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?
Best Answer
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.