Undergrad Supermassive black hole, surface gravity and tidal forces

[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.

 

[Orodruin]

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.

It should not be surprising that such coordinates do exist. However, it should be noted that these time slices are not orthogonal to the time coordinate lines, which is a bit dissatisfying in itself.

 

[PAllen]

It should not be surprising that such coordinates do exist. However, it should be noted that these time slices are not orthogonal to the time coordinate lines, which is a bit dissatisfying in itself.

But these slices are orthogonal to the congruence of inertial free fallers from infinity (which is different from the Gullestrand-Painleve t coordinate lines; to see this you write out the rain drop world line tangents in GP coordinates). Who says the congruence of stationary observers is fundamentally preferred over the ‘rain drop’ congruence?

 

[Orodruin]

Who says the congruence of stationary observers is fundamentally preferred over the ‘rain drop’ congruence?

That depends to some extent on your preference. In a general spacetime there would be no preference for one set of observers over the other. However, in the case of a stationary spacetime, such as the exterior Schwarzschild solution, the existence of a timelike Killing field is related to a time translation symmetry of the spacetime itself, independent of any actual observers. That stationary observers are then defined based on this time translation symmetry is a different issue.

So I would say that if you are going to call any convention ”preferred”, the it better be based on the symmetries of the spacetime than an arbitrarily chosen set of observers. Thereby not saying that you definitely should do so.

 

[PAllen]

What the discussion of Lemaitre and Gullestrand-Painleve coordinates leads to is the following conclusion:

From any event outside an non-rotating uncharged BH, if you define distance to the horizon by the proper distance along the radial spacetime geodesic [edit: path of Gullestrand-Painleve simultaneity] through that event 4-orthogonal to a free faller with escape velocity toward the BH, then that distance is just r - R, where r is the areal radius, and R is the SC radius.

In contrast, the radial lines in SC coordinates are spacelike spacetime geodesics orthogonal to a hovering observer at a given event. Unfortunately, for a BH from collapse, these don’t intersect the horizon at all, as discussed in recent posts. Thus, IMO, it is useless to use them to discuss possible definitions of distance to the BH.

Of course, among all the radial spacelike geodesics from an event that do reach the horizon, there is no reason to favor one orthogonal to an escape velocity free faller. Thus you can justify a wide range of values greater than zero using proper distance along geodesics.

 

[Orodruin]

The question is how to relate all of this to the question of surface gravity as posed in the OP.

In respect to this I’d argue that sufficiently locally, so locally that tidal effects may be neglected, there is nothing separating the two black holes of different masses at the horizon. For the stationary observer hovering just above the horizon with constant proper acceleration, the event horizon is mapped to its Rindler horizon in the local Minkowski frame. Regardless of the size of the black hole.

 

[PeterDonis]

the radial spacetime geodesic through that event

I don't think those radial lines are spacetime geodesics. They are geodesics of the 3-surface of constant Painleve coordinate time that passes through the chosen event, but I believe those surfaces are not generated by radial spacetime geodesics. The radial spacetime geodesics lie in the surfaces of constant Schwarzschild coordinate time, which are different from the surfaces of constant Painleve coordinate time.

 

[PeterDonis]

In respect to this I’d argue that sufficiently locally, so locally that tidal effects may be neglected, there is nothing separating the two black holes of different masses at the horizon.

But if we adopt this approximation then there is no difference in their surface gravities either. The "surface gravity" of a Rindler horizon is zero.

In order to differentiate between surface gravities, we have to include the black hole masses in our model somehow.

 

[PAllen]

I don't think those radial lines are spacetime geodesics. They are geodesics of the 3-surface of constant Painleve coordinate time that passes through the chosen event, but I believe those surfaces are not generated by radial spacetime geodesics. The radial spacetime geodesics lie in the surfaces of constant Schwarzschild coordinate time, which are different from the surfaces of constant Painleve coordinate time.

They are geodesics. This is most easily seen by looking at the variation in Lemaitre coordinates, where the result is almost obvious. Note that through a point there are infinite radially directed geodesics, each with a different tangent. Similar to in x t Minkowsky plane, at any point there are infinite space like geodesics with -x component. Thus, the SC radial geodesic from an event is just one of infinitely many radial geodesics from that event.
[see later discussion].

 

[Orodruin]

But if we adopt this approximation then there is no difference in their surface gravities either. The "surface gravity" of a Rindler horizon is zero.

No, it is infinite as the proper acceleration of a hovering observer tends to infinity as you get closer to the horizon. My point is that locally there is no difference until you start feeling the tidal effects.

In order to differentiate between surface gravities, we have to include the black hole masses in our model somehow.

That’s assuming that whatever we want to call ”surface gravity” is different in the cases of different mass black holes.

 

[PeterDonis]

That’s assuming that whatever we want to call ”surface gravity” is different in the cases of different mass black holes.

The standard definition of surface gravity for a black hole (and indeed for a Killing horizon in general) is, as I said earlier in this thread, the redshifted proper acceleration at the horizon (or more precisely the appropriate limit as the horizon is approached). For a good (although A-level) treatment, see Wald's monograph Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. This surface gravity is different for different black holes.