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## Main Question or Discussion Point

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:

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:

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.

[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*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?**mass**
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