I Is a Manifold with a Boundary Considered a True Manifold?

[Nullstein]

But such cases typically have unnessessary geodesic incompleteness. So, for physical plausibility (removal of inessential singularities), you would extend the manifold.

While this is a reasonable physical assumption, it needn't be true in general. There are two ways in which an inextendible geodesic $\gamma:(-\infty,t_0)\rightarrow M$ (affinely parametrized) can fail to be complete (assuming a Lipschitz condition):

1. The solution of the geodesic equation blows up (in a coordinate system) as $t\rightarrow t_0$. In that case, one can't extend the spacetime.
2. The solution converges as $t\rightarrow t_0$.

In the second case, one may add the limit $\gamma(t_0) := \lim_{t\rightarrow t_0}\gamma(t)$ (yes, I'm being sloppy with the math here) to the spacetime manifold. The resulting set may have the structure of a manifold with boundary. In other words, geodesics may come to an end after finite proper time and end on the boundary of the manifold. I don't know any examples though. While such situations may be pathological, they are not strictly excluded.

 

[PAllen]

While this is a reasonable physical assumption, it needn't be true in general. There are two ways in which an inextendible geodesic $\gamma:(-\infty,t_0)\rightarrow M$ (affinely parametrized) can fail to be complete (assuming a Lipschitz condition):

1. The solution of the geodesic equation blows up (in a coordinate system) as $t\rightarrow t_0$. In that case, one can't extend the spacetime.
2. The solution converges as $t\rightarrow t_0$.

In the second case, one may add the limit $\gamma(t_0) := \lim_{t\rightarrow t_0}\gamma(t)$ (yes, I'm being sloppy with the math here) to the spacetime manifold. The resulting set may have the structure of a manifold with boundary. In other words, geodesics may come to an end after finite proper time and end on the boundary of the manifold. I don't know any examples though. While such situations may be pathological, they are not strictly excluded

What I was referring to was a manifold with boundary not at conformal infinity, equipped with a pseudo-Reimannian metric, defined on the boundary. In this case, every geodesic reaching the boundary is incomplete, but there is no physical reason for this. That the geodesic is defined on the boundary rules out any "true" singularity. Thus the manifold with metric can be extended. And this would be necessary for physical plausibility.

 

[Nullstein]

What I was referring to was a manifold with boundary not at conformal infinity, equipped with a pseudo-Reimannian metric, defined on the boundary. In this case, every geodesic reaching the boundary is incomplete, but there is no physical reason for this. That the geodesic is defined on the boundary rules out any "true" singularity. Thus the manifold with metric can be extended. And this would be necessary for physical plausibility.

But it's not guaranteed that the geodesic will be extendible. The geodesic may end on the boundary, yet the solution to the geodesic equation may not be extendible beyond the boundary. A compact interval $[t_1,t_2]$ may already be the maximal interval of existence of the solution.

 

[PAllen]

But it's not guaranteed that the geodesic will be extendible. The geodesic may end on the boundary, yet the solution to the geodesic equation may not be extendible beyond the boundary. A compact interval $[t_1,t_2]$ may already be the maximal interval of existence of the solution.

How is this possible? For example, if a continuous function of reals to reals is defined on a closed interval, it is always possible to continuously extend it. This is trivially false, in general, for an open interval. To my knowledge, the same would be true for geodesics in GR, excluding boundaries at conformal infinity. Can you provide any example of what you claim, or a clear argument or link to proof that it’s possible.

 

[Nullstein]

How is this possible? For example, if a continuous function of reals to reals is defined on a closed interval, it is always possible to continuously extend it. This is trivially false, in general, for an open interval. To my knowledge, the same would be true for geodesics in GR, excluding boundaries at conformal infinity. Can you provide any example of what you claim, or a clear argument or link to proof that it’s possible.

You can always extend a curve, but the extension may no longer solve a particular differential equation, in this case the geodesic equation. There are existence and uniqueness theorems that guarantee the existence of solutions of differential equations. In particular, the Picard-Lindelöf theorem is relevant here and it guarantees the existence of a local solution of an ODE. Once you have a local solution, you can try to extend it further, but this may not always be possible. In particular, the solution may blow up after finite time or it may just not be extendible beyond a finite time, because no extension of the solution beyond a particular finite time $t_0$ may be able to solve the geodesic equation. The first case corresponds to a singularity and the second case corresponds to a boundary. The geodesic ends without developing a singularity. It ends, because no continuous extension satisfies the geodesic equation.

 

[PAllen]

You can always extend a curve, but the extension may no longer solve a particular differential equation, in this case the geodesic equation. There are existence and uniqueness theorems that guarantee the existence of solutions of differential equations. In particular, the Picard-Lindelöf theorem is relevant here and it guarantees the existence of a local solution of an ODE. Once you have a local solution, you can try to extend it further, but this may not always be possible. In particular, the solution may blow up after finite time or it may just not be extendible beyond a finite time, because no extension of the solution beyond a particular finite time $t_0$ may be able to solve the geodesic equation. The first case corresponds to a singularity and the second case corresponds to a boundary. The geodesic ends without developing a singularity. It ends, because no continuous extension satisfies the geodesic equation.

I still don't buy it. You can extend manifold and metric such that the geodesic is extensible. I don't find this believable without an example or an existence proof. I have seen no hint that such a thing is possible in any GR literature I have seen. (Also, the most common definition of singularity in GR is the presence of incomplete inextensible geodesics, so either case would be a singularity by the standard definition).

 

[Nullstein]

I still don't buy it. You can extend manifold and metric such that the geodesic is extensible. I don't find this believable without an example or an existence proof. I have seen no hint that such a thing is possible in any GR literature I have seen.

It's just simple ODE theory, e.g. see this Wikipedia link:

In the case that $x_\pm \neq \pm\infty$, there are exactly two possibilities
- explosion in finite time: $\limsup_{x \to x_\pm} \|y(x)\| \to \infty$
- leaves domain of definition: $\lim_{x \to x_\pm} y(x)\ \in \partial \bar{\Omega}$

 

[PeterDonis]

It's just simple ODE theory, e.g. see this Wikipedia link:

The second alternative requires that the curve leaves the domain in which $F$ is defined. But for the geodesic equation, $F$ is a function composed entirely of the metric and its derivatives, and any such function is defined everywhere the metric and its derivatives are defined, i.e., everywhere in the spacetime. So it is impossible for this case to be realized for the geodesic equation.

 

[Nullstein]

The second alternative requires that the curve leaves the domain in which $F$ is defined. But for the geodesic equation, $F$ is a function composed entirely of the metric and its derivatives, and any such function is defined everywhere the metric and its derivatives are defined, i.e., everywhere in the spacetime. So it is impossible for this case to be realized for the geodesic equation.

No, we're exactly in the situation of a manifold with boundary. $\bar\Omega$ is the manifold with boundary, $\Omega$ is its interior and $\partial\bar\Omega$ is the boundary. ODE theory requires $F$ to be defined on an open set, in this case $\Omega$, not $\bar\Omega$. You solve the ODE there and then study its behavior on $\bar\Omega$ by taking limits.

 

[PeterDonis]

No, we're exactly in the situation of a manifold with boundary. $\bar\Omega$ is the manifold with boundary, $\Omega$ is its interior and $\partial\bar\Omega$ is the boundary. ODE theory requires $F$ to be defined on an open set, in this case $\Omega$, not $\bar\Omega$. You solve the ODE there and then study its behavior on $\bar\Omega$ by taking limits.

You are not responding to the actual issue I raised. Go back and read my post again. Responding by simply repeating your position is not sufficient.

 

[PAllen]

No, we're exactly in the situation of a manifold with boundary. $\bar\Omega$ is the manifold with boundary, $\Omega$ is its interior and $\partial\bar\Omega$ is the boundary. ODE theory requires $F$ to be defined on an open set, in this case $\Omega$, not $\bar\Omega$. You solve the ODE there and then study its behavior on $\bar\Omega$ by taking limits.

But the theory of singularities in GR is different. You ask whether there exists a manifold without boundary within which the manifold with boundary is a submanifold, with the overall metric only needing to satisfy junction conditions at the boundary, such that the geodesics defined on each side of the boundary meet. In many cases, e.g the extension of exterior Schwarzschild manifold to Kruskal, not only smoothness but analyticity is possible. But as long as the first limited condition is met, we say the geodesic incompleteness is removable, and does not constitute a singularity. Further, the physical principle of equivalence requires that some such extension be made. So far as I know, if the geodesic is defined on the boundary of a manifold with boundary, then such an extension is possible. Nothing you have said so far seems relevant to this.

 

[Nullstein]

You are not responding to the actual issue I raised. Go back and read my post again. Responding by simply repeating your position is not sufficient.

The post is addressing what you said. It doesn't matter whether $F$ is defined on $\partial\bar\Omega$. "Leaving the domain of definition" in the second case doesn't refer to leaving $\bar\Omega$, but leaving $\Omega$.

 

[PeterDonis]

The post is addressing what you said.

No, it isn't. See below.

The post is addressing what you said. It doesn't matter whether $F$ is defined on $\partial\bar\Omega$. "Leaving the domain of definition" in the second case doesn't refer to leaving $\bar\Omega$, but leaving $\Omega$.

The spacetime is $\Omega$, not $\bar{\Omega}$. I explained why, for the particular case of the geodesic equation, $F$ is guaranteed to be defined on $\Omega$. You have not responded to what I said about that at all.

 

[PAllen]

The post is addressing what you said. It doesn't matter whether $F$ is defined on $\partial\bar\Omega$. "Leaving the domain of definition" in the second case doesn't refer to leaving $\bar\Omega$, but leaving $\Omega$.

But that is then not relevant to the question essential geodesic incompleteness as used in GR, nor is relevant to the question of extending a manifold with boundary to satisfy the POE (which is the post of mine you originally responded to). If the geodesics don't reach the boundary, then physically, the boundary is not part of manifold for GR purposes. If the geodesics do reach the boundary, then this is simply to be considered a submanifold of some larger manifold without boundary.

 

[Nullstein]

But the theory of singularities in GR is different. You ask whether there exists a manifold without boundary within which the manifold with boundary is a submanifold

No, you don't necessarily ask this. You can also just have a manifold with boundary, which isn't embedded into a larger manifold without boundary. It may not be possible to extend the manifold beyond the boundary.

with the overall metric only needing to satisfy junction conditions at the boundary, such that the geodesics defined on each side of the boundary meet.

In the second case I discussed, it is not possible for two geodesics to meet on this boundary, because then the full curve would constitute extensions of the original solutions, which is in contradiction to the assumption that the original solutions were already maximal.

So far as I know, if the geodesic is defined on the boundary of a manifold with boundary, then such an extension is possible. Nothing you have said so far seems relevant to this.

So can you provide a reference that proves that such an extension is always possible? So far, you have only made a claim while I have provided literature.

If the geodesics don't reach the boundary, then physically, the boundary is not part of manifold for GR purposes. If the geodesics do reach the boundary, then this is simply to be considered a submanifold of some larger manifold without boundary.

But that just isn't possible in general if we're in the second case. It is possible for a geodesic to reach a boundary and not possesses an extension. That's what the theorem in Wikipedia says.

 

[Nullstein]

The spacetime is $\Omega$, not $\bar{\Omega}$. I explained why, for the particular case of the geodesic equation, $F$ is guaranteed to be defined on $\Omega$. You have not responded to what I said about that at all.

No, if we're talking about a spacetime manifold with boundary, then the spacetime is $\bar\Omega$. $\Omega$ doesn't have a boundary, so it's not the spacetime manifold we're talking about.

 

[Nullstein]

Every ODE $\ddot x(t) + f(x(t))=0$ can be transformed into a geodesic equation by introducing a new function $t(s)$ (reparametrizing time) and writing $\ddot x(s)+\dot t(s) \dot t(s) f(x(s))=0$ and $\ddot t(s) = 0$ (without loss of generality, we choose inital conditions that include $\dot t(0) = 1$). Then we just have $\Gamma^x_{tt}=f(x)$ and all other Christoffel symbols are $0$. So if the second case can never occur for a geodesic equation, is can also automatically never occur for the initial ODE in the first place and then the Wikipedia article would be lying.

 

[PAllen]

I really don't see how this theorem is applicable. If we start by assuming a manifold with boundary equipped with a metric, that means metric is defined on the boundary. That is, in the context of the theorem, F is defined on a closed interval rather than an open interval. Thus, I reject the applicability of this theorem to the case at hand. As I understand it, @PeterDonis has also tried to make this same point, without getting a satisfactory answer.

 

[Nullstein]

I really don't see how this theorem is applicable. If we start by assuming a manifold with boundary equipped with a metric, that means metric is defined on the boundary. That is, in the context of the theorem, F is defined on a closed interval rather than an open interval. Thus, I reject the applicability of this theorem to the case at hand. As I understand it, @PeterDonis has also tried to make this same point, without getting a satisfactory answer.

It is not necessary for the metric not to be defined on a boundary for the theorem to apply. Whether the metric is defined on the boundary or not is not relevant for the applicability of the theorem.

Suppose we chose initial conditions $x(0)=x_0\in\Omega$. Then the Picard-Lindelöf theorem guarantees the existence of a solution to the geodesic equation on a maximal open interval: $x:(t_-,t_+)\rightarrow\Omega$. It may then be the case that $\lim_{t\rightarrow t_+} x(t) = x_+$ exists and that $x_+\in\partial\bar\Omega$. In this case, the solution ends on the boundary and can't be extended beyond it. It's not relevant whether the metric is defined on $\partial\bar\Omega$ for this situation to be possible.

 

[PeterDonis]

No, if we're talking about a spacetime manifold with boundary, then the spacetime is $\bar\Omega$. $\Omega$ doesn't have a boundary, so it's not the spacetime manifold we're talking about.

You evidently failed to comprehend the Wikipedia article you yourself linked to.

The article explicitly says (right after it gives the two cases you quoted):

$\Omega$ is the open set in which F is defined, and $\bar{\Omega}$ is its boundary.

For a spacetime, $F$, as I said, is a function built entirely from the metric and its derivatives, and $F$ must be defined on whatever the entire spacetime is. That must at least include $\Omega$. If the spacetime has a boundary $\bar{\Omega}$ that is part of the spacetime--i.e., if the spacetime is a manifold with boundary, which is the case you are making claims about--then $F$ must also be defined on $\bar{\Omega}$. If it isn't, then we are talking about the first possibility ("explosion in finite time"), not the second. And if it is, then the second possibility cannot happen, because evaluating the geodesic equation on $\bar{\Omega}$ is not "leaving the domain of definition" of $F$.

It may then be the case that $\lim_{t\rightarrow t_+} x(t) = x_+$ exists and that $x_+\in\partial\bar\Omega$.

Yes, this part is fine, and in fact it will always be true in the case of a spacetime where the boundary is included in the manifold, so the metric is defined on the boundary.

In this case, the solution ends on the boundary and can't be extended beyond it.

This does not follow; the premises you have stated are not sufficient to establish this conclusion. In fact, for a spacetime, I believe the opposite conclusion will hold: if the affine parameter along a geodesic has a finite value at the boundary, and all geometric invariants are finite (so the first case, "explosion in finite time", does not apply), then there must exist a further extension of the spacetime beyond the boundary. (I think this may be proved in Hawking & Ellis.)

It's not relevant whether the metric is defined on $\partial\bar\Omega$ for this situation to be possible.

Yes, it is, because the metric being defined on the boundary, and all geometric invariants being finite there (which will always be true if the metric is defined on the boundary), is what grounds the conclusion that there must be an extension of the spacetime beyond the boundary.

 

[PeterDonis]

I really don't see how this theorem is applicable.

The theorem is perfectly valid as far as it goes; it just doesn't go far enough to make the claims about inextendibility that @Nullstein is making.

For example, consider the exterior patch of Schwarzschild spacetime, i.e., the portion outside the horizon, and suppose we include the horizon itself as a boundary and view this patch as a manifold with boundary. It is straightforward to compute that every geodesic that intersects the boundary does so at a finite value of its affine parameter, and that all geometric invariants are finite at the horizon. So we have a case where, in the notation of the Wikipedia article, $x_\pm \neq \pm \infty$ and the "explosion in finite time" possibility does not apply. So the other possibility must be true. That is what the theorem says, and, as noted above, it is perfectly correct as far as it goes.

However, the Wikipedia article misdescribes this second possibility when it uses the phrase "leaves the domain of definition of $F$", because all of the possible functions $F$ that we could be using, which must be built from the metric and its derivatives, are perfectly well-defined on the boundary $\bar{\Omega}$ (the horizon) as well as on the open set $\Omega$. So the theorem is not telling us that the curve leaves the "domain of definition" of whatever differential equation it is an integral curve of (in this case, the geodesic equation). It is merely telling us that the curve leaves the "domain of definition" of the open interval that describes its affine parameter on the open set $\Omega$. Which tells us precisely nothing, in itself, about whether or not the curve is extendible past the boundary $\bar{\Omega}$. And in fact we know that all geodesics that intersect the horizon are extendible past the horizon, because the spacetime itself is.

 

[vanhees71]

But the theory of singularities in GR is different. You ask whether there exists a manifold without boundary within which the manifold with boundary is a submanifold, with the overall metric only needing to satisfy junction conditions at the boundary, such that the geodesics defined on each side of the boundary meet. In many cases, e.g the extension of exterior Schwarzschild manifold to Kruskal, not only smoothness but analyticity is possible. But as long as the first limited condition is met, we say the geodesic incompleteness is removable, and does not constitute a singularity. Further, the physical principle of equivalence requires that some such extension be made. So far as I know, if the geodesic is defined on the boundary of a manifold with boundary, then such an extension is possible. Nothing you have said so far seems relevant to this.

But aren't there these theorems saying that there are non-extensible singularities in the solutions of Einstein's field equations? E.g., isn't it true that the Schwarzschild manifold has a true singularity at the origin $r=0$, for which there is no extension? Of course, as @PeterDonis writes in #71, the singularities at the event horizon are coordinate singularities, which can by overcome just by choosing other coordinates like the Kruskal coordinates. So here the singularities are singularities of the coordinate chart defined by Schwarzschild coordinates, but you can simply use another chart to cover a "completed" Schwarzschild spacetime with, e.g., Kruskal coordinates.

 

[PeterDonis]

isn't it true that the Schwarzschild manifold has a true singularity at the origin $r = 0$, for which there is no extension?

Yes, but $r = 0$ itself is not part of the manifold, so the manifold itself is a manifold without boundary. $r = 0$ is called a singularity because geodesics approach a finite value of their affine parameter as a limit as they approach $r = 0$ as a limit.

 

[jbergman]

I need to look at the arguments more closely to refute Peter's claim, but my gut instinct is to side with Nullstein, here. As he stated, these are standard results in manifold and ODE theory, i.e., the existence of flows that can't be extended. I believe if your Manifold is compact then you can always extend. But for the non-compact case that isn't the case.

 

[martinbn]

I need to look at the arguments more closely to refute Peter's claim, but my gut instinct is to side with Nullstein, here. As he stated, these are standard results in manifold and ODE theory, i.e., the existence of flows that can't be extended. I believe if your Manifold is compact then you can always extend. But for the non-compact case that isn't the case.

No, if the maifold has a boundary and geodesics reach it in finite afine parameter, with all relevant quantities bounded, the manifold can be extended.

 

[jbergman]

No, if the maifold has a boundary and geodesics reach it in finite afine parameter, with all relevant quantities bounded, the manifold can be extended.

Did, i say otherwise?

 

[PeterDonis]

these are standard results in manifold and ODE theory, i.e., the existence of flows that can't be extended.

Nobody is saying that these standard results are wrong. We are simply explaining why those results don't help in assessing extensibility for the particular case of geodesics in a Lorentzian spacetime.

I believe if your Manifold is compact then you can always extend.

This is obviously false as you state it: any n-sphere for n >= 1 is a counterexample.

 

[jbergman]

Nobody is saying that these standard results are wrong. We are simply explaining why those results don't help in assessing extensibility for the particular case of geodesics in a Lorentzian spacetime.This is obviously false as you state it: any n-sphere for n >= 1 is a counterexample.

It's a theorem in Lee's Smooth Manifolds book. Can you more precisely state your counterexample? I think you are probably getting hung up on my choice of the wording. The technical definition is that one can extend a flow such that it is parametrized by a parameter such that the parameters domain is $[-\infty, \infty]$.

 

[martinbn]

Did, i say otherwise?

Yes, that is what Nullstein said and you agree with him.

 

[jbergman]

Yes, that is what Nullstein said and you agree with him.

I'll have to go back and look at his claims more closely. I agree with your statement here.

 

[PeterDonis]

Can you more precisely state your counterexample?

You said any manifold that is compact can always be extended. Any n-sphere for n >=1 is a compact manifold but cannot be extended.

The technical definition is that one can extend a flow such that it is parametrized by a parameter such that the parameters domain is $[-\infty, \infty]$.

Suppose I have a geodesic on a 2-sphere, a great circle. For concreteness, say it's the equator. It is parameterized by longitude $\varphi$, with parameters $0 \le \varphi \lt 2 \pi$ denoting distinct points. I can of course extend this parameterization to cover $- \infty < \varphi < \infty$, but only if I am ok with having multiple values of $\varphi$ denoting the same points. I cannot extend the geodesic itself.

 

[jbergman]

You said any manifold that is compact can always be extended. Any n-sphere for n >=1 is a compact manifold but cannot be extended.Suppose I have a geodesic on a 2-sphere, a great circle. For concreteness, say it's the equator. It is parameterized by longitude $\varphi$, with parameters $0 \le \varphi \lt 2 \pi$ denoting distinct points. I can of course extend this parameterization to cover $- \infty < \varphi < \infty$, but only if I am ok with having multiple values of $\varphi$ denoting the same points. I cannot extend the geodesic itself.

Of course. I am talking about the domain. What you are describing is a periodic orbit in the flow domain. You can also have stationary points that never move but have a solution parametrized for all of $$\mathbb{R}$$.

When talking about extending a flow, I'm talking about it's domain, not whether the resulting path is injective.

 

[PeterDonis]

When talking about extending a flow, I'm talking about it's domain, not whether the resulting path is injective.

Then what you are saying is irrelevant to the discussion in this thread, which is about whether and under what conditions a manifold with boundary can be extended. You seem to be saying that theorems about whether a parameterization can be extended do not say anything about whether the underlying manifold can be extended, which I agree with and I think @PAllen would agree with. But @Nullstein has been claiming the opposite.

 

[jbergman]

Then what you are saying is irrelevant to the discussion in this thread, which is about whether and under what conditions a manifold with boundary can be extended. You seem to be saying that theorems about whether a parameterization can be extended do not say anything about whether the underlying manifold can be extended, which I agree with and I think @PAllen would agree with. But @Nullstein has been claiming the opposite.

Well, there are only three kinds of flows, stationary, periodic and, aperiodic. For the periodic and constant cases you can always extend your domain. For the aperiodic case, whether or not you can extend your domain does say something about the extension of the image.

The case I'm thinking of that might cause trouble would be if you can't extend to the boundary of the manifold your flow. For example, if the flow on the boundary is either stationary or only lives in the boundary this might be the case.

 

[PeterDonis]

For the periodic and constant cases you can always extend your domain.

Yes, but without extending the image (to use your terminology), so these cases are irrelevant to the discussion in this thread.

For the aperiodic case, whether or not you can extend your domain does say something about the extension of the image.

Ok, then what, specifically, does it say?

The case I'm thinking of that might cause trouble would be if you can't extend to the boundary of the manifold your flow.

This is also irrelevant to the discussion in this thread, because we are talking about cases where we know this potential problem does not occur. The theorem @Nullstein referenced explicitly has this as one of its assumptions.

 

[PeterDonis]

Every ODE $\ddot x(t) + f(x(t))=0$ can be transformed into a geodesic equation by introducing a new function $t(s)$ (reparametrizing time) and writing $\ddot x(s)+\dot t(s) \dot t(s) f(x(s))=0$ and $\ddot t(s) = 0$ (without loss of generality, we choose inital conditions that include $\dot t(0) = 1$). Then we just have $\Gamma^x_{tt}=f(x)$ and all other Christoffel symbols are $0$. So if the second case can never occur for a geodesic equation, is can also automatically never occur for the initial ODE in the first place and then the Wikipedia article would be lying.

Once again you are failing to address what we are actually saying. The issue is not the form of the equation but the domain of definition of what you are here calling $f$, and the Wikipedia article calls $F$. The second case is described in the Wikipedia article as the curve "leaving the domain of definition" of $F$. But in the cases we have been discussing, of spacetime regions that are manifolds with boundary, $F$ is defined on the boundary, so the curve reaching the boundary at a finite value of its curve parameter is not "leaving the domain of definition" of $F$.

Here's another way of putting it. The theorem says that there exists a maximal extension of a curve given certain assumptions, and that if that maximal extension stops at a finite value of its parameter, either case 1 or case 2 must apply. But it does not tell you how to prove that some particular parameterization you have that covers the curve in some region is the maximal extension of the curve. In particular, the theorem does not say that any parameterization that reaches the boundary of some region at a finite value of the parameter and does not "blow up" there, so it can't be an instance of case 1 of the theorem, must therefore be an instance of case 2 of the theorem. But that is what you are claiming.

 

[jbergman]

Yes, but without extending the image (to use your terminology), so these cases are irrelevant to the discussion in this thread.Ok, then what, specifically, does it say?

Basically there is a limit point which the curve cannot reach. Typically it's speed would also get slower and slower as it approaches that point.

 

[PeterDonis]

there is a limit point which the curve cannot reach. Typically it's speed would also get slower and slower as it approaches that point.

How does this relate to whether the image can be extended or not?

Please bear in mind that this is the relativity forum, and we are talking about spacetime, not general abstract math. For example, look at my post #71 and the curve I described there, a geodesic that approaches the horizon. In Schwarzschild coordinates, your verbal description, quoted above, appears to apply to this curve: the Schwarzschild coordinate $t$ goes to infinity as the horizon is approached, and the "speed" of the curve, $dr / dt$, goes to zero. But this curve can be extended to and beyond the horizon; the appearance of "not being able to reach the horizon" is an artifact of a particular choice of coordinates. (And we have threads all the time in this subforum where we have to explain this to newbies who misunderstand it.) So, again, what exactly does whatever math you are relying on say about what you call the aperiodic case and what it says about whether or not the image can be extended?

 

[jbergman]

How does this relate to whether the image can be extended or not?

Please bear in mind that this is the relativity forum, and we are talking about spacetime, not general abstract math. For example, look at my post #71 and the curve I described there, a geodesic that approaches the horizon. In Schwarzschild coordinates, your verbal description, quoted above, appears to apply to this curve: the Schwarzschild coordinate $t$ goes to infinity as the horizon is approached, and the "speed" of the curve, $dr / dt$, goes to zero. But this curve can be extended to and beyond the horizon; the appearance of "not being able to reach the horizon" is an artifact of a particular choice of coordinates. (And we have threads all the time in this subforum where we have to explain this to newbies who misunderstand it.) So, again, what exactly does whatever math you are relying on say about what you call the aperiodic case and what it says about whether or not the image can be extended?

I'll need more time to more carefully look at the particular case you mention. I'll try and look at this evening.

 

[jbergman]

How does this relate to whether the image can be extended or not?

Please bear in mind that this is the relativity forum, and we are talking about spacetime, not general abstract math. For example, look at my post #71 and the curve I described there, a geodesic that approaches the horizon. In Schwarzschild coordinates, your verbal description, quoted above, appears to apply to this curve: the Schwarzschild coordinate $t$ goes to infinity as the horizon is approached, and the "speed" of the curve, $dr / dt$, goes to zero. But this curve can be extended to and beyond the horizon; the appearance of "not being able to reach the horizon" is an artifact of a particular choice of coordinates. (And we have threads all the time in this subforum where we have to explain this to newbies who misunderstand it.) So, again, what exactly does whatever math you are relying on say about what you call the aperiodic case and what it says about whether or not the image can be extended?

This question turned out to be harder than I expected as I wasn't familiar with coordinate singularities. There is a nice discussion about this at https://math.stackexchange.com/questions/852678/known-methods-to-detect-coordinate-singularities

I particularly like Jack Lee's answer on this question, https://math.stackexchange.com/a/853519

The TLDR I am taking away from this discussion is that coordinate singularities are hard to detect, but if you can and can remove them then what I said above, I believe is correct.

 

[PeterDonis]

The TLDR I am taking away from this discussion is that coordinate singularities are hard to detect, but if you can and can remove them then what I said above, I believe is correct.

What you said above where?

Note that, for the particular example I gave, of the exterior patch of Schwarzschild spacetime in Schwarzschild coordinates, we know what the maximal analytic extension is, and it is a manifold without boundary.

The question is whether there are any solutions of the Einstein Field Equation which are (a) manifolds with boundary, and (b) not extendible.

If we assume that we can always discover and remove coordinate singularities, how does what you said wherever you said it bear on this question?

 

[PeterDonis]

I particularly like Jack Lee's answer on this question, https://math.stackexchange.com/a/853519

His answer is irrelevant to this discussion because he is talking about a case where we have an open manifold, which will not include any boundary. The claim at issue in this thread concerns a manifold with boundary.

 

[jbergman]

What you said above where?

Note that, for the particular example I gave, of the exterior patch of Schwarzschild spacetime in Schwarzschild coordinates, we know what the maximal analytic extension is, and it is a manifold without boundary.

The question is whether there are any solutions of the Einstein Field Equation which are (a) manifolds with boundary, and (b) not extendible.

If we assume that we can always discover and remove coordinate singularities, how does what you said wherever you said it bear on this question?

I wasn't trying to answer the full question yet. I don't know if there is a solution to Einstein Field Equations that is a manifold with boundary. I was just trying to address the question of extendability of the metric.

What I've concluded so far is the same as most others in the thread. That is if we can extend to a boundary we can always extend it further.

I was trying to ascertain if you could have a weird case where geodesics would remain on the boundary or in the interior but never cross between these regions. It seems unlikely, but personally, I still want to think about that case.

 

[WWGD]

Just curious here, hope not to derail the post: what are the differences from Euclidean 4-space and Space-time, being a product of 3-space and (1-dimensional) time?

 

[PeterDonis]

Just curious here, hope not to derail the post: what are the differences from Euclidean 4-space and Space-time, being a product of 3-space and (1-dimensional) time?

Euclidean 4-space has a flat Riemannian metric, with signature (+, +, +, +). Minkowski spacetime has a flat Lorentzian metric (strictly speaking a pseudo-metric since it is not positive definite) with signature (-, +, +, +) (using the spacelike signature convention) or (+, -, -, -) (using the timelike signature convention). Both of them have the same underlying manifold, $\mathbb{R}^4$.

 

[sysprog]

What I've concluded so far is the same as most others in the thread. That is if we can extend to a boundary we can always extend it further.

Are you contending that a majority of those who have posted in this thread have in it concluded that in the case of every 'manifold with a boundary', extensibility of the manifold to its boundary entails that the manifold or the boundary thereto is ispo facto further extensible? I don't see where anyone else in this thread has exposited such a conclusion.

 

[PAllen]

Are you contending that a majority of those who have posted in this thread have in it concluded that in the case of every 'manifold with a boundary', extensibility of the manifold to its boundary entails that the manifold or the boundary thereto is ispo facto further extensible? I don't see where anyone else in this thread has exposited such a conclusion.

What most on this thread claim is that a pseudo-Riemannian manifold with boundary (excluding boundaries at conformal infinity), such that the metric is well defined everywhere including the boundary, and curvature scalars don't blow up on approach to the boundary, then:

- there is inessential geodesic incompleteness at the boundary, which, in GR terms, is an inessential singularity. It is physically implausible to accept this as a complete spacetime.
- under these circumstances, it is always possible to extend the manifold and metric (typically, in an infinite number of different ways), to become an manifold without boundary, such that any remaining geodesic incompleteness is irremovable (no further exension that would remove the icnompleteness is possible).

 

[jbergman]

Euclidean 4-space has a flat Riemannian metric, with signature (+, +, +, +). Minkowski spacetime has a flat Lorentzian metric (strictly speaking a pseudo-metric since it is not positive definite) with signature (-, +, +, +) (using the spacelike signature convention) or (+, -, -, -) (using the timelike signature convention). Both of them have the same underlying manifold, $\mathbb{R}^4$.

It might be also worth mentioning that in Minkowski spacetime we are typically interested only in the transformations that preserve the direction of time (along with orientation) which is different from a generic Euclidean 4-space.

 

[WWGD]

It might be also worth mentioning that in Minkowski spacetime we are typically interested only in the transformations that preserve the direction of time (along with orientation) which is different from a generic Euclidean 4-space.

Is there a name for these transformations that preserve the directiin, orientation of space, time in Minkowski Space? Is it even a group; i.e., are these transformations even invertible? Sorry for my ignorance on this topic?

 

[PAllen]

Is there a name for these transformations that preserve the directiin, orientation of space, time in Minkowski Space? Is it even a group; i.e., are these transformations even invertible? Sorry for my ignorance on this topic?

General coordinate transforms form a group. There is really no restriction on them, because pullbacks associated with the transform ensure all tensors (including the metric tensor) represent the same geometric object. That is, in the framework of GR, there is no such thing as an excluded coordinate transform. While I mention GR, the same is true in SR. Absolutely coordinate transform may is allowed, but the pullbacks ensure the spacetime has zero curvature and all invariants are preserved.

I like to give the following example for SR:

Consider the metric: $ds^2 = du dv + du dw + du da + dv dw + dv da + dw da$. This is just standard Minkowski spacetime with timelike signature in 4-lightlike coordinates. All of SR physics can be done in these coordinate with no problem