Hawking radiation and black hole evaporation

[Mister Earl]

Really? So the matter and antimatter have no way of "cancelling" out? No annihilation at all?

#EDIT: Kindly ignore my incredulity. It's not that I think you're wrong, but rather am startled to due an unexpected result, and cannot imagine how that would work.

Stupid singularities!

 

[jasonpatterson]

Yes, energy is conserved, but why? Shouldn't the infalling radiation be exactly the same as the outgoing Hawking radiation?

Yep, it is.

Someone said that the infalling particle will annihilate with a particle inside the black hole. Whatever allows that to happen should allow the outgoing particle to annihilate with a particle outside the black hole, since there's no difference between the particles.

They misunderstood the nature of antimatter. In a nutshell, antimatter is exactly like regular matter, but the charges are all swapped around. Whether they encounter another particle and are annihilated or not doesn't make any difference to your question.

What happens is that the particles exist first as virtual objects. That is, the energy density of the universe has some uncertainty (look up Heisenberg's Uncertainty Principle for more info) and that uncertainty is greater and greater on smaller and smaller scales. On sufficiently small scales it's great enough that there ought to be swarms of particles popping in and out of existence. Hawking radiation is the idea that if things work out just so, a virtual particle pair could pop into existence but one of the pair gets trapped in a black hole while the other escapes. Once that happens and the particles exceed the limits of uncertainty, their energy cost has to be paid for, and it is taken out of the mass of the black hole via E = mc². One particle is recovered but the other zips away, taking a teeny bit of mass with it.

For a half-assed real world analogy, it's like a person going clothes shopping. They can try on clothes all day, but when they finally want to escape the store, bringing any clothes with them requires payment.

Why doesn't the same logic apply to real particles? In which case it seems that if I shine a light into a black hole the energy could become negative momentum and the black hole could grow smaller, rather than larger?

Nope, light has positive energy and would add to the total energy/mass of the black hole. None of the particles (whether they're your typical, massive particles or only photons carrying mass via E=mc²) have negative energy in this scenario. The total energy of the system is conserved in the end because the total mass-energy of the black hole+escaped particle is equal to the original mass-energy of the black hole.

 

[rjh01]

Yep, it is.



They misunderstood the nature of antimatter. In a nutshell, antimatter is exactly like regular matter, but the charges are all swapped around. Whether they encounter another particle and are annihilated or not doesn't make any difference to your question.

What happens is that the particles exist first as virtual objects. That is, the energy density of the universe has some uncertainty (look up Heisenberg's Uncertainty Principle for more info) and that uncertainty is greater and greater on smaller and smaller scales. On sufficiently small scales it's great enough that there ought to be swarms of particles popping in and out of existence. Hawking radiation is the idea that if things work out just so, a virtual particle pair could pop into existence but one of the pair gets trapped in a black hole while the other escapes. Once that happens and the particles exceed the limits of uncertainty, their energy cost has to be paid for, and it is taken out of the mass of the black hole via E = mc². One particle is recovered but the other zips away, taking a teeny bit of mass with it.

<snip>

Here is a problem for you. How can we create virtual particles? Then maybe use them as energy source or to wipe out an approaching black hole?

Edit. My question might be unanswerable.

 

[Toontown]

Particle-antiparticle pairs are popping into existence all the time, but the energy of the universe as a whole is conserved because they annihilate each other quickly. When this occurs near the event horizon of a black hole, one particle can be sucked into the black hole while the other escapes as Hawking radiation.

This is somewhat OT, but I don't get this.

The universe is very large. At any instant there is a very large number of virtual particles in existence, if only for that instant. The next instant that batch is gone, only to be replaced by yet another multitude of seemingly illegitimate account-skimmers.

Q: How is this not an energy imbalance? By what right do these multitudes of virtual particles exist at every instant?

 

[jasonpatterson]

Here is a problem for you. How can we create virtual particles? Then maybe use them as energy source or to wipe out an approaching black hole?

Edit. My question might be unanswerable.

We don't need to create them; they're all around us all the time. Unfortunately it appears that making use of them is more or less impossible, since the moment we actually use them they stop being virtual and start being real and their mass-energy has to come from somewhere. It appears that they aren't just a bookkeeping measure though (check out the Casimir Effect for an example.)

Q: How is this not an energy imbalance? By what right do these multitudes of virtual particles exist at every instant?

It is an energy imbalance, but it's on a time scale that is so short that it doesn't violate conservation of energy when uncertainty is taken into consideration. There are numerous other examples of energy-time uncertainty and some have been put to good use (quantum tunneling, for instance.)

 

[sol invictus]

I can understand the fact that energy is given off. I don't get how it's conserved. Why, when a particle falls into the black hole, does the mass of the black hole decrease. I still don't see it.

If you want to take virtual particle pairs seriously as a concept, then - since energy is exactly conserved by every interaction, including virtual pair production - one particle must have negative energy and one positive. They are, after all, a vacuum fluctuation - there is no non-zero energy to be "borrowed" from. With that in mind, if the negative energy particle falls in, the black hole will lose mass.

Now you may ask why the negative energy particle falls in more often than the positive energy one. I don't have a fully satisfying glib answer. One answer is that if you detect a virtual particle it must be "on shell" (i.e. have the correct, positive energy); hence (since you can detect the outgoing flux) the ingoing particles must be the ones with negative energy.

A better answer perhaps is that these words are describing a physical process the black hole must be evaporating. The hole cannot spontaneously gain mass - that would violate conservation of energy. So all it can do that is consistent with COE is to evaporate; hence, negative energy must be going in, and positive energy going out.

 

[lenny]

I think this may be one of those cases where trying too hard to treat virtual particles like they are particles is going to fail.

i'd support that thought...

If you want to take virtual particle pairs seriously as a concept, then - since energy is exactly conserved by every interaction, including virtual pair production - one particle must have negative energy and one positive.

... so i am not at all sure that i want to take virtual particle pairs seriously as a concept here; but if i were too then i would have thought of them as a normal ol' particle antiparticle pair, which violate conservation of energy by an amount delta energy (about twice the rest mass) for a time delta t such that the product is less than the bound given by the indeterminacy relations.

CoE is violated, just not for very long. at least that is how i recall TD Lee discussing it decades ago. have things changed since then? (?got a modern review paper cite?)

perhaps as sol and edd suggest we should simply not try and picture the details of the process in this way?

 

[Senex]

I brought this problem up at my local cosmology club meeting. It turns out my neighbor Bob has been doing some sort of optical light pulse experiments in his basement and he says he can recreate Hawking radiation. I asked him to bring a video with him next week because ever since Hal died doing his superfluid helium experiments I keep away from some of the sketchier members basement and garage experiments.

Because I believe Plank mass is not the lower limit to the mass of Black Holes I think we have large extra dimensions at play but those European rascals won't even give me a couple of days with their precious collider. I knew we should have built it here in the USA. I may have more insight next week when I see Bob's video.

 

[sol invictus]

i
CoE is violated, just not for very long. at least that is how i recall TD Lee discussing it decades ago. have things changed since then? (?got a modern review paper cite?)

perhaps as sol and edd suggest we should simply not try and picture the details of the process in this way?

Well, virtual particles are a way of describing quantum fluctuations. In quantum mechanics and quantum field theory, energy is strictly conserved, with no exceptions. If TD Lee says energy isn't conserved for even a short while, he's wrong. That's not what the energy-time uncertainty principle says.

 

[edd]

I've seen things phrased as violating conservation of energy temporarily, I think. I doubt I could find one of those examples now. Without a doubt I would not like it.

 

[RussDill]

For me its enough just to realize that black holes must both follow the laws of thermodynamics and any behavior that can happen forward in time, can happen in reverse.

 

[Roboramma]

If you want to take virtual particle pairs seriously as a concept, then - since energy is exactly conserved by every interaction, including virtual pair production - one particle must have negative energy and one positive. They are, after all, a vacuum fluctuation - there is no non-zero energy to be "borrowed" from. With that in mind, if the negative energy particle falls in, the black hole will lose mass.

Now you may ask why the negative energy particle falls in more often than the positive energy one. I don't have a fully satisfying glib answer. One answer is that if you detect a virtual particle it must be "on shell" (i.e. have the correct, positive energy); hence (since you can detect the outgoing flux) the ingoing particles must be the ones with negative energy.

A better answer perhaps is that these words are describing a physical process the black hole must be evaporating. The hole cannot spontaneously gain mass - that would violate conservation of energy. So all it can do that is consistent with COE is to evaporate; hence, negative energy must be going in, and positive energy going out.

Hey Sol, thanks for all that, it all makes sense to me. I guess I was looking for some understanding of a mechanism that leads to the conservation of energy in this case, rather than simply taking the conservation of energy as given and then seeing what follows from that... because it still leaves me wondering how energy is conserved.

Obviously the above conclusions are all going to be correct and this is the easiest way to think about the problem. It just leaves me scratching my head. But maybe it's the best I can do without actually understanding the physics.

 

[sol invictus]

Hey Sol, thanks for all that, it all makes sense to me. I guess I was looking for some understanding of a mechanism that leads to the conservation of energy in this case, rather than simply taking the conservation of energy as given and then seeing what follows from that... because it still leaves me wondering how energy is conserved.

Obviously the above conclusions are all going to be correct and this is the easiest way to think about the problem. It just leaves me scratching my head. But maybe it's the best I can do without actually understanding the physics.

Try this. Instead of a strong gravity field like a black hole, consider a strong electric field (inside an idealized capacitor, for example). There's a quantum process that's similar to Hawking radiation where a charged particle/anti-particle pair nucleates in the field, then accelerates apart. The net effect is a spontaneous, vacuum-generated electric current that will tend to discharge the field. Obviously, the current flows in the direction that reduces the field - going the other way would increase the field and violate COE, and plus, that's not the direction the field pushes the charges after they appear.

Hawking radiation is analogous in many ways. In the electric field case, the field is converted into charged particles. In Hawking radiation, the gravitational field is converted into energetic particles. In both cases, since the fundamental equations governing all of this conserve energy, the forces act in the way that's consistent with COE.

 

[sol invictus]

I've seen things phrased as violating conservation of energy temporarily, I think. I doubt I could find one of those examples now. Without a doubt I would not like it.

Here's what Griffiths has to say:

It is often said that the uncertainty principle means energy is not strictly conserved in quantum mechanics - that you're allowed to "borrow" energy $\Delta E$, as long as you "pay it back" in a time $\Delta t \approx \hbar/(2 \Delta E)$; the greater the violation, the briefer the period over which it can occur. Now, there are many legitimate readings of the energy-time uncertainty principle, but this is not one of them.

 

[arthwollipot]

Warning: handwavy yet technical explanation below.

Background: whenever definable, energy is the time-component of the energy-momentum four-vector, while (more ordinary) momentum consists of the spatial components.

For the case of the Schwarzschild spacetime, the Schwarzschild time is a Killing vector field ∂_t, which is what provides energy conservation for particle orbits: if the energy of the outgoing particle at infinity is ε, then for any part of its orbit with four-velocity vector u, ε = -∂_t·u = (1-2m/r)(dt/dτ).

Only that's not quite right across the horizon, because the Schwarzschild time is actually spacelike within, with Schwarzschild radial coordinate being timelike instead. This is obvious from the switcheroo the signs of the metric coefficients do at the horizon for the Schwarzschild metric at r = 2m. Thus, for a particle inside the horizon, what a far-away stationary observer considers energy is 'actually' spatial momentum, and thus has no particular problem with being negative.

This didn't help at all.

What happens is that the particles exist first as virtual objects. That is, the energy density of the universe has some uncertainty (look up Heisenberg's Uncertainty Principle for more info) and that uncertainty is greater and greater on smaller and smaller scales. On sufficiently small scales it's great enough that there ought to be swarms of particles popping in and out of existence. Hawking radiation is the idea that if things work out just so, a virtual particle pair could pop into existence but one of the pair gets trapped in a black hole while the other escapes. Once that happens and the particles exceed the limits of uncertainty, their energy cost has to be paid for, and it is taken out of the mass of the black hole via E = mc². One particle is recovered but the other zips away, taking a teeny bit of mass with it.

This helped a little.

If you want to take virtual particle pairs seriously as a concept, then - since energy is exactly conserved by every interaction, including virtual pair production - one particle must have negative energy and one positive. They are, after all, a vacuum fluctuation - there is no non-zero energy to be "borrowed" from. With that in mind, if the negative energy particle falls in, the black hole will lose mass.

Now you may ask why the negative energy particle falls in more often than the positive energy one. I don't have a fully satisfying glib answer. One answer is that if you detect a virtual particle it must be "on shell" (i.e. have the correct, positive energy); hence (since you can detect the outgoing flux) the ingoing particles must be the ones with negative energy.

A better answer perhaps is that these words are describing a physical process the black hole must be evaporating. The hole cannot spontaneously gain mass - that would violate conservation of energy. So all it can do that is consistent with COE is to evaporate; hence, negative energy must be going in, and positive energy going out.

And this helped a lot.

Thanks, all.

 

[Senex]

Posted by Vorpa:
Yadda, yadda, yadda...

This didn't help at all.


Posted by jasonpatterson:
Yadda,yadda,yadda

This helped a little.

Posted by Sol Invictus:
Yadda, yadda, yadda

And this helped a lot.

Thanks, all.

Why so stingy with the thanks? Bob's going to drink all my beer while I watch his video not to mention Hal made the ultimate sacrifice for knowledge trying to save a few bucks and use vortex sheets (notoriously unstable in a basement environment) to make superfluid helium-4. Those guys you mentioned just pasted some mumbo jumbo while the real heroes are on the front lines in their basements and garages. Where is the love for the amateur cosmologist and particle physicist?