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A Interpretation of [itex]\partial_\nu T^{\mu \nu}[/itex]

  1. May 24, 2016 #1
    I am working on Sean Carroll's problem 1.8.

    If [itex]\partial_\nu T^{\mu \nu}=Q^\mu[/itex], what physically does the spatial vector [itex]Q^i[/itex] represent? Use the dust energy momentum tensor to make your case.

    The dust energy momentum tensor is

    [tex]T^{\mu \nu}= \rho U^\mu U^\nu,[/tex]

    where [itex]U[/itex] is the four-velocity and [itex]\rho[/itex] the energy density in the rest frame.

    Trying to manipulate [itex]\partial_\nu \rho U^\mu U^\nu[/itex] directly didn't give me anything I could reasonably interpret. I then tried to look at [itex]U_\mu Q^\mu[/itex] following the example in the textbook to see if I could possibly factor the final expression into giving me an answer I could interpret. What I got was

    [tex]U_\mu \partial_\nu \rho U^\mu U^\nu=-U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu[/tex].

    I can't factor four-velocity out of the second term since it is inside the derivative, so I'm not too sure that I can simplify this into something that I can interpret for Q.

    I also tried to consider the part of this equation that was orthogonal to the four velocity. To do this, I multiplied the expression by the following projection vector:

    [tex]P^\sigma_\nu=\delta^\mu_\nu+U^\sigma U_\nu.[/tex]

    This gives

    [tex]P^\sigma_\mu Q^\mu = P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu),[/tex]

    [tex]P^\sigma_\mu \partial_\nu (\rho U^\mu U^\nu) =U^\nu \rho \partial_\nu U^\sigma, [/tex]

    [tex]P^\sigma_\mu Q^\mu =Q^\sigma + U^\sigma U_\mu Q^\mu,[/tex]

    [tex]U^\nu \rho \partial_\nu U^\sigma = Q^\sigma + U^\sigma U_\mu Q^\mu.[/tex]

    From here, if we plug in [itex] -U^\nu \partial_\nu \rho-\rho \partial_\nu U^\nu=U_\mu Q^\mu[/itex] into the final expression above, we get the original expression: [itex]\partial_\nu T^{\mu \nu}=Q^\mu[/itex]. This indicates that I have done all of the algebra correctly, however it doesn't help me interpret what the spatial part of Q is physically. Can anyone help me out on where to go from here?
    Last edited: May 24, 2016
  2. jcsd
  3. May 24, 2016 #2


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    really how did you end up [in your 2nd equation] with somethin with no free-index when you began with something with 1 free index?
  4. May 24, 2016 #3
    Not sure what you mean when you say second equation. If it's the one I think you mean, it's because I'm projecting the equation along the four-velocity.
  5. May 24, 2016 #4

    George Jones

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    But that doesn't allow you to change the number of free indices in an equation.
  6. May 24, 2016 #5
    Ahh, I forgot to write a factor up there.
    Ahh, I forgot to write a factor up there. Is it better now?
  7. May 24, 2016 #6


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    Well, have you been introduced to what physically a energy-momentum tensor [itex]T^{\mu \nu}[/itex] stands for?
    Like what its components are...[I guess that's a better way to make use of the dust E-M tensor].
    If you get what the [itex]T[/itex] stands for, it's straightforward to get a physical interpretation of its derivative [which will give you something like a "continuity equation": a relation of the flow with time and space of the T quantitiy].
    In particular you want to look at:
    [itex]Q^i = \partial_\mu T^{\mu i}[/itex]
    So you'll only need to know what's [itex]T^{i0}[/itex] and [itex]T^{ij}[/itex].
    Last edited: May 24, 2016
  8. Jun 7, 2016 #7
    The problem is that I don't know how to isolate Q in my equations above when trying to look at the dust energy-momentum tensor.
  9. Jun 7, 2016 #8


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    what do you mean by isolate?
  10. Jun 7, 2016 #9


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    I mean you have:
    [itex] Q^i = \partial_\mu T^{\mu i}[/itex]
    [itex] Q^i = \partial_0 T^{0 i } + \partial_j T^{ji}[/itex]
    which is 3 equations:
    [itex] Q^1 = \partial_0 T^{0 1} + \partial_j T^{j1}[/itex]
    [itex] Q^2 = \partial_0 T^{0 2} + \partial_j T^{j2}[/itex]
    [itex] Q^3 = \partial_0 T^{0 3} + \partial_j T^{j3}[/itex]
    I gave you hints when I refered to a continuity equation... Qs are supposed to play the role of sources/sinks in the equation... the only thing you need to interpret then are the Energy-Momentum components [which you can obtain directly from the dust equation you gave]...as an example I tell you that the [itex]T^{01}= \rho U^1 U^0 = \gamma^2 \rho u^1 = \rho' u^1[/itex] is the flux of energy density along the 1-axis [pretty much proportional to the 1-momentum]. T[ij] can be a little more tricky at least if you haven't seen a stress tensor from your mechanics course (or in general the flux of a vector along the 3 directions which would be the gradF for the whole derivative quantity).
    Last edited: Jun 7, 2016
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