Invariant divergence and christoffel symbols

[Felicity]

Homework Statement

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/∂x^k + Гⁱ_klA^l

Г^k_ij = g^kl/2 (∂g_il/∂x^j+∂g_lj/∂xⁱ-∂g_ij/∂x^l)

Homework Equations

g is the metric tensor

hints from class:
you will have to check that Г^l_il = 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

The Attempt at a Solution

so far I have

divA = ∂Aⁱ/∂x^k + (∂g_il/∂x^j+∂g_lj/∂xⁱ-∂g_ij/∂x^l)A^l

but i don't 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

 

[tiny-tim]

so far I have

divA = ∂Aⁱ/∂x^k + (∂g_il/∂x^j+∂g_lj/∂xⁱ-∂g_ij/∂x^l)A^l

but i don't 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?

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 + … ?

 

[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 )A^l

which brings me to the hint

Г^l_il = 1/√g ∂i√g

which must be equal to

Г^k_ij = g^kl/2 (∂g_il/∂x^j+∂g_lj/∂xⁱ-∂g_ij/∂x^l)

when k is the same as j

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

Г^l_il = 1/√g ∂i√g = g^im/2 (∂g_im/∂x^l+∂g_ml/∂xⁱ-∂g_il/∂x^m)

I also learned that ∂g_ml/∂xⁱ=∂g_il/∂x^m) so those terms cancle to give

1/√g ∂i√g = g^im/2 (∂g_im/∂x^l)

since

∂g/∂x_i = gg_im(∂g_im/∂x∂g/∂x^l)


1/√g ∂i√g = 1/2g (∂g/∂x^l)

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

 

[tiny-tim]

Hi Felicity!

The question requires you to prove that

Г^l_il = 1/√g ∂i√g

Hint: g is just a number, gⁱ_i,

so ∂_i√g = … ?