Would the disco ball really still end up inside a black hole?
Moderators: gmalivuk, Moderators General, Prelates
 Shufflepants
 Posts: 16
 Joined: Wed Jul 20, 2016 3:12 pm UTC
Would the disco ball really still end up inside a black hole?
PBS Space Time recently had a challenge question involving two different proposed solutions to saving the earth from being trapped inside a Kugelblitz. https://www.youtube.com/watch?v=v3hd3AI2CAA
They also released the solution video https://www.youtube.com/watch?v=q_oHv6HCMX4
In it they claim that the disco ball solution would not work because the event horizon would still form once the light shell reached a radius of 1 light second. But I'm unconvinced that the disco ball wouldn't work. I feel like the reasoning around it doesn't take into account that changes in spacetime geometry only propagate at the speed of light.
While all of the energy is at the distance of the alien ships before they fire, it's not concentrated enough to make a black hole yet.
And then for all incident photons on their entire journey inward to the disco ball, the space that they are traveling through is only warped to the exact same extent as it was at the beginning of their journey (which is to say not at all except for whatever warping occurs normally outside of this wacky scenario).
On their return trip back out they immediately run into warped space caused by some cone of the other nearby photons whose effects of their inward trip has had time to reach the space that the photon is now traveling through, but it's not the full effect because the effects of nearly 3/4 of the sphere still hasn't reached the space the photon is traveling through on it's return trip.
It seems to me that the event horizon would not form because there isn't enough time for all the gravity waves to propagate to all parts of the 1/2 light second radius sphere.
Now, I believe there is some radius at which if you built your disco ball, enough of the effects would have had time to propagate to to form a black hole anyway, but I'm not at all sure how to calculate that. But I feel like you'd need to calculate that radius to be sure 1/2 light second is inside it.
Am I wrong about this? Would the disco ball really not work at any radius less than 1 light second? Or are they just ignoring the finite speeds of gravity waves in their explanation and Penrose diagram?
They also released the solution video https://www.youtube.com/watch?v=q_oHv6HCMX4
In it they claim that the disco ball solution would not work because the event horizon would still form once the light shell reached a radius of 1 light second. But I'm unconvinced that the disco ball wouldn't work. I feel like the reasoning around it doesn't take into account that changes in spacetime geometry only propagate at the speed of light.
While all of the energy is at the distance of the alien ships before they fire, it's not concentrated enough to make a black hole yet.
And then for all incident photons on their entire journey inward to the disco ball, the space that they are traveling through is only warped to the exact same extent as it was at the beginning of their journey (which is to say not at all except for whatever warping occurs normally outside of this wacky scenario).
On their return trip back out they immediately run into warped space caused by some cone of the other nearby photons whose effects of their inward trip has had time to reach the space that the photon is now traveling through, but it's not the full effect because the effects of nearly 3/4 of the sphere still hasn't reached the space the photon is traveling through on it's return trip.
It seems to me that the event horizon would not form because there isn't enough time for all the gravity waves to propagate to all parts of the 1/2 light second radius sphere.
Now, I believe there is some radius at which if you built your disco ball, enough of the effects would have had time to propagate to to form a black hole anyway, but I'm not at all sure how to calculate that. But I feel like you'd need to calculate that radius to be sure 1/2 light second is inside it.
Am I wrong about this? Would the disco ball really not work at any radius less than 1 light second? Or are they just ignoring the finite speeds of gravity waves in their explanation and Penrose diagram?
 gmalivuk
 GNU Terry Pratchett
 Posts: 25789
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Would the disco ball really still end up inside a black hole?
For those who lack the time/inclination to watch the 8minute video:
Scenario: Aliens want to destroy the Earth by surrounding the solar system with their ships and sending a pulse of light toward us that has enough energy to create a black hole with a Schwarzschild radius of 1 lightsecond.
Disco Ball Solution: We build a perfectly reflective surface at a radius of 1/2 lightsecond, to send the light back out toward the offending ships.
Scenario: Aliens want to destroy the Earth by surrounding the solar system with their ships and sending a pulse of light toward us that has enough energy to create a black hole with a Schwarzschild radius of 1 lightsecond.
Disco Ball Solution: We build a perfectly reflective surface at a radius of 1/2 lightsecond, to send the light back out toward the offending ships.
Re: Would the disco ball really still end up inside a black hole?
spherical shells behave externally as if all their mass is at the centre. This is just as true in GR as it is in Newtonian gravity. Once the light pulse sphere is within the 1 lightsecond radius it will therefore behave as a black hole externally and thus won't be able to escape.
my pronouns are they
Magnanimous wrote:(fuck the macrons)
 Shufflepants
 Posts: 16
 Joined: Wed Jul 20, 2016 3:12 pm UTC
Re: Would the disco ball really still end up inside a black hole?
eSOANEM wrote:spherical shells behave externally as if all their mass is at the centre.
I know this is true of static systems and systems where you don't take into account the fact that gravity only propagates at the speed of causality, but I'm fairly certain that this is not necessarily true of dynamic systems where gravity propagates at the speed of causality.
And I know how to prove that fact for Newtonian systems with gravity propagating at infinite speed, but it seems like to show that in GR in a dynamic system with finite propagation speed is completely nontrivial.
Re: Would the disco ball really still end up inside a black hole?
eSOANEM wrote:spherical shells behave externally as if all their mass is at the centre. This is just as true in GR as it is in Newtonian gravity. Once the light pulse sphere is within the 1 lightsecond radius it will therefore behave as a black hole externally and thus won't be able to escape.
But which observers would be able to observe a sphere? An observer on earth certainly could (as the rays are timed to arrive simultaneously on earth). But what do the light rays observe? I think Shufflepants made the argument that the light rays aren't even aware of the existence of the far side of the light sphere, hence they do not observe a sufficiently dense sphere forming, hence they observe no event horizon, hence they do not need to care and will escape.
But, like Shufflepants, I lack the math to figure out the actual required radius.
And while we're nitpicking, their preferred solution won't save humanity. The energy is getting absorbed, the dyson sphere heats up, the energy is reemitted as blackbody radiation. A good portion of the energy will end up inside the sphere. Even if the energy getting inside is not enough to form a black hole, it's still enough to remove earth in all ways imagineable.
Re: Would the disco ball really still end up inside a black hole?
It doesn't matter who sees it as a sphere, just that there exists some frame where it is spherically symmetric (in which we can do our normal Schwarzschildy stuff). Also, the light definitely can't see anything because light doesn't really have a reference frame, it's kind of nonsensical.
Anyway, as for the nonstatic objection, in the system in question we have some frame where the stressenergy tensor is spherically symmetric (and so depends only on our radial and time coordinates). The Einstein Field Equations tell us that the curvature tensor is proportional to the stressenergy tensor and so must also be spherically symmetric.
We know that the Schwarzschild metric is the unique static, spherically symmetric vacuum solution (so outside a spherically symmetric matter distribution). You are right that the metric here is not quite the Schwarzschild one because of the fact it isn't static but in any sufficiently thin shell around this body, the collapse is slow enough that it will look locally like the Schwarzschild metric. The only parameter in the metric is the "mass" of the Schwarzschild body which we could upgrade to a function dependent on our r and t coordinates and, well, we already know what it should look like; it should be the sums of the energies including the gravitational binding energy. This last bit, together with the standard retarded potentials we expect should at least give us a very good first order solution (and the kugelblitz being proposed here is large enough that these approximate solutions should be valid for the entire experiment).
So yeah, to work out the actual point at which the kugelblitz forms, you can use the gravitational binding energy of a spherically symmetric shell and, well, if you do, you find that it forms once the kugelblitz collapses within the expected Schwarzschild radius.
For smaller kugelblitzen where this first order solution is no longer accurate at the horizon, this may break down a bit, but in this case, it's 100% valid.
Anyway, as for the nonstatic objection, in the system in question we have some frame where the stressenergy tensor is spherically symmetric (and so depends only on our radial and time coordinates). The Einstein Field Equations tell us that the curvature tensor is proportional to the stressenergy tensor and so must also be spherically symmetric.
We know that the Schwarzschild metric is the unique static, spherically symmetric vacuum solution (so outside a spherically symmetric matter distribution). You are right that the metric here is not quite the Schwarzschild one because of the fact it isn't static but in any sufficiently thin shell around this body, the collapse is slow enough that it will look locally like the Schwarzschild metric. The only parameter in the metric is the "mass" of the Schwarzschild body which we could upgrade to a function dependent on our r and t coordinates and, well, we already know what it should look like; it should be the sums of the energies including the gravitational binding energy. This last bit, together with the standard retarded potentials we expect should at least give us a very good first order solution (and the kugelblitz being proposed here is large enough that these approximate solutions should be valid for the entire experiment).
So yeah, to work out the actual point at which the kugelblitz forms, you can use the gravitational binding energy of a spherically symmetric shell and, well, if you do, you find that it forms once the kugelblitz collapses within the expected Schwarzschild radius.
For smaller kugelblitzen where this first order solution is no longer accurate at the horizon, this may break down a bit, but in this case, it's 100% valid.
my pronouns are they
Magnanimous wrote:(fuck the macrons)

 Posts: 660
 Joined: Fri Feb 07, 2014 3:15 pm UTC
Re: Would the disco ball really still end up inside a black hole?
I am fairly certain that I know the answer to the following question. I just want to double check, because GR is complex. If the aliens were 2 lightsecond away and the sphere had a diameter of 1.5 lightseconds, then would a black hole form?
Remember that the shell is shaped like a discoball (a.k.a. buckyball), so the light will reach the shell at different times and distances from the shell's center. Although treating the shell as a sphere is good enough to get an approximate solution, I do not want to rule out the possibility that the discoball's exact shape is a factor until some evidence is provided.
eSOANEM wrote:spherical shells behave externally as if all their mass is at the centre.
Remember that the shell is shaped like a discoball (a.k.a. buckyball), so the light will reach the shell at different times and distances from the shell's center. Although treating the shell as a sphere is good enough to get an approximate solution, I do not want to rule out the possibility that the discoball's exact shape is a factor until some evidence is provided.
 doogly
 Dr. The Juggernaut of Touching Himself
 Posts: 5212
 Joined: Mon Oct 23, 2006 2:31 am UTC
 Location: Somerville, MA
 Contact:
Re: Would the disco ball really still end up inside a black hole?
eSOANEM wrote:the collapse is slow enough
It's not going to be at c?
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Re: Would the disco ball really still end up inside a black hole?
Shit. Yeah. That's a goooood point
my pronouns are they
Magnanimous wrote:(fuck the macrons)
 Shufflepants
 Posts: 16
 Joined: Wed Jul 20, 2016 3:12 pm UTC
Re: Would the disco ball really still end up inside a black hole?
eSOANEM wrote:Shit. Yeah. That's a goooood point
I assume you're referring to the previous post pointing out that the collapse is happening at c, and that's exactly what led me down this path of not being convinced. If instead of a light shell, it was a bunch of subatomic bouncy balls that all some how fall in at some slow constant speed and then bounce away, I totally believe that everything would have time to equalize and it would form an event horizon almost immediately upon reaching a radius of 1 light second, well before the bouncy balls could bounce away.
But the fact that the light is infalling at the speed of causality, and the ricocheting back out at the speed of light, and this should be the same result even if we start the light shell at arbitrarily far away, it doesn't seem like any of the simple approximations would necessarily work here. After all, weird things can happen if a black hole were charged or spinning that results in horizons that form in different shapes and distances with the same mass.
 doogly
 Dr. The Juggernaut of Touching Himself
 Posts: 5212
 Joined: Mon Oct 23, 2006 2:31 am UTC
 Location: Somerville, MA
 Contact:
Re: Would the disco ball really still end up inside a black hole?
I have a strong feeling it's still stuck though.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?

 Posts: 63
 Joined: Mon Jun 15, 2009 9:56 pm UTC
 Location: Prague, Czech Republic
Re: Would the disco ball really still end up inside a black hole?
jewish_scientist wrote:Remember that the shell is shaped like a discoball (a.k.a. buckyball), so the light will reach the shell at different times and distances from the shell's center. Although treating the shell as a sphere is good enough to get an approximate solution, I do not want to rule out the possibility that the discoball's exact shape is a factor until some evidence is provided.
I think that the unsaid assumption is that the disco ball is also spherically symmetric.
Re: Would the disco ball really still end up inside a black hole?
Yeah, if there are enough satellites in the disco ball it's pretty much spherical. Watching the video, I assumed that they intended it to be spherically symmetric (or rather, approximately so) and the disco ball and the pic they flashed up were just illustrations
my pronouns are they
Magnanimous wrote:(fuck the macrons)
Who is online
Users browsing this forum: Yahoo [Bot] and 11 guests