Defining the surface gravity of a black hole Consider the following two definitions of the surface gravity of a black hole (taken from page 23 of Thomas Hartman's lecture notes on Quantum Gravity):


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*The surface gravity is the acceleration due to gravity near the horizon (which goes to infinity) times the redshift factor (which goes to zero).

*If you stand far away from the black hole holding a fishing pole, and dangle an object on your fishing line near so it hovers near the horizon, then you will measure the tension in your fishing line to be $\kappa M_{\text{object}}$.

What does it mean for the acceleration due to gravity near the horizon to go to infinity?
Why do we multiply by the redshift factor and what does it mean for the redshift factor to go to zero?
 A: 
What does it mean for the acceleration due to gravity near the horizon to go to infinity?

It means that a person dangling near the horizon feels an almost infinite pull of gravity, and he observes that anything dropped almost immediately reaches almost the speed of light. 

Why do we multiply by the redshift factor and what does it mean for the redshift factor to go to zero?

The force felt at the fishing pole should somehow depend on the force felt by the thing dangling near the horizon, right? As we noted earlier that force 'goes to infinity'. But the force on the pole actually stays about the same when the force on the dangling thing is 'going to infinity'. So to get the right force at the fishing pole, we must multiply the large force felt by the dangling thing by a small number. The  number must 'go to zero' near the horizon. We call that number 'redshift factor'.
If gravity field is uniform, then we feel a constant force on our fishing pole, when we dangle an object at different altitudes in that field.
