Hawking Radiation and negative energy

[Sacroiliac]

Most of the explanations I have read on Hawking radiation have certain things in common.

1.) Virtual pair creation via borrowing energy from the vacuum (HUP).
2.) One of the virtual particles has positive energy and the other has negative energy.
3.) The negative energy particle is capture by the hole and the positive energy particle escapes to infinity.

I’m having a lot of trouble understanding the above. If one of the particles has negative energy then why does energy have to be borrowed from the vacuum? It would seem that +E -E equals a net zero energy requirement. Why is any borrowing of energy via the HUP necessary?

Why does the negative energy particle necessarily get captured by the BH? It would seem that either particle would have an equal probability of being captured.

Thanks

[Sacroiliac]

How come no one will answer my question? Have I done something wrong, or is it just a hopelessly confused question?

[Glom]

It's a valid question. I instantly thought of that when I read about positive energy and negative energy creation in the seconds before reading the very question you asked.

Clearly the experts are threatened by the intelligence of your question and have decided to let the thread fade into the wastes of the archives. :P :wink:

kpi? JS? BA? dgruss, you know you want a go. :P

One question I have for you, what is this hole that sucks up the negative energy particle?

[Chuck]

What happens if the positive energy particle is captured by the black hold and the negative one escapes? Shouldn't there be some negative particles floating around from when this happens?

[Zathras]

Most of the explanations I have read on Hawking radiation have certain things in common.

1.) Virtual pair creation via borrowing energy from the vacuum (HUP).

correct.

2.) One of the virtual particles has positive energy and the other has negative energy.

Not quite. See below.

3.) The negative energy particle is capture by the hole and the positive energy particle escapes to infinity.

I’m having a lot of trouble understanding the above. If one of the particles has negative energy then why does energy have to be borrowed from the vacuum? It would seem that +E -E equals a net zero energy requirement. Why is any borrowing of energy via the HUP necessary?

Why does the negative energy particle necessarily get captured by the BH? It would seem that either particle would have an equal probability of being captured.

Thanks.

When they say "negative energy" rest energy is not considered. They are just considering the sum of kinetic and potential energy for the particle in question. Because this sum is less than zero, the particle does not have enough energy to get away from the black hole, and thus it fall in. Meanwhile, the positive energy particle goes off because it does have enough energy to get away.

[Glom]

So it's a datum thing?

[Sacroiliac]

When they say "negative energy" rest energy is not considered. They are just considering the sum of kinetic and potential energy for the particle in question. Because this sum is less than zero, the particle does not have enough energy to get away from the black hole, and thus it fall in. Meanwhile, the positive energy particle goes off because it does have enough energy to get away.

Thanks, Zathras now I got it. And thanks for responding Glom.

[heusdens]

Most of the explanations I have read on Hawking radiation have certain things in common.

1.) Virtual pair creation via borrowing energy from the vacuum (HUP).
2.) One of the virtual particles has positive energy and the other has negative energy.
3.) The negative energy particle is capture by the hole and the positive energy particle escapes to infinity.

I’m having a lot of trouble understanding the above. If one of the particles has negative energy then why does energy have to be borrowed from the vacuum? It would seem that +E -E equals a net zero energy requirement. Why is any borrowing of energy via the HUP necessary?

Why does the negative energy particle necessarily get captured by the BH? It would seem that either particle would have an equal probability of being captured.

Thanks

You have some errors.

The virtual particle creation creates two particles. A particle and it's anti-particle. This costs net energy, borrowed from the vacuum, and if the two annihilate, the energy is paid back.
But if this occurs near the event horizon of a Black hole, one can fall in, while the other can escape. This will cost nett energy then, which is subtracted from the energy of the Black hole.

That is what happens.

[Glom]

Is that something to do with black holes evaporating?

[Avatar28]

Yes. If enough particles do that, the black hole eventually evaporates. As I understand it, in the early moments of the universe, tiny black holes were created with a minuscle amount of mass (some probably as light as you or me even). Most of them would have evaporated by now, but there may be a few of the larger ones still floating around. Don't ask me how a black hole with the mass of, say, an asteroid can exist, but that's my understanding of it.

[Glom]

I think it's a case of density. They always say that if Earth was compressed into the size of a pin, it would be a black hole.

[DStahl]

huesdens, thanks for clearing that up. I was wondering about that negative energy bit--my understanding is that Hawking radiation can be thought of as involving a matter-antimatter particle pair, but since antimatter does not have negative energy that was leaving me a bit puzzled.

It's my understanding that a stellar-mass black hole would last hundreds of billions of years before evaporating via the Hawking radiation.

[Zathras]

huesdens, thanks for clearing that up. I was wondering about that negative energy bit--my understanding is that Hawking radiation can be thought of as involving a matter-antimatter particle pair, but since antimatter does not have negative energy that was leaving me a bit puzzled.

It's my understanding that a stellar-mass black hole would last hundreds of billions of years before evaporating via the Hawking radiation.

Yes, this is true. A back of the envelope calculation can show this. The only assumption made is that the wavelength(w) is about the same size as the Schwartzschild radius:

w=2Gm/c^2=c/f.

E=hf, and in a blackbody spectrum, the mean energy of a photon is about =kT, so

f=kT/h,

so combining the first and last equations give:

2Gm/c^2=hc/kT,
or
T=hc^3/(2Gkm), so the "effective temperature" of a black hole due to evaporation is inversely proportional to its mass.

Furthermore, we know that for a blackbody (such as a black hole, of course), E per unit area is equal to sigma*T^4, or total E radiated per unit time is:

dE/dt=sigma*T^4*(4*pi*r^2)

But r again =2Gm/c^2, so combining the various equations, one gets:

dE/dt=sigma*[hc^3/(2Gkm)]^4*4*pi*(2Gm/c^2)^2

After you clean up this mess, you get

dE/dt=sigma*h/[(c*G^2*k^4]*m^-2

Since E=mc^2, you get

dm/dt=sigma*h/[(c^3*G^2*k^4]*m^-2

Let s=sigma*h/[(c^3*G^2*k^4]

Then dm/dt=s/m^2

so m^3=m0^3-3st

In short, the math shows that the smaller a black hole is, the faster it radiates. Conversely, a supermassive black hole will take eons of eons to radiate away.

[kilopi]

kpi? JS? BA? dgruss, you know you want a go.

I answer to a lot of things, but--OK, I admit, since the PX infusion, I haven't had time to read every post on every thread thoroughly, so I cheat a little and search for my name. If you abbreviate it, I may miss it on the search. Or else get here late for the party. Is there any beer left?

[Glom]

All the beer's disappeared down the black hole.