Covariant derivative of a contravariant vector

[nrqed]

Fine, I will delete it then. He (or she)was not doing it the way the initial question was asking because he/she was doing it your way, so I don't see the harm in showing a different approach. But fine, I won't try to help anymore. (Note that you did not try to answer his/her initial question about his/her initial approach. I guess I did worse than not answering by giving too many details). But no worry, there are many other websites where I can help people so no problem! :-)

You know, it's a bit funny after all. The OP asked a question about why a specific calculation was not working and you ignored that initial question completely. And yet, what I did was apparently less useful to the OP than what you did.

Anyway, I will get out of your hair. Enjoy policing the forums.

 

[BiGyElLoWhAt]

Well, first off, I did specifically ask to not be given spoilers, that in mind, I read about the first line of your post then immediately closed my app, since you said you were going to provide a solution. That in mind, I started over from scratch, and feel as though I am closer, but I am not sure what to do from here:
$D_m T^k \to D_m g^{\mu k} T_{\mu}$
$D_m g^{\mu k} T_{\mu} = T_{\mu}D_m g^{\mu k} + g^{\mu k} D_m T_{\mu}$
$= 0 + g^{\mu k} [ \partial_m T_{\mu} - \Gamma_{m \mu}^z T_z ]$
$= g^{\mu k} [ \partial_m T_{\mu} - g^{zr}(\partial_m g_{\mu r} + \partial_{\mu} g_{mr} - \partial_r g_{m \mu})T_z]$
$= "..." + (-g^{zr}g^{\mu k} \partial_m g_{\mu r} - g^{zr}g^{\mu k} \partial_{\mu} g_{mr} + g^{zr}g^{\mu k}\partial_r g_{m \mu}) T_z$
$= "..." + (-\delta_r^k \partial_m g^{zr} - \delta_m^z \partial_{\mu} g^{\mu k} + \delta_m^k \partial_r g^{zr})T_z$
$= "..." + (-\delta_r^k \partial_m g^{zr} - g^{zk}g_{mk} \partial_{\mu} g^{\mu k} + \delta_m^k \partial_r g^{zr})T_z$
$= "..." + (-\delta_r^k \partial_m g^{zr} - \delta_m^{\mu} \partial_{\mu} g^{zk} + \delta_m^k \partial_r g^{zr})T_z$

The middle term obviously stands out, you can see it in the partial_mu g^mu k. I'm just not sure what to do about it. The last couple steps might even be redundant. My thinking was I shouldn't have a partial of g^mu k w.r.t. the muth component. That just seems weird, and it doesn't happen in the christoffel symbols. However, I now have two upper and 1 lower components...
I'm still of the idea that I should have partial_something g^zr for each term, as I need to transform T_z to T^r. This warrants a lot of these steps useless, and leaves me at the 4th from last line:
$= "..." + (-g^{zr}g^{\mu k} \partial_m g_{\mu r} - g^{zr}g^{\mu k} \partial_{\mu} g_{mr} + g^{zr}g^{\mu k}\partial_r g_{m \mu}) T_z$
Which can be simplified to:
$= "..." + (-\delta_r^k \partial_m g^{zr} - g_{mr}g^{\mu k} \partial_{\mu} g^{zr} + \delta_m^k\partial_r g^{zr}) T_z$
and the g_zr can be factored out and contracted with T_z to give
$= "..." + (-\delta_r^k \partial_m - g_{mr}g^{\mu k} \partial_{\mu} + \delta_m^k\partial_r ) T^r$
My thinking is that I need to use the "..." and rearrange, but I'm having trouble seeing how I can get rid of the T_mu in the ...
So, I am, again, at another unique impasse. Fun.
I honestly feel like I'm doing something completely idiotic, I'm just not sure what, when, or where. I expanded the deltas as sums over mu to get a g^mu k attached to every term, then I can factor that out, but that seem to get me anywhere useful, either. The closest that gets me to anything (from what I can see) is:
$g^{\mu k}\partial_m T_{\mu} + g^{\mu k}(\partial_r g_{\mu m} - \partial_{\mu} g_{mr} - \partial_m g_{\mu r}) T^r$
Which almost looks like $\Gamma_{m r}^k$ but I have a floating negative sign somewhere. If I could get $\partial_m g_{\mu r}$ to be positive, then I would have the christoffel, but it's not. However, the other partial term (the one not nested in the christoffel) is a partial w.r.t. the same variable, the only problem is the term is the partial of T_mu, and not T^r. What's even more peculiar is that if you distribute the T^r to the -partial_m, and group it with the other term, you can collapse T^r to T_mu and then you have $g^{\mu k} \partial_m(T_{\mu} - T_{\mu})$
I'm either making mistakes, doing stupid things, or both. Can someone point out where I've went astray?

 

[BiGyElLoWhAt]

Well, I thought I figured it out, but I seem to have overlooked something. Ok so my question is this:
Can I just make both of those T_mu terms positive? I mean, based on what you said earlier with the identities, $g^{\mu k}\partial_m g_{\mu r} T^r = 0$, which means I should just be able to change the signs willy nilly.
So we have $g^{\mu k}\partial_m(T_{\mu} - g_{\mu r} T^r) = g^{\mu k}\partial_m(T_{\mu} + g_{\mu r} T^r)$ which allows us to get the positive m partial term.
Distributing and regrouping with the other two terms gives:
$g^{\mu k}\partial_m (T_{\mu}) + g^{\mu k}(\partial_r g_{\mu m} +\partial_m g_{\mu r} - \partial_{\mu} g_{mr} )T^r$
But according to one poster above, I can't contract the metric in the first term with the T_mu. I'm not sure why. Assuming I can, then I have the answer, but it was very round-about. If I multiply the first term by the kroenecker delta, the term should still be unchanged, but then I can expand the delta in terms of a metric and inverse, and I can't seem to find the correct delta(s) as I need both a mu and a k in the upper index, meaning I need more than one, but then I have introduced issues with the lower indices. This doesn't seem like the way, unless I can multiply by $\delta^{\mu k} = g^{\mu k}g_{\mu k}$, which should be right, considering that the metrics are inverses of each other. Then I can pull out the contravariant metric, which then collapses with the original one to the kroenecker delta, and leaves the covariant inside the partial with the T_mu, allowing me to raise it to T^k.
Is this correct?
Assuming it is, with that solved, I have a question:
You have to write a term that is zero, and carry that through in order to get the relationship. Without knowing what the end is supposed to look like, how was I supposed to find that out? My intuition would have been to just cancel the two terms.
Also, that almost feels like cheating.

 

[Orodruin]

First of all, you are missing a factor 1/2 in the definition of the Christoffel symbols.

The closest that gets me to anything (from what I can see) is:
$g^{\mu k}\partial_m T_{\mu} + g^{\mu k}(\partial_r g_{\mu m} - \partial_{\mu} g_{mr} - \partial_m g_{\mu r}) T^r$

Note that the first term is also not on the form that you would want it to be, you want it to be on the form $\partial_m T^k = \partial_m g^{\mu k} T_\mu$. How can you rewrite this term so that it is the first term you have plus something else? How does this affect what is left for the second term? (The factor of 1/2 is important here!)

 

[BiGyElLoWhAt]

Ok, so I came here because woe is me, and I couldn't get it, as I ended up here:
$g^{\rho \nu} \partial_{\mu} V_{\rho} - 1/2 g^{\rho \nu}\partial_{\mu} g^{\lambda \sigma} g_{\rho \sigma} V_{\lambda} + 1/2 (g^{\nu \sigma}\partial_{\sigma} g_{\mu \rho}V^{\rho} - g^{\nu \sigma}\partial_{\rho}g_{\mu \ sigma}V^{\rho}$
the first term can be expanded to be identical to the second, and you can add them together.
I'm only ending up with 3 terms...
$g^{\rho \nu} \partial_{\mu} V_{\rho}$
$=(1-1/2)g^{\rho \nu} \partial_{\mu}g^{\lambda \sigma}g_{\rho \sigma} V_{\lambda}$
So I'm literally just ending up with the christoffel symbol, and I lost the ordinary derivative term.

 

[bolbteppa]

An easier way to remember these calculations is to include the basis vectors and to treat this as a basic calculus problem, then for a contravariant vector field you have

\begin{align}
\vec{A}_{;j} &= \partial_j (A^i \vec{e}_i) = A^i_{,j}\vec{e}_i + A^i \vec{e}_{i,j} = A^i_{,j} \vec{e}_i + A^i \vec{e}_{i,j} \vec{e}^k \cdot \vec{e}_k \\

&= A^i_{,j} \vec{e}_i + A^i (\vec{e}_{i,j} \cdot \vec{e}^k) \vec{e}_k = A^i_{,j} \vec{e}_i + A^k (\vec{e}_{k,j} \cdot \vec{e}^i) \vec{e}_i = A^i_{,j} \vec{e}_i + A^k \Gamma^i_{kj} \vec{e}_i \\

& = (A^i_{,j} + \Gamma^i_{kj} A^k) \vec{e}_i = A^i_{;j} \vec{e}_i
\end{align}

while for a covariant vector field we have

\begin{align}
\vec{A}_{;j} &= \partial_j (A_i \vec{e}^i) = A_{i,j}\vec{e}^i + A_i \vec{e}^i_{,j} = A_{i,j} \vec{e}^i + A_i \vec{e}^i_{,j} \vec{e}_k \cdot \vec{e}^k \\

&= A_{i,j} \vec{e}^i + A_i (\vec{e}^i_{,j} \cdot \vec{e}_k) \vec{e}^k = A_{i,j} \vec{e}^i - A_i (\vec{e}^i \cdot \vec{e}_{k,j}) \vec{e}^k = A_{i,j} \vec{e}^i - A_k (\vec{e}^k \cdot \vec{e}_{i,j}) \vec{e}^i \\

&= A_{i,j} \vec{e}^i - A_k \Gamma^k_{ij} \vec{e}^i = (A_{i,j} - \Gamma^k_{ij} A_k) \vec{e}_i = A_{i;j} \vec{e}_i

\end{align}

Where I used $$(\vec{e}_i \cdot \vec{e}^k)_{,j} = (\delta_i^k)_{,j} = 0 \rightarrow \vec{e}_{i,j} \cdot \vec{e}^k = - \vec{e}_i \cdot \vec{e}^k_{,j}$$

Note you can do both of these without randomly adding $$1 = \vec{e}_k \cdot \vec{e}^k$$ in the calculation, that just helps me get the indices right, you can ignore them and switch the indices yourself to get the answer, as is done here

http://www.physicspages.com/2013/02/16/covariant-derivative-and-connections/