### Transformation of Christoffel Symbol

We have the metric transformations between the two different coordinate systems as;

$$g_{\mu '\nu '}'=\frac{\partial x^{\mu}}{\partial y^{\mu '} }\frac{\partial x^{\nu}}{\partial y^{\nu '} }g_{\mu \nu}$$ We also know that the Christoffel symbol in terms of the metric tensors is as follows $$\Gamma_{\mu \nu}^{\lambda}=\frac{1}{2} g^{\lambda \rho}\left(\partial_{\mu}g_{\nu \rho} +\partial_{\nu}g_{\rho \mu}-\partial_{\rho}g_{\mu \nu}\right)$$ This then implies that the christoffel symbol in the primed coordinate system is then; $$\Gamma_{\mu ' \nu '}^{\lambda '} =\frac{1}{2} g^{\lambda ' \rho '}\left(\partial_{\mu '}g_{\nu ' \rho '} +\partial_{\nu '}g_{\rho ' \mu '}-\partial_{\rho '}g_{\mu ' \nu '}\right)$$ Our aim here, is to find the transformation relation between  these christoffel symbols which are in different coordinate system. We first have to find the derivative of the metric tensor in the primed coordinate system. Let us differentiate with respect to $\lambda '$ $$\partial_{\lambda '}g_{\mu ' \nu '}'=\frac{\partial x^{\lambda } }{\partial y^{\lambda '} }\frac{\partial x^{\mu } }{\partial y^{\mu '} }\frac{\partial x^{\nu } }{\partial y^{\nu '} } \partial_\lambda g_{\mu \nu}+g_{\mu \nu}\left(\frac{\partial x^{\nu } }{\partial y^{\nu '} }\frac{\partial^{2} x^{\mu } }{\partial y^{\lambda '}\partial y^{\mu '} }+ \frac{\partial x^{\nu } }{\partial y^{\mu '} }\frac{\partial^{2} x^{\mu } }{\partial y^{\lambda '}\partial y^{\nu '} } \right)$$ We know that the christoeffel symbol in the primed coordinate system has the derivatives of metric of the cycle $\rho ' ,\mu ' ,\nu '$.

Using this expression three times and relabeling the indices, one can write;

$$\partial_{\mu '}g_{\nu ' \rho '} +\partial_{\nu '}g_{\rho ' \mu '}-\partial_{\rho '}g_{\mu ' \nu '}=\frac{\partial x^{\lambda } }{\partial y^{\lambda '}}\frac{\partial x^{\rho } }{\partial y^{\rho '}}\frac{\partial x^{\nu } }{\partial y^{\nu '} }\left(\partial_{\lambda}g_{\rho \nu}+\partial_{\nu}g_{\rho \lambda}-\partial_{\rho}g_{\nu \lambda}\right)+$$ $$g_{\rho \nu}\left(\frac{\partial x^{\nu } }{\partial y^{\nu '} }\frac{\partial^{2} x^{\rho } }{\partial y^{\lambda '}\partial y^{\rho '} }+\frac{\partial x^{\nu } }{\partial y^{\lambda '} }\frac{\partial^{2} x^{\rho } }{\partial y^{\nu '}\partial y^{\rho '} }+ 2\frac{\partial x^{\nu } }{\partial y^{\rho '} }\frac{\partial^{2} x^{\rho } }{\partial y^{\lambda '}\partial y^{\nu '} }-\frac{\partial x^{\rho } }{\partial y^{\lambda '} }\frac{\partial^{2} x^{\nu } }{\partial y^{\rho '}\partial y^{\nu '} }-\frac{\partial x^{\rho } }{\partial y^{\nu '} }\frac{\partial^{2} x^{\nu } }{\partial y^{\rho '}\partial y^{\lambda '}}\right)$$ Because, the metric is symmetric in $\rho$ and $\nu$ we are just left with:

$$\partial_{\mu '}g_{\nu ' \rho '} +\partial_{\nu '}g_{\rho ' \mu '}-\partial_{\rho '}g_{\mu ' \nu '}=\frac{\partial x^{\lambda } }{\partial y^{\lambda '}}\frac{\partial x^{\rho } }{\partial y^{\rho '}}\frac{\partial x^{\nu } }{\partial y^{\nu '} }\left(\partial_{\lambda}g_{\rho \nu}+\partial_{\nu}g_{\rho \lambda}-\partial_{\rho}g_{\nu \lambda}\right)+$$ $$2g_{\rho \nu}\frac{\partial x^{\nu } }{\partial y^{\rho '} }\frac{\partial^{2} x^{\rho } }{\partial y^{\lambda '}\partial y^{\nu '} }$$ Now substituting the result in the primed christoffel symbol we have the following:
$$\Gamma_{\mu ' \nu '}^{\lambda '}=\frac {1}{2}\frac{\partial y^{\mu '}}{\partial x^{\mu} }\frac{\partial y^{\nu '}}{\partial y^{\nu} }g^{\mu \rho}\left(\frac{\partial x^{\lambda } }{\partial y^{\lambda '}}\frac{\partial x^{\rho } }{\partial y^{\rho '}}\frac{\partial x^{\nu } }{\partial y^{\nu '} }\left(\partial_{\lambda}g_{\rho \nu}+\partial_{\nu}g_{\rho \lambda}-\partial_{\rho}g_{\nu \lambda}\right)+2g_{\rho \nu}\frac{\partial x^{\nu } }{\partial y^{\rho '} }\frac{\partial^{2} x^{\rho } }{\partial y^{\lambda '}\partial y^{\nu '} }\right)$$
$$=\frac{\partial y^{\mu '}}{\partial x^{\mu} }\frac{\partial x^{\lambda}}{\partial y^{\lambda '} }\frac{\partial x^{\nu }}{\partial y^{\nu '} }\frac {1}{2}g^{\mu \rho}\left(\partial_{\lambda}g_{\rho \nu}+\partial_{\nu}g_{\rho \lambda}-\partial_{\rho}g_{\nu \lambda}\right)+\frac{\partial y^{\mu '} }{\partial x^{\mu} }\delta_{\rho}^{\nu}\delta_{\nu}^{\mu}\frac{\partial^{2} x^{\rho } }{\partial y^{\lambda '}\partial y^{\nu '} }$$ Thus, we have the transformation relation between the christoffel symbol as follows;
$$\Gamma_{\mu ' \nu '}^{\lambda '}=\frac{\partial y^{\mu '}}{\partial x^{\mu} }\frac{\partial x^{\lambda}}{\partial y^{\lambda '} }\frac{\partial x^{\nu }}{\partial y^{\nu '} }\Gamma_{ \nu \lambda}^{\mu}+\frac{\partial y^{\mu '} }{\partial x^{\mu} }\frac{\partial^{2} x^{\mu} }{\partial y^{\lambda '}\partial y^{\nu '} }$$

#### 3 comments:

Unknown said...

While relabeling tags you provide the negative term with a common index with the contravariant metric which wasn't the case originally, ie you change an index to another currently in use but different from the one you changed. Why can you do this?

George said...

I agree with Unknown who commented December 2018. One thing is that The indices μ,ν in the denominator of the last term of the fourth equation are the wrong way round. Another is that the indices on unprimed and primed Christoffel symbol in the last equation have moved around in a very odd way. It is not right!

I have written out the correct derivation here. (https://www.general-relativity.net/2019/03/transformation-of-christoffel-symbol.html)

George said...

In addition this proof is for a torsion-free metric-compatible connection. It would be better to have a proof for any type of connection. I learnt this in Spacetime and Geometry: An Introduction to General Relativity by Sean M Carroll. He also had an error in the transformation! The correct proof and transformation are also now on www.general-relativity.net HERE.