How Do Black Holes Emit Radio Waves if They Only Suck Things In?

[Siadeug]

From my limited understanding, black holes supposedly have "event horizons" where everything near them just falls inside the black hole and becomes nothingness, totally anhilated. If this is the case then how is it that black holes emit stuff, like radio waves? If they suck things in and then spit things back out, isn't that the opposite of what a black hole is really supposed to do? I would really like to gain a better understanding of black holes. Thank you!

 

[mgb_phys]

Anything that crosses the event horizon of a black hole is gone for good.
However all the material that is attracted to a black hole can't immediately fall in. Because of conservation of angular momentum it forms a disk around the black hole at large a distance from the event horizon. It is the heating of the material in the disk from the attraction of the black hole and the acceleration of new material arriving that creates the x-rays and radio waves

But yes it is a little odd that 'black' holes are the among the brightest sources of energy in the sky!

 

[Siadeug]

Well that makes sense.. I have some more newbie questions. I've read that black holes grow larger as they consume more things. Doesn't that mean that the matter is not getting anhilated? Is a black hole actually made of some sort of matter then? I don't see how "nothingness" can really grow. And, of course, a most basic question: why does a black hole even happen? I understand that the gravity of a black hole is very strong and that it's basically an imploded star but why do black holes form when stars die?

 

[DaveC426913]

Well that makes sense.. I have some more newbie questions. I've read that black holes grow larger as they consume more things. Doesn't that mean that the matter is not getting anhilated?

The matter degenerates to a form that we don't yet understand. But it is still there. As is its mass.

And, of course, a most basic question: why does a black hole even happen? I understand that the gravity of a black hole is very strong and that it's basically an imploded star but why do black holes form when stars die?

It is quite simply gravity gone wild.

There are various lesser forms of degenerate stars: white dwarf, neutron star.

A neutron star is gravity gone almost wild. There is just enough gravity to crush the matter of the star to the point where the electron and protons combine into neutrons and the neutrons get packed almost edge-to-edge.

 

[Ultrastar 1]

First, black holes don't 'spit stuff out". They only emit waves (radio,infared etc.) The reson they do that is because of their intense gravity. I suggest that you google black holes and see what you find out about them.

 

[Nabeshin]

First, black holes don't 'spit stuff out". They only emit waves (radio,infared etc.) The reson they do that is because of their intense gravity. I suggest that you google black holes and see what you find out about them.

Be cautious with the word emit. Most people define the object "a black hole" to be everything contained within the event horizon, and this region certainly doesn't emit anything. The only candidate would be the proposed Hawking radiation that happens at the event horizon. The common x-ray emissions are, like mgb says, created by accreting gas, which really doesn't have anything to do with the black hole itself. The same happens any compact object accreting gas, say, a neutron star.

 

[protonchain]

As others have already mentioned, the black hole itself doesn't emit anything (except perhaps hawking radiation).

The radio jets and x-ray/gamma ray emissions are due to infall of material from nearby stars and clusters of stars as it forms a massive accretion disk of gas around the black hole. This matter, as it slowly gets pulled in by gravity is obviously increasing in temperature due to the acting gravitational force from the black hole. In turn, the increase in energy of these atoms and subatomic particles results in the emission of high energy radiation. Radio jets are also formed from similar reactions but that radiation "travels" along the magnetic poles of the BH (hence the jet).

 

[Unit]

i am venturing a guess that anything that radiates from the surface of the event horizon must do so perfectly perpendicular to the surface? at the speed of light? (the escape velocity AT the event horizon is the speed of light, yes?) what would happen if the radiated wave or particle had a velocity that was not perpendicular? would it "fall" back in?

also, this might be another topic, but since when is gravity so "intense"? it's the weakest of the four (or three) forces; would it ever be possible to have a large gathering of, let's say, electrons, all negatively charged, so collectively together as if in a black hole?

 

[DaveC426913]

i am venturing a guess that anything that radiates from the surface of the event horizon must do so perfectly perpendicular to the surface? at the speed of light? (the escape velocity AT the event horizon is the speed of light, yes?) what would happen if the radiated wave or particle had a velocity that was not perpendicular? would it "fall" back in?

You're looking at it backwards. The radius within which light cannot escape defines the event horizon.

If some photons are on a trajectory [say, that is not directly away from the centre of the BH] and they fall back in, then the EH is above that. If, OTOH, there are photons that are able to escape, that the EH is below that.

So, the question is: looking at a given photon, did it escape or did it fall back in?

 

[Unit]

to be honest i don't quite understand how I'm seeing it backwards, but maybe it's because I'm tired and sleepy. either way, i hope I'm not being redundant in asking this:

i imagine the event horizon to be a sphere, a very large sphere, without a tangible surface of course. let's say it is 10 km from the centre of the black hole. if a photon, whose velocity is radially outward away from the centre of the black hole, is placed just outside the edge of this event horizon (let's say, 10.00001 km from the centre of the black hole), will it escape? what would happen if the photon's velocity was NOT perpendicular to the "surface" of the event horizon sphere, and instead had an angle of 80 degrees to it. that would mean that the vertical, "escaping" component would be 0.98c. would it escape then?

also,

The radius beyond which light cannot escape defines the event horizon.

By this, I assume you mean the radius within which, yes?

 

[DaveC426913]

also, By this, I assume you mean the radius within which, yes?

Why, whatever are you talking about?

 

[DaveC426913]

to be honest i don't quite understand how I'm seeing it backwards, but maybe it's because I'm tired and sleepy.

Because you're defining the EH first, then asking what the photons will do second.
You can only define the EH by determining what the photons did.

either way, i hope I'm not being redundant in asking this:

i imagine the event horizon to be a sphere, a very large sphere, without a tangible surface of course. let's say it is 10 km from the centre of the black hole. if a photon, whose velocity is radially outward away from the centre of the black hole, is placed just outside the edge of this event horizon (let's say, 10.00001 km from the centre of the black hole), will it escape? what would happen if the photon's velocity was NOT perpendicular to the "surface" of the event horizon sphere, and instead had an angle of 80 degrees to it. that would mean that the vertical, "escaping" component would be 0.98c. would it escape then?

also, By this, I assume you mean the radius within which, yes?

Observe your photon. Did it escape? Then the EH is defined as below that point. If it didn't, escape, the EH is defined as above that point.


Imagine the black hole is not symmetrical, even for a moment, say while it's absorbing an extra helping of matter (this may or may not be realistic, let's just pretend). The point is, the EH is not mathematically defined as a radius or any other predetermined shape or formula. At this or that point, photons may or may not escape. The EH depends on whether photons escaped or not.

 

[Unit]

The point is, the EH is not mathematically defined as a radius or any other predetermined shape or formula. At this or that point, photons may or may not escape. The EH depends on whether photons escaped or not.

isn't the Schwarzschild radius defined as r_s = \frac{2GM}{c^2}? i think I'm wrong, though, because of general relativity and the way how spacetime bends in the presence of gravity. the radius cannot be thought of like a pretty little sphere in plain, 3d cartesian coordinates. plus, what if the black hole is spinning? then it gets messier.

i totally understand your explanation though, thanks a bundle