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Given the standard geodesic equation: $$\frac{d^2 x^\mu}{d\lambda^2}+\Gamma ^\mu _{\sigma \rho}\frac{d x^\sigma}{d \lambda}\frac{d x^\rho}{d \lambda}=0$$ we want to apply it to the Schwarzschild metric; conceptually this simply means using the proper $\Gamma$ coefficients, but in practice this leads to a gargantuan calculation. But fortunately we can apply some simplification to get the following equation: $$\frac{1}{2}\left(\frac{dr}{d\lambda}\right)^2+V(r)=\mathcal{E} \ \ \ \ \ \ \ \ \ \ (1)$$ where: $$V(r)=\frac{1}{2}\epsilon-\epsilon\frac{GM}{r}+\frac{L^2}{2r^2}-\frac{GML^2}{r^3}$$ $$\mathcal{E}=\frac{1}{2}E^2$$ we should also specify that:

  1. This equations should hold for massive and massless particles alike, so $\lambda$ is in one case the proper time and in the other an affine parameter of the proper time.
  2. $r \equiv r(\lambda)$ which is the radius of the orbit
  3. $\epsilon := -g_{\mu\nu}\frac{d x^\mu}{d\lambda}\frac{d x^\nu}{d\lambda}$
  4. $E=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\lambda}$
  5. $L=r^2\frac{d \phi}{d\lambda}$ (where $\phi$ is the fourth coordinate in our Schwarzschild's spherical coordinates)

End of the setup; my question is: I followed the proof of this statement, involving killing vectors ecc., and I get the mathematical gist of it, but I have problems interpreting the result. Is there an intuitive way to read statement (1)?

(For example: One problem is that it is said that $V(\lambda)$ should represent the potential and $\mathcal{E}$ should represent the energy, but I don't see how.)

Noumeno
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  • I do not understand precisely what intuition you are trying to gain here. – Prahar Jul 21 '20 at 14:29
  • From what I understand now equation (1) just pops out of the calculations. I am interested in a more physical and less mathematical way to see equation (1). This equation seems to me like some sort of energy conservation, but I don't get how to relate the quantities to potential or kinetical energy for example. – Noumeno Jul 21 '20 at 14:36
  • Have a look at this page, particularly between equations (7.32) and (7.48). The discussion on orbits continues further after this, but the derivation is in the range I give here. https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html There is also stuff in MTW if you have access to that ("the pit in the potential"). – m4r35n357 Jul 21 '20 at 16:29
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    When you took classical mechanics, did you encounter the concept of the effective potential for Keplerian orbits? – G. Smith Jul 21 '20 at 16:55
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    Is there an intuitive way to read statement (1)? Since you used the word “statement”, some readers will think you are talking about the numbered paragraph rather than the numbered equation. – G. Smith Jul 21 '20 at 17:01
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    this leads to a gargantuan calculation No, it doesn’t, because the Schwarzschild metric has only five nonzero metric derivatives ($g_{tt,r}$, $g_{rr,r}$, $g_{\theta\theta,r}$, $g_{\phi\phi,r}$, and $g_{\phi\phi,\theta}$) and the Christoffel symbols are symmetric. Only 9 of the 40 Christoffels require any algebra. – G. Smith Jul 21 '20 at 17:16
  • Are you familiar with the concept of an effective potential? It appears in many more situations than this one. – Javier Jul 21 '20 at 20:30

1 Answers1

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  1. OP's first-order ODE (3) involves 4 spacetime coordinates, but after using various Killing symmetries of the Schwarzschild solution, we eliminate$^1$ 3 coordinates, and we are left with OP's first-order ODE (1) in a single radial coordinate $r$, which readily has an interpretation as a non-relativistic-looking$^2$ 1D energy conservation law. See also my related Phys.SE answer here.

  2. An additional existence argument comes from the fact that we know from the Kepler problem that some version of OP's eq. (1) must exist in the non-relativistic limit $c\to \infty$.

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$^1$ Explicitly this elimination is done by (i) going to equatorial plane, and by using that (ii) specific energy & (iii) angular momentum are conserved.

$^2$ While the "kinetic term" $\dot{r}^2$ naively looks non-relativistic, everything is of course fully relativistic!

Qmechanic
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