# Clocks inside a cavity inside a collapsing dust shell

1. Nov 27, 2013

### yuiop

Consider a perfectly spherical dust shell with a large internal cavity. Imagine there is a moment when the dust is momentarily stationary and there is a string of clocks strung out along a radial axis. At this instant, I will assume that the spacetime inside the cavity is flat and all the clocks inside the cavity are synchronised with each other and ticking at the same rate as each other, but ticking slower than clocks outside the cavity.

If we take a 'snapshots' at later stages of the collapse, and consider only the ticking rate of a clock that happens to be at the inner surface of the shell (before it gets swept along by the dust) then the ticking rate of of each representative clock will be even slower than the previous representative clock, relative to the one at infinity. This slightly elaborate set up of representative clocks rather than a single clock co-moving with the inner surface, is to avoid the complications due to the motion of such a clock.

Since the outermost clock of the cavity at any given stage of the collapse is ticking slower than the outermost clocks at earlier stages, this change has to propagate into the cavity if we want the spacetime to remain flat inside the cavity and have all the clocks inside the cavity ticking at the same rate.

Now the difficulty is that this propagation of changes in the field, into the cavity can only happen at a finite speed, so it appears there will inevitably be a gravitational gradient within the cavity. Clocks at outermost part of the cavity will be ticking slower than clocks near the centre of the cavity, because the innermost clocks are not yet 'aware' of the changes. This gradient would be from a high potential near the centre to a lower potential near the inner surface of the cavity. This suggests test particles inside the shell might start falling outwards towards the collapsing shell due the gravitational gradient. Now I am aware that there will be issues due to the various notions of simultaneity, but I am not sure how this is resolved and would be interested in any opinions on this.

Another approach to eliminating gradient issues within the cavity is to assume the velocity of the collapsing dust conspires with the mass of the dust in such a way that the time dilation at the inner surface of the cavity remains constant at any stage of the collapse, but I am not sure that is what actually happens.

I am also aware that the Oppenheimer-Snyder pressureless dust solution, sort of addresses this situation, but it is from the point of view of an observer co-falling with the dust and it's only applicable to a sphere of dust, rather than a sphere with a cavity.

Any thoughts on what happens inside the cavity in this dynamic situation?

P.S. To quote Wikipedia Birkhoff's theorem implies:

That statement does not make it clear if that conclusion only applies to a static shell or not.

Last edited: Nov 27, 2013
2. Nov 27, 2013

### PAllen

Synge covers this in his book, in his chapter on spherical symmetry. He considers even vibrating matter fields as long as they maintain perfect spherical symmetry. Birkhoff still applies to any vacuum region. GW are impossible - thus metric propagation of any type through vacuum is impossible.

I don't think there is any paradox - a contracting shell's mass measured at infinity cannot change for spherical symmetry. Further, EM radiation is impossible with formal spherical symmetry. You can add the idea of null dust emanating from the collapse radially to model loss of energy due to radiation, but then, again, the comparison between an interior clock and one at infinity remains constant.

3. Nov 27, 2013

### Staff: Mentor

At this instant, this seems OK. But once the motion of the shell is significant, you can no longer use your intuitions about things like the "field", "potential", or "ticking rate of clocks relative to infinity", because those concepts only work for a static scenario.

I don't think this works, as I commented above. You can't handwave away the effects of the motion of the dust this way. See below.

No, what will happen (I think--I haven't done the detailed math) is that there will be a sort of "shock front" on the inner surface of the shell, where the relative "ticking rate" of adjacent clocks gets more and more different. This corresponds to the dust at the inner surface of the shell moving faster and faster, relative to an observer who is stationary within the cavity. Nothing has to "propagate" into the cavity; the cavity just gradually gets replaced by infalling dust, and observers who start out static within the cavity get swept along with the infalling dust with a more and more sudden and sharp inward acceleration. The result will be that adjacent infalling observers *within the shell* will have experienced more and more sharply different elapsed times since the starting instant (the instant where everything was at rest).

I don't think so; I think the only gradient will be at the inner surface of the shell, as above. Observers within the cavity will remain in free fall and at rest relative to each other until the inner surface of the shell overtakes them.

I think this illustrates why the concept of "gravitational potential" doesn't work in a non-static scenario. Spacetime in a vacuum surrounded by a spherically symmetric mass distribution is flat, regardless of whether the mass distribution is static or dynamic; hence, as I said above, test particles inside the cavity will remain in free fall and at rest relative to each other until the infalling dust overtakes them.

4. Nov 27, 2013

### WannabeNewton

See section 3.9 of "A Relativist's Toolkit: The Mathematics of Black Hole Mechanics"-Poisson.

EDIT: Also see section 7.6 of "Gravitation: Foundations and Frontiers"-Padmanabhan, page 336 in particular.

Last edited: Nov 27, 2013
5. Nov 27, 2013

### yuiop

Thanks for that. I have never before seen it explicitly stated that Birkhoff's theorem implies that in the spherically symmetrical situation, that gravitational waves are impossible in the vacuum outside AND inside a collapsing shell.

Now consider a thin dust shell that has a cavity with a huge internal radius. let's say for the sake of argument that the time dilation factor is 0.98 at the inner surface and 0.99 at the outer surface of the shell. Imagine the dust is 'sticky' but due to the initial spatial separation of the particles it behaves initially like a normal dust with me internal pressure or stress and is free to fall.

At a later stage, the dust shell has collapsed significantly and the dust particles start glueing and interlocking together with the increasing density of the dust, until eventually they eventually form a solid shell that resists the collapse and this occurs before the cavity radius is zero. Now if there can be no propagation to the interior of the shell, the time dilation factor everywhere within the shell (and by continuity at the inner surface of the shell) must still be 0.98, while at the outside surface of the shell the time dilation factor will be <0.98 with sufficient collapse. Yes, there will be a build up of stress and pressure in the shell, but by Birkoff' theorem as stated above, this can have no influence on the time dilation within the cavity. This contradicts the assumption that the lowest time dilation factor must occur at the inner surface of the shell.

Note that we started with a momentarily stationary shell and ended up with a stationary but compacted shell so we can compare two static spacetimes.

Last edited: Nov 27, 2013
6. Nov 27, 2013

### PAllen

Good point, I had a mistake in my prior discussion (operating from memory of the chapter).

Everything I said was correct except the following: "the comparison between an interior clock and one at infinity remains constant. " It is interesting to see how this could be. First, note that physically, any notion 'clocks ticking slower here' is wrong. Instead, you can compare a clock here to one there, via signals. Thus, change in matter distribution between the interior and infinity can and does change the clock comparison, but this does not imply a metric change to the interior. It is still flat Minkowski space. It would seem that there is a problem because the interior metric, while flat, has g00 changing with time. However, this is purely a coordinate expression feature from from fitting the interior metric to SC coordinates. One can equally use coordinate patch whose 3-volumes are shrinking, but with diag(1,-1,-1,-1) metric everywhere/when. Scaling g00 to match the inner shell only comes into play to produce a global coordinate system, or (more physically) to represent world lines or null paths crossing from inside through the shell. Remember, all coordinates, even ones that produce time changing metric components, are not changing the interior geometry at all as long as they preserve that the curvature tensor is zero. Think of arbitrary, non-inertial coordinates in SR. Nothing has changed about the geometry - Rijkl is still 0.

7. Nov 27, 2013

### Staff: Mentor

The spacetime inside the cavity is flat and it remains flat. There is nothing physical propagating. The only change in curvature happens at the location of the shell, and "propagates" only at the speed of the shell if you can even call that "propagating".

The "ticking rate wrt infinity" is a coordinate artifact, not anything physical. So it is not limited to any finite propagation speed and can change instantaneously according to your changing synchronization convention. Any redshift or other physical observable at infinity changes smoothly and causally.

The time dilation factor is not something physical which needs to propagate. Nor are changes to it prevented by Birkhoff's theorem which applies only to changes in curvature.

Last edited: Nov 27, 2013
8. Nov 27, 2013

### Staff: Mentor

Actually, for the final state to be static, it will have to occur before the outer radius of the shell reaches 9/4 M, where M is the total mass of the system as measured at infinity (which is constant throughout the scenario). Otherwise no static equilibrium is possible. If the shell remains thin throughout, this means the cavity radius (i.e., the inner radius of the shell) will also have to be greater than 9/4 M.

No, this is wrong, and it's a good example of why "time dilation factor" (or "gravitational potential") doesn't work well for non-static scenarios. See further comments below.

This is correct; but note that it requires that the final time dilation factor inside the shell is also <0.98, since it must be less than the time dilation factor at the outer surface. That is obvious from the equations in my blog post, which apply to both the initial and final static configurations (but do *not* apply to the dynamic region in between--see below).

Birkhoff's Theorem doesn't say this. See below.

No, it contradicts the assumption that you can model the dynamic portion of this scenario using "time dilation factor" and "propagation". See below.

First, a comment: if the initial state is truly dust with zero pressure, it isn't static; if you evolve that state backwards in time, it would have to be expanding and decelerating, to momentarily "hover" at the initial state you defined. For the initial state to really be static, the shell would have to be supported by something that is suddenly taken away at time t = 0. For concreteness, I'll assume that there is some sort of EM field generator inside the shell that was generating a field to keep the shell static, and which is rigged to turn itself off everywhere in the shell at time t = 0 (the timing is all set up in advance to ensure this).

That said, the comparison you describe requires, as I noted above, that the time dilation factor inside the cavity is *smaller* in the final configuration than in the initial configuration, so it *must* change during the collapse process. Physically, we could verify this by putting a static observer, O (outside), very far away, who "hovers" at a constant, very large radius, and exchanges light signals with another static observer, I (inside), who is far enough inside the cavity that the shell stops collapsing before it reaches him. Observer O will find that his proper time elapsed between successive light signals is constant; observer I will find that his proper time elapsed starts out at a constant value smaller than that of observer O, then *decreases* during the collapse, then remains constant (and smaller than before) in the final static configuration.

How can this be? I haven't done the detailed math, but the key is that the equations we have been throwing around in other threads, such as the ones in my blog post, *only* apply in static regions of spacetime. They do *not* apply in a region of spacetime which is dynamic, which the "collapse" region is (see further clarification of this below). So we can't deduce the properties of the dynamic region by using concepts like "time dilation factor" that arise from equations that are only valid in the static region. However, we *can*, as I noted above, compute the properties of both static regions; and that computation alone is sufficient to show that the "time dilation factor" inside the shell must be smaller after the collapse than before. Whatever happens in the dynamic region must be consistent with that.

Furthermore, while Birkhoff's Theorem applies to the vacuum regions of the spacetime, it does *not* apply to the non-vacuum region containing the shell. (This is explicit in the theorem itself: one of its assumptions is that the SET is zero, i.e., vacuum; the proof of the theorem makes essential use of the vacuum EFE, with the RHS zero.) That means that, whatever Birkhoff's Theorem says about "propagation" of gravity in a vacuum region, it does *not* say anything about how spacetime curvature "propagates" through the shell. So the fact that the "time dilation factor" of an observer inside the shell, relative to infinity, is smaller after the collapse than before, in no way contradicts Birkhoff's Theorem.

So what *is* going on in the dynamic region? As I said above, I haven't done the detailed math, but here's what I think is happening. If we just consider the region of spacetime outside the shell, it is obvious that a light signal traveling from the shell's outer radius to observer O will redshift, and the amount of redshift will be inversely proportional to the shell's outer radius (larger radius -> smaller redshift). This is guaranteed by Birkhoff's Theorem. It is equally obvious that light traveling in the vacuum region inside the shell experiences no redshift at all; spacetime there is flat. So if the shell itself has negligible effect on the light, the redshift of a light signal traveling from I to O will gradually increase as the shell collapses, because the regions of spacetime outside and inside the shell remain static the whole time (i.e., the *only* dynamic region of the spacetime is that occupied by the shell itself).

Now, what effect will the shell itself have on the light? In the static configurations before and after, we know that $J(r)$ decreases by some amount (which may be small) as we go from the outside to the inside of the shell. This corresponds to some amount of redshift of light going outwards. But the properties of the shell in the dynamic region should change continuously from the initial to the final configuration: so if the only effect of the shell on outgoing light before and after the collapse is a small redshift, the same should be true during the collapse. That means the net effect of the collapse process on light signals going from I to O is just to increase their redshift in proportion to the decrease in the shell's outer radius.

This process could be described as the "time dilation factor" inside the cavity getting smaller: but note that nothing has to "propagate" to make this happen; it's just a simple consequence of the effect of the shell's collapse on the redshift of light signals. You can describe the whole process without ever thinking about "time dilation" (I basically just did so); the only reason to bring "time dilation" into it at all is our intuitions. But our intuitions aren't suited for this kind of dynamic scenario.

Last edited: Nov 27, 2013
9. Nov 27, 2013

### PAllen

Peter, I'm confused by your last argument. I would think that if a shell contracts, the time dilation factor for the inside compared to infinity increases rather than decreases. The mass measured at infinity must stay the same. The exterior vacuum becomes a larger part of the exterior SC geometry with the same M. Thus there is higher proper acceleration for static exterior surface observers than before contraction. At then, the time dilation factor here is greater than before contraction. It can only increase on the way through the shell to the interior vacuum. Maybe it's just terminology: smaller g00 in the final static static state (for the interior), which I would call larger time dilation factor.

10. Nov 27, 2013

### Staff: Mentor

Yes, it's just terminology. What decreases is $J$ inside the shell (the negative of $g_{00}$). What increases is the ratio of $J$ inside the shell to $J$ at infinity. Either one could be called the "time dilation factor", but I've been using the term to denote $J$ itself, which decreases, not the ratio of $J$ values. You are using it to denote the ratio, which does increase, yes.

11. Nov 27, 2013

### yuiop

Thanks all for taking the time to investigate my concerns. It all still seems a bit mysterious how the gravitational potential changes *simultaneously* everywhere inside the cavity in response to physical changes outside the cavity. I have highlighted the word 'simultaneously' because as we all know that is a relative concept, and even if the potential is changing simultaneously according to one set of observers inside the cavity, it will not be simultaneous to another set of observers with motion relative to the first set, and this second set will therefore see a gravitational potential gradient inside the cavity during the dynamical stage, (or so it seems to me).

I have found some texts on the subject of collapsing dust spheres, but they are a bit over my head. I am listing them here in the hope that someone may be able to glean something from them.

THE GENERAL EXACT SOLUTION FOR RELATIVISTIC SPHERICAL SHELLS

Last edited: Nov 27, 2013
12. Nov 27, 2013

### PAllen

The gravitational potential (when you can introduce it) is a global feature - a comparison between here and infinity, affected by all the matter. It is not something that propagates. No local geometry of the interior is changing at all. Remember, there is nothing wrong with describing the interior in its own coordinate patch, consisting of shrinking 3-ballsXt, in which the metric is diag(1,-1,-1,-1). From the complete manifold (dynamic and static phases), represented in a complete metric on global coordinates, you can define this patch and do a coordinate transform to get this Minkowski metric. This underscores that within the interior, the geometry is absolutely nothing but ordinary Minkowski. The potential is a global relationship between this volume and the whole rest of the manifold. For emphasis: it is a feature of all the rest, that we assign here for bookkeeping. In no conceivable way at all is it a feature of the physics of the interior.

13. Nov 27, 2013

### PAllen

Only the third of these is relevant to the case under discussion: a shell surrounding flat spacetime. The first two have a black hole inside the the shell. The last two only cover balls, not shells.

14. Nov 27, 2013

### yuiop

OK, I will delete the others to avoid confusion.

15. Nov 27, 2013

### Staff: Mentor

That's *not* what's happening; that's the whole point of my posts. You are not showing that the gravitational potential "changes" during the collapse; you are showing why the concept of "gravitational potential" breaks down in a dynamic spacetime.

It's interesting that you did, since you are making the same sort of error that a poster in another recent thread made when he claimed that an observer who accelerates can make someone else's time "flow backward" according to him, just by changing the tilt of his lines of simultaneity. All the poster was really showing was why the concept of "simultaneity" is limited, and breaks down if you try to push it too far. "Gravitational potential" is limited in the same way.

No, they will not. Let's add another observer inside the cavity, I3, who is rotating at a constant angular velocity around I1. I3 and I1 can exchange light signals between themselves and I2, and can verify that, for example, I3 experiences less elapsed proper time per revolution than I1 (they can both judge a "revolution" by marking successive events at which I3 passes the line between I1 and I2) by some fixed ratio. The key point then is that *nothing* in the relationship between I3 and I1 changes during or after the collapse; neither one sees any change at all in their local region of spacetime. The only change is in the redshift that observer O sees in the light signals coming from them.

Once again, the fact that there is no locally observed change in the cavity at all means that there can't possibly be any change in the "gravitational potential" there--at least not if you want to tie the term "gravitational potential" to anything locally observable. If you want to tie it to the redshift of the light signals, that's fine, but then it's clear that changes in that redshift are purely due to changes in the radius of the shell, and don't affect the spacetime inside the cavity at all.

(You could, in fact, construct an alternate assignment of "gravitational potential", which would be more natural for the observers inside the cavity, and according to which the potential *outside* the cavity changes during the collapse. The observations justifying this are simple: just observe the blueshift in light signals coming from observer O, and confirm that the blueshift increases during the collapse. But this interpretation has the same problem as the other: there's no "propagation" of the potential change in the vacuum region outside the shell; the change in blueshift is purely due to the change in the shell's radius.)