackrel{·}{\overline{r}}}{{\overline{r}}^2})\\ \phantom{\frac{d^{2}R}{d{T}^2}}=-\frac{2}{c^{2}r{\overline{r}}^3}\left({\omega }^2\right|r{|}^2+\left|\stackrel{·}{r}{|}^2\right)\\ \frac{d^{2}R}{d{T}^2}=-\frac{4r^2}{m{c}^2|r^2{|}^3}\left[\frac{m}{2}\left({\omega }^2\right|r{|}^2+\left|\stackrel{·}{r}{|}^2\right)\right]\end{array}$

The quantity in brackets is the energy of the simple harmonic oscillator, so the constant can be used to scale the right-hand side as one pleases. In terms of the squared variable one then has

$\frac{d^{2}R}{d{T}^2}=-\gamma \frac{R}{|R{|}^3}c=\sqrt{\frac{4E_{\mathrm{SHO}}}{m\gamma }}$

which is the differential equation for the Cartesian coordinates of the radial variable in a gravitational system. This was expected from the simple geometrical analysis above, but it is quite nice to see the details fall into place.

Lest one think that the Kepler problem is completely solved as an extension of the simple harmonic oscillator, one must now evaluate the explicit relationship between the two temporal variables,

$\begin{array}{l}\frac{dT}{dt}=c|r{|}^2=c(a^2{cos}^2\omega t+b^2{sin}^2\omega t)\\ \phantom{\frac{dT}{dt}}=c[\frac{a^2}{2}({cos}^2\omega t-{sin}^2\omega t+1)+\frac{b^2}{2}({sin}^2\omega t-co{s}^2\omega t+1)]\\ \frac{dT}{dt}=c[\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2}cos2\omega t]\end{array}$

which has the solution

$T\left(t\right)=c[At+\frac{C}{2\omega }sin2\omega t]=\frac{cA}{2\omega }[2\omega t+Esin2\omega t]$

This is essentially the Kepler equation, apart from a scaling of the temporal variable. The sign on the sine function can be adjusted to the usual negative by shifting this same variable, with the modification to the linear term absorbed into a constant of integration.

The Kepler equation cannot be inverted in practice in terms of known special functions, so that its appearance means that this relationship between the two potentials cannot be used to solve the Kepler problem any more simply than by methods already available. Schade!

While the simple harmonic oscillator is easily soluble, most power potentials are not. In these cases on has to perform a direct integration of pairs of differential equations and compare the results. The critical step in this process is choosing correct initial conditions for the second set of equations compared to the first. Consider the process first for the known exact solution.

For the simple harmonic oscillator, the initial conditions being used are

$x\left(0\right)=a\stackrel{·}{x}\left(0\right)=0y\left(0\right)=0\stackrel{·}{y}\left(0\right)=\omega b$

Two of the conditions for the corresponding Kepler potential come directly from the squared ellipse,

$X\left(0\right)=a^{2}Y\left(0\right)=0$

but care must be taken with the final two conditions, since there are two different time scales in the integration:

$\frac{dR}{dT}=\frac{dt}{dT}\frac{dR}{dt}=\frac{1}{c|r{|}^2}\frac{dR}{dt}=\sqrt{\frac{m\gamma }{4E_{\mathrm{SHO}}}}\frac{1}{|r{|}^2}\frac{dR}{dt}$

This leads to the final two initial conditions

$\stackrel{·}{X}\left(0\right)=0\stackrel{·}{Y}\left(0\right)=\sqrt{\frac{2\gamma }{a^2+b^2}}\frac{b}{a}$

The domain of integration for the second equation must also take account of the separate time scale and its own doubled frequency. The endpoint of the second integration is thus

$T\left(\frac{\pi }{\omega }\right)=\frac{\pi }{\omega }cA=\frac{\pi }{\sqrt{2\gamma }}(a^2+b^2{)}^{3/2}$

Here is the integration of the differential equations for the two potentials, with the option to square that of the simple harmonic oscillator to verify the comparison. This graphic takes $\omega =\gamma =1$ for convenience, and limits the range of input variables to avoid accumulated error in the integrations:

The square of the first integration matches the second integration as expected. Pretty cool!

To extend the relationship between power potentials, assume one has a solution $r\left(t\right)$ for one potential and consider a more general power $R=r^p$ between radial variables. The differential equation linking the two statements of constancy of angular momentum becomes

$\frac{dT}{dt}=c\frac{|R{|}^2}{|r{|}^2}=c|r{|}^{2p-2}$

Again evaluate the second derivative of the squared variable with respect to this new temporal variable:

$\begin{array}{l}\frac{d^{2}R}{d{T}^2}=\frac{1}{c|r{|}^{2p-2}}\frac{d}{dt}\frac{1}{c|r{|}^{2p-2}}\frac{d}{dt}{r}^p\\ \phantom{\frac{d^{2}R}{d{T}^2}}=\frac{p}{c^2|r{|}^{2p-2}}\frac{d}{dt}\frac{\stackrel{·}{r}}{{\overline{r}}^{p-1}}\\ \phantom{\frac{d^{2}R}{d{T}^2}}=\frac{p}{c^2|r{|}^{2p-2}}[\frac{\stackrel{··}{r}}{{\overline{r}}^{p-1}}-(p-1)\frac{\stackrel{·}{r}\stackrel{·}{\overline{r}}}{{\overline{r}}^p}]\\ \phantom{\frac{d^{2}R}{d{T}^2}}=-\frac{p}{c^2{r}^{p-1}{\overline{r}}^{2p-1}}[\frac{k\alpha }{m}|r{|}^k+(p-1)|\stackrel{·}{r}{|}^2]\\ \frac{d^{2}R}{d{T}^2}=-\frac{2p(p-1)}{m{c}^2}\frac{r^p}{|r^p{|}^{(4p-2)/p}}[\frac{k\alpha }{2(p-1)}|r{|}^k+\frac{m}{2}|\stackrel{·}{r}{|}^2]\end{array}$

The quantity in brackets is the energy in the first potential if one chooses

$2(p-1)=k\to p=\frac{k+2}{2}$

Again scaling the right-hand side with the constant, in terms of the squared variable one then has

$\frac{d^{2}R}{d{T}^2}=-\gamma \frac{R}{|R{|}^{4(k+1)/(k+2)}}c=\sqrt{\frac{k(k+2)E_k}{2m\gamma }}$

Referencing the differential equations near the beginning of this presentation, one should have

$\kappa -2=-\frac{4(k+1)}{k+2}\to \kappa =-\frac{2k}{k+2}$

which is precisely the relationship given at the outset. Thus if one has a solution to a potential with the exponent k as a complex variable and raises it to the power p, one will have the corresponding solution for a potential with exponent κ. Stupendous!

Note that the choice for p corresponds exactly to the exponent chosen in the previous presentation. Apparently the scaling of classical variables matches the scaling of quantum mechanical variables.

For the given choice of the power of the radial variable, the equation relating the two temporal variables becomes

$\frac{dT}{dt}=c|r{|}^k$

which curiously enough is proportional to the first potential. The power of the radial mapping and this power of the potential are only equal for $p=k=2$ , the simple harmonic oscillator.

In order to present comparisons of integrated differential equations, one must determine appropriate corresponding initial conditions. To match the previous graphic as much as possible, for the first potential with positive exponent choose

$x\left(0\right)=a\stackrel{·}{x}\left(0\right)=0y\left(0\right)=0\stackrel{·}{y}\left(0\right)=b$

since the frequency was set to unity for convenience. Two initial conditions for the second potential are clearly then

$X\left(0\right)=a^{(k+\mathrm{2<mo>)</mo>\; <mo>/</mo><mn>2</mn>}}Y\left(0\right)=0$

The connection between the two temporal variables is now

$\frac{dR}{dT}=\frac{dt}{dT}\frac{dR}{dt}=\frac{1}{c|r{|}^k}\frac{dR}{dt}=\sqrt{\frac{2m\gamma }{k(k+2)E_k}}\frac{1}{|r{|}^k}\frac{dR}{dt}$

If one approximates the initial temporal behavior of the first radial variable as $r\approx a+ibt$ , then one can take

$\frac{dR}{dt}\approx \frac{d}{dt}(a+ibt{)}^{(k+2)/2}=\frac{k+2}{2}b{a}^{k/2}+O\left(t\right)$

ignoring the complex unit in the spirit of the entire presentation. The final two initial conditions are

$\stackrel{·}{X}\left(0\right)=0\stackrel{·}{Y}\left(0\right)=\sqrt{\frac{k+2}{k}\frac{\gamma }{b^2+\frac{2\alpha }{m}{a}^k}}\frac{b}{a^{k/2}}$

Taking $\frac{2\alpha }{m}=1$ and $\gamma =1$ will then produce a result consistent with the previous graphic. Comparing the transformed equation here to the general equations at the outset indicates that $\gamma =\frac{\kappa \alpha }{m}$ , so that the extra physical factors are being absorbed into the temporal variable.

One could at this point calculate the period of each motion from closest approach to the center of force to farthest approach and use multiples of these values as domains of integration. Since the focus of this presentation is a comparison of the orbit paths, not their complete temporal evolution, instead just use $2\pi$ for the domain of the first potential and $\frac{\pi }{\sqrt{2\gamma }}(a^k+b^2{)}^{3/2}$ for the domain of the second potential. This allows enough difference between the two to see the comparison a bit more clearly.

There is one final twist to the comparison: fractional powers of a complex variable have multiple sheets. This can be understood from the polar form of any complex number,

$z^p=(r{e}^{i\phi }{)}^p=(r{e}^{i\phi +2n\pi i}{)}^p=e^{2np\pi i}{z}^p$

since the exponential function is unchanged by the additional term the exponent. This means that on passing to a different complex sheet on needs an extra factor of $e^{2p\pi i}$ : this is exactly the plus and minus signs of a square root generalized to an arbitrary exponent. Different sheets begin along the branch cut on the negative real axis, and since this has nothing to do with the physics one can just run through the integrated data and determine when the imaginary part of the solution to the differential equation becomes negative.

And here we go! This is how the two integrated systems appear, with the option to raise the first to the appropriate power:

The power of the solution in the first potential matches the solution in its corresponding second potential as expected. After all of the math, seeing it in action is still quite something.

While the orbits in general are not closed, one can still see a distinct difference between the two cases: the orbits for positive exponents are centered around the origin, while those for negative exponents are centered around what corresponds to a focus of a closed orbit. Taking the appropriate complex power of the first clearly shifts its center from the origin to an elliptic focus. This shifting of the center of the orbit is also clearly part and parcel of the relationship at the outset between pairs of exponents.

As a final note, it is interesting to compare the simple complex power here to a transformation group introduced by Lynden-Bell in the context of approximations to power potential orbits. His approach relates derivatives of radial variables but reaches the same relationship between pairs of potential exponents, albeit with extra multiplicative factors in the transformation (although this occurs in the quantum mechanical context as well). He also relates radial variables using the same power found here, but in his context the variables are purely real. Curiouser and curiouser...

Uploaded 2022.10.15 analyticphysics.com