# Covariant derivative

1. Feb 21, 2008

### blorp

1. The problem statement, all variables and given/known data

Help! I wish to prove the following important statements:

(1) The presence of Christoffel symbols in the covariant derivative of a tensor assures that this covariant derivative can transform like a tensor.

(2) The reason for this is because, under transformation, the Christoffel symbol picks up a term that cancels out a problematic term which would prevent the covariant derivative from transforming as a tensor. That term arises from the transformation of the other part of the covariant derivative, which is the partial derivative of the tensor. (This is why the partial derivative of a tensor does not in general transform as a tensor.)

These statements come from 'Relativity Demystified' (McMahon, p. 68; you can look it up in Amazon book search). The book provides some supporting equations for the assertions, but not a full proof. I would like to derive a proof. Here is what I've gotten so far.

3. The attempt at a solution

Let's start with the covariant derivative of of a vector . The components of the covariant derivative are denoted as . The various books establish (without getting into the proofs & definitions behind this) that this also =

, where the Г is the Christoffel symbol.

Our task is to show that this transforms as a tensor; and that the reason it does, is because the Г term transformation cancels out otherwise problematic quantities arising from transformation of the other term. This is what the book says.

A way to do this is presumably as follows:

(a) Write out an equation showing how the covariant derivative would transform if it transformed under ordinary tensor rules.

(b) Transform each term of the covariant derivative separately. Show that this leads to the same result as in (a). And show that it does so because the transformation of the Г term cancels out "problems" arising from transformation of the other other term.

I am now going to put primes after all the indices used above, because this will more closely follow the relevant equations in the book, and I already did it this way so it's easier for me. Transformation will then be to an uprimed coordinate system. First (a), we write an equation showing how the covariant derivative (the one above but primed) would transform if it did so under ordinary tensor rules.

(Eq. 1)

Above the first two terms show the basic transformation equation. Subsequent terms show its expansion in terms of the two known parts of the unprimed covariant derivative.

Our next task, as stated in (b), is to recover the same equation by transforming each of the two parts of the original (primed) covariant derivative separately. i.e., transform each of the two halves of

.

This transformation leads to the following results. For the first half:

(Eq. 2)

[Note that the 2nd order derivative term in Eq. 2 that makes it impossible for the partial derivative of a tensor -- alone -- to transform as a tensor. It would clearly do so without that term.]

For the second half:

. (Eq. 3)

This skips over quite a bit of calculation, but it's all shown in McMahon and the final answer is lifted straight from there. Note that I have left in some terms that apparently would be easily canceled out. This again follows McMahon's presentation; I think some of these extra terms may be useful for conversions that would complete the proof.

In fact, our proof would in fact be complete if we could show that

(everything after the last = in Eq. 2) +
(everything after the last = in Eq. 3) =
(everything after the last = in Eq. 1).

It looks promising: the terms look vaguely similar, and presumably the second-order derivative terms cancel out. Unfortunately, even the most elaborate index gymnastics that I can think of fail to deliver me the conclusion for which I yearn. The terms just don't seem to add up.

Any help would be greatly appreciated!

2. Feb 22, 2008

### Ben Niehoff

In EQ2, you have a factor of

$$\frac{\partial A^n}{\partial x^{b\,'}}$$

This isn't fully transformed. You need to apply the chain rule to get:

$$\frac{\partial A^n}{\partial x^m} \frac{\partial x^m}{\partial x^{b\,'}}$$

That should help.

In EQ3, you also need to transform

$$A^{c\;'}$$

to

$$\frac{\partial x^{c\;'}}{\partial x^k}A^k$$

this should cancel one of the factors on $\Gamma^d_{mn}$ in that equation.

Then, you have to play with the partial derivatives a bit to get the second-order terms to cancel.

Also, remember that

$$\frac{\partial x^{\mu}}{\partial x^{\lambda\;'}}\frac{\partial x^{\lambda\;'}}{\partial x^{\nu}} = \delta^{\mu}_{\nu}$$

I'm not sure if that will help.

Last edited: Feb 22, 2008
3. Feb 24, 2008

### blorp

Thanks for your reply. I suspect you are right that the transformation in Eq. 2 is wrong. However, the correction you suggest still doesn't seem to lead to the correct result. Implementing both your suggested changes, the full equation (that is, the addition of eq's 2&3 above) becomes:

As mentioned earlier, this should add up to my Eq. 1 above, in particular the part after the last "=" sign. Unfortunately, it doesn't seem to work out. Let me focus only on the fact that the terms containing second derivatives (essentially the first & third terms) should cancel out. This would mean that

There is certainly plenty of simplification and relabeling that can be done here to approach our desired result. For one, we relabel k to n. Then, we can simplify using the kronecker delta notation that Ben noted. The dx's with indices n, b' and d can be subsumed into one kronecker delta as follows:

[Broken] or equivalently . [Broken]

If I'm not mistaken we can also cancel out from the right side any dx's with the same index on top and bottom (this in fact leads to the same result as using kronecker deltas) so here the c' indices can come out from the right side. Additionally we can eliminate the (1/dxn)An from both sides of the equation.

These two simplifications lead to the following highly questionable (to me) proposition, which is in fact the result I've been getting all along:

[Broken]

I have never heard of such a result [although strangely a somewhat similar equation seems true in certain cases: dxnd2xm = - dxmd2xn. This is derivable from the product rule if one assumes d(dxmdxn)=0. This equation raised my hopes for awhile but it doesn't actually seem applicable here.]

Another troubling issue, quite apart from this, is that my original derivation of a transformation for covariant derivative -- the one directly using tensor transformation rules (Eq. 1) -- seems fishy. In the second, Christoffel term, the indices don't 'pair up' properly by the Einstein convention. The n's are both on top and the m's both below. Something looks wrong there. I suspect this is contributing to the problem somehow.

Last edited by a moderator: Apr 23, 2017 at 11:03 AM