# Different approaches to calculating the Christoffel symbols

I would be very grateful to whoever can debug the following calculations...

We have the metric for static spacetime: $$ds^2 = -\exp(2U(\vec x))dt^2+h_{ij}(\vec x) d x^i d x^j$$ I want to find the associated Christoffel symbols.

I know of 2 ways:

The first is to use the formula $$\Gamma^a_{bc}={1\over 2}g^{ad}(g_{dc,b}+g_{db,c}-g_{bc,d})$$ Using this formula, I get $$\Gamma^0_{0i}={\partial U\over \partial x^i}$$ for $i=1,2,3$.

However, by using a second way -- by reading off the Euler-Lagrange equation ,I got an extra factor of $2$ and I can't spot where I have made a mistake. Here is what I did:

The Lagrangian $$L=\exp(2U(\vec x))\dot t^2-h_{ij}(\vec x) \dot x^i \dot x^j$$ So by the Euler-Lagrange equation for the $t$ component, $${d\over d\tau}(\exp(2U(\vec x))\dot t)=0$$ which leads to $$\ddot t +2{\partial U\over \partial x^i} \dot x^i \dot t=0$$ Therefore giving $$\Gamma^0_{0i}=2{\partial U\over \partial x^i}$$

What went wrong, where? Are my approaches and formulae correct?

Thanks.

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You get the factor 2 because the Christoffel symbols are symmetric in their lower indices. Therefore, you read off $\Gamma^0_{0i} + \Gamma^0_{i0}$ from the geodesic equation. –  queueoverflow Jul 12 '14 at 15:15

The first one is probably correct, and you are likely not counting correctly in the second method. You should compare you result

$\ddot{t}+ 2 \frac{\partial U}{\partial x^{i}} \dot{x}_i \dot{t}=0$

to

$\ddot{t}+ \Gamma^0_{\mu \nu} \dot{x}^\mu \dot{x}^{\nu}=0$

or

$\ddot{t}+ \Gamma^0_{i 0} \dot{x}^i \dot{x}^{0}+ \Gamma^0_{0 i} \dot{x}^0 \dot{x}^{i}=0$

which is

$\ddot{t}+ \Gamma^0_{i 0} \dot{x}^i \dot{t} + \Gamma^0_{0 i} \dot{t} \dot{x}^{i}= \ddot{t}+ 2\Gamma^0_{i 0} \dot{x}^i \dot{t} = 0$.

Which yields

$\Gamma^0_{i 0} = \frac{\partial U}{\partial x^{i}}$.

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Thanks, DJBunk! –  Gene Apr 8 '13 at 18:26
+1: Thanks^2 DJBunk! –  joshphysics Apr 8 '13 at 20:05
@Gene - no problem. I am pretty sure I lost an afternoon of my life to these factors of 2 at one point, so I am happy to help. –  DJBunk Apr 8 '13 at 20:40

Couple mistakes:

1. The second term in the metric should read $h_{ij}(\vec x)dx^idx^j$ not $h_{ij}(\vec x) \dot x^i\dot x^j$

2. The first term in the Langrangian should have $\dot t^2$ instead of $dt^2$ which means that in the equation after the Lagrangian you should have a $\dot t^2$, and (I redid the calc.) this gives an answer that matches the first method without the extra factor of $2$.

Edit. My #2 doesn't solve the problem of the factor of two; BJBunk's answer does.

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Thanks, Josh! Indeed, I should have $$L=\exp(2U(\vec x))\dot t^2-h_{ij}(\vec x) \dot x^i \dot x^j$$. But doesn't $${d\over d\tau}\left({\partial L\over \partial \dot t}\right)=0$$ and here $${\partial L\over \partial \dot t}=2\exp(2U(\vec x))\dot t$$ thus still giving $${d\over d\tau}\left(\exp(2U(\vec x))\dot t\right)=0$$? (Or am I just being stupid again? :/ ) –  Gene Apr 8 '13 at 18:08
@Gene No, you're right, I was being careless my apologies. Now this is bothering me; I'll try to think about this more. –  joshphysics Apr 8 '13 at 18:54
@Gene I just read DJBunk's answer and realized he resolves the issue. Sorry if I caused any confusion. –  joshphysics Apr 8 '13 at 20:03