Divergence of (covaraint) energymomentum tensor

[Torg]

whyT^[ab][;b]≠T_[ab][;b] for spatially flat FLWR cosmology ((ds)^2=(c^2)* (dt)^2-a(t)^2[(dx)^2+(dy)^2+(dz)^2])?
τ[ab][/;b] gives the right answer, but not τ[ab][/;b].

(T^(ab) or T_(ab)) contra-variant and co-variant energy momentum tensor of perfect fluid
(;) covariant derivative,
(c) spped of light in vacuum

 

[Orodruin]

Because covariant components generally are not the same as contravariant components. This should not come as a surprise.

τ[ab][/;b] gives the right answer, but not τ[ab][/;b].

Here you just wrote the same thing twice. I suggest using LaTeX to better transmit the meaning of your post.

 

[Torg]

$${T}_{ab;b}=0$$ \ne $${T}^{ab}_{;b}=0$$ for flat FLWR cosmology line element $${ds}^{2}=(c^{2}(dt)^{2}-a(t)^{2}[(dx)^{2}+(dy)^{2}+(dz)^{2}])$$

 

[Stavros Kiri]

Symbols need fixing.

Because covariant components generally are not the same as contravariant components. This should not come as a surprise.Here you just wrote the same thing twice. I suggest using LaTeX to better transmit the meaning of your post.

I agree

 

[Torg]

But the equations o motion should be the same whether i use covariant or contrvariant. They mathematical
construct!

 

[Stavros Kiri]

But the equations o motion should be the same whether i use covariant or contrvariant. They mathematical
construct!

The equations of motion yes, but not every tensor ... etc.

 

[Orodruin]

${T}_{ab;b}=0$ \ne ${T}^{ab}_{;b}=0$ for flat FLWR cosmology line element ${ds}^{2}=(c^{2}(dt)^{2}-a(t)^{2}[(dx)^{2}+(dy)^{2}+(dz)^{2}])$

See https://www.physicsforums.com/help/latexhelp/

Regardless, it does not change the answer. You have not explained why you would expect contravariant and covariant components to be the same.

 

[Orodruin]

But the equations o motion should be the same whether i use covariant or contrvariant. They mathematical
construct!

##\nabla_b T^{ab}## is a nice expression that transforms as a contravariant vector. ##\nabla_b T_{ab}## is not and it makes no sense to write it down as it does not transform covariantly. You could write ##\nabla^b T_{ab}=0##, which would be equivalent to ##\nabla_b T^{ab} = 0##, but the left-hand sides would be different (although the entire system of equations would be equivalent).

 

[Torg]

\[\begin{array}{l}

{T^{ab}}_{;b} = {T^{ab}}_{,b} + {\Gamma _{bc}}^a{T^{bc}} + {\Gamma _{bd}}^b{T^{ad}} \\

{T_{ab;b}} = {T_{ab;b}} - {\Gamma _{ab}}^c{T_{bc}} - {\Gamma _{bb}}^d{T_{ad}} \\

\end{array}\]



The zero components give

\[\begin{array}{l}

{T^{0b}}_{;b} = {T^{0b}}_{,b} + {\Gamma _{bc}}^0{T^{bc}} + {\Gamma _{bd}}^b{T^{0d}} = {T^{00}}_{,0} + {\Gamma _{11}}^0{T^{11}} + {\Gamma _{22}}^0{T^{22}} + {\Gamma _{33}}^0{T^{33}} + {\Gamma _{01}}^0{T^{00}} + {\Gamma _{02}}^0{T^{00}} + {\Gamma _{03}}^0{T^{00}} \\

{T_{0b;b}} = {T_{0b;b}} - {\Gamma _{0b}}^c{T_{bc}} - {\Gamma _{bb}}^d{T_{0d}} = {T_{00}}_{,0} - {\Gamma _{01}}^1{T_{11}} - {\Gamma _{02}}^2{T_{22}} - {\Gamma _{03}}^3{T_{33}} - {\Gamma _{11}}^0{T_{00}} - {\Gamma _{22}}^0{T_{00}} - {\Gamma _{33}}^0{T_{00}} \\

\end{array}\]
\[{T^{ab}}_{;b} \ne {T_{ab;b}}\]

 

[Torg]

they don't give the same answer

 

[Stavros Kiri]

they don't give the same answer

They are not supposed to.
[Why would they? Explain]

 

[Orodruin]

they don't give the same answer

As you have already been told, one of the expressions is fine and the other is essentially nonsense.

 

[Torg]

Because both are zero when energymomentum is conserved in General Relativity and they should give the same answer for equation of motion.

 

[Torg]

The upper expression gives the true Frieman's equations, but the lower one doesn't. That what amazes me. I have been working on it for days couldn't get right.

 

[Ibix]

Not true. ##T^{ab}{}_{;b}## represents conservation of energy, yes. ##T_{ab;b}## doesn't mean anything. You can't contract a lower index with a lower index - it doesn't mean anything.

 

[Torg]

shall i just forget about it? I use the lower expression a lot.

 

[Stavros Kiri]

but the lower one doesn't.

It's not supposed to.

shall i just forget about it? I use the lower expression a lot.

I agree with the others. It means nothing

 

[Orodruin]

shall i just forget about it? I use the lower expression a lot.

You should forget about using that expression. It is just wrong.

 

[Torg]

Why do I need to contact it? when i use EFEs in their covariant form and when I differentiate both sides covariantly and put the divergence of the energymomentum tensor equals to zero the left hand side should give the same set of equation of motion.

 

[Orodruin]

Taking the divergence is a contraction.

 

[Ibix]

A lower and upper index can be contracted over because their product is a Lorentz scalar. A vector is a map from the space of co-vectors to the reals.

But a co-vector does not map a co-vector to a real. So contracting them isn't expected to produce a consistent result.

Ben Crowell gives an example of a frequency (a covariant quantity) and a time (a contravariant quantity). The product is the number of cycles in the time period, independent of the units used (a scalar). You can convert a frequency to a period by taking the reciprocal (lower an index). And you can multiply that period by the time. But the result is unit-dependent and physically meaningless.

 

[Torg]

I am really sad! to give up using EFEs in their covariant form when it comes to apply the conservation of energymomentum tensor.

 

[Ibix]

You can always raise an index on the covariant derivative operator, as Orodruin noted. ##\nabla^bT_{ab}## is fine and each element is a linear combination of the elements of ##\nabla_bT^{ab}##.

 

[Torg]

I got it, but I will take it with a grain of salt.

 

[Torg]

I have to checked that if it doesn't affect the Bianchi second identity.

 

[Torg]

I understand it now. Thank you all very much for the help. You have been excellent in explaining it.

 

[Orodruin]

If you struggle with this kind of manipulations I would strongly suggest that you read up on differential geometry using a dedicated text that is more detailed than the summary you would typically find in a GR textbook.

 

[Stavros Kiri]

I understand it now. Thank you all very much for the help. You have been excellent in explaining it.

And again welcome to PF! I hope it works for you now. I'll go back to the chat later (it's still there).

Note: within 4hrs you can edit your last post (in general) instead of making many consecutive ones, if no one is in between.

 

[Dale]

shall i just forget about it? I use the lower expression a lot.

Younshould not use the lower expression ever. It is syntactically wrong. Never contract a lower index with a lower index.

 

[Torg]

Thank you all again.
I will ask for much today :-)
I would like to know name of a good reference in introductory Differential Geometry.
I struggled whole night with the second expression below, I couldn't figure out the positions of the indices

\[\begin{array}{l}
{T^{ab}}_{;b} = {T^{ab}}_{,b} + {\Gamma _{bc}}^a{T^{bc}} + {\Gamma _{bd}}^b{T^{ad}} \\
{T_{ab}}^{;b} = {T_{ab}}^{,b} - {\Gamma _{..}}^{.}{T_{..}} - {\Gamma _{..}}^{.}{T_{..}} \\
\end{array}\]
but happily I could write latex expression in PF :-)

I need a paper which I don't have and I couldn't afford and my institution too! one of these is about non-conserved of energy momentum tensor by Rastall.

 

[Stavros Kiri]

I will address just the first part (as I am currently ~ in bed and my Diff. Geometry won't be that walken up yet).
In college, to get ready for General Relativity, I used:
D. F. Lawden "Introduction to Tensor Calculus, Relativity and Cosmology" (e.g. Dover publ.)
https://www.amazon.com/dp/0486425401/?tag=pfamazon01-20
It's kind of educational and affordable.
I believe you can also find it on line (e.g. pdf)
https://www.google.gr/search?ei=lvB...67j0i7i30j0i22i10i30j33i160j0i13.ide/B7FO8fE=

 

[Torg]

Sorry, there a little mistake in the second expression. The first term in the RHS should be the ordinary partial derivative instead of the covariant derivative.

 

[Orodruin]

I struggled whole night with the second expression below, I couldn't figure out the positions of the indices

That would be because you raised the index of the covariant derivative and you simply cannot do that without involving the metric tensor in your expression.
$$
T_{ab}^{;b} = \nabla^{b} T_{ab} = g^{bc} \nabla_c T_{ab} = g^{bc} (T_{ab,c} - \Gamma_{ca}^d T_{db} - \Gamma_{cb}^d T_{ad})
= T_{ab}^{,c} - g^{bc} \Gamma_{ca}^d T_{db} - g^{bc} \Gamma_{cb}^d T_{ad}
$$

 

[Torg]

I thought the metric goes immediately to work into the tensor indices to give a mixed tensor not to the differentiation index itself. I am learning a lot.
Thank you very much Orodruin.

 

[Orodruin]

Of course you can absorb the metric tensor into the energy momentum tensor with lower indices and obtain a mixed tensor, but that did not seem to be what you were after as your template expression included ##T_{\cdot\cdot}##.

Edit: Note that in the term on the form ##\Gamma_{\cdot\cdot}^\cdot T_{\cdot\cdot}## there is no way to place indices such that there is only one free covariant index left, since there are 4 covariant and only one contravariant index and the left-hand side has only one free covariant index. This should immediately tell you that it is impossible to write the term on that form.