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What energy term determines gravity in GR?

  1. Nov 4, 2011 #1
    The relativistic relativistic energy–momentum relationship is:

    [tex]E^2 = \left(pc\right)^2 + \left(m_0c^2\right)^2[/tex]

    The relativistic energy (for a mass) is:

    [tex]E = \gamma m_0c^2[/tex]

    It has been said that the curvature of space is contributed to by photons, not just mass. However:

    • The curvature of space produced by a photon field is virtually undetectable as photons are simply not packed dense enough.
    • The ability for a photon to follow a curvs ed path is based primarily on the mass of the celestial objects. Doubling the energy of a photon only does not double this curvature and indeed no detectable diffraction has occurred as a result of curvature of space.

    If it is still true that photons themselves produce some amount of curvature, then the invariant mass of system [itex]m_0[/itex] cannot account for this curvature. However, neither can the following:

    • [itex]E^2 = \left(pc\right)^2 + \left(m_0c^2\right)^2[/itex] is relative to the observer because [itex]m_0[/itex] is fixed according to SR, not just [itex]c[/itex], while [itex]p[/itex] is relative to the observer, requiring [itex]E^2[/itex] to be relative to the observer.
    • [itex]E = \gamma m_0c^2[/itex] is relative to the observer because [itex]\gamma[/itex] is defined relative to the observer, requiring [itex]E[/itex] to be relative to the observer.

    B.T.W. In SR, [itex]m_0[/itex] is not a time-invariant mass, but a frame-invariant mass. Photons colliding head-on have been shown to produce additional [itex]m_0[/itex].

    So I ask, "What energy term determines gravity in GR?" My basic premise of course is that this quantity should not be dependent on the observer. But, maybe velocity time dilation relative to each mass has something to do with the apparent difference in curvature (more velocity time dilation would mean greater perceived acceleration of bodies in orbit around the mass). Does [itex]\sqrt{\left(pc\right)^2 + \left(m_0c^2\right)^2}[/itex] give us the amount of energy involved in this curvature? :confused:
    Last edited: Nov 4, 2011
  2. jcsd
  3. Nov 4, 2011 #2


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    Please remember that Maxwell's Equations have not been repealed, and it is still perfectly permissible to describe electromagnetic waves in terms of ... electromagnetic fields! Photons are appropriate for very low intensity or very high frequency, but otherwise they are not the best way to go.
    But energy IS dependent on the observer. Energy is the 00 component of the stress energy tensor. For electromagnetism,

    Tμν = Fμα Fνα - 1/4 gμν Fαβ Fαβ

    and this is the source of the gravity.
  4. Nov 4, 2011 #3


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    In special relativity to conserve energy for fields it is not sufficient to assign fields energy and momentum, but also "pressure" via a stress-energy-momentum tensor (which may be called the "energy-momentum tensor" or "stress-energy tensor" for convenience).

    In general relativity, the energy-momentum tensor is the generalization of Newtonian gravitational mass, and is the source of the gravitational field (the metric field).
  5. Nov 4, 2011 #4
    Is this "pressure" the key component in causing gravity (with energy and momentum simply being things which help to define it)? In other words, can the curvature be largely expressed in terms of pressure (as soon it has been calculated), or do energy and momentum make significant contributions separate from pressure?
  6. Nov 5, 2011 #5
    Among other mistakes, your «relativistic energy–momentum» is valid in SR not in GR when gravity is present. The energy of a photon with momentum p traveling near Sun is not given by your square root.
  7. Nov 5, 2011 #6


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    They all make significant contributions. Here is a Wikipedia link about the tensor Bill_K mentioned which is the source of gravity for an EM field:

    And some further details about the components of stress-energy tensors in general:
  8. Nov 5, 2011 #7
    Since we are on the topic of light waves, I asked a question a year or so ago and never got any answers... ( i figured it was kind of a dumb question) but now a year later, it still bugs me if it still applies.... can I get a hand with this?

    One way to "build" the SEM tensor is to associate a 3-volume with the energy-momentum 4-vector
    \Delta p^{\alpha\beta}=T^{\alpha\beta}n_\beta \Delta V
    This can get one the energy density, momentum density, and pressure by studying the different spacelike/timelike surfaces that comprise the volume. So my question is: is it appropriate to substitute [itex]p^\alpha =\hbar k^\alpha [/itex] for the momentum 4-vector? This would give the SEM tensor in terms of wave quantities. I find this interesting since it phrases SEM in terms of k and ω and the fluxes of these quantities.

  9. Nov 5, 2011 #8


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    My guess would be almost, but one would need one more parameter for the amplitude of the wave. Classically, the energy depends on the amplitude (not frequency) of the wave, so I would expect that to show up in the SEM tensor.

    You could check it by computing the SEM tensor the usual way (via the Lagrangian), then substituting the plave wave solution.
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