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[The reference for this question is the book Gravitation by Misner,Thorne, & Wheeler.]

The trajectories of massive particles around a spherically symmetric body is governed by the effective potential $$ \tilde{V}(\tilde{L}, r) = \left( 1 - \frac{2M}{r}\right) \left( 1 + \frac{\tilde{L}^2}{r^2}\right) $$ where the orbital invariant $\tilde{L}$ is the angular momentum per unit rest mass $\frac{L}{\mu}$.

The question is, Is there any upper limit to $\tilde{L}$ for bound orbits?

I am thinking the following: In Newtonian physics, the angular momentum is unconstrained. However, we know from Special relativity that there is an upper limit to the velocity of any particle. Since angular momentum is related to $\dfrac{d\phi}{dt}$, naively using

$$r \frac{d\phi}{dt} \leq c$$

gives me,

$$ \tilde{L} \leq \tilde{E} \frac{r^2}{r- 2M} $$

How to proceed from here?

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    $\begingroup$ For unbound trajectories the angular momentum can obviously be arbitrarily high since $p$ can be arbitrarily high. Are you specifically thinking of bound orbits? $\endgroup$ – John Rennie Jun 29 '16 at 7:15
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    $\begingroup$ @JohnRennie Yes. Thank you . I have made the appropriate Edit. $\endgroup$ – negligible_singularity Jun 29 '16 at 9:57
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This turns out to have a really boring answer. We can find the two circular orbits by finding the maxima and minima of the effective potential, and we get:

$$ r = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - \frac{12M^2}{L^2}}\right) \tag{1} $$

where the $+$ gives the outer stable orbit and the $-$ gives the inner unstable orbit. Note that both orbits exist only for $L^2 \ge 12M$, which gives the expected result for the innermost stable orbit $r = 6M$.

Anyhow, if we take the limit of large $L$ we find:

$$ r \propto L^2 $$

So we can make $L$ as large as we want by making the orbital radius as large as we want.

This is easily understood by looking at the classical orbit. The orbital velocity falls with orbital radius as $v = \sqrt{GM/r}$ so $\omega=\sqrt{GM/r^3}$, but the moment of inertia rises as $I=mr^2$. Put it all together and you get $L=I\omega=m\sqrt{GMr}$.

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