Contracting dummy indices tensors

[whatisreality]

Homework Statement

Write out this covariant derivative in terms of partial derivatives and Christoffel symbols:

$\nabla_{\mu} S^{\nu}_{\nu \rho}$

Homework Equations

The Attempt at a Solution

I think you can contract that so it reads
$\nabla_{\mu} S_{\rho}$, in which case the solution would be $\partial _{\mu} S_{\rho} - \Gamma^{\varepsilon}_{\mu \rho} S_{\varepsilon}$. But I thought I'd just check it without contracting, fully expecting them to be equal, but sadly it isn't obvious to me that they are. If I don't contract before taking the covariant derivative then I get:

$\partial_{\mu} S^{\nu}_{\nu \rho} + \Gamma^{\nu}_{\mu\varepsilon} S^{\varepsilon}_{\nu\rho} - \Gamma^{\varepsilon}_{\mu\nu} S^{\nu}_{\varepsilon\rho} - \Gamma^{\varepsilon}_{\mu\rho} S^{\nu}_{\nu\varepsilon}$

So I just want the middle two terms to cancel and then it would be fine, but I'm not sure I can do that. Are both my solutions the same or are they different? If they're different then which one is valid?

Any help is much appreciated, thank you. :)

 

[Orodruin]

They are the same. You can rename dummy indices without changing the expression (it does not matter what you call your summation indices) so make the exchange $\varepsilon \leftrightarrow \nu$ in the second term.

 

[whatisreality]

They are the same. You can rename dummy indices without changing the expression (it does not matter what you call your summation indices) so make the exchange $\varepsilon \leftrightarrow \nu$ in the second term.

I don't understand why it doesn't matter. Ok, so yes the symbol I initially choose is random and I could swap out every epsilon for a theta, for example. What I don't understand is why I can relabel an index in just one term, without also making the swap in all the rest.

If the epsilon in the third term is completely unrelated to the epsilon in the second term, then maybe I can see why they can just be changed. But if they're completely different, then why choose the same symbol for them? Isn't that like me writing $x + y + x^2 = 9$ and then saying the first $x$ and the $x$ in $x^2$ are completely different?

Potentially that's a silly question... sorry if it is, I'm struggling to get my head round tensors in general, and this notation!

 

[Orodruin]

What I don't understand is why I can relabel an index in just one term, without also making the swap in all the rest.

Because summation is linear. Writing out the sums explicitly, you would have (only writing out the two relevant terms)
$$
\Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon - \Gamma_{\mu\varepsilon}^\nu S^\varepsilon_\nu
= \sum_\nu \sum_\varepsilon (\Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon - \Gamma_{\mu\varepsilon}^\nu S^\varepsilon_\nu)
= \sum_\nu \sum_\varepsilon \Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon - \sum_\nu \sum_\varepsilon \Gamma_{\mu\varepsilon}^\nu S^\varepsilon_\nu.
$$

If the epsilon in the third term is completely unrelated to the epsilon in the second term, then maybe I can see why they can just be changed. But if they're completely different, then why choose the same symbol for them? Isn't that like me writing $x + y + x^2 = 9$ and then saying the first $x$ and the $x$ in $x^2$ are completely different?

No, it is not the same. Your $x$ is not a dummy variable, it is an actual variable of your equation. If you integrated over $x$ it would be a dummy variable and you could change it for another variable $t$ in one of the terms (assuming you also change the integration in that term to be over $t$ instead of $x$, just as here you change the sum over one dummy index to the sum over the dummy index you replace it by - the difference here is that you have sums over two dummy indices and you change them so you get the same sums in the end).

If you feel uncomfortable changing both directly. Start by changing one in the second term to a completely unrelated index. Note that, if the indices run over $n$ values, you would have (skipping some indices for brevity)
$$
\Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon = \Gamma_{\mu 1}^\varepsilon S^1_\varepsilon + \Gamma_{\mu 2}^\varepsilon S^2_\varepsilon + \ldots + \Gamma_{\mu n}^\varepsilon S^n_\varepsilon,
$$
which is exactly the same thing you would get if you expanded the sum over $\rho$ in $\Gamma_{\mu\rho}^\varepsilon S^\rho_\varepsilon$ in the same fashion and so you can replace $\nu\to \rho$. Then change $\varepsilon \to \nu$ in the same manner and finally $\rho \to \varepsilon$.

 

[whatisreality]

Because summation is linear. Writing out the sums explicitly, you would have (only writing out the two relevant terms)
$$
\Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon - \Gamma_{\mu\varepsilon}^\nu S^\varepsilon_\nu
= \sum_\nu \sum_\varepsilon (\Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon - \Gamma_{\mu\varepsilon}^\nu S^\varepsilon_\nu)
= \sum_\nu \sum_\varepsilon \Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon - \sum_\nu \sum_\varepsilon \Gamma_{\mu\varepsilon}^\nu S^\varepsilon_\nu.
$$No, it is not the same. Your $x$ is not a dummy variable, it is an actual variable of your equation. If you integrated over $x$ it would be a dummy variable and you could change it for another variable $t$ in one of the terms (assuming you also change the integration in that term to be over $t$ instead of $x$, just as here you change the sum over one dummy index to the sum over the dummy index you replace it by - the difference here is that you have sums over two dummy indices and you change them so you get the same sums in the end).

If you feel uncomfortable changing both directly. Start by changing one in the second term to a completely unrelated index. Note that, if the indices run over $n$ values, you would have (skipping some indices for brevity)
$$
\Gamma_{\mu\nu}^\varepsilon S^\nu_\varepsilon = \Gamma_{\mu 1}^\varepsilon S^1_\varepsilon + \Gamma_{\mu 2}^\varepsilon S^2_\varepsilon + \ldots + \Gamma_{\mu n}^\varepsilon S^n_\varepsilon,
$$
which is exactly the same thing you would get if you expanded the sum over $\rho$ in $\Gamma_{\mu\rho}^\varepsilon S^\rho_\varepsilon$ in the same fashion and so you can replace $\nu\to \rho$. Then change $\varepsilon \to \nu$ in the same manner and finally $\rho \to \varepsilon$.

Oh, I get it! Thank you very much for your help, and for taking the time to write such a detailed answer. I really appreciate it!