Will a disco ball really end up inside a black hole?

Question

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?

Answer 1

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!

Answer 2

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.