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Energy-stress tensor integration proof (from schutz ch.4)

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Use the identity [tex]T^{\mu \nu}_{ ,\nu} = 0[/tex] to prove the following results for a bounded system (ie a system for which [tex]T^{\mu \nu} = 0[/tex]
    outside a bounded region of space),

    [tex]\frac{\partial}{\partial t}\int T^{0\alpha}d^{3}x = 0[/tex]

    2. Relevant equations

    3. The attempt at a solution

    The integral obviously gives 4 equations (one for each [tex]\alpha[/tex]) which must all be = 0.
    I tried just working on the first and passing the partial derivative into the integral and writing [tex]\frac{\partial}{\partial t} T^{0 0} = -\left(\frac{\partial}{\partial x} T^{0 1} +\frac{\partial}{\partial y} T^{0 2} + \frac{\partial}{\partial z} T^{0 3} \right)[/tex]

    This gives
    [tex]- \int T^{0 i}_{,i}d^{3}x[/tex]

    which doeesnt really seem to be getting me anywhere - also the same reasoning wouldn't work for the other equations because they must all be partially differentiated wrt t and this is only relevant for [tex]T^{0 0}[/tex] in [tex]T^{\mu \nu}_{ ,\nu} = 0[/tex]

    Another direction I thought of was to use Gauss' law but then there is no outward normal one-form and so maybe not....

    There are another two parts to this question but I thought that if I had an idea of how to do the first part I could figure the others out by myself.

    Thanks for any replies :)
    Last edited: Nov 2, 2008
  2. jcsd
  3. Nov 2, 2008 #2


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    Since the system is bounded, why not use [itex]T^{0 \alpha}=T^{\alpha 0}[/itex]? Then [tex] \frac{\partial}{\partial t}T^{0 \alpha}=\frac{\partial}{\partial t}T^{\alpha 0}=T^{\alpha 0}_{,0}[/tex]....but what is this last quantity if [itex]T^{\mu \nu}_{ ,\nu} = 0[/itex]? ;0)
    Last edited: Nov 2, 2008
  4. Nov 2, 2008 #3
    Wow, thanks, that is a nifty little trick.

    [tex]T^{\mu \nu}_{ ,\nu} = 0[/tex]
    [tex]\Rightarrow T^{\alpha 0}_{,0}= - \left( \frac{\partial}{\partial x} T^{\alpha 1} + \frac{\partial}{\partial y} T^{\alpha 2} + \frac{\partial}{\partial z} T^{\alpha 3} \right)[/tex]

    [tex] T^{\alpha 0}_{,0}= - T^{\alpha i}_{,i}[/tex] where [itex] i = 1,2,3[/itex] and [itex]\alpha = 0,1,2,3.[/itex]

    and so
    [tex]\frac{\partial}{\partial t} \int T^{0 \alpha} d^{3}x = \int T^{\alpha i}_{,i} d^{3}x[/tex]

    which can be written as
    [tex]\int \int T^{\alpha 1} (x_{1}) - T^{\alpha 1} (x_{2}) dydz + \int \int T^{\alpha 2} (y_{1}) - T^{\alpha 2} (y_{2}) dxdz + \int \int T^{\alpha 3} (z_{1}) - T^{\alpha 3} (z_{2}) dxdy[/tex]

    and I'm afraid that I have come to that dead end again - but this time all four equations are included which is nice... Is the last step right? why is this 0?
  5. Nov 3, 2008 #4


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    Well, what do the components [tex]T^{\alpha i}[/tex] represent physically?
    Last edited: Nov 3, 2008
  6. Nov 3, 2008 #5
    The [itex]\alpha[/itex] momentum per second crossing a unit area of surface of constant [itex]x^{i}[/itex]

    So does the " constant [itex]x^{i}[/itex] " mean that [itex]x_{1}=x_{2}[/itex], [itex]y_{1}=y_{2}[/itex] and [itex]z_{1}=z_{2}[/itex] and therefore the integrands in the above integrals are all zero and so the integrals are also zero?

    I feel a bit like I'm clutching at straws here...
    Is that right or am I just making stuff up? :D
    Last edited: Nov 4, 2008
  7. Nov 5, 2008 #6
    I have managed to complete the proof of the second question (Tensor Virial Theorem) but it requires that

    [tex]\int \left[T^{0k}x^{i}x^{j} \right] ^{x^{k}_2}_{x^{k}_1} d^{2}x = 0[/tex]

    Which makes me rethink my previous post

    I said that [itex]x_{1}=x_{2}[/itex] etc...

    But the second proof made me realise that this doesnt really make sense.... why should [itex]x_{1}=x_{2}[/itex]? these values are unrelated to [itex]T^{\alpha 1}[/itex]
    What would make more sense (and satisfy both proofs) is that [itex]T^{\alpha 1} (x_{1}) = T^{\alpha 1}(x_{2}) = 0[/itex]
    Then both
    [tex] \left[T^{0k}x^{i}x^{j} \right] ^{x^{k}_2}_{x^{k}_1} = 0[/tex] (satisfying the second proof)
    [tex]T^{\alpha 1} (x_{1}) - T^{\alpha 1} (x_{2}) = 0[/tex] (satisfying the first)

    But I'm not sure why [itex]T^{\alpha 1} (x_{1}) = 0[/itex].
    My logic says this is incorrect - if we have a constant x surface, then [itex]T^{\alpha 1}[/itex] doesn't depend on x and so [itex]T^{\alpha 1} (x_{1}) = T^{\alpha 1}(x_{2}) = T^{\alpha 1}[/itex] but not necessarily 0. So I have a problem with my second proof.

    Am i thinking in the right direction anywhere here?
    Last edited: Nov 5, 2008
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