I Covariant Derivative Rank 2 Contravariant Tensor

[Ibix]

Agreed.

The presence of a $\mathrm{cosec}^2$ in OP's expressions (plus previous confusing threads about whether the Schwarzschild $g^{ab}$ was different from a $g_{ab}$ whose components happen to be equal to the components of the Schwarzschild $g^{ab}$) makes me think OP has actually supplied $T^{ab}$ mis-labelled as $T_{ab}$. I've closed Maxima but can check later, but would his expressions be correct for $T^{ab}$?

 

[PeterDonis]

would his expressions be correct for $T^{ab}$?

Possibly, the $\csc^2 \theta$ factor would be right, but I haven't checked it in detail since the OP's expressions are in terms of the "potential" $\phi$ instead of $M$.

 

[PeterDonis]

$$g_{tt}=-(1+2GM/r)^{-1}, g_{rr}=(1+2GM/r), g_{\theta\theta}=r^2 and g_{\phi\phi}=r^2{sin^2\theta}$$

Where does this metric come from?

 

[Ibix]

Possibly, the $\csc^2 \theta$ factor would be right, but I haven't checked it in detail since the OP's expressions are in terms of the "potential" $\phi$ instead of $M$.

It kind of makes sense as a mishmash of mixed-index and upper-index components if you assume that $\Phi=GM/r$, rather than $2GM/r$ as originally stated and that there has been some carelessness about signs.

Eh. Either OP is being careless about presentation or needs to revisit his calculations. Either way I think I've done enough algebra for now.

 

[Bishal Banjara]

@lbix, I have not stated that $${\phi=GM/r}$$ and the correct one is $${\phi=2GM/r}..$$

 

[PeterDonis]

@lbix, I have not stated that $${\phi=GM/r}$$ and the correct one is $${\phi=2GM/r}..$$

You should be aware that this is not standard notation. The standard Newtonian potential, which is what $\phi$ normally refers to, is $- GM / r$ (note both the minus sign and the absence of the factor 2). Making up your own idiosyncratic notation is generally not a good idea.

 

[Bishal Banjara]

sorry, $$\phi=GM/r$$...not $$\phi=-GM/r$$

 

[PeterDonis]

sorry, $$\phi=GM/r$$...not $$\phi=-GM/r$$

I know that is your notation. I am telling you that it's not standard notation. Standard notation has a minus sign in front, for good reasons.

 

[PeterDonis]

Where does this metric come from?

@Bishal Banjara you still have not answered this question. You need to.

 

[Bishal Banjara]

I took it heuristically..

 

[PeterDonis]

I took it heuristically..

That's not a helpful response. Did you find this metric in a textbook? Or derive it from something in a textbook? What is its physical meaning supposed to be? Please give more details.

 

[Ibix]

@lbix, I have not stated that $${\phi=GM/r}$$ and the correct one is $${\phi=2GM/r}..$$

Then you definitely need to recheck your maths.

 

[Bishal Banjara]

I find it helpful to write out the sums explicitly. Thus$$\begin{eqnarray*}
0&=&\partial_aT^{ab}+\Gamma^a_{ad}T^{db}+\Gamma^b_{ad}T^{da}\\
&=&\sum_a\partial_aT^{ab}\\
&&+\sum_a\sum_d\Gamma^a_{ad}T^{db}\\
&&+\sum_a\sum_d\Gamma^b_{ad}T^{da}\end{eqnarray*}$$where there is no implied summation over repeated indices in the second line (there is in the first). There is one free index, so this us four equations, one for each $b$. Doing the $b=r$ case, you have$$\begin{eqnarray*}
0&=&\sum_a\partial_aT^{ar}\\
&&+\sum_a\sum_d\Gamma^a_{ad}T^{dr}\\
&&+\sum_a\sum_d\Gamma^r_{ad}T^{da}\\
&=&\partial_rT^{rr}\\
&&+\sum_a\Gamma^a_{ar}T^{rr}\\
&&+\sum_a\Gamma^r_{aa}T^{aa}\end{eqnarray*}$$Again there is no implied summation. We have used the fact that $T^{ab}=0\ \forall\ a\neq b$ to drop summation terms that are obviously zero in order to go from the first line to the second.

I should note that putting the summation signs in has absolutely no effect except that I personally find it useful as a way of tracking what is being summed over and what is not.

Why can't I write the covariant derivative of $$T_{tt}$$ components as $${\nabla}_{t}T_{tt}=\partial_{t}T_{tt}-\Gamma^t_{t\gamma}T_{t\gamma}-\Gamma^t_{r\gamma}T_{t\gamma}$$ where, $$\gamma$$ runs from $$t$$ to $$\phi$$?

 

[Ibix]

You can't take the covariant derivative of a component. A component is defined with respect to a coordinate system - how could it vary covariantly?

Furthermore your expression involves summing over two lower indices, which is not legal.

 

[Bishal Banjara]

We could find numerous example following the covariant type as what I have done. My question is not exactly for that. I am expecting the difference in the result that I mentioned and what you mentioned (you could assume as what you took of contravariant type). I have mentioned the same way you expressed formerly.

 

[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 $${\nabla}_{t}T^{tt}=\partial_{t}T^{tt}+\Gamma^t_{tt}T^{tt}+\Gamma^t_{tr}T^{tr}+\Gamma^t_{t\theta}T^{t\theta}+\Gamma^t_{t\phi}T^{t\phi}+\Gamma^t_{tt}T^{tt}+\Gamma^t_{rt}T^{tr}+\Gamma^t_{r\theta}T^{t\theta}+\Gamma^t_{r\phi}T^{t\phi}$$.?

 

[PeterDonis]

@Bishal Banjara you do not seem to understand what the covariant divergence is. It is a tensor. If you have a tensor $T^{ab}$, its covariant divergence is $\nabla_a T^{ab}$. Since this expression has one free upper index, it is a vector; the "covariant divergence" is the entire vector, the geometric object, not just one component of it.

 

[Bishal Banjara]

In the lecture of Leonardo Susskind, he writes the same form I mentioned when deriving the Christoffels from metric tensors. There are other also, Iets not make this as an issue but my core quest is other as I mentioned above.

 

[PeterDonis]

In the lecture of Leonardo Susskind

Reference, please?

 

[PeterDonis]

Iets not make this as an issue

If you are doing wrong things, how can we not make it an issue?

my core quest is other as I mentioned above.

I'm afraid I still don't understand what your "core quest" is, if it isn't trying to understand the proper way to compute a covariant divergence.

 

[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.