# I Covariant derivative of a contravariant vector

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1. Jun 1, 2016

### BiGyElLoWhAt

This is (should be) a simple question, but I'm lost on a negative sign.
So you have $D_m V_n = \partial_m V_n - \Gamma_{mn}^t V_t$ with D_m the covariant derivative.
When trying to deduce the rule for a contravariant vector, however, apparently you end up with a plus sign on the gamma, and I'm not sure how to get there. I think I'm missing a property of christoffel symbols or something.
$D_m V^n = D_m (g^{np}V_p$
$= V_p D_m g^{np} + g^{np} D_m V_p$
$= 0 + g^{np} [\partial_m V_p - \Gamma_{mp}^z V_z]$
$= \partial_m V^n - g^{np} \Gamma_{mp}^z V_z$
The last term, I wouldn't think the negative sign cancelled out, but apparently it does. How?

2. Jun 1, 2016

### Orodruin

Staff Emeritus
You are not using the easiest approach and end up with an expression containing the metric and it is not directly clear how what you obtained relates to the Christoffel symbols and the contravariant components of V. Instead, start from the inner product between a tangent vector field and a dual vector field. This is a scalar field and its derivative is just the partial derivative, but you can also compute it using the product rule for the connection.

This is a general property which is not related to the metric tensor at all, it holds for all connections.

3. Jun 1, 2016

### BiGyElLoWhAt

In the video series, he worked out explicitly and solved for the covariant derivative of a covector, which involved the christoffel symbol. His hint for the derivative of a contravariant vector was to rewrite it as a covector contracted with the metric tensor and solve from there. If I contract the last term, I get exactly what he has except the minus sign, which I assumed came out in some property of contracting christoffel symbols or something.

Is using the tangent space really easier? And for the connection, do you mean the geodesic equation with the second derivative? Both of those seem rather in depth on the surface, but I haven't worked out many things explicitly, so I may be mistaken.

4. Jun 1, 2016

### Orodruin

Staff Emeritus
Who is "he"?

Are you writing in terms of the covariant or contravariant components. Make sure that all components of V which appear are contravariant.

Much easier. It does not even rely on the connection being metric compatible - or even the existence of a metric.

No, I mean a general affine connection.

5. Jun 2, 2016

### BiGyElLoWhAt

He is Leonard Susskind ( sussikind? ) from stanford. He has approx 20 hours of GR Lecture up. the first few episodes are all geometry after you gwt past the equivalence principle.

As of right now, V only has 1 index. Is that what you mean?

Is a form that you're referring to different than an n-form?

I'll have to look into the affine connection. The only thing I know of from the lecture series is the tangent definition of a geodesic, which involves the second derivative, and with connections, just things I've read on the internet in random places at random times, namely the levi-cevita connection, which I believe is also affine and a geodesic. That may kr may not be wrong, though.

6. Jun 2, 2016

### BiGyElLoWhAt

Here, for reference. It comes in at about the 9minute mark.
Also, when he refers to "the best frame" he is referring to what he was previously referring to as gaussian coordinates, but people were confused and at the beginning of 4 he changed the terminology to be more general.

7. Jun 2, 2016

### Orodruin

Staff Emeritus
No, I am referring to whether the components of $V$ which are referred to are covariant or contravariant.

The correct form for the covariant derivative of the contravariant components is $\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\sigma} V^\sigma$ (note that the index on the $V$ is contravariant!). This is generally true for any affine connection, but you can show it explicitly for the case of the Levi-Civita connection. You will likely need the relation $0 = \partial_\mu \delta^\nu_\sigma = \partial_\mu g^{\nu\rho}g_{\sigma\rho} = g^{\nu\rho}\partial_\mu g_{\sigma\rho} + g_{\sigma\rho}\partial_\mu g^{\nu\rho}$ in order to play with the expression of the Levi-Civita connection in terms of the metric.

8. Jun 2, 2016

### BiGyElLoWhAt

That is the same expression that is presented. I'll give the levi civita connection a shot and post back if I have more questions.
Thanks.

9. Jun 2, 2016

### Orodruin

Staff Emeritus
It is only the same expression as you gave in #1 if you can show that $g^{ab} \Gamma^c_{mb} V_c = -\Gamma^a_{mb}V^b$. However, I do not understand why Susskind would take a detour via the metric tensor when it is a general property for any affine connection.

10. Jun 2, 2016

### BiGyElLoWhAt

My guess is because we're working with limited differential geometry. It's 20 hours of GR, but with (probably) the bare essentials of the diff geom. We haven't talked about nor defined a connection, let alone an affine connection, and Levi-Civita hasn't been mentioned once.

When I said "presented", I meant by Susskind. That's the expression he presents on the board, just not sure how to get there.

11. Jun 2, 2016

### Orodruin

Staff Emeritus
Well, in reality you are doing the Levi-Civita connection. It is the unique metric compatible and torsion free connection. (Torsion free is equivalent to the Christoffel symbols being symmetric in the lower indices.) In my opinion, it actually helps later understanding to do the more general case first, but it might be personal preference.

Well, try to do it by manipulating the expression you got using the identity I mentioned. Once you are done we can discuss the general case.

12. Jun 3, 2016

### nrqed

Well, you still have to go from $g^{np} \Gamma_{mp}^z V_z$ to $\Gamma^n_{m r} V^r$, right? So there is still some work needed to bring it in the form he gives. Do you see how to do that?

13. Jun 3, 2016

### BiGyElLoWhAt

Not immediately, no. Currently, I'm still exploring the identity that Orodruin said might be useful. I'm trying to add zero to one side, but I ended up having to introduce a second partial in order to keep the indices correct. (Partial of zero w.r.t. anything is zero).
Don't spoil it for me, but if you want to give me a hint, I would appreciate that.

Oh, wait, do you mean just the index contraction? I know how to do that. The issue is getting the negative sign to come out.

14. Jun 3, 2016

### Orodruin

Staff Emeritus
Don't do that, use it to rewrite terms. Anyway, that you do what was suggested in #12 was the intention.

15. Jun 3, 2016

### wrobel

axiomatic definition of covariant derivative includes the following axiom
If $f, v$ are the covector and vector fields respectively then $\nabla_u\langle f,v\rangle=\langle\nabla_u f,v\rangle+\langle f,\nabla_u v\rangle$. From this formula it follows that if
$\nabla_i v^k=\frac{\partial v^k}{\partial x^i}+\Gamma_{si}^kv^s$ then $\nabla_i f_k=\frac{\partial f_k}{\partial x^i}-\Gamma_{ki}^sf_s$ Metric tensor does not relate here

16. Jun 3, 2016

### BiGyElLoWhAt

You mean there is a change of index that I was missing?
I thought that it was simply choice of running index.

17. Jun 3, 2016

### Orodruin

Staff Emeritus
Start with the expression of the Christoffel symbols in terms of the metric. You should then be able to rewrite the term you have in the form required, as described in #12.

18. Jun 3, 2016

### BiGyElLoWhAt

Oh man, I feel dumb lol. Thanks for bearing with me. "Expand the christoffell" is what I needed. Let me continue working through it, and I'll post my work just because.

19. Jun 3, 2016

### BiGyElLoWhAt

Ok, well, I think I'm on the right track, but I seem to have encountered a road block.
$g^{np} \Gamma_{mp}^z V_z = g^{np} [1/2 g^{zr}(\partial_m g_{pr} + \partial_p g_{mr} - \partial_r g_{mp})]$
$=1/2 [g^{np}g_{pr}\partial_m +g^{np} g_{mr} \partial_p - g^{np}g_{mp}\partial_r ]g^{zr}V_z$
$=1/2 [ \delta^n_r\partial_m + g^{np}g_{mr} \partial_p - \delta^n_m\partial_r ] V^r$
And I'm not sure where to go from here. Is this the right path? I'm pretty confident I need the $g^{zr}$ inside the partials, so I can raise the index on V, and if I keep them on the outside, I get deltas with an upper z index.
I'm not sure exactly what I'm looking for. I know I need z and r on a metric, so I can raise V_z to V^r. If I don't collapse the metrics to the delta's, I get
$1/2 g^{np}[\partial_m g_{pr} + \partial_p g_{mr} - \partial_r g_{mp} ]V^r$
which looks like $\Gamma_{mp}^n V^r$, but that still doesn't do much, I think...
***
Wait a second. So I'm a little confused, all of my indices still add up, but this doesn't seem to make sense, at least to me. The christoffel should be intrinsically summed over p (within itself) because of the g^np, however, the partial term, -partial_r , should have the intrinsic summation index, implying that it should be summing over r within the christoffel, but it only sums r when multiplied by V^r.

Last edited: Jun 3, 2016
20. Jun 3, 2016

### Orodruin

Staff Emeritus
The p on the LHS of your expression is not a free index ...