# Formula for gravitational acceleration as you approach a black hole's event horizon

With a classical point particle we have $$Gm/r^2$$ acceleration, but with a massive object such as a neutron star or black hole we have additional geometrical and time distortions (radial distance increases and local time slows down relative to a distant observer).

What is the formula for gravitational acceleration around a super massive object as a function of distance from its center (defined as equal to circumference/$$2\pi$$) for an object hovering at rest (relative to the massive body)?

I should be able to convert this myself to a formula of acceleration according to a distant observer but I would still like to know what this is so that I can use it to double check my understanding of how space and time is warped. Since descriptions of how to calculate when something is dense enough to become a black hole use the classical escape velocity, I am guessing that somehow everything will cancel out and it will still end up being $$Gm/r^2$$.

• Possible duplicate. The answer is there anyway. physics.stackexchange.com/q/47379 Commented Feb 24, 2020 at 21:21
• That answered most of my question and had enough information for me to see the answer to the rest. I'm surprised this didn't come up in the search. Commented Feb 24, 2020 at 22:58
• I might be wrong but for a Schwarzchild black hole we can use the classic acceleration formula. It is when we have rotating BH we need to include effects such as gravitational frame-dragging (en.wikipedia.org/wiki/Frame-dragging) Commented Feb 24, 2020 at 23:04

Rob Jeffries found a post that answered most of my question (what is the weight equation through general relativity).

$$accel=\frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$

where $$r = \frac{circumferenc}{2 \pi}$$

As I guessed, this is just the Newtonian equation divided by time dilation $$\sqrt{1-\frac{2GM}{c^2r}}$$ and assuming that space coordinates are adjusted for radial space expansion. All hovering observers agree on circumference but you have to integrate the following for total (non-relativistic) radial distance traveled: $$\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$

One way to look at it is that it is the same acceleration as for the classical case (as pointed out by pavel) except that time runs slower for someone closer to the event horizon so it seems like a higher acceleration to them.

• the same acceleration as for the classical case” - Not really. In the Schwarzschild frame, acceleration at the horizon is asymptotically zero. All motion freezes there. Commented Feb 25, 2020 at 15:40
• By "same" I meant when you calculate it relative to local geometry of space (but not adjusting for time dilation). Space expansion (which looks like compression from the outside) near the event horizon goes to infinity both in d(radius)/d(circumference) and total integral so infalling objects never get there relative to the outside reference frame. The way I understand it, observers hovering gradually closer to the event horizon see acceleration approach infinity but will never actually reach it according to any frame. Free-falling ones will reach it in their own frame due to SR effects. Commented Mar 1, 2020 at 22:24
• I suspected from some thought experiments that everything freezes there according to the outside reference frame but it was really hard to figure out the correct way to ask the question. The main issue is that "according to outside observer" can mean either "according to distant observer's calculations in their coordinate system" or "what the distant observer actually sees". I've noticed that physicists tend to freely switch between these definitions, leaving it up to the audience to determine from context what it means. Commented Mar 1, 2020 at 22:38
• "According to [some] observer" always means "expressed in terms of his coordinates" (e.g. predicted or calculated). It never means what this observer actually sees. To figure out what he actually sees (or doesn't) requires additional calculations. For example, when things freeze forever at the horizon, a distant observer doesn't actually see anything. He may see some redshfted photons for a while, but their energy drops to near zero almost instantly. Commented Mar 2, 2020 at 4:32