- #1
- 750
- 41
From wikipedia 'http://en.wikipedia.org/wiki/Surface_gravity" '-
For a static black hole, this is-
[tex]\kappa=\frac{c^2}{4M}=\frac{c^2}{2r_s}[/tex]
where [itex]M=Gm/c^2[/itex] and rs is the Schwarzschild radius (2M)
this is synonymous to Newtons equation for gravity [itex]g=Gm/r^2[/itex] at the event horizon where r=2M (the Schwarzschild radius), the results being the same.
When looking at Kerr black holes, the Killing surface gravity is-
[tex]\kappa=c^2\frac{(r_+-r_-)}{2(r_+^2+a^2)}[/tex]
where r+ is the outer event horizon [itex]r_+=M+\sqrt(m^2-a^2)[/itex], r- is the inner event horizon [itex]r_+=M+\sqrt(m^2-a^2)[/itex] and a is the spin parameter [itex]a=J/mc[/itex].
which reduces to-
[tex]\kappa=c^2\frac{r_s}{2r_s^2}[/tex]
when a=0 which is equivalent to the first equation for a Schwarzschild black hole.
The Killing surface gravity works out less for the Kerr black hole than that for the static solution. Obviously this has something to do with spin and the Newtonian equation for gravity is 'modified' but what is this based on? I would have thought the reduction of gravity would be a consequence of the vacuum solution (Kerr metric) as apposed to something that seems related to the equivalent of Newtonian gravity. I understand that κ is abstract but it seems odd that it should reduce the higher the spin when normally it's considered that rotational kinetic energy may contribute to the stress energy tensor.
In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which can only be truly treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity.
Therefore, when one talks about the surface gravity of a black hole, one is defining a notion that behaves analogously to the Newtonian surface gravity, but is not the same thing. In fact, the surface gravity of a general black hole is not well defined. However, one can define the surface gravity for a black hole whose event horizon is a Killing horizon.
For a static black hole, this is-
[tex]\kappa=\frac{c^2}{4M}=\frac{c^2}{2r_s}[/tex]
where [itex]M=Gm/c^2[/itex] and rs is the Schwarzschild radius (2M)
this is synonymous to Newtons equation for gravity [itex]g=Gm/r^2[/itex] at the event horizon where r=2M (the Schwarzschild radius), the results being the same.
When looking at Kerr black holes, the Killing surface gravity is-
[tex]\kappa=c^2\frac{(r_+-r_-)}{2(r_+^2+a^2)}[/tex]
where r+ is the outer event horizon [itex]r_+=M+\sqrt(m^2-a^2)[/itex], r- is the inner event horizon [itex]r_+=M+\sqrt(m^2-a^2)[/itex] and a is the spin parameter [itex]a=J/mc[/itex].
which reduces to-
[tex]\kappa=c^2\frac{r_s}{2r_s^2}[/tex]
when a=0 which is equivalent to the first equation for a Schwarzschild black hole.
The Killing surface gravity works out less for the Kerr black hole than that for the static solution. Obviously this has something to do with spin and the Newtonian equation for gravity is 'modified' but what is this based on? I would have thought the reduction of gravity would be a consequence of the vacuum solution (Kerr metric) as apposed to something that seems related to the equivalent of Newtonian gravity. I understand that κ is abstract but it seems odd that it should reduce the higher the spin when normally it's considered that rotational kinetic energy may contribute to the stress energy tensor.
Last edited by a moderator: