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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?

Shufflepants
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    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. – Peter Shor Jan 06 '17 at 17:05
  • 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? – Michael Seifert Jan 06 '17 at 17:51
  • The conservation of momentum is just for simplicity, the real question is if the event horizon forms fast enough to include the reflected photon – Yukterez Jan 06 '17 at 18:09
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    @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. – Shufflepants Jan 06 '17 at 18:23
  • 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. – Michael Seifert Jan 06 '17 at 18:56
  • Perhaps I'm making implicit assumptions about the disco ball being perfectly rigid, though? – Michael Seifert Jan 06 '17 at 19:05
  • @MichaelSeifert yeah, that would work too, it's just that in the original formulation of the disco ball it was a network of satellites rather than some actual rigid ball. And somehow I feel more comfortable with fiating this magic ignoring conservation of momentum than in assuming some surface that can transfer forces across it faster than the speed of light :) – Shufflepants Jan 06 '17 at 20:08
  • I don't think this question can have a reasonable answer. The conservation of energy and momentum are inextricably linked in relativity. Consider the following thought experiment. You have just fallen into a black hole. But you know how to violate the conservation of energy. You produce a negative energy object with energy equal to the mass of the black hole. There now is no black hole, and you can escape. Does this work? I don't know. Producing energy from nothing gives a space-time which is inconsistent with Einstein's equations. Does GR even give a prediction for this? – Peter Shor Jan 07 '17 at 20:34
  • Rigidity aside, the problem remains: how can the photon hitting the left mirror curve the spacetime on the right side of the reflecor soon enough to capture the other photon and vice versa, if at the time of the reflection the changes in the gravitational field produced by the other photon is still 2 lightseconds (the diameter of the ball) and therefore 2 seconds away? The changes in the field propagate with c, so the right photon does not know about the left photon and neither does the spacetime (yet). That would take D/c = 2 sec, and 2 sec later the reflected photons are already gone – Yukterez Jan 08 '17 at 07:29
  • So, the last thing you see after you plunge through an event horizon is Jon Travolta in a white suit. – Selene Routley Jan 11 '17 at 23:24

2 Answers2

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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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John Rennie
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  • 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 – Yukterez Jan 06 '17 at 18:35
  • ...should not feel any effects, and therefore also no gravitational effects, or do they? – Yukterez Jan 06 '17 at 18:36
  • 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. – Shufflepants Jan 06 '17 at 18:37
  • 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. – Shufflepants Jan 06 '17 at 18:40
  • 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) – Yukterez Jan 06 '17 at 18:54
  • I think Gullstrand-Painlevé coordinates should come with a health warning. Einstein was not fond of them, see Wikipedia, for good reason. The black hole is black because the "coordinate" speed of light goes to zero. And it can't go lower than that. So to claim that the outward-directed light beam is "falling inward" like the waterfall analogy is IMHO badly wrong. – John Duffield Jan 07 '17 at 13:57
  • I think your answer is on the order of Assume 0 = 1. Then 13 is not prime. Violations of the conservation of momentum are inconsistent with general relativity, so you can prove anything if you assume them. – Peter Shor Jan 07 '17 at 20:40
  • Conservation of momentum is not really the point of the question. The real problem is: how can the photon hitting the left side of the ball already attract the photon hitting the right side of the ball if the speed of changes is the field is only c, and the left photon is still 2 lightseconds away from the right side of the ball at the moment of reflection - the gravitational force of the left photon should not have any effects on the right side until 2 seconds after the reflection, but at this time the reflected and now outgoing photon is already 2 lightseconds away from the reflecting point – Yukterez Jan 08 '17 at 00:54
  • I'm not sure if Schwarzschild or Gullstrand coordinates describe this problem correctly since they assume a static spacetime while in this scenario we have an n-body (or n-photon) problem where the propagation speed of the changes in the gravitational field might be essential. I still don't get how the photon approaching from the left could change the gravitational field on the right when the seperation between the two points is still 2 lightseconds. 2 sec later when the curvature changes have propagated across the diameter the reflected photons should already be outside the critical radius! – Yukterez Jan 08 '17 at 07:15
  • @СимонТыран: GR ios a local theory so it relates changes in the local curvature to changes in the local stress-energy tensor. But in this case we have imposed a global spherical symmetry, and this causes the geometry to be Schwarzschild. Photons don't need to know what photons on the opposite side of the shell are doing. The local behaviour plus the global symmetry is enough to make the geometry Schwarzshild. If the system was less symmetric that would be a very different matter and we would get an initially time dependent oscillating geometry. – John Rennie Jan 09 '17 at 06:28
  • I see, I also also found out that Birkhoff's theorem applies in this scenario. I'll correct my own answear and upvote yours, thanks for the explanation. – Yukterez Jan 11 '17 at 23:18
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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.

Yukterez
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  • 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. – Shufflepants Jun 28 '17 at 15:00