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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

The short of it: Aliens fire a spherical shell of photons that have a total mass energy sufficient to create a black hole with a Schwarzschild radius of 1 light second at the Earth. The people of earth can build a perfectly reflective barrier at a radius of 1/2 light seconds that can reflect all of the incoming light and are able to violate the conservation of momentum so that the barrier will not be imparted with all of that momentum that a reflective surface normally would be.

Will the earth be saved? Or will it still be trapped in this massive Kugelblitz?

The answer provided by the solution video is that this disco ball solution will 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?

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    $\begingroup$ In general relativity, momentum is conserved. So the disco ball solution breaks general relativity, which means that there may be no consistent theory that will let us figure out what would happen. $\endgroup$ – Peter Shor Jan 6 '17 at 17:05
  • $\begingroup$ Why is violation of conservation of momentum required? If everything is perfectly spherically symmetric, the disco ball wouldn't have any momentum imparted to it. Or am I missing something here? $\endgroup$ – Michael Seifert Jan 6 '17 at 17:51
  • $\begingroup$ The conservation of momentum is just for simplicity, the real question is if the event horizon forms fast enough to include the reflected photon $\endgroup$ – Yukterez Jan 6 '17 at 18:09
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    $\begingroup$ @MichaelSeifert The violation of conservation of momentum is necessary because a photon reflecting off of a mirror imparts momentum on the mirror, and even if all the reflected photons didn't form a black hole, all the momentum imparted on the reflective surface would be enough to send the surface flying at the earth at an appreciable fraction of the speed of light and might be enough energy to form a black on its own. $\endgroup$ – Shufflepants Jan 6 '17 at 18:23
  • $\begingroup$ Hmm... I suspect I haven't explained myself very well, or perhaps I've misunderstood the situation. The photon impinging on the sphere from the $+z$ direction will push the disco ball in the $-z$ direction with a particular impulse; but the identical photon impinging on the sphere from the $-z$ direction will impart exactly the opposite impulse on the sphere. Because of spherical symmetry, all of the impulses delivered by the photons will cancel out. The net impulse delivered to the sphere should therefore be zero, and the sphere will not move. $\endgroup$ – Michael Seifert Jan 6 '17 at 18:56
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Our starting point is that the ingoing light is a spherical shell. Anything that maintains spherical symmetry does not radiate gravitational waves, so the geometry does not oscillate in any sense. As the shell sweeps inwards it leaves a static Schwarzschild geometry behind it, and that means by the time the light reaches your disco ball the light is already inside the event horizon and doomed to collapse to a singularity.

Describing what happens as the collapse proceeds turns out to be a complicated business, because you get different behaviour for different choices of the time coordinate. For the aliens watching outside the light shell slows to a halt as it approaches the Schwarzschild radius (yes, for external observers the speed of light changes near a black hole) and the event horizon never forms. The aliens would have to wait an infinite time before light even reached the horizon let alone crossed it.

For the light itself, well light has no rest frame so we can't ask what the light itself observes. And as discussed in the video for the observers inside the shell the radius decreases at a velocity of $c$.

The best approach I've seen for describing how light propagates inside an event horizon is a calculation done using the Gullstrand-Painleve coordinates, and indeed this is what I did in my answer to Why is a black hole black?. Rather than go thorugh the calculation again I'll just quote the results:

$$ \frac{dr}{dt_r} = c\left(-\sqrt{\frac{r_s}{r}} \pm 1\right) \tag{1} $$

where the $+1$ gives us the outbound velocity and the $-1$ gives us the inbound velocity. You need to be a bit careful with the Gullstrand-Painleve coordinates as this calculated velocity is not something any human observer could actually measure. However it is still physically relevant in the sense that zero veocity means the radial coordinate of the light is not changing while a negative velocity means the radial coordinate is decreasing.

And it should be obvious from equation (1) that for $r \lt r_s$ both the ingoing and outgoing velocities are negative, which means:

Even the ray of light directed outwards is still moving inwards towards the singularity

So while your disco ball might reflect the incoming light shell it would make no difference to your ultimate fate - it would only serve to delay it slightly. The light that is reflected outwards is still actuall moving inwards and will sweep your disco ball and you to an inevitable if spectacularly bright death!

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  • $\begingroup$ So photon A, which is approaching the ball from the right, does change the gravitational field on the left of the ball, where photon B approaches? The question is, how does the gravitational attraction of the photon make it to the opposite (left) side of the ball when the photon itself is stuck on the right side, and nothing, not even gravity, can travel faster than the photon itself? Since photon B is in front of photon A and vice versa they should only feel each others gravity as soon as they would have had the time to cross each other's paths, as long as they couldn't "see" each other they $\endgroup$ – Yukterez Jan 6 '17 at 18:35
  • $\begingroup$ ...should not feel any effects, and therefore also no gravitational effects, or do they? $\endgroup$ – Yukterez Jan 6 '17 at 18:36
  • $\begingroup$ It's that first paragraph I'm having trouble accepting when accounting for the finite speed propagation of gravity and the fact that the shell itself is collapsing at the speed of causality. It's my understanding that spacetime at any given point is shaped by exactly all of the mass/energy inside that point's past light cone. But for a point on the surface of the mirror at the moment that side of the light shell hits it, its past light cone doesn't include the entire shell at a distance smaller than 1 light second. $\endgroup$ – Shufflepants Jan 6 '17 at 18:37
  • $\begingroup$ Because of the relativity of simultaneity, there exist reference frames and locations from which the entire photon is never inside the 1 light second radius at the same time. Only times when part of it is inside. Once the rest of the shell has made it inside, parts of the shell will have already made it back outside. $\endgroup$ – Shufflepants Jan 6 '17 at 18:40
  • $\begingroup$ Not because of relativity of simultaneity (in the frame of the earth they are indeed inside the Schwarzschildradius at the same time), but because of the finite propagation speed of light and gravity it is not so clear if photon A could effect photon B before their paths could actually meet, and therefore photon A should act as if photon B would not exist, so it should be reflected since the only gravity it should experience would be that of the earth and the ball which were already there long enough to change the field at the point where the photon is now (but I'm not all sure about that) $\endgroup$ – Yukterez Jan 6 '17 at 18:54
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Shufflepants wrote: "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."

The answear lies in the Birkhoff's theorem, which states that as long as the situation is spherically symmetric the outside metric is reduced to the Schwarzschild solution, so it works with collapsing stellar matter as well as with the radially infalling photons. As soon as the photons get reflected they already get pulled back, even if the photons on the left did not have the time yet to influence the photons on the right.

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  • $\begingroup$ But Birkhoff's theorem only applies to an external observer who is and has been external to the mass for all time doesn't it? Here we're talking about an observer who starts out inside the the mass shell, and then transitions to outside the mass shell as it passes. $\endgroup$ – Shufflepants Jun 28 '17 at 15:00

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