Under what circumstances do black holes evaporate? I think it has something to do with the kind of black hole and also the role of ‘virtual particles’ but I’m very sketchy. Any info welcome.
All black holes are thought to emit radiation due to really weird quantum effects. If they emit more radiation than they absorb energy/mass they will be losing energy/mass, i.e. evaporate.
The Wikipedia article briefly describes several ways to explain the, as of yet unobserved, Hawking radiation.
There are thousands of webpages out there explaining these topics, and I assume you’re reading lots of them. It would possibly be helpful if you referred to whichever resource you’ve used and specified which part confused you instead of just opening with a completely open question on a huge subject.
Not that these open threads aren’t interesting, mind you.
Any object at any temperature above absolute zero (that is to say, any object at all) is continually radiating energy due to its temperature. If there’s anything else in its environment (which there always is), then those other things are also radiating, and so our original object is also absorbing energy from those other things. Depending on the other things in the environment, how they’re arranged, and their temperature, our original object might be gaining energy on net, losing energy on net, or neither, and be in equilibrium.
Since everything I just said applies to everything, that includes black holes. Black holes are continually emitting Hawking radiation, according to their temperature. But the temperature of a black hole depends on its size, and a stellar-mass black hole would have a temperature of only a millionth of a degree above absolute zero, and hence be radiating only extremely slowly (a larger hole would be even colder). In the present Universe, in fact, a black hole will be gaining energy much faster from the microwave background radiation which pervades space than it would be losing it to Hawking radiation, so if you want to radiate a black hole away to nothing, the first step is to wait hundreds of billions of years for the Universe to cool down enough. Not that that matters much, though: Even without the background radiation, Hawking radiation from a stellar-mass hole is still slow enough that it would take far longer than that to completely radiate away.
Now, a sufficiently small black hole could radiate much quicker, and we can’t completely rule out the possibility of those existing. But we don’t have any concrete reason to believe they do exist, either.
Are larger black holes colder because it is harder for the thermal radiation to escape the greater pull of the black hole?
Oh, sorry I asked. :rolleyes:
A very clear, helpful reply, as usual, Chronos. Thank you very much. 
Just one point, if you don’t mind me asking: What exactly is Hawking radiation?
Short answer is, Hawking radiation is the radiation emitted by the black hole as a result of its temperature.
If you mean “what is it made of”, then the answer is “probably mostly photons, some gravitons, and maybe some neutrinos, plus other particles for a sufficiently-hot hole”.
I would hesitate to say anything so simplified. The temperature of a black hole is proportional to its surface gravity, which (for a non-rotating hole) is inversely proportional to its mass. In other words, in some sense larger holes are colder because they have less gravity, not more.
Just a piece of advice for getting more relevant answers. If for instance you’d written “I read the Wikipedia article on Hawking radiation and have some questions”, I would have known you wouldn’t appreciate an answer at that level.
This, I’m afraid, totally evades the point (or what I think is the point) of OP’s question. We’ve all been conditioned to believe that a black hole doesn’t emit anything because it’s gravitation field is so intense that even light itself can’t make it out.
So the question becomes: Given that popular understanding of a black hole, what is this contradictory Hawking Radiation, and how does it happen at all?
The explanation I’ve seen (but, like any explanations of Quantum Anything) seem, at first glance at least, to be unlike any “normal” kind of heat radiation coming from “normal” objects. Rather than just a stream of photons coming from the object, Hawking Radiation is described as having to do with a random creation of particle/antiparticle pairs near (just outside) the event horizon. In some cases, the anti-particle will randomly fall into the black hole (reducing the mass therein because of its anti-ness?) while the particle will shoot off in a different direction, increasing the mass of the outside universe.
Or is that actually how ALL radiation works, and we just were never taught that?
And of course, if Chronos or anyone else wants to respond that my explanation here is total ignorant bullshit – well, that’s entirely possible. Absent a detailed level of physics and mathematical understanding, it seems that ANY layman-accessible explanation of Quantum Anything must necessarily be a very poor actual explanation. Am I at least right in this?
I am not a physicist but an engineer and my understanding is limited to what my Thermodynamics professor said :
Max Planck came out with black body radiation. Basically, it says that every thing in the universe emits radiation that is a function of its temperature. For a body to emit no radiation, it has to be at absolute zero.
Thermodynamics on the other hand says absolute zero is impossible (entropy).
Black holes were initially theorized to absorb everything and emit nothing - thereby implying absolute zero temperatures leading to a thermodynamic contradiction.
Hence, Hawking radiation was theorized as a means to satisfy thermodynamics. I believe it was later proven to be correct.
Interesting, but why is absolute zero impossible?
In classical physics - heat flows from a hot object to a cold one. How do you get all the heat to flow out of an object? You need something colder than it. So basically there is no mechanism to move all the heat out of something. Even given infinite time to do so.
After that we get to problems of quantum effects, and an intrinsic bit of energy (zero point energy*). This also prevents you ever getting all the energy out of something.
So the temperature can never be exactly zero.
Some tricks allow us to get ridiculously cold, in the low milli-Kelvins. But actually precisely zero. Nope.
*Related to the Heisenberg Uncertainly principle.
Because of the third law of thermodynamics.
An object at absolute zero would have to have zero entropy and that is not possible. No system can have zero entropy.
Err…I do not think this is true.
They use lasers and magnetic traps and evaporative cooling to cool things near absolute zero. They do not use something “colder”.
A question about the wavelength of hawking radiation. I am not very great at math which is why while physics has always fascinated me, I could never get too much further than a pretty good layman’s understanding, but one thing I’ve tried to work out would be the wavelength of a photon emitted by hawking radiation, and its relation to the schwarzschild radius.
So, real quick, if you happen to know the answer there, or are able to work it out with more ease than I, is there actually a relation, and if so, at what proportion?
I suspect, but am probably wrong, that the wavelength of a photon emitted by a black hole would be equal to twice the schwarzschild radius, or the diameter of the black hole’s event horizon. The reason for me to think this is that I am pretty sure I’ve come across in susskind’s lectures that a black hole cannot absorb a photon of greater wavelength that its diameter, and it would make sense to me that an emitted photon would be at a similarly low energy level. I was expecting him to be leading down the path of showing that the wavelength of hawking ration would be related in the same way, but went onto a different subject.
It’s a more complicated way of cooling, but ultimately, yes the heat is rejected into a colder space.
Just like your air conditioning at home is not “creating” cold, but is instead rejecting the excess heat to a place you don’t mind getting hotter, the same is true with the magnetic traps and such. The very simple idea is that most of the momentum of the stem gets transferred into one particle, and that particle leaves, carrying that energy off with it. That particle will be warmer than the environment it enters.
I’m thinking the physicists on the board will hate my handwaving non-mathematical model (some of which comes from reading pop-sci articles, some made up of whole cloth), BUT:
Virtual particle pairs are constantly forming and self-annihilating throughout space. I imagine that virtual particle pairs that form near the horizon are torn apart by the tidal forces there. If one half of the pair falls in, the other half becomes “real” (somehow) and may then fly out as radiation. (The other particle is apparently imaginary and has ‘negative’ mass- somehow.)
The tides at the surface are greater the smaller the black hole. Therefore smaller holes can rip apart more pairs and more energetic pairs (somehow) than larger holes, therefore leading to more and higher energy radiation. This corresponds to the mathematical predictions that smaller black holes are hotter than large black holes.
This model is made of hand-wavium. It leads me to wonder if there is radiation originating in the space at some distance from a very small black hole (where the tides may still be more intense than that seen at the surface of a larger hole).
So when NASA cools something to a billionth of a degree above absolute zero they had to make something colder than that first?
Yeah, the empty vacuum in which the higher energy particles are ejected into.
There are two things here, heat, and temperature.
Simplified, temperature is the amount of momentum a particle has, and heat is how much momentum energy the group of particles has.
The temperature of the vacuum may be higher, but it contains less heat, so the heat of the sample can be rejected into it.
You cannot “make cold” without violating at least one of the laws of thermodynamics, you can only move heat from one place to another.
The wavelength of Hawking radiation is certainly comparable to the Schwarzschild radius, but it’s not all at the same wavelength: It has a thermal spectrum. You could still calculate the peak wavelength, though I’m not sure precisely what it comes out to (and it’ll depend on how you define “peak”).