3
$\begingroup$

https://en.wikipedia.org/wiki/Observable_universe
Mass (ordinary matter) 10^53 kg
Ordinary (baryonic) matter (4.9%)
Diameter 8.8×10^26 m


OU=observable universe
SR=Schwarzschild radius

Mass including dark matter/energy = 10^53 kg / 4.9% = 2E54 kg

SR = 2*G*m/c^2 = 3E27 m = 7 times the radius of the OU.

Anything wrong so far? I want to write that in wikipedia. Any unaccounted effect of matter expanding thus gravity at a distance not reflecting the position of where matter is now or so? I read the SR of the OU is equal to the radius of the OU, making the OU "flat".

Normally I would call a mass in a sphere 1/7 of it's SR a black hole. But here I'm assuming that the universe is filled with matter fairly evenly (despite irregularities like galaxies, supermassive black holes, stars or voids on a small scale :), so gravity generally cancels and everything is free to go wherever it wants and cross Schwarzschild radii which are everywhere depending on where you pick the center of the OU.

Now, would it be possible to get stuff out of a black hole if it's gravity was cancelled by nearby matter similar to the OU? Maybe a black hole cluster with overlapping event horizons. Would singularities remain (naked?) singularities and stuff could escape the singularities it was originally bound to? Could stuff eventually escape the black hole cluster, perhaps if there were also non singularities like neutron stars and normal stars in the cluster?

$\endgroup$

2 Answers 2

1
$\begingroup$

Because the universe is expanding faster than light. This allows it to escape its own Schwarzschild radius. Dark energy also helps but that is not strictly necessary. Also, the "observable universe" could be much smaller than the entire universe.

But wait, how is faster-than-light allowed!?

Special relativity ensures two particles can't exceed the speed of light relative to each other when they are close to each other. Expansion doesn't break this rule because the velocity-difference of nearby co-moving points is tiny.

General relativity with realistic matter/energy and boundary conditions ensures that there can be no time travel. This also rules out warp drives. A pair of faster-than-light travel tubes, for example, can be used to make a time-machine: moving faster-than-light in one reference frame is traveling back-in-time in some other frame. A stationary tube and a (slower-than-light) moving tube, if set up properly, can be used to travel back in time. But again, there is no way to do this using the expansion of the universe.

Apart from the universe itself, there is no way to create a non-blackhole that is smaller than its Schwarzschild radius. An object that is expanding faster than light cannot have an outside. A stationary particle outside the expanding object would either collide with or be pushed away from it faster than light, which would violate one of the two rules above.

$\endgroup$
4
  • $\begingroup$ But even if the OU were not expanding stuff could get out. Imagine a static universe filled with iron at 300 K. Every point is the center of a mass large enough to be within a SR, but since gravity cancels there is no force on particles and they can travel anywhere, even across Schwarzschild spheres, which are not event horizons in this case. $\endgroup$
    – darsie
    Aug 20, 2016 at 8:47
  • $\begingroup$ @darsie no it would collapse. Each atom would see all the other atoms going toward them. See the en.wikipedia.org/wiki/Friedmann_equations. $\endgroup$ Aug 20, 2016 at 12:14
  • $\begingroup$ the OP says that the expected radius of the universe is strictly lower than its Schwarzschild radius. Meaning we are in a super BH. $\endgroup$
    – user46925
    Aug 20, 2016 at 15:18
  • $\begingroup$ @kevin You mean it would collapse similar as it's expanding now? Ok, but still there would be no one way event horizons. Thermal electrons could go both ways. $\endgroup$
    – darsie
    Aug 20, 2016 at 16:17
1
$\begingroup$

A cluster of black holes with overlapping event horizons has a much simpler name: one black hole.

Sending two black holes towards each other causes some very strange effects, but matter escaping from inside isn't one of them.

As two black holes approach, both deform, allowing their mutual equipotential surface to remain outside the event horizon until the moment of impact, whereupon they merge and form a sphere. (Until then it's possible to follow the equipotential plane between them without falling into either one, even as their centres are separated by less than the sum of their Schwarzchild radii.)

Of course the "moment of impact" isn't a "moment" in the classical sense because one runs up against the limits of simultaneity under relativity. Instead the two hemispheres appear to zip together starting at the point nearest the observer and proceeding away around the circumference at the speed of light. So an observer momentarily sees a "dumbell" simply because they're seeing different parts at different times.

There are several theories of what the interior of a black hole is like. Some figure it's essentially an infinitely deep hole into which matter falls forever; or perhaps stops due to quantum effects before the diameter shrinks to the planck length. Some figure that normal space simply stops at the event horizon, and anything falling in is essentially squished into a 2-D shadow on impact, and "interior" is a meaningless concept.

Whilst the event horizon briefly recedes from an approaching massive object, it's at a lower velocity than the approaching object, and so can't overtake any matter already inside the black hole (if indeed "interior" actually has any meaning.)

The stupendous energy released goes mostly into gravity waves, which briefly causes ripples in the new event horizon

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.