0
$\begingroup$

When employing the Schwarzschild metric, I understand there are nine non-vanishing Christoffel symbols:

$ \Gamma^t_{rt} = -\Gamma^r_{rr} = \frac{r_{\rm s}}{2r(r - r_{\rm s})} \\[3pt] \Gamma^r_{tt} = \frac{r_{\rm s}(r - r_{\rm s})}{2r^3} \\[3pt] \Gamma^r_{\phi\phi} = (r_{\rm s} - r)\sin^2(\theta) \\[3pt] \Gamma^r_{\theta\theta} = r_{\rm s} - r \\[3pt] \Gamma^\theta_{r\theta} = \Gamma^\phi_{r\phi} = \frac{1}{r} \\[3pt] \Gamma^\theta_{\phi\phi} = -\sin(\theta)\cos(\theta) \\[3pt] \Gamma^\phi_{\theta\phi} = \cot(\theta) $

What I am confused about, is that all textbooks appear to be "selective" about which Christoffel symbols to use, when deriving the Geodesic equations via:

$\frac{d^2x^{\lambda}}{d q^2} + \Gamma^{\lambda}_{\mu\nu} \frac{dx^{\mu}}{d q} \frac{dx^{\nu}}{dq} = 0$

While $\Gamma^{\lambda}_{\mu\nu}$ apparently can be substituted with each one of the nine Christoffel Symbols, the sources I encounter always go on to obtain only four Geodesic equations, with $\lambda$ (the upper index of the Christoffel symbol) being equal to: $t,r,\theta,\phi$. This is understandable as far as being sufficient to find a solution, and yet I am not very clear on whether:

  1. Is it correct that an additional five geodesic equations can be derived but are just redundant / over-complicated?
  2. The other five equations for some reason are not interesting, such as being trivial (in the sense $0=0$)?
$\endgroup$

2 Answers 2

4
$\begingroup$

The expression $$\frac{d^2 x^\lambda}{dq^2} + \Gamma^\lambda_{\mu \nu} \frac{dx^\mu}{dq}\frac{dx^\nu}{dq} = 0$$ yields one equation for each value of $\lambda$. The dummy indices $\mu$ and $\nu$ are repeated, and therefore summed over. Performing the sums explicitly leads to the following expanded form of the geodesic equation:

$$\ddot{x}^\color{red}{\lambda} + \Gamma^\color{red}{\lambda}_{tt} \dot t \dot t+ \Gamma^\color{red}{\lambda}_{tr} \dot t \dot r+ \Gamma^\color{red}{\lambda}_{t\theta} \dot t \dot \theta+ \Gamma^\color{red}{\lambda}_{t\phi} \dot t \dot \phi$$ $$+ \Gamma^\color{red}{\lambda}_{rt} \dot r \dot t+ \Gamma^\color{red}{\lambda}_{rr} \dot r \dot r+ \Gamma^\color{red}{\lambda}_{r\theta} \dot r \dot \theta+ \Gamma^\color{red}{\lambda}_{r\phi} \dot r \dot \phi$$ $$+ \Gamma^\color{red}{\lambda}_{\theta t} \dot \theta \dot t+ \Gamma^\color{red}{\lambda}_{\theta r} \dot \theta \dot r+ \Gamma^\color{red}{\lambda}_{\theta \theta} \dot \theta \dot \theta+ \Gamma^\color{red}{\lambda}_{\theta \phi} \dot \theta \dot \phi$$ $$+ \Gamma^\color{red}{\lambda}_{\phi t} \dot \phi \dot t+ \Gamma^\color{red}{\lambda}_{\phi r} \dot \phi \dot r+ \Gamma^\color{red}{\lambda}_{\phi \theta} \dot \phi \dot \theta+ \Gamma^\color{red}{\lambda}_{\phi \phi} \dot \phi \dot \phi = 0$$

where the dot denotes differentiation with respect to $q$. This is obviously a mess, but the point I mean to emphasize is that there is one geodesic equation for each value of the upper index $\lambda$, and every nonzero $\Gamma$ appears in one of those four.

$\endgroup$
1
  • $\begingroup$ Oh no, I should have seen this! Thank you so much for your answer! $\endgroup$
    – Amit
    Dec 30, 2022 at 21:27
2
$\begingroup$

The questions seems to stem from a misunderstanding of the Einstein notation. There are four and not nine equations, because there are four, not nine, components of the tensor $x^\lambda$. Nevertheless, all nine non-zero values of the Christoffel symbol occur in the four equations, because the indices appearing twice in each term are summed over.

In more detail

The geodesic equation as written is a system of four equations for the four $x^\lambda(\tau)$ (where I use $\tau$ for the proper time, instead of $q$ which is used in the question), labelled by the values of the free index $\lambda$ which runs over the values $0, 1, 2, 3$ (where $0$ corresponds to the time coordinate and $1,2,3$ to the spatial coordinates – of course you can also write $t, r, \theta, \phi$ instead of $0,1,2,3$ when working in spherical coordinates).

The Einstein sum convention is, that there is implicit summation over every pair of indices appearing once in upper (contravariant) and once in lower (covariant) position in each term of the sum. So written in full the four geodesic equations read: $$ \frac{d^2x^\lambda}{d\tau^2} + \sum_{\mu=0}^3 \sum_{\nu=0}^3 \Gamma^{\lambda}_{\mu\nu} \frac{x^\mu}{d\tau}\frac{x^\nu}{d\tau} \quad\text{ for} \lambda = 0, 1, 2, 3$$ As you can see, all the nine non-zero values of the Christoffel symbols are taken into account in these equations, as $\mu$ and $\nu$ run over all indices.

$\endgroup$
1
  • $\begingroup$ You are very kind to provide such a detailed explanation and I thank you! $\endgroup$
    – Amit
    Dec 30, 2022 at 21:28

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.