Built on Facts

More Potential Universes

Posted by Matt Springer on September 10, 2009
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All right, time for an actual example of this gravitational force law I’ve been ruminating on for the last two days. Today we’ll look at an alternate version of the gravitational potential that’s truly screwy:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

The first thing to notice is that it’s not finite as r becomes very large. Among other things, that means there’s no such thing as an escape velocity in a universe with this kind of gravity. What goes up really must come down.

To say more, we ought to take a look at the effective potential, which takes into account the angular momentum of the orbiting object. We’ll write that effective potential down and call it V:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

Here the letter “l” is the angular momentum, which is the orbiting body’s mass times its velocity times its distance from the center of the orbit. A circular orbit will occur if the orbital radius r is such that the particle is held right at the bottom of the effective potential well. If there is such a location, it’ll have to be where the derivative of the potential is zero.

In our case we can find that location pretty easily:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

Which implies that the radius which produces a circular orbit is:

i-8d7070ce28f0af28047c7a5fb8246d53-4.png

The condition that the orbit be a stable one is that this location is actually a local minimum of the potential. A theorem from calculus tells us that the test for this is that the second derivative of the potential is positive at that location. We can do that pretty easily too:

i-bb5c2c6b0452df43a61e3974bd9b473f-5.png

Substituting in the particular r that we found for the radius of the circular orbit gives:

i-80f9713f6d11461837a9f9b540684e36-6.png

(My notation’s a little sloppy, generally you should be sure to indicate that the derivative is being evaluated at a specific point.)

Anyway. We know that m is positive and both squared terms are positive, so the second derivative at that point is certainly positive. Therefore that point is a true minimum of the potential energy and the orbit is stable. Who’d have thunk it?

I’m not prepared to say that it’s a realistic possibility for the law of gravitation in some alternate universe. The bare fact that orbits are stable doesn’t say much about much. We don’t know if stars would be stable, and that’s certainly a bare minimum requirement for an interesting universe with life as we know it. But it is quite interesting to see that such a wild variation of gravity does in fact result in at least a few features of normal gravity in our own universe.

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Comments

  1. #1 Uncle Al
    September 10, 2009

    What about the near case? As r becomes very small we revisit the beauty of Yukawa potentials – nothing is ever decided. What of supernovae and black holes? What is the Schwarzschild radius expression? And what happens if the log goes negative?

  2. #2 Crux Australis
    September 10, 2009

    I’ve wondered this for a while (at least a week); what do you use to import the equations into your post?

  3. #3 Alex
    September 11, 2009

    Seems like you’ve broken Bertrand’s theorem. The only potentials that are supposed to admit stable closed orbits are the 1/r and r^2 potentials. In fact, numerical simulation seems to bear out that the orbits for this potential are not in fact stable, which can also be seen by considering a perturbation in the orbit.

  4. #4 andy
    September 11, 2009

    Alex: who said anything about the orbit being closed?

  5. #5 Alex
    September 12, 2009

    A circle is closed. Matt found that a circular (i.e. closed orbit) in a logarithmic potential was stable. This isn’t true. I simulated the orbit numerically and found that it isn’t stable, and if you try and do the problem out fully, and look at the solutions to the Lagrangian and consider perturbations to the orbit, these are also unstable. Something has gone horribly wrong in the post, and I’m not sure what.

    Plotting the effective potential does seem to yield a minimum for the circular radius, but as I said the resulting dynamics seems to suggest that the orbit is not in fact stable. I don’t think it is enough to consider the shape of the effective potential, for that ignores the r dependence of l.

  6. #6 Alex
    September 12, 2009

    I meant to say that I don’t think taking the second partial derivative of the effective potential with respect to r serves as a proper test for stability anymore. In that, I know the orbits are not stable. I’m not sure what’s gone wrong, I haven’t taken a lot of time to think about it, but I think that since Matt ignored the r dependence of l, the partial derivative we took no longer speaks to the stability of orbits, but something more akin to the fact that the effective potential has a minimum at the circular radius, which has nothing to say about the stability of the orbit, and only something to say about the class of circular solutions to a particular choice of l.

    A mistake has been made, and some investigation and correction are in order.

  7. #7 Matt Springer
    September 12, 2009

    I think it warrants further hashing out before saying anything definitive. Specifically with respect to Bertrand’s theorem though: we know that in this case a circular orbit is closed but not stable under perturbation. But once perturbed, there’s nothing stopping the orbit from being stable but not closed. I expect this is what happens, though it will take a little more poking around to verify.

    Also: L doesn’t have an r-dependence. It’s angular momentum, and thus has to be fixed by the conservation law. The specific value it takes for a circular orbit is a function of r, but once you set L and start the system moving L is strictly constant.

  8. #8 andy
    September 15, 2009

    Alex stated that he has numerically integrated orbits in this potential and found them to be unstable. I have tried writing a Runge-Kutta integrator in Perl and find stable orbits. If you perturb a circular orbit (e.g. varying the initial velocity) then you end up with a particle on a stable but non-closed orbit.

  9. #9 Matt Springer
    September 16, 2009

    Excellent work, Andy. I had Mathematica integrate the explicit equation of motion numerically, with the result that there is apparently a numerical instability in whatever method Mathematica uses to perform the integral. The instability is not in the physics itself. I think that’s probably the source of the problem.

    My previous comment is a little unclear, by the way. A circular orbit is not “unstable” as such, rather, Bertrand’s theorem simply does not apply to exactly circular orbits. A circular orbit is possible in any attractive central potential, but once perturbed out of circularity it will be either unstable or open except in the special cases of Coulomb and harmonic oscillator potentials.