Proving Covariant Derivative Transforms as Tensor

[blorp]

Homework Statement

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.

The Attempt at a Solution

Let's start with the covariant derivative of of a vector http://members.aol.com/mlucen/2.bmp...up. Any help would be greatly appreciated!

 

[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.

 

[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:

http://members.aol.com/mlucen/8.bmp

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

http://members.aol.com/mlucen/9.bmp

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:

http://members.aol.com/mlucen/10.bmp or equivalently .http://members.aol.com/mlucen/11.bmp

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:

http://members.aol.com/mlucen/12.bmp

I have never heard of such a result [although strangely a somewhat similar equation seems true in certain cases: dxⁿd²x^m = - dx^md²xⁿ. This is derivable from the product rule if one assumes d(dx^mdxⁿ)=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.