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So I was toying around with the idea of blackholes,

What I'm consider a spherical structure $M$, consisting of an extremely rigid material forming it's surface, that is otherwise hollow. So now suppose $M$ is so dense that it is contained in its Schwarzschild radius, but the material is so rigid that it doesn't actually collapse.

Then at the absolute center of $M$ the gravitational force one would experience is 0, (in fact there exists a sphere of non-zero volume around the center for which one doesn't experience significant amounts of gravitational pull)

In that case, $M$ is a black hole, but it doesn't carry a singularity, i.e. it's an event horizon enveloping what is otherwise a pretty normal piece of space.

Now, I don't understand them too well, but it appears that all black holes are believed to carry singularities, does that imply that this type of structure is impossible? If so, can one then derive relativistic-bounds on rigidity of a material using this?

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  • $\begingroup$ Layers of molecules would simply be stripped off. $\endgroup$
    – Jimmy360
    Aug 1 '16 at 2:33
  • $\begingroup$ As a thought experiment, try working this out for a structural ring, not a sphere. The math is quite a bit simpler and the advantage of a rigid ring is that you can rotate it, and the rotation would help resist the gravity. It still doesn't work. It stops working well before it gets close to a black hole's necessary mass to radius ratio. The larger you build the less of a problem tidal forces are, but it's never possible to achieve your goal of turning the object into a stable structure that's also a black hole. $\endgroup$
    – userLTK
    Aug 1 '16 at 3:13
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Indeed no such structure is possible, and one can look at this as saying something about rigidity.


First some calculations

Consider a part of such a structure. It's Schwarzschild coordinates $(t,r,\theta,\phi)_\mathrm{S}$ will be $x^\mu = (t,r_0,\theta_0,\phi_0)_\mathrm{S}$, where $t$ is an arbitrary parameter that I've chosen to agree with the standard time coordinate, and the other values are fixed. That is, the point doesn't move except through time.

Let's now work in Kruskal–Szekeres coordinates $(T,X,\theta,\phi)_\mathrm{KS}$, which carry through the event horizon without any issue. (We could work in Schwarzschild too and get the same answer, but this is just to be on the safe side.) The transformation is given by \begin{align} T & = \begin{cases} \sqrt{\frac{r}{2M}-1} \mathrm{e}^{r/4M} \sinh\Big(\frac{t}{4M}\Big), & r > 2M; \\ \sqrt{1-\frac{r}{2M}} \mathrm{e}^{r/4M} \cosh\Big(\frac{t}{4M}\Big), & r < 2M; \end{cases} \\ X & = \begin{cases} \sqrt{\frac{r}{2M}-1} \mathrm{e}^{r/4M} \cosh\Big(\frac{t}{4M}\Big), & r > 2M; \\ \sqrt{1-\frac{r}{2M}} \mathrm{e}^{r/4M} \sinh\Big(\frac{t}{4M}\Big), & r < 2M; \end{cases} \\ \theta & = \theta; \\ \phi & = \phi. \end{align}

In these coordinates we have $x^\mu = (T,X,\theta_0,\phi_0)_\mathrm{KS}$, where $T$ and $X$ both depend on our worldline parameter $t$ (and fixed parameter $r$). Letting dots denote differentiation with respect to $t$, we have $\dot{x}^\mu = (\dot{T},\dot{X},0,0)_\mathrm{KS}$, where $$ \dot{T} = \begin{cases} \frac{1}{4M} \sqrt{\frac{r}{2M}-1} \mathrm{e}^{r/4M} \cosh\Big(\frac{t}{4M}\Big), & r > 2M; \\ \frac{1}{4M} \sqrt{1-\frac{r}{2M}} \mathrm{e}^{r/4M} \sinh\Big(\frac{t}{4M}\Big), & r < 2M; \end{cases} $$ and $$ \dot{X} = \begin{cases} \frac{1}{4M} \sqrt{\frac{r}{2M}-1} \mathrm{e}^{r/4M} \sinh\Big(\frac{t}{4M}\Big), & r > 2M; \\ \frac{1}{4M} \sqrt{1-\frac{r}{2M}} \mathrm{e}^{r/4M} \cosh\Big(\frac{t}{4M}\Big), & r < 2M. \end{cases} $$

The metric is given by $$ ds^2 = \frac{32M^3}{r} \mathrm{e}^{-r/2M} (-\mathrm{d}T^2 + \mathrm{d}X^2) + r^2 (\mathrm{d}\theta^2 + \sin^2\theta \mathrm{d}\phi^2). $$ With this we can evaluate for our worldline the quantity $$ g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu = \frac{2M}{r} - 1, $$ which holds for all $r$ both inside and outside the horizon.


Now some analysis

The quantity $g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu$ is the inner product of the tangent to the worldline with itself. In particular, a particle's $4$-velocity $u^\mu$ obeys $$ g_{\mu\nu} u^\mu u^\nu = g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu \Big(\frac{\mathrm{d}\tau}{\mathrm{d}t}\Big)^{-2}, $$ where $\tau$ is the particle's proper time. But for a massive particle, $g_{\mu\nu} u^\mu u^\nu = -1$ always. Thus we can solve to find $$ \frac{\mathrm{d}\tau}{\mathrm{d}t} = \pm \sqrt{1-\frac{2M}{r}}. $$

The sign determines whether or not we are moving forward or backward in time. But note that something terrible happens for $r < 2M$: the right-hand side becomes imaginary. That is, there is no solution for a massive particle traveling along the worldline we proposed, which remember was defined to be constant in radius and angular coordinates.


What this says about the scenario

No massive particle can resist moving toward the center of the black hole. No acceleration (and it is finite intermolecular forces that apply finite accelerations to prevent particles in rigid objects from free falling) will make that worldline timelike. Indeed, once inside the event horizon, moving toward the singularity is moving into the future.

Material properties are subservient to the structure of spacetime. Just as you could not design a material that was so strong it can avoid moving forward in time, you cannot design a material that can hold up against gravity inside a black hole.

You can do a more general analysis allowing for $\theta$ and $\phi$ to vary, and the result will be the same. Moreover, you can show that any timelike worldline inside the event horizon will hit the singularity in finite proper time -- you cannot even let you structure continuously shrink but never hit the center. If the black hole is spinning things get... interesting... but that's an analysis for another time.

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  • $\begingroup$ I like the fact that there are some proper calculations here! I was actually wondering about whether this analysis is correct when the particle is sitting exactly at the boundary of the spherical shell (not inside, not outside). According to Birkhoff's theorem, the solution is Schwarzschild outside the shell. I remember having a heated debate about this with a fellow student during an EM course some 3 years ago on a similar problem concerning a charged spherical shell. $\endgroup$
    – OTH
    Aug 1 '16 at 14:30
  • $\begingroup$ From what I remember, it was argued that the case is different when we take a thin charge layer and take the limit of infinitely thin layer. $\endgroup$
    – OTH
    Aug 1 '16 at 14:37
  • $\begingroup$ There is no "moving forward or backward in time". A clock features some kind of regular cyclical motion, time is a measure of motion, you can't move through it. "Something terrible" happens at $r = 2M$ which corresponds to the coordinate speed of light being zero at the event horizon. It can't go lower than that. And please do note that a wordline represents the motion of a particle through space over time in the static 3+1 dimensional arena called spacetime. This models space at all times, there is no motion in this static arena, the particle does not move along its worldline. $\endgroup$ Aug 4 '16 at 15:22
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What I'm consider a spherical structure $M$, consisting of an extremely rigid material forming it's surface, that is otherwise hollow. So now suppose $M$ is so dense that it is contained in it's Schwarzschild radius, but the material is so rigid that it doesn't actually collapse.

It's not realistic, because in practice you wouldn't be able to form such a structure. But never mind, it's good to consider such things.

Then at the absolute center of $M$ the gravitational force one would experience is 0, (in fact there exists a sphere of non-zero volume around the center for which one doesn't experience significant amounts of gravitational pull)

Fair enough. Note that for an ordinary gravitating body the force of gravity at some location is related to the the local gradient in gravitational potential. This in turn relates to the local gradient in the "coordinate" speed of light. You can effectively measure this using optical clocks at different elevations. Clocks run slower when they're lower. If you're at a location such as the centre of the Earth where all the clocks around you are running at the same rate, you are at a location where there is no gravitational force.

In that case, $M$ is a blackhole, but it doesn't carry a singularity, ie its an event horizon enveloping what is otherwise a pretty normal piece of space.

It isn't particularly normal, but nevermind. We could think of some gedanken location within the event horizon of a supermassive black hole.

Now, I don't understand them too well, but it appears that all blackholes are believed to carry singularities

That's what people say. But if you were to read the Einstein digital papers I suspect you would not agree with the claim that there's a point-singularity at the centre of every black hole. The crucial point is this: "the curvature of light rays occurs only in spaces where the speed of light is spatially variable".

does that imply that this type of structure is impossible?

No. You will find people claiming that once a body is within the event horizon, it is going to end up in the central singularity. But note that at the event horizon, the coordinate speed of light is zero, and it can't go lower than that. This means there is no force of gravity within the event horizon. So your structure will not fall down.

If so, can one then derive relativistic-bounds on rigidity of a material using this?

No. The scenario is too unrealistic for that. But what you can derive is some bounds on abstraction. For example, if anybody tells you about Kruskal-Szekeres coordinates, take it with a pinch of salt, because they contain a schoolboy error. They replace the t coordinate with a new time coordinate which totally ignores the infinite gravitational time dilation at the event horizon. At this location your optical clock stops. Kruskal-Szekeres coordinates effectively sit a stopped observer in front of a stopped clock and claim that he sees it ticking normally. He doesn't, because the clock is stopped, and so is he. You can spot a similar issue on the MTW page below. The Schwarzschild chart on the left features a chopped-off time axis and an infalling body that a) goes to the end of time and back and b) is in two places at once. The Kruskal-Szekeres chart on the right makes this "well behaved", but it's a mathematical fairy tale I'm afraid.

enter image description here

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