Then I probably always misunderstood the Hawking radiation.
I have always thought that this is caused by virtual particles that emerge outside the black hole.
Join the club! There are a lot of confusing things about black holes, but the mechanism behind Hawking radiation might be one of the most commonly misunderstood ones, largely because it is too often oversimplified. The idea that the radiation is particles of matter and antimatter popping into existence and annihilating, except for a few pairs being torn apart and then one particle escapes, which somehow reduces the black hole's mass... is very common and also very wrong.
The correct explanation requires some quantum field theory, and it is probably not very accessible to most people. PBS Space Time made a couple of videos explaining it, but they may be among their more difficult videos to follow. I'll link them at the end here in case you want to check them out. Otherwise, I can present a simpler approach that, while not totally rigorous, still gives a lot of insight and almost correct answers.
In fact, let me answer the question of what the radiation is right up front. It is mostly photons, with a blackbody spectrum. In other words,
[center]
black holes emit light, exactly like an ideal black body with a particular temperature. [/center]
This may seem extremely surprising. Why is it true?
The way I like to think about about this is through the original argument proposed by Jacob Bekenstein, who figured out a lot about black hole thermodynamics even before Hawking did. (Hawking just made it much more rigorous.) I like Bekenstein's approach because it is very simple and beautiful, and it gets the right answers save for a few constants.
Bekenstein started by thinking about the entropy of the black hole. For a long time this was a big mystery. Entropy is related to the information content of a system, but what happens to the information of whatever falls into a black hole? If it is lost, then the entropy of the universe decreased. But that seems to contradict the 2nd law of thermodynamics. Bekenstein quickly realized that the entropy must somehow be preserved in the black hole. When he went through this simple argument, he discovered the entropy is actually related to the black hole's horizon.
So let's follow Bekenstein's argument and see this for ourselves, which will then lead us to realizing that a black hole has a temperature, and therefore also emits radiation. It won't clearly explain how that radiation is emitted, but we'll see that it must be emitted, find approximately correct formulas describing it, and see that it is mostly photons with a blackbody spectrum.
What is the entropy of a black hole? That is, how much hidden information does it contain? A simple way to calculate it is to imagine building a black hole out of photons. To maximize the number of photons (and thus the entropy), let each photon have as little energy, and therefore as large of a wavelength, as possible. That would be a wavelength equal to the diameter of the black hole.
The diameter of the black hole is twice the Schwarzschild radius of 2GM/c
2, so we'll let each photon have a wavelength of λ=4GM/c
2. Then because the photon energy is E=hf = hc/λ, each photon contributed an energy of hc
3/4GM. But the total energy of the black hole is Mc
2, so the number of photons that went into it is Mc
2/(hc
3/4GM) = 4GM
2/hc.
The entropy of the black hole should then be that number times the Boltzmann constant, k. So we've derived through very simple means the entropy of the black hole as 4GM
2k/hc. In fact, this is very close to the correct formula derived by Hawking. The only difference is that the factor of 4 becomes an 8pi
2:
[center]

[/center]
The hidden information (the entropy) of the black hole is proportional to its mass squared. But hold on. The radius of a black hole is proportional to its mass, and so its surface area is proportional to mass squared.
This equation is telling us that the black hole's entropy is proportional to its surface area! This is very surprising. Usually the entropy of something is proportional in some way to its volume, not surface area. Now if you've ever heard of the "Holographic Principle", this result here is essentially what sparked the idea... but I digress, that's a subject for another day. Getting back on track:
We've found the entropy of the black hole. Why is it useful for understanding Hawking radiation? The reason is because there is a fundamental relationship between the entropy of an object, the energy of an object, and its temperature. The definition of temperature is how quickly the energy changes when the entropy changes. Formally,
[center]

[/center]
[left]In words, what this says is that temperature is the change in energy U with respect to entropy S. But I wrote it upside-down (the change of entropy with respect to energy, to the -1 power) just because it is usually more easy to find how entropy changes with energy than the other way around, which then makes calculating the partial derivative (that curly ∂ symbol) easier.[/left]
Aside: this definition of temperature which comes from statistical mechanics is probably unfamiliar. Isn't temperature just the "average kinetic energy per particle"? Well, I guarantee this definition is completely consistent with that one, and you can check by applying it so something more familiar like an ideal gas (use the
Sackur-Tetrode equation). But this definition is also more general, and will work for systems where the more common definition
won't work. Earlier I used this to show why a 2-state paramagnet can have a
negative temperature.
[left]So, if we write the formula for the entropy of the black hole in terms of its energy (which is just Mc
2), and then differentiate with respect to energy and invert it, we'll get the black hole's temperature! Unfortunately this means a bit of calculus is required, which I'll go through quickly:[/left]
[left]First writing the entropy in terms of total black hole energy U,[/left]
[left]

[/left]
[left]Then differentiating with respect to U,[/left]
[left]

[/left]
[left]Finally, by inverting that, we get the black hole's temperature:[/left]
[center]

[/center]
[left]Does it seem totally weird that a black hole should have a temperature at all? It must be true, because it has both energy and entropy. And because it has a temperature, it will radiate blackbody radiation! By the
Stefan-Boltzmann law, it will radiate a blackbody spectrum with a power proportional to its surface area times the 4th power of its temperature. If you like you can try doing some algebra from here and calculate what that radiated power is, or the peak wavelength of the radiation. You'll find that
the power radiated is inversely proportional to the black hole's mass, and
the peak wavelength of the radiation is approximately 8 times the black hole's diameter. (If you could image a black hole by its Hawking radiation, it would look extremely fuzzy. Fortunately they tend to be surrounded and lit up by very bright accretion disks, so as EHT proved we can image them that way instead.)

[/left]
Let's apply these results to a particular case. How about the black hole Sagittarius A* at the center of our galaxy? It weighs about 4 million solar masses.
Temperature: About 1.5x10
-14 Kelvin.
Very cold.
Rate of energy emitted: 6x10
-42 Watts.
Peak wavelength: 2x10
11 meters or slightly over 1AU!
With an average photon energy of 10
-36 Joules, this means SgrA* emits about one photon per
day, and most of the photons have huge wavelengths comparable to the Earth-Sun distance. These are some pretty crazy numbers to contemplate. The effective horizon temperature is very low for most astrophysical black holes, and the Hawking radiation is correspondingly extremely weak and long in wavelength.
So that's about as complete an explanation for why black holes radiate and what that radiation is made of as I can give, without going into quantum field theory. If you do want to go deeper, then you can give these PBS Spacetime videos a try:
[youtube]bG-xu5H6plk[/youtube]
[youtube]qPKj0YnKANw[/youtube]