# Consequence of approaching infinite mass for near-light speed particles

Considering that the inertial mass of an object approaches infinity as the speed of the object approaches $c$, and that inertial mass equals gravitational mass, does this not imply that particles nearing $c$ would have gravitational mass approaching infinity?

Postulating that in fact gravitational mass nears infinity, would an object moving at near $c$, due to approaching infinite gravitational mass, cause profound gravitational acceleration on very distant objects? Infinity is a big value, even after decaying at $1/d^2$.

• Careful here. "Inertial mass" is not a relativistic invariant, since it is observer dependent. The concept of "inertial mass w.r.t. to a moving observer" is also called "relativistic mass", and is effectively considered obsolete, cf. physics.stackexchange.com/a/133395/50583. It's not clear that "inertial mass = gravitational mass" is meant to hold in frames other than the rest frame, especially given that I don't know what "gravitational mass" is supposed to mean outside of the Newtonian limit. The second question is a duplicate of physics.stackexchange.com/q/3436/50583. – ACuriousMind Sep 30 '16 at 11:05
• @ACuriousMind: Thank you for the link. In fact the second query was a dupe of that, so I've removed it from this question. Now this question is more focused, thank you. – dotancohen Sep 30 '16 at 11:57
• This kind of question is exactly why "relativistic mass" it not a mere harmless-but-obsolete curiosity, but a notion that should be actively avoided in elementary instruction and popularization: it enables people to ask questions which sound reasonable despite being obviously un-reasonable when asked in better language. Do you expect that a particle passing the Earth near light-speed should experience a near infinite gravitational attraction from Earth? If not (the right answer) then why would you expect the answer to change simply because you changed reference frame? – dmckee --- ex-moderator kitten Sep 30 '16 at 17:20
• @dmckee: Actually, if in fact mass approaches infinity as an object approaches $c$ then I would expect that object to experience a near infinite gravitational attraction from the Earth. However, that same near infinite mass would also provide it near infinite resistance to the applied force, that is what inertial mass does. So the net effect on the particle would be minor, but the net effect on the Earth, without its near infinite inertial mass, would be major. – dotancohen Sep 30 '16 at 20:35
• @dotancohen You haven't thought that through: in the frame of the passing particle the Earth would have increased mass and the particle would not, which leads you to a silly prediction. – dmckee --- ex-moderator kitten Sep 30 '16 at 21:11

As @ACuriousMind pointed out: it is the particles momentum that goes against infinity its rest mass $m$ is a scalar: a fixed particle property that does not change.

To your question about "gravitational mass": in General Relativity (GR) this is a very very delicate point. The concept of a "gravitational mass" is not trivial in GR and one needs to specify very precisely what one understands as gravitational mass. First of all "mass" is not governing the curvature of space: the energy momentum tensor is.

So if you are interested in the gravitational field of a massive particle moving very fast one would need to solve the field equations with the energy momentum tensor of this particle. You can NOT use lets say the Schwarzschild metric (SSM) and put in the "relativistic mass" of your particle. First the concept of relativistic mass is not a good one 2. the SSM does not hold at all for moving particles/sources. It holds only outside a static, spherical symmetric object.

I did not find a reference for the gravitational field of a relativistic moving particle but one can for sure not use the SSM. The gravitational field will depend on the particles four-momentum, it will be non static and not spherical symmetric, since the direction of motion is an axis of anisotropy. Maybe one can work out some spatial axis-symmetric solution for that problem choosing a coordinate system in which the four momentum becomes $p^\alpha=(p^0,0,0,p^3)$ but I have not found/ calculated one. Apart from that I would assume that even at relativistic speeds the curvature caused by single massive particle will be relatively small, since even at very high speeds the energy/momentum of such a particle would be rather small.

This gravitational mass against infinity picture does not hold: its all about the energy momentum tensor and you will not get this one to infinity with just relativistic speeds.

That being said one more point towards "gravitational mass" in GR, to be specific in the SSM. In the SSM the "gravitational mass" comes up in form of a integration constant and from the Newtonian limit we see that this constant can be identified with the classical/Newtonian gravitational mass. This is one special case where classical gravitational mass has a meaning/equivalent in GR. It is one of very few cases where the concept holds some meaning in GR. In general this is not the case and one can define a multitude of "masses" (integrals over very different objects).

the whole reason that the inertial mass increases as an object approaches C is that it requires an increasing amount of energy to be able to accelerate faster, hence to accelerate an object (that has mass) to C would require infinite energy (and break a LOT of physics)

• Keep in mind that the "inertial mass" of a particle in motion relative you depends on which direction you push it. There is more (anti-)parallel to the relative velocity vector than transverse to it. Search terms "longitudinal mass" and "transverse mass". – dmckee --- ex-moderator kitten Sep 30 '16 at 17:22

Some time before the mass of the particle approaches infinity, surely its gravitational field must make it collapse into a micro-black hole. Or not, the whole thing sounds basically flawed the more you extrapolate its implications.