Is the Schwarzchild Radius Relativistic or Newtonian?

[PeterDonis]

In GR $\dot r$ cannot be initially positive below the horizon.

It can if you are in the white hole region of the maximally extended spacetime, which @stevendaryl said he was including in his analysis.

 

[Dale]

It can if you are in the white hole region of the maximally extended spacetime, which @stevendaryl said he was including in his analysis.

Which is not only considered unphysical but also has no Newtonian counterpart. It makes the GR solution even more different from the Newtonian solution.

The fact that there exists any region of spacetime where $\dot r$ cannot be positive is distinct from Newtonian gravity. The existence of a region where $\dot r$ cannot be negative is even weirder. The event horizon (either one) simply is not equivalent to the Newtonian surface where the escape velocity is c.

 

[PeterDonis]

The fact that there exists any region of spacetime where $\dot r$ cannot be positive is distinct from Newtonian gravity. The existence of a region where $\dot r$ cannot be negative is even weirder.

Ah, I see. Yes, I agree that the lack of any region where there are restrictions on $\dot r$ at all in the Newtonian case is a key difference.

 

[A.T.]

Yes, which means that if we are discussing a comparison between Newtonian mechanics and GR, which we are in this thread, we have to differentiate between the two.

For GR we have to differentiate between the two, while in Newtonian mechanics it's impossible to differentiate them.

$r$ is actually defined as the physical distance from the center in Newtonian mechanics, so that's the relevant definition for this discussion.

Since the definitions are equivalent in Newtonian mechanics, you can pick either of them for comparison.

 

[PeterDonis]

in Newtonian mechanics it's impossible to differentiate them

No, it isn't. One is the physical distance from the center, the other is the areal radius. Both of these are perfectly good different definitions. The fact that they happen to have the same value does not mean they're indistinguishable; it just means they have the same value.

the definitions are equivalent in Newtonian mechanics

No, the definitions are not equivalent. The values are the same. The definitions are still distinct, because the equality (or inequality) of the values is experimentally testable; but to even formulate such a test you need to understand the distinct definitions of the two quantities, and the distinct ways of measuring them, so you can make the measurements and compare their results. Even in a case like a planet, where there is no event horizon and the physical distance from the center is well-defined, GR still predicts different values for "physical distance from the center" and "areal radius", whereas Newtonian mechanics predicts they are the same; and this could be tested.

 

[A.T.]

No, the definitions are not equivalent. The values are the same. The definitions are still distinct, because the equality (or inequality) of the values is experimentally testable; ...

Newtonian mechanics assumes Euclidean geometry, in which the two definitions of "r" are interchangeable. Whether that assumption matches experiments is a whole different question.

 

[Dale]

For GR we have to differentiate between the two, while in Newtonian mechanics it's impossible to differentiate them.

Which is a clear distinction between the theories. In the region of interest (around the EH) GR makes a distinction between the radial and areal coordinate whereas there is no such distinction in the region of the Newtonian surface.

 

[PeterDonis]

Newtonian mechanics assumes Euclidean geometry, in which the two definitions of "r" are interchangeable.

They are "interchangeable" in the sense that their numerical values being the same is a theorem of Euclidean geometry. But that does not mean their definitions are the same. Their definitions involve obviously different physical procedures: one has you extending a ruler from the center, the other has you measuring the area of a 2-sphere and then calculating the square root of the area divided by $4 \pi$. These are different definitions. And Newtonian mechanics defines the gravitational force as depending on the first thing--the distance from the center.

 

[PeterDonis]

GR makes a distinction between the radial and areal coordinate

I would not put it this way. I would put it that GR predicts that the physical distance from the center is not the same as the square root of the area divided by $4 \pi$. This is a coordinate-independent fact. And for a comparison with Newtonian physics, the coordinate-independent fact is what's relevant, since Newtonian physics does not say the gravitational force depends on a radial coordinate: it says the force depends on the physical distance from the center.

 

[A.T.]

And Newtonian mechanics defines the gravitational force as depending on the first thing--the distance from the center.

Yes, but it is a definition in the context of Euclidean geometry, where picking "distance from the center" instead of "areal radius" has no significance, other than going with the more convenient, straightforward one.

 

[PeterDonis]

it is a definition in the context of Euclidean geometry, where picking "distance from the center" instead of "areal radius" has no significance

Yes, it does--the two definitions describe two different concepts with two different methods of measuring them. "Euclidean geometry" just means we've adopted a model in which you can prove as a theorem that the values measured by those two different methods give the same answer. It doesn't make the two different methods the same.

 

[A.T.]

Yes, it does--the two definitions describe two different concepts with two different methods of measuring them.

The definitions don't describe any measuring methods.

 

[bbbl67]

Typically in studying black holes, the portion of the Schwarzschild geometry with $t < 0$ is considered an unphysical "white hole", and the real black hole geometry only includes the section $t > 0$. That makes sense, because black holes aren't actually eternal, but are formed from the collapse of massive stars. So there is no actual black hole prior to that, and so no white hole geometry in the limit $t \rightarrow -\infty$. But as far as the Schwarzschild metric is concerned, you can include the white hole region for the sake of studying solutions of the field equations and what the paths of test particles look like. In the complete spacetime that includes both the $t >0$ and $t < 0$ regions, the meaning of escape velocity for a test particle is very much analogous to what it is in Newtonian physics.

If you extend the black hole case out to its nearest analogy, the birth of the universe, wouldn't a white hole simply be all of the energy coming into a black hole being released into a white hole inside a new spacetime? Therefore the white hole happens, just not in the same universe as the black hole? So the white hole will act as the Big Bang of a new universe?

 

[bbbl67]

Yes, but it is a definition in the context of Euclidean geometry, where picking "distance from the center" instead of "areal radius" has no significance, other than going with the more convenient, straightforward one.

Yes, it's pedantics the two distances are describing the same thing, whether you're talking about areal radius vs. distance from the center radius.

 

[PeterDonis]

The definitions don't describe any measuring methods.

Sure they do. "Physical distance from the center" isn't just an abstraction; it implies a certain measurement method. So does "square root of the area of a 2-sphere divided by $4 \pi$". And the two methods are different. The fact that the measurement methods aren't explicitly spelled out in the words used for the definitions doesn't mean they aren't there.

 

[A.T.]

Sure they do. "Physical distance from the center" isn't just an abstraction; it implies a certain measurement method.

But it doesn't imply one unique measurement method in the context of Euclidean geometry.

 

[PeterDonis]

it doesn't imply one unique measurement method in the context of Euclidean geometry

Yes, it does. You extend rulers from the center to the point of interest. The fact that doing this gives the same numerical answer as measuring the area of a 2-sphere, dividing by $4 \pi$, and taking the square root, does not mean the physical distance from the center is defined to be measured that way. A theorem is not the same as a definition.

 

[A.T.]

You extend rulers from the center to the point of interest. The fact that doing this gives the same numerical answer as measuring the area of a 2-sphere, dividing by $4 \pi$, and taking the square root, does not mean the physical distance from the center is defined to be measured that way.

Again, the definitions don't mention any specific measuring methods. You claim they "imply" one method, but admit there are multiple methods that would give the same answer (assuming Euclidean geometry). With multiple equivalent possibilities I don't see the implication of one method.

 

[PeterDonis]

You claim they "imply" one method

If you honestly think that the words "physical distance from the center" imply measurement by the method of "measure the area of a 2-sphere, divide by $4 \pi$, then take the square root" instead of "lay rulers end to end from the center to the point of interest", then we are speaking different languages and there is really no point in further discussion. I don't see how the theorem that, given Euclidean geometry, the two methods give the same answer means that they are the same method or that the words that plainly (to me) describe one method can be taken to describe the other.

 

[PeterDonis]

You claim they "imply" one method, but admit there are multiple methods that would give the same answer (assuming Euclidean geometry).

Newtonian physics "assumed" Euclidean geometry as a model. If you conflate the two different measurement methods, you are eliminating the possibility of testing the model by actual measurements.

 

[Dale]

If you extend the black hole case out to its nearest analogy, the birth of the universe,

This is not a good analogy. The Schwarzschild spacetime is a vacuum spacetime, but the FLRW spacetime is not. Also, the FLRW metric is homogenous while Schwarzschild is not. Also, Schwarzschild is static but FLRW is not. And Schwarzschild is asymptotically flat and FLRW is not. The analogy is really bad.

 

[Dale]

I would not put it this way. I would put it that GR predicts that the physical distance from the center is not the same as the square root of the area divided by 4π. This is a coordinate-independent fact.

Yes, I like that way better.

 

[bbbl67]

This is not a good analogy. The Schwarzschild spacetime is a vacuum spacetime, but the FLRW spacetime is not. Also, the FLRW metric is homogenous while Schwarzschild is not. Also, Schwarzschild is static but FLRW is not. And Schwarzschild is asymptotically flat and FLRW is not. The analogy is really bad.

How do we know the inside of a black hole is a vacuum? How do we know the inside of a black hole is static?

 

[Matterwave]

How do we know the inside of a black hole is a vacuum? How do we know the inside of a black hole is static?

Note that Dale is talking about the Schwarzschild spacetime - which is a mathematical model (the one from which the concept of "black holes" arose) - and not physical black holes. The Schwarzschild spacetime is built upon the static and vacuum assumptions - it is everywhere static and everywhere a vacuum. The interior of the "black hole" in Schwarzschild spacetime is by definition vacuum and static.

A physical black hole need not necessarily have a vacuum and static interior. A Kerr black hole for example is not static (it is - however - stationary). The interior of the black hole can't be material in (hydrostatic) equilibrium because all world lines for all physical particles must reach the singularity in finite proper time - but there could be material inside a "black hole" that's just on its way towards the singularity. I guess in that case it wouldn't be technically a "vacuum" inside there.

 

[PeterDonis]

How do we know the inside of a black hole is a vacuum?

Because it is impossible for it not to be (except for the small scattered regions of spacetime occupied by matter that falls in).

How do we know the inside of a black hole is static?

The interior of a black hole is not static. That's why it has to be (almost all) vacuum: because matter that falls in can't just stay in place--it has to keep falling until it hits the singularity.

 

[PeterDonis]

A Kerr black hole for example is not static (it is - however - stationary).

The exterior (outside the horizon) of Kerr spacetime is stationary. The interior (inside the horizon) is not. (Nor is the interior of Schwarzschild spacetime.)

The interior of the black hole can't be material in (hydrostatic) equilibrium because all world lines for all physical particles must reach the singularity in finite proper time

"All worldlines must hit the singularity in finite proper time" is only true for Schwarzschild spacetime. It is not true for Kerr spacetime; there are worldlines in maximally extended Kerr spacetime that never hit the singularity at all (they eventually emerge into another stationary exterior region that is disconnected from the original one that the material fell in from). Also, Kerr spacetime, unlike Schwarzschild spacetime, has an inner Cauchy horizon that makes any kind of extrapolation beyond it problematic. And also again, the interior regions of both Schwarzschild and Kerr spacetime are unstable against small perturbations, which means they probably don't actually describe the interiors of physically realistic black holes, at least not well inside the horizon. (Look up "BKL singularity" for an example of a more realistic possibility.)

None of this changes the key points for this discussion: black hole interiors are not static, and they must be almost all vacuum because material falling in doesn't stay in place.

 

[Matterwave]

The exterior (outside the horizon) of Kerr spacetime is stationary. The interior (inside the horizon) is not. (Nor is the interior of Schwarzschild spacetime.)

There is no time-like killing field for the interior of a Schwarzschild space time? Somehow this point has been glossed over in my readings on the subject. I assumed that since the "static" and "vacuum" assumptions are built into finding the Schwarzschild solution itself, that it would be valid everywhere. My mistake.

 

[Dale]

How do we know the inside of a black hole is a vacuum? How do we know the inside of a black hole is static?

We don’t have to go inside a black hole. The analogy fails well outside of the horizon, where we have plenty of experimental evidence.

If you still think it is a valid analogy despite the contradictions indicated above then it is up to you to provide some support for your claim.

 

[PeterDonis]

There is no time-like killing field for the interior of a Schwarzschild space time?

No, there is not. Only in the exterior. Inside the horizon, the Killing field is still there, but it's spacelike, not timelike. (On the horizon, it's null; the horizon is actually generated by the null integral curves of the Killing field at $r = 2M$.)

I assumed that since the "static" and "vacuum" assumptions are built into finding the Schwarzschild solution itself, that it would be valid everywhere.

As it happens, I wrote an Insights article a while back about the derivation of Birkhoff's theorem which discusses this.

https://www.physicsforums.com/insights/short-proof-birkhoffs-theorem/

The key is that "static" is not an assumption; it's a result of the derivation, but the result isn't actually "static" (though it's often misstated that way, even in textbooks, unfortunately), it's "there's a fourth Killing field in addition to the three that are there because of spherical symmetry". The theorem does not prove, or require, that the fourth Killing field is timelike everywhere. And indeed, you can see from the line element that it is only timelike for $r > 2M$; it's null at $r = 2M$ and spacelike for $r < 2M$.

 

[Matterwave]

No, there is not. Only in the exterior. Inside the horizon, the Killing field is still there, but it's spacelike, not timelike. (On the horizon, it's null; the horizon is actually generated by the null integral curves of the Killing field at $r = 2M$.)
As it happens, I wrote an Insights article a while back about the derivation of Birkhoff's theorem which discusses this.

https://www.physicsforums.com/insights/short-proof-birkhoffs-theorem/

The key is that "static" is not an assumption; it's a result of the derivation, but the result isn't actually "static" (though it's often misstated that way, even in textbooks, unfortunately), it's "there's a fourth Killing field in addition to the three that are there because of spherical symmetry". The theorem does not prove, or require, that the fourth Killing field is timelike everywhere. And indeed, you can see from the line element that it is only timelike for $r > 2M$; it's null at $r = 2M$ and spacelike for $r < 2M$.

Thanks for this! I will read it when I get the time.

 

[PAllen]

There is no time-like killing field for the interior of a Schwarzschild space time? Somehow this point has been glossed over in my readings on the subject. I assumed that since the "static" and "vacuum" assumptions are built into finding the Schwarzschild solution itself, that it would be valid everywhere. My mistake.

Precise statement is that vacuum plus spherical symmetry imply an additional killing vector field besides those for spherical symmetry. Outside the horizon, it is timelike. Inside it is spacelike, and may be considered axial. If you allow nonanalytic manifolds, there are many different topologies than Kruskal that are possible. I believe some have no timelike killing vector field at all. I will try to dig up a paper I’ve linked in the past that explores this.

Addendum: I found the papers:

https://arxiv.org/abs/0908.4110
https://arxiv.org/abs/0910.5194

See top of page 6 of second paper which gives vacuum spherically symmetric manifolds that are nowhere static, nor are they asymptotically flat. They are also not extensible, thus not a cheat by just chopping a manifold at some boundary, e.g. a piece of the interior.

 

[bbbl67]

We don’t have to go inside a black hole. The analogy fails well outside of the horizon, where we have plenty of experimental evidence.

If you still think it is a valid analogy despite the contradictions indicated above then it is up to you to provide some support for your claim.

What experimental evidence is that?

This paper, by Carlo Rovelli & Marios Christodoulou, shows that the inside of a black hole is bigger than the outside, if you use certain coordinate systems! The volume grows bigger with time, and even with a stellar mass black hole, the interior volume can become bigger than the current observable universe. This sounds pretty much like a universe to me (also sounds like a Tardis)!

Paper: [1411.2854] How big is a black hole?

 

[Dale]

This paper, by Carlo Rovelli & Marios Christodoulou, shows that the inside of a black hole is bigger than the outside, if you use certain coordinate systems! The volume grows bigger with time, and even with a stellar mass black hole, the interior volume can become bigger than the current observable universe.

First, to use your own words, how do you know the inside of a black hole is bigger than the outside? Second, how is that any kind of analogy with the universe since the Big Bang (FLRW) universe has no outside? Third, that does not address the vacuum objection. Fourth, that does not address the homogeneity objection. Fifth, that does not address the asymptotic flatness objection.

Note that at no point in that paper does Rovelli make any claim that a Schwarzschild spacetime is analogous to a FLRW spacetime, as you did. While being interesting, the paper does not support your specific objectionable claim.

Regarding the evidence that we have outside of the EH you can consider all of the solar system tests of the PPN that show the PPN parameters match those of GR. All of the solar system tests are based on the Schwarzschild spacetime.

 

[martinbn]

Just a pedantic comment about Schwartschild being static. To show that the interior is not even stationary one needs to prove that there aren't any timelike Killing fields there. The fact that the Killing field that is timelike outside becomes spacelike inside, by itself is not enough, there still could be another timelike Killing field there. Of course that is not true, but it needs to be shown.

 

[PeterDonis]

To show that the interior is not even stationary one needs to prove that there aren't any timelike Killing fields there. The fact that the Killing field that is timelike outside becomes spacelike inside, by itself is not enough, there still could be another timelike Killing field there.

No, there couldn't, because the metric inside is still Schwarzschild (by Birkhoff's theorem), and we already know Schwarzschild has no other Killing fields.