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Speed of mass-energy flux, derived from stress-energy?

  1. Sep 23, 2012 #1

    bcrowell

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    The dominant energy condition is interpreted as meaning that no observer will see a flux of energy flowing at speeds greater than c. In this post I'm basically asking how to relate this concept of "speed of the energy flux" more precisely to the stress-energy tensor.

    If we start with dust, described in its rest frame, we have a stress-energy tensor that looks like this:

    [tex]
    \left(\begin{matrix}
    1 & 0 \\
    0 & 0
    \end{matrix}\right)
    [/tex]

    This is in coordinates (t,x), ignoring an over-all multiplicative constant.

    Boosting along x gives a stress-energy tensor that looks like this:

    [tex]
    \left(\begin{matrix}
    1 & v \\
    v & v^2
    \end{matrix}\right)
    [/tex]

    This is exact, not approximate, although I'm leaving out a factor of [itex]\gamma^2[/itex] in front, since I don't care about multiplicative constants.

    The DEC says that if I multiply this matrix by a vector that lies in or on the future light-cone, I get another vector that lies in or on the cone...but this doesn't seem to me to connect in any obvious way to the concept of a flux of mass-energy flowing with some velocity.

    If I let [itex]v=1-\epsilon[/itex], with [itex]|\epsilon| \ll 1[/itex], then [itex]\epsilon=0[/itex] gives the correct stress-energy tensor for, e.g., an electromagnetic wave, and [itex]\epsilon<0[/itex] gives one that violates the DEC.

    In the above example, [itex]T_{xt}/T_{tt}[/itex] is the velocity at which mass-energy flows in the x-t frame. However, this doesn't seem to be a valid interpretation in general. For example, if I start with


    [tex]
    \left(\begin{matrix}
    1 & p \\
    p & q
    \end{matrix}\right) ,
    [/tex]

    and I want to find a velocity v such that a boost eliminates the off-diagonal component, then I only get [itex]v \approx -q[/itex] in the case where both p and q are small.

    In terms of old-fashioned E&M, the existence of a nonzero Poynting vector doesn't indicate that there is a flow of energy; that requires a nonzero divergence for the Poynting vector. By analogy, it seems unlikely to me that one could conclude the "speed of energy flux" from knowing T at a point, without knowing T's derivatives. If I put together a static EM field with a nonzero Poynting vector, then I get a stress-energy tensor where basically [itex]T_{tt}[/itex] is the energy density and [itex]T_{tx}[/itex] is the Poynting vector. (See http://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor .) So clearly the ratio [itex]T_{xt}/T_{tt}[/itex] doesn't tell me the "speed of the energy flux."

    So it seems that it is not possible at all to read off the "speed of the energy flux" from T. In that case, what is the justification for this interpretation of the DEC? The example above with the epsilons makes it plausible to me, but it only seems like a plausibility argument.
     
    Last edited: Sep 23, 2012
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  3. Sep 23, 2012 #2

    Mentz114

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    Interesting. The EMT of a null dust takes the form
    [tex]
    A\left(\begin{array}{cc}
    1 & 1 \\
    1 & 1
    \end{array}\right)
    [/tex]

    in regular (t,x) coords. In this case [itex]T_{xt}/T_{tt}[/itex] =1 is giving the speed.

    The transformation from the dust EMT is
    [tex]L=\left(\begin{matrix}
    -1/\sqrt{2} & 1/\sqrt{2} \\
    1/\sqrt{2} & 1/\sqrt{2}
    \end{matrix}\right)
    [/tex]

    which is not a boost.
     
  4. Sep 23, 2012 #3

    bcrowell

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    Null dust...that's a new one on me. WP has this: http://en.wikipedia.org/wiki/Null_dust_solution Written by CH, who was formerly a heavy on PF. It's too bad he's left -- we're poorer for it.

    Since the divergence of the Poynting vector gives an indication of whether an EM field is transmitting energy, I wonder whether there is some GR analog constructed from the stress-energy tensor and its derivatives. The simplest thing I can think of is [itex]\nabla_a T^{ab}[/itex], but that vanishes identically. Presumably if energy is being transmitted, the flux is dependent on the observer, so we need to introduce some vector [itex]U^a[/itex] that represents the observer's velocity vector. Then the recipe for the flux would have to involve a derivative, U, and T.
     
    Last edited: Sep 23, 2012
  5. Sep 23, 2012 #4

    Mentz114

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    The transformation that takes dust to null dust is its own inverse so it takes null dust to dust. It also swaps between Brinkmann and Minkowski coords.

    (I agree about CH. His Wiki archive is till available).

    How about ∇aPb where P is the top row of the EMT ? This would be the divergence of the Poynting vector ( is it a 4-vector ?)
     
    Last edited: Sep 23, 2012
  6. Sep 23, 2012 #5

    bcrowell

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    No, the Poynting vector is just a 3-vector. To get the divergence, you'd have to add the three derivatives. Making that into valid index gymnastics in 3+1 dimensions would involve contracting the indices a and b, which would give the divergence of T, which vanishes.

    One way that I can think of to quantify the "speed of flux" is this. Given a frame (t,x,y,z), we could define the speed of flux to be the smallest boost that gives a stress-energy tensor T with [itex]\nabla_{t'} T^{t't'}=0[/itex]. But I don't know how to connect this to the DEC.
     
  7. Sep 23, 2012 #6

    George Jones

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    See 2.1.5 from Eric Poisson's notes,

    http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

    which evolved into the excellent book, A Relativist's Toolkit: The Mathematics of black hole Mechanics.
     
  8. Sep 24, 2012 #7

    bcrowell

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    Thanks, George!

    Poisson says this:

    This helps, but there are a couple of things still bugging me.

    It does manage to phrase the whole thing without referring to the "speed of flux" or similar concepts, as other treatments do. (E.g., WP has: "That is, mass-energy can never be observed to be flowing faster than light."). That suggests that the "speed of flux" concept could just be dispensed with. One thing that seems odd to me is that with a static field, I can saturate the DEC. For example, if I take a uniform E field and superimpose on it an equal-strength, uniform B field in the perpendicular direction, then in that frame, the DEC holds only by the skin of its teeth. I suppose this is reasonable, since there is no lower limit on the frequency of an EM wave, so there is no empirically observable distinction between this example, with no flow, and an EM wave propagating in the direction orthogonal to E and B.

    What still kind of bugs me is that there seems to be a big logical leap between (a) observing whether the first row of T is timelike and (b) inferring whether matter has a "flow along timelike or null world lines." It's certainly true that if we take dust and boost it, we have a iff b. It's also true that if I superimpose timelike energy-momentum densities that are in different directions, their vector sum is also timelike, so b implies a in that special case. What is far less obvious to me is that b is a valid interpretation of a in all cases.

    In the old-fashioned E&M case, we have two separate criteria, [itex]E\times B\ne 0[/itex] and [itex]\nabla \cdot (E\times B)\ne 0[/itex]. The latter works better than the former. I have the feeling that the DEC is the moral equivalent of the former, and the verbal interpretations people give of it are not really valid, for the same reasons.
     
  9. Sep 24, 2012 #8

    bcrowell

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    I think I've finally made some progress in understanding this.

    You can have flow without accumulation or depletion, e.g., when a river flows by but always stays at the same water level.

    In some physical examples, flow without accumulation (I'll call it FWA) has physical significance, and in others it doesn't.

    When charge flows, FWA is detectable from the magnetic field.

    When mass flows, FWA is detectable from the momentum it carries.

    When energy flows in Newtonian mechanics, FWA is undetectable. For example, if I superimpose a point charge and a magnetic dipole at the origin, then the Poynting vector in the surrounding space circulates. Its effect is undetectable.

    In relativity, we can't have the Newtonian picture where FWA of mass is detectable and FWA of energy isn't. Mass and energy are equivalent.

    If I show you a stress-energy tensor like the one Lut wrote,
    [tex]
    A\left(\begin{array}{cc}
    1 & 1 \\
    1 & 1
    \end{array}\right) ,
    [/tex]
    how do you know whether the off-diagonal elements can really be interpreted as a flow? If it was really some perfectly uniform dust moving ultrarelativistically in the positive x direction (so that the four elements are only approximately equal), you'd say it's really a flow. If it was really due to a uniform, static pattern of E and B (pointing along the suppressed y and z axes), you'd say it wasn't really a flow, because the fields were static. But either way, those off-diagonal elements play the same role in the Einstein field equations, and the distinction becomes meaningless.

    What this means, I think, is that the interpretation of the time-space components of T as fluxes of momentum is only valid (a) because it's clearly the right interpretation for dust, so we need to interpret it that way in the Newtonian limit, and (b) if we put aside the Newtonian idea that FWA of mass is detectable, while FWA of energy isn't.

    In the example of the superimposed dipole and point charge, the circulation of energy has no effect on the far gravitational field, because when you average T over the whole region, the time-space parts cancel out. But I think it does have an effect on the near gravitational field. (Maybe we'd see something analogous in the Kerr-Newman metric of a charged, rotating black hole?)

    In E&M, we can test for accumulation (as opposed to FWA) by checking whether the Poynting vector has a divergence. I suggested before that we could do the same thing by looking for a frame in which [itex]\nabla_t T^{tt}[/itex] vanishes. But I think this is basically not what we *want* to test for, since we actually don't want to reject FWA in GR.

    By the way, that test seem to turn out as sort of a mess. Suppose I have a T of this form

    [tex]
    A\left(\begin{array}{cc}
    q+at+bx & ... \\
    ... & r+ct+dx
    \end{array}\right) .
    [/tex]

    The ...'s represent elements whose derivatives we can infer from the ones that are explicitly shown, since T is divergenceless. If you then apply a boost v along the x axis and impose the non-accumulation condition I suggested above, you get the following (probably with some sign mistakes): [itex]a+(-b-2d)v+(c+2a)v^2-dv^3[/itex]. In general this is going to have up to 3 distinct solutions for v. I don't see any particular reason to expect that exactly one solution will be real and lie in the interval [itex]-1\le v \le 1[/itex], regardless of whether you impose the DEC (which really just restricts the constant terms q and r).
     
    Last edited: Sep 24, 2012
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