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.)

  • $\begingroup$ I do not understand precisely what intuition you are trying to gain here. $\endgroup$
    – Prahar
    Jul 21, 2020 at 14:29
  • $\begingroup$ 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. $\endgroup$
    – Noumeno
    Jul 21, 2020 at 14:36
  • 1
    $\begingroup$ When you took classical mechanics, did you encounter the concept of the effective potential for Keplerian orbits? $\endgroup$
    – G. Smith
    Jul 21, 2020 at 16:55
  • 1
    $\begingroup$ 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. $\endgroup$
    – G. Smith
    Jul 21, 2020 at 17:01
  • 1
    $\begingroup$ 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. $\endgroup$
    – G. Smith
    Jul 21, 2020 at 17:16

1 Answer 1

  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 $c\to \infty$.


$^1$ 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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.