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Given the line element $$ds^2 = e^v dt^2 - e^{\lambda} dr^2 - r^2 d \theta^2 - r^2 \sin^2 \theta d \phi^2$$

we wish to compute the Christoffel symbols $\Gamma^{a}_{bc}$ using the geodesic equation.

Our Lagrangian is $$L = g_{ab} {\dot{x}}^a {\dot{x}}^b = e^v \dot{t}^2 - e^{\lambda} \dot{r}^2 - r^2 \dot{\theta}^2 - r^2 \sin^2 \theta \dot{\phi}^2.$$ Applying the Euler-Lagrange equations and dividing by $-2e^v$ yields $$\ddot{t} - \frac{1}{2} \dot{v} \dot{t}^2 + v' \dot{t} \dot{r} + \frac{1}{2} \dot{\lambda} e^{\lambda - v} \dot{r}^2 = 0.$$

I incorrectly conclude that $\Gamma^{0}_{00} = - \frac{1}{2} \dot{v}$. Why? I have a feeling this has something to do with the difference between the geodesic equations for space-like and time-like geodesics.

Ivo Terek
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Andrew Whelan
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  • $v$ and $\lambda$ are functions of what, exactly? The equations are the same, independently of the geodesic's causal type. – Ivo Terek Mar 12 '16 at 16:28
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    They are both functions of $t$ and $r$. – Andrew Whelan Mar 12 '16 at 17:26
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    Is the correct result $\Gamma_{00}^0 = \frac{1}{2}\dot{v}$? I did the computations here and I got $\Gamma_{00}^0 = \frac{1}{2}\dot{v}$. I think you probably messed up a sign. Also, I assume that you're using the dot at the same time in $v$ and $\lambda$ to denote derivatives wrt to $t$, and the dot in $t$ and $r$ to denote derivative wrt to the parameter of the curve. It is easy to get something wrong like this. If you can check that the result I got is okay, I'll put my calculations in an answer here. – Ivo Terek Mar 12 '16 at 23:16
  • Yes, that is correct. Would be great to see your calculation – Andrew Whelan Mar 12 '16 at 23:47
  • Ok. I'm on it. ${}{}$ – Ivo Terek Mar 12 '16 at 23:48
  • Hope it helps. Sorry for the delay (I still haven't mastered the skill of eating pizza and typing in tex quickly at the same time :D) – Ivo Terek Mar 13 '16 at 00:17
  • Haha, thanks! Yeah that helps a lot (: – Andrew Whelan Mar 13 '16 at 17:33

1 Answers1

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Let $\xi$ be the parameter of the geodesic. I'll introduce a factor of $\frac{1}{2}$ in your Lagrangian for psychological reasons. Then my Lagrangian will be $$L = \frac{1}{2}\left(e^{v(t(\xi),r(\xi))}t'(\xi)^2 - e^{\lambda(t(\xi),r(\xi))}r'(\xi)^2 - r(\xi)^2\theta'(\xi)^2 - r(\xi)^2\sin^2\theta(\xi)\phi'(\xi)^2\right),$$where $\xi$ is the parameter of the curve. From here on I'll drop the point of application - I just wrote everything so we can keep track of chain rules. I won't use dot notation to avoid any confusion, and prime will denote derivatives in $\xi$. We compute the Euler-Lagrange equation $$\frac{\partial L}{\partial t}-\frac{{\rm d}}{{\rm d}\xi}\left(\frac{\partial L}{\partial t'}\right)=0$$First step: $$\frac{1}{2}\left(\frac{\partial v}{\partial t}e^v t'^2 - \frac{\partial \lambda}{\partial t}e^\lambda r'^2\right)-\frac{{\rm d}}{{\rm d}\xi}\left(e^vt'\right)=0.$$Now: $$\frac{1}{2}\frac{\partial v}{\partial t}e^vt'^2 - \frac{1}{2}\frac{\partial \lambda}{\partial t}e^\lambda r'^2 - t'e^v\left(t'\frac{\partial v}{\partial t}+r'\frac{\partial v}{\partial r}\right)-e^vt'' = 0$$ Dividing out $e^v$, writing $t''$ first and distributing: $$-t''+\frac{1}{2}\frac{\partial v}{\partial t}t'^2 - \frac{1}{2}\frac{\partial \lambda}{\partial t}e^{\lambda-v} r'^2 - \frac{\partial v}{\partial t}t'^2-\frac{\partial v}{\partial r}t'r'= 0$$

Adding the terms with $t'^2$ and multiplying everything by $-1$ gives: $$t''+\frac{1}{2}\frac{\partial v}{\partial t}t'^2 + \frac{1}{2}\frac{\partial \lambda}{\partial t}e^{\lambda-v} r'^2 +\frac{\partial v}{\partial r}t'r'= 0.$$So we conclude that $$\Gamma^0_{00} = \frac{1}{2}\frac{\partial v}{\partial t}, \quad \Gamma^0_{11} = \frac{1}{2}\frac{\partial \lambda}{\partial t}e^{\lambda -v}, \quad \Gamma_{01}^1 = \frac{1}{2}\frac{\partial v}{\partial r}.$$

Ivo Terek
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