Is the Schwarzchild Radius Relativistic or Newtonian?

[A.T.]

No, because the fact that the physical distance from the center happens to be the same in Newtonian mechanics as the areal radius does not mean that r is defined as the areal radius. r is defined as the physical distance from the center;

Newtonian mechanics assumes flat space, so in its context the two definitions are equivalent and interchangeable. GR doesn't assume flat space, so it has to be more specific and differentiate between the two.

 

[Dale]

Well, the issue was whether it's a "coincidence" that you can use Newtonian physics to calculate the radius rrr such that not even light can escape if it starts at a smaller radius than that.

But that is not what a surface with escape velocity c is. Light can easily escape that surface, as can material objects.

Escape velocity is not the velocity required to escape that radius, it is the velocity required to ballistically escape to infinity from that radius. Light could escape the Newtonian radius as could rockets and even bullets. Light and bullets wouldn’t make it to infinity, they would orbit, but a rocket could even make it to infinity.

The Newtonian surface has none of the interesting properties of an event horizon. The form of the equation is coincidentally the same. That’s it.

 

[stevendaryl]

But that is not what a surface with escape velocity c is. Light can easily escape that surface, as can material objects.

Escape velocity is not the velocity required to escape that radius, it is the velocity required to ballistically escape to infinity from that radius. Light could escape the Newtonian radius as could rockets and even bullets. Light and bullets wouldn’t make it to infinity, they would orbit, but a rocket could even make it to infinity.

The Newtonian surface has none of the interesting properties of an event horizon. The form of the equation is coincidentally the same. That’s it.

I disagree completely. If we have the equations of motion:

$E = \frac{1}{2} m (\dot{r})^2 - \frac{GMm}{r}$

we can easily sketch the motion, qualitatively: If $E < 0$ and $\dot{r}$ is initially positive, then $r$ will increase up until the point where

$r_{max} = -\frac{GMm}{E}$ (that's a positive number, since $E < 0$.

Then it will turn around, and $\dot{r}$ will be negative until it falls into singularity at $r = 0$. Note that $r_{max}$ will typically be greater than the Schwarzschild radius, $\frac{2GM}{c^2}$. So it's not true, at least not based on the Schwarzschild geometry alone, that nothing can rise above the event horizon.

Now, if you study this "orbit" from the point of view of the Schwarzschild geometry, what you'll see is that, in terms of the Schwarzschild coordinates $r, t$, the initial rise above the event horizon occurs at a time that is not covered by the Schwarzschild coordinates, as does the fall into the event horizon. But speaking informally, the rise happens at an event that has Schwarzschild time $t = - \infty$, while the fall happens at an even that has Schwarzschild time $t = + \infty$. From the point of view of an observer stationary from the black hole, the test particle takes forever to reach its maximum height, then turns around and takes forever to dip below the event horizon again (although both events happen at finite proper time).

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.

 

[PeterDonis]

Newtonian mechanics assumes flat space, so in its context the two definitions are equivalent and interchangeable. GR doesn't assume flat space, so it has to be more specific and differentiate between the two.

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

 

[Dale]

If E<0E<0E < 0 and ˙rr˙\dot{r} is initially positive

This is another difference. In GR $\dot r$ cannot be initially positive below the horizon. In Newtonian gravity it can.

 

[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.