What gives a Black Hole its tremendous gravity?

by DHF
Tags: black holes
 P: 65 I have been fascinated by stars and Black holes for most of my life but one of the things I cant get a satisfactory explanation on is what gives a Black Hole it's mind boggling gravity. We know that a Black Hole cannot be seen because it's gravity is so great that not even light can escape it but if it is formed from a massive star, wouldn't it have a mass equal to that star (if not less if some of the mass was lost during the collapse) so why would its gravity change? Or am I chasing a red herring because this would be the part where the physics break down? Don
 P: 740 You're quite right that its gravity is pretty much the same as the gravity of the core that had collapsed - if viewed from far away. The difference is only noticeable when one gets very close to the black hole(closer than the surface of the stellar core). The reason is the density. Black hole has got the same mass crammed into a much tighter space, so the gravity at what you might call its "surface" is much stronger than at the surface of the stellar core that begat it. You probably know the equation of gravitational acceleration: $g=G\frac{M}{R^2}$ One can treat all the mass of a sphere as being simply concentrated in its centre, so R is the distance from the surface to the centre of mass M. With regular bodies, getting closer to the centre than R means that you're leaving some mass behind(or, "above" you), and that mass can be shown to contribute exactly zero to the forces acting on you. The result is that gravity raises until you hit the surface, and then falls down as you approach the centre. Black holes really have all of their masses concentrated at their centres, so as you get closer to the centre the gravity keeps raising until it reaches the levels sufficient for trapping the light within its event horizon.
 C. Spirit Sci Advisor Thanks P: 5,661 A black hole whose event horizon happens to be a killing horizon has defined over its event horizon a quantity called the surface gravity ##\kappa## given by ##\kappa^{2} = -\frac{1}{2}\nabla^{a}\xi^{b}\nabla_{a}\xi_{b}## where ##\xi^{a}## is the killing vector field normal to the event horizon (which is a null surface). Now, I don't know if any of the above makes sense to you or not but I am writing all this just to be complete in case other people happen to land on this thread. You can, under the above conditions, interpret ##\kappa## as the force that would need to be exerted at infinity in order to hold in place a unit test mass, in the limit as one approaches the event horizon. Take the simplest case of a Schwarzschild black hole, which is the one you are probably thinking of. We can then calculate the surface gravity across the event horizon using appropriate coordinates (e.g. Eddington - Finklestein coordinates) and we will find that ##\kappa = \frac{c^{4}}{4GM}## where ##M## is the Schwarzschild mass-parameter. Consider a Schwarzschild black hole of one solar mass. If you plug in the values in the newly calculated surface gravity, you will find that ##\kappa = 1.5\times 10^{13}\frac{m}{s^{2}}## whereas the surface gravity of the sun itself is about ##274\frac{m}{s^{2}}##!
 P: 65 What gives a Black Hole its tremendous gravity? Thank you Bandersnatch that makes perfect sense. and thank you Wannebenewton, that was very detailed although I am afraid a little over my head.
C. Spirit