Supermassive black hole, surface gravity and tidal forces

[haushofer]

TL;DR Summary

A light black hole has stronger surface gravity and tidal forces just outside the horizon than a supermassive black hole. Why can't light nevertheless escape just inside the horizon of the supermassive black hole?

As the summary says: a light black hole has stronger surface gravity and tidal forces just outside the horizon than a supermassive black hole. So if you want to hoover just outside the horizon of a black hole and care about your well-being it better be supermassive. I understand this perfectly from a GR-point of view and a Newtonian analogy. But somehow my intuition clashes with the fact that also for this supermassive black hole light can't escape from inside the horizon to the outside. In the large M limit you can make the surface gravity arbitrarily weak just outside the horizon while the horizon still prevents light to escape. Is my confusion the fact that the surface gravity is defined for an observer at infinity, and the amount of required thrust for a stationary observer right outside the horizon becomes incredibly high as is mentioned e.g. here,

https://www.mathpages.com/rr/s7-03/7-03.htm

?

So, to give something concrete: is the following statement true? "A supermassive black hole has a smaller surface gravity right outside its horizon than a light black hole as measured by an observer far away, but the required thrust for an observer to stay stationary just outside the horizon is for a supermassive black hole much bigger than for a light black hole"

 

[Orodruin]

First of all, the concept of a “surface” of the black hole is rather ambiguous. There is no surface as such. The event horizon is locally no different from any other null surface.

That in mind, loosely the null geodesics (light paths in spacetime) are determined by the metric while what is closer to the gravitational acceleration of Newtonian gravity are the Christoffel symbols, which are derivatives of the metric. The tidal forces correspond to the curvature tensor, which is in essence consisting of derivatives of the Christoffel symbols.

(Note: I want to point out again that this is an extremely heuristic argument)

 

[Dale]

TL;DR Summary: A light black hole has stronger surface gravity and tidal forces just outside the horizon than a supermassive black hole. Why can't light nevertheless escape just inside the horizon of the supermassive black hole?

a light black hole has stronger surface gravity and tidal forces just outside the horizon than a supermassive black hole. So if you want to hoover just outside the horizon of a black hole and care about your well-being it better be supermassive.

I disagree with this premise. The acceleration required to hover outside of any size black hole goes to infinity. The tidal forces are finite and are smaller at the horizon of a supermassive black hole, but not the “surface gravity”.

 

[PeterDonis]

a light black hole has stronger surface gravity

But "surface gravity" in this context does not mean what you think it means. It does not mean the actual proper acceleration of an object hovering at epsilon above the horizon; as @Dale has pointed out, that goes to infinity for any black hole. The "surface gravity" is the redshifted proper acceleration; the physical interpretation of this, heuristically, is the force that an observer at rest at infinity would have to exert on a rope connected to an object at epsilon above the horizon to get it to hover. (The more technical statements of the above involve appropriate limits taken as the horizon is approached.)

 

[haushofer]

But "surface gravity" in this context does not mean what you think it means. It does not mean the actual proper acceleration of an object hovering at epsilon above the horizon; as @Dale has pointed out, that goes to infinity for any black hole. The "surface gravity" is the redshifted proper acceleration; the physical interpretation of this, heuristically, is the force that an observer at rest at infinity would have to exert on a rope connected to an object at epsilon above the horizon to get it to hover. (The more technical statements of the above involve appropriate limits taken as the horizon is approached.)

Yes, I do understand that. It's completely different from the local acceleration needed to hoover just outside the black hole, which becomes infinite at r=2M.

I guess I'm struggling to reconcile the fact that an outside far away observer measures this surface gravity to be weak for supermassive black holes, yet also will agree that no light can escape from that same surface.

I've found Wald's treatment on the surface gravity and exercise 4 in ch.6. Maybe that also helps.

 

[Ibix]

I guess I'm struggling to reconcile the fact that an outside far away observer measures this surface gravity to be weak for supermassive black holes, yet also will agree that no light can escape from that same surface.

Is the question here something like "if I were half way between a stellar mass BH and an SMBH, 1,000ly from each, and holding things above the event horizons on long ropes, which way do I get pulled"?

If so, is it worth thinking about what "close to" and "far away" mean? You're evaluating them in limits, but is that what you are intuitively thinking about? I mean, an object 1km above the event horizon is at about $3M$ for a small stellar mass BH, but is at $2M$ to 6+ significant figures for an SMBH. Similarly, "far away" means $2M\ll r$ in practice, but you can easily find an $r$ such that $2M_{Sun}\ll r\ll 2M_{SMBH}$. What does the ratio of surface gravities look like if you evaluate them between $r=2M+l$ and $r=R_0$, where $l\ll M_{Sun}$ and $R_0\gg M_{SMBH}$, but both quantities are finite and neither is expressed in terms of the mass of either hole?

 

[PeterDonis]

I guess I'm struggling to reconcile the fact that an outside far away observer measures this surface gravity to be weak for supermassive black holes, yet also will agree that no light can escape from that same surface.

When you say "measuring the surface gravity", you have to remember that the name "surface gravity" is misleading. The measurement you're making is not at the "surface" or just above it. It is at infinity. That's why the measurement can be finite, and gets smaller as the hole gets more massive--because it's not actually measuring anything at the "surface". It's measuring a "property at infinity"--in fact it's indirectly measuring the mass of the hole, which is a global geometric property of the spacetime.

 

[Dale]

I guess I'm struggling to reconcile the fact that an outside far away observer measures this surface gravity to be weak for supermassive black holes, yet also will agree that no light can escape from that same surface

Nobody measures this “surface gravity”. It is a very artificial concept. If I am standing on Earth and I want to measure a more natural concept of surface gravity I simply look at what the accelerometer app on my cellphone says. I don’t even think about correcting it for the gravitational time dilation that someone in intergalactic space would see from me. None of the physics depends on that corrected value.

 

[PeterDonis]

Nobody measures this “surface gravity”. It is a very artificial concept.

Actually, it does have a physical meaning in black hole thermodynamics, but that meaning has nothing to do with "gravity". It is actually the temperature of the black hole (in "natural" units).

 

[haushofer]

Is the question here something like "if I were half way between a stellar mass BH and an SMBH, 1,000ly from each, and holding things above the event horizons on long ropes, which way do I get pulled"?

If so, is it worth thinking about what "close to" and "far away" mean? You're evaluating them in limits, but is that what you are intuitively thinking about? I mean, an object 1km above the event horizon is at about $3M$ for a small stellar mass BH, but is at $2M$ to 6+ significant figures for an SMBH. Similarly, "far away" means $2M\ll r$ in practice, but you can easily find an $r$ such that $2M_{Sun}\ll r\ll 2M_{SMBH}$. What does the ratio of surface gravities look like if you evaluate them between $r=2M+l$ and $r=R_0$, where $l\ll M_{Sun}$ and $R_0\gg M_{SMBH}$, but both quantities are finite and neither is expressed in terms of the mass of either hole?

That's a beautiful scenario to think about, thanks! I'm not sure I'm getting the point you make after that, so let me ask this: if I'm at a coordinate distance of 500 ly from both black holes (a distance of 1000 ly apart; I'm right in the middle) and I lower a 1 kg mass at a rope just outside both horizons, which black hole will pull the most? According to the definition of surface gravity (which I can roughly apply here even though I'm not at 'infinity') I'd say the stellar BH will exert the greatest 'force'. Am I right?

 

[haushofer]

Nobody measures this “surface gravity”. It is a very artificial concept.

How about this observer lowering a mass attached to a rope from a great distance from the BH?

Or Earth, in your example.

 

[haushofer]

When you say "measuring the surface gravity", you have to remember that the name "surface gravity" is misleading. The measurement you're making is not at the "surface" or just above it. It is at infinity. That's why the measurement can be finite, and gets smaller as the hole gets more massive--because it's not actually measuring anything at the "surface". It's measuring a "property at infinity"--in fact it's indirectly measuring the mass of the hole, which is a global geometric property of the spacetime.

Yes, but in the thought experiment of post #10 the surface gravity determines the tension in the string, right?

 

[timmdeeg]

TL;DR Summary: A light black hole has stronger surface gravity and tidal forces just outside the horizon than a supermassive black hole. Why can't light nevertheless escape just inside the horizon of the supermassive black hole?

Because according to the metric the r-coordinate of anything (including light) inside is decreasing.

 

[PeterDonis]

in the thought experiment of post #10 the surface gravity determines the tension in the string, right?

Only at infinity. As you go down the string from infinity towards the object that is being held stationary just above the horizon, the tension in the string increases without bound.

 

[PeterDonis]

if I'm at a coordinate distance of 500 ly from both black holes (a distance of 1000 ly apart; I'm right in the middle) and I lower a 1 kg mass at a rope just outside both horizons, which black hole will pull the most?

It depends on how you define "just outside". The surface gravity is the limit of the pull on a rope at infinity as the object approaches the horizon; but you can't actually have an object hover at the horizon. It has to be some distance above. How is that distance determined, and does it depend on the mass of the hole?

Also, 500 ly is not infinity; for a SMBH it might not even be close enough to infinity to be treated as infinity, depending on how supermassive the SMBH is.

 

[jartsa]

That's a beautiful scenario to think about, thanks! I'm not sure I'm getting the point you make after that, so let me ask this: if I'm at a coordinate distance of 500 ly from both black holes (a distance of 1000 ly apart; I'm right in the middle) and I lower a 1 kg mass at a rope just outside both horizons, which black hole will pull the most? According to the definition of surface gravity (which I can roughly apply here even though I'm not at 'infinity') I'd say the stellar BH will exert the greatest 'force'. Am I right?

Force is arbitrarily close to mass multiplied by surface gravity, when a test mass is hanged arbitrarily close to the horizon using a rope, the other end of which is at infinity.

By force I mean the force felt at the upper end of the rope The rope is of course massless. I mean small rockets along the rope cancel out its weight.

 

[jartsa]

I disagree with this premise. The acceleration required to hover outside of any size black hole goes to infinity. The tidal forces are finite and are smaller at the horizon of a supermassive black hole, but not the “surface gravity”.

If the left side of your body is pulled with infinite force to one direction, and the right side of your body is pulled with infinite force to slightly different direction, then is there not an infinite tidal force?

 

[jartsa]

Yes, I do understand that. It's completely different from the local acceleration needed to hoover just outside the black hole, which becomes infinite at r=2M.

I guess I'm struggling to reconcile the fact that an outside far away observer measures this surface gravity to be weak for supermassive black holes, yet also will agree that no light can escape from that same surface.

I've found Wald's treatment on the surface gravity and exercise 4 in ch.6. Maybe that also helps.

If an observer lowers a box that contains light, using a rope, towards a super massive black hole, from infinity, we know that the trajectories of light beams inside the box become similar to trajectories of light beams in a box in a very strong gravity field. We know this, because we know that according to an observer inside the box the proper acceleration goes towards infinity, as the box is lowered.

But the observer only feels a force that is not larger than mass of box multiplied by the surface gravity of the black hole, which is kind of weird.

Well, this could be explained by the observer by light propagating slowly according to the observer. Trajectories of light beams can have tight curves because the beam propagates slowly. He could even see the beams propagating slowly with some suitable equipment.

 

[Dale]

If the left side of your body is pulled with infinite force to one direction, and the right side of your body is pulled with infinite force to slightly different direction, then is there not an infinite tidal force?

No. Tidal forces specifically refer to the gravitational interaction. If you are drawn and quartered by infinitely strong ropes attached to infinitely strong horses that is not an infinite tidal force. And being pulled up by an infinitely strong rope accelerating you at an infinite rate is also not a tidal force. It is non-gravitational.

 

[Dale]

How about this observer lowering a mass attached to a rope from a great distance from the BH?

Or Earth, in your example.

Any such rope will break:

https://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html

 

[jartsa]

No. Tidal forces specifically refer to the gravitational interaction. If you are drawn and quartered by infinitely strong ropes attached to infinitely strong horses that is not an infinite tidal force. And being pulled up by an infinitely strong rope accelerating you at an infinite rate is also not a tidal force. It is non-gravitational.

You said pull of gravity goes to infinity, but not tidal forces.

I asked how can pull of gravity go to infinity without tidal forces going to infinity. Seems impossible to me. I mean, I meant to ask that.

I used as an example a body experiencing an infinite pull of gravity, said pull not being to exactly to the same direction all over the body, but instead towards the center point of the gravitating body.

 

[Dale]

I asked how can pull of gravity go to infinity without tidal forces going to infinity. Seems impossible to me. I mean, I meant to ask that

Gravity doesn’t pull in GR. The force of gravity is zero. What goes to infinity at the horizon is the force that must be exerted by a rope or a rocket to make the object hover.

The hovering force is simply an artifact of the coordinates. You can get the same thing in flat spacetime using Rindler coordinates. A rocket must use an infinite thrust to hover at the Rindler horizon, but a free falling object feels nothing at all.

 

[timmdeeg]

I asked how can pull of gravity go to infinity without tidal forces going to infinity.

GR requires that the Newtonian gravitational acceleration has to be multiplied with 1/sqrt(1-r_S/r, whereby r_S is the Schwarzschild radius. So the force approaches infinity if r goes to r_S.
Whereas the tidal force at the event horizon is proportional to 1/M².

 

[jartsa]

GR requires that the Newtonian gravitational acceleration has to be multiplied with 1/sqrt(1-r_S/r, whereby r_S is the Schwarzschild radius. So the force approaches infinity if r goes to r_S.
Whereas the tidal force at the event horizon is proportional to 1/M².

Well, hmm, tidal force on rocket fleet placed so that the rockets are symmetrically at the opposite sides of a black hole, experiences a tidal force that compresses the fleet at force that goes to infinity, as the distance between rockets approaches 2r_s.

I used a very large object, the fleet, Is it not allowed?

 

[PeterDonis]

tidal force on rocket fleet placed so that the rockets are symmetrically at the opposite sides of a black hole, experiences a tidal force that compresses the fleet at force that goes to infinity, as the distance between rockets approaches 2r_s.

No. What you are calling the "tidal force" in this case is not even well defined, because the "distance between rockets" you describe is not well defined (because this "distance" has to go through the interior of the hole, and the spacetime geometry in the interior of the hole doesn't work that way).

I used a very large object, the fleet, Is it not allowed?

Not if the "object" has to span the interior of the hole. See above.

In addition, "tidal force" means the Riemann curvature tensor, which can vary over a large region, so using objects that span large regions is not a good way to assess "tidal force". You should use objects that are small enough that you are not seeing significant changes in the curvature tensor over the region spanned by the object.

 

[Dale]

I used a very large object, the fleet, Is it not allowed?

In GR what we are talking about when we discuss "tidal forces" is more technically called geodesic deviation. This is defined for "neighboring geodesics" or "a family of closely spaced geodesics". I don't know if there is a term for the summed or integrated geodesic deviation over some finite volume. Such things tend to be difficult to define in GR.

 

[Dale]

Well, hmm, tidal force on rocket fleet placed so that the rockets are symmetrically at the opposite sides of a black hole, experiences a tidal force that compresses the fleet at force that goes to infinity, as the distance between rockets approaches 2r_s.

I used a very large object, the fleet, Is it not allowed?

This is essentially the idea behind the Ricci curvature scalar. Of course, except for the fact that it is defined for a small fleet of very nearby ships. It measures the change in the volume of a small cluster of nearby geodesic test particles that start out at rest wrt each other.

 

[PeterDonis]

This is essentially the idea behind the Ricci curvature scalar. Of course, except for the fact that it is defined for a small fleet of very nearby ships. It measures the change in the volume of a small cluster of nearby geodesic test particles that start out at rest wrt each other.

More precisely, it measures that assuming that the small cluster of particles is in the interior of a massive object, i.e., a region with nonzero stress-energy.

However, in this case we are talking about a vacuum solution (the black hole). In case the Ricci scalar (and indeed the Ricci tensor) is zero. The "tidal force" in the vacuum region is due to the Weyl tensor, which unfortunately does not have a simple scalar that captures anything useful about it.

 

[Dale]

More precisely, it measures that assuming that the small cluster of particles is in the interior of a massive object, i.e., a region with nonzero stress-energy.

However, in this case we are talking about a vacuum solution (the black hole). In case the Ricci scalar (and indeed the Ricci tensor) is zero. The "tidal force" in the vacuum region is due to the Weyl tensor, which unfortunately does not have a simple scalar that captures anything useful about it.

I would say that it still has the same geometrical meaning whether it is in vacuum or in matter. It is just that the value is 0 in vacuum.

 

[PeterDonis]

I would say that it still has the same geometrical meaning whether it is in vacuum or in matter. It is just that the value is 0 in vacuum.

Yes, but the value being 0 in vacuum means that there is no geodesic deviation in vacuum due to Ricci curvature. So you can't explain things like what @jartsa was describing using Ricci curvature. You have to look at the Weyl curvature in the vacuum region, and how it changes as you go from one side of the black hole to the other.

 

[Dale]

So you can't explain things like what @jartsa was describing using Ricci curvature.

Right, which is why I put the “except for the fact …” part of my post to them. If what he were describing were an infinitesimal fleet then the change in the fleet’s volume would be the Ricci scalar

 

[PeterDonis]

Right, which is why I put the “except for the fact …” part of my post to them. If what he were describing were an infinitesimal fleet then the change in the fleet’s volume would be the Ricci scalar

Which would be zero in vacuum, yes.

 

[haushofer]

It depends on how you define "just outside". The surface gravity is the limit of the pull on a rope at infinity as the object approaches the horizon; but you can't actually have an object hover at the horizon. It has to be some distance above. How is that distance determined, and does it depend on the mass of the hole?

Also, 500 ly is not infinity; for a SMBH it might not even be close enough to infinity to be treated as infinity, depending on how supermassive the SMBH is.

Let's say for both black holes 1 meter outside the horizon in the Schwarzschild coordinates (of course, that means a different proper distance to the horizon in both cases). And if you're worried about whether we can treat these distances to be "infinite", let's take a black hole with 1 solar mass and 10 solar masses.

Then what?

 

[haushofer]

Only at infinity. As you go down the string from infinity towards the object that is being held stationary just above the horizon, the tension in the string increases without bound.

Yes, but in our thought experiment we measure the tension far away from both black holes (at "infinity" to a certain degree of approximation).

 

[PeterDonis]

Let's say for both black holes 1 meter outside the horizon in the Schwarzschild coordinates (of course, that means a different proper distance to the horizon in both cases).

In other words, the areal radius $r$ in both cases is $2M$ + 1 meter. Ok.

And if you're worried about whether we can treat these distances to be "infinite", let's take a black hole with 1 solar mass and 10 solar masses.

10 solar masses is not "supermassive", but basically you're asking how things vary with the mass of the hole, so ok.

I'll use SI units throughout.

Some useful constants:

One solar mass: $1.9885 \times 10^{30}$
Newton's gravitational constant $G$: $6.6732 \times 10^{-11}$
Speed of light $c$: $299792458$
1 Year: $31558118.4$

$r = 2GM/c^2$ plus 1 meter for

One solar mass hole: $2953.897$
Ten solar mass hole: $29529.97$

Redshift factor at $r$ for

One solar mass hole: $0.01839761$
Ten solar mass hole: $0.005813797$

Proper acceleration at 1 m above horizon for

One solar mass hole: $8.266231 \times 10^{15}$
Ten solar mass hole: $2.617423 \times 10^{15}$

$2GM / c^2 r$ at 500 ly for

One solar mass hole: $6.242327 \times 10^{-16}$
Ten solar mass hole: $6.242327 \times 10^{-15}$

So we can treat this as "infinity" for both holes. That means the redshift factor above is the factor by which we multiply the proper acceleration at 1 meter above the horizon, to get the force exerted at infinity. So...

Force exerted at infinity for

One solar mass hole: $1.520789 \times 10^{14}$
Ten solar mass hole: $1.521716 \times 10^{13}$

Both of these are quite close to the surface gravity, which is the limit of the redshifted proper acceleration at the horizon, and which in SI units is $c^4 / 4 G M$. So...

Surface gravity for

One solar mass hole: $1.521819 \times 10^{14}$
Ten solar mass hole: $1.521819 \times 10^{13}$

Notice how the ten solar mass value 1 meter above the horizon is closer to the limiting surface gravity; that is because 1 meter is a smaller increment in terms of the mass of the hole.