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 Jul 12 '14 at 15:46
  • 1
    $\begingroup$ Also by $\Gamma_{\mu\nu}^{4}$ you mean $\Gamma_{\mu\nu}^{3}$ right? $\endgroup$ – PhotonBoom Jul 12 '14 at 15:52
  • $\begingroup$ Do they have dependence on $t = x^0$ or the proper time $\tau$ (or written as $s$)? $\endgroup$ – Martin Ueding Jul 12 '14 at 16:15
  • $\begingroup$ On the $t$ used in the Euler-Lagrange equation i.e $x^0$ $\endgroup$ – PhotonBoom Jul 12 '14 at 16:24
  • 1
    $\begingroup$ The main trouble is that your first equation is actually incorrect, because of product rule and differentiation of $\nu$. $\endgroup$ – Stan Liou Jul 12 '14 at 17:01

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.

| cite | improve this answer | |
  • $\begingroup$ 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! $\endgroup$ – Martin Ueding Jul 13 '14 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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