I 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.

 

[haushofer]

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

Yes, this seems right (as a check for myself ;) ). So this is one example of a general relativistic problem in which a naive intuition gives the wrong answer (the most massive black hole pulls the least), while a Newtonian intuition gives the right answer (the Schwarzschild radius combined with the fact that gravity falls of as one over r squared gives a surface gravity inversely proportional to mass, so the lightest black hole will pull the most). This is a really nice scenario to show different aspects of general relativity (surface gravity as measured at "infinity", surface gravity as measured near the horizon, etc).

Thanks (and sorry for the late reply)!

 

[Orodruin]

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

”1 meter outside” does not sound to me like it is referring to the areal radius. The first thing to do should be to clarify what ”1 meter outside” means. If we are taking slices of constant t, then it would make sense to define “1 meter outside” as the distance along that spatial hypersurface. Since the metric has a singularity, this involves doing the integral
$$
\int_{r_*}^{r_* + \delta} \frac{\sqrt r \, dr}{\sqrt{r - r_*}}
\int_0^\delta \frac{\sqrt{x + r_*}) dx}{\sqrt x}
\simeq 2 \sqrt{r_* \delta} = 1\, \textrm m
$$
as long as $r_*$ >> 1 m. Consequently the difference in r coordinate would be much smaller for a bigger mass black hole.

 

[haushofer]

”1 meter outside” does not sound to me like it is referring to the areal radius. The first thing to do should be to clarify what ”1 meter outside” means. If we are taking slices of constant t, then it would make sense to define “1 meter outside” as the distance along that spatial hypersurface. Since the metric has a singularity, this involves doing the integral
$$
\int_{r_*}^{r_* + \delta} \frac{\sqrt r \, dr}{\sqrt{r - r_*}}
\int_0^\delta \frac{\sqrt{x + r_*}) dx}{\sqrt x}
\simeq 2 \sqrt{r_* \delta} = 1\, \textrm m
$$
as long as $r_*$ >> 1 m. Consequently the difference in r coordinate would be much smaller for a bigger mass black hole.

I don't get this. In my setup, I stated that the 1 m was in the Schwarzschild coordinates I used. So if the one black hole with mass M is situated at r = R, its horizon is situated at r = R - 2M, and the mass is hanging at r = R- 2M - 1.

 

[Ibix]

@Orodruin is arguing that a Schwarzschild $r$ coordinate of $R_S+1\mathrm{m}$ is not what you actually want for a fair comparison.

If you hover somewhere near a black hole and slowly wind out a string, it'll extend radially downward towards the hole. It'll snap under its own weight eventually, and even a string whose tensile strength tends towards infinity will break at the horizon. So one measure of "how far above the horizon am I" is "what is the longest length of string I can get back after lowering it out, assuming an arbitrarily strong string is available". The answer to that is Orodruin's integral, the integral of $\sqrt{|g_{rr}|}dr$.

 

[Orodruin]

I don't get this. In my setup, I stated that the 1 m was in the Schwarzschild coordinates I used. So if the one black hole with mass M is situated at r = R, its horizon is situated at r = R - 2M, and the mass is hanging at r = R- 2M - 1.

You can of course specify that, but it is not a physical distance - just a random coordinate difference. (Distances also are arbitrary to some extent, but slightly less so.)

 

[PeterDonis]

”1 meter outside” does not sound to me like it is referring to the areal radius.

I was interpreting it that way because that's the way I think @haushofer stated it. But I agree that is not the only possible or relevant way to do the comparison.

 

[PAllen]

I would like to add to the discussion of distance versus coordinate difference described recently in this thread, motivating just how ambiguous it is to talk about distance to a BH.

As with so many GR questions, it helps to look at SR first. There are, in fact, many ambiguities defining distance in SR.

Consider first, a definition of distance between two timelike world lines, W1 and W2. Consider the problem only in one plane, for simplicity. Consider using proper distance computed along a spacelike geodesic 4-orthogonal to one W1 at some point P1. This definition corresponds physically to a ruler distance measured in a global inertial frame in which W1 happens to be at rest at P1.

As long as W1 and W2 are both inertial, this definition seems fine - no unexpected anomalies whether you measure from W1 or W2 (though the results are typically different for each choice). However, simply adding the feature W1 is performing a small amplitude, slow zig zag, and that W2 is very far away (as defined by the definition), you can have the bizarre result that each event on W2 is used as the end point for multiple different points on the history of W1 (using W1 as the reference for the definition).

Note that this definition corresponds to the one @Orodruin proposed, and that in the case of Schwarzschild geometry, it has the anomaly that distances for all times along a stationary world line are computed to the same point on the horizon history, and further, that this horizon event does not even exist for a BH resulting from collapse (rather than an idealized eternal one).

Another issue with this definition (at least as used in a single plane) is that the reference world line cannot be light like. That is, distance from a lightlike world line cannot be defined at all, because there are no spacelike othogonals.

An alternative is to try to define distance between an event and a world line. One simple geometric idea has similar issues to the prior defintion: proper distance along a spacelike geodesic from an event to a world line 4-orthogonal to the world line at intersection. In particular, this cannot be defined at all for distance to a light like world line if the light is moving towards or away from the point in a plane defined by the light and the point. Also, a point can have multiple distances to a non-inertial world line, per this definition.

A third definition sometimes used for a point to a world line is the maximum proper distance over spacelike geodesics connecting the point and the world line. This avoids several anomalies of the prior definitions while agreeing with them for simple cases. It helps to work a simple example to see how maximum is what you really want. However, for a point to a light like world line (in x-t plane, for example) the set of spacelike geodesic distances span 0 to inifnity. Thus, the point is considered to be infinitely far away.

For these reasons, I consider the distance from a BH horizon to be fundamentally undefinable by simple geometric definitions.

@Ibix definition may coincide in result with one of the defintions above, but it doesn't use a simple geometric definition. Instead, it has the limitation that it only makes sense for stationary observers in a stationary geometry. It would be at least a little surprising to me if it coincides with @Orodruin's because it would imply that the strings lowered and returned from any event along a stantionary world line approach the same event on the horizon's history.

 

[Ibix]

@Ibix definition may coincide in result with one of the defintions above, but it doesn't use a simple geometric definition. Instead, it has the limitation that it only makes sense for stationary observers in a stationary geometry. It would be at least a little surprising to me if it coincides with @Orodruin's because it would imply that the strings lowered and returned from any event along a stantionary world line approach the same event on the horizon's history.

I admit I was assuming a Schwarzschild geometry. I think if I add a requirement of slow lowering that helps, doesn't it? The idea being that each point on the string is, in the limit of slowness, a Schwarzschild hovering observer, the last one of which can be arbitrarily close to the horizon.

I note that my proposed integral is the same as Orodruin's.

 

[PeterDonis]

each point on the string is, in the limit of slowness, a Schwarzschild hovering observer, the last one of which can be arbitrarily close to the horizon.

Yes, and if you take the common lines of simultaneity of all of these observers, all along their worldlines (these are the surfaces of constant $t$ in Schwarzschild coordinates), they all intersect at the horizon, or more precisely at a single point (event) on the horizon. This is the bifurcation point of the horizon (in a Kruskal diagram, it's the center point of the diagram--just as all of the common lines of simultaneity of Rindler observers in Minkowski spacetime intersect at the bifurcation point of the Rindler horizon, which is the center point of a Minkowski diagram).

But if you look at the worldline of the highest point on the string that falls through the horizon after it breaks (or for that matter any point on the string that falls through the horizon, but the highest point is the simplest one to consider because it marks the "break point" of the string), that worldline does not intersect the horizon at the bifurcation point. (It can't possibly, because the bifurcation point is on the white hole horizon as well as the black hole horizon, and is therefore unreachable by any timelike or null curve from the exterior region.) It intersects the horizon at some event to the future of the bifurcation point.

So to properly ground the physical interpretation you are proposing of the integral you and @Orodruin are using, you have to explain how the upper limit of the length of string you can get back after the break, which involves events to the future of the bifurcation point on the horizon, ends up being the same as the length measured along a spacelike line that passes through the bifurcation point.

 

[Orodruin]

I’d like to make it clear that I am not suggesting the proper distance the Schwarzschild radius along surfaces of constant t is unique or even physical (a physical black hole will not have that event). All I am saying is that there are more physical definitions of distance than staring oneself blind on the r coordinate. Typically we should do what we do in SR, define distance as distance along some spacelike hypersurface we define as a simultaneity. Just as in SR, this is not going to be unique, but that is simply something we’ll have to live with.

I will say this though: Outside of the horizon, the surfaces of constant t are obviously orthogonal to the time-like Killing field and therefore a natural candidate for the definition of simultaneity. Then it of course does not extend all the way to the horizon.

 

[PeterDonis]

a physical black hole will not have that event

Yes, in a physical black hole spacetime, the surfaces of Schwarzschild coordinate simultaneity will enter the region of spacetime occupied by the collapsing matter, and they will do so outside the horizon. At least some large fraction of them will then never intersect the horizon at all (because they will pass through the collapsing matter region to the past of the event at the center of that matter where the horizon first forms).

 

[PAllen]

Having discussed above some peculiarities of Schwarzschild coordinate simultaneity, one of many alternatives that actually include horizon events and cover the case of BH from collapse, is Lemaitre coordinates (https://en.wikipedia.org/wiki/Lemaître_coordinates). (Others are Kruskal, Eddington-Finkelstein, Gullestrand-Panlieve etc.). Note, each event on an external stationary world line measures to a different event on the horizon in these coordinates. For Lemaitre, a horizon distance takes a very simple form.

Given some chosen Schwarzschild r coordinate (better, areal radius) and a BH Schwarzschild radius R, define $\rho_0={\frac 2 3}R$, $\rho={\frac 2 3}r^{3/2}/\sqrt R$, then the distance between r and the horizon is given by:
$$\int_{\rho_0}^{\rho} \sqrt[3]{\rho_0/\rho}\,d\rho$$ which is easily integrated.

[added: Performing the integration and a bunch of algebra leads simply to r - R as the result, the same as Gullestrand-Painleve coordinates discussed below. This is because Lemaitre coordinates share the same foliation as GP; they just put a different embedded coordinate system on each spatial slice. This difference, of course, doesn’t change invariants like radial proper distance within a slice.]

 

[haushofer]

You can of course specify that, but it is not a physical distance - just a random coordinate difference. (Distances also are arbitrary to some extent, but slightly less so.)

I know. I just wanted a concrete example to calculate, as observed by someone far away from both black holes.

 

[Orodruin]

I know. I just wanted a concrete example to calculate, as observed by someone far away from both black holes.

Well, yes, but my point is to question whether that computation holds any actual physical relevance or not.

 

[PAllen]

Gullestrand-Painleve coordinates use a time slicing good from approaching the singularity to infinity. Each slice includes different horizon events as desired. In this time slicing, any difference in areal radius r, is, in fact, a proper distance per the metric along the time slice.