At a very basic level I know that gravity isn't generated by mass but rather the stress-energy tensor and when I wave my hands a lot it seems like that implies that energy in $E^2 = (pc)^2 + (mc^2)^2$ is the source of gravity.

If the total energy of a particle contributes to the generation of a gravitational field, that seems to imply some strange things. For example, you could accelerate a particle like an electron to such a high velocity that it has enough momentum / kinetic energy to collapse into a black hole. An observer traveling with the electron would see an electron and stationary observers would see an arbitrarily massive black hole flying by. These observations seem to be contradictory. The same would be true of an extremely high energy photon.

Another consequence is that a stationary particle could appear to have so much mass that it could attract / deflect the trajectory of a rapidly moving observer. It seems odd that at low velocities you wouldn't even notice a stationary electron as you pass it by but at high velocities the gravitational field you observe could be significant enough to alter your path.

Am I correct that non-mass-energy generates a gravitational field? If it does, how are these strange observations like a particle looking like a black hole in one reference frame but not another reconciled?


A black hole won't form. The reason why is that the boosted particle is equivalent by a boost to a reference frame where there is no black hole, and the presence/abscence of a black hole is coordinate-independent. While the energy of, say, an object with Earth's density profile can be made arbitrarily large through a boost, the boosted Earth will still have a distinguishable stress-energy tensor from a highly compact object that is not boosted. And, it will be distinguishable from the stress-energy tensor of a boosted black hole, which can be defined by noting that the Kerr metric can be written in the form:

$$g_{ab} = \eta_{ab} + C \ell_{a}\ell_{b}$$

where $\ell_{a}$ is a particular null vector relative to both $\eta_{ab}$ and $g_{ab}$ and $C$ is an exactly specified function. In this form, we can define boosts relative to the background Minkowski metric, and find out what the spacetime of moving black holes is.

Which is a long way of saying that the math encodes the difference between an earth boosted to $.999999999c$ and something that is natively super-dense and is moving that fast. You really have to consider the whole stress-energy tensor, which doesn't just include energy density, but includes momentum and all of the internal pressures in the object.

  • $\begingroup$ Okay so "the math encodes the difference..." takes care of the black hole formation. Does that also mean that since the particle isn't getting bigger, it puts an upper limit on the the gravitational field that can be formed from boosting alone? It seems like allowing a boosted particle to form any sort of extra gravitational field causes issues. In the reference frame moving with the boosted object, surrounding particle would be attracted to the moving particle as though it has a gravitational field that the observer can't see. $\endgroup$ – Brandon Enright Apr 26 '13 at 21:19
  • $\begingroup$ @BrandonEnright: I've never calculated the force on a stationary object at a finite distance due to a boosted black hole. I would assume that there would be some sort of upper limit to the force, but I'd have to do the calculation to know the answer for sure, though you'll get other effects that are as potentially as big as the boosted energy creating more apparent mass, such as gravetomagnetism. $\endgroup$ – Jerry Schirmer Apr 26 '13 at 21:24
  • $\begingroup$ @JerrySchirmer Should be trivial to pull it out, at least for a spherical central body. Just take the Schwarschild solution and boost it. In the frame of the black hole objects follow (approximately) conic section orbits, so you can just boost those to find the trajectories in the other frame. $\endgroup$ – Michael Brown Apr 27 '13 at 1:01
  • $\begingroup$ @MichaelBrown: sure. I tell you above how to do it. I just haven't done the calculation. $\endgroup$ – Jerry Schirmer Apr 27 '13 at 4:24

As you know, gravity isn't generated by mass but rather the stress-energy tensor and a property of tensors is that they're co-ordinate invariant. So it is same in all the reference frames.


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