This is actually one of the problems with black holes, the event horizon is listed as an impenetrable barrier, but if you emitted light (which is in principle no different from an electrical impulse) from just under the barrier what prevents it from traveling past the barrier and away?
And don't tell me that it curves away because that would be a violation of conservation of momentum.
It also can't travel directly up then back down, because light can't do that - at the moment of reversal it's not traveling at the speed of light.
And the light particle can't evaporate (i.e. red shift into nothingness) by loosing all its energy to the blackhole because that would be a violation of conservation of angular momentum (every photon caries some angular momentum).
So what happens to it?
I should mention that after studying all the physics that I could understand I do not believe black holes exist. (Super massive objects might, like the observed objects in the center of galaxies, but not as black holes.)
Gravity causes or is (depending on how you look at it) curvature of space so the light goes "straight" but doesnt leave. Sorry if that means it "curves away" but that's what having an escape velocity greater than the speed of light means.
And what does that have to do with conservaton of momentum? If i throw a rock upwards it curves away rather than leaving the solar system — momentum is still conserved and the rock has much more than a photon.
The object can not suddenly switch from moving straight to a curve. That's what I mean. In order to curve something must cause it to do so, but there are no lateral forces on it (or more accurately the forces in all directions are exactly balanced).
It is still moving in a straight line. Large amounts of gravity tend to distort spacetime, such that a "straight line" appears bent to a distant observer.
It's the same phenomenon as what causes gravitational lensing.
Different effect. A rock is actually following a curved trajectory through ordinary space. Light always goes straight - but in these extreme circumstances space itself is curved. You can easily verify this by looking at said curving rock: the fact that you can see it without distortions (or at all) implies that its region of space has insignificant curvature and the light from the event is reaching your eyes is an easy indication of it.
Ordinary masses bend space nowhere nearly enough. The Sun bends space just barely enough for the effect to be observable in a solar eclipse (it was used as a verification of Einstein's theories) by causing stars behind it to appear ever so slightly out of place when its disc is just about to overlap them. The light from these stars is NOT being bent - it keeps going in a straight line, but the Sun's gravity slightly alters the very idea of straight in its immediate vicinity. The effect is a tiny distorsion, about 0.00048 degrees.
And the Sun is not exactly small.... but it's not dense enough to have a strong effect. Nothing in our everyday experience is. If you could stuff all of its mass into a tiny volume then in its immediate vicinity you could experience this.
if you emitted light (which is in principle no different from an electrical impulse) from just under the barrier what prevents it from traveling past the barrier and away?
The "barrier" you speak of (the event horizon) is already traveling outward at the speed of light. Gravity curves spacetime so much at the horizon that a light beam traveling radially outward stays at the same radius forever. Inside the horizon, spacetime is curved even more strongly, so that a light beam traveling radially outward still gets pulled inward towards the singularity.
Which way? To the left or to the right? There is always a point where the curvature is exactly balanced.
The Hairy ball theorem requires it.
Are you somehow claiming that traveling in a straight line away from the black hole leads toward the black hole? If that were the case then gravity would be pulling objects away from the black hole, and toward it, at the same time, which makes little sense.
And if it existed anyway, despite making no sense, that would imply there is a point away from the black hole where the gravity of the black hole is exactly balanced (i.e. you could hover there forever), which would a stunningly huge thing to say.
If you mean which way is the event horizon traveling, it's traveling radially outward. That is a well-defined direction all by itself; there is no "left or right" involved.
The Hairy ball theorem requires it.
If you mean the theorem that says there can't be an everywhere non-vanishing vector field on a sphere, that theorem doesn't apply in this case because the manifold of the black hole spacetime is not a sphere (meaning, topologically it's not a sphere).
Are you somehow claiming that traveling in a straight line away from the black hole leads toward the black hole?
No. I'm saying that inside the horizon, even something that is moving as fast as it can away from the black hole still has a radial coordinate that decreases with time. That's because of the way the spacetime is curved; your intuitions about how "space" works don't work in a region of spacetime that is curved as strongly as the region inside the black hole's horizon.
that would imply there is a point away from the black hole where the gravity of the black hole is exactly balanced
How does it imply this? I don't understand your argument here.
No, I mean which way is the photon traveling - left or right? (Relative to a line pointing "down".) You can't just decide to arbitrarily curve - you have to curve in a particular direction.
> How does it imply this? I don't understand your argument here.
Emit a photon from just inside the event horizon, traveling directly away from "down".
What does the photon do?
Does it go into orbit? (What gave it the angular momentum to do that?) Does it hit the black hole? (How did it turbn around without ever curving?) Does it just travel forever thinking it's moving away from the black hole, but not actually going anywhere? (i.e. redshift into nothingness, since it's constantly fighting gravity)
Something else?
Tell me what it does from the POV of the photon, not the black hole.
Everything we "know" about the inside of a black hole is speculation - there is no way we can observe what's going on inside.
That said, from just inside the black hole - what is "down"? Towards the singularity? Does that even make sense when you'll hit the singularity no matter what direction you travel?
From the POV of the photon, nothing special happens. You get emitted, you move towards the singularity, and then we have no idea what happens next. You might happen to take a longer path, but that's it.
No, I mean which way is the photon traveling - left or right? (Relative to a line pointing "down".)
The photon is traveling radially outward; that means it is traveling in a direction opposite to a line pointing "down". At least, that's the way it's traveling spatially. In spacetime, a photon at the horizon is traveling along a curve of constant radial coordinate r (and constant angular coordinates theta, phi if you include them). A photon inside the horizon is traveling along a curve of decreasing radial coordinate r; inside the horizon even outgoing null curves (the worldlines of outgoing light beams) have decreasing r.
Emit a photon from just inside the event horizon, traveling directly away from "down".
What does the photon do?
According to an observer that is falling inward just inside the event horizon, the photon moves radially outward at the speed of light.
Does it go into orbit?
No. As you say, it has no angular momentum, but that's not the primary reason; the primary reason is that there are no "orbits" inside the horizon. In fact, there are no "orbits" inside a radius of 3/2 the horizon radius; at that radius, a photon can orbit the hole in a circular orbit.
Does it hit the black hole?
Eventually, yes.
(How did it turn around without ever curving?)
It didn't. The spacetime itself is curved inside the horizon to such an extent that even a photon traveling radially outward ends up hitting the singularity.
Once again, your intuitions about how "space" works and how things "travel in space" break down inside the horizon. You are thinking of a point with radius r < 2M, i.e., just inside the horizon, as a "place in space". It isn't. It's more correct to think of it as a "moment of time". The horizon itself is also not a place in space; it's more correct to think of it as an outgoing light beam.
Does it just travel forever thinking it's moving away from the black hole, but not actually going anywhere?
This is one way (but probably not the best way) to visualize what a photon exactly at the horizon does; it is moving radially outward at the speed of light, but because of the curvature of spacetime at the horizon it stays at the horizon.
(i.e. redshift into nothingness, since it's constantly fighting gravity)
"Redshift" is relative; you have to specify what observer is receiving the photon and measuring its frequency.
Tell me what it does from the POV of the photon, not the black hole.
There is no such thing as "the POV of the photon"; photons don't have "rest frames" in the usual sense of that term. I said above what the photon does from the POV of an observer falling into the hole.
My suspicion (ie no facts) would be that either the light just does make it to the event horizon to leave because of time dialation effects, or we can't answer it because our math breaks down at the horizon so we can't say anything about the universe past it.
And don't tell me that it curves away because that would be a violation of conservation of momentum.
It also can't travel directly up then back down, because light can't do that - at the moment of reversal it's not traveling at the speed of light.
And the light particle can't evaporate (i.e. red shift into nothingness) by loosing all its energy to the blackhole because that would be a violation of conservation of angular momentum (every photon caries some angular momentum).
So what happens to it?
I should mention that after studying all the physics that I could understand I do not believe black holes exist. (Super massive objects might, like the observed objects in the center of galaxies, but not as black holes.)