# Help understand article conclusion for thin shell separating flat and Schwarzschild spacetimes

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1. Oct 6, 2014

### PLuz

I've been learning the Israel formalism (see original article here) for thin shells. I think I understand the formalism well and how to do the matching given two manifolds (that are solutions of the Einstein's field equations - EFE).

I've been studying several articles that use the matching formalism to construct new solutions of the EFE and I've stumbled on the article: "Tension shells and tension stars" by D. Lynden-Bell and J. Katz (see original article here). The article is only accessible through payment so I'll try to explain my doubt.

The authors consider the following setup: A static spherical shell of surface $4\pi R^2$ placed between two Schwarzschild spacetimes, with mass $m$ and $\bar{m}<m$. The metrics on each side of the shell, in Schwarzschild coordinates, are
$$ds^{2}=-\left(1-\frac{2\, m}{r}\right)dt^{2}+\left(1-\frac{2\, m}{r}\right)^{-1}dr^{2}+r^{2}d\Omega^{2},$$
where $d\Omega ^2 =d\theta^2+\sin(\theta)d\varphi^2$, and
$$ds^{2}=-\left(1-\frac{2\, \bar{m}}{\bar{r}}\right)dt^{2}+\left(1-\frac{2\, \bar{m}}{\bar{r}}\right)^{-1}dr^{2}+r^{2}d\Omega^{2}.$$

Following Israel formalism the authors find the following expression for the surface energy density of the shell:
$$8\pi \sigma=-\frac{2}{R}\left(\bar{\epsilon}\bar{A}+\epsilon A \right),$$
where
$$A^2=1-\frac{2\, m}{r}$$
and
$$\bar{A}^2=1-\frac{2\, \bar{m}}{\bar{r}}.$$

As seen from the $\bar{m}$-side, if $\bar{\epsilon}=1$: $\bar{r}>R$ near the shell; or $\bar{\epsilon}=-1$: $\bar{r}<R$ near the shell. Similarly, $\epsilon=1$: $r>R$ near the shell; or $\epsilon=-1$: $r<R$ near the shell.

Now the authors claim that in order to have a positive surface energy density $\bar{\epsilon}=-1$. If they consider that "inside" the shell the spacetime is flat - $\bar{A}=1$ then by considering $\epsilon=-1$ they will have a shell on the other side of the Einstein-Rosen bridge. Now, I understand they argument and I accept it. My problem is, how can they relate the sign of the $\bar{\epsilon}$ and $\epsilon$ with the condition that the Schwarzschild radial coordinate is greater or smaller than the radial coordinate of the shell.

From my calculations their $\epsilon$ variables appear in the expression for the normal to the shell, this is, for example, as seen from the $m$-side I get:
$$n_\mu = \epsilon (0, u^0),$$
where $u^0$ is the time coordinate of the velocity of a particle comoving (well in this case just chilling out while time passes) with the shell.

Can somebody help me?

2. Oct 6, 2014

### Staff: Mentor

I'm not sure I understand. Wouldn't this just be a thin spherical shell of matter with positive mass, and vacuum inside?

3. Oct 6, 2014

### George Jones

Staff Emeritus
No, this paper uses the GR junction conditions to "glue" two different spacetimes together. See, e.g., section 3.7 from Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

which evolved into the excellent book, "A Relativist's Toolkit: The Mathematics of black hole Mechanics".

Sorry, I don't have time to write more.

4. Oct 6, 2014

### Staff: Mentor

I understand the basic methodology (I agree Poisson's notes are a great reference, btw). I'm just not seeing how applying it with the given conditions in the OP leads to an Einstein-Rosen bridge. However, after looking again I'm not sure it leads to a shell with flat spacetime inside and Schwarzschild spacetime outside either.

The conditions as I understand them are:

(1) On the "unbarred" side of the shell, $m > 0$ and $\epsilon = -1$, meaning that $r < R$ as the shell is approached from this side.

(2) On the "barred" side, $\bar{m} = 0$ (i.e., spacetime is flat) and $\bar{\epsilon} = -1$, meaning that $\bar{r} < R$ as the shell is approached from this side.

The barred side makes sense: it looks like a flat Minkowski region inside a spherical shell. However, the unbarred side does not: it looks like a Schwarzschild region *interior* to the shell (because $r < R$), not exterior to it. So this looks like a shell with two interiors but no exterior (no region with $r > R$).

5. Oct 6, 2014

### pervect

Staff Emeritus
I'm not familiar enough with the Israel formalism to use it, but I've used the Einstein equations to do glue metrics together in a few special cases.

I'm left with some questions looking at your post. I'm not sure how much I can help, but perhaps the questions themselves will help you.

If $\sigma$ is the surface energy density of the shell, what is $\epsilon$ and $\bar{\epsilon}$?
What is the surface tension in the shell, do you calculate that?
What basis vectors do you use for calculating the energy density? (And the tension, if you calculate that).

I.e. are you using the cobasis vector sqrt(1-2m/r) as one of your orthonormal cobasis vactors, and are you assuming that dt is timelike, when you talk about the "surface energy density" of the shell?

6. Oct 6, 2014

### PLuz

You are confusing what is the Schwarzschild radial coordinate. It's possible to have both the radial coordinates of the inside and outside smaller than the Schwarzschild radius of the shell, I called it $R$, notice: near the shell. The radial coordinate of the outside solution grows after some minimum and it gets greater then $R$. They way I visualize this is by thinking in the embedding diagram and indeed for this to be possible the shell must be on the other side of the Einstein -Rosen bridge.

7. Oct 6, 2014

### PLuz

The $\epsilon$ and $\bar{\epsilon}$ appear when I compute the normal to the shell - please see the final part of my question.

The authors did compute the tension of the shell, it's given by:
$$T=\frac{(1+A)^2}{16 R A}.$$

I chose the usual coordinate basis $y^a=(\tau, \theta, \varphi)$. I assume that $\partial_t$ is timelike because I'm assuming that the shell is outside the horizon - otherwise the shell couldn't be considered static since it would always infall to the singularity.

8. Oct 6, 2014

### Staff: Mentor

This makes sense for the unbarred side of the shell, yes; but then the barred side wouldn't be flat; it would be the "outside" of the shell in the other "universe" and would be a Schwarzschild geometry with mass equal to $m$ plus the mass of the shell (and with $\bar{r} > R$ as the shell was approached). It seems like there must be a missing piece to this model that's not described in your OP.

Last edited: Oct 6, 2014
9. Oct 6, 2014

### pervect

Staff Emeritus
You specified two line elements, but not their domains. If we assume that we use the first line element, expressed in terms of r, in the domain r<R, and the second line element, expressed in terms of $\bar{r}$, in the domain $\bar{r} > R$, things start to make sense to me. Though I'm not sure that you are doing anything so simple. If you are doing a different "gluing", or trying to consider all possible gluings, I'd consider each gluing separately. The three possible sorts of gluing I see are gluing together cases where $r, \bar{r} < R$, cases where $r, \bar{r} >R$ and the mixed case where one of the two is less than R and the other is greater than R (as above).

Using the "mixed" gluing, with r<R and $\bar{r}>R$ we simply have one physical space with a spherical shell , the Schwarzschild mass parameter outside the spherical shell is $\bar{m}$, the mass parameter inside the spherical shell is m, and if the shell has no energy and pressure then $m = \bar{m}$ which seems consistent with your equations.

To glue together the other cases, we need to insist that $r=\bar{r}=R$ is represented by a single point in the gluing. The resulting domain will be a line segment in one case, the real line in the other.

Surely if you were using a coordinate basis, then you'd have $\sigma$ be a function of $\theta$? I suspect you're using an orthonormal basis, similar to for instance Wald pg 121, or what I wrote earlier.

10. Oct 13, 2014

### PLuz

I start by apologizing for the very late reply, I had a problem with my computer.

Now, perhaps it's easier for me to just post some images of the article. I hope I'm not breaking any copyright rule since it's only a small part of the article.

I don't understand what you mean with:
Why would the surface energy density depend on $\theta$ because I choose to not use an orthonormal coordinate system? The spacetime is spherically symmetric.

Check the images below with parts of the article:

the next page:

And here are the - mentioned - figures 1 and 2:

11. Oct 13, 2014

### Staff: Mentor

Something still doesn't add up for me. First, in the "normal shell" diagram, there should not be horizons (the lines r = 2m). The spacetime inside the normal shell is flat, so the left side of the Penrose diagram (i.e., the boundary to the left of the shell) should just be a vertical line $r = 0$ between $i^-$ and $i^+$ (i.e., the points at the bottom and top ends of the shell), as in ordinary Minkowski spacetime. So that diagram doesn't make sense.

The "tension shell" diagram doesn't make sense either. The implication appears to be that the shell is now the left boundary of the diagram--i.e., there is no spacetime to the left of the shell. That means the region to the right of the shell, or at least a portion of it, is supposed to be the "inside" of the shell (the barred region) as well as the "outside" (the unbarred region). That seems inconsistent: the barred region is flat ($\bar{m} = 0$) and the unbarred region is curved ($m > 0$). The same region of spacetime can't be both flat and curved.

So ISTM that either the paper is making some kind of error somewhere, or there's still a missing piece that isn't in the parts you've quoted so far.

12. Oct 13, 2014

### PLuz

I Agree with you. I thought the same thing but either the authors were incorrect and weren't very good physicists (nor were the peers that reviewed the article) or the the authors were just lazy with the figure and threw a Carter-Penrose for the whole Schwarzschild spacetime, drew a line a stated in that line we have a normal shell.

On the tension shell case, I thought that figure. 1. b) is also incomplete and it is only showing the Schwarzschild spacetime outside the shell. I don't understand your doubt in here, looking at the embedding diagram of the tension shell it seems correct to me, although I still don't see their relation between the sign of the $\epsilon$s and the comparison with the Schwarzschild radius near the shell.

There's nothing else to quote from the article. The images that I have uploaded are all there is.

13. Oct 13, 2014

### Staff: Mentor

Penrose diagrams are not supposed to be incomplete; they are supposed to show the entire spacetime. If there is a region of spacetime "inside" the shell (i.e., to the left of it on the diagram), it should be shown.

However, another reason why I don't think there is such a region is that $\bar{r} < R$ near the shell. On the Penrose diagram, that means the "barred" region should be to the right of the shell, since in that region the radial coordinate is decreasing to the right (i.e., toward the $r = 2m$ lines). The region to the left of the shell, if there were one, would, it seems to me, have to have $\bar{r} > R$.

It may be "correct" in the sense that it's consistent with the other descriptions in the paper, but I don't think it's consistent in the sense of being a valid solution. The "flat" region (the rectangular box) should not be able to overlap the curved region (the paraboloids), because, once again, the same region of spacetime (or the same portion of a spacelike slice) cannot be both flat and curved.

If we suppose for the sake of argument that the equations in the paper do in fact form a valid solution (which, for the reasons I've given, I don't think is the case), then their equation (3) makes this clear. That equation (omitting unit conversion factors) is

$$\sigma = - \frac{2}{R} \left( \bar{\epsilon} \bar{A} + \epsilon A \right)$$

We are also given that $\bar{A} > A$, which follows from $\bar{m} < m$ and the equations for $\bar{A}$ and $A$ in terms of $\bar{m}$ and $m$. (There is also an unstated assumption that $\bar{A}$ and $A$ are both positive, but that's a standard assumption in these types of scenarios, though the notation is a bit different here.)

Now if $\sigma > 0$ (positive energy density), we must have $\bar{\epsilon} \bar{A} + \epsilon A < 0$, which implies $\bar{\epsilon} \bar{A} < - \epsilon A$. Now suppose that $\bar{\epsilon} = 1$. We then have two possibilities:

(1) If $\epsilon = 1$, we have $\bar{A} < - A$, which is impossible since $\bar{A}$ and $A$ are both positive.

(2) If $\epsilon = -1$, we have $\bar{A} < A$, which contradicts the given condition that $\bar{A} > A$.

So if $\sigma > 0$, $\bar{\epsilon} = 1$ is impossible and we must have $\bar{\epsilon} = -1$. That gives either $- \bar{A} < - A$ or $- \bar{A} < A$, depending on whether $\epsilon = 1$ or $\epsilon = -1$, both of which are obviously true, so both possibilities for $\epsilon$ need to be considered (which they are in the paper).

14. Oct 14, 2014

### PLuz

I still can't follow your reasoning on this. Your reason for not be able to have a tension shell on the other side of the bridge is because in the Penrose diagram the region inside the shell (flat) and some region on the $m$-side would overlap. But, for me that was the reason that convinced me that the diagram might be incomplete: because it would be weird to have the diagram for the whole space - outside and inside - drawn, they would have overlapping regions. However, that overlap is only apparent, the radial Schwarzschild coordinate is not a good coordinate for this. When we look at the embedding diagram we see that the construction of a shell on the other side is possible if we consider a tension shell (the Israel junction formalism allows it, thence it's a solution to the Einstein field equations), and there's no overlapping, although, inside the shell there's a region that has the same radial Schwarzschild coordinate than other region on the outside, but so what? If we consider a better coordinate, like the the proper distance measured by an observer, to parameterize a curve that started on a point on the outside, entered the inside region and left again to the outside, the curve would be continuous, would it not?

For the second part of your answer, you misunderstood my doubt. I can follow their equations and reasoning, thence I understand why $\bar{\epsilon}=-1$ and so on. My problem is the relation between the $\epsilon$s and the comparison between the radial coordinate near the shell. I have done the calculations and for me the $\epsilon$s appear when we compute the normal to the junction surface. So the sign appears when we have to choose the direction of the normal as seen from each side - barred and unbarred. I agree with their conclusion for a normal shell: $\bar{\epsilon}=-1$ and $\epsilon=1$; but I can't follow when they claim they can take the same sign for $\bar{\epsilon}$ and $\epsilon$, that doesn't make sense to me - thinking in the direction of the normal as seen from each side. I would like for that to appear naturally from the mathematics because conceptually it doesn't make sense to me. Perhaps the best way is to work on the Kruskal-Szekeres coordinates but that isn't simple and perhaps the authors used some heuristic argument - that they don't explain in the article - that would spare me the work of doing the calculations using the other, more appropriate, coordinates.

Last edited: Oct 14, 2014
15. Oct 14, 2014

### Staff: Mentor

I agree the Schwarzschild radial coordinate is not a good coordinate, but I'm not using it as a coordinate here. I'm using it as a function of the Penrose diagram coordinates, which is how the authors of the paper used it, for example to derive the curves that mark the position of the shell in their diagrams (which are curves of constant $r = R$). What I'm having trouble seeing is how there can be a valid coordinate transformation into the Penrose diagram coordinates (which there must be if the diagram is valid) that has the region "inside" the shell being to the left of the shell's position in the "tension shell" diagram (which would mean the diagram is incomplete), but that also has $r$ in that region decreasing to the left (as opposed to decreasing to the right on the right side of the shell). Does the paper give the explicit transformation that generates the Penrose diagrams they use?

If the model is valid, yes, it should be. But I'm not convinced the model is valid; or at least, if it is, there's still a missing piece somewhere.

Ah, ok. I don't think there's enough information in what you posted for me to see how they derived that, but I'll take another look. I suspect they were viewing it more as a definition than as something that could be derived.

16. Oct 15, 2014

### PLuz

Unfortunately they don't. I also stumbled on that, the authors don't give any hints on which transformation they used to draw the Penrose diagram.

The authors don't explain anything about the formulas on Eq. $(3)$. They do, however, claim that they have studied in detail tension shells on a previous work but I can't seem to find any link to that work, not even in the journal's website.

In this same article they also find solutions of tension stars (assuming a Schwarzschild star model - constant density) and they find that such stars also must be on the other side of the Einstein-Rosen bridge, so it's somehow consistent with their claim on tension stars.