- #1
binbagsss
- 1,254
- 11
So Einstein Equation: ##G_{uv}= 8 \pi G T_{uv} ##,
Justifying the cosmological constant can be included is done by noting that ## \bigtriangledown^{a}g_{ab} =0 ## and so including it on the LHS, conservation of energy-momentum tensor still holds.
I'm not sure why ## \bigtriangledown^{a}g_{ab} =0 ##. The source I'm using says to 'recall' this, and it is talking about the FRW tensor.
The only thing I can think of is the fundamental theorem of Riemannian geometry : ## \bigtriangledown_{a}g_{bc}= 0 ##. But this doesn't does look right as it has 3 free indicies, not 1, and a lower indice instead of a upper on the ## \bigtriangledown ##
Thanks for your help in advance.
On a side note, I think I am confused between 'divergence' and 'covariant derivative', when we say ## \bigtriangledown_{a} T^{ab} = 0 ##, conservation of energy-momentum tensor that its 'divergence' is zero, is this saying it's convariant derivative is zero?
Justifying the cosmological constant can be included is done by noting that ## \bigtriangledown^{a}g_{ab} =0 ## and so including it on the LHS, conservation of energy-momentum tensor still holds.
I'm not sure why ## \bigtriangledown^{a}g_{ab} =0 ##. The source I'm using says to 'recall' this, and it is talking about the FRW tensor.
The only thing I can think of is the fundamental theorem of Riemannian geometry : ## \bigtriangledown_{a}g_{bc}= 0 ##. But this doesn't does look right as it has 3 free indicies, not 1, and a lower indice instead of a upper on the ## \bigtriangledown ##
Thanks for your help in advance.
On a side note, I think I am confused between 'divergence' and 'covariant derivative', when we say ## \bigtriangledown_{a} T^{ab} = 0 ##, conservation of energy-momentum tensor that its 'divergence' is zero, is this saying it's convariant derivative is zero?