Black Hole: Knock on the horizon

[mef]

TL;DR Summary

What is the meaning of the coordinates of R with a spherical Schwarzschild metric?

First of all, I want to note that geometry is being discussed, which in fact is the General Theory of Relativity. And in any geometry, there are infinitely thin, weightless, etc. lines, rulers, and so on. In the future I will remind you about this.

The system of units is meters.
There is a Schwarzschild metric, with a horizon radius of r (meters).
The observer is at rest at the point R (meters). This is the coordinate.

Next, the observer wants to "knock on the horizon."

He takes a ruler (weightless, infinitely thin, etc., etc.) with a length of 1 meter and applies it from himself in the opposite direction from the origin (center).
Shifts it 1 meter towards the center.

At the nth step, the ruler N is applied and shifted together with the N-1 rulers placed earlier.

Mathematically, this corresponds to adding 2 vectors (-1, 0) and (0, N-1) and shifting the total vector (-1, N-1) by 1 meter to the center and obtaining the final vector (0, N), where N is at least the number of rulers.

The procedure is repeated only R-r times, and after R-r steps the total ruler (vector) should hit the firmament, the event horizon.
If it does not work out to "reach out", due to the fact that the far end of the vector moves in a "special" way during the shift (for example, at a shorter distance, when "looking" from point R), and moreover, it may never reach the horizon at all, then what is the meaning of the coordinate R (meters)?

I note that the described procedure has a deep physical meaning.

If you start "walking" with a step length of 1 meter according to such a constructed ruler, then no matter how the position of EACH gluing of meter rulers (we will call them serifs) is transformed, the "steps" will fall into these serifs, since the step length is transformed in the same way as the length of the rulers at each N step.

That is, it is a clear and BASIC concept of the distance between the beginning and the end of the vector

 

[Nugatory]

Summary:: What is the meaning of the coordinates of R with a spherical Schwarzschild metric?

That’s actually not such a hard question, although the implications of the answer for the discussion in the rest of your post are not so easy.

The set of all points in space with a given $R>R_S$ ($R_S$ is the Schwarzschild radius) form a sphere with surface area $4\pi R^2$; the analogy with Euclidean geometry is so strong that we cannot resist the temptation to call $R$ a “radius” although it is not in fact the distance from the center of anything.

Note that the distance between the sphere with “radius” $R$ and the sphere with “radius” $R+\Delta R$ is not $\Delta R$.

Note also that this area of the sphere interpretation works only in the vacuum region outside of a spherically symmetric mass distribution, and then only for $r\gt R_S$.

 

[mef]

You have strange temptations ... but let's still rewrite in coordinates when R is the usual radius vector. And then what will we get? That the horizon is infinitely distant from any observer?

 

[Ibix]

He takes a ruler (weightless, infinitely thin, etc., etc.) with a length of 1 meter and applies it from himself in the opposite direction from the origin (center).
Shifts it 1 meter towards the center.

Which direction is "towards the center"? This is 4d spacetime, so keeping your $\theta$ and $\phi$ coordinates fixed leaves you two directions to play with. The answer depends which one you pick.

The obvious choice is to pick the direction orthogonal to the timelike Killing vector field. In that case, the distance from the horizon at $r=R_S$ to $r=R$ is $$\int_{R_S}^R\sqrt{g_{rr}}dr=\sqrt{R(R-R_S)}$$where $g_{rr}$ is specified in Schwarzschild coordinates. Other values are possible with different choices.

 

[mef]

we take only the spatial part of the coordinates , the coordinate (0,0,0) is the center. The observer is located at the coordinate (R,0,0) When we switch to the reference frame associated with the observers (the coordinate of the center will change, but its name is not so clear?

 

[Ibix]

You are still thinking three dimensionally. How are you defining "space"? Note that $r=0$ is not a place in space in Schwarzschild spacetime. The changing $r$ direction (keeping everything else fixed) is timelike inside the horizon.

 

[mef]

You are still thinking three dimensionally. How are you defining "space"?

just like a plane in three-dimensional space - I fix one of the coordinates
If you like, this is a set of three-dimensional spaces for different t and they are the same since the metric does not depend on t
Maybe it will be easier for you and me to switch to the R-t plane altogether?

 

[PeroK]

And then what will we get? That the horizon is infinitely distant from any observer?

It's pointless to insist on applying classical (Newtonian) physics to GR. What you are really saying is: Newtonian physics cannot explain a Schwarzschild black hole. And that's true. You need the concepts and mathematical framework of GR.

It's difficult because Newtonian ideas are so well established that they can be seen as obviously correct - which is why they are called "laws" rather than "theory".

But, Newton's laws are not universal and you need to completely rethink your ideas about spacetime if you want to understand 20th century physics.

 

[mef]

Note that $r=0$ is not a place in space in Schwarzschild spacetime.

what prevents you from using not r=0 but r->0 as in a mathematical analyzer?)
so do you understand what a center is?))

 

[Ibix]

just like a plane in three-dimensional space - I fix one of the coordinates

Fix a coordinate in which coordinate system? Schwarzschild's? Eddington-Finkelstein? Gullstrand-Painleve? Kruskal-Szekeres? You'll get different answers in each case because they define different spatial planes.

 

[Ibix]

what prevents you from using not r=0 but r->0 as in a mathematical analyzer?)
so do you understand what a center is?))

Go look up a Kruskal diagram. $r=0$ is a line, not a point, and corresponds more closely to a time than a place.

 

[PeroK]

so do you understand what a center is?))

A spherical surface, for example, has no center. In this sense, $r = 0$ is not the center of a black hole. It's actually a singularity in the spacetime model. It's not a center.

 

[mef]

The changing direction (keeping everything else fixed) is timelike inside the horizon.

I don't understand what you want to say.
Let the bh in the center of the galaxy have a Schwarzschild metric.
being in her field (her curved space) on Earth, can we fix t, phi and theta?
But if we want to consider the movement in time to fix only phi and theta?

 

[Nugatory]

but let's still rewrite in coordinates when R is the usual radius vector.

The problem is there is no such coordinate system - the vectors you are looking for do not exist. It will be easier to see this if you look at the spacetime using Kruskal coordinates (Google will find some good references) which avoid the coordinate singularity at the Schwarzschild radius and cover the entire spacetime.

 

[Nugatory]

I don't understand what you want to say.

That is your cue to back up and read the Insights articles linked by @PeterDonis in one of your other threads - there’s a lot of background about the Schwarzschild spacetime that you need before the answers to your questions will make sense.

Also follow up the recommendation to look at the Kruskal coordinate description of the Schwarzschild spacetime (using Kruskal coordinates instead of Schwarzschild coordinates is like using a globe instead of a Mercator map to study the surface of the Earth - it’s the same physical thing presented in a different way).

Another reference you may find helpful is https://arxiv.org/abs/0804.3619 and (because you are trying to apply flat-space concepts like radius vectors to curved spacetimes) https://preposterousuniverse.com/wp-content/uploads/2015/08/grtinypdf.pdf will be essential background.

 

[Ibix]

I don't understand what you want to say.

There are four linearly independent directions in spacetime. One is timelike, three spacelike (usually - more complicated choices are possible). The direction of changing $r$ is one of the spacelike directions outside the horizon, but is the timelike direction inside it. That's why you can't avoid the singularity once inside the horizon: $r=0$ is literally in your future.

 

[Dale]

what is the meaning of the coordinate R (meters)?

The r coordinate is sometimes called the areal radius. The spacetime has spherical symmetry, so it is foliated into spheres. The r coordinate is is defined such that the area of the area of the foliated sphere is $4\pi r^2$. The purpose of doing that is so that the angular coordinates will have the standard familiar spherical metric $r^2(d\theta^2 + \sin^2(\theta)d\phi^2)$

He takes a ruler (weightless, infinitely thin, etc., etc.) with a length of 1 meter and applies it from himself in the opposite direction from the origin (center).
Shifts it 1 meter towards the center.

To measure r they should actually shift it horizontally and join them each at an angle such that eventually the first stick pokes them from the other side. Then they can just use the usual circumference formula to find r

 

[Dale]

A spherical surface, for example, has no center. In this sense, $r = 0$ is not the center of a black hole. It's actually a singularity in the spacetime model. It's not a center.

And, in fact, $r=0$ is not even part of the spacetime manifold.

 

[Dale]

what prevents you from using not r=0 but r->0 as in a mathematical analyzer?)
so do you understand what a center is?))

You can pick any spacelike line from any event at some $r$ to some event $\epsilon$ and take the limit as $\epsilon$ approaches the singularity.

What you appear to be missing is that such a spacelike line cannot be described simply as an integral along the Schwarzschild $r$ coordinate. This is not because of the geometry, but because the Schwarzschild $r$ coordinate is badly behaved at the horizon and is timelike inside the horizon.

 

[Ibix]

What you appear to be missing is that such a spacelike line cannot be described simply as an integral along the Schwarzschild $r$ coordinate. This is not because of the geometry, but because the Schwarzschild $r$ coordinate is badly behaved at the horizon and is timelike inside the horizon.

The $r$ coordinate is fine (it's $2GM/c^2$), it's the $t$ one that doesn't work, essentially because it's a radar coordinate which is ill-defined when the pulse never comes back. $r$ is defined geometrically and is fine except at the singularity.

 

[Dale]

The $r$ coordinate is fine (it's $2GM/c^2$), it's the $t$ one that doesn't work, essentially because it's a radar coordinate which is ill-defined when the pulse never comes back. $r$ is defined geometrically and is fine except at the singularity.

The $r$ coordinate itself is still problematic for defining a spacelike path because it is not spacelike at or below the horizon.

 

[Dale]

In that case, the distance from the horizon at $r=R_S$ to $r=R$ is $$\int_{R_S}^R\sqrt{g_{rr}}dr=\sqrt{R(R-R_S)}$$where $g_{rr}$ is specified in Schwarzschild coordinates. Other values are possible with different choices.

I just tried to work this out myself and got a slightly different answer. So I have $$g_{rr}=\frac{1}{1-\frac{R_S}{r}}$$ which integrates (according to Mathematica) to $$\int_{R_S}^R \sqrt{g_{rr}} \ dr = \sqrt{R \left(R-R_S\right)}+\frac{1}{2} R_S \log \left(\frac{-R_S+2 \sqrt{R \left(R-R_S\right)}+2
R}{R_S}\right) $$
Are you thinking that second term evaluates to 0 or otherwise goes away? Or did I or Mathematica make a mistake somewhere?

 

[PeterDonis]

Note also that this area of the sphere interpretation works only in the vacuum region outside of a spherically symmetric mass distribution

Yes. But see below.

and then only for $r\gt R_S$.

Actually, no. It still works at and inside the horizon, as long as we are still in a spherically symmetric vacuum region. The spherically symmetric mass distribution can be the one that originally collapsed to form the black hole; it doesn't have to be static.

 

[PeterDonis]

The $r$ coordinate itself is still problematic for defining a spacelike path because it is not spacelike at or below the horizon.

More than that: the line that the OP is implicitly imagining, which is a line in a surface of constant $t$ (and constant angular coordinates, but we can leave those out of the discussion since ony radial paths are being considered), bends at the horizon. In other words, it is not a single straight line from some value of $r$ outside the horizon all the way down to $r \to 0$. At $r = 2M$ it changes direction. (Note that here we are talking about an idealized "eternal" black hole, not one formed from an actual gravitational collapse; in the latter spacetime the line can't even reach $r = 2M$ in the first place, it will hit the collapsing matter first.)

 

[Nugatory]

Actually, no. It still works at and inside the horizon, as long as we are still in a spherically symmetric vacuum region.

Even though inside the horizon $r$ is timelike not spacelike I can still identify a spatial three-sphere with area $4\pi r^2$?

Although I must confess that I didn't really think about it, I just threw the $\gt R_S$ qualifier in there so as not to have to think about it.

(Also I generally think that the world would be a less confusing place if we didn't use the same labels for the timelike(spacelike) coordinate outside the horizon and the spacelike(timelike) coordinate in the different patch inside the horizon - even if something applies for $r$ outside the horizon and $r$ inside the horizon we're talking about different things.)

 

[Dale]

Even though inside the horizon r is timelike not spacelike I can still identify a spatial three-sphere with area 4πr2?

Yes. $\theta$ and $\phi$ are spacelike inside and outside the horizon and they still have the metric of a 2-sphere of radius $r$ even when $r$ itself is timelike.

 

[PeterDonis]

Even though inside the horizon $r$ is timelike not spacelike I can still identify a spatial three-sphere with area $4 \pi r^2$?

Not a spacelike 3-sphere, a spacelike 2-sphere. Every such 2-sphere anywhere in the spacetime is spacelike.

If you mean a spacelike 3-surface formed by a continuous series of such spacelike 2-spheres, yes, those also exist inside the horizon. They just aren't surfaces of constant Schwarzschild $t$ coordinate, since those are timelike (more precisely, they have one timelike and 2 spacelike linearly independent tangent vectors) inside the horizon.

 

[Nugatory]

Yes. $\theta$ and $\phi$ are spacelike inside and outside the horizon and they still have the metric of a 2-sphere of radius $r$ even when $r$ itself is timelike.

Ah - right, of course.

 

[mef]

Thank you all, but I would like to receive in response not reasoning, but CALCULATION...I think I can understand him

 

[Dale]

Please don’t shout, you have already been told about that.

@Ibix and I have provided calculations. We agree on the setup, although I got an extra term in the final answer compared to his. Importantly, both of us found that the length is finite. You should do the integral yourself and see what you get.

By the way, it is more than a little counterproductive to shout a demand for a calculation when you have, in fact, already received calculations which you have not even responded to.

 

[Ibix]

Are you thinking that second term evaluates to 0 or otherwise goes away? Or did I or Mathematica make a mistake somewhere?

No, you're right. I did think the log term went away, but I'd misread the result on my tiny phone screen.

 

[Dale]

I'd misread the result on my tiny phone screen.

That has happened to me more times than I would care to admit.

 

[Dale]

I just tried to work this out myself and got a slightly different answer. So I have $$g_{rr}=\frac{1}{1-\frac{R_S}{r}}$$ which integrates (according to Mathematica) to $$\int_{R_S}^R \sqrt{g_{rr}} \ dr = \sqrt{R \left(R-R_S\right)}+\frac{1}{2} R_S \log \left(\frac{-R_S+2 \sqrt{R \left(R-R_S\right)}+2
R}{R_S}\right) $$
Are you thinking that second term evaluates to 0 or otherwise goes away? Or did I or Mathematica make a mistake somewhere?

@mef to illustrate the issue about the coordinate system, I performed the same calculation in isotropic coordinates. We have $$g_{rr}=\frac{\left(2 r+\frac{R_S}{2}\right){}^4}{16 r^4}$$ which integrates to $$\int_{R_S}^R \sqrt{g_{rr}} \ dr = R-R_S+\frac{\left(R-R_S\right) R_S}{16 R}+\frac{1}{2} R_S \log \left(\frac{R}{R_S}\right)$$

Note that the results are different, but still finite.

 

[cianfa72]

If you mean a spacelike 3-surface formed by a continuous series of such spacelike 2-spheres, yes, those also exist inside the horizon. They just aren't surfaces of constant Schwarzschild $t$ coordinate, since those are timelike (more precisely, they have one timelike and 2 spacelike linearly independent tangent vectors) inside the horizon.

Since the $t$ coordinate is spacelike inside the horizon and the Schwarzschild metric in Schwarzschild coordinate chart is diagonal, then any 3-surface of constant coordinate time $t= \text{const}$ inside the horizon cannot be spacelike (there are not 3 linearly independent spacelike 4-vectors part of the orthogonal complement of a spacelike vector in the tangent space at each point).

 

[PeterDonis]

Since the $t$ coordinate is spacelike inside the horizon and the Schwarzschild metric in Schwarzschild coordinate chart is diagonal, then any 3-surface of constant coordinate time $t= \text{const}$ inside the horizon cannot be spacelike

Yes, you are agreeing with what I said.