How to Simplify the Covariant Derivative Transformation?

[JTFreitas]

Homework Statement

I need to prove that the covariant derivative of a vector is a tensor. In other words, I need to show that $\nabla_{\mu} V^{\nu}$ is a tensor.

Relevant Equations

I know by definition that $\nabla_{\mu} V^{\nu} = \frac{\partial}{\partial x^{\mu}} V^{\nu} +\Gamma^{\nu}_{\mu \sigma} V^{\sigma}$

Apologies in advance if I mess up the LaTeX. If that happens I'll be editing it right away.

By starting off with $\nabla^{'}_{\mu} V^{'\nu}$ and applying multiple transformation laws, I arrive at the following expression
$$ \frac{\partial x^{\lambda}}{\partial x'^{\mu}} \frac{\partial x'^{\nu}}{\partial x^{\alpha}} (\nabla_{\lambda} V^{\alpha}) + \frac{\partial x^{\lambda}}{\partial x'^{\mu}} \frac{\partial^{2} x'^{\nu}}{\partial x^{\lambda} \partial x^{\alpha}} V^{\alpha} + \frac{\partial^{2} x^{\alpha}}{\partial x'^{\mu} \partial x'^{\sigma}} \frac{\partial x'^{\nu}}{\partial x^{\alpha}} \frac{\partial x'^{\sigma}}{\partial x^{\beta}} V^{\beta} $$

I know my next step is to prove that the last two factors in the expression add up to zero. I know I need to "factor out" the partials, but I'm unsure which indices I should do that in terms of, or in what way I can relabel the indices in my favor. (I'm not very clear on when I can/cannot relabel indices in general, which has become problematic.)

Thanks for any help in advance!

 

[PeroK]

Yes, you just need a bit of index manipulation. First, if you swap the dummy indices $\alpha$ and $\beta$ in the second expression, you should get a common factor of $V^{\alpha}$.

Then, in the second expression, you can take out the $'\sigma$ derivative:
$$\frac{\partial^{2} x^{\beta}}{\partial x'^{\mu} \partial x'^{\sigma}} \frac{\partial x'^{\nu}}{\partial x^{\beta}} \frac{\partial x'^{\sigma}}{\partial x^{\alpha}} V^{\alpha} = \frac{\partial x'^{\sigma}}{\partial x^{\alpha}}\frac{\partial}{\partial x'^{\sigma}}(\frac{\partial x^{\beta}}{\partial x'^{\mu}})\frac{\partial x'^{\nu}}{\partial x^{\beta}}V^{\alpha} = \frac{\partial}{\partial x^{\alpha}}(\frac{\partial x^{\beta}}{\partial x'^{\mu}})\frac{\partial x'^{\nu}}{\partial x^{\beta}}V^{\alpha} $$
Now, think about using the product rule on your alpha derivative.

 

[PeroK]

PS all I really used above was the chain rule: $$\frac{\partial}{\partial x^{\alpha}} = \frac{\partial x'^{\sigma}}{\partial x^{\alpha}}\frac{\partial}{\partial x'^{\sigma}}$$

 

[JTFreitas]

Yes, you just need a bit of index manipulation. First, if you swap the dummy indices $\alpha$ and $\beta$ in the second expression, you should get a common factor of $V^{\alpha}$.

Then, in the second expression, you can take out the $'\sigma$ derivative:
$$\frac{\partial^{2} x^{\beta}}{\partial x'^{\mu} \partial x'^{\sigma}} \frac{\partial x'^{\nu}}{\partial x^{\beta}} \frac{\partial x'^{\sigma}}{\partial x^{\alpha}} V^{\alpha} = \frac{\partial x'^{\sigma}}{\partial x^{\alpha}}\frac{\partial}{\partial x'^{\sigma}}(\frac{\partial x^{\beta}}{\partial x'^{\mu}})\frac{\partial x'^{\nu}}{\partial x^{\beta}}V^{\alpha} = \frac{\partial}{\partial x^{\alpha}}(\frac{\partial x^{\beta}}{\partial x'^{\mu}})\frac{\partial x'^{\nu}}{\partial x^{\beta}}V^{\alpha} $$
Now, think about using the product rule on your alpha derivative.

God it makes complete sense now. I've been at this for days. My professor's notes weren't exactly very clear, but pulling out the $\partial x'^{\sigma}$ made everything work out nicely and made complete sense.

Thank you so much for the response, that was really helpful!

A quick follow-up question about relabelling: correct me if I'm wrong, but we're allowed to relabel indices if they're dummy indices, or we can also relabel true indices if we don't relabel them as some index that already exists in the expression? Is this a good way of going about it?

I try my best not to reuse indices in case I get them mixed up, and later I have to relabel them back into indices that make sense. Is this good practice?

 

[PeroK]

A quick follow-up question about relabelling: correct me if I'm wrong, but we're allowed to relabel indices if they're dummy indices, or we can also relabel true indices if we don't relabel them as some index that already exists in the expression? Is this a good way of going about it?

Yes, dummy indices tend to come out any old way and generally you need to sort them out in one or more of your terms at some stage to make the terms manifestly equal. It's really just $$\sum_k a_kv^k - \sum_m a_mv^m = 0$$.
Your indexing looked good. You should always have the same free indices in each term. You can change these, of course, as long as you change them in every term. For example: $$A^k_m = B^k_m \equiv A^n_r = B^n_r$$ As they mean for all $k, m = 0, 1, 2, 3$ and all $n, r = 0, 1, 2, 3$ (which is the same thing).

You'll see quite a bit of that as well, when an equation has one set of free indices when first stated - but then when it's used later a different set of free indices is used.

 

[JTFreitas]

Perfect, thank you so much!

We can close the thread now