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- TL;DR Summary
- 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.

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- Thread starter Bishal Banjara
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M \right)^2}$$As you can see, these are not the same as your equations. I have no idea what you have done to get such a result, so I can't tell you what is wrong with your computation. If you want to check this, I suggest you post your derivation.You are right; the stress-energy tensor that Bishal Banjara posted is the diagonal part of a more general stress-energy tensor that has all off-diagonal terms equal to zero. This more general form is the standard form for the stress-energy tensor in GR, and it is what is used in most GR textbooks. In the general form, the off-diagonal terms do not vanish, but the diagonal terms are still as given

- #1

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- TL;DR Summary
- 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.

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$$\mathrm{D}_j A^{kl}=\partial_j A^{kl} + {\Gamma^k}_{ij} A^{il} + {\Gamma^l}_{il} A^{ki}.$$

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Having text and equations in images is not permitted here at PF. Please type your text directly in the post and use the LaTeX support to type equations directly. (A LaTeX Guide link appears at the bottom left of the post window.)Bishal Banjara said:

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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 $${\tau_{rr}}^r$$?

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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.

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

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yes, all others off diagonal components are zero.

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

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In order to compute the Christoffel symbols, you need the metric. What is the metric?Bishal Banjara said:I felt a bit confusion here, regarding the expression of Christoffel's and other components.

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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.

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That formula doesn't tell you theBishal Banjara said:$$\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.

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

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I find it helpful to write out the sums explicitly. Thus$$\begin{eqnarray*}Bishal Banjara said:I felt a bit confusion here, regarding the expression of Christoffel's and other components.

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.

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Ibix said: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.

??

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

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To be frank it is truly accidental post but roaming here and there I came to same point.

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

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yes..

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

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&&\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.

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

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Bishal Banjara said:Please specify...what have you obtained?

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}

$$

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Are you sure? I get$$\begin{eqnarray*}PeterDonis said: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}

$$

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.

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

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Yes, with the correct metric from #14, that's what I get from Maxima. Lowering an index on each then givesIbix said: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*}$$

$$

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.

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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}##?

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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##.Ibix said:would his expressions be correct for ##T^{ab}##?

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Where does this metric come from?Bishal Banjara said:$$g_{tt}=-(1+2GM/r)^{-1}, g_{rr}=(1+2GM/r), g_{\theta\theta}=r^2 and g_{\phi\phi}=r^2{sin^2\theta}$$

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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.PeterDonis said: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##.

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.

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

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