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A Divergence of (covaraint) energymomentum tensor

  1. Dec 25, 2017 #1
    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
     
  2. jcsd
  3. Dec 25, 2017 #2

    Orodruin

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    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.
     
  4. Dec 25, 2017 #3
    $${T}_{ab;b}=0$$ \ne $${T}^{ab}_{;b}=0$$ for flat FLWR cosmology line element $${ds}^{2}=(c^{2}(dt)^{2}-a(t)^{2}[(dx)^{2}+(dy)^{2}+(dz)^{2}])$$
     
    Last edited by a moderator: Dec 25, 2017
  5. Dec 25, 2017 #4
    Symbols need fixing.
    I agree
     
  6. Dec 25, 2017 #5
    But the equations o motion should be the same whether i use covariant or contrvariant. They mathematical
    construct!
     
  7. Dec 25, 2017 #6
    The equations of motion yes, but not every tensor ... etc.
     
  8. Dec 25, 2017 #7

    Orodruin

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    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.
     
  9. Dec 25, 2017 #8

    Orodruin

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    ##\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).
     
  10. Dec 25, 2017 #9
    \[\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}\]



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

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  11. Dec 25, 2017 #10
    they don't give the same answer
     
  12. Dec 25, 2017 #11
    They are not supposed to.
    [Why would they? Explain]
     
  13. Dec 25, 2017 #12

    Orodruin

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    As you have already been told, one of the expressions is fine and the other is essentially nonsense.
     
  14. Dec 25, 2017 #13
    Because both are zero when energymomentum is conserved in General Relativity and they should give the same answer for equation of motion.
     
  15. Dec 25, 2017 #14
    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.
     
  16. Dec 25, 2017 #15

    Ibix

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    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.
     
  17. Dec 25, 2017 #16
    shall i just forget about it? I use the lower expression a lot.
     
  18. Dec 25, 2017 #17
    It's not supposed to.
    I agree with the others. It means nothing
     
  19. Dec 25, 2017 #18

    Orodruin

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    You should forget about using that expression. It is just wrong.
     
  20. Dec 25, 2017 #19
    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.
     
  21. Dec 25, 2017 #20

    Orodruin

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    Taking the divergence is a contraction.
     
  22. Dec 25, 2017 #21

    Ibix

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    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.
     
  23. Dec 25, 2017 #22
    I am really sad! to give up using EFEs in their covariant form when it comes to apply the conservation of energymomentum tensor.
     
  24. Dec 25, 2017 #23

    Ibix

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    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}##.
     
  25. Dec 25, 2017 #24
    I got it, but I will take it with a grain of salt.
     
  26. Dec 25, 2017 #25
    I have to checked that if it doesn't affect the Bianchi second identity.
     
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