Why In 4D, the four-divergence of the four-curl is not zero, for ∂νGμν

[physicsgirl07]

1. prove in the 4-dimensional Riemannian space, the 4-divergence of the 4-curl is not zero that is
where is the 2d’Alembertian operator



2.∂νGμν = ∂μ∂νaν(xκ)−2aμ(xκ) = 0

 

[dextercioby]

So you're asked to prove that in 4D Riemann space (no torsion, connection is symmetric and metric compatible)

$$\nabla_{\mu}\left(\nabla^{\mu}T^{\nu} -\nabla^{\nu}T^{\mu}\right) \neq 0$$

Do you know which formulas you need to use ?

 

[physicsgirl07]

yes iknow ∇· [∇×a(x)] = ∂i[∇×a(x)]i = i jk∂i∂jak(x)

but i want to prove it in 4-d

becouse in 3-d equal to zero

see this link

http://www.scribd.com/doc/19388495/152/The-curl

see the page that have title curl
becouse i don't know how to wright the formula

thanx

 

[dextercioby]

Ok, I would choose a free component (let's say in my notation $\nu$) and make the additions involved. What would you get, if you did the same ?