Vector identity proof in general curvilinear coordinates

[Earthland]

Homework Statement

Need to prove that:

,b means partial differentation with respect to b, G is the metric tensor and Γ is Christoffel symbol.

I think I could proceed with this quite well if I could understand the hint given, that I should lower the index j.

Homework Equations

a_m=G_mja^j

The Attempt at a Solution

For the first member on the left I could write

G_mjG^jk∂_k=δ^k_m∂_k=∂_k

and else than that it should stay the same

For members on the right, lowering j would give me

v^α(v_j,α+Γ_jαβv^β)

However, what should I do with the second member on the left? And I am told that this one is the important one.

 

[mighty2000]

First, let's rewrite the given equation in a more clear form:

∂b(Gam)=∂b(GmjGaj)

Now, we can use the product rule for differentiation to expand the left side:

∂b(Gam)=∂b(Gmj)Gaj+Gmj∂b(Gaj)

Next, we can use the definition of the metric tensor to rewrite the first term on the right side:

∂b(Gam)=∂b(Gmj)Gaj+δmj∂b(Gaj)

Now, we can use the definition of the Christoffel symbol to rewrite the second term on the right side:

∂b(Gam)=∂b(Gmj)Gaj+δmj(∂bGaj+ΓajbGab)

Next, we can use the definition of the metric tensor again to rewrite the first term on the right side:

∂b(Gam)=∂b(Gmj)Gaj+δmj(∂bGaj+ΓajbGab)

=∂b(Gmj)Gaj+δmj(∂b(Gmj)Gab+ΓajbGab)

Now, we can use the product rule again to expand the first term on the right side:

∂b(Gam)=∂b(Gmj)Gaj+δmj(∂b(Gmj)Gab+ΓajbGab)

=∂b(Gmj)Gaj+δmj(∂b(Gmj))Gab+δmjGmj(∂bGab)+δmjΓajbGab

Finally, we can use the definition of the metric tensor one more time to simplify the second term on the right side:

∂b(Gam)=∂b(Gmj)Gaj+δmj(∂b(Gmj))Gab+∂bGab+δmjΓajbGab

=∂b(Gmj)Gaj+δmj(∂b(Gmj))Gab+∂b(Gmj)Gab+δmjΓajbGab

Now, we can see that the first and third terms on the right side are equivalent, and the second and fourth terms are equivalent. Therefore, we can rewrite the equation as:

∂b(Gam)=∂b(Gmj)Gaj+