I would like to compute the Christoffel symbols of the second kind using the geodesic equation. To practice, I have tried the Schwarzschild Ansatz $$ g_{00} = \mathrm e^\nu,\quad g_{11} = - \mathrm e^\lambda,\quad g_{22} = -r^2,\quad g_{33} = -r^2 \sin(\theta)^2, $$ where $\nu$ and $\lambda$ are functions of $r$.

The Lagrangian is $$ L = \mathrm e^\nu \dot t^2 - \mathrm e^\lambda \dot r^2 - r^2 \dot \theta^2 - r^2 \sin(\theta)^2 \dot \phi^2.$$

From this, I have computed for Euler Lagrange equations: $$ 0 = 2 \mathrm e^\nu \ddot t $$ $$ \mathrm e^\nu \nu' \dot t^2 - \mathrm e^\lambda \lambda' \dot r^2 - 2 r \dot \theta^2 - 2 r \sin(\theta)^2 \dot\phi^2 = -2 \mathrm e^\lambda \ddot r $$ $$ - 2 r^2 \sin(\theta) \cos(\theta) \dot \phi^2 = - 2r^2 \ddot \theta $$ $$ 0 = - 2 r^2 \sin(\theta)^2 \ddot \phi $$

With the second I got: $$ \Gamma^1_{00} = \frac{\nu'}2 \mathrm e^{\nu-\lambda}, \quad \Gamma^1_{11} = - \frac{\lambda'}2, \quad \Gamma^1_{22} = - r \mathrm e^{-\lambda}, \quad \Gamma^1_{33} = - r\sin(\theta)^2 \mathrm e^{-\lambda}$$

And the third one: $$ \Gamma^2_{33} = - \sin(\theta)\cos(\theta) $$

From the first and fourth equation I would deduce that any $\Gamma^0_{\mu\nu} = 0$ as well as $\Gamma^3_{\mu\nu} = 0$. The solution says that this is not the case. How can I obtain the other nonzero Christoffel symbols?

  • $\begingroup$ Hint: $\theta$ and $r$ have time dependence $\endgroup$
    – PhotonBoom
    Commented Jul 12, 2014 at 15:46
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    $\begingroup$ Also by $\Gamma_{\mu\nu}^{4}$ you mean $\Gamma_{\mu\nu}^{3}$ right? $\endgroup$
    – PhotonBoom
    Commented Jul 12, 2014 at 15:52
  • $\begingroup$ Do they have dependence on $t = x^0$ or the proper time $\tau$ (or written as $s$)? $\endgroup$ Commented Jul 12, 2014 at 16:15
  • $\begingroup$ On the $t$ used in the Euler-Lagrange equation i.e $x^0$ $\endgroup$
    – PhotonBoom
    Commented Jul 12, 2014 at 16:24
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    $\begingroup$ The main trouble is that your first equation is actually incorrect, because of product rule and differentiation of $\nu$. $\endgroup$
    – Stan Liou
    Commented Jul 12, 2014 at 17:01

1 Answer 1


Notation: I will use overdot for differentiation with respect to $\tau$, overtilde for partial differentiation with respect to $x^0 = t$, and prime for partial differentiation with respect to $x^1 = r$. (Edit: removed overloading of $\lambda$, sorry.)

I assumed a general $\nu = \nu(t,r)$; reading the question more carefully, they're functions of $r$ only, which makes $\tilde\nu = \tilde\lambda = 0$, but the rest applies equally well.

From the Euler-Lagrange equation for $x^0 = t$: $$\frac{\mathrm{d}}{\mathrm{d}\tau}\left(2e^{\nu}\dot{t}\right) = \frac{\partial L}{\partial t} = e^\nu\tilde\nu\dot{t}^2 - e^\lambda\tilde\lambda\dot{r}^2\text{.}$$ Remember that $$\frac{\mathrm{d}\nu}{\mathrm{d}\tau} = \frac{\partial\nu}{\partial x^\alpha}\frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} = \tilde\nu\dot{t}+\nu'\dot{r}\text{,}$$ and you should be able to complete the calculation correctly.

  • $\begingroup$ The $\mathrm e^\nu$ looks really nice, but it actually is $$\exp\left(\nu\left(r(\tau)\right)\right).$$ If you write it as such, the chain rule becomes obvious. I tried it out, and it works out correctly! $\endgroup$ Commented Jul 13, 2014 at 10:35

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