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Carroll gives the potential energy in general relativity by

$$ V(r)=\frac{1}{2}\epsilon-\epsilon\frac{G\,M}{r}+\frac{L^{2}}{2r^{2}}-\frac{G M L^{2}}{r^{3}} $$

My first question is does $V(r)$ have the right units because the second term must be multiple by the mass of the test particle for $V(r)$ to have units of energy i.e. the second term should be $-\epsilon\frac{G M m}{r}$.

My Second question: Why is the last term given by $-\frac{G M L^{2}}{r^{3}}$ while the same term is given by $-\frac{G (M+m) L^{2}}{r^{3}}$ in Wikipedia. Why was $m$ not included in Carroll's derivation?

Thanks.

user3741635
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  • That's the effective potential, and Carroll is giving you the potential per unit mass. See the discussion in Could we send a man safely to the Moon in a rocket without knowledge of general relativity?. – John Rennie Jun 05 '16 at 16:34
  • @JohnRennie, But that still doesn't make sense because if Wikipedia derivation is right then the last term should be $-\frac{G(M+m)L^{2}}{m,r^{3}}$ – user3741635 Jun 05 '16 at 16:38
  • @JohnRennie, I am just not sure why there is $-\frac{G(M+m)L^{2}}{r^{3}}$ term on wikipedia while in your derivation you have $-\frac{GM L^{2}}{m r^{3}}$ Why are the masses arranged differently. Could you explain this please. Thanks. – user3741635 Jun 05 '16 at 16:48
  • From a quick glance, Carroll and I are assuming an infinitely small test mass so the central mass can be taken as stationary. The Wikipedia calculation assumes an orbiting mass high enough to be significant. Note that Wikipedia uses the reduced mass $\mu$ in the equation. – John Rennie Jun 05 '16 at 16:54
  • @JohnRennie, Thanks for help. By "infinitely small test mass" do you mean the mass of the moon compared to the mass of the earth? – user3741635 Jun 05 '16 at 17:03
  • The effective potential is derived from the Schwarzschild metric. The assumption is that the test mass $m$ is too small to cause any significant perturbation to this metric. – John Rennie Jun 05 '16 at 17:08
  • Mass of the Moon doesn't sound infinitesimal does it? A baseball would infinitesimal for example. The Moon is more than 1% of the Earth's mass which will have an influence on the calculation. The baseball would not have any significant impact. – Peter R Jun 05 '16 at 18:17
  • @PeterR, Do you mean that John's assumption is wrong? – user3741635 Jun 05 '16 at 18:27
  • John is right on. The nice thing about mathematics is that I can let values go to Zero. They don't have to be zero but they can approach zero so that the equation that remains is the one without small m since its impact will be negligable. – Peter R Jun 05 '16 at 18:37

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Since no answers have been forthcoming I will summarise what has been discussed in the comments.

The usual analysis of geodesic motion around a spherical mass assumes that the spacetime geometry is described by the Schwarzschild metric, and that the test mass is too small to perturb this metric to any significant extent. In that case the motion can be analysed using an effective potential that includes the gravitational forces and the fictitious centrifugal force. As described in the link you cite this produces an effective potential:

$$ V_{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3} \tag{1} $$

Note that Carroll gives the energy for unit mass, i.e. $m=1$:

In (7.47) we have precisely the equation for a classical particle of unit mass and "energy" $ {1\over 2}$E2 moving in a one-dimensional potential given by V(r).

and he uses the usual relativist's units of $c = 1$.

In a two body problem where the test mass is large enough to significantly perturb the system it's usual to describe the motion using the reduced mass $\mu$ given by:

$$ \mu = \frac{Mm}{M+m} $$

In the limit of $M \gg m$ this simply reduces to $\mu \approx m$. If you look at the effective potential described in the Wikipedia article it gives the potential without the assumption that $m$ is negligably small and uses the reduced mass $\mu$:

$$ V_{wiki}(r) = -\frac{GMm}{r} + \frac{L^2}{2\mu r^2} - \frac{G(M+m)L^2}{c^2\mu r^3} \tag{2} $$

If we take the limit $M \gg m$, so that $\mu\approx m$ and $M+m\approx m$, the equation (1) from Carroll and (2) from Wikipedia become identical.

I have to confess I'm not sure how the equation Wikipedia cite is derived, other than possibly as a heuristic approach. I don't think there is an analytical approach that takes into account the perturbation of the Schwarschild metric by the test mass.

John Rennie
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  • Sorry to bother you again John but you said that Carroll gives the potential per unit mass, so to get the potential I need to multiply by $m$ but doing that I get the effective potential as $$ V_{eff}(r) = \frac{1}{2}\epsilon,m-\frac{GMm}{r} + \frac{m L^2}{2r^2} - \frac{GmML^2}{c^2r^3} \tag{1} $$ If you notice the last two terms are different from the last two term in Eq $(1)$ in you answer. Could you explain this please . – user3741635 Jun 07 '16 at 04:59
  • Hmm, yes, OK. I'll have to see if I can dig out a copy of Carroll's book to see exactly what he has derived. The form of $V_\text{eff}$ I've given is the one I was taught, and the one that matches Wikipedia in the limit of $M\gg m$. – John Rennie Jun 07 '16 at 05:01
  • @user3741635: aha, Carroll derives the expression for a unit mass, i.e. $m=1$, so $m$ does not appear in his equation. – John Rennie Jun 07 '16 at 17:54