Is spacetime curved by relativistic mass?

The top answer to this question asserts that only rest mass curves spacetime. My poor understanding of the EFEs & special relativity is that $$T_{\mu\nu}$$ doesn't care about what whether energy density includes by rest mass or relativistic mass.

The linked author says that "relativistic mass is not "real" mass. The gravity of an object travelling at relativistic speeds does not increase, because its mass does not increase" and goes on to talk some stuff about frames of reference that I think is probably irrelevant.

I always thought that a black hole does not form because you reach a certain rest mass energy, it forms because a region of space reaches a certain energy density... and that accelerating a mass to a sufficiently high velocity that the region of space it occupies achieves this energy density would form a black hole.

In short, I think that guy's totally wrong. Are they? Why or why not?

• Relativistic mass is a concept that is kinda outdated. At my university it is not taught anymore at all. Commented Oct 30, 2021 at 23:16
• Except for a units-conversion factor $c^2$, the "relativistic mass" of an isolated object is a synonym for the energy of the object (which is relative because velocity is relative). I don't know why anybody ever thought introducing that kind of redundant language would be a good idea, but in any case, the question is closely related to this one: Does kinetic energy warp spacetime? Commented Oct 30, 2021 at 23:40
• Commented Oct 31, 2021 at 6:49
• $T_{\mu\nu}$ is "relativistic". Think about Maxwell's field or any massless field. The rest mass is zero, hence only relativistic mass/energy contributes to T00 component. Commented Oct 31, 2021 at 9:54

The Oh My God particle was a cosmic ray...perhaps a proton...that struck Earth with about 50 joules of kinetic energy. In its own frame, it was a perfectly ordinary proton at rest, and it is Earth that had a Lorentz factor of 3.2e11, which would be enough "relativistic mass" to make a black hole 18 light seconds across. Notably, we're all still here.

Energy and mass are equivalent, but kinetic energy is frame dependent, and has no physical meaning in isolation. It only means something in relation to some larger system. So no, just moving fast is not enough to form a black hole.

• Double-checked your calculations & I got the same thing. I still doubt your conclusion. The stress–energy tensor of course includes momentum. Thus, I suspect that under Lorentz boosts the curvature of spacetime is variant and Earth was indeed a supermassive black hole in the OMG particle's frame of reference (although I doubt one can derive its Schwarzschild radius as simply as just plugging it into the equation in this high-velocity frame).
– neph
Commented Oct 31, 2021 at 0:55
• "Earth was indeed a supermassive black hole in the OMG particle's frame of reference": that would imply that one would see things turn into black holes and back into planets and stars just by accelerating. You could pass through the solar system and see the planets fully illuminated by light escaping the black hole at the center of the system, etc. The suggestion raises all sorts of contradictions. An object or system is always what it is in its own center of momentum frame, whether that's the OMG particle, Earth, or the system consisting of both of them together. Commented Oct 31, 2021 at 1:26
• @safesphere Thank you! Is this answer incorrect, then? physics.stackexchange.com/q/479299
– neph
Commented Oct 31, 2021 at 4:56
• @safesphere I think your comment about gravitational effect attributed to momentum (gravitomagnetism) is well worth entering as an answer. There is the article (that you may well already know) by Steve Carlip: 'Aberration and the speed of Gravity', which does a lot to show how the concept of gravitomagnetism is integral to relativistic theory of gravity. Carlip devotes a section to the question 'Is the cancellation a miracle?' Carlip discusses: there is a connection between the cancellation and conservation laws. Commented Oct 31, 2021 at 12:12
• The Wikipedia page on Mass in special relativity says that "The measurable inertia and the warping of spacetime by a body in a given frame of reference is determined by its relativistic mass, not merely its invariant mass.". Is this incorrect? Commented Dec 18, 2023 at 13:36

Let me discuss this question in terms of the principle of equivalence.

Thought experiment:
A chamber, with walls that reflect light perfectly. That is, light that has entered that chamber is never absorbed by the walls; the light persists in that chamber just as a gas would persist in that chamber.

The energy of the light in that chamber contributes to the inertial mass of the assembly.

Next we consider a thought experiment where the chamber is in free fall. We compare the free fall of a chamber that is holding a significant amount of light energy to the motion of an empty chamber.

According to the principle of equivalence the free fall of the chamber-with-light-energy will be indistinguishable from the free fall of the empty chamber. It follows logically that if the principle of equivalence holds good the chamber-with-light-energy must have a larger gravitational mass than the empty chamber, matching the ratio of inertial masses of the two chambers.

Next we consider a thought experiment where the chamber is a housing for a setup for flywheel energy storage

We have that the spinning flywheel will have a larger inertial mass than a non-spinning flywheel, in accordance with the difference in accumulated kinetic energy. The kinetic energy of a spinning flywheel is confined to a finite volume of space. That is: the kinetic energy of a spinning flywheel has a definable energy density.

(Of course, in an actual setup this additional inertial mass is too small to be actually measured. An actual flywheel will disintegrate long before it reaches relativistic velocity.)

A celestial counterpart of a spinning flywheel is a spinning neutron star. Some neutron stars have an extremely large rate of spin.

The logic of the principle of equivalence implies that the kinetic energy of the neutron star spin must be contributing to the gravitational mass of that neutron star.