Undergrad Transformation of connection coefficients

[accdd]

I don't understand why the highlighted term is there.
This image was taken from Sean Carroll's notes available here: preposterousuniverse.com/wp-content/uploads/grnotes-three.pdf

 

[Orodruin]

This follows directly (together with the first term) from writing out the transformation relation $V^{\nu’} = \frac{\partial x^{\nu’}}{\partial x^\nu} V^\nu$ and applying the Leibniz rule.

 

[accdd]

Thanks for the answer, could you show me the steps to get to the result?

 

[Orodruin]

I think it would be more instructive if you do what I proposed starting from the first line in (3.3) and show us the steps until you get stuck.

 

[accdd]

$$V^{\nu'}=\frac{\partial x^{\nu'}}{\partial x^{\nu}}V^{\nu}$$
$$
\nabla_{\mu'}V^{\nu'}=\partial_{\mu'}(\frac{\partial x^{\nu'}}{\partial x^{\nu}}V^{\nu})+\Gamma^{\nu'}_{\mu'\lambda'}(\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}) =\partial_{\mu'}(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \partial_{\mu'}V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}=
$$

 

[Orodruin]

Ok. So the piece that you're missing is expressing $\partial_{\mu'}$ in terms of $\partial_\mu$. Are you familiar with any way to relate those two partial derivatives?

 

[accdd]

Is this the correct transformation?
$$
\partial_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu
$$
$$
\partial_{\mu'}(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \partial_{\mu'}V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}= \frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu(\frac{\partial x^{\nu'}}{\partial x^{\nu}})V^{\nu}+\frac{\partial x^{\nu'}}{\partial x^{\nu}} \frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu V^{\nu}+\Gamma^{\nu'}_{\mu'\lambda'}\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda}
$$

so the first line of 3.3 should be more clearly:
$$
\nabla_{\mu'}V^{\nu'}=(\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu)(\frac{\partial x^{\nu'}}{\partial x^{\nu}}V^{\nu})+\Gamma^{\nu'}_{\mu'\lambda'}(\frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda})
$$

 

[Orodruin]

Is this the correct transformation?
$$
\partial_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\partial_\mu
$$

Yes. That is the chain rule for partial derivatives.