# Shrinking event horizon to point singularity

Varying coordinate systems in GR has given me a new perspective that may help to resolve a problem that has been nagging at me ever since I began working with GR. In every problem I've ever dealt with, a complex mathematical result describes an impossible scenario, something that cannot occur. Yet in the Schwarzschild coordinate system, we have a coordinate radius beyond which all sorts of complex results arise, below the event horizon. This says nothing should exist within this space, since we cannot have clocks that tick at a complex rate and we cannot have complex lengths. This space below 2m, then, really shouldn't exist imo. Yet using Schwarzschild coordinates, we can still plot coordinate distances between zero and 2 m according to a distant observer, that a freefaller falls into within finite proper time, although we can't describe what happens below 2 m.

I had always thought of Schwarzschild coordinates as the "real" coordinate system, while all others were psuedo-systems for ease of derivation that must be switched back to Schwarzschild to gain the "actual" results. But if Schwarzschild is just as arbitrary a coordinate system as any other, then we can just as easily make a different coordinate choice. Of course, even in deriving Schwarzschild, we must follow certain rules: the locally measured speed of light is always c, SR is valid locally, the equivalence principle holds, the EFE equations are valid, locally measured angular momentum is constant, the locally measured energy of a particle per local time dilation is constant, etc. I will add one more which will limit my choice of coordinate systems even further. It is simply that we will only measure "real" space with our coordinate system.

My choice of coordinate systems, at least the simplest I've found so far, then, is

r1 = r (1 - 2 m / r), r = r1 (1 + 2 m / r1) which transforms the metric to

ds^2 = c^2 dt^2 / (1 + 2 m / r1) - dr1^2 (1 + 2 m / r1) - dθ^2 r1^2 (1 + 2 m / r1)^2

I have now turned the event horizon at r = 2 m in Schwarzschild into a point singularity at r1 = 0. Nothing below r1 = 0 exists within this coordinate system, so complex space is eliminated. Problem solved. This is the way I believe it should be, and if I am allowed to use any coordinate system I choose, this is definitely the one that makes sense to me. If a freefalling observer were to fall all the way to the point singularity in this coordinate system, in a finite time according to his own watch, then since that is also where the point mass must exist, the freefaller will immediately strike the mass rather than falling further into some complex space. If the mass were not struck somehow, and the freefaller could keep going, he should simply pass through the point in real space and begin decelerating as he now travels away from the point while his watch continues to tick at a gravitationally time dilated rate, although to the distant observer, it would take him an infinite amount of time to reach the point singularity.

The transformations are now

dt' = dt / sqrt(1 + 2 m / r1)

dr' = dr sqrt(1 + 2 m / r1)

with tangent distances measured as (1 + 2 m / r1) greater locally. Locally, tangent and radial speeds of particles would be measured as

v_t' = v_t (1 + 2 m / r1)^(3/2)

v_r' = v_r (1 + 2 m / r1)

Of course, there are still plenty of other coordinate systems that create a point singularity, r1 = r sqrt(1 - 2 m / r) for instance. The only other rule that must apply is that r1 approximates r as r goes to infinity. If we want to describe -r1 as the same as r1 but in the opposite direction, so that r1 and -r1 must give the same time dilations and length contractions, then we could use r1 = sqrt(r^2 - (2 m)^2). Even if others might not agree, since there is theoretically no right or wrong in GR coordinate choices, I truly believe this condition of only allowing real space should be included in the coordinate choice. Other conditions might narrow it down further as well. Does anyone have any other conditions for the coordinate choice in mind that might help things make more sense to you?

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PeterDonis
Mentor
I had always thought of Schwarzschild coordinates as the "real" coordinate system, while all others were psuedo-systems for ease of derivation that must be switched back to Schwarzschild to gain the "actual" results.

You realize, I trust, that many of us on PF will be tempted to stop reading right here, because of the ten to the umpteenth power previous threads centering around this same (erroneous) belief. However, you do raise some points that are worth discussing.

But if Schwarzschild is just as arbitrary a coordinate system as any other, then we can just as easily make a different coordinate choice.

Yes, you can. But in choosing *any* coordinate chart, including Schwarzschild, you have to remember two things:

(1) There is no guarantee that a given single coordinate chart will cover the entire spacetime. To check that, you have to compute invariants (numbers that don't depend on which chart you use to compute them) and see how they behave. You can't make any assertions about what portions of a spacetime "exist" or "are real" based on what happens in a particular coordinate chart.

For example, what people usually mean when they say "Schwarzschild coordinates" is *exterior* Schwarzschild coordinates (see item #2 below for more on this), which only cover the portion of the spacetime with r > 2m. The fact that these coordinates become singular at r = 2m does not, in itself, say anything about the *physics* at r = 2m. To find out about that, you have to compute invariants, and when you do, you find that they are all finite and well-behaved at r = 2m (just like all the invariants describing the geometry of the Earth's surface are finite and well-behaved at the North Pole, even though the standard latitude-longitude coordinates are singular there). That means there is a portion of spacetime with r < 2m that exterior Schwarzschild coordinates do *not* cover, but which still "exists".

(2) A single coordinate chart, strictly speaking, can only cover a portion of spacetime in which it has no singularities. So, for example, when you say this:

Yet in the Schwarzschild coordinate system, we have a coordinate radius beyond which all sorts of complex results arise, below the event horizon...

What you are calling "the Schwarzschild coordinate system" here, covering the region 0 < r < 2m, is a *different* chart than the one that covers the region 2m < r < infinity. The coordinate singularity at r = 2m separates the two; they are not continuous with each other, and so all the issues you raise about how the coordinate labeled "t" behaves when r < 2m are irrelevant to the actual physics, because in the interior Schwarzschild chart (0 < r < 2m), the coordinate labeled "t" is not the "time" coordinate. (Yes, that means the labeling is confusing. Sorry, but it's so entrenched now that we'll just have to deal.)

Of course, even in deriving Schwarzschild, we must follow certain rules: the locally measured speed of light is always c, SR is valid locally, the equivalence principle holds, the EFE equations are valid

Yes, these are all good things to keep in mind for *any* coordinate chart on *any* spacetime.

locally measured angular momentum is constant, the locally measured energy of a particle per local time dilation is constant, etc.

These are more particular to Schwarzschild spacetime (which is *not* the same as "Schwarzschild coordinates"--the spacetime is the underlying geometric object), and I'm not quite sure what you mean by them. But I think this is a minor issue and probably not worth delving into; you basically appear to be saying that local physics should look like we expect it to look, which is true.

I will add one more which will limit my choice of coordinate systems even further. It is simply that we will only measure "real" space with our coordinate system.

How do you define "real" space? Remember that, since you are using it to restrict your choice of coordinates, you have to define it *without* using any coordinate-dependent concepts. Can you do that?

I have now turned the event horizon at r = 2 m in Schwarzschild into a point singularity at r1 = 0. Nothing below r1 = 0 exists within this coordinate system, so complex space is eliminated.

Nope. All you have done is ensure that your coordinates only cover the portion of the spacetime outside the event horizon. (I think Schwarzschild himself came up with a coordinate chart that was somewhat similar in his original paper.) You can use any coordinates you choose, but you can't make invariant physical assertions like what parts of spacetime "exist" or "are real" based on what happens in a given coordinate chart. You have to, as I said above, compute invariants and see how they behave as r1 -> 0 in your chart.

I haven't bothered to do that computation because, if your coordinate transformation is correct, the invariants must all come out the same as they do in the standard Schwarzschild chart, or indeed in *any* chart on the same spacetime, such as the Painleve chart, which is singularity-free all the way down to r = 0, or the Kruskal chart--you would do well to look at them. And I already know how the invariants behave in those other charts: they clearly show that there *is* a region of spacetime with r < 2m, with a real singularity--one where the curvature invariants become infinite--at r = 0. That's the physics, and you can't change the physics by changing coordinate charts.

pervect
Staff Emeritus
As others have posted, this will work fine outside the event horizon, but inside the event horizon, you're coordinate transform r' = r(1-2m/r) = r-2m assigns a circle of radius 2m to a single point - and it doesn't assign any valid coordinates at all for r < 2m.

So you've basically tried to get rid of the event horizon and the interior region by refusing to assign coordinates to the points inside it.

If you're actually interested in the event horizon and the interior region of the black hole, one of the first things you'll have to do is to reconsider your initial decision to look at things from the viewpoint of a static observer. The reason is simple - static observers don't exist at or inside the event horizon.

Trying to analyze what the black hole looks like from the standpoint of observers who don't actually exist is not going to be terribly successful.

There are successful coordinate systems that DO cover the inside of the black hole. They're known as Kruskal-Szerkes coordinates, see for instance http://en.wikipedia.org/w/index.php?title=Kruskal–Szekeres_coordinates&oldid=506720782

Not accidentally, the K-S coordinates respect the fact that the black hole event horizon is not like a stationary point in space, but rather a null surface, it can be considered to be "trapped light". Which makes it more or less obvious why one can't have a "static observer" on the event horizon, any more than you can sensibly have a "static observer" on a light beam.

The full K-S solution has some rather interesting features, including a whie hole region, that are rather intersting, but in the interests of time and not hijacking the thread I won't go into them. But I'd encourage you to read some actual textbooks and get in touch with what is already known if you have the necessary background.

Thank you PeterDonis.

How do you define "real" space? Remember that, since you are using it to restrict your choice of coordinates, you have to define it *without* using any coordinate-dependent concepts. Can you do that?
By "real" space, I don't really mean the space itself so much as the effects within that space. The Schwarzschild coordinates say that any observer at r > 2m will infer that rulers within r < 2m become contracted to a complex length and clocks tick at a complex rate. Generally in mathematics, from any example I've ever seen, that means that it is impossible for clocks and rulers or anything physical to exist within that space. If nothing can exist within that space, nothing can enter that space, so it is unreachable. So more than just being some extra dimension we cannot fathom below 2m, it simply doesn't exist as part of our universe.

Otherwise, to any observers, depending upon the coordinate system, the space itself smoothly transitions from r > 2m to r < 2m, but for that same reason, we shouldn't have two different equations that describe what is taking place inside and outside of 2m. There is no physical boundary there, nothing physical that changes the physics, so precisely the same equation that describe the effects outside of 2m should describe the inside of 2m. For an object falling past the surface of the Earth, for example, the surface would describe a boundary where the physics changes, and we would require two equations, one which applies outside the body and another that applies internally. But for a point mass with only a vacuum outside to infinity, only a single equation should be necessary to describe what is taking place at all r.

Nope. All you have done is ensure that your coordinates only cover the portion of the spacetime outside the event horizon. (I think Schwarzschild himself came up with a coordinate chart that was somewhat similar in his original paper.) You can use any coordinates you choose, but you can't make invariant physical assertions like what parts of spacetime "exist" or "are real" based on what happens in a given coordinate chart. You have to, as I said above, compute invariants and see how they behave as r1 -> 0 in your chart.

I haven't bothered to do that computation because, if your coordinate transformation is correct, the invariants must all come out the same as they do in the standard Schwarzschild chart, or indeed in *any* chart on the same spacetime, such as the Painleve chart, which is singularity-free all the way down to r = 0, or the Kruskal chart--you would do well to look at them. And I already know how the invariants behave in those other charts: they clearly show that there *is* a region of spacetime with r < 2m, with a real singularity--one where the curvature invariants become infinite--at r = 0. That's the physics, and you can't change the physics by changing coordinate charts.
Right. I am not changing the physics, but changing coordinates to eliminate complex events which cannot take place according to our "external" universe. What if instead of the Schwarzschild coordinates we're all used to, Schwarzschild had come up with this coordinate system I have introduced in its place? Think about it. We are considering a point mass and want a single equation that will tell us what is taking place in a vacuum at any distance r from that point mass. This coordinate system does that, all results are real (not complex), and the singularity lies at the point mass in the same way that it would for Newton, although relativistic.

If this coordinate system had been the original, then if after getting used to it, someone comes along and says "I've got a different coordinate system that will open up the point singularity to a perimeter at r = m or r = 2m or r = 100m, etc., although we cannot describe what takes place within that boundary using this coordinate system, unless perhaps we generate a separate coordinate system that applies inside the perimeter and matches the exterior coordinate system at the boundary." We might reply with something like "What would be the purpose of that? We have a point singularity at a point mass. What happens when we open the point singularity? Is the point mass expanded to fill the space within 2m also or does it remain a point mass? Why should we need a separate coordinate system to describe that internal space? Why should we open a point and attempt to describe its internal physics when that physics cannot even exist as part of our ordinary universe, except as it is described as a point in space?" Something like that. Opening up a point singularity this way probably wouldn't even be seriously considered except by "out-of-the-box" enthusiasts. Wouldn't you agree?

Bill_K
The Schwarzschild coordinates say that any observer at r > 2m will infer that rulers within r < 2m become contracted to a complex length and clocks tick at a complex rate.
No indeed, not at all. There is nothing complex in the Schwarschild coordinate system. What happens for r < 2m is that gtt and grr both change sign. This means that t becomes a spacelike coordinate and r becomes a timelike coordinate. They basically trade places.

As others have posted, this will work fine outside the event horizon, but inside the event horizon, you're coordinate transform r' = r(1-2m/r) = r-2m assigns a circle of radius 2m to a single point - and it doesn't assign any valid coordinates at all for r < 2m.

So you've basically tried to get rid of the event horizon and the interior region by refusing to assign coordinates to the points inside it.
Yes, exactly. :)

If you're actually interested in the event horizon and the interior region of the black hole, one of the first things you'll have to do is to reconsider your initial decision to look at things from the viewpoint of a static observer. The reason is simple - static observers don't exist at or inside the event horizon.

Trying to analyze what the black hole looks like from the standpoint of observers who don't actually exist is not going to be terribly successful.
Right. Under this coordinate system, with the point singularity at a point mass, no observers would exist within that point in space. If an observer were to fall all the way down to r1 = 0, he would simply strike the point mass and that would be that. I am considering the viewpoint of observers that are external to the singularity, although I suppose you mean that external observers would only infer clocks rates and ruler lengths for observers falling within the singularity, since internal observers cannot remain static.

There are successful coordinate systems that DO cover the inside of the black hole. They're known as Kruskal-Szerkes coordinates, see for instance http://en.wikipedia.org/w/index.php?title=Kruskal–Szekeres_coordinates&oldid=506720782

Not accidentally, the K-S coordinates respect the fact that the black hole event horizon is not like a stationary point in space, but rather a null surface, it can be considered to be "trapped light". Which makes it more or less obvious why one can't have a "static observer" on the event horizon, any more than you can sensibly have a "static observer" on a light beam.

The full K-S solution has some rather interesting features, including a whie hole region, that are rather intersting, but in the interests of time and not hijacking the thread I won't go into them. But I'd encourage you to read some actual textbooks and get in touch with what is already known if you have the necessary background.
Thank you. I've been looking into those.

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No indeed, not at all. There is nothing complex in the Schwarschild coordinate system. What happens for r < 2m is that gtt and grr both change sign. This means that t becomes a spacelike coordinate and r becomes a timelike coordinate. They basically trade places.
They wouldn't be measured as complex locally, no, since they are within the same space, and we always assume the local space to be Minkowski regardless of where that space resides according to external observers. They would only be complex as inferred by observers exterior to the singularity according to their coordinate system. As has been pointed out to me, though, the complex time dilation and length contraction I have been referring to is taken directly between static observers, so one way we can define the complex result is simply to imply that no static observers exist within the singularity, that they must be falling. Fair enough. But then, we have kinetic effects on top of the gravitational effects, which should make things worse, not better I think. And of course the metric changes from timelike to spacelike as you stated, but I am referring to the individual transformations of length and time. But apart from this and other reasons I have given within this thread, wouldn't it just make more sense that the singularity be described at a point for a point mass rather than stretched out over an arbitrary volume?

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pervect
Staff Emeritus
Basically, not assigning coordinates to something so you can then argue it doesn't exist is inventive, but not particular fruitful or sensible.

IT seems to me like this thread is re-inventing the wheel ,improving it by making it square rather than round. (Excpet that in this case it's not the wheel that's being re-invented, but the Schwarzschild geometry).

I suppose that means I've been spending too much time on the thread, and it's time for me to move on to other things.

Basically, not assigning coordinates to something so you can then argue it doesn't exist is inventive, but not particular fruitful or sensible.

IT seems to me like this thread is re-inventing the wheel ,improving it by making it square rather than round. (Excpet that in this case it's not the wheel that's being re-invented, but the Schwarzschild geometry).

I suppose that means I've been spending too much time on the thread, and it's time for me to move on to other things.
Okay thanks. I'm actually looking at it the other way around, but anyway, before you go, answer me this one last post please. I'm cutting out the space below the singularity, saying it doesn't exist within our universe, really just an arbitrarily expanded point singularity, right, but if you disagree and you think it's important that it be considered, then let's take things from the other direction. If I were to adopt a coordinate system that opens up the singularity even more, say at r = 20 m, then a singularity at r = 2m cuts out 99.9% of that space, shrinking the inner 18 m to the center point. So why isn't that extra space important, that we still consider it to be at r = 2m? Or r = 200 m from which r = 20 m cuts 99.9% of that? What's the minimum limit for how much space should be accounted for within a singularity? Surely not infinite, right? Of course, I'm taking it the other way, down to zero. But if Schwarzschild is arbitrary, and some amount of space should really be considered within the singularity, then how far should we take it?

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PeterDonis
Mentor
The Schwarzschild coordinates say that any observer at r > 2m will infer that rulers within r < 2m become contracted to a complex length and clocks tick at a complex rate.

No, they don't. The coordinates don't tell you how clocks tick; you have to compute physical invariants to see that. The "t" coordinate in the region r < 2m is *not* the "time" coordinate, and it doesn't tell you how clocks tick.

Otherwise, to any observers, depending upon the coordinate system, the space itself smoothly transitions from r > 2m to r < 2m

This doesn't depend on the coordinate system. How the space "transitions" is an invariant, and can be seen by computing invariants. It is true that some coordinate charts (like the Painleve chart) do a much better job of *capturing* how the space transitions than other charts (like the Schwarzschild charts--remember that there are *two* such charts, one for r > 2m and one for r < 2m, and they are disconnected from each other).

but for that same reason, we shouldn't have two different equations that describe what is taking place inside and outside of 2m.

We don't. The Einstein Field Equation holds at every event in the spacetime, including the region where r < 2m.

If you are trying to claim that we shouldn't need two different coordinate charts to describe the two regions, that's not a valid claim. There is no guarantee that a spacetime can be described in its entirety by a single chart.

If you are trying to claim that any coordinate chart you pick should be able to describe any feature you choose about a spacetime, that's not valid either. You can pick whatever chart you like, but you don't get to pick how well your chosen chart describes the spacetime.

There is no physical boundary there, nothing physical that changes the physics

This is true.

so precisely the same equation that describe the effects outside of 2m should describe the inside of 2m.

Only if you pick the right "equation". See above.

What if instead of the Schwarzschild coordinates we're all used to, Schwarzschild had come up with this coordinate system I have introduced in its place?

Then someone would compute invariants using this chart and, from those invariants, would be able to show that the chart could not possibly be describing the entire spacetime. Coordinate charts don't determine the physics. They are useful in *describing* the physics, but to determine how well they describe the physics, you have to look at things that are independent of the particular chart.

Opening up a point singularity this way probably wouldn't even be seriously considered except by "out-of-the-box" enthusiasts. Wouldn't you agree?

No. See above.

No, they don't. The coordinates don't tell you how clocks tick; you have to compute physical invariants to see that. The "t" coordinate in the region r < 2m is *not* the "time" coordinate, and it doesn't tell you how clocks tick.
Okay thanks, but I would want to know how a clock physically ticks compared to dt on our own external clock, and how a ruler is contracted. As far as I know, the external coordinate system cannot describe it, but I'll save that for another thread.

This doesn't depend on the coordinate system. How the space "transitions" is an invariant, and can be seen by computing invariants. It is true that some coordinate charts (like the Painleve chart) do a much better job of *capturing* how the space transitions than other charts (like the Schwarzschild charts--remember that there are *two* such charts, one for r > 2m and one for r < 2m, and they are disconnected from each other).
Right. I just added the statement about coordinate dependence because with the one I have, if an observer could fall all the way down to the point singularity, he shouldn't pass through smoothly, but should just immediately strike the point mass there, so there is no "falling through" the boundary for him, unless he somehow passes through or doesn't interact with the mass, but then he would just be falling through a point and simply begin to decelerate again as he travels away from the point mass.

We don't. The Einstein Field Equation holds at every event in the spacetime, including the region where r < 2m.

If you are trying to claim that we shouldn't need two different coordinate charts to describe the two regions, that's not a valid claim. There is no guarantee that a spacetime can be described in its entirety by a single chart.

If you are trying to claim that any coordinate chart you pick should be able to describe any feature you choose about a spacetime, that's not valid either. You can pick whatever chart you like, but you don't get to pick how well your chosen chart describes the spacetime.
I still don't see why would need two equations. To me, it seems that the mathematics of the external equation cannot describe what occurs below a singularity, so we are basically trying to change the mathematics around in such a way below the singularity as to make more sense of it. But if there is nothing physically different at the boundary that would change the physics according to either an external observer or a local observer that is falling through it, then I don't see why the mathematics should suddenly be applied differently there.

Then someone would compute invariants using this chart and, from those invariants, would be able to show that the chart could not possibly be describing the entire spacetime. Coordinate charts don't determine the physics. They are useful in *describing* the physics, but to determine how well they describe the physics, you have to look at things that are independent of the particular chart.
Okay, this is the question that the whole thread boils down to at this point (pun intended :) ). If you're saying that a point singularity could not possibly describe the entire spacetime, then as also asked in my last post, how far should we open the singularity so that the entire space-time is accounted for? If Schwarzschild coordinates truly are arbitrary, then it could just as easily be 20 m or 200 m, whereby 2m would be cutting much of it out. Or it might be zero. Surely not infinite, though, right? What is the minimum amount we should expand the space within a singularity to fully account for the space-time?

PeterDonis
Mentor
I would want to know how a clock physically ticks compared to dt on our own external clock, and how a ruler is contracted. As far as I know, the external coordinate system cannot describe it

You're right, the external coordinate system, the exterior Schwarzschild chart, can't. So what? As I said before, there's no guarantee that the chart you prefer even covers the entire spacetime, let alone describes it in a way you find intuitively plausible.

The *interior* Schwarzschild coordinates can, but using the "r" coordinate, *not* the "t" coordinate. (Yes, as I said before, that means the coordinate names are confusing for the interior chart.) Also, there are other coordinate charts, such as Painleve, Eddington-Finkelstein, or Kruskal, which can.

if an observer could fall all the way down to the point singularity, he shouldn't pass through smoothly, but should just immediately strike the point mass there, so there is no "falling through" the boundary for him

If you mean the singularity at r = 0, you are correct, the observer just "hits" it and stops. (Technically, there is not a "point mass" at r = 0, but a singularity of infinite curvature, so the observer doesn't "hit" the point mass and bounce, he is destroyed by the infinite curvature and ceases to exist.)

If you mean the "singularity" that occurs at r = 2m in the standard Schwarzschild charts, but at r1 = 0 in your transformed chart, no, the observer does not "strike" anything there. He continues to fall, into a region which simply does not appear, either in the exterior Schwarzschild chart or in your transformed chart. In other words, there has to be a portion of the spacetime where r < 2m. This can be shown by computing invariants, and it holds regardless of which chart you use. You can't change the physics by changing coordinate charts.

I still don't see why would need two equations.

We don't. There is only one "equation", the Einstein Field Equation. The coordinate charts are just ways of describing one particular solution to the EFE. Everything comes back to that one equation.

If you mean you don't understand why we need two coordinate charts, we only do, in this particular case, if you insist on using the Schwarzschild charts. You can pick other charts that are not singular at r = 2m, so a single chart covers the whole range of r from 0 to infinity.

To me, it seems that the mathematics of the external equation cannot describe what occurs below a singularity

If you mean the exterior Schwarzschild chart, yes, that's correct; it can't describe the region with r < 2m. But that doesn't mean the region isn't there.

so we are basically trying to change the mathematics around in such a way below the singularity as to make more sense of it.

Your transformation changes the "mathematics", but it doesn't change the physics. You can't change the physics by changing coordinate charts. The region with r < 2m is there, physically, whether it appears in your chart or not.

But if there is nothing physically different at the boundary that would change the physics according to either an external observer or a local observer that is falling through it, then I don't see why the mathematics should suddenly be applied differently there.

Because the mathematics of the exterior Schwarzschild chart doesn't describe the physics well as r -> 2m. Again, this can be shown by computing invariants; all the invariants are finite and well-defined at r = 2m. So the apparent singularity of Schwarzschild coordinates there is only apparent; it's not real. It's an artifact of those coordinates, just as the apparent singularity of longitude at the North Pole is only apparent, not real; there is no actual physical singularity at the Earth's North Pole.

If you're saying that a point singularity could not possibly describe the entire spacetime

I'm not saying that at all. I'm saying that the *exterior Schwarzschild coordinate chart* does not, as a matter of physical fact, describe the entire spacetime. And since your transformed chart covers the same spacetime region (the transformation is one-to-one), your transformed chart does not, as a matter of physical fact, describe the entire spacetime either.

how far should we open the singularity so that the entire space-time is accounted for?

I'm not sure I understand what this means.

If you mean, how can we find a chart that covers the entire spacetime (or at least the entire region that an infalling observer can explore), I and others have already given you several: the Painleve chart, the Eddington-Finkelstein chart, and the Kruskal chart. I strongly suggest that you look at them.

If you mean, how can we "adjust" either the exterior Schwarzschild chart, or your transformed chart, to "make" it cover the entire spacetime, well, coordinate transformations exist between that chart and the other charts I named.

Thanks again PeterDonis, good post.

In other words, there has to be a portion of the spacetime where r < 2m. This can be shown by computing invariants, and it holds regardless of which chart you use.

The region with r < 2m is there, physically, whether it appears in your chart or not.

Because the mathematics of the exterior Schwarzschild chart doesn't describe the physics well as r -> 2m. Again, this can be shown by computing invariants; all the invariants are finite and well-defined at r = 2m. So the apparent singularity of Schwarzschild coordinates there is only apparent; it's not real.
What do you mean by these statements? What invariants show that a portion of the spacetime must exist below 2m?

If you mean you don't understand why we need two coordinate charts, we only do, in this particular case, if you insist on using the Schwarzschild charts. You can pick other charts that are not singular at r = 2m, so a single chart covers the whole range of r from 0 to infinity.
By this, are you referring to charts that place the singularity at r1 = 0 like the one I have, which only require the external coordinate system, although it would not map any internal coordinates?

I'm not sure I understand what this means.

If you mean, how can we find a chart that covers the entire spacetime (or at least the entire region that an infalling observer can explore), I and others have already given you several: the Painleve chart, the Eddington-Finkelstein chart, and the Kruskal chart. I strongly suggest that you look at them.

If you mean, how can we "adjust" either the exterior Schwarzschild chart, or your transformed chart, to "make" it cover the entire spacetime, well, coordinate transformations exist between that chart and the other charts I named.
I am referring to the amount of spacetime that must be included below the singularity in order to be fully accounted for. Schwarzschild says 2m, others may give more or less. Mine brings it all the way down to zero. Potentially we could stretch the singularity all the way to infinity. The point at r = 0 in Schwarzschild would be opened up in the same way my point singularity opens up when transforming to Schwarzschild, or it would close further, taking along some of the surrounding spacetime depending upon which coordinate system we use, as mine does. If Schwarzschild is an arbitrary coordinate system, then 2m is most probably not the absolute amount of spacetime that must be accounted for within a singularity, or the amount of spacetime that is transformed 1 to 1 between r=0 and r = 2m in Schwarzschild to another coordinate system, so now that I have finally managed to close the singularity in a way that makes sense to me, I would not want to open it again until I see some demonstration or proof that determines how much spacetime should be accounted for within the singularity and why.

Here's something interesting. I am still looking into the coordinate systems you have been referring to, but I know Eddington's isotropic coordinates, so I was going to use them to show how much space is cut out when switching from Schwarzschild to Eddington, but found something I didn't expect. The distance transformation for Eddington is

r = r1 (1 + m / (2 r1))^2, r1 = (r - m + r sqrt(1 - 2m/r))/2

The coordinate system for Eddington's isotropic coordinates places the singularity at m/2. The interesting thing is that if we try to transform r<2m from Schwarzschild to Eddington, we get complex results. That space doesn't transform, so r<2m in Schwarzschild doesn't transform directly to Eddington at all. However, some space must exist in Eddington below m/2, but if we transform that space back from Eddington to Schwarzschild, it places those points outside the Schwarzschild singularity, in the exterior region. For instance, r1 = m/10 in Eddington, which lies below the singularity in Eddington's coordinate system, transforms to r = 3.6m in Schwarzschild, so we have a "doubling back" effect over the exterior region of Schwarzschild like a mirror.

I suppose this why two separate coordinate systems, one which applies above and the other below the singularity, would be required, to make the mathematics work out as we think they should when transforming between them. I would still say that the mathematics of a single coordinate system should apply equally in both cases, however, and with a point singularity, we would only have to transform the exterior region. But anyway, getting back to the main topic, if I were to start with Schwarzschild, say, and transform all of those same coordinates above and below the singularity to a new coordinate system, but opening up the point at the center of the Schwarzschild interior chart as well to include even more spacetime than Schwarzschild does, then since the new system accounts for even more spacetime, wouldn't that be saying that Schwarzschild cuts some of it out, so that Schwarzschild, being arbitrary to begin with, so making the spacetime covered within 2m arbitrary as well, does not fully describe the spacetime either in that case, since the new coordinate system includes more spacetime and so can describe it even more fully? In other words, how much spacetime must be included below the singularity before we can say we have covered "the entire spacetime"?

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Okay well, upon further reflection, I suppose technically I can only really say that I have mapped the exterior coordinates and closed up the interior coordinates, which is fine. My only argument there, then, is that since events that occur there cannot be mapped, measured, or even inferred according to the coordinate system of an external observer, why should external observers measure a space there to begin with? Better to just close it up to them altogether, rather than to consider that some coordinate space (not the spacetime, but actual space) exists there that contains events which can never be observed or mapped according to their coordinate system.

All observations are still the same externally. Particles follow the same paths and upon falling to the singularity, are lost forever. Nothing can ever escape from there, so there is no conflict between coordinate systems. My coordinate system says particles immediately strike the point singularity and point mass at the same time and that is all, but again, that is only as observed by external observers and according to their external time, and they can only map the proper time upon clocks all the way until they strike the singularity. If we consider that the point mass lies at the center of the singularity in Schwarzschild, falling observers must still fall in and strike the point mass, so we are still in agreement there, especially since external observers cannot really map what might take place locally within a singularity anyway.

I have stated that if the observer could pass through the point mass somehow with this new coordinate system, he would just pass through the point singularity and begin decelerating again on the other side, but even if he could pass through the mass, the gravity at that point is still infinite, so he could never really escape again anyway, so the point (again, pun intended :) ) is moot. Really, according to the external coordinate system, a star will not collapse all the way to the singularity and observers never fall all the way to it in finite external time anyway, so external observers don't even need to consider that a true singularity exists in the first place, or what would happen locally if it did, since it never fully forms according to the external coordinate system, so there is no need for them to attempt to map it.

I suppose now I should just turn my efforts toward finding solutions to the metric according to this new coordinate system. In the meantime, however, I am still curious about how much spacetime should be accounted for internally. I'm sure I could find a coordinate system that would open up the point at the center of the singularity, allowing for more spacetime than Schwarzschild does, but I suppose I would have to transform the internal Schwarzschild coordinate system in order to do so in that case, as all of the external coordinates have already been accounted for with this new coordinate system.

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PeterDonis
Mentor
What invariants show that a portion of the spacetime must exist below 2m?

(1) We can compute the proper time it takes for a freely falling observer, starting at any finite radius r > 2m, to reach r = 2m. This proper time is finite.

(2) We can compute the tidal gravity acting on the freely falling observer at r = 2m, and it is finite; in fact, it decreases with the mass of the black hole, so for a large enough black hole it is negligible.

(3) The Schwarzschild spacetime as a whole is a vacuum--that is, there is no matter or energy anywhere (we are assuming that the freely falling observer himself is too small to have any effect on the spacetime geometry). Therefore, there is nothing at r = 2m that could affect the motion of the freely falling observer.

(4) Combining the above makes it clear that the freely falling observer, after he reaches r = 2m in a finite proper time, *must* continue to fall, and therefore must pass into a region where r < 2m, and therefore there must *be* a region where r < 2m. If anything else were to happen, it would show up in one of the items above--the proper time would be infinite (but it isn't), or there would be infinite tidal gravity (but there isn't), or there would be something there to stop the motion (but there isn't).

Similar arguments show that the freely falling observer must continue all the way down to r = 0; only there does an invariant (the tidal gravity felt by the observer) become infinite. (He reaches r = 0 in a finite proper time as well, btw; only the infinite tidal gravity signals an actual physical singularity there.)

By this, are you referring to charts that place the singularity at r1 = 0 like the one I have, which only require the external coordinate system, although it would not map any internal coordinates?

No, I'm referring to the charts I named: Painleve, Eddington-Finkelstein, and Kruskal. All of them cover the entire region from r = 0 to infinity with a single chart. References for these charts:

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

http://en.wikipedia.org/wiki/Eddington–Finkelstein_coordinates

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

For the Painleve and Eddington-Finkelstein charts, I have been referring to the "ingoing" versions; there are also "outgoing" versions.

(Btw, I don't think the "Eddington isotropic chart" you refer to later on is the same as the Eddington-Finkelstein chart I'm referring to. See further comments below.)

I am referring to the amount of spacetime that must be included below the singularity in order to be fully accounted for.

The spacetime, physically, extends all the way down to r = 0. Btw, this is not a statement that depends on the Schwarzschild chart; the "r" as it's defined here appears in *all* the charts I mentioned (though not always as a coordinate in itself--in the Kruskal chart, for example, "r" is a function of the coordinates), and it can be given a coordinate-free, invariant definition. So the statement I just made is a genuine physical statement; it doesn't depend on which coordinates you use.

Schwarzschild says 2m, others may give more or less. Mine brings it all the way down to zero. Potentially we could stretch the singularity all the way to infinity.

You could come up with a coordinate chart that made it *look* like the "singularity" stretched all the way to infinity, but that wouldn't change the physics. You are spending a lot of time talking about things that don't prove anything about the physics.

I would not want to open it again until I see some demonstration or proof that determines how much spacetime should be accounted for within the singularity and why.

See the argument I gave above for why a region with r < 2m (with "r" defined as I referred to above) must exist, and why it must extend all the way down to r = 0. No amount of gymnastics with coordinate charts will change that fact. If you want more elaboration on what I said, feel free to ask questions about it.

Eddington's isotropic coordinates

Do you have a reference for these? As I said above, it doesn't look like they're the same as the Eddington-Finkelstein coordinates I referred to.

In other words, how much spacetime must be included below the singularity before we can say we have covered "the entire spacetime"?

Again, see above; physically, the spacetime extends all the way down to r = 0 (with the standard definition of "r").

PeterDonis
Mentor
the spacetime extends all the way down to r = 0 (with the standard definition of "r").

I suppose I should add something to this about what the standard definition of "r" is. The Schwarzschild spacetime is spherically symmetric, which means that a spacelike slice through it is composed of a family of nested 2-spheres, each with a slightly different area. It is extremely helpful in analyzing the physics to give each such 2-sphere a convenient label; one way to do that is to label each such 2-sphere with a number $r$, such that the physical area of the 2-sphere is equal to $4 \pi r^2$. The number $r$, so defined, is the "radial coordinate" that appears in the Schwarzschild charts (exterior and interior). It also appears as the "radial coordinate" in the Painleve and Eddington-Finkelstein charts. In the Kruskal chart, it is not a coordinate, but it is a well-defined function of the coordinates, so we can tell which 2-sphere any point in the Kruskal chart lies on by evaluating the function.

The reason I go into this, other than just for clarity, is that it shows one other important fact that I forgot to mention in my last post: the physical area of the 2-sphere at r = 2m is, by the formula above, $16 \pi m^2$. That is, *not* zero. So r = 2m is *not* a "point"; it's a 2-sphere (technically, r = 2m at a given "instant of time" is a 2-sphere). That's a coordinate-independent, physical fact. That being the case, one can't evade the conclusion that a freely falling observer must fall into a region where r < 2m, by saying that somehow the locus r = 2m is "only a point", or that "the spacetime ends there", or something like that. The locus r = 2m is a 2-sphere, and it smoothly connects to its neighbors on both sides in the family of nested 2-spheres.

(1) We can compute the proper time it takes for a freely falling observer, starting at any finite radius r > 2m, to reach r = 2m. This proper time is finite.
I don't see that this one demonstrates anything. We wouldn't expect that an observer falling from a finite radius would take an infinite amount of proper time to reach a singularity to begin with, not even in Newton, especially as his locally measured speed increases. What is surprising, however, is that it takes an infinite amount of external time as the externally measured coordinate speed slows to zero. But after realizing that external observers measure an infinite time, and knowing that rulers also shrink to zero at the singularity according to the external observers, then similarly, looking at it from this perspective, we might then expect to measure an infinite ruler distance from a finite radius to the singularity, using local rulers laid end to end from the singularity to r, but that is also finite, as only the very end of the last ruler shrinks to zero. There are potentially many measurements we might expect to be infinite or finite, depending upon who's perspective we take, but they don't necessarily directly reflect upon spacetime below the singularity.

(2) We can compute the tidal gravity acting on the freely falling observer at r = 2m, and it is finite; in fact, it decreases with the mass of the black hole, so for a large enough black hole it is negligible.
I don't see how a radial tidal gradient at the singularity would tell us anything about below it. But are you referring to a radial or tangent tidal gradient? A finite tangent tidal gradient might say something, since it might infer that a 2-sphere exists there. I'm not sure, though, I would have to see it. Could you clarify the tidal gravity you are referring to, describe how you are thinking about it, and show the calculation for that?

(3) The Schwarzschild spacetime as a whole is a vacuum--that is, there is no matter or energy anywhere (we are assuming that the freely falling observer himself is too small to have any effect on the spacetime geometry). Therefore, there is nothing at r = 2m that could affect the motion of the freely falling observer.
Well, we would already have to have assumed that more spacetime exists there in order to apply it as a vacuum, so the reasoning here is somewhat circular.

Similar arguments show that the freely falling observer must continue all the way down to r = 0; only there does an invariant (the tidal gravity felt by the observer) become infinite. (He reaches r = 0 in a finite proper time as well, btw; only the infinite tidal gravity signals an actual physical singularity there.)
Right, if we already assume that more spacetime exists there, and that the star has collapsed to a point mass at the center, then the freefaller will continue falling to r=0 in the same way that the surface of the star did. If the proper time is still finite there, though, then by your reasoning for #1, the full spacetime has still not yet been accounted for, so the central point should be opened up to include even more spacetime as well, using a different coordinate system, such that the freefaller's time becomes infinite upon reaching the center, right? Also, at what point does a singularity at 2m become a singularity at the central point, and how does that look to the freefaller?

The spacetime, physically, extends all the way down to r = 0. Btw, this is not a statement that depends on the Schwarzschild chart; the "r" as it's defined here appears in *all* the charts I mentioned (though not always as a coordinate in itself--in the Kruskal chart, for example, "r" is a function of the coordinates), and it can be given a coordinate-free, invariant definition. So the statement I just made is a genuine physical statement; it doesn't depend on which coordinates you use.
I thought about trying to work out a coordinate free system for the spacetime, but even then, it would still involve two separate and distinct spacetime systems, one above and one below the singularity, so I would still have to work through assumptions about the spacetime below a singularity (like that it exists for starters ;) )in order to apply a coordinate system there in the first place.

You could come up with a coordinate chart that made it *look* like the "singularity" stretched all the way to infinity, but that wouldn't change the physics. You are spending a lot of time talking about things that don't prove anything about the physics.
Right. The physics is what is important, not coordinate effects. That's why I am interested in the tidal gravity and what that implies. What I meant about opening up the singularity to infinity, though, was not so much the singularity itself, but opening up the point at its center, at r = 0, to include more spacetime. For instance, my coordinate system only includes the external coordinate system with a point singularity at r1=0. The spacetime appears to stop there in that particular coordinate system. But I could apply a different coordinate system that opens that point up so that the singularity lies at 2m, regaining Schwarzschild. I could then apply another internal coordinate system that opens up the point at the center to say, a radius of m/10, allowing even more spacetime to exist within the singularity. But likewise, I could also close the point in my coordinate system further by applying the coordinate system r1 = r (1 - 3m/ r) to Schwarzschild instead of r1 = r (1 - 2m/r), where r1 = 0 now lies at r = 3m instead of 2m, cutting out some of the external spacetime as well. This ability actually works for and against what I have been arguing about whether internal spacetime exists, since according to these coordinates, the external spacetime from 2m to 3m no longer exists either. But looking at it the other way, I shouldn't be able to just keep opening up the point at the center of the singularity by changing the coordinate system either, it should end somewhere, surely. But as I said, I realize now that I can really only say that I have mapped the external spacetime and only cut out what cannot be mapped by external observers.

I suppose to better demonstrate what I am asking about with opening up the central point to add more spacetime, I'll need to show an example. What is the internal Schwarzschild metric? I always have trouble finding it. I can see how the spacetime metric changes from timelike to spacelike below the singularity, but I'm not sure what is meant by the statements that t and r themselves changes places. In the internal Schwarzschild metric, though, t is still time and r distance, so I can just apply them directly from the internal metric, right?

Do you have a reference for these? As I said above, it doesn't look like they're the same as the Eddington-Finkelstein coordinates I referred to.
Sure. Here is a Wiki article on the Schwarzschild metric. They discuss Eddington's isotropic coordinates about half way down.

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PeterDonis
Mentor
I don't see that this one demonstrates anything.

I'm not claiming that any of the factors I gave are sufficient by themselves; it's the *combination* of them that is crucial.

A general comment, which I've made before, but maybe not quite in this form: You are approaching this whole question backwards. You are starting from a particular coordinate chart, *assuming* that it has to describe all the physics, and then using that assumption to determine what the physics is like.

But the physics is not governed by coordinates; it's governed by the Einstein Field Equation, which is an invariant, coordinate-free equation. The right way to approach this whole question is to first figure out the physics--what solution of the EFE corresponds to the physical scenario you are interested in--and *then* see how well any particular coordinate chart describes that physics.

Both. The tidal gravity at r = 2m is finite, and becomes negligible for a large enough black hole, in *all* directions, not just radial. The calculation is simple: it just amounts to computing the appropriate components of the Riemann curvature tensor, which is straightforward but tedious (and these days, is best done by computer, for better accuracy with the algebra).

The result is that, for an object which extends for a proper length $L$ in the appropriate direction, the radial tidal gravity is, in general, $- \left( 2M / r^3 \right) L$, and the tangential tidal gravity is, in general, $\left( M / r^3 \right) L$. The negative sign means tidal tension--the tidal gravity tries to pull things apart--and the positive sign means tidal compression--the tidal gravity tries to push things together.

At r = 2m, these become $- L / 4 M^2$ and $L / 8 M^2$, which are obviously finite and get smaller as M increases. (I'm using units where c = G = 1. The above also assumes that L is small compared to M, so that the curvature components can be assumed constant over the length of the object.)

Well, we would already have to have assumed that more spacetime exists there in order to apply it as a vacuum

You're missing the point. We don't *assume* that more spacetime exists in the region r < 2m; we *find it out* it by looking at the solution of the EFE. The only assumptions (aside from the EFE itself) are that the spacetime is a vacuum, and that it is spherically symmetric. All the rest is derived as theorems from those starting assumptions--*including* the fact that the spacetime must extend all the way down to the curvature singularity at r = 0. Once again, no amount of gymnastics with coordinates will change that, because you don't solve the EFE by assuming that a particular coordinate chart covers all the spacetime or describes all the physics; that's backwards, as I noted above.

Right, if we already assume that more spacetime exists there...

Same comment as above. We don't *assume* this; we *derive* it as a theorem from the assumptions I gave above, and *only* those assumptions.

If the proper time is still finite there, though, then by your reasoning for #1, the full spacetime has still not yet been accounted for

Nope. At r= 0, the curvature is infinite, not finite. It's the infinite curvature there that signals that there isn't "any more spacetime". Once again, the deduction of "more spacetime" is only valid when the *combination* of all the factors I gave holds true; as soon as one is violated (which it is at r = 0, since the curvature becomes infinite there), the deduction no longer holds.

I thought about trying to work out a coordinate free system for the spacetime, but even then, it would still involve two separate and distinct spacetime systems, one above and one below the singularity

No, it doesn't. In fact, it isn't even required for a non-coordinate-free system, since, as I've noted several times now, there are charts that cover the entire region from r = 0 to infinity with no singularities (except of course at r = 0 itself).

Right. The physics is what is important, not the coordinates.

And yet you keep on drawing deductions that only apply in your particular set of coordinates, and claiming that they tell us something about the actual physics.

What I meant about opening up the singularity to infinity, though, was not so much the singularity itself, but opening up the point at its center, at r = 0, to include more spacetime.

See above; the curvature becomes infinite at r = 0, and that's what tells us that there is no more spacetime "beyond" that point.

Both. The tidal gravity at r = 2m is finite, and becomes negligible for a large enough black hole, in *all* directions, not just radial. The calculation is simple: it just amounts to computing the appropriate components of the Riemann curvature tensor, which is straightforward but tedious (and these days, is best done by computer, for better accuracy with the algebra).

The result is that, for an object which extends for a proper length $L$ in the appropriate direction, the radial tidal gravity is, in general, $- \left( 2M / r^3 \right) L$, and the tangential tidal gravity is, in general, $\left( M / r^3 \right) L$. The negative sign means tidal tension--the tidal gravity tries to pull things apart--and the positive sign means tidal compression--the tidal gravity tries to push things together.

At r = 2m, these become $- L / 4 M^2$ and $L / 8 M^2$, which are obviously finite and get smaller as M increases. (I'm using units where c = G = 1. The above also assumes that L is small compared to M, so that the curvature components can be assumed constant over the length of the object.)
This is the part that I'm interested in. It demonstrates the actual local physics that takes place there, more than just coordinate transformations. If I use a local Newtonian gravity of a = G M / r^2 that acts upon each end of a static radial ruler of length dr, where the acceleration would be as measured upon an accelerometer, then I get a radial tidal gradient of - 2 G M dr / r^3 as you've got, right. Of course, in Newton, the singularity for a point mass only exists at the central point anyway, at r = 0, and would be finite everywhere else. But if I use the locally measured GR gravity of a = (G M / r^2) / sqrt(1 - 2 m / r), I get a radial tidal gradient of G M (3 m - 2 r) dr/ (r^3 sqrt(1 - 2 m / r) (r - 2 m)), which becomes infinite at r = 2m, so a singularity physically exists there, even according to the local static observers. They can physically place rulers locally end to end radially from the singularity, which will give a finite ruler distance, although the ruler must be infinitely strong to remain intact, so they can physically measure their ruler distance from the singularity at 2m. They can physically measure the acceleration there with an accelerometer, or by how much constant acceleration must be applied to remain static. They can physically measure the stress along a local rod or spring. Now that I think about it, though, just measuring an infinite local acceleration there indicates a singularity to begin with, being the reason nothing can escape from there, as infinite acceleration would need to be applied in order to escape, but it still would have been interesting if the tidal effect turned out to be finite.

EDIT - I just realized this tidal effect I found is for a rod that is released at local r while in freefall, not held static, since that would produce a positive tidal compression, as one end of the rod would have to be accelerated to hold it in place. What I posted is the differential between the local accelerations that act upon each end of a rod that has just been let go at r, a moment after it has regained its normal non-compressed inertial length.

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PeterDonis
Mentor
This is the part that I'm interested in. It demonstrates the actual local physics that takes place there, more than just coordinate transformations.

Good.

if I use the locally measured GR gravity of a = (G M / r^2) / sqrt(1 - 2 m / r), I get a radial tidal gradient of G M (3 m - 2 r) dr/ (r^3 sqrt(1 - 2 m / r) (r - 2 m)), which becomes infinite at r = 2m

You're assuming that the tidal gravity [Edit: meaning "tidal gravity acting on a freely falling object"] is the gradient of the "acceleration due to gravity" [meaning "proper acceleration required to hold an object static"]. That's only true in the Newtonian approximation (i.e., very far away from the source of gravity). It's not true in general in GR. The tidal gravity is given by the Riemann curvature tensor, as I said; and computing the components of that gives the result I gave. It is true that this result "looks like" the gradient of the *Newtonian* "acceleration due to gravity"; that's an interesting mathematical fact about Schwarzschild spacetime, but it's not how the result is actually derived.

They can physically place rulers locally end to end radially from the singularity, which will give a finite ruler distance, although the ruler must be infinitely strong to remain intact, so they can physically measure their ruler distance from the singularity at 2m.

This is true, except for the incorrect statement that r = 2m is a physical singularity. See above.

They can physically measure the acceleration there with an accelerometer, or by how much constant acceleration must be applied to remain static.

Yes, and as you say, this acceleration goes to infinity as r -> 2m.

though, just measuring an infinite local acceleration there indicates a singularity to begin with, being the reason nothing can escape from there, as infinite acceleration would need to be applied in order to escape

The infinite acceleration does tell you something important about the location r = 2m, but the important thing it tells you is *not* that r = 2m is a "physical singularity".

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PeterDonis
Mentor
Cleaning up a couple of things I forgot to respond to...

What is the internal Schwarzschild metric? I always have trouble finding it. I can see how the spacetime metric changes from timelike to spacelike below the singularity, but I'm not sure what is meant by the statements that t and r themselves changes places. In the internal Schwarzschild metric, though, t is still time and r distance, so I can just apply them directly from the internal metric, right?

The line element in the interior Schwarzschild chart looks the same as it does in the exterior chart, if we keep the standard names for the coordinates:

$$ds^2 = - \left( 1 - \frac{2m}{r} \right) dt^2 + \frac{1}{1 - \frac{2m}{r}} dr^2 + r^2 d\Omega^2$$

However, since r < 2m, the signs of the first two terms are switched from what they are in the exterior chart. So a less confusing way to write this line element would be to define $T = r$ and $R = t$, to obtain

$$ds^2 = - \frac{1}{\frac{2m}{T} - 1} dT^2 + \left(\frac{2m}{T} - 1 \right) dR^2 + T^2 d\Omega^2$$

which at least makes the signs of the terms explicit. However, you still have to remember that the T coordinate, *not* the R coordinate, is "radial", in the sense that the area of a 2-sphere at T is $4 \pi T^2$, whereas R has no connection at all with the areas of 2-spheres. In other words, T is both a kind of "time coordinate" *and* a kind of "radial coordinate". One way of stating what this is telling you is to say that, in the region r < 2m, falling radially inward is the same as moving into the future; that's why, once you are in the region r < 2m, you can't avoid the singularity at r = 0, any more than you can avoid tomorrow.

Of course, you could avoid all of this confusion by just using a chart such as the Painleve chart (which is my personal preference), which is not singular at r = 2m, so that the same form of the line element works the same way all the way down to r = 0. However, even that chart has some counterintuitive properties; the most important one is that, in the region r < 2m, *all four* Painleve coordinates are spacelike; radial timelike lines (such as the worldline of a freely falling observer) can only be described by line elements with nonzero dT *and* dr (where T here is now the Painleve time coordinate).

Here is a Wiki article on the Schwarzschild metric. They discuss Eddington's isotropic coordinates about half way down.

Ah, I see. These are usually just called "isotropic coordinates" (or "isotropic coordinates on Schwarzschild spacetime" if more precision is needed). I wasn't aware that Eddington was the one who originally proposed them; interesting.

As far as the coverage of these coordinates, yes, they only cover the exterior portion of the spacetime, i.e., the portion with r > 2m in the standard Schwarzschild coordinates. The region R > m/2 (where R is the isotropic radial coordinate) and the region R < m/2 *both* cover the *same* exterior region of the spacetime; that is, this coordinate chart maps *two* values of the R coordinate to each spacetime point. (One of the exercises in MTW is to find an explicit transformation that shows this, by mapping R -> 0 to R -> infinity and vice versa, while leaving the line element unchanged.)

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PeterDonis
Mentor
What is surprising, however, is that it takes an infinite amount of external time as the externally measured coordinate speed slows to zero.

One other comment I meant to make about this but forgot to earlier: the "externally measured coordinate speed" here is *not* the same as "the speed that the infalling observer would be observed to have, by static observers that he passes as he approaches r = 2m". If we imagine a family of static observers, all using rocket thrust to "hover" at values of r closer and closer to 2m (and requiring increasing rocket thrust to do so), these observers will see a freely falling observer pass them, falling inward, at speeds approaching the speed of light; i.e., the freely falling observer is *not* "slowing down" according to local static observers; he is "speeding up", just as you would expect from the fact that the black hole's gravity pulls inward. This is another reason, btw, why the freely falling observer can't just "stop" at r = 2m.

PeterDonis
Mentor
You're assuming that the tidal gravity [Edit: meaning "tidal gravity acting on a freely falling object"] is the gradient of the "acceleration due to gravity" [meaning "proper acceleration required to hold an object static"].

On thinking it over, the above may not be the best way to state it. A better way might be: you're assuming that the gradient of the "acceleration due to gravity" is just the partial derivative with respect to r. In a curved spacetime, that's not correct; you have to include extra terms involving the Christoffel symbols. In other words, you have to take a *covariant* gradient, not an ordinary gradient. (Some people don't like using the term "gradient" for the covariant version; that's a matter of terminology, not physics.)

You *can* view the tidal gravity as a *covariant* gradient of the "acceleration due to gravity"; that's what the Riemann tensor is. The general formula for the Riemann tensor is:

$$R^a_{bcd} = \partial_c \Gamma^a_{db} - \partial_d \Gamma^a_{cb} + \Gamma^a_{ce} \Gamma^e_{db} - \Gamma^a_{de} \Gamma^e_{cb}$$

The radial tidal gravity on an object that is momentarily static (i.e., momentarily "hovering" at radius r) is $R^r_{trt} u^t u^t$, where $u^t$ is the only nonzero component of the object's 4-velocity (because it is "hovering") and is given by $u^t = 1 / \sqrt{|g_{tt}|}$. So the radial tidal gravity is $R^r_{trt} / \left( 1 - 2m / r \right)$.

The Riemann tensor component is (substituting appropriately into the above formula)

$$R^r_{trt} = \partial_r \Gamma^r_{tt} - \partial_t \Gamma^r_{rt} + \Gamma^r_{re} \Gamma^e_{tt} - \Gamma^r_{te} \Gamma^e_{rt}$$

The first term, $\partial_r \Gamma^r_{tt}$, more or less corresponds to what you calculated. But the correct calculation has to include the other terms as well.

The second term is zero because the spacetime is static, so there are no nonzero time derivatives; and the sums over the e index in the third and fourth terms collapse to one term each because only certain Christoffel symbols are nonzero in this spacetime, so we have

$$R^r_{trt} = \partial_r \Gamma^r_{tt} + \Gamma^r_{rr} \Gamma^r_{tt} - \Gamma^r_{tt} \Gamma^t_{rt}$$

If you calculate all this out, it comes to

$$R^r_{trt} = - \frac{2m}{r^3} \left( 1 - \frac{2m}{r} \right)$$

so the final tidal gravity comes out to be

$$R^r_{trt} u^t u^t = - \frac{2m}{r^3}$$

as I said.

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The line element in the interior Schwarzschild chart looks the same as it does in the exterior chart, if we keep the standard names for the coordinates:

$$ds^2 = - \left( 1 - \frac{2m}{r} \right) dt^2 + \frac{1}{1 - \frac{2m}{r}} dr^2 + r^2 d\Omega^2$$

However, since r < 2m, the signs of the first two terms are switched from what they are in the exterior chart. So a less confusing way to write this line element would be to define $T = r$ and $R = t$, to obtain

$$ds^2 = - \frac{1}{\frac{2m}{T} - 1} dT^2 + \left(\frac{2m}{T} - 1 \right) dR^2 + T^2 d\Omega^2$$

which at least makes the signs of the terms explicit. However, you still have to remember that the T coordinate, *not* the R coordinate, is "radial", in the sense that the area of a 2-sphere at T is $4 \pi T^2$, whereas R has no connection at all with the areas of 2-spheres. In other words, T is both a kind of "time coordinate" *and* a kind of "radial coordinate". One way of stating what this is telling you is to say that, in the region r < 2m, falling radially inward is the same as moving into the future; that's why, once you are in the region r < 2m, you can't avoid the singularity at r = 0, any more than you can avoid tomorrow.
I've been staring at this post for a couple of days now and I'm still not sure how I should be thinking about it. I was thinking about the internal Schwarzschild solution before, but I forgot that is for beneath the surface of a body, completely different from this. I've never seen this before and I have a couple of questions. For one, this metric is now based upon T, so dT and dR now have co-efficients that are functions of time, right? But I thought the metric was supposed to be time independent. Second, who's time is this T? In the external metric, we can find the tangent and radial components of the scalar speed of a photon according to an arbitrary static observer that happens to coincide with the photon, and transform those coordinates to those of the distant observer to get the metric. So dr and dt in the external metric are the change in radius and time according to the distant observer. But who's are they in the internal metric? There aren't even any static observers internally, so we could only transform the internal metric between freefallers.

Of course, you could avoid all of this confusion by just using a chart such as the Painleve chart (which is my personal preference), which is not singular at r = 2m, so that the same form of the line element works the same way all the way down to r = 0. However, even that chart has some counterintuitive properties; the most important one is that, in the region r < 2m, *all four* Painleve coordinates are spacelike; radial timelike lines (such as the worldline of a freely falling observer) can only be described by line elements with nonzero dT *and* dr (where T here is now the Painleve time coordinate).
Yes, that would be better I think. I've still barely started studying those other coordinate systems though. I need to get on that.

As far as the coverage of these coordinates, yes, they only cover the exterior portion of the spacetime, i.e., the portion with r > 2m in the standard Schwarzschild coordinates. The region R > m/2 (where R is the isotropic radial coordinate) and the region R < m/2 *both* cover the *same* exterior region of the spacetime; that is, this coordinate chart maps *two* values of the R coordinate to each spacetime point. (One of the exercises in MTW is to find an explicit transformation that shows this, by mapping R -> 0 to R -> infinity and vice versa, while leaving the line element unchanged.)
Would I just perform precisely the same couple of operations on this exterior coordinate system as you did for Schwarzschild to gain the interior metric for the isotropic coordinates?

One other comment I meant to make about this but forgot to earlier: the "externally measured coordinate speed" here is *not* the same as "the speed that the infalling observer would be observed to have, by static observers that he passes as he approaches r = 2m". If we imagine a family of static observers, all using rocket thrust to "hover" at values of r closer and closer to 2m (and requiring increasing rocket thrust to do so), these observers will see a freely falling observer pass them, falling inward, at speeds approaching the speed of light; i.e., the freely falling observer is *not* "slowing down" according to local static observers; he is "speeding up", just as you would expect from the fact that the black hole's gravity pulls inward. This is another reason, btw, why the freely falling observer can't just "stop" at r = 2m.
Right, the local static observer measures a greater speed. One thing I just realized, though, is that conservation of energy, from what I've been able to work out anyway, shows that static observers that coincide at the same place as a particle travelling past them would measure precisely c for the scalar speed of the particle right at 2m, regardless of whether the particle is massive or massless, and regardless of its original speed or direction of travel when the partilce started to freefall. So if all particles travel at c as they approach the event horizon according to a static observer there, then so would a freefaller in the same way, and the passing of the freefaller's proper time there would fall to zero according to both static observers, distant and local, and probably any static observers in between. Both would say that the ticking of his clock stops at the event horizon.

If you calculate all this out, it comes to

$$R^r_{trt} = - \frac{2m}{r^3} \left( 1 - \frac{2m}{r} \right)$$

so the final tidal gravity comes out to be

$$R^r_{trt} u^t u^t = - \frac{2m}{r^3}$$
I don't know the notations you were using, but can understand the algebraic form. The second to last there looks like the gravity gradient the distant observer might measure. I haven't worked that out though. I'm not sure how you go from there to the last. The gradient I calculated was just finding the local gravity gradient according to local static observers. The acceleration of gravity locally is just a = ( 1 - (v_r'/c)^2)(G M / r^2) / sqrt(1 - 2 m / r), where v_r' is the locally measured radial speed, again worked out according to conservation of energy. Upon releasing a rod of length dr at r, v_r' = 0, and after giving the rod a moment to regain its inertial length since it was compressed when static (or we could just throw it upward from a slightly lower radius such that it comes to rest, then turns around to fall again at r), we should be able to just compute the differential between the local accelerations acting upon each end of the rod since SR is valid locally. Rindler would probably be more accurate, actually, but starting from rest with an infinitesimal length of the rod, I don't see why we couldn't just take the difference directly within the local spacetime using the local measurements.

PeterDonis
Mentor
I was thinking about the internal Schwarzschild solution before, but I forgot that is for beneath the surface of a body, completely different from this.

Sorry, I misinterpreted the term "interior Schwarzschild solution". I always take that to mean what I wrote, the interior *vacuum* solution for the region r < 2m of a black hole; but I should have remembered that it can also mean the solution for the interior of a massive gravitating body, like a planet or star.

I've never seen this before and I have a couple of questions. For one, this metric is now based upon T, so dT and dR now have co-efficients that are functions of time, right?

Yes, in the sense that they are functions of "T", which is the renamed "r" coordinate. But remember that T is not just a "time" coordinate, it's also a "radial" coordinate. See further comments below.

But I thought the metric was supposed to be time independent.

That's actually *not* an initial assumption that is used to derive the Schwarzschild metric. The only initial assumptions (as I think I've posted before in this thread) are that the spacetime is vacuum and spherically symmetric. The fact that the metric turns out to be time-independent, at least in the region r > 2m, is a *consequence* of the assumptions plus the EFE. And you are right that it is *not* true in the region r < 2m; the metric in that region is *not* time-independent in the relevant sense.

It's worth expanding a bit on what "in the relevant sense" means. The following Wikipedia pages defines what a "static" spacetime is:

http://en.wikipedia.org/wiki/Static_spacetime

I won't go into too much technical detail, but the key point about Schwarzschild spacetime is that the vector field $\partial / \partial t$ is a Killing vector field on the entire spacetime (i.e., from r = 0 to r = infinity). Here $t$ is the standard Schwarzschild "time" coordinate (the one that I redefined as "R" in my restatement of the interior metric). However, the vector field $\partial / \partial t$ is only timelike in the region r > 2m. At r = 2m, it is null; and in the region r < 2m, it is spacelike. (In that region we would call it $\partial / \partial R$ in my revised coordinate nomenclature; but it's still not a "radial" coordinate there.) The spacetime is only static in the region where the Killing vector field is timelike, i.e., in the region r > 2m.

What this means physically is that, for r <= 2m, there is no such thing as a "hovering" observer--i.e., it is not possible to have an observer that stays at a constant r (or T in my revised nomenclature for the interior region) for any finite length of time. You appear to realize this, since you say so further on in your post. But you may not have realized that this is equivalent to saying that the metric is no longer static; having a static (time-independent) metric is equivalent to having a family of observers who are static.

Second, who's time is this T?

There is no obvious observer whose proper time is given by T in the interior region. An ordinary infalling observer can have a worldline where only T (i.e., r) varies, since such a curve is timelike; but if that observer is freely falling, his proper time will not be given by T (that's obvious from the metric I wrote down). I suppose one could construct an accelerated radial worldline such that an observer traveling on it would have his proper time given directly by T; but I haven't tried to do that explicitly.

So dr and dt in the external metric are the change in radius and time according to the distant observer. But who's are they in the internal metric? There aren't even any static observers internally, so we could only transform the internal metric between freefallers.

That's correct. But you can still construct a local inertial frame about any event on any free-faller's worldline, and transform the coordinates in such a frame to and from the global (t, r) or (R, T) coordinates using the metric. You don't need static observers to do that.

Would I just perform precisely the same couple of operations on this exterior coordinate system as you did for Schwarzschild to gain the interior metric for the isotropic coordinates?

There is no "interior metric" for the isotropic coordinates; that's the point. There is a region 0 < R < m/2 in isotropic coordinates (R = m/2 is the black hole horizon, corresponding to r = 2m in the standard Schwarzschild coordinates), but it doesn't cover the region 0 < r < 2m in the standard Schwarzschild coordinates; it covers the region r > 2m. In other words, the region 0 < R < m/2 and the region m/2 < R < infinity in isotropic coordinates cover the *same* region of the spacetime; there are *two* values of the isotropic R coordinate assigned to each spacetime point in that region (i.e., to each single value of r > 2m in the standard Schwarzschild coordinates).

The coordinate transformation I referred to when talking about the MTW homework problem is a transformation that takes the region m/2 < R < infinity to the region 0 < R < m/2, and vice versa, in isotropic coordinates. It's not a transformation between isotropic and any other kind of coordinates. (Of course those exist too, for both regions of the isotropic coordinates.)

One thing I just realized, though, is that conservation of energy, from what I've been able to work out anyway, shows that static observers that coincide at the same place as a particle travelling past them would measure precisely c for the scalar speed of the particle right at 2m, regardless of whether the particle is massive or massless, and regardless of its original speed or direction of travel when the partilce started to freefall.

If there *were* such static observers. But there aren't. What this reasoning actually shows is that, as any free-falling object passes the horizon, regardless of its mass or intial speed, it will see the horizon itself moving outward past it at the speed of light.

the passing of the freefaller's proper time there would fall to zero according to both static observers, distant and local, and probably any static observers in between. Both would say that the ticking of his clock stops at the event horizon.

No, this does not follow. Timelike worldlines (which do not have zero lapse of "proper time" along them) can still pass through r = 2m on the way down. However, it is impossible for Schwarzschild coordinates to show this directly, since they are singular at r = 2m. That's one reason why the other coordinate charts I mentioned are useful; they *can* show it directly.

The gradient I calculated was just finding the local gravity gradient according to local static observers.

No, it wasn't. What you calculated was the *partial derivative with respect to r* of the proper acceleration felt by static observers. But in a curved spacetime, the partial derivative is *not* the same as the physical gradient. That's what all the extra terms in the equations I wrote down are for; to correct for the fact that the spacetime is curved and therefore has to be described by curvilinear coordinates.

If you want to see how this works in a simpler case, try computing partial derivatives with respect to polar coordinates on a flat plane. You will find that partial derivatives with respect to the angular coordinate, theta, are *not* the same as gradients with respect to tangential distance. What's going on in curved spacetime is the same sort of thing, but it can happen even with ordinary "distance" coordinates, not just angular ones.