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

  • #1
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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
dextercioby
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So you're asked to prove that in 4D Riemann space (no torsion, connection is symmetric and metric compatible)

[tex] \nabla_{\mu}\left(\nabla^{\mu}T^{\nu} -\nabla^{\nu}T^{\mu}\right) \neq 0 [/tex]

Do you know which formulas you need to use ?
 
  • #3
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 dont know how to wright the formula

thanx
 
  • #4
dextercioby
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Ok, I would choose a free component (let's say in my notation [itex] \nu [/itex]) and make the additions involved. What would you get, if you did the same ?
 

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