# Derivation of the value of christoffel symbol

1. Apr 19, 2012

### Boltzmann2012

Hi,
I am new to general relativity and as I would like to find out how we could derive the value of christoffel symbol in terms of the metric tensor.
I have also heard that it was given as a definition for the christoffel symbol and would like a clarification on that.

Regards
Bltzmn2012

2. Apr 19, 2012

### Staff: Mentor

3. Apr 19, 2012

### Boltzmann2012

By permuting the indices, and resumming, one can solve explicitly for the Christoffel symbols as a function of the metric tensor:

I didnt quite get that(although I was given the formula underneath).could you please explain what is permuting the indices. I am really new to this stuff.

Regards
Bltzmnn2012

4. Apr 19, 2012

### elfmotat

$$\Gamma^\rho_{~\mu \nu}=\frac{1}{2} g^{\rho \lambda} (\partial_\mu g_{\nu \lambda}+\partial_\nu g_{\mu \lambda}-\partial_\lambda g_{\mu \nu})$$

It's relatively straightforward to calculate the components of the Christoffel symbols from the components of the metric. It can get pretty tedious though.

Last edited: Apr 19, 2012
5. Apr 19, 2012

### pervect

Staff Emeritus
Yes - a few comments on notation for the OP
$$\partial_\mu = \frac{\partial}{\partial_\mu}$$

Summation over lambda is implied by the Einstein convention, (which is that you sum over repeated indices), i.e: [correction to index]

$$\Gamma^\rho_{~\mu \nu}=\sum_{\lambda=0}^{\lambda=3} \frac{1}{2} g^{\rho \lambda} (\partial_\mu g_{\nu \lambda}+\partial_\nu g_{\mu \lambda}-\partial_\lambda g_{\mu \nu})$$

Last edited: Apr 19, 2012
6. Apr 19, 2012

### Nabeshin

The sum should extend from 0-3, not 0-4 :)

7. Apr 19, 2012

### pervect

Staff Emeritus
Ooops - yes, good point, space-time has 4 dimensions, not 5. I corrected the original, for whatever it's worth.