Exploring a New Type of Black Hole?

[yuiop]

I was playing around with the Schwarzschild interior solution when I came up with this interesting solution that I think would be fun exploring.

The interior solution for the uniform density case is given as:

$$\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M r^{2}}{R^{3}}}$$

where M is the mass of the gravitational object enclosed within a surface radius of R,
r is the radial location of a clock with proper time $$\tau$$
and t is the coordinate time.

Now if I multiply the top and bottom of the last fraction by $$4/3\pi r p$$ where p is the density I get:

$$\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M}{r}\frac{4/3 \pi r^3 p}{4/3 \pi R^3 p}}$$

Now density times enclosing volume equals mass, so the above equation can be expressed as:

$$\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M_r}{r}}$$

where $$M_r$$ is the mass enclosed within a spherical volume of radius r (where the proper time is measured).

Now the interesting bit. If we have a hollow shell with an outer radius of 9M/4 and an inner radius of 2M so that the mass enclosed within r is zero, the time ratio measured by an observer at 2M is:

$$\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1-\frac{8}{9}}-\frac{1}{2}\sqrt{1-\frac{0}{r}} = 0$$

This means the ratio of proper to coordinate time everywhere inside the cavity will be zero making the interior one large event horizon. My first reaction was that this shell would immediately collapse but on closer inspection, the inward gravitational force at the inner surface is zero and the gravitational force at the surface of the shell is finite so in principle such a shell could exist.

This artificial “black hole” contains no singularity at the centre and in principle can be explored. If holes are drilled through the shell, probes can be launched into the cavity and can be expected to pop out the other side, because there is no gravitational force pulling everything back towards the centre. Clocks can be placed at the “event horizon” to see if they really stop. This raises the question as to whether clocks at the “event horizon” can be seen by outside observers, as I have yet to determine whether light can escape the cavity or not.

 

[Q-reeus]

I think the primary concern with this interesting setup is whether such a shell could actually resist collapse. Best I could do is find the following article "Newtonian and General Relativistic Models of Spherical Shells - II" at http://arxiv.org/pdf/1006.2280" , Section 6 p11-14, esp eqn's 37, 38. Not easy going but not an easy problem!

 

[Mentz114]

yuiop, can you give the metric you worked from ?

Q-reeus, thanks for the reference, which I failed to find somehow in my searches.

Their metric ( equation (29) ) which covers all the cases that one can find a Newtonian potential for looks very well behaved, except when f=-1. However it's not clear to me if this can happen.

This is equation 29

$$ds^2 = c^2\left( \frac{1-f}{1+f} \right)^2 dt^2 - (1+f)^4(dr^2+r^2d\theta^2+r^2\sin(\theta)^2d\phi^2 )$$

where f is related to the Newtonian potential by

$$f=-\frac{\Phi}{2c^2}$$

If the potential has a singularity then this could cause a singularity in the metric, maybe.

This is avery cool approach, because all that's required is to find the Newtonian potential of interest , plug it into the metric and analyse the spacetime.

 

[yuiop]

yuiop, can you give the metric you worked from ?

The full form of the interior Schwarzschild metric for the constant density case is:

$$c^2d\tau^2 = \left(\frac{3}{2}\sqrt{1-\frac{r_s}{R}}-\frac{1}{2}\sqrt{1-\frac{r_s r^{2}}{R^{3}}}\right)^2 c^2 dt^2 - \left(1-\frac{r_s r^2}{R^3}\right)^{-1}dr^2 -r^2(d\theta^2+r^2\sin(\theta)^2d\phi^2)$$

where
$r_s = 2GM/c^2$ is the Schwarzschild radius,
R is the surface radius of the spherical gravitational body and
r is the radius at which a clock that measures the proper time $\tau$ is located.

I posted a link for non-uniform density version of the interior solution in a very old thread but the link is broken now and information on the interior solution is hard to find. There is some information on the uniform density version here in this book by Hobson & Lasenby: http://books.google.co.uk/books?id=xma1QuTJphYC&pg=PA295#v=onepage&q&f=false Page 295 equation 12.29 and in this book by Gron http://books.google.co.uk/books?id=IyJhCHAryuUC&pg=PA255#v=onepage&q&f=false Page 255, equation 10.266.

Q-reeusTheir metric ( equation (29) ) which covers all the cases that one can find a Newtonian potential for looks very well behaved, except when f=-1. However it's not clear to me if this can happen.

This is equation 29

$$ds^2 = c^2\left( \frac{1-f}{1+f} \right)^2 dt^2 - (1+f)^4(dr^2+r^2d\theta^2+r^2\sin(\theta)^2d\phi^2 )$$

where f is related to the Newtonian potential by

$$f=-\frac{\Phi}{2c^2}$$

If the potential has a singularity then this could cause a singularity in the metric, maybe.

This is avery cool approach, because all that's required is to find the Newtonian potential of interest , plug it into the metric and analyse the spacetime.

It is not well behaved. For f<+1 the proper time becomes imaginary for $dr=d\theta=d\phi=0$. The equation for f=GM/(c^2r) is just the isotropic Schwarzschild solution which in turn is just the exterior vacuum Schwarzschild solution with the radial coordinate relabelled. The isotropic version of the Schwarzschild version is obtained by replacing r with $r_i * (1+GM/(2c^2))$ where $r_i$ is the radial coordinate used in the isotropic metric so to avoid confusion the equation you gave above should be written as:

$$ds^2 = c^2\left( \frac{1-f}{1+f} \right)^2 dt^2 - (1+f)^4(dr_i^2+r_i^2d\theta^2+r_i^2\sin(\theta)^2d\phi^2 )$$

While the isotropic version is useful in allowing the metric to be analysed in rectangular coordinates rather than polar coordinates it does not change the physics. It can also be noted that in regular Schwarzschild coordinates local observers as well as distant coordinate observers all agree that the relationship of the circumference of a shell to the radius is $2*\pi*r$ when r is defined in the usual way, i.e. Euclidean, but in the isotropic coordinates the circumference of a shell is not equal to $2*\pi*r_i$

Your quoted equation is based on the vacuum solution and so it is not valid for analysing shells that contain mass.

 

[yuiop]

I think the primary concern with this interesting setup is whether such a shell could actually resist collapse.

For an arbitrarily large shell the finite forces can be made arbitrarily small, so in principle such an artificial black hole shell that does not collapse could be constructed without requiring material of infinite strength.

 

[yuiop]

I did the analysis in the wrong order in #1 but the results still stand. I should have started with:

$$\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M}{r}\frac{ r^3 p(r)}{ R^3 p(R)}}$$

where p(r) is the average density inside the sphere of radius r and p(R) is the average density of the whole body within a sphere of radius R. In the uniform density case we can set p(r)=p(R) and the whole equation simplifies to the more familiar:

$$\frac{d\tau}{dt}= \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M r^{2}}{R^{3}}}$$

In the constant density case the pressure P within a sphere of perfect liquid is not constant and the pressure at a given radius P(r) is given by:

$$P(r) = \frac{2}{3} p \frac{\sqrt{1-R_s r^2/R^3} - \sqrt{1-R_s/R}}{3\sqrt{1-R_s/R}-\sqrt{1-R_sr^2/R^3}}$$

When the surface radius is $$9Rs/8$$ the pressure at the centre becomes infinite. Buchdahl's theorem states that for any distribution of matter or equation of state, that the pressure term becomes infinite somewhere within the sphere for $$R<=9Rs/8$$ and that the sphere should collapse due to infinite gravitational force. However it is obviously not true for a sphere with a vacuum cavity, but we can note that whenever the pressure becomes infinite at a point in the liquid that the time dilation $$\sqrt{1-r_s/r}$$ term also goes to zero at that point. A perhaps better way of stating Buchdahl's theorem, that can cope with cavity case, is that any gravitational body that has a surface radius of $$R_s<R<9Rs/8$$ will contain an absolute horizon where the time dilation ratio goes to zero. Another interesting aspect of the equations is that for any $$R_s<R<9Rs/8$$ the pressure terms below the absolute horizon become negative but still real! Negative pressure means repulsive gravity. The time dilation terms below the absolute horizon also become negative but also remain real so we do not have to worry about awkward imaginary or complex values. Negative pressure at the centre of the forming black hole suggests that the singularity at the centre may have difficulty forming because of the repulsive gravity.

 

[pervect]

I haven't had time to look at this - I'd want to make sure that it satisfies the hydrodynamic equilibrium equation

http://en.wikipedia.org/w/index.php?title=Tolman–Oppenheimer–Volkoff_equation&oldid=379273952

and Einstein's field equations. I think that you shouldn't get a metric coefficeint of zero with a finite presure.

 

[Q-reeus]

For an arbitrarily large shell the finite forces can be made arbitrarily small, so in principle such an artificial black hole shell that does not collapse could be constructed without requiring material of infinite strength.

I'll conveniently leave the detailed maths to folks like yourself that are far more adept mathematically than myself. One thing is evident - stress distribution will be highly non-uniform as a function of radius, going obviously from zero at the outer surface and rising steeply at the inner surface - just touching the Schwarzschild radius 2M. A very delicate balance one would think! Perhaps you may have to nudge it ever so slightly out.

 

[yuiop]

I have noticed something interesting about the pressure equation for constant density perfect fluid given in #6:

$$P(r) = \frac{2}{3} p \frac{\sqrt{1-R_s r^2/R^3} - \sqrt{1-R_s/R}}{3\sqrt{1-R_s/R}-\sqrt{1-R_sr^2/R^3}}$$

If the outer surface is set to $$R=R_s$$ the equation becomes:

$$P(r) = -\frac{2}{3}p = \frac{-Mc^2}{2\pi R^3}$$

and the pressure is everywhere negative and independent of r so is uniform throughout the fluid. Can the negative gravity effect of this negative pressure offset the positive gravity of the mass of the fluid?

 

[PeterDonis]

When the surface radius is $$9Rs/8$$ the pressure at the centre becomes infinite. Buchdahl's theorem states that for any distribution of matter or equation of state, that the pressure term becomes infinite somewhere within the sphere for $$R<=9Rs/8$$ and that the sphere should collapse due to infinite gravitational force. However it is obviously not true for a sphere with a vacuum cavity, but we can note that whenever the pressure becomes infinite at a point in the liquid that the time dilation $$\sqrt{1-r_s/r}$$ term also goes to zero at that point. A perhaps better way of stating Buchdahl's theorem, that can cope with cavity case, is that any gravitational body that has a surface radius of $$R_s<R<9Rs/8$$ will contain an absolute horizon where the time dilation ratio goes to zero. Another interesting aspect of the equations is that for any $$R_s<R<9Rs/8$$ the pressure terms below the absolute horizon become negative but still real! Negative pressure means repulsive gravity. The time dilation terms below the absolute horizon also become negative but also remain real so we do not have to worry about awkward imaginary or complex values. Negative pressure at the centre of the forming black hole suggests that the singularity at the centre may have difficulty forming because of the repulsive gravity.

I'm confused--are you claiming that, even if there is a vacuum cavity inside the shell, the pressure can be non-zero inside the vacuum cavity? That's obviously wrong: pressure can only be nonzero where the stress-energy tensor is nonzero. For the case of a hollow shell, the pressure (in the radial direction) has to be zero on both the outer and inner surface of the shell; otherwise the shell will not be in equilibrium.

 

[yuiop]

I'll conveniently leave the detailed maths to folks like yourself that are far more adept mathematically than myself. One thing is evident - stress distribution will be highly non-uniform as a function of radius, going obviously from zero at the outer surface and rising steeply at the inner surface - just touching the Schwarzschild radius 2M. A very delicate balance one would think! Perhaps you may have to nudge it ever so slightly out.

As long as the outer surface is at 2.25M, moving the inner surface of the shell does not change anything because there is always a time (and pressure) singularity at the inner surface under those conditions. Nudging the outer surface out beyond 2.25 loses the event horizon which is the item of interest and the object just becomes something more ordinary like a neutron star.

I'm confused--are you claiming that, even if there is a vacuum cavity inside the shell, the pressure can be non-zero inside the vacuum cavity? That's obviously wrong: pressure can only be nonzero where the stress-energy tensor is nonzero. For the case of a hollow shell, the pressure (in the radial direction) has to be zero on both the outer and inner surface of the shell; otherwise the shell will not be in equilibrium.

At first the equations seemed to show changes in the pressure of the vacuum in the cavity and I was wondering if the vacuum can have a pressure of anything other than zero in the extreme conditions inside a black hole possibly as a result of vacuum energy manifesting itself or simple as a result of gravitational energy density. Does anyone have any references on this aspect? Obviously a vacuum with anything other than absolute zero pressure goes against common expectations so for now I will assume the pressure in the cavity vacuum is zero and assume the pressure equations are not valid in that region.

I have done some graphs showing the pressure profiles of the shell material as it collapses. The horizontal axis in the diagrams is the spatial axis extending from 0M on left to just over 2.5M on the right. The filled curve represents pressure in the shell material and the blue curve represents proper time rate relative to coordinate time. The dashed vertical line to the right represents the outer radius of the shell (R) and the solid black vertical line to the left of the yellow shaded region represents the inner radius (r) of the shell material. Everything to the left of that is the vacuum cavity. The yellow area represents a region of positive pressure.

In this first diagram the outer shell is at 2.25M and the pressure at inner surface has become infinite and and a absolute horizon (time singularity) has also formed there, despite the fact that the inner shell surface is still outside r=2M. This is basically Buchdahl's theorem that predicts that a pressure singularity (and time singularity) will form somewhere inside a spherical object no matter what the mass distribution or equation of state is inside the the object, when the outer surface is at 2.25M. At this point there is nothing that can prevent the shell collapsing further.

In the second diagram, after further collapse, it can be seen that a "pressure discontinuity" has formed inside the shell material with a surface that changes instantaneously from plus infinite pressure to minus pressure. The negative pressure is represented by the green shaded area. The negative pressure regions also have proper time running in the reverse direction to coordinate time. Since a positive pressure at the inner surface boundary requires that the inner surface should move inwards, a negative pressure suggests the inner surface should move outwards. Negative pressure also implies a negative force of gravity. For the sake of argument let's say these effects are insufficient to overcome the positive gravity of the outer shell material and the shell continues to collapse further.

In the final diagram the inner and outer surfaces have moved further inwards but as result in the change of mass distribution the pressure discontinuity moves rapidly towards the outer surface. As a result the negative pressure of the bulk of the shell material dominates the positive pressure of the outer skin of the shell which is vanishing and there is suggestion that this may cause a rebound of the shell outwards. Another result of the pressure singularity is that there is never a point when the inner shell surface is at zero pressure so that surface can never reach equilibrium and it raises the intriguing possibility that the shell would oscillate with a characteristic frequency that is a function of its mass. Could there be any connection between such an oscillation frequency and the characteristic thermodynamic temperature of a black hole?

Of course I recognise that the equations I have used are for static geometries with uniform mass density and the sequence I have described is very dynamic. I am hoping that the tensor experts on this forum may be able to come up with some answers for a similar thought experiment with more realistic parameters. Possibly such an analysis would require a finite element simulation on a university computer, but the spherical symmetry should ease that task.

 

[grav-universe]

The equation is determined by the gravitational potentials M/R and M_r/r with uniform density only. If you take away all of the mass within radius r, the gravity drops to zero within the cavity, but the gravitational potential does not, so you can no longer use just M_r/r = 0, but must apply something else, based upon the gravitational potential that still exists from the mass above r. The equation only stands as it is with uniform density, so if one is to vary that density by creating a cavity, then M/R and M_r/r must vary according to the gravitational potentials that still apply, producing an entirely different equation.

 

[PeterDonis]

At first the equations seemed to show changes in the pressure of the vacuum in the cavity and I was wondering if the vacuum can have a pressure of anything other than zero in the extreme conditions inside a black hole possibly as a result of vacuum energy manifesting itself or simple as a result of gravitational energy density. Does anyone have any references on this aspect? Obviously a vacuum with anything other than absolute zero pressure goes against common expectations so for now I will assume the pressure in the cavity vacuum is zero and assume the pressure equations are not valid in that region.

If you're using classical general relativity (i.e, no quantum physics), the only way to have a non-zero energy or pressure in the vacuum is to have a cosmological constant that's non-zero. If you're assuming a zero cosmological constant (which I assume you are), the vacuum must have zero density and zero pressure.

Of course I recognise that the equations I have used are for static geometries with uniform mass density and the sequence I have described is very dynamic.

That's the problem: if the shell outer radius is 9/8 times the Schwarzschild radius (2M) or less, I believe there is *no* static solution; the spacetime *has* to be dynamic (because the shell has to collapse into a black hole), because any static solution would have to have infinite pressure somewhere inside the material. It doesn't matter whether there's a vacuum cavity inside the shell or whether the shell contains material all the way down to the center. I think there's a general theorem that covers this case; I'll try to find a reference.

 

[pervect]

Your equation for pressure in #6 _almost_ matches up with MTW's on pg 610, but there may be a constant factor missing in yours:

MTW has rho_0 where you have (2/3) rho.

I'm going to assume that this is some very minor typo or miscalculation somewhere, but at this point it doesn't seem to have any major significance.

The pressure as defined by MTW is the pressure in a frame-field, as is the "constant density", which is also defined by MTW as the density in a frame field. And I assume that this is also the pressure you calculated.

In coordinate terms that means that the pressure, and density, are defined by some scaled local coordinates chosen to make the metric diagonal, giving the "density" and "pressure" an intuitive physical significance (one that would be lacking if we kept to strictly using the Schwarzschild coordinates). It also means that one does NOT need, or want , to consider the local time dilation factors when interpreting these pressures and densities - this has already been factored in.

So, when you look at the assumptions behind your solution, you are assuming that you have some physical substance that's so strong that it's density remains constant even as the pressure approaches infinity, as measured in some locally Minkowskian frame. This is very non-physical behavior for an equation of state, one would expect infinite pressure to cause any material substance to become compacted - offhand, I can't think of any limit that would cause the density to be less than infinite under infinite pressure - but perhaps this is too speculative. What isn't speculative is to say that it's very unrealistic and unphysical to consider a solution where the density remains constant under infinite pressure.