# Spherical GR dust shell with vacuum inside and outside

1. Nov 29, 2013

### yuiop

In various other threads we have been kicking around various equations for a spherical shell and discussing the implications. In this thread I would like to present what (I think) I have worked out about how the shell metric relates to the vacuum metric inside and outside the shell.

I hope to show here that $g_{tt}$ or $J(r)$ inside the vacuum cavity must always be unity and that the 'fishy' equation 7.32 given by Sean Carroll can be derived from the shell equation derived by PeterDonnis and that the metric for the shell can connect the metric inside and outside without any 'patching' factors.

For reference I am assuming spherical symmetry with no rotation, no charge and a momentarily stationary dust shell.

First consider this post by WannabeNewton
that concludes that $g_{tt} = g_{rr}^{-1}$ inside the cavity and this quote by PeterDonnis
.. concludes that $g_{rr}=1$ inside the cavity. If we accept the above conclusions by WBN and Peter, then we must accept that $g_{tt}=J(r) =1$ inside the cavity and cannot take any value, other than unity, as long as the vacuum cavity exists. <EDIT: Now I have looked again, I see that WBN had already concluded that $g_{tt}$ must be unity inside the vacuum cavity, in his elegant post above>.

If we require the metric of the shell to smoothly connect $g_{tt}=1$ at the inside surface to $g_{tt} = (1-2M/r)$ at the outside surface of the shell then that requirement is met by Sean Carroll's metric that has $g_{tt} = J(r) = (1-2m(r)/r)$. I know this is not a popular idea but I will try to show here that it can be obtained from Peter's equation. We start with:

$\frac{1}{J} \frac{dJ}{dr} = \frac{2(m(r) - 4 \pi r^3 \rho (r))}{r \left( r - 2 m(r) \right)}$

Initially I will consider the constant density case so $\rho = \frac{M}{(4/3)*\pi(R2^3-R1^3)}$ where M is the total mass of the shell, R1 is the inner radius of the shell and R2 is the outer radius. Similarly we can replace $m(r)$ with $M*(r^3-R1^3)/(R2^3-R1^3)$ as shown by pervect here. When these substitutions are made, the equation becomes:

$\frac{1}{J} \frac{dJ}{dr} = \frac{2M(-2r^3-R1^3)}{r^2(R2^3-R1^3)-r2M(r^3-R1^3)}$

Now when we carry out the definite integration from r=R1 to r=R0 where R0 is the location where we wish to determine $g_{tt}$, we get a closed form solution:

${J(R0)} = \left(\frac{R1^3(1-2M/R0)+2MR0^2-R2^3}{R1^3(1-2M/R1)-2MR1^2-R2^3}\right)$

The online calculation of the indefinite integral can be seen here.

This algebraically simplifies to:

$J(R0) = \left(1-\frac{2M}{R0}\left(\frac{R0^3-R1^3}{R3^3-R1^3}\right)\right)$

The algebraic manipulation is not obvious, but the equivalence can be confirmed by subtracting the expressions from each other and getting zero, as shown here.

It is easy to see that the last term is the enclosed volume and that the equation can be expressed as:

$J(r) = \left(1-\frac{2M(r)}{r}\right)$

where it should be understood that M(r) is the mass enclosed within a sphere of radius r. This is the result quoted by Sean Carroll and has been obtained directly from Peter's equation. It also has the pleasant result that the integral has a closed form and directly connects $g{tt}=1$ of the vacuum cavity with $g_{tt}=(1-2M/r)$ of the Schwarzschild metric outside the shell without any scaling factors.

The above derivation is for the constant density case. It turns out there is alternative way to obtain the same result that implies the result has a more general application. This time I start with Peter's equation and replace $(m(r) - 4 \pi r^3 \rho (r))$with M(x), where M(x) is a constant that represents the total effective gravitational mass enclosed by a volume with r=x, taking into account all factors such as energy, stress, pressure that contribute to the gravitational mass, just as M does the same job in the external Schwarzschild metric. By this definition we have

$\frac{1}{J(r)}\frac{dJ(r)}{dr} = \frac{2M(x)}{r \left( r - 2M(x) \right)}$

The indefinite integral is then:

$J(r) = (1-2M(x)/r) +C$

This is the same result as obtained earlier if the constant of integration is discarded. Normally an indefinite integral does not have defined boundaries but it works here as the boundaries are implicitly defined when M(x) is calculated. A more general form of M(x) can given as:

$M(x) = M\frac{x^d-R1^d}{R2^d-R1^d}$

For constant density, d=3 but other density profiles can be defined by using different values of d. For example when d=(-1), the density near the inner surface of the shell is considerably higher than near the outer surface. Changing the density profile change the shape of the potential gradient within the shell, but the curve always neatly connects the inner Minkowski metric to the outer Schwarzschild metric.

See these plots for J(r) where d=1,-1,3 with M=G=c=1 and R1=1/2 and R2=7.

$g_{tt}$ or $J(r)$ can represent the gravitational potential and examination of the slope of the potential in the graphs linked above, produces some surprising conclusions that might offend our Newtonian based intuitions, but we should not be too surprised that GR can differ significantly.

An interesting aspect of this solution is that one metric (with no 'patching' factors) can continuously cover the entire region inside the vacuum of the cavity, through the shell and on through the external vacuum, if it is clearly understood that M(x) is the gravitational mass enclosed within a volume of radius r=x.

Last edited: Nov 29, 2013
2. Nov 30, 2013

### pervect

Staff Emeritus
I haven't read the long threads in enough detail to comment on all of your questions (but I think other posters have been doing a good job from the little I have followed).

But there is one obvious point here that I think should be made.

If J(r) is a (part of) a solution to Einstein's field equations, $\alpha \, J(r)$, where alpha is an arbitrary constant, is also solution to Einstein's field equations. Therefore there is no reason derivable from Einsteins field equations that J(0) "needs to be unity".

In generalized coordinates, J(0) can have any value you like.

What I suspect is happening is that yuiop has some personal view on what coordinates "should be", based on previous experience with his posts, where I have noticed he tends to give them a lot of physical significance.

Mathematically speaking, though, with truly general coordinates, there is no reason that J(0) "has to have" any particular value.

3. Nov 30, 2013

### yuiop

After reviewing Peters notes, I now see that I chose the equation for (1/J)(dJ/dr) that included a density term that was derived by assuming J(r) had the form (1-2m(r)/r), so it is not surprising I got that result. Obviously I need to think about this some more.

I can see there is an arbitrary constant and in the external vacuum coordinates this must be 1 if the speed of light at infinity is c. Still, if we take WBN's calculations at face value it seems this arbitrary constant must be unity inside the cavity also, if $g_{tt} = g_{rr}^{-1}$. Possibly WBN assumed a constant of $\alpha =1$. I think that needs to be clarified.

Last edited: Nov 30, 2013
4. Nov 30, 2013

### Staff: Mentor

I think that's effectively what he did, yes.

Look at the EFE components in my blog post: for a static, spherically symmetric spacetime we always have $B(r) = 1 / \left(1 - 2m(r) / r \right)$, whether there is vacuum or not. Then, if the region is vacuum (so the pressure $p$ is zero and the mass $m$ is a constant), we have (using my notation $J(r)$ instead of $A(r)$)

$$\frac{1}{J} \frac{dJ}{dr} = \frac{2 m}{r^2 \left( 1 - 2 m / r \right)}$$

This is solved by the general ansatz

$$J(r) = \alpha \left( 1 - \frac{2m}{r} \right)$$

where $\alpha$ is an arbitrary constant, as pervect said; $\alpha$ cancels out when we evaluate $(1 / J) dJ / dr$, so the equation above can't constrain its value.

So in general we have $A(r) = \alpha B(r)^{-1}$. However, in a region of spacetime where we have a boundary condition on $A(r)$, we can obtain a definite value for $\alpha$. For example, if we are in a vacuum region that is asymptotically flat, so $A(r) \rightarrow 1$ as $r \rightarrow \infty$, then we must have $\alpha = 1$.

But if we are in the interior of a spherically symmetric cavity inside a spherically symmetric shell, then the boundary condition on $A(r)$ is determined by its value on the inner surface of the shell, which must be determined by integrating inward from the outer surface, using the value of $A(r)$ on the outer surface that is determined by its behavior in the outer vacuum region. This will make the value of $\alpha$ inside the cavity *less* than 1 (and since $m = 0$ inside the cavity, $A(r)$ will just be equal to $\alpha$ everywhere inside the cavity).

5. Dec 6, 2013

### yuiop

Let's say the metric inside a cavity is described by $ds^2 = t*dt^2 - dx^2-dy^2-dz^2$ where the coordinate time (t) is measured by an observer outside the cavity. Is there any possible measurement that observers inside the cavity can make, to determine they are not in ordinary Minowski spacetime $(ds^2 = dt^2 - dx^2-dy^2-dz^2)$, without reference to measurements outside the cavity?

6. Dec 6, 2013

### Staff: Mentor

This doesn't make sense; this metric isn't flat, but the metric inside the cavity has to be flat. You can't just arbitrarily choose it.

Perhaps I should clarify what the metric will be inside and outside the cavity, based on what I said in previous posts. In the global chart in which $g_{\mu \nu} \rightarrow \eta_{\mu \nu}$ at infinity, the metric outside the cavity is the normal Schwarzschild metric, and the metric inside the cavity is

$$ds^2 = - \alpha dt^2 + dr^2 + r^2 d\Omega^2$$

where $\alpha < 1$ (I've written this in spherical oordinates to make clear the continuity with the normal way of writing the metric in the exterior vacuum region). This metric is the ordinary Minkowski metric, but the time coordinate has a non-standard scaling in this chart; however, a simple coordinate transformation $dT = \sqrt{\alpha} dt$ puts the metric into the standard Minkowski form:

$$ds^2 = - dT^2 + dr^2 + r^2 d\Omega^2$$

Now, what does the *exterior* metric look like in this new chart? That's easy; it looks like this:

$$ds^2 = - \frac{1}{\alpha} \left( 1 - \frac{2M}{r} \right) dT^2 + \frac{1}{1 - 2M / r} dr^2 + r^2 d\Omega^2$$

In other words, in this chart, the metric at *infinity* is the Minkowski metric with a non-standard scaling of the time coordinate. But any physical invariants will still be the same, because any invariant in which a time interval $dt$ appears in the standard chart will be expressed in terms of a time interval $dT / \sqrt{\alpha}$ in the new chart, which will make the invariant's value the same (for example, try it with a specified value of $ds^2$ using both formulas).

7. Dec 7, 2013

### yuiop

Imagine we have a self supporting spherical shell with inner radius $R1_0$ and outer radius $R2_0$ at time t0. The shell is designed so that the size can be mechanically changed at will. We reduce the size of the shell so that the new outside radius $R2_1$ at time t1 is less than $R1_0$. Assume initially the metric is described by: $ds^2 = \alpha_0 dt^2 - dr^2 - r^2 d\Omega^2$

After reducing the size of the shell, the metric is now described by: $ds^2 = \alpha_1 dt^2 - dr^2 - r^2 d\Omega^2$

If we assume the redshift factor at the inner surface of the shell is always less than the same factor at the outer surface of the shell, then we can expect that $\alpha_1<\alpha_0$. If that is the case, we can imagine reducing the size of the shell very slowly in very small spatial and temporal steps approaching infinitesimal, is such a way that the velocity of the shell components is insignificant and then to a reasonable approximation the metric describing the inner vacuum cavity could be described by: $ds^2 = \alpha(t) dt^2 - dr^2 - r^2 d\Omega^2$

where $\alpha(t)$ is a function of time. Signals sent from a source inside the cavity would be observed to have increasing redshift over time according to a stationary observer outside the shell. If the source inside the cavity is stationary so that $dr=d\Omega=0$ then the redshift factor is $\frac{ds}{dt} = \sqrt{\alpha(t)}$.

This is the ratio of the redshift factor inside the cavity relative to the redshift factor outside the cavity. Now if we restrict ourselves to measurements made by a stationary observer inside the cavity without reference to anything outside the cavity, can we reasonably expect this observer to consider the metric inside the cavity to described by: $ds^2 = dT^2 - dr^2 - r^2 d\Omega^2$

where the coordinate time T, is the proper time of our stationary observer inside the cavity?

Offhand, I do not think the two metrics contradict each other, they are just different points of view. One has coordinate time defined as he proper time of an observer at infinity and the other has coordinate time defined as the proper time of an observer inside the cavity.

8. Dec 7, 2013

### Staff: Mentor

Yes; this is a similar coordinate transformation to the one I described in an earlier post, just with a transformation factor $\alpha$ that depends on time. In this chart, the metric inside the cavity is constant, but the metric *outside* the cavity varies with time (the same $\alpha$ factor will appear in the exterior metric).

I agree.