The surface gravity of a Schwarzschild black hole is said to be inversely proportional to the mass of the black hole. But if the event horizon represents the "point of no return" even for light, then I would have thought that the surface gravity must have a fixed relationship to the speed of light, and hence should be the same for all black holes regardless of mass. Why am I wrong?
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1$\begingroup$ No contradiction, just different frames. For a remote observer the speed of light at the horizon is zero and gravity is infinite. For an infalling observer the local speed of light is constant and the surface gravity is less for a bigger black hole, because it's horizon is farther from the center. $\endgroup$– safesphereCommented Jul 11, 2018 at 15:30
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$\begingroup$ Interestingly black hole surface gravity is independent on gravitational constant G. In SI units it is given by $$ g_{BH} = \frac{ G M_{BH} }{ R_{BH}^2 } = \frac{c^2}{D_{BH}} $$ where $ D_{BH} $ is a black hole diameter. $\endgroup$– SzymonCommented Dec 6, 2021 at 17:52
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1 Answer
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As with so many things in GR the answer depends on exactly what you mean and which observer you are considering.
There is a property called the surface gravity, $\kappa$, that is defined in a rather technical way but is kind of the gravitational acceleration at the event horizon in the reference frame of a distant observer. This is given (in geometrical units) by:
$$ \kappa = \frac{1}{4M} = \frac{1}{2r_s} $$
so this is indeed inversely proportional to the black hole mass/radius. This surface gravity is not something that can be directly observed i.e. it is not something any observer could measure with their accelerometer. However it is an important property of a black hole. For example the temperature of the Hawking radiation is proportional to the surface gravity, as discussed in my answer to Why do larger black holes emit less Hawking Radiation than smaller black holes?.
However this is not the gravity someone hovering near the event horizon feels. If you hover at a distance $r$ from the centre of the black hole the gravity you feel is:
$$ a=\frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} $$
and as you approach the event horizon, i.e. as $r \to r_s$, this gravitational acceleration goes to infinity. This is true for any size of black hole.
The reason the surface gravity is finite while the acceleration felt by a hovering observer goes to infinity is because the hovering observer's time is dilated. If you are hovering close to the event horizon while I am far from the black hole your clock will be running more slowly than mine. Acceleration has units of metres per second squared, and it's because the two of us disagree about the length of one second that we disagree abut the gravitational acceleration.
answered Jul 11, 2018 at 15:21
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$\begingroup$ Sorry, but is not acceleration from the point of view of distant observer is negative as the infalling body will slow down as it approaches the horizon?There is no need to hold it, it will never fall through. $\endgroup$– AnixxCommented Jun 27, 2021 at 20:37
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$\begingroup$ @Anixx the Schwarzschild coordinate acceleration $\ddot r$ does indeed go to zero then become positive (it is initially negative i.e. inwards) for a freely falling object. However this is not a physically meaningful quantity as it depends on the coordinate system chosen. $\endgroup$ Commented Jun 28, 2021 at 4:38
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$\begingroup$ @John Rennie Is this equation applicable for a Kerr Black Hole as well? $\endgroup$– VickCommented Aug 30, 2023 at 7:06