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Derivation of energy-stress tensor in GR

  1. Dec 4, 2013 #1
    Would there be a direct proof of the energy-stress tensor of general relativity? My lecturer only provides me with a simplified proof -
    1. Guess the form of the tensor in special relativity in co-moving frame

    (ρ+p)uμuv+pημv

    Note that the pη00 term cancels the p in u0u0, to simplify the understanding of the tensor.

    2. Transform the tensor to arbitrary frame in special relativity

    3. The η is replaced by g in general relativity

    Problem -

    1. In step 1, the pη could be 'plugged into' the uμuv, why do we need to extract it out for the sake of fitting the form of GR tensor? It seems that it knows the answer before head and construct it. Is there any more vigorous proof?

    2. You could replace g by η as a special case of g. But, replacing all η by g means "special relativity implies general relativity" which is not valid in a logical sense.

    Could anyone please suggest a better derivation, or help me understand the proof better?
    Thank you very much. =]
     
  2. jcsd
  3. Dec 4, 2013 #2

    WannabeNewton

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    First this is only for a perfect fluid so I assumed you meant as much. Here we have a fluid with 4-velocity field ##u^{\mu}## with rest mass density ##\rho## and isotropic pressure ##p##. Go to the instantaneous rest frame of the fluid. In this frame we know that the stress-energy tensor has to have the form ##T^{\mu}{}{}_{\nu} = \text{diag}(\rho,p,p,p)## (you can easily convince yourself of this using basic physical arguments since ##p## is isotropic and we are in the instantaneous rest frame of the fluid).

    Now comes the important part: we want to rewrite this in a Lorentz covariant form. This is easy to do because ##u^{\mu} = \delta^{\mu}_{t}## in the instantaneous rest frame of the fluid hence ##T^{\mu}{}{}_{\nu} = \rho u^{\mu}u_{\nu} + p(\delta^{\mu}_{\nu} + u^{\mu}u_{\nu})## where ##\delta^{\mu}_{\nu} + u^{\mu}u_{\nu}## projects onto the orthogonal complement of ##u^{\mu}##. Therefore ##T^{\mu\nu} = \rho u^{\mu}u^{\nu} + p(\eta^{\mu\nu} + u^{\mu}u^{\nu})##. This is a Lorentz covariant expression so it must hold in all Lorentz frames, not just the instantaneous rest frame of the fluid.

    In order to generalize to GR we just have to apply the Einstein equivalence principle which states that in a local inertial frame, all measurements must reduce to measurements in SR so in a local inertial frame the stress-energy tensor for a perfect fluid in curved space-time must reduce to the stress-energy tensor for a perfect fluid in flat space-time. Thus the prescription ##\eta^{\mu\nu} \rightarrow g^{\mu\nu}## is the natural one and we get ##T^{\mu\nu} = \rho u^{\mu}u^{\nu} + p(g^{\mu\nu} + u^{\mu}u^{\nu})##. This is also sometimes called "minimal coupling": http://people.sissa.it/~percacci/lectures/genrel/08-matter.pdf
     
  4. Dec 4, 2013 #3

    atyy

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    One of the assumptions of GR is the equivalence principle, also called the "minimal coupling". This principle says that spacetime should look "locally" like SR, which is why it is ok to use SR to derive the stress energy tensor.

    GR uses a symmetric stress tensor rather than the canonical stress energy tensor. If one defines the gravitational and matter fields with an action, the symmetric stress-energy tensor can be gotten from the action via Eq 6 of http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf or in the section on the Hilbert stress-energy in http://en.wikipedia.org/wiki/Stress–energy_tensor .
     
  5. Dec 5, 2013 #4
    You can get the S-E tensor directly by variaring the Ricci curvature scalar with respect to the metric.
     
  6. Dec 5, 2013 #5

    Bill_K

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    Not true, of course. The variation of the Ricci scalar gives you the left side of Einstein's equations, namely the Einstein tensor. To get the other side, the stress energy tensor, you must vary the matter Lagrangian with respect to the metric.
     
  7. Dec 5, 2013 #6
    I am not sure whether the local spacetime reducing to flat one is an approximation, a just a result of different observers.
    Because the SE tensor is first derived in SR with the assumption of flat spacetime. Then shall I regard all purely SR equations are still valid under GR conditions, since the SE tensor is now used in GR situations.
     
  8. Dec 5, 2013 #7

    WannabeNewton

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    It's not an approximation: all (pseudo) Riemannian manifolds are locally flat i.e. if ##M## is a (pseudo) Riemannian manifold then given any ##p\in M## there exists a neighborhood ##U## of ##p## and a coordinate chart ##\{x^{\mu}\}## such that ##g_{\mu\nu}(p) = \eta_{\mu\nu}## and ##\Gamma^{\mu}_{\nu\gamma}(p) = 0##. SR equations are valid in GR, as I have already said, because of the Einstein equivalence principle; all you have to do is make sure you lift Lorentz covariant equations to generally covariant equations by replacing the Minkowski metric with a general metric and partial derivatives with covariant derivatives (this works almost all the time but there are instances where this simple prescription fails-however the failure is due to things related to curvature and self-evident on a case by case basis).
     
  9. Dec 5, 2013 #8

    George Jones

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    Yea! You're a physicist!:biggrin: :wink:

    See

    https://www.physicsforums.com/showthread.php?t=374403
     
  10. Dec 5, 2013 #9

    WannabeNewton

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