Covariant Derivative Rank 2 Contravariant Tensor

[Bishal Banjara]

TL;DR

Though, we are familiar with covariant derivative of contravariant tensor of index 1 it is perhaps that we don't have it with index 2. I want a clarification on it.

 

[Orodruin]

I am not sure I understand your question, this should be covered in any basic text on GR or differential geometry. See also, eg, https://en.wikipedia.org/wiki/Covariant_derivative#Covariant_derivative_by_field_type

 

[vanhees71]

You simply have to put the "correction" with the Christoffel symbol for each index on the tensor components:
$$\mathrm{D}_j A^{kl}=\partial_j A^{kl} + {\Gamma^k}_{ij} A^{il} + {\Gamma^l}_{il} A^{ki}.$$

 

[PeterDonis]

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[Bishal Banjara]

I have got one of the stress energy momentum tensor component $$T_{rr}=-\frac{\Phi^2}{2\pi\mathcal{G}r^2 \left(1+2\Phi\right)^2,} where, \phi=2GM/r$$, how could I solve the covariant derivative following this rule to check whether that tensor is conserved? or, how to abstract/specify the Christoffel's other than $${\tau_{rr}}^r$$?

 

[Ibix]

Without specifying the other components of $T_{ab}$? You can't, since you need to show that$$\begin{eqnarray*}
0&=&\nabla_aT^{ab}\\
&=&\partial_aT^{ab}+\Gamma^a_{ad}T^{db}+\Gamma^b_{ad}T^{da}\end{eqnarray*}$$for all $b$, which ends up requiring you to know all the components of $T$, or else that all Christoffel symbols that multiply an unknown element are zero. I tend to suspect the latter would lead you into a contradiction if you only know one element of $T$, although I haven't tried to prove it.

 

[Bishal Banjara]

@lbix, sure, the rest components are $$T_{tt}=\frac{\phi^2}{2\pi{G}r^2(1+2\phi)^2}, T_{\theta\theta}=\frac{\phi^2}{2\pi{G}r^2(1+2\phi)^3} and T_{\phi\phi}=\frac{\phi^2 cosec^2\theta}{2\pi{G}r^2(1+2\phi)^3}$$

 

[Ibix]

And the off diagonals are zero? Then just grind through the maths.

 

[Bishal Banjara]

yes, all others off diagonal components are zero.

 

[Bishal Banjara]

I felt a bit confusion here, regarding the expression of Christoffel's and other components.

 

[PeterDonis]

I felt a bit confusion here, regarding the expression of Christoffel's and other components.

In order to compute the Christoffel symbols, you need the metric. What is the metric?

 

[Bishal Banjara]

$$\begin{eqnarray*}
0&=&\nabla_aT^{ab}\\
&=&\partial_aT^{ab}+\Gamma^a_{ad}T^{db}+\Gamma^b_{ad}T^{da}\end{eqnarray*}$$. I meant for these christoffels and components.

 

[PeterDonis]

$$\begin{eqnarray*}
0&=&\nabla_aT^{ab}\\
&=&\partial_aT^{ab}+\Gamma^a_{ad}T^{db}+\Gamma^b_{ad}T^{da}\end{eqnarray*}$$. I mean these christoffels and components.

That formula doesn't tell you the values of the Christoffel symbols. You need their values in order to compute the covariant derivative of $T_{ab}$.

 

[Bishal Banjara]

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

 

[Ibix]

I felt a bit confusion here, regarding the expression of Christoffel's and other components.

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.

 

[Bishal Banjara]

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.
??

 

[Ibix]

Is there a question in your last post @Bishal Banjara, or was that an accidental post?

 

[Bishal Banjara]

how to define $${\tau^r}_{aa}, T^{aa}$$?
To be frank it is truly accidental post but roaming here and there I came to same point.

 

[Ibix]

What's $\tau^r_{aa}$? Do you mean the Christoffel symbol usually denoted $\Gamma^r_{aa}$?

 

[Bishal Banjara]

yes..

 

[Ibix]

Then I don't understand what the problem is. You know how to calculate the Christoffel symbols, correct? And you've already stated the components of the stress-energy tensor. You have everything you need...

 

[Bishal Banjara]

According to the information I had given, what are those values I have to insert for those christoffels and components you just mentioned?

 

[Bishal Banjara]

Does this $$T_{aa}=T_{tt}$$ and so on..?

 

[Ibix]

No - $a$ is a dummy index so you will never see it outside a summation. Thus$$\begin{eqnarray*}
&&\sum_a\Gamma^a_{ar}T^{rr}\\
&=&\Gamma^t_{tr}T^{rr}+\Gamma^r_{rr}T^{rr}+\Gamma^\theta_{\theta r}T^{rr}+\Gamma^\phi_{\phi r}T^{rr}
\end{eqnarray*}$$and similarly for the other sum.

 

[PeterDonis]

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

You need to check your computations; this metric does not give an Einstein tensor that corresponds to the stress-energy tensor you have posted.

 

[Bishal Banjara]

Please specify...what have you obtained? So that I could recheck mine. I had followed the same procedure as Sean Carroll's book.

 

[PeterDonis]

Please specify...what have you obtained?

[Edit--this was wrong, see post #30 for correction.]

In terms of $M$ instead of $\phi$, which is the more usual notation, I get the following from Maxima for the Einstein tensor (note that these are the mixed components):

$$
G^t{}_t = - \frac{4M^2}{r^2 \left( r + 2M \right)^2}
$$

$$
G^r{}_r = \frac{4M^2}{r^2 \left( r - 2M \right) \left( r + 2M \right)}
$$

$$
G^\theta{}_\theta = G^\phi{}_\phi = - \frac{4M^2 \left( r - M \right)}{r \left( r - 2M \right)^2 \left( r + 2M \right)^2}
$$

Lowering an index on each component (which is simple because the metric is diagonal) then gives

$$
G_{tt}= \frac{4M^2 \left( r - 2M \right)}{r^3 \left( r + 2M \right)^2}
$$

$$
G_{rr} = \frac{4M^2}{r^3 \left( r - 2M \right)}
$$

$$
G_{\theta \theta} = - \frac{4M^2 r \left( r - M \right)}{\left( r - 2M \right)^2 \left( r + 2M \right)^2}
$$

$$
G_{\phi \phi} = - \frac{4M^2 r \left( r - M \right) \sin^2 \theta}{\left( r - 2M \right)^2 \left( r + 2M \right)^2}
$$

 

[Ibix]

In terms of $M$ instead of $\phi$, which is the more usual notation, I get the following from Maxima for the Einstein tensor (note that these are the mixed components):

$$
G^t{}_t = - \frac{4M^2}{r^2 \left( r + 2M \right)^2}
$$

$$
G^r{}_r = \frac{4M^2}{r^2 \left( r - 2M \right) \left( r + 2M \right)}
$$

$$
G^\theta{}_\theta = G^\phi{}_\phi = - \frac{4M^2 \left( r - M \right)}{r \left( r - 2M \right)^2 \left( r + 2M \right)^2}
$$

Are you sure? I get$$\begin{eqnarray*}
G^t{}_t=G^r{}_r&=&-\frac{4M^2}{r^2(r+2M)^2}\\
G^\theta{}_\theta=G^\phi{}_\phi&=&\frac{4M^2}{r(r+2M)^3}
\end{eqnarray*}$$From your index lowering, it looks to me like your $g_{tt}$ is $-(1-2M/r)$ where I believe @Bishal Banjara specified $-(1+2M/r)^{-1}$ in #14.

 

[PeterDonis]

From your index lowering, it looks to me like your $g_{tt}$ is $-(1-2M/r)$ where I believe @Bishal Banjara specified $-(1+2M/r)^{-1}$ in #14.

Ah, you're right, I misread the metric. I'll have to re-do the computation in Maxima.

 

[PeterDonis]

Are you sure? I get$$\begin{eqnarray*}
G^t{}_t=G^r{}_r&=&-\frac{4M^2}{r^2(r+2M)^2}\\
G^\theta{}_\theta=G^\phi{}_\phi&=&\frac{4M^2}{r(r+2M)^3}
\end{eqnarray*}$$

Yes, with the correct metric from #14, that's what I get from Maxima. Lowering an index on each then gives

$$
G_{tt}= \frac{4M^2}{r \left( r + 2M \right)^3}
$$

$$
G_{rr}= - \frac{4M^2}{r^3 \left( r + 2M \right)}
$$

$$
G_{\theta \theta} = \frac{4M^2 r}{\left( r + 2M \right)^3}
$$

$$
G_{\phi \phi} = \frac{4M^2 r \sin^2 \theta}{\left( r + 2M \right)^3}
$$

This still doesn't look like what the OP posted.