The covariant derivative of a contravariant vector

1. Jan 21, 2005

tennishaha

Since there are some equations in my question. I write my question in the following attachment. It is about the covariant derivative of a contravariant vector.

Thank you so much!

Attached Files:

• The covariant derivative of a contravariant vector is.doc
File size:
34.5 KB
Views:
154
Last edited: Jan 22, 2005
2. Jan 22, 2005

tennishaha

Can anyone give me some guidance?
Thanks

3. Jan 22, 2005

dextercioby

Okay.
1.Those Gamma's components are not zero...Not in the general case,anyway...

2.I'll use the column-semicolumn notation (though we physicst are not really fond of it...)
In the following,"g" is the determinant of the metric tensor:
$$g_{,i}=g \ g^{kl} g_{kl,i}$$ (1)
(1):This is the rule as how to differentiate the determinant of a matrix...

$$A^{i}_{;i}=A^{i}_{,i}+\Gamma^{i} \ _{ij}A^{j}$$(2)

(2):The covariant divergence (the one u're interested in).

$$\Gamma^{i} \ _{ij} =\frac{1}{2}g^{ki}(g_{kj,i}+g_{ik,j}-g_{ji,k}) =\frac{1}{2}g^{ki}g_{ki,j}=\frac{1}{2g}g_{,j}$$ (3)

In getting (3) I made use of the definition of 2-nd rank Christoffel symbols (mannifold with both connection & metric) and of relation (1).

Use (3) and (2) and the fact that:
$$g=(\sqrt{g})^{2}$$ (4)

Report any problems...

Daniel.

Last edited: Jan 22, 2005
4. Jan 24, 2005

tennishaha

But from your result (3), there should be 3 terms in the following equation's second part of right hand side.
$$A^{i}_{;i}=A^{i}_{,i}+\Gamma^{i} \ _{ij}A^{j}$$
Then the result is not the same with my results.

5. Jan 24, 2005

dextercioby

What 3 terms are u talking about??The ones in the definition of Christoffel's symbols...??

Daniel.