Paradox in special and general relativity - a thought experiment?

Question

Consider a neutron star - say 13 solar masses that is in a state of collapse such that its radius is JUST larger than the schwarzschild radius i.e. it is on the verge of becoming a black hole. Assume further that this star is in a state of equilibrium i.e. energy radiated away = mass arriving on the star.

Now consider two observers - one is stationary relative to the star and can make measurements as to the radius and mass of said star - and agrees with the above situation. 

Now consider a second observer moving at a relativistic speed relative to the neutron star. Applying the Lorentz transformations then the radius of the neutron star as measured is reduced and the mass of the neutron star is increased. Both of these factors would seem to indicate that the radius of the neutron star is now LESS than the schwarzschild radius and that the star should now become a black hole.

Now the second observer changes their frame of reference to be the same as the first observer i.e. stationary relative to the stellar object.

Does the second observer still 'see' the black hole or has the black hole changed back to a neutron star and, if so, how?

Lola F · Accepted Answer

The second observer never sees a black hole. A body is either a black hole or it isn't, and changing reference frames doesn't change that.

Not all bodies observed to be smaller than their Schwarzschild radii are black holes. You remembered to apply the Lorentz transformation to the mass of the star but neglected to do so for the metric. The standard Schwarzshild metric is only one particular solution (the "static solution") of the black hole problem for bodies at rest relative to distant observers. For bodies in motion (or bodies with residual internal structure, i.e. black hole "ringdown"), that solution itself must be transformed.

Cisco · Answer

The whole Schwazschild analysis assumes the star is at rest with respect to the person measuring the radius. Also, for a moving observer, the star does not contract equally, only along the axis of motion, so its radius in the perpendicular directions remains the same. If you do the calculations making all the corrections necessary for the moving observer, the results are the same.

A P · Answer

Surely no observer outside a forming black hole can see its transition into a black hole. Not even if that observer is falling into the black hole. Time freezes at the event horizon. Let's assume the black hole is truly massive, say one billion solar masses, such that an in falling astronaut can survive the tidal forces at the event horizon as measured by a second and distant observer. Our in falling astronaut would have his own version of the event horizon which as he approaches would shrink inward towards the singularity, and the subatomic particles of his mass would be destroyed by tidal forces as he catches up with all the material that has already fallen in, and meet all of the subatomic particles which in the future will fall in, all coming together at the singularity for a final big crunch.

Our outside observer will simply see the in falling astronaut falling ever more slowly as he approaches the event horizon and is red shifted to ever longer electromagnetic wave lengths until disappearing from view.

It remains paradoxical, but I think your scenario is somewhere inside my extreme scenario. The paradox appears to be, how can a black hole form in the observable universe if time stands still at its event horizon as measured in the observable universe?

Paradox in special and general relativity - a thought experiment?

3 Answers