Gravity at Schwarzschild Radius of a Black Hole

[Roberto Pavani]

TL;DR Summary

It would be possible to have the gravity acceleration experienced by an observer close to an event horizon so "little" to let a spaceship to stay above the event horizon?

From "standard" formula we have that the gravity acceleration a = GM/r^2 and that the Schwarzschild radius rs = 2 GM / c^2

Is it possible to compute the gravity acceleration at Schwarzschild radius putting r = rs?
In this case we will have a = c^4 / (4GM) This mean that a very very ultramassive black hole can have a gravity acceleration for an observer O1 close the event horizon, similar to the Earth gravity acceleration ?
If the above formula is wrong (a = GM/r^2) , what is the gravity acceleration formula for a black hole?

From a very far observer O2 the observer O1 is seen slowed down.

Is O1 able to hover the event horizon because of the "relative small" gravity ?
If yes can that observe send a probe "below" the event horizon and using the probe engines, retrieve it ?

Another question is, when the probe will be below the event horizon, the radio signals from the probe aren't able to "escape" (e.g. will never reach O2) because the escape velocity is greater than the speed of light.
Would be possible for the light however to reach O1 ?

Probably most of the assumption above are wrong, otherwise it would be possible for a light ray from the probe below the event horizon to hit O1 and by scattering or reflection, "escape" from the black hole.
There should be some error on all the description, because also without a spaceship at O1, a photon could exit the event horizon, hit an incoming particle outside the event horizon and "flee" away by scattering or reflection.

 

[Ibix]

Summary:: It would be possible to have the gravity acceleration experienced by an observer close to an event horizon so "little" to let a spaceship to stay above the event horizon?

From "standard" formula we have that the gravity acceleration a = GM/r^2 and that the Schwarzschild radius rs = 2 GM / c^2

Is it possible to compute the gravity acceleration at Schwarzschild radius putting r = rs?

Your formula is the Newtonian approximation. The nearest equivalent relativistic formula is the proper acceleration of a hovering observer, which is $$\frac{1}{\sqrt{1-r_S/r}}\frac{GM}{r^2}$$which diverges as $r\rightarrow r_S$. Note that this is not strictly the acceleration due to gravity since there is no such thing in relativity, but it is the best equivalent.

Strictly speaking the answer to your question is no, we cannot calculate the acceleration at the event horizon. This is because gravity doesn't cause acceleration in relativity - you only feel acceleration when you are being pushed out of your free fall path. Also, note that the formula I gave above is not valid at or below the horizon, since one cannot hover there by definition.

Would be possible for the light however to reach O1 ?

Light from outside can reach a probe inside the horizon, although there is very little time before the probe is destroyed (15$\mu$s is the maximum survival time inside a solar mass black hole). No reply is possible.

 

[pervect]

One can approximate the proper acceleration required to be hold station at a certain distance d from the event horizon of a black hole by a = c^2 / d, d being the distance from the event horizon as measured by a stationary observer. This works only for d << r_s, where r_s is the Schwarzschild radius of the black hole. The formula itself is independent of the mass or size of the black hole, but is only valid for a black hole that's large enough so that the schwarzschild radius >> the distance d.

See for instance https://www.physicsforums.com/threads/jetpacking-above-a-black-hole.914198/#post-5761517 for more details.

To put some numbers into the formula, to hover 1km away from a black hole would require a proper acceleration of 9*10^13 m/s^2, about 9 trillion times the proper acceleration one needs to "hold station" on the surface of the Earth.

 

[Roberto Pavani]

Thanks all for the answers.
However, looking around, I've found on Wikipedia a "solution" to my first answer that seems the same as my solution. Probably, because of your answer, the Wikipedia page is wrong: https://en.wikipedia.org/wiki/Surface_gravity

From that equation it seems that the acceleration needed for O1to stand still "just above" the event horizon, measured by O1 (and because of this generated by O1 engines) would be $\frac {c^4}{4GM}$
But probably from the point of view on an Observer O3 very far from O1, the O1 acceleration will be different (because of different time dilatation).

Is this correct?

 

[Will Learn]

Hi.

However, looking around, I've found on Wikipedia a "solution" to my first answer that seems the same as my solution.

That Wikipedia article states the following:
...the Newtonian concept of acceleration turns out not to be clear cut...
...the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity. Because of this, a renormalized value is used...

To say this another way: $\frac {c^4} {4GM}$ is NOT the Newtonian acceleration of the object O1 at the event horizon. It is a strange quantity that I've not seen before and I'm not convinced that it is all that useful (but my experience is limited).

 

[Roberto Pavani]

Would it be possible to consider the object O1 at a very small distance ε and to compute the limit with ε -> 0 ?
Could be possible that the equation $\frac {c^4} {4GM}$ is coming from that limit?

 

[jbriggs444]

Would it be possible to consider the object O1 at a very small distance ε and to compute the limit with ε -> 0 ?
Could be possible that the equation $\frac {c^4} {4GM}$ is coming from that limit?

No.

That limit is infinite. As one approaches the horizon from the outside, the proper acceleration that would be required to hover at the current radius increases without bound.

 

[jartsa]

Thanks all for the answers.
However, looking around, I've found on Wikipedia a "solution" to my first answer that seems the same as my solution. Probably, because of your answer, the Wikipedia page is wrong: https://en.wikipedia.org/wiki/Surface_gravity

From that equation it seems that the acceleration needed for O1to stand still "just above" the event horizon, measured by O1 (and because of this generated by O1 engines) would be $\frac {c^4}{4GM}$
But probably from the point of view on an Observer O3 very far from O1, the O1 acceleration will be different (because of different time dilatation).

Is this correct?

If O3 sees dust particles land on O1 at rate one per second, then O1's experience is to be crushed by zillion dust particles landing on him every second.

So the time dilation factor seems to be 1/zillion in this case, in which O1 is very close to the surface.

Proper gravity at surface = surface gravity / time dilation factor = infinity

Proper gravity close to the surface ≈ surface gravity / time dilation factor

(The horizon is the 'surface')
('Proper' means according to someone that is personally right there at the location )

 

[jbriggs444]

But probably from the point of view on an Observer O3 very far from O1, the O1 acceleration will be different (because of different time dilatation).

The coordinate acceleration of a particular object will depend on the coordinates you choose. Obviously.

In General Relativity, choosing an observer is not equivalent to choosing a set of coordinates. The "point of view of an observer O3" is not defined well enough to decide what is meant by the acceleration of O1 from the point of view of O3.

However, the Wiki article does appear to be taking this sort of a point of view:

"The surface gravity

of a static Killing horizon is the acceleration, as exerted at infinity"

So they are not considering proper acceleration but are somehow ascribing an acceleration figure at infinity corresponding to the acceleration figure near the event horizon. When you take the limit as the test object O1 approaches the horizon, the local proper acceleration increases without bound. But the transformation to the corresponding figure at infinity skews the result more and more so that the result remains finite.

 

[Will Learn]

Sorry, @Roberto Pavani I have to agree with jbriggs (reply #7) earlier. The limit is undefined.
The "surface gravity" that your wikipedia article is describing does seem to be a product of the conventional Newtonian acceleration and the time dilation. This is something like being the distant observer (O3 I think you called it) and asking "what sort of acceleration would be there (at the event horizon) IF the rate of flow of time there was identical to my rate of flow time here (far away from the event horizon)". You seem to be able to get an answer but the question just isn't all that sensible to start with. The local geometry is very different near the event horizon compared to far away from it and it seems quite strange to try and find accelerations with a time co-ordinate that is not the local time co-ordinate at that place. I have to ask why would you continue to use the local space co-ordinates but NOT the local time co-ordinates?
If I've understood the Wikipedia correctly then the quantity they describe is something like the rate of change of my local space with respect to someone else's non-local time. It's a quantity and it seems to be well defined, I'm just not sure it's useful.

 

[Ibix]

Probably, because of your answer, the Wikipedia page is wrong:

I think it is misapplying something valid.

The formula I gave is actually the modulus of $k^a\nabla_ak^b$, the left hand side of their expression, and is the acceleration you would feel if you tried to stay at a constant altitude. It diverges at the event horizon, which is the whole point of an event horizon. It's too late there, you are going into the hole and there is no acceleration that can stop you even in principle.

The formula they are quoting is, I think, the force per unit of your mass that I would need to provide if you were hanging from a rope and I were at infinity holding the other end. That apparently doesn't diverge as you approach the event horizon - but the time taken for you to reach the horizon from my perspective does (edit: their Killing horizon based construction won't let them calculate a value as you approach the event horizon because it's only definable on the Killing horizon, but proper acceleration divided by time dilation factor is certainly definable elsewhere). So there's no way to set up the situation where you are at the horizon and hanging by a rope and I feel that force. So it's a meaningless number in this case.

It would make sense for black hole types where the Killing horizon is not at the event horizon, but then it would be one measure of "gravitational force" at the Killing horizon, not the event horizon.

 

[Roberto Pavani]

Inverting the formula we have: $M = \frac {c^4}{4aG}$ that as an example to have a=g (gravity on Earth surface) the mass would be very very huge (something around 1558 billion solar mass) with a radius close to 1LY.
Now if we suppose that the Rs is just one meter below the surface, we should be able to walk to this surface "feeling just" 1g as gravity acceleration.

Is this "teorically" possible?
Do I have missed some calculation or did some wrong calculation?

If the calculation are correct, the surprising effect is that we have an object very massive with the event horizon just 1m below the surface. However, the gravitational acceleration (perceived from an astronaut walking on the surface) is "small" (1g)

Now if everything above is correct what happens if someone on the surface, starts "digging" ?
Will never be able to reach the event horizon because approaching it the time is slowing down approaching the event horizon?

 

[Vanadium 50]

You can't stand on the surface of a black hole. There's no matter there to hold you up.

 

[Ibix]

Inverting the formula we have: $M = \frac {c^4}{4aG}$ that as an example to have a=g (gravity on Earth surface) the mass would be very very huge (something around 1558 billion solar mass) with a radius close to 1LY.
Now if we suppose that the Rs is just one meter below the surface, we should be able to walk to this surface "feeling just" 1g as gravity acceleration.

No. You need my formula for the acceleration the astronaut feels. Wikipedia's formula is not appropriate for this.

 

[DrGreg]

Inverting the formula we have: $M = \frac {c^4}{4aG}$ that as an example to have a=g (gravity on Earth surface) the mass would be very very huge (something around 1558 billion solar mass) with a radius close to 1LY.
Now if we suppose that the Rs is just one meter below the surface, we should be able to walk to this surface "feeling just" 1g as gravity acceleration.

No. You need my formula for the acceleration the astronaut feels. Wikipedia's formula is not appropriate for this.

Or, easier, use the approximate formula given in post #3. With $d$ = 1 metre, that gives approximately 10¹⁷ m/s², or 10 quadrillion $g$.

 

[pervect]

Would it be possible to consider the object O1 at a very small distance ε and to compute the limit with ε -> 0 ?
Could be possible that the equation $\frac {c^4} {4GM}$ is coming from that limit?

Yes, it's possible to compute the acceleration of an object at a very small distance $\epsilon$ from the event horiozon. That's what I did in the earlier post #3, for instance. However, when you do it that way, you do not get the wiki formula you quote, which appears to be for what is called "surface gravity".

The key to understanding the difference is to understand the specifics of the language being used. A careful reading of wiki will demonstrate that while it's telling you that the "surface gravity" of a black hole is finite, what they call "surface gravity" is not the sort of gravity one might measure in a hovering spaceship with a test mass and a calibrated scale.

Wiki also says something equivalent to this, as was quoted in post #5. "...the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity. "

The sort of gravity one would measure with a test mass and a scale (or more generally, with a gravimeter) is called "proper accleration", and, for a stationary observer increases without limit as one approaches the event horizon.

I'm a bit too lazy to dig it out, but Kip Thorne had a good popularized discussion about this early on in

 

[pervect]

I thought I'd say a few words about one way of understanding physical significance of what wiki terms the "surface gravity" of a black hole. Imagine that you have a very strong string that also had a negligible mass (as measured by a static observer), and you used this string to lowered a test mass of some known value towards the surface of the black hole.

If you measure the tension in the string via an in-line scale at or near the point where the string attaches to the object , i.e. at a point near the black hole, the scale reading approaches infinity as the object approaches the black hole. And as a consequence, the string (and/or scale) must eventually break.

However, if you measure the tension in the string "at infinity" in a similar manner, you find the scale reading is different (lower) at infinity than it is near the black hole. It turns out that the force exerted by the string, approaches a finite value when measured at infinity as one lowers the object towards the black hole. The ratio of force at infinity to mass of the lowered object is one way of understanding the "surface gravity" of the black hole.

This is related to the phenomenon called "gravitational time dilation", and the conservation of energy in GR for static space-times. There is a rather technical discussion in Wald's textbook, "General Relativity", about the "force at infinity" in the section where Wald talks about energy conservation. I have not even nearly done this topic justice, but the intent is to write a B-level description of what is going on.

 

[Roberto Pavani]

Thanks you all for the answers.