How can particles escape from a black hole?

[Aaron Maiwald]

In simplified terms Hawking Radiation exists, because in the vacuum surrounding a black hole these subatomic-particle-pairs pop into existence and one of these particles manages to escape from the black hole. This stream of escaping particles is called Hawking Radiation, right?(Please correct me) Now my questions:
1. Why can it escape? Wouldn't that mean that the escaping particles velocity is above the speed of light?
2. How does the gravitation of a black hole affect this production of particle-pairs? How is it different there compared to other vacuums?

 

[PeterDonis]

In simplified terms Hawking Radiation exists, because in the vacuum surrounding a black hole these subatomic-particle-pairs pop into existence and one of these particles manages to escape from the black hole.

This is a heuristic description that is often given in pop science sources, but it's worth noting that it doesn't really correspond to anything in the actual underlying math.

Wouldn't that mean that the escaping particles velocity is above the speed of light?

No. In this heuristic model, the pairs pop into existence just above the hole's horizon. One falls into the hole, the other escapes. The one that falls into the hole has negative energy, so it removes a small bit of energy from the hole. The one that escapes has positive energy, equal to the energy removed from the hole.

How does the gravitation of a black hole affect this production of particle-pairs?

It doesn't affect the pair production itself; a small patch of vacuum just above a black hole's horizon is no different from a small patch of vacuum anywhere else.

The difference the presence of the hole makes is its tidal gravity, which (in the heuristic viewpoint we're using here) pulls apart the pair of particles before then can annihilate each other again.

 

[Orodruin]

In simplified terms Hawking Radiation exists, because in the vacuum surrounding a black hole these subatomic-particle-pairs pop into existence and one of these particles manages to escape from the black hole.

These are very simplified terms. Hawking himself said he used it only because it was the closest he could get using regular language, but let us assume the premise. These "virtual particle-antiparticle" pairs would not be inside the event horizon.

Why can it explain? Wouldn't that mean that the escaping particles velocity is above the speed of light?

No. Again, the pairs would not appear inside the horizon and therefore not need to be escaping faster than light.

2. How does the gravitation of a black hole affect this production of particle-pairs? How is it different there compared to other vacuums?

To understand this you need to understand that, in regular flat space-time, the vacuum looks different in an accelerated reference frame than it does in an inertial frame. An accelerated observer would observe a background radiation that depends on their acceleration. This is known as the Unruh effect and is very similar to how Hawking radiation works.

Also, your profile says that you have completed high-school. This means you likely should not be opening A-level threads as putting A-level indicates that you have an understanding of the subject equivalent to that of a graduate student in the subject or better and expect answers aimed at that level.

 

[kimbyd]

In simplified terms Hawking Radiation exists, because in the vacuum surrounding a black hole these subatomic-particle-pairs pop into existence and one of these particles manages to escape from the black hole. This stream of escaping particles is called Hawking Radiation, right?(Please correct me) Now my questions:
1. Why can it explain? Wouldn't that mean that the escaping particles velocity is above the speed of light?

In the semi-classical description, the particles do not come from inside the black hole. They are produced outside the event horizon.

In the full quantum description, it may be possible for particles to come from inside the black hole, but only if the black hole doesn't actually have a real event horizon (the event horizon would have to be a classical approximation to the true behavior).

2. How does the gravitation of a black hole affect this production of particle-pairs? How is it different there compared to other vacuums?

It's the horizon itself that causes the effect. Any such horizon will create similar radiation. This has been tested and verified using sound horizons in fluids: it's possible to create a system where a fluid is forced through a pipe at a rate faster than the speed of sound in the fluid, which can be used to create a "sound horizon". These horizons emit sound waves in a way that is directly analogous to Hawking Radiation from black holes.

To the extent that gravity impacts the outgoing radiation, there will naturally be some gravitational redshift as the radiation climbs out of the black hole's gravitational potential well. This redshift is taken into account: the black hole temperature is that observed from far away, after the outgoing radiation has redshifted.

 

[Grinkle]

Read this -

https://www.physicsforums.com/insights/misconceptions-virtual-particles/

 

[newjerseyrunner]

Or watch this:

 

[Aaron Maiwald]

Thank you all for your responses!
I think one of the things that led me of track was a misconception about black holes. So please let my know, whether my current understanding of black holes(alltough massively simplified) is correct: A black hole is a region in space, in which gravity is so strong, that nothing can escape from inside it. In space black holes look like black spheres. In the centre of these Spheres is a singularity, an object that has infinite mass-density, infinitesimal volume and infinite spacetime curvature. The surface of the sphere is called the eventhorizon. Nothing can, once it has passed the eventhorizon, escape. That's because gravity is so strong that something what have to move faster than light to escape, which is impossible. The size of the black hole is proportional to the mass of the singularity/the mass of the planet that collapsed into the black hole.
Now to Hawking Radiation:
Like anywhere in space, there are virtualphotons-pairs popping into existence close to the eventhorizon of a black hole all the time. But since gravity is so strong close to the eventhorizon, something similar to the dynamical casimireffect happens. Through gravity the virtualphotons-pairs become real (can somebody explain the causality here?) Now my knowledge becomes even more vague: Sometimes it happens that one of the photons from the pairs fall into the black hole.(why?) Now the created energy has to be "paid back"(i have no idea what that means) and this is done by decreasing the mass of the black hole.

Please keep in mind that I am (obviously) not a physicist and that my goal here is to get a picture of these concepts that is correct on a very simplified level.
It would be great to get some Feedback :)

 

[PeterDonis]

A black hole is a region in space, in which gravity is so strong, that nothing can escape from inside it.

A region of spacetime, not space. "Space" inside a black hole doesn't work the way you would think.

In space black holes look like black spheres.

Black holes, strictly speaking, don't "look" like anything, since no light can escape from them. But you can tell they are there because there will be a region from which no light is coming. That region is spherical, yes.

In the centre of these Spheres is a singularity, an object that has infinite mass-density, infinitesimal volume and infinite spacetime curvature.

The singularity is not really at the "center" of the hole; it's a moment of time, not a place in space. This moment of time is to the future of every event inside the hole's horizon. That's why the singularity is unavoidable once you're inside the horizon: because you can't avoid moving into the future.

Also, strictly speaking, the singularity itself--the point (actually it's not a point, it's a spacelike curve) where spacetime curvature is infinite--is not part of spacetime at all. What "singularity" really means is that there are freely falling worldlines which come to an end in a finite time, and in that finite time, spacetime curvature increases without bound. Note that an idealized black hole is a vacuum; there is no stress-energy inside it. So thinking of the singularity as an "object" isn't really correct.

The size of the black hole is proportional to the mass of the singularity/the mass of the planet that collapsed into the black hole

It depends on what you mean by the "size" of the hole. The only real invariant measure of "size" is the area of the hole's horizon, which is proportional to the square of its mass.

Now to Hawking Radiation

As I have already said, the picture you give of this, while common in pop science treatments, doesn't really correspond to anything in the actual math. So it really isn't a good idea to try to reason from it as though it described what's actually happening.

 

[PeterDonis]

my goal here is to get a picture of these concepts that is correct on a very simplified level.

As far as Hawking radiation is concerned, I would say there isn't one beyond the bare fact that, when quantum effects are taken into account, black holes can radiate and thereby lose energy (and mass). Anything beyond that really requires a serious investment of time and energy to learn the underlying concepts in more detail.

 

[Aaron Maiwald]

Black holes, strictly speaking, don't "look" like anything, since no light can escape from them. But you can tell they are there because there will be a region from which no light is coming. That region is spherical, yes.

Would you say the relationship between the black sphere and the actual black hole is analogous to a footprint and the actual foot?

Also, strictly speaking, the singularity itself--the point (actually it's not a point, it's a spacelike curve) where spacetime curvature is infinite--is not part of spacetime at all. What "singularity" really means is that there are freely falling worldlines which come to an end in a finite time, and in that finite time, spacetime curvature increases without bound.

I'm afraid I do not understand. What do you mean by "freely falling worldlines which come to an end in a finite time, and in that finite time, spacetime curvature increases without bound"? What concepts do I need to understand in order to understand that?

 

[jbriggs444]

I'm afraid I do not understand. What do you mean by "freely falling worldlines which come to an end in a finite time, and in that finite time, spacetime curvature increases without bound"? What concepts do I need to understand in order to understand that?

Here is how I understand it:

A "freely falling worldline" means an unaccelerated trajectory. Like a satellite in orbit. Each point on the world line has a specific position and time, that is, an "event". A worldline can be thought of as a continuous sequence of events.

The fact that it is freely falling or unaccelerated means that it is (locally) straight. Every incremental segment on the trajectory is extended in the same direction as the previous. Another term for freely falling worldline is "geodesic". A geodesic in curved space-time is analogous to a straight line in flat space-time.

[We often imagine that a freely falling object is following a curved path in flat space-time, but the description in general relativity is that it is following a straight path in curved space-time]

Let us suppose that we have a massive particle on a freely falling world line (aka freefall trajectory, aka geodesic). Time passes for that particle on its trip. We can label each event on the world line with a time stamp according to the elapsed time measured by a hypothetical clock attached to the particle.

Now let us imagine a world line that approaches a black hole. It passes through the event horizon surrounding the black hole. The crossing is uneventful. There is no local feature of the horizon that makes the crossing remarkable or even noticeable. The important feature of the event horizon is that from a global perspective, it is a "point of no return". No trajectory, accelerated or not, can pass back out through the event horizon. Locally, the event horizon seems to sweep past our free-falling particle at the speed of light.

The particle keeps falling. As it falls, it gets closer and closer to the "singularity" and space-time curvature in its neighborhood gets more and more extreme. As the particle continues falling, time is still passing for the it at a rate of one second per second, as it always does.

The space-time curvature at the particle's location increases without bound. The particle approaches the singularity and its clock approaches a limiting value.

Instead of having our mathematical description cover the singularity, we stop just short. Our mathematical description covers everything up to, but not including the singularity. We do not model the particle's clock striking a metaphorical midnight. We do not model curvature actually being infinite. We can not and do not extend the trajectory through and past the singularity. Every point on the world line has a finite time tag.

A pithy term for this is "geodesic incompleteness".

 

[PeterDonis]

Would you say the relationship between the black sphere and the actual black hole is analogous to a footprint and the actual foot?

I'm not sure. I don't think this comparison is useful, but I can't say it's wrong.

What do you mean by "freely falling worldlines which come to an end in a finite time, and in that finite time, spacetime curvature increases without bound"?

I mean that it is possible for you, or any observer, to freely fall (which just means falling with zero proper acceleration--i.e., you are weightless) into a black hole and find tidal gravity in your vicinity increasing without bound (meaning, it would tear you apart) in a finite time by your own clock.

 

[Aaron Maiwald]

Thank you for the great explanations! What do you think which of hawkings other scientific contributions can be explained as intuitiv as hawking radiation?