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Is the vacuum stress energy tensor Lorentz invariant ?

  1. Jul 11, 2006 #1
    In many text books on relativity, one finds at some point a statement that the vacuum stress energy tensor should be Lorentz invariant, from which it then follows that the vacuum pressure is minus the vacuum energy density.
    However, the vacuum energy density (or stress tensor) is not an observable as such, so it is not clear to me why it should be Lorentz invariant. Only when it is included in some physical law, then this law should be Lorentz invariant (or so called covariant). An invariant law does not have to imply the invariance of all its components. Is there a flaw in my reasoning ?
     
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  3. Jul 11, 2006 #2

    selfAdjoint

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    Tensors in general are not invariant but tensor equations are covariant. The equation remains true even though the components of the tensors in it change with viewpoint. So if the vacuum stress energy tensor equals zero in one frame then it equals zero in all frames. Perhaps this is what you read?
     
  4. Jul 11, 2006 #3
    Yes, that is indeed what I thought. There seems not to exist a separate equation for the stress-energy tensor of the vacuum(for instance that it is zero). One considers the vacuum to be a fluid with no viscosity, such that it has only diagonal components. And then one insists that this tensor should be invariant under a Lorentz boost. I do not see the reason for this requirement.
     
  5. Jul 11, 2006 #4

    George Jones

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    Consider two observers coincident at the spacetime event p. Each observer can set up orthonormal set (tetrad) of (tangent) 4-vectors with respect to which the observers make measurements. If the observers are in relative motion, then the two tetrads are different, but even in general relativity, the tetrads are related by a Lorentz transformation.

    Now consider an electromagnetic field also present at p. Each observer measures the components of the field with respect to the appropriate tetrad, and components are related in the standard way by Lorentz transformations. This is covariance, not invariance. Also, energies are related in the standard way.

    Now consider the vacuum. Lorentz invariance of the vacuum energy-momentum tensor means that the components are the *same* with respect to both (actually, all) tetrads at (every) p. The only second-rank tensor that has the same components with respect to all tetrads is g.
     
  6. Jul 11, 2006 #5
    Thanks for this explanation, but I don't see how this justifies why the vacuum energy-momentum tensor must be Lorentz invariant as nobody has the ability to measure and compare its components in different frames.
     
  7. Jul 11, 2006 #6

    George Jones

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    Yes, I carefully avoided giving a reason. :biggrin:

    I think the hope is that someday a theory will predict the values of the components, i.e., the propotionality to g, and that this will correspond to the measured value of the cosmological constant/dark energy.
     
  8. Jul 11, 2006 #7

    pervect

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    If we write the stress-energy tensor of the vacuum-with-a-cosmological constant it in cartesian coordinates (t,x,y,z)

    [itex]T^{ij}[/itex] = diag(k,-k,-k,-k)

    then boost it in the x direction

    [tex]T'^{00} = (\Lambda^0{}_0) ^2 T^{00} + (\Lambda^0{}_1)^2 T^{11}[/tex]

    [itex]T'^{11} = (\Lambda^1{}_1)^2 T^{11} + (\Lambda^1{}_0)^2 T^{00}
    [/tex]

    With [itex]\Lambda^0{}_0 = \Lambda^1{}_1 =\gamma [/itex] and [itex]\Lambda^0{}_1 = \Lambda^1{}_0 = -\beta \gamma[/itex]

    we get

    [tex]T'^{00} = \gamma^2 T^{00} + \beta^2 \gamma^2 T^{11} = \gamma^2 k - \beta^2 \gamma^2 k = k[/tex]

    [tex]T'{11} = \gamma^2 T^{11} + \beta^2 \gamma^2 T^{00} = -k \gamma^2 + \beta^2 \gamma^2 k = -k[/tex]

    which is the same. Unless I've made a silly sign error somewhere?
     
    Last edited: Jul 11, 2006
  9. Jul 11, 2006 #8
    Yes, you calculated correctly but it only shows that if rho = - pressure one obtains a Lorentz invariant tensor (or vice versa). So it does not answer why the vacuum energy-stress tensor should be Lorentz invariant in the first place (or should have rho = - p in the first place)
     
  10. Feb 2, 2012 #9
    Although vacuum does not mean containing nothing, there should be no preferred coordinate system in a vacuum. But if its stress-energy tensor (T) is not Lorentz-invariant, then there would be one. Therefore, T must be Lorentz-invariant.
    It turns out that we have only one choice for such a tensor: the Minkowski metric (up to some constant).
     
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