3 ms·
Indeed, in their local coordinate system, their mass hasn't changed, while from a still observer's point of view, such a star gains that extra relativistic mass
by akomtu 6y ago
Indeed, in their local coordinate system, their mass hasn't changed, while from a still observer's point of view, such a star gains that extra relativistic mass and should in theory collapse once it reaches a certain speed. Of course, it would be strange if the same star would be collapsed or normal in different corrdinate systems. I guess the solution is that the extra relativistic mass isn't real, but a curious artifact that manifests as the higher than normal kinetic energy.
- XorNot 6y agoBlack hole's aren't really defined by mass though - they're defined by spacetime curvature. Specifically, inside the event horizon of a blackhole, every possible vector winds up at the singularity. A very speedy object however doesn't look like this: objects it passes have vectors which take them away from it. So while it can warp spacetime with that mass, it can't form an event horizon - there are vectors which lead things away from it and thus can escape. [EDIT: I had to go do some reading to check up on this and it actually has an analogue to electromagnetism: moving objects develop gravitational effects perpendicular to their direction of motion, essentially this is one of the origins of gravitational waves - but it's also important to the overall point, which is that even if you had infinite gravitational acceleration perpendicular to your direction of travel, you've got numerous other degrees of freedom where this doesn't happen and thus escape vectors - so no event horizon though you might suffer a lot of damage and be spagettified anyway] Black holes only work when you have a point force exerting uniform acceleration in all directions towards itself - beyond a key point, no attainable velocity ever results in an object not moving towards you. Whereas a speedy object has an obvious one: just jump off the back of it (i.e. accelerate slightly against it's vector) and your trajectory now points you away from it.