# Homework Help: Invariant divergence and christoffel symbols

1. Sep 27, 2008

### Felicity

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

show that the definition of the invariant divergence

divA = 1/√g ∂i (√g Ai)

is equivalent to the other invariant definition

divA = Ai;i

Ai;k = ∂Ai/∂xk + ГiklAl

Гkij = gkl/2 (∂gil/∂xj+∂glj/∂xi-∂gij/∂xl)

2. Relevant equations

g is the metric tensor

hints from class:
you will have to check that Гlil = 1/√g ∂i√g

you will have to differentiate determinants

write the answer in terms of the derivative of the metric tensor and inverse metric tensor

3. The attempt at a solution

so far I have

divA = ∂Ai/∂xk + (∂gil/∂xj+∂glj/∂xi-∂gij/∂xl)Al

but i dont know if this is correct or where to go from here. Christoffel symbols are new to me, can anyone point me in the right direction?

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

2. Relevant equations

3. The attempt at a solution

2. Sep 27, 2008

### tiny-tim

Hi Felicity!

You haven't got this index summation thing, have you?

Hint: divA is a scalar, so that line must have no uncancelled indices (every index up must be paired with itself down) …

divA = ∂Ai/∂xi + … ?

3. Sep 29, 2008

### Felicity

Thank you so much for the reply! I see my mistake now but I still have not solved the problem

I see now that

divA = 1/√g ∂i (√g Ai) = ∂Ai/∂xi + (1/√g ∂i √g )Al

which brings me to the hint

Гlil = 1/√g ∂i√g

which must be equal to

Гkij = gkl/2 (∂gil/∂xj+∂glj/∂xi-∂gij/∂xl)

when k is the same as j

So if I relabel my indices to be consistent I get the equation

Гlil = 1/√g ∂i√g = gim/2 (∂gim/∂xl+∂gml/∂xi-∂gil/∂xm)

I also learned that ∂gml/∂xi=∂gil/∂xm) so those terms cancle to give

1/√g ∂i√g = gim/2 (∂gim/∂xl)

since

∂g/∂xi = ggim(∂gim/∂x∂g/∂xl)

1/√g ∂i√g = 1/2g (∂g/∂xl)

which requires me to find the derivative of the determinant of the metric tensor, how do I do that?

Last edited: Sep 29, 2008
4. Sep 29, 2008

### tiny-tim

Hi Felicity!

The question requires you to prove that

Гlil = 1/√g ∂i√g

Hint: g is just a number, gii,

so ∂i√g = … ?