Source of the post I wonder if we can simulate the long times of these prehistoric eclipses in modern times during flight (that is, artificially extending the duration of a total eclipse on an airplane by following the path of the moon's shadow.)

Average speed of a commercial jet is about 250m/s, which could extend a theoretical 7.5 minute totality to just over 13 minutes.

To extend a typical 4 minute totality to over an hour, one would need to fly at least 550m/s (Mach 1.6). No commercial airline lets you do this today, though before its retirement the Concorde jet could (and often was) used for this purpose. Nowadays NASA flies some research aircraft into eclipses to allow longer and better observations above the clouds:

But Wat, protons and neutrons aren't really particles are they? They are the union of three quarks each, bound together by the strong force (gluons). So how and why do they get treated like single masses? Is it analogous to the sun being composed of trillions of particles and yet being treated as a single mass when computing its gravitational force? Also, is gravity a factor when dealing with such small particles? As I recall, the strong force governs most interactions at that level and ironically, the further away the quarks get, the stronger it is (like a rubber band.)

Also, fascinating stuff about noble gas compounds. They also react well with Fluorine. The only noble gases that are still considered to be completely inert are Helium and Neon- the ones with the smallest atomic numbers!

Source of the post What if there was an atom with electrons in the center, and the protons and neutrons surrounding the electron nucleus, in a shell or a housing.

Fascinating question.

First off, the reason that the protons and neutrons are in the central nucleus surrounded by electrons, and not vice versa, is because the protons and neutrons are much heavier, so the center of mass is essentially fixed inside the nucleus. Electrons feel an electric attraction to the protons in the nucleus, and vice versa, but the massive nucleus mostly stays still and the much lighter electrons "orbit" around it. (I emphasize the quotes around "orbit", since the electrons really don't orbit at all like planets orbiting a star. What they really do is deeply quantum mechanical.)

If we instead place the electrons in the center of a spherical shell containing an equal number of protons (and maybe we throw in some neutrons as well because why not?), then what happens? Well, the neutrons might actually prove to be irrelevant, since they are electrically neutral. In a nucleus they serve to help bind the protons together, preventing them from flying apart. But spread that same number of protons and neutrons across a spherical shell around the electrons, and most of the time they will never be in contact with each other. If the protons are free to move throughout the shell, then they will quickly spread out as far as they possibly can from one another.

Also, neutrons that are not confined within an atomic nucleus are unstable. They decay with a half-life of about 10 minutes. In a few hours, all of the neutrons in this housing would decay into more protons, electrons, and neutrinos. Maybe in this hypothetical scenario we get around this problem by constructing the housing in such a way that the neutrons inside are kept stable -- although I have no idea how that would be possible.

So, it sounds like we have a spherical shell will some protons that quickly become evenly spread around it, with an equal number of electrons stuck inside. How would this "inverted atom" behave?

It would be surprisingly boring.

If the atomic number for this atom is large (a large number of protons), then the shell of them more closely approximates a uniform spherical shell of positive charge. But the electric field inside a hollow sphere of charge is zero, everywhere. So the electrons inside it will feel basically no net electric force from the protons at all. But they will feel electric repulsion from the other electrons. So they will fly outward away from each other and distribute themselves evenly along inner surface of the shell of protons. Our inverted atom would just be a thin shell of electrons stuck against the inside wall of a shell of protons.

Everywhere inside this shell, the electric field is zero. And if the number of protons equals the number of electrons, then the electric field is zero everywhere outside as well. Our inverted atom is tiny, electrically neutral, and doesn't form chemical bonds to anything. It would act a lot like a noble gas like neon or xenon -- single atoms that don't do anything interesting chemically.^{1}

So, basically, we've built a weird, inside out version of a noble gas atom.

1. Actually this is not entirely true -- noble gases can bond with other elements under certain conditions. For example, very short-lived bonds between xenon and oxygen can occur when the gas is electrically excited, which adds an eerie green glow to some plasma globes. In fact there are a fairly large number of noble gas compounds that have been synthesized.

Very interesting response. Not what I predicted. I had though that an inverted atom would be rather dangerous and unstable, and when it would fall apart it'd be quite dangerous. I wonder if these types of atoms would be possible. Going a step further, would this inverted atom behave in a similar form for it's antimatter form? If that would even be possible.

Is there a periodic table for antimatter and other types of atoms? If these atoms and other types of things are possible then the possibilities of existence are far greater, we could have entire regions of space consisted of them. Anti-matter planets. In versed atomic stars, inversed antimatter asteroids, and all sorts of things. How would radiation function in antimatter or these inversed atomic forms? Never thought about these things before, it's quite interesting to think about. Not sure how possible life with function, possibly with inversed processes or similar processes that are inversed from ours to create something similar to ours but for them. Not sure if emitting light for inversed atoms and inversed antimatter would be possible, would it?

Source of the post But Wat, protons and neutrons aren't really particles are they? They are the union of three quarks each, bound together by the strong force (gluons). So how and why do they get treated like single masses? Is it analogous to the sun being composed of trillions of particles and yet being treated as a single mass when computing its gravitational force?

Yes, it is very analogous. The Sun is a big ball of a huge number of particles, but you don't need to know which particles are where within the Sun to compute how the planets move. All you need to know is the Sun's mass and use the fact that it is approximately spherically symmetric. Similarly, a proton is made up of quarks, but you can compute how the electron in a hydrogen atom behaves without any reference to the quarks in that proton at all. All you need to know is the proton's mass and charge.

Where the existence of quarks becomes important is in the physics of the nucleus and at very high energies. If you shoot electrons into an atom with enough energy, then their de Broglie wavelengths will be short enough to interact with the quarks in observable ways. In fact this is how the existence of quarks was first demonstrated. :)

A-L-E-X wrote:

Source of the post Also, is gravity a factor when dealing with such small particles?

On a per-particle basis, gravity is so weak compared to the other forces that we can very safely ignore it. Let's do a simple calculation to convince ourselves of this. What is the ratio of the strength of the gravitational force to the electrical force between two protons?

The gravitational force between them is F_{g}=Gm_{p}^{2}/r^{2} where of course G is the gravitational constant, m_{p} is the proton mass, and r is the distance between them.

The electrical force is F_{e} = ke^{2}/r^{2}, where k=1/(4πε_{0}), e is the charge of a proton and ε_{0} is the electric constant.

Their ratio is

In other words, the role of gravitation in nuclear physics is completely unmeasurable, unless you had more than 30 decimal places of precision (which we never do)! So you can absolutely ignore it. The only time you can't ignore it is if there are so many particles that their combined mass produces a lot of gravity, such as in extremely compact objects like neutron stars or black holes.

Source of the post Is there a periodic table for antimatter and other types of atoms?

Yes, but it looks exactly like the regular periodic table. The chemical properties of antimatter atoms should be exactly the same as the regular matter atoms. The antimatter particles themselves have the same properties as their matter particle twins except for the opposite electric charge. So you can imagine building up an anti-hydrogen atom by putting an antielectron (positron) around an antiproton. In fact, this has regularly been done in the lab, and its properties are identical to regular hydrogen. (Except the bit about annihilating on contact with regular matter, and also being the most expensive material ever made.)

It is surprisingly common for antimatter to be produced in nature (even bananas emit about 1 antiparticle per hour, as part of the radioactive decay of potassium-40), but they are still so little in abundance compared to regular matter that you don't need to worry much about them. We can also be sure that there are not regions of the universe full of antimatter, and there are no antimatter planets or stars (because they would have annihilated with regular matter, especially earlier on in the universe when conditions were very dense.)

The Big Bang itself "should have" generated equal numbers of matter and antimatter, which if true would have resulted in all of it annihilating, leaving only radiation behind. So there's a bit of a mystery as to why the universe is full of matter at all. Apparently, they weren't produced in exactly equal numbers, and a little bit of excess matter won out.

Cantra wrote:

Source of the post Not sure if emitting light for inversed atoms and inversed antimatter would be possible, would it?

This question actually goes down a really deep and interesting digression of quantum mechanics, and it gets pretty complicated. To understand how your inverted atom would interact with light, we would need to know what the wave functions of the electrons inside it would be, and from them calculate their allowed energy levels.

Solving this for multiple electrons is very difficult, since we have to account for the interactions between them. Fortunately it is not very bad at all for a single electron, which would be very similar to the classic problem of "a particle in a box" which I described in gruesome detail here. In this case we have a spherical box instead of a square one, and I suppose we shall assume the walls are very strong (the potential energy barrier very high and wide) so that the electrons don't tunnel out too quickly.

As abstract as all of this may seem, there is a very important real-world use of it: the "quantum dot". Basically it's a tiny cavity that holds an electron inside. The trapped electron has specific allowed energy levels (this is a very fundamental result for any particle that is "trapped" in quantum mechanics), and it can absorb and emit photons whose energies are equal to some transition between those energy levels (provided the transition also obeys some selection rules). Quantum dots are very interesting because the allowed energy levels depend largely on the geometry of the container holding the electron inside. So they can be specifically tailored to absorb and emit light in all sorts of ways, a lot like a custom artificial atom, with a wide range of applications.

Could not it be that when inflation set in, not all matter and antimatter had annihilated each other? And that the remaining antimatter would have been widely separated from matter?

Or that matter and antimatter would not have annihilated each other at all, because antimatter is developing backwards in time? Then there would be an antimatter universe at the 'other end of time'! (Only a few very amateurish thoughts. )

Source of the post But Wat, protons and neutrons aren't really particles are they? They are the union of three quarks each, bound together by the strong force (gluons). So how and why do they get treated like single masses? Is it analogous to the sun being composed of trillions of particles and yet being treated as a single mass when computing its gravitational force?

Yes, it is very analogous. The Sun is a big ball of a huge number of particles, but you don't need to know which particles are where within the Sun to compute how the planets move. All you need to know is the Sun's mass and use the fact that it is approximately spherically symmetric. Similarly, a proton is made up of quarks, but you can compute how the electron in a hydrogen atom behaves without any reference to the quarks in that proton at all. All you need to know is the proton's mass and charge.

Where the existence of quarks becomes important is in the physics of the nucleus and at very high energies. If you shoot electrons into an atom with enough energy, then their de Broglie wavelengths will be short enough to interact with the quarks in observable ways. In fact this is how the existence of quarks was first demonstrated. :)

A-L-E-X wrote:

Source of the post Also, is gravity a factor when dealing with such small particles?

On a per-particle basis, gravity is so weak compared to the other forces that we can very safely ignore it. Let's do a simple calculation to convince ourselves of this. What is the ratio of the strength of the gravitational force to the electrical force between two protons?

The gravitational force between them is F_{g}=Gm_{p}^{2}/r^{2} where of course G is the gravitational constant, m_{p} is the proton mass, and r is the distance between them.

The electrical force is F_{e} = ke^{2}/r^{2}, where k=1/(4πε_{0}), e is the charge of a proton and ε_{0} is the electric constant.

Their ratio is

In other words, the role of gravitation in nuclear physics is completely unmeasurable, unless you had more than 30 decimal places of precision (which we never do)! So you can absolutely ignore it. The only time you can't ignore it is if there are so many particles that their combined mass produces a lot of gravity, such as in extremely compact objects like neutron stars or black holes.

One of the most fascinating and somewhat paradoxical qualities of quarks (and something which makes them quite fascinating to me) is that the mass of each individual quark is MUCH more than the mass of a proton or a neutron..... Something happens to all that extra mass when the three of them combine, but what? Does the mass get bound up in energy and is that why it's so difficult to break a proton or neutron down to its constituent quarks? Also is a neutron heavier than a proton because a neutron's mass = the mass of a proton + electron?

I've also read of the possibility of such things as quark stars or strange stars and perhaps some evidence has been found for them- it would be fascinating if they actually existed. I am a big believer in the idea that anything humankind can create already exists out in the universe, somewhere.

JD- I've written about the idea of an antiverse as you describe it (also a pair of mirrorverses, one of mirror matter, one of antimirror matter) creating its own dimensions, called complementary dimensions. Time would go backwards relative to our universe, but would be forward relative to any denizens of that universe. It's an exercise in mathematics but I find it logical and self-consistent.

Last edited by A-L-E-X on 13 Oct 2019 04:04, edited 1 time in total.

Source of the post One of the most fascinating and somewhat paradoxical qualities of quarks (and something which makes them quite fascinating to me) is that the mass of each individual quark is MUCH more than the mass of a proton or a neutron..... Something happens to all that extra mass when the three of them combine, but what?

You have it backwards! Individual quark masses are much less than the proton mass (and also much less than one would expect by just by dividing a proton mass by 3, which is what makes it surprising).

In other words, the mass of a proton is much more than the sum of the masses of its parts. Why is that the case? Because the mass of a system is more than the sum of its parts. It also depends on how those parts are arranged and interacting with one another.

Examples:

1) A stretched spring weighs more than it does when relaxed. You can think of the extra mass as being associated with the potential energy of being stretched. Release it, and it will collapse back toward its natural length, exchanging that potential energy for kinetic energy. If you then damp out that motion (halt the spring at its natural length), it will weigh less, because now it has lost that energy. (This effect is so small it is unmeasurable, but it is easy to calculate by the formulas m=E/c^{2} and the spring potential energy U=1/2kx^{2}.)

2) The total mass of a black hole plus a 1kg mass far away from it is greater than if you slowly lowered the same mass down to the event horizon.

For quarks, one way to think about the extra mass is in terms of the strong force that is binding them together. The strong force involves a lot of gluons being exchanged, which also transfer a lot of momentum (like in the box filled with photons example). So much of the mass of a proton is not just the quarks, but also the gluon exchange, even though the gluons themselves are massless.

Source of the post One of the most fascinating and somewhat paradoxical qualities of quarks (and something which makes them quite fascinating to me) is that the mass of each individual quark is MUCH more than the mass of a proton or a neutron..... Something happens to all that extra mass when the three of them combine, but what?

You have it backwards! Individual quark masses are much less than the proton mass (and also much less than one would expect by just by dividing a proton mass by 3, which is what makes it surprising).

In other words, the mass of a proton is much more than the sum of the masses of its parts. Why is that the case? Because the mass of a system is more than the sum of its parts. It also depends on how those parts are arranged and interacting with one another.

Examples:

1) A stretched spring weighs more than it does when relaxed. You can think of the extra mass as being associated with the potential energy of being stretched. Release it, and it will collapse back toward its natural length, exchanging that potential energy for kinetic energy. If you then damp out that kinetic energy (halt the spring at its natural length), it will weigh less, because now it has lost that energy. (This effect is so small it is unmeasurable, but it is easy to calculate by the formulas m=E/c^{2} and the spring potential energy U=1/2kx^{2}.)

2) The total mass of a black hole plus a 1kg mass far away from it is greater than if you slowly lowered the same mass down to the event horizon.

For quarks, one way to think about the extra mass is in terms of the strong force that is binding them together. The strong force involves a lot of gluons being exchanged, which also transfer a lot of momentum (like in the box filled with photons example). So much of the mass of a proton is not just the quarks, but also the gluon exchange, even though the gluons themselves are massless.

Wow, that solves the paradox! I think I got the idea that quarks are much more massive from this article: https://phys.org/news/2010-05-masses-common-quarks-revealed.html Quarks have an astonishingly wide range of masses. The lightest is the up quark, which is 470 times lighter than a proton. The heaviest, the t quark, is 180 times heavier than a proton -- or almost as heavy as an entire atom of lead.

"So why these huge ratios between masses? This is one of the big mysteries in theoretical physics right now," Lepage said. "Indeed it is unclear why quarks have mass at all." He added that the new Large Hadron Collider in Geneva was built to address this question.

According to their results, the up quark weighs approximately 2 mega electron volts (MeV), which is a unit of energy, the down quark weighs approximately 4.8 MeV, and the strange quark weighs in at about 92 MeV.

What you said and illustrated with the examples makes a lot of sense considering the fact that mass is just another form of energy. I think that's why with regards to particles, when referring to mass, it's usually called "resting mass."

An interesting quality about the box of massless photons not being massless if the photons are bouncing around is that, since, to move at the speed of light, the particle must be massless, but since the box of photons isn't massless it means the box isn't moving at the speed of light (the individual motions of the photons must be canceling each other out or be deducted from the individual motions of each photon.)

Source of the post Quarks have an astonishingly wide range of masses. The lightest is the up quark, which is 470 times lighter than a proton. The heaviest, the t quark, is 180 times heavier than a proton -- or almost as heavy as an entire atom of lead.

Oh yes, the different types of quarks have a wide range of masses. But a proton is two up quarks (2.3 +/- 0.6 MeV each) and a down quark (4.8 +/- 0.5 MeV), for a total of 9.4 +/- 1 MeV. That's about 1% of the mass of a proton (938MeV).

A-L-E-X wrote:

Source of the post I think that's why with regards to particles, when referring to mass, it's usually called "resting mass."

Yes, though I think it really should be called rest energy, and the mass is invariant and doesn't depend on how fast the particle is moving. A proton moving at 99.9% of c still has a mass of 1.67x10^{-27} kg. But its energy will be 20986MeV, of which 20047MeV is kinetic energy. A common temptation here would be to calculate the energy and then say that m=E/c^{2} to get 22.37 proton masses, but that is really misleading. That famous formula E=mc^{2} is more famous than it has any right to be, and it is only valid as a relation between total energy and mass of a particle in the rest frame. The correct formula in any frame is E^{2} = m^{2}c^{4} + p^{2}c^{2}, with m invariant. This formula also works equally well for massive or massless particles.

I actually prefer to write m^{2} = E^{2} - p^{2} (in units where c=1), to emphasize that m is invariant while total energy E and momentum p vary depending on reference frame. This reminds (and not by coincidence) of the relativity of space and time (where the distance in spacetime, or spacetime interval, is s^{2} = x^{2} - t^{2}. The time intervals t and spatial distances x can vary, but in such a way as to keep the spacetime interval s the same.)

A-L-E-X wrote:

Source of the post An interesting quality about the box of massless photons not being massless if the photons are bouncing around is that, since, to move at the speed of light, the particle must be massless, but since the box of photons isn't massless it means the box isn't moving at the speed of light.

Indeed! In fact we can always choose a frame where the box is at rest, and no matter what frame we choose the photons are all always moving at c.

Source of the post I do not know if that has been asked before, but could one distinguish a black hole made of antimatter from a black hole made of matter?

Nope! Not in any way whatsoever.

The easiest but least satisfying way to reach this conclusion is by the No Hair Theorem, which says a black hole's only properties are mass, spin, and charge, so whether it was built from matter or antimatter is irrelevant. But it can be more fun to think about why certain common ideas for how you might distinguish them would fail.

For example, what if you made a black hole out of antimatter, and then add matter to it? The results will be the same as if the black hole were made of matter. You might think that matter meeting antimatter would annihilate and make the black hole shrink. But not so. Matter-antimatter annihilation releases energy, and energy is mass. So in either case the black hole grows by the same amount, and this method fails.

What if instead of making observations from outside, you dive in yourself to see what it is made of? Surely if you sacrifice yourself, you discover the true answer, right? No. If you made a black hole, and then dive in to it, then you never catch up to the material that made it -- it is always ahead of you, all the way to the singularity, before which you are destroyed.

The smartest idea for how to distinguish a matter from antimatter black hole might be to build a very giant telescope and observe the horizon, hoping to catch the occasional random photon from the material that built it which, from your frame of reference, is frozen in time very close to the horizon. Surely if your telescope is sensitive enough, and you wait long enough, you should see something that would answer the question, right? No. The idea is ruined by the quantization of light. There are only so many photons that will ever escape, and most escape within a very short time -- about the same time as it took the material to fall through the horizon according to its frame of reference. After that, the black hole's horizon is absolutely black, save for Hawking radiation which would not betray any useful information.

The inescapable conclusion is that a black hole really isn't "made of" anything except curved spacetime. Whatever information about what went into creating it is effectively lost and scrambled into its Hawking radiation, leaving the only measurable properties of mass, spin, and charge.

Interesting question. Assuming that the Big Bang produced exactly equal amounts of matter and anti-matter, could the creation of anti-matter black holes be what shifted the balance and prevented everything from just annihilate into pure energy?

Source of the post For example, what if you made a black hole out of antimatter, and then add matter to it? The results will be the same as if the black hole were made of matter. You might think that matter meeting antimatter would annihilate and make the black hole shrink. But not so. Matter-antimatter annihilation releases energy, and energy is mass. So in either case the black hole grows by the same amount, and this method fails.

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. And that these virtual particles consist of matter and antimatter. But if I think about it now, that would not work, the probability that the particle of matter falls into the black hole is as great as the probability that the antimatter particle will fall into the black hole. But if the virtual particles are not matter and antimatter, what do they consist of? Or if the Hawking radiation is not caused by virtual particles, what causes it?

Source of the post 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,

black holes emit light, exactly like an ideal black body with a particular temperature.

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^{2}k/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}:

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,

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.

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.

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:

First writing the entropy in terms of total black hole energy U,

Then differentiating with respect to U,

Finally, by inverting that, we get the black hole's temperature:

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.)

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: