Physics 321 

Prof. Dale E. Gary
NJIT 
Black Holes
When a stellar corpse is between 0.08 and 1.4 solar masses, the state of matter in the star is electron degenerate matter. Between 1.4 and 3 solar masses it is neutron degenerate. What happens when it is greater than 3 solar masses? At such a mass we know of no force that can support the star against the crush of gravity. As far as we know, it collapses to a "singularity." What really happens we do not know, but we do know that when the object gets small enough the surface gravity gets so high that even light cannot escape. The object becomes a black hole.
We can calculate the escape velocity (the outward velocity that an object would have at the surface of an star or planet in order to reach infinity with no velocity)
1/2 mv^{2} = GMm/RThis means that the deeper one goes into a gravitational well, the more energy (velocity) it takes to get out again. At some radius, the escape velocity reaches the speed of light! That radius is found by setting v = c and solving for R:
v = [2GM/R]^{1/2}
R = 2GM/c^{2} = (2GM_{o}/c^{2})(M/M_{o}) = 3 (M/M_{o}) km Schwarzschild RadiusThis is the radius of a nonrotating black holealso called the Event Horizon. Thus, the smallest stellar corpse black hole should be about 9 km in size (since M/M_{o} = ~3). Unlike the case for white dwarfs and neutron stars, the event horizon grows as mass increases. However, the event horizon is merely the location of the point where the escape speed reaches the speed of light. The object itself has shrunk to a singularity.
From the expression for Gravitational Redshift from the previous lecture,
l_{f}/ l_{i} = [1  2GM/Rc^{2}]^{1/2} Gravitational Redshift (relativistic)we find that the photon wavelength goes to zero when
l_{f}/ l_{i} = [1  2GM/Rc^{2}]^{1/2} = 0or
2GM/Rc^{2} = 1Solving for the radius, we find
R = 2GM/c^{2}which is the same radius (the Schwarzschild Radius) that we got from the escape speed argument.

Getting Close to a Black Hole
Distortion due to bending of light High orbital velocityWhat do you feel?
Tidal Forces (gravity gradient in curved spacetime)Trip to a Black Hole
Normally we are so far from the center of a gravitating body (say the Sun or the Earth) that our local space is flat, and only large objects feel any difference in the force of gravity from one side to the other. Because a black hole can be so tiny and yet so massive, we can get very close to it and even small objects like ourselves can feel a difference in the force of gravity between our feet and our heads, for example. This difference in force can literally be enough to rip us apart.
Say we are in a ship, in orbit at 1 AU around a block hole of mass 10 M_{o}. Our orbit is normal, and is not affected by the black hole any differently than any other star of 10 M_{o}. We now let you drop into the black hole with a laser and a stop watch.Observational evidenceFor most of the trip, nothing much happens, but as you get very close to the event horizon your laser signal gets progressively redshifted, as seen by me, still on the ship. You also feel progressively stronger tidal forces that stretch you head to toe. At about 3000 km radius, you would be pulled apart, unfortunately. The pieces, however, continue on through the event horizon rather uneventfully.
Back at the ship, your laser signal is redshifted, and your stop watch appears to tick more and more slowly due to time dilation. These two things can be considered the same phenomenon, in fact. Just as you cross the event horizon, and signal you send back takes forever to arrive (and it is red shifted to zero energy anyway). As seen from the ship, in fact, you appear to move more and more slowly and freeze at the event horizon, never crossing.
What happens on the ship as seen by you? Is there a paradox?
Accretion disk and Xray emission
Doppler shifts of visible companion and orbital velocity
Total mass of system
Energy output of system and jets