In classical mechanics one typically specifies a potential of interaction and solves the Lagrange or Hamilton equations for the orbits determined by the potential. The process is easily reversed for planar orbits: given an invertible polar graph relating radial distance to an angular variable, one can always find the corresponding spherically symmetric potential that generates that orbit.

Begin with a spherically symmetric Lagrangian in two dimensions:

$ℒ=\frac{1}{2}m({\stackrel{·}{x}}^2+{\stackrel{·}{y}}^2)-V\left(r\right)=\frac{1}{2}m({\stackrel{·}{r}}^2+r^2{\stackrel{·}{\phi }}^2)-V\left(r\right)$

The method holds more generally for an arbitrary number of dimensions greater than or equal to two, but two is sufficient for orbit calculations in an invariant plane. Since the angular variable does not appear in the Lagrangian, its corresponding conjugate momentum is conserved:

$p_{\phi }=\frac{\partial ℒ}{\partial \stackrel{·}{\phi }}=m{r}^2\stackrel{·}{\phi }=\mathrm{constant}=L$

This is the statement of conservation of angular momentum for the spherically symmetric system. It can be used to determine the temporal derivatives of the angular variable in terms of the radial variable:

$\stackrel{·}{\phi }=\frac{L}{m{r}^2}\stackrel{··}{\phi }=-\frac{2L}{m{r}^3}\stackrel{·}{r}$

The Lagrange equation for the radial variable is

$\begin{array}{c}\frac{d}{dt}\frac{\partial ℒ}{\partial \stackrel{·}{r}}=\frac{\partial ℒ}{\partial r}\\ m\stackrel{··}{r}=mr{\stackrel{·}{\phi }}^2-\frac{dV}{dr}=\frac{L^2}{m{r}^3}-\frac{dV}{dr}\end{array}$

where the statement of angular momentum conservation has been used to write the right-hand side of the equation entirely in terms of the radial variable.

Now let the orbiting body follow an arbitrary invertible curve in polar coordinates:

$r=f\left(\phi \right)\leftrightarrow \phi =f^{-1}\left(r\right)$

Noting that $\stackrel{·}{r}=\stackrel{·}{f}$ by definition, evaluate the second temporal derivative of the orbit:

$\begin{array}{l}\stackrel{··}{r}=\frac{d}{dt}\left[\frac{df}{d\phi }\stackrel{·}{\phi }\right]\\ \phantom{\stackrel{··}{r}}=\frac{d^{2}f}{d{\phi }^2}{\stackrel{·}{\phi }}^2+\frac{df}{d\phi }\stackrel{··}{\phi }\\ \phantom{\stackrel{··}{r}}=\frac{d^{2}f}{d{\phi }^2}{\stackrel{·}{\phi }}^2-\frac{df}{d\phi }\frac{2L}{m{f}^3}\left[\frac{df}{d\phi }\stackrel{·}{\phi }\right]\\ \stackrel{··}{r}=\frac{L^2}{m^2{f}^4}[\frac{d^{2}f}{d{\phi }^2}-\frac{2}{f}(\frac{df}{d\phi }{)}^2]\end{array}$

Putting this into the Lagrange equation for the radial variable, one has

$\frac{dV}{dr}=-\frac{L^2}{m{f}^4}[\frac{d^{2}f}{d{\phi }^2}-\frac{2}{f}(\frac{df}{d\phi }{)}^2-f]$

While the two symbols r and f are interchangeable, it is useful to keep each side of this equation in separate variable labels to make the method explicit. The derivatives on the right-hand side are evaluated with respect to the angular variable, which after simplification is replaced with the radial variable. One more integration then gives the potential that generates the arbitrary orbit used as input to the method.

Before considering explicit examples, noting that

$\frac{d^{2}f}{d{\phi }^2}-\frac{2}{f}(\frac{df}{d\phi }{)}^2=-f^2\frac{d^2}{d{\phi }^2}\left(\frac{1}{f}\right)$

the equation determining the potential can be written more simply as

$\frac{dV}{dr}=\frac{L^2}{m{f}^3}[f\frac{d^2}{d{\phi }^2}\left(\frac{1}{f}\right)+1]$

The form of this equation indicates that a multiplicative constant factor for f merely rescales the potential, and so can be ignored for the sake of simplicity. It is even simpler to work with the equation

$\frac{d𝒱}{dr}=\frac{1}{f^3}[f\frac{d^2}{d{\phi }^2}\left(\frac{1}{f}\right)+1]$

for determining the potential, and then add back the angular momentum and mass factors at the end as needed.

Now the fun can begin! For a first case, consider a straight line in polar coordinates:

$\begin{array}{c}x=rcos\phi =\mathrm{constant}=1\\ r=\frac{1}{cos\phi }\end{array}\to \begin{array}{l}\frac{dV}{dr}=0\\ V=\mathrm{constant}\end{array}$

which is exactly what one expects for a free particle. An initially curious result is that for a perfectly circular orbit,

$r=\mathrm{constant}=1\to \begin{array}{l}\frac{dV}{dr}=\frac{L^2}{m{f}^3}=\frac{L^2}{m{r}^3}\\ V=-\frac{L^2}{2m{r}^2}+\mathrm{constant}\end{array}$

until one remembers that perfectly circular orbits exist for any attractive central force with an extremum in its effective potential energy. The result here represents the balancing of the value of the actual potential energy with the angular momentum term in the effective potential energy.

Now consider an elliptical orbit described with respect to one of its foci:

$r=\frac{1}{1-ecos\phi }$

The corresponding potential (without scale factors) is

$\begin{array}{l}\frac{d𝒱}{dr}=\frac{1}{f^3}[\frac{ecos\phi }{1-ecos\phi }+1]=\frac{1}{f^2}=\frac{1}{r^2}\\ 𝒱=-\frac{1}{r}\end{array}$

which is of course the Kepler potential when all necessary constant factors are included. If the ellipse is described with respect to its geometric center,

$r=\frac{1}{\sqrt{1-e^2{cos}^2\phi }}$

then the corresponding potential is

$\begin{array}{l}\frac{d𝒱}{dr}=\frac{1}{f^3}[\frac{e^2({cos}^2\phi -{sin}^2\phi )}{1-e^2{cos}^2\phi }-\frac{e^4{cos}^2\phi {sin}^2\phi }{(1-e^2{cos}^2\phi {)}^2}+1]\\ \phantom{\frac{d𝒱}{dr}}=(1-e^2{cos}^2\phi {)}^{3/2}\frac{1-e^2}{(1-e^2{cos}^2\phi {)}^2}\\ \frac{d𝒱}{dr}=(1-e^2)r\\ \\ 𝒱=\frac{1-e^2}{2}{r}^2\end{array}$

This is a wonderful result: shifting the source of the attractive potential from the focus to the center of the ellipse turns the Kepler potential into that of the simple harmonic oscillator. This result was already known by Newton and included in the Principia. It represents a deep mathematical relationship between pairs of power potentials.

Now try something more exotic: an Archimedes spiral

$r=\phi$

for which the corresponding potential is

$\begin{array}{l}\frac{d𝒱}{dr}=\frac{1}{{\phi }^3}[\frac{2}{{\phi }^2}+1]=\frac{2}{r^5}+\frac{1}{r^3}\\ \\ 𝒱=-\frac{1}{2r^4}-\frac{1}{2r^2}\end{array}$

This result is more meaningful if the missing factors are restored in writing the Hamiltonian

$H=\frac{p_r}{2m}+\frac{L^2}{2m{r}^2}+\frac{L^2}{m}𝒱=\frac{p_r}{2m}-\frac{L^2}{2m{r}^4}$

The effective potential does not have an inverse-squared angular momentum term, so that there is no well in the potential and thus no extremum. This is of course because the Archimedes spiral continues to infinity and cannot represent a bound orbit.

Continuing with bound orbits, consider the polar equation

$r=1-ecos\phi$

which is a cardioid for e = 1 , but is more generally a limaçon. The corresponding potential is

$\begin{array}{l}\frac{d𝒱}{dr}=\frac{1}{f^3}[-\frac{ecos\phi }{1-ecos\phi }+\frac{2e^2{sin}^2\phi }{(1-ecos\phi {)}^2}+1]\\ \phantom{\frac{d𝒱}{dr}}=\frac{2(e^2-1)}{r^5}+\frac{3}{r^4}\\ \\ 𝒱=\frac{1-e^2}{2r^4}-\frac{1}{r^3}\\ \end{array}$

For 0 < e < 1 the effective potential here does indeed have a well in which closed orbits can occur. This does not necessarily mean that such orbits would be stable: that is a separate question that will not be investigated in this short presentation.

Rose-type curves, which are named for their lobes like the petals of a flower, can be described in two different ways. One is with a direct trigonometric function,

$r=cos\left(n\phi \right)$

where n is rational for closed orbits. The potential corresponding to this description is

$\begin{array}{l}\frac{d𝒱}{dr}=\frac{1}{f^3}[n^2+\frac{2n^2{sin}^{2}n\phi }{{cos}^{2}n\phi }+1]\\ \phantom{\frac{d𝒱}{dr}}=\frac{2n^2}{r^5}+\frac{1-n^2}{r^3}\\ \\ V=-\frac{n^2}{2r^4}+\frac{n^2-1}{2r^2}\end{array}$

The effective potential for this case does not have a well because the orbits necessarily pass through the center of the attractive force. The potential reduces to one term for n = 1 , and the corresponding orbit is a circle with the origin on its circumference.

The other description for rose-curves is as a square root of a trigonometric function:

$r=\sqrt{cos\left(n\phi \right)}$

where n is again rational for closed orbits. The potential corresponding to this description is

$\begin{array}{l}\frac{d𝒱}{dr}=\frac{1}{f^3}[\frac{n^2}{2}+\frac{3n^2{sin}^{2}n\phi }{4{cos}^{2}n\phi }+1]\\ \phantom{\frac{d𝒱}{dr}}=\frac{3n^2}{4r^7}+\frac{4-n^2}{4r^3}\\ \\ 𝒱=-\frac{n^2}{8r^6}+\frac{n^2-4}{8r^2}\end{array}$

The effective potential for this case again does not have a well because the orbits necessarily pass through the center of the attractive force. The potential reduces to one term for n = 2 , and the corresponding orbit is a lemniscate.

These evaluations of potentials are primarily meant as illustrations of the mathematical structure of classical mechanics, but there is a temptation to ask the question as to whether one could realize these orbits in some sense. It is well known that the gravitational potential of ordinary matter outside of a spherically symmetric source is always a Kepler potential: this result allowed Newton to treat planets as point particles in the Principia. It is also well known that the potential inside a uniform spherical mass is that of the simple harmonic oscillator, which allows one to assert that a body moving in a tunnel bored through a spherical mass will execute simple harmonic motion.

In the same idealization, one can think of a spherical distribution of ordinary matter with a narrow tunnel in the shape of an arbitrary planar orbit, where the removal of the tunnel material is small enough not to appreciably alter the overall gravitational field of the mass. One can then ask what radial distribution of matter will produce the potential corresponding to the arbitrary planar orbit. In that idealized sense one can attempt to realize the orbit physically using ordinary matter.

In classical Newtonian gravity, the potential inside a matter distribution is related to the matter density by Poisson’s equation,

$\rho =\frac{1}{4\pi G}{\nabla }^{2}V$

where the equation can in general depend upon all spatial variables. For spherically symmetric potentials, the Laplacian in n dimensions reduces to a single term, and Poisson’s equation becomes

$\rho \left(r\right)=\frac{r^{1-n}}{4\pi G}\frac{d^2}{d{r}^2}\left[r^{n-1}V\left(r\right)\right]$

where n = 3 corresponds to our physical reality. For a potential containing a term proportional to r^k, the contribution to the mass density is proportional to

$r^{1-n}\frac{d^2}{d{r}^2}\left[r^{k+n-1}\right]=(k+n-1)(k+n-2)r^{k-2}$

A realistic mass density cannot have a singularity at the origin, which means that the exponent of the term in the potential must be two or greater. Looking back on the potentials evaluated for arbitrary planar orbits, the only one that would not correspond to a central singularity in the matter density is the ellipse with the gravitational source at its center, i.e. the well-known simple harmonic oscillator potential. The other orbits considered must remain mathematical curiosities.

As a final comment, the n-dimensional Poisson equation provides one way to extrapolate classical gravity to higher dimensions, leading to a higher-dimensional Kepler potential of the form

$V(r,n)=\frac{GMm}{r^{n-2}}$

This same form arises in the weak-field limit of the n-dimensional Schwarzschild solution in general relativity. The reason for this is that gravity via Poisson’s equation is equivalent to equal-time slices of a spatially flat general relativistic metric.

The effective potential for the extrapolated Kepler potential only has a well in the potential for n = 3 , so that closed orbits not passing through the center of attraction can only exist for this particular dimension in classical mechanics. Paul Ehrenfest [Proc. Amsterdam Acad. 20, 200-209 (1917)] took the existence of stable planetary orbits as an indication that our physical space cannot have more than three dimensions. His conclusion provides a significant boundary condition to speculation about higher-dimensional physics.

Uploaded 2014.10.19 — Updated 2015.06.11 analyticphysics.com