If we assume there is a singularity at the centre of a black hole:
For an observer at the centre of a black hole (of mass M) the force on a small mass (m0) approaching from infinity when at distance r from the centre
|Where||m||=||mo||by special relativity, which is towards the centre.|
|(1 - v2/c2)½|
|and||dr||=||-||GM||- as ro o|
The kinetic energy =
and the added mass = /c + mo
so that the black hole has infinite mass, which is clearly not true by observation
If on the other hand the mass of the black hole is contained in a neutron star at absolute zero, with radius ro;
If a particle of rest mass mo arrives from infinity
|Where||m||=||mo||invoking special relativity, and|
|(1 - v2/c2)½|
d r/dt = -GMm/r m/r m
Particle arrives at r0 with kinetic energy moGM/2ro and the added mass is moGM/2ro c + mo
Note that dr/dt can exceed the velocity of light if M is large enough
And -dr/dt = vo say.
For example, for G = 6.67X10-11 , M = 4×1033 kg and ro= 1.343×106m the particle arrives at the neutron surface at c
The value for r0 requires some justification.
If a neutron has a cross section of one barn its radius is 5.64×10-5m assuming it is incompressible.
A spherical neutron star of mass M with hexagonal close packing has a radius of (5.656 M / mo)1/3×5.64×10-5 = 1.343×106m.
An alternative way of calculating the escape velocity from ro allowing for r inside the neutron star is as follows:
F = -GMm/r .r /ro when r < ro
and d r/dt = -GM/r r /ro
|Therefore||-||dr||=||GM||=||- vo||giving the same value for vo as before.|
J F Fleming 24 February 2011
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