I Covariant Derivative Rank 2 Contravariant Tensor

[Bishal Banjara]

after 1 hr onwards

 

[Bishal Banjara]

Particularly, lets take as as what you mentioned, $${\nabla}_{t}T^{tt}=\partial_{t}T^{tt}+\Gamma^t_{t\gamma}T^{t\gamma}+\Gamma^t_{r\gamma}T^{t\gamma}$$. Is this possible or not, if we extend the equation taking all summation where $$\gamma$$ runs from t to $$\phi$$ as $${\nabla}_{t}T^{tt}=\partial_{t}T^{tt}+\Gamma^t_{tt}T^{tt}+\Gamma^t_{tr}T^{tr}+\Gamma^t_{t\theta}T^{t\theta}+\Gamma^t_{t\phi}T^{t\phi}+\Gamma^t_{tt}T^{tt}+\Gamma^t_{rt}T^{tr}+\Gamma^t_{r\theta}T^{t\theta}+\Gamma^t_{r\phi}T^{t\phi}$$.?

my core quest is this...

 

[PeterDonis]

after 1 hr onwards

In other words, the same material you would find in any GR textbook. As @Orodruin already pointed out way back in post #2. And as he said, that material will of course tell you how to take the covariant divergence of a tensor of any rank whatever.

my core quest is this...

"This" makes no sense. It looks like you are trying to take the covariant divergence of one component of a tensor. That is nonsense.

If that is the best you can do at explaining what your "core quest" is, we might as well close this thread. Can you do any better?

 

[Ibix]

Particularly, lets take as as what you mentioned, $${\nabla}_{t}T^{tt}=\partial_{t}T^{tt}+\Gamma^t_{t\gamma}T^{t\gamma}+\Gamma^t_{r\gamma}T^{t\gamma}$$. Is this possible or not, if we extend the equation taking all summation where $$\gamma$$ runs from t to $$\phi$$ as $${\nabla}_{t}T^{tt}=\partial_{t}T^{tt}+\Gamma^t_{tt}T^{tt}+\Gamma^t_{tr}T^{tr}+\Gamma^t_{t\theta}T^{t\theta}+\Gamma^t_{t\phi}T^{t\phi}+\Gamma^t_{tt}T^{tt}+\Gamma^t_{rt}T^{tr}+\Gamma^t_{r\theta}T^{t\theta}+\Gamma^t_{r\phi}T^{t\phi}$$.?

No. This:$${\nabla}_{t}T^{tt}=\partial_{t}T^{tt}+\Gamma^t_{t\gamma}T^{t\gamma}+\Gamma^t_{r\gamma}T^{t\gamma}$$does not make sense. The general equation is $${\nabla}_{a}T^{ab}=\partial_{a}T^{ab}+\Gamma^b_{ad}T^{ad}+\Gamma^d_{ad}T^{ba}$$It is actually four equations, which you can tell because there is one free index, one that is not repeated and hence not summed over. You can specify one of the equations by setting $b=t$, for example, in which case that one equation is $${\nabla}_{a}T^{at}=\partial_{a}T^{at}+\Gamma^t_{ad}T^{ad}+\Gamma^d_{ad}T^{ta}$$I think what you are trying to do is ask if the above means that $${\nabla}_{t}T^{tt}=\partial_{t}T^{tt}+\Gamma^t_{td}T^{ad}+\Gamma^d_{td}T^{tt}$$(where no summation is implied over $t$). It does not, any more than 1+2+3=2+2+2 implies that 1=2, 2=2, and 3=2.

 

[Bishal Banjara]

In other words, the same material you would find in any GR textbook. As @Orodruin already pointed out way back in post #2. And as he said, that material will of course tell you how to take the covariant divergence of a tensor of any rank whatever."This" makes no sense. It looks like you are trying to take the covariant divergence of one component of a tensor. That is nonsense.

If that is the best you can do at explaining what your "core quest" is, we might as well close this thread. Can you do any better?

yes, @Orodruin had already mentioned. "If that is the best you can do at explaining what your "core quest" is, we might as well close this thread. Can you do any better?" but why?

 

[PeterDonis]

yes, @Orodruin had already mentioned. "If that is the best you can do at explaining what your "core quest" is, we might as well close this thread. Can you do any better?" but why?

Because, as has been said multiple times now, what you appear to be doing makes no sense. Since you are apparently unable to explain what you are doing any better, this thread is now closed. References have been given to the correct way of computing a covariant divergence.