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Stress–energy tensorT^uv in theory and in practice

  1. Aug 31, 2013 #1
    The observer and the object has the same four-velocity U^u as (1,0,0,0),which means they are relatively static.Then the the observer measures a point A of the object's enery density,which is ρ.
    My question is,the ρ is T^00 or T_00 at the point A?
     
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  3. Aug 31, 2013 #2

    PAllen

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    It is the contravariant 00 component. The covariant could be different if the metric differs from Minkowski at point A. I'm hoping this is what you were asking. Also, the density would have to be as measured at point A.

    A further clarification is that in GR, the component expression of the 4-velocities being the same at two different points doesn't mean much (unless they are close together). You can't really say anything about their relative motion.
     
    Last edited: Aug 31, 2013
  4. Aug 31, 2013 #3

    PeterDonis

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    Actually, if the metric differs from Minkowski, I believe neither the covariant nor the contravariant 00 component will be the measured energy density; that will be the mixed component ##T_0{}^0##.

    More precisely, if the metric coefficient ##g_{00}## is not +/- 1 (+/- depending on the choice of metric signature convention), then we have ##u^0 = 1 / \sqrt{| g_{00} |}## (because we must have ##| g_{00} | u^0 u^0 = 1##), and therefore

    $$
    T^{00} = \rho u^0 u^0 = \frac{1}{g_{00}} \rho
    $$

    We lower one index to get

    $$
    T_0{}^0 = g_{00} T^{00} = \rho
    $$

    We lower the other index to get

    $$
    T_{00} = g_{00} T_0{}^0 = g_{00} \rho
    $$
     
  5. Aug 31, 2013 #4

    Dale

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    If the metric differs from Minkowski then in general the 00 component has no specific significance. E.g. if we allow general metrics then the 0 component could be spacelike or null. I think that the OP's question only makes sense if we assume a (locally) Minkowski metric.
     
  6. Aug 31, 2013 #5

    WannabeNewton

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    Perhaps the OP meant for it to be evaluated in a momentarily comoving locally inertial frame at event ##A## wherein ##u^{\mu} = (e_0)^{\mu}## so ##\rho = T^{\mu\nu}u_{\mu}u_{\nu} = T^{00}##.
     
  7. Aug 31, 2013 #6

    PAllen

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    Oops, you are right. Put another way, the statement that U=(1,0,0,0) with 'reasonable' coordinates, means g00=1, therefore covariant, contravariant and mixed T00 are all the same, and all equal to the locally measured density.

    [Edit] It seems to me that Peter's analysis would extend to any coordinates where the components 0,1,2,3 are respectively timelike, spacelike,.... (and, I think, orthogonality = diagonal metric is needed as well, for the overall argument). Then we assume a single isolated particle with U = (a,0,0,0) [to generalize from (1,0,0,0)]
     
    Last edited: Aug 31, 2013
  8. Aug 31, 2013 #7
    Yeah,I agree the observer should measure it locally and so on,but I'm afraid this generalization is a matter of unitarity.e.g.guvUuUv=+/-1.I think another generalization is Uu=(1,a,0,0),that is to say,orthogonality or diagonal metric is not needed...In that case,ρ is not necessarily equal to T00 or T00
     
    Last edited: Aug 31, 2013
  9. Aug 31, 2013 #8
    That's it!
     
  10. Aug 31, 2013 #9
    I'm confused about the equation I quoted...Why?
     
  11. Aug 31, 2013 #10

    PeterDonis

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    Yes, I was implicitly assuming that; but as DaleSpam pointed out, there are plenty of coordinate charts where that doesn't hold.

    A diagonal metric isn't required, as long as the 0 coordinate is timelike (more precisely, as long as integral curves of of ##\partial_0## are timelike). For example, an observer who remains at a constant ##r, \theta, \phi## in Painleve coordinates has u = (a, 0, 0, 0), where ##a = 1 / \sqrt{1 - 2M / r}##, but the metric is not diagonal in the Painleve chart.

    If ##g_{00} = +/- 1 / a^2##, then ##u^0 = a##, ##u^i = 0## satisfies this equation. That's what PAllen was referring to.

    Have you plugged this into the equation ##g_{uv} u^u u^v = +/- 1##? What do you get?
     
  12. Aug 31, 2013 #11

    PeterDonis

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    Do you mean, why is ##T^{00} = \rho u^0 u^0##? In general it isn't, but it is for the special case a perfect fluid with 4-velocity ##u## and energy density ##\rho## as measured by an observer comoving with the fluid. Your OP pretty much implicitly assumed this special case.
     
    Last edited: Aug 31, 2013
  13. Aug 31, 2013 #12

    PAllen

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    But if the metric isn't diagonal, then ρ is, in some sense, split between T00 and other components of T. Consider that T00 (mixed indices) = ρ in a local frame where the metric is Minkowski (and all other T components are zero). Then transform to coordinates where the metric is non-diagonal (without changing the timelike/spacelike character of any basis vector). It seems to me that such a transform will distribute locally measured ρ in several components of T.
     
  14. Aug 31, 2013 #13

    PeterDonis

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    This depends on which "version" of T you are looking at. ##T^{00} = \rho u^0 u^0## still holds, and if ##u## has only a 0 component, that will be the only component of T that has ##\rho## in it. But yes, if you lower one or both of the indexes on ##T^{00}##, you will get multiple components if the metric is not diagonal. But the 4-vector ##u^a## will still only have one component (though the 1-form ##u_a## corresponding to it will not if the metric is not diagonal).
     
  15. Aug 31, 2013 #14

    PAllen

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    I must be missing something. Let me restate my argument: you have T00 (top indices, if you will) = ρ in a locally Minkowski frame, where U has only time components. Consider a general coordinate transform (that still leaves each basis having the same timelike/spacelike character, but the new basis is not orhogonal). Apply the tensor transform rule. It seems to me, that, in general, this will distribute the ρ to several components of T.
     
  16. Aug 31, 2013 #15

    PeterDonis

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    The question (at least for the special case we're discussing, where in general ##T = \rho u u##) is not how it distributes the components of T, but how it distributes the components of u.

    Consider a "hovering" observer in Schwarzschild spacetime, and take a local inertial frame centered on some event on the observer's worldline. At the origin of that LIF, the observer's 4-velocity is (1, 0, 0, 0). If we transform to global Painleve coordinates, the observer's 4-velocity at the same event is ##(1 / \sqrt{1 - 2M / r}, 0, 0, 0)##. So at that event, u still has only one component, ##u^0##, and therefore ##T^{ab} = \rho u^a u^b## will also have only one component, ##T^{00}##. But the mixed tensor ##T_a{}^b## and the covariant tensor ##T_{ab}## will have more than one component at the same event in Painleve coordinates, because the metric is not diagonal.

    In the completely general case, transforming from a LIF to a general coordinate chart can cause ##u^a## to have multiple components; and in *that* case, yes, even the contravariant tensor ##T^{ab} = \rho u^a u^b## will have multiple components. But there are cases, as the above shows, where that doesn't happen.
     
  17. Aug 31, 2013 #16

    PAllen

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    Thanks. What I was missing was that if the coordinate transform is such as to map U=(1,0,0,0) to U=(a,0,0,0), then it will leave contravariant T with only one nonzero component as also having only one nonzero component. But if mixed and covariant T have other nonzero components, only the contravariant part of your argument remains valid. For the whole argument (in post #3) to work, it seems to me you need to assume a diagonal metric.
     
  18. Aug 31, 2013 #17

    PeterDonis

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    Yes, you're right, I was assuming a diagonal metric in post #3. But that may not make post #3 wrong, just incomplete.

    Let's take the Painleve case as an example. We have ##u^a = (1 / \sqrt{1 - 2M / r}, 0, 0, 0)##, so

    $$
    T^{00} = \rho u^0 u^0 = \frac{\rho}{1 - 2M / r}
    $$

    is the only nonzero component of the contravariant tensor. Then we lower one index:

    $$
    T_a{}^b = g_{ac} T^{cb}
    $$

    which gives

    $$
    T_0{}^0 = g_{00} T^{00} = \rho
    $$

    $$
    T_1{}^0 = g_{10} T^{00} = \frac{\rho}{1 - 2M / r} \sqrt{\frac{2M}{r}}
    $$

    Now lower the second index:

    $$
    T_{ab} = g_{cb} T_a{}^c
    $$

    which gives

    $$
    T_{00} = g_{00} T_0{}^0 = \rho \left( 1 - \frac{2M}{r} \right)
    $$

    $$
    T_{10} = g_{00} T_1{}^0 = \rho \sqrt{\frac{2M}{r}}
    $$

    $$
    T_{11} = g_{01} T_1{}^0 = \frac{\rho}{1 - 2M / r} \frac{2M}{r}
    $$

    So ##T_0{}^0## and ##T_{00}## don't change from what I wrote in post #3; what changes is that we now have additional mixed and covariant components of T.
     
  19. Sep 1, 2013 #18
    I just want this equation to be the unitarity(normalization) condition.E.g. Uu=(1,2,0,0) and (C,2C,0,0) are different in mathematics,but they have the similar physical meaning.So in order to make work easy,we use (1,2,0,0) to represent (C,2C,0,0)
     
  20. Sep 1, 2013 #19
    But for perfect fluid ,T00=ρU0U0+p(g00+U0U0) and according to 3#,then it=ρ/g00+p(g00+1/g00),where p is presure.
     
  21. Sep 1, 2013 #20

    PeterDonis

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    Not quite, because ##g^{00} + u^0 u^0 = 0## for any frame in which the fluid is at rest, so the term in ##p## vanishes for the 0-0 component of the contravariant tensor.

    It is true that for a perfect fluid, ##T^{ab} = \rho u^a u^b + p \left( g^{ab} + u^a u^b \right)##. So if ##p## is nonzero, there will be other terms besides ##T^{00}## that are nonzero in a frame in which the fluid is at rest. But those terms won't have ##\rho## in them.
     
    Last edited: Sep 1, 2013
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