# 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!

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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.