# What's the G-forces in a black hole event horizon?

As the tile says, anyone's got an idea what the g-forces might be at the boundary of a black hole's event horizon? I got a formula to calculate the event horizon radius but not the the gravitational forces. Equivalence to Earth G's would be nice or in m/s2 = meter per second squared!

Is it also true that if I accelerate a craft to that equivalent g-force, will it produce an artificial event horizon? I'm familiar with that theory, but I've lost track if it's ever tested already.

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cristo
Staff Emeritus
As the tile says, anyone's got an idea what the g-forces might be at the boundary of a black hole's event horizon? I got a formula to calculate the event horizon radius but not the the gravitational forces. Equivalence to Earth G's would be nice or in m/s2 = meter per second squared!
Well, it firstly depends on what you mean by force. I presume that you mean gravitational tidal forces. They are very large past the event horizon of a black hole, but I cannot provide you with exact numbers.

Is it also true that if I accelerate a craft to that equivalent g-force, will it produce an artificial event horizon?
Again, what do you mean here by "g-force"? It is true that any accelerating observer will have a sort of event horizon, however it is not the same as an event horizon around a black hole. The event horizon that the accelerating observer experiences is only a one way event horizon-- there exists regions of spacetime from which causal rays cannot reach the observer, but not vice-versa.

mgb_phys
Homework Helper
If you mean what is the gravitational force (eg. relative to 'g' at the earths surface) this is easy to work out.
F = Gm1m2/r^2 and F=m1a so a=Gm/r^2 where m is the mass of the black hole and r is it's radius.
You have the formula for the radius of the event horizon for a given mass of black hole so you can work out how the 'surface gravity' varies with radius.
You will see that as a black hole gets bigger it's surface gravity gets less - work out how big a black hole would be to have 1g of surface gravity.

The second part of your question misunderstands acceleration and gravity. We use 'g' as a convenient measure of the strength of a gravitational field because it produces the same apparent force as needed to accelarate the mass. This doesn't mean that acceleration causes gravity.

pervect
Staff Emeritus
As the tile says, anyone's got an idea what the g-forces might be at the boundary of a black hole's event horizon? I got a formula to calculate the event horizon radius but not the the gravitational forces. Equivalence to Earth G's would be nice or in m/s2 = meter per second squared!

Is it also true that if I accelerate a craft to that equivalent g-force, will it produce an artificial event horizon? I'm familiar with that theory, but I've lost track if it's ever tested already.
It depends on what it is you want to measure.

The proper acceleration measured by an accelerometer mounted on an observer who is hovering above the event horizon at a constant Schwarzschild coordinate r is

[corrected]

$$a = \frac{GM}{r^2 \sqrt{1 - \frac{r_s}{r}}}$$

where $$r_s = \frac{2GM}{ c^2} =$$ the Schwarzschild radius
M = mass of black hole
G = gravitatioanl constant
c = speed of light

if r >> $r_s$ this approximates the Newtonian equation for gravitational acceleration when the Schwarzschild coordinate r is approximated by the Newtonian radial distance.

The above equation represents what someone in a spaceship hovering at a constant distance from the black hole would feel "on the seat of their pants", or with any other onboard acceleration measuring instrument.

Note that this is not the same as what is called "the surface gravity of the black hole" in black hole thermodynamics, which is finite. (The above expression goes to infinity at the event horizon, which is located at r=$r_s$.

This is also not the same as the coordinate acceleration that someone at infinity would see - that would be zero.

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Jorrie
Gold Member
Equation right?

It depends on what it is you want to measure.

The proper acceleration measured by an accelerometer mounted on an observer who is hovering above the event horizon at a constant Schwarzschild coordinate r is

$$a = \frac{GM}{r^2 \sqrt{1 - \frac{r}{r_s}}}$$

where $$r_s = \frac{2GM}{ c^2} =$$ the Schwarzschild radius
M = mass of black hole
G = gravitatioanl constant
c = speed of light
Sorry to dig up a 'dead thread', but I just noticed something here: haven't you got r / r_s the wrong way around under the square root? Should it not have been:

$$a = \frac{GM}{r^2 \sqrt{1 - \frac{r_s}{r}}}$$ ?

I'm asking this because I've seen the same thing in the notes of Kip Thorne's "Black Holes and Time Warps", note 40. So I'm wondering if I simply understand it wrongly.

pervect
Staff Emeritus
Sorry to dig up a 'dead thread', but I just noticed something here: haven't you got r / r_s the wrong way around under the square root? Should it not have been:

$$a = \frac{GM}{r^2 \sqrt{1 - \frac{r_s}{r}}}$$ ?

I'm asking this because I've seen the same thing in the notes of Kip Thorne's "Black Holes and Time Warps", note 40. So I'm wondering if I simply understand it wrongly.
Your understanding is fine, I made an error in converting from geometric units. It should indeed be r_s / r, which is equivalent to 2m / r. I thought it would be better not to use geometric units, but unfortunately I didn't do the conversion from geometric to ordinary units right.

I took the liberty of correcting the original.

Thought this was an interesting question to play with the math.... See attached.

As expected, the G forces at the event horizon are pretty fierce, but the bigger the mass of the black hole, the bigger the Swartchild radius gets and the lower the g-force at the event horizon.

I don't suppose our hypothetical spaceman would be all that worried about the g-forces at the event horizon though. I don't know anything about the dynamics of any gravitational waves, but the tidal forces anywhere near a small black hole would tear our poor spaceman to pieces long before he got there unless it was a real monster. I'll have a bash at working that out when I get another spare hour or so Much more interesting then compiling reports too.

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work out how big a black hole would be to have 1g of surface gravity.
That works out to be about 150 Billion Solar Masses or 2.98E+42Kg I think. Pretty big then.

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