Black hole horizon for different observers

[shinobi20]

TL;DR Summary

I'm reading a paper about the volume inside a black hole. Reading the abstract and the early part of the introduction, I already have questions regarding the event horizon as viewed by different observers and the uniqueness of the black hole area.

The paper is The Volume Inside a Black Hole (0801.1734)

Looking at the abstract, I have a question already.

It is stated: Because the light rays are orthogonal to the spatial 2-dimensional surface at one instant of time, the surface of the black hole is the same for all observers (i.e. the same for all coordinate definitions of "instant of time")

I don't understand this statement, as I know the horizon $r_H$ is found by calculating the norm of the null normal vector to the event horizon (a null hypersurface) which then leads to the radial component of the metric to be $g^{rr} = 0 \rightarrow r_H = 2GM$.

• How can light rays (null) be orthogonal to a spatial surface (spacelike tangent)?
• What is this spatial 2-dimensional surface being talked about?
• What does a light ray being orthogonal to the spatial surface have to do with the black hole surface being the same for all observers?

Looking at the first paragraph of the Introduction,

It is stated: Because the black hole is spherical, we simply need to measure the area in a transverse direction. This produces the unique result (Area $= 16 \pi m^2$). The uniqueness follows because if we consider a different definition of the 3-space in which we measure the area, we just shift our points in null directions along the (null) horizon. Null directions have zero length and cannot contribute to (or change) the area.

• What does "the black hole is spherical, we simply need to measure the area in a transverse direction" mean?
• Can anyone explain more about the statement on the uniqueness of the derived black hole area? If there's a different 3-space definition, why should we just shift in the null direction?

 

[martinbn]

Think about it this way. Consider Minkowski space-time and start with a space-like vector $s$ and a null vector $n$ perpendicular to it with respect to the metric, which I will call $g(\cdot,\cdot)$. The the length of $s$ is the same as the length of $s+n$ because $g(s+n,s+n)=g(s,s)+2g(s,n)+g(n,n)=g(s,s)$, because the last terms are zero $n$ being null and orthogonal to $s$.

This can be rephrased as follows. Imagine a strip of parallel null rays. Cut it orthogonaly by a space-like line. Then the length of this cross section is independent of the choice of a space-like line by the argument above.

Go one more demension and take three parallel light rays and the triangular prism they bound. Cut it orthogonaly by a spacelike surface. The areas of the triangles, the cross sections, si independent of the choice of a space-like surface. Same argument as above.

If you are in a general space-time it is in a way the same.

 

[shinobi20]

Think about it this way. Consider Minkowski space-time and start with a space-like vector $s$ and a null vector $n$ perpendicular to it with respect to the metric, which I will call $g(\cdot,\cdot)$. The the length of $s$ is the same as the length of $s+n$ because $g(s+n,s+n)=g(s,s)+2g(s,n)+g(n,n)=g(s,s)$, because the last terms are zero $n$ being null and orthogonal to $s$.

This can be rephrased as follows. Imagine a strip of parallel null rays. Cut it orthogonaly by a space-like line. Then the length of this cross section is independent of the choice of a space-like line by the argument above.

Go one more demension and take three parallel light rays and the triangular prism they bound. Cut it orthogonaly by a spacelike surface. The areas of the triangles, the cross sections, si independent of the choice of a space-like surface. Same argument as above.

If you are in a general space-time it is in a way the same.

But your statement is exactly why I'm confused in the first place, an orthogonal (Lorentzian sense) vector to a null vector is null itself, so there can be no spacelike vector orthogonal to a null vector, at least in the Lorentzian sense.

 

[vanhees71]

Of course, take $(1,0,0,1)$ (light-like) and $(0,0,1,0)$ (space-like).

 

[martinbn]

But your statement is exactly why I'm confused in the first place, an orthogonal (Lorentzian sense) vector to a null vector is null itself, so there can be no spacelike vector orthogonal to a null vector, at least in the Lorentzian sense.

Why not? Take Minkowski space with the usual coordinates $(t,x,y,z)$, the the vector $n=(1,1,0,0)$ is null, the vector $(0,0,1,0)$ is space-like, and they are orthogonal to each other.

 

[shinobi20]

Why not? Take Minkowski space with the usual coordinates $(t,x,y,z)$, the the vector $n=(1,1,0,0)$ is null, the vector $(0,0,1,0)$ is space-like, and they are orthogonal to each other.

Ah, I'm sorry, for the null normal vector of the event horizon, indeed one of the orthogonal vectors to it is the null tangent vector but there are two additional orthogonal vectors for which one of them is spacelike.

So what does the orthogonality of the light ray to the spatial surface have to do with the black hole surface being the same for all observers?

 

[martinbn]

So what does the orthogonality of the light ray to the spatial surface have to do with the black hole surface being the same for all observers?

The black hole area is the same for every one. That was in my first post.

 

[shinobi20]

The black hole area is the same for every one. That was in my first post.

I think I might be confusing the 2-dimensional surface which should be the black hole area that the paper is talking about (?) and the event horizon represented by $r_H$ for which the area is proportional to $r_H^2$.

 

[PeterDonis]

an orthogonal (Lorentzian sense) vector to a null vector is null itself

No. A null vector is orthogonal to itself, but that does not mean that all vectors orthogonal to a given null vector are null. Other than itself (or multiples of itself), the vectors orthogonal to a given null vector are spacelike.

 

[PeterDonis]

for the null normal vector of the event horizon, indeed one of the orthogonal vectors to it is the null tangent vector

Which are both the same vector.

there are two additional orthogonal vectors for which one of them is spacelike.

No, both of the other orthogonal vectors are spacelike.

 

[Ibix]

I think the language used isn't particularly clear, if I'm following correctly. The intersection between my choice of spacelike surface "now" and the event horizon is a set of events. The intersection between your choice of "now" and the event horizon is (in general) a different set of events. In that sense, then, the surfaces are not the same surface. However, martinbn's argument shows that the areas are the same.

 

[shinobi20]

I think the language used isn't particularly clear, if I'm following correctly. The intersection between my choice of spacelike surface "now" and the event horizon is a set of events. The intersection between your choice of "now" and the event horizon is (in general) a different set of events. In that sense, then, the surfaces are not the same surface. However, martinbn's argument shows that the areas are the same.

Exactly where part of my confusion comes in. From the statement in the abstract, it means that the 2-dimensional surface orthogonal to the light ray IS the black hole surface. But shouldn't the black hole surface be the same as the event horizon surface (if you could call it that). I'm confused since it is as if we're talking about two different surfaces pertaining to the same black hole.

 

[PeterDonis]

the 2-dimensional surface orthogonal to the light ray IS the black hole surface.

No. The event horizon (which I assume is what you mean by "the black hole surface") is not a single 2-sphere. It is an infinite, continuous series of 2-spheres parameterized by a null generator. In the case of a stationary black hole (i.e., one that never absorbs any mass), the areas of all the 2-spheres are the same.

If you pick some spacelike surface, it will intersect the horizon in just one of the 2-spheres. Different spacelike surfaces can intersect the horizon in different 2-spheres.

 

[shinobi20]

No. The event horizon (which I assume is what you mean by "the black hole surface") is not a single 2-sphere. It is an infinite, continuous series of 2-spheres parameterized by a null generator. In the case of a stationary black hole (i.e., one that never absorbs any mass), the areas of all the 2-spheres are the same.

If you pick some spacelike surface, it will intersect the horizon in just one of the 2-spheres. Different spacelike surfaces can intersect the horizon in different 2-spheres.

I understand, however, that is why I wanted to clarify the 2-dimensional surface that the paper is pertaining to which states that the surface of the black hole is the same for all observers. What is this surface that is being talked about?

 

[martinbn]

I understand, however, that is why I wanted to clarify the 2-dimensional surface that the paper is pertaining to which states that the surface of the black hole is the same for all observers. What is this surface that is being talked about?

I think this was already said, and is usually said in the texts. The event horizon is a three dimensional null surface, which is the boundary of the events from which signals can go to infinity and those from which signals cannot go to infinity.. The black hole region is the bounded component the horizon divides the spacetime in. By black hole often (if not always) is meant the intersection of the horizon with a space-like surface. The surface area is always the area of such a section.

By the way did you look into any of the standard references on black holes, say those that are cited in the paper?

 

[PeterDonis]

the 2-dimensional surface that the paper is pertaining to which states that the surface of the black hole is the same for all observers. What is this surface that is being talked about?

The surface that is the same for all observers is not 2-dimensional, it's 3-dimensional; it's the event horizon as I described it in post #13.

The language you quoted from the abstract of the paper is saying the same thing I said in post #13, though in language that is meant for a technical audience, not for a lay person.

 

[shinobi20]

The surface that is the same for all observers is not 2-dimensional, it's 3-dimensional; it's the event horizon as I described it in post #13.

So you are saying that the paper is wrong for saying that it is 2-dimensional, right?

No. The event horizon (which I assume is what you mean by "the black hole surface") is not a single 2-sphere. It is an infinite, continuous series of 2-spheres parameterized by a null generator. In the case of a stationary black hole (i.e., one that never absorbs any mass), the areas of all the 2-spheres are the same.

If you pick some spacelike surface, it will intersect the horizon in just one of the 2-spheres. Different spacelike surfaces can intersect the horizon in different 2-spheres.

What you are saying can be seen in the Kruskal diagram (courtesy of Wikipedia: Kruskal-Szekeres coordinates), right?

The dashed line represents the horizon $r=r_H$, and that is a null line which of course is generated by a null generator. In the Kruskal diagram, the angular coordinates are suppressed so that each point/dot on the diagram represents a 2-sphere. Therefore, each yellow dot represents a 2-sphere of which the event horizon is composed of. This is what you're saying that the event horizon is an infinite, continuous series of 2-spheres parameterized by a null generator.

Knowing this, let's suppose I can draw a yellow line across the diagram to represent a spacelike surface (pardon the inaccuracy of the yellow line but I know you know what I'm saying, just wanted to point out something on the horizon), it will intersect the horizon at the yellow point. This is what you're saying that different spacelike surfaces can intersect the horizon in different 2-spheres (different yellow dots).

Correct me if I'm wrong here.

 

[PeterDonis]

So you are saying that the paper is wrong for saying that it is 2-dimensional, right?

No, I'm saying you are misunderstanding the paper. The paper, or at least what you quoted from it, does not say the event horizon as a whole is 2-dimensional. It only says that the intersection of the event horizon with "space at an instant of time" is 2-dimensional.

What you are saying can be seen in the Kruskal diagram (courtesy of Wikipedia: Kruskal-Szekeres coordinates), right?

Yes. Your description of this is correct. Note, though, that the Kruskal spacelike surfaces are not the only possible set of spacelike surfaces you could use.

 

[PAllen]

At least part of the paper linked in the OP is total BS. They consider a surface in the interior of t=constant. But this is not a spacelike surface. It has only two spatial dimensions, considered as a submanifold (the other dimension is timelike). Unsurprisingly, they find a ‘volume’ for such a slice to be zero. But this is utter nonsense. To discuss interior volume in SC coordinates you want an r=constant slice (since r is the timelike coordinate in the interior), and this does not have zero volume. Given this totally egregious error, I see little point in reading the rest of the paper.

 

[PeterDonis]

To discuss interior volume in SC coordinates you want an r=0 slice (since r is the timelike coordinate in the interior), and this does not have zero volume.

I think you mean an $r =$ constant slice, where the constant is greater than zero. The $r = 0$ "slice" would be the singularity. (There are also other ways to specify a spacelike slice in the interior--but of course as you note specifying $t = 0$, or indeed $t =$ any constant whatever, is not one of them.)

 

[PAllen]

I think you mean an $r =$ constant slice, where the constant is greater than zero. The $r = 0$ "slice" would be the singularity. (There are also other ways to specify a spacelike slice in the interior--but of course as you note specifying $t = 0$, or indeed $t =$ any constant whatever, is not one of them.)

Yes, typo corrected. Also, corrected is that they assume t = constant, i.e. dt=0

 

[PeterDonis]

they assume t = constant, i.e. dt=0

Looking at section 3, they also aren't even using the t = constant slices inside the horizon; they are using the ones outside the horizon. Then they claim that their volume integral in Schwarzschild coordinates must vanish because the "inner" and "outer" radial coordinates, the limits of the integral, must both be the same--both must be $r = 2m$. They even mention explicitly that slices of constant $r$ are spacelike inside the horizon, but fail to consider the implications (which are that the volume of each $r =$ constant spacelike slice inside the horizon is infinite).

They also fail to realize that Schwarzschild spacetime at and inside the horizon is not static. Apparently they think that if a coordinate is called $t$ it must be a "time" coordinate.

 

[PeterDonis]

They also fail to realize that Schwarzschild spacetime at and inside the horizon is not static.

In fact, they appear to think that "static" is a coordinate-dependent property of a metric.

 

[PAllen]

What is unbelievable to me is that this paper was published in a respectable peer reviewed journal. How is this possible ?!

https://arxiv.org/abs/0801.1734

 

[vanhees71]

Well, some years ago there was even a paper published, which claimed that the Lienard-Wiechert potentials were no solutions of the Maxwell equations. Then Jackson wrote a comment showing that the authors simply forgot the derivatives wrt. to $\vec{x}$ due to the dependence of the retarded time on $\vec{x}$. It's unbelievable that any referee could let such a blunder through since every physicist should have done this calculation since more than 100 years which shows that the retarded potentials indeed are a solution of Maxwell's equations for a given charge-current density ;-).