Frontiers and Controversies in Astrophysics

ASTR 160 - Lecture 8 - Introduction to Black Holes

Chapter 1. Introduction [00:00:00]

Professor Charles Bailyn: Okay, welcome to the second part of Astro 160. This is going to be about black holes and relativity. And just to give you kind of a preview, the whole point of black holes is that, of course, they emit no light, so you can’t see them directly. And so, the question arises, “How do you know that they’re there?” And the reason you can demonstrate that black holes exist is because they’re in orbit around other things and you can see the motion of the other things that interact gravitationally with the black hole.

This concept should be familiar, to a certain extent, because it’s exactly the same thing we’ve been doing for discovering exoplanets. You don’t see the exoplanet directly. What happens is that there’s something else that you can see that’s affected by the presence of the exoplanet. So, exactly the same thing happens with black holes. And so, we’re going to use the same equations, the same concepts, to explore this very different context. So, black holes can’t be seen directly. And so, instead of detecting them directly, you use this combination of orbital dynamics and things like the Doppler shift to infer their presence, and more than just inferring their presence, to infer their properties.

Now, the context is more complicated. And, in particular, we’re no longer going to be using Newton’s laws ‒ Newton’s Law of Gravity, Newton’s Laws of Motion–because there is a more comprehensive theory that replaced Newton, which is necessary to understand these things. That more complex theory is Einstein’s Theory of Relativity. So, we’re going to be using some relativity rather than Newtonian physics. This gets weird very fast, okay? And so, I’m not going to start there. I’m going to start with a kind of Newtonian explanation for what black holes are, we’ll do that this time, and then the weirdness will start on Thursday.

Chapter 2. Escape Velocity [00:02:38]

So, the first concept and the easiest way, I think, to understand black holes is the concept of the escape velocity. This is a piece of high school physics. Some of you may have encountered it before. And it just means how fast you have to go to escape from the gravitational field of a given object. If you go outside and you shoot up a rocket ship or something like that, how fast do you have to shoot it up so that it doesn’t fall back to the Earth? And so, you can define an escape velocity for the Earth, or for any other object for that matter, which is just how fast you have to go to escape its gravitational field.

There is an equation associated with this. It looks like this, V_escape, that’s the escape velocity, 2GM / R all to the ½ power. And this is the speed required to escape the gravitational field of an object; supposing that the object has mass equal toM and radius equal to R. Oh, one other assumption, I’m assuming here that you’re standing on the–that you start from standing on the surface of the object. If you are on the surface. Okay.

This equation should look vaguely, but not a hundred percent familiar to you, because you derived something that looked a lot like it on the second problem set, where you worked out the relationship between the semi-major axis of an orbit and the speed and object had to go in to be in that orbit, and what–the way that calculation worked out it was the velocity equals GM over the semi-major axis, a. So that 2 wasn’t there in that derivation, but otherwise, the form of this is actually quite similar to that.

And so, let me explain why that is. Here’s some object. It’s got a radius of R and a mass of M. And imagine that you’ve got something that’s in orbit around this object, but it’s in orbit right above the surface. It’s just skimming the surface of the thing. That would be impossible in the case of Earth, because the friction from the atmosphere would slow you down, but in a planet without an atmosphere, this would be possible. Imagine you’re just sort of skimming over the tops of the mountains, or whatever. So here you are in orbit, this is something in a nice circular orbit right above the surface of this object, and it’s going around. And it’s got some velocity, which I’m going to call the circular velocity. And as we calculated in the previous section of the course, that’s GM / a, which is, in this case, GM / R, because you’re skimming the surface, to the ½ power. So that’s how fast you have to go to stay in orbit. If you go slower than that–let’s take a different color, here. If you go slower than that, you crash into the surface right away, because you don’t–you’re not going fast enough to stay in orbit. So, that’s what happens in the case–if you have a velocity less than the circular velocity.

What happens if you’re going faster than the circular velocity? Well, you’re going so fast that you move further away from this object. You don’t stay skimming the surface. So, here you are, going a little bit faster. And now you’re not in a circular orbit and your orbit ends up looking something like this–nice elliptical orbit. And this is a velocity greater than the circular velocity, but still less than the escape velocity because you haven’t escaped the gravitational field, because you’re still in an orbit. It’s an elliptical orbit now, but you’re still in an orbit, and it’ll come back to the same place.

Now, if you’re going at somewhat faster than that at the escape velocity, then you never come back. You continue all the way out to infinity. Your orbit continues to change direction a little bit due to the gravitational force of this thing. You continue to slow down, but in the end you never come back. So, it’s a non-repeating orbit. And, if you’re going even faster than the escape velocity, then you get to infinity even faster. And, more importantly, when you get there, you’re still moving pretty fast. This, you gradually slow down.

So, all of these different kinds of orbits–the one that crashes into the planet, the one that skims the planet, the one that’s an elliptical orbit, the one that escapes altogether–those are all within a factor of the square root of 2 of each other. Because, remember, the escape velocity here is equal to (2GM / R)^½. So, just by increasing your orbital speed by a factor of the square root of 2, you go from being in a nice circular orbit to escaping the gravitational field of the object altogether.

So, you can calculate escape velocities of things. Let’s do that once. The escape velocity of Earth–that would be–the escape is (2GM / R)^½.

2G = 7 x 10^-11

M is, for the Earth, is 6 x 10²⁴

R for the Earth is something like 7 x 10⁶. This all has to be taken to the ½ power. The 7s cancel. What do we have here? Let’s see, 12 x 10¹³.

24 - 11 over 10⁶

½–let’s see, that’s 1.2.

14 - 6 = 8, times 10⁸, to the ½. That’s something like 10⁴ meters per second, because I’ve used the value of G that’s appropriate for meters per second. So, that’s about 10 kilometers a second. So, that’s the escape velocity of the Earth. If you go outside, you throw a football up into the air at 10 kilometers per second, it’s not going to come back down. Try it–good exercise.

Okay, you could calculate the escape velocity of any particular object. You could calculate the escape velocity of the human being–of a human. A human has a mass of 100 kilograms, a radius of about 1 meter; this is kind of a big human, but we’ll go with it. Let’s see, V_escape is equal to 2GM over–a half–there’s a very famous book about physics with the title, “Assume A Spherical Cow,” and this–so, this is the kind of thing that physicists like to do. It’s an idealized situation. We have this perfectly spherical human being and–okay so let’s do the calculation, 2 x 7 x 10^-11 x 100, that’s 10², over 1, to the square root. That’s 14 x 10^-9. Call it (1.4 x 10^-8)^½. That’s something like 1.2, let’s call it 1.

1 x 10^-4 meters per second. Or–let’s see, that’s 1/10 of a millimeter per second. That’s less than 1 meter per hour. And that’s why we don’t go into orbit around each other. Okay? Because you’re always moving way, way, way faster than the escape velocity of the people you’re interacting with. The escape velocity of the Earth is 10 kilometers a second. So, you’re not moving that fast. So, you’re bound to the Earth. But when you’re hanging around with your friends, you’re probably going faster than 1 meter per hour, and therefore you’re moving faster than the escape velocity. You don’t feel a great effect of the gravitational force of other people. This is kind of anti-Valentine’s Day calculation, right? Because it proves that human beings are not attracted to each other; at least not by the force of gravity. And so, you can do this calculation for any given object.

Chapter 3. Defining Black Holes and the Schwarzschild Radius [00:12:12]

Fine, so what’s a black hole? Black hole has, actually, a very simple definition. A black hole is simply something in which the escape velocity is greater than or equal to the speed of light. c is the expression for the speed of light. This is 3 x 10⁸meters per second. And, if you’ve got something where the escape velocity is greater than the speed of light, that’s what a black hole is. And it makes a certain amount of sense. If the escape velocity’s greater than the speed of light, then light won’t escape this object, so you won’t be able to see it, right? Because, the way you see something is you see the photons that come off it.

And, there’s nothing particularly extraordinary about this, as far as it goes. In fact, the middle of–this was already talked about and worked out in some detail in the middle of the eighteenth century by an otherwise obscure English clergyman named, I think, John Michell, and he did the following calculation. He said, okay now, if this is going to be true, how big is such an object? How dense does it have to be? And, you can work this out. Supposing the escape velocity is equal to the speed of light–so c = (2GM / R)^½. You can square both sides and regroup, and you have R = 2GM / c². This is now–wasn’t called this in the eighteenth century, but it’s now called the Schwarzschild radius. Schwarzschild was a contemporary of Einstein. And, this is the size something has to be in order for its escape velocity to be equal to the speed of light. And a black hole–another definition of a black hole is something in which the radius of the object is less than the Schwarzschild radius–because if the radius is less, then the escape velocity will be even greater.

So as I say, this was worked out in obscurity in the middle of the eighteenth century, and nobody thought anything much about it. Michell sort of pondered for a little bit what such an object would look like, and he decided it would look dark. And true enough, but who cares. And nobody thought anything further about it for 150 years until Einstein came along at the start of the twentieth century, and came up with the Theory of Relativity. And one of the pieces, important pieces of relativity is that the speed of light, c is a very, very special velocity, and has rather bizarre and extremely profound properties. And it was only at that point that the concept of the black hole was recognized as something that was in any way out of the ordinary. And so, for 150 years this idea kind of lay dormant until it was resurrected by a profound change in the thinking about the underlying physics.

So, this is one of these kinds of historical fables that we’ve hit a couple of times. This is the fable of Michell’s discovery of black holes. And the moral of this little fable might be that the importance of a result changes depending on the context; result changes, sometimes dramatically, with context. So, something that looks quite unimportant for a long time can all of a sudden become a very big deal. And so, it happened in this case.

Okay, so, one question you can ask is that’s all fine, how big is the Schwarzschild radius for some bunch of objects? Let’s try the Sun; here, let me start a new piece of paper. So, how big is the Schwarzschild radius of the Sun? Schwarzschild radius, again, 2GM / c² = 2 x 7 x 10^-11. Mass of the Sun, now, 2 x 10³⁰. Divided by c². c is 3 x 10⁸ and we’ll square it. Okay, so let’s see, 2 x 2 = 4, times 7 is 30.

30 x 10¹⁹

30 - 11 = 19

Divided by–3 x 3 = 10.

(10⁸)² = 10¹⁶. So, that’s 3 x 10³, because 19 - 16 = 3–oh, in meters, because we’re still in MKS units, because that what we’re using. That’s the units of G that we’ve chosen.

3 x 10³; that’s 3 kilometers. That’s pretty small for the Sun. And so, you might imagine that such objects are rare, or perhaps even non-existent, because it would have to be incredibly small, and therefore, incredibly dense, to have a strong enough gravitational field to prevent light from escaping. And so, you might have thought that this is an entirely theoretical concept, and that no matter how interesting it is, there’s no point in really studying it further, because you’re unlikely to ever encounter one of these things in real life.

But one of the strange things that was–has been known for quite some time is that black holes really should exist, and that they are predicted to exist. This has been known for quite a while–known for at 70 years, that black holes should exist. And the reason they should is that they are one of the possible end points of stellar evolution–the evolution of stars.

Chapter 4. Gravity and Pressure in the Evolution of Stars [00:18:50]

And so, now I want to summarize how stars evolve. I should say, this is the subject of whole courses. You can take Astro 350 and they will talk about this for the entire semester. You can take Astro 110 and then they’ll talk about it for a month. But we’re going to do it in about twenty minutes, save you time and effort, here.

So, a star’s life is determined by the competition between two forces; so, star’s lifetime and evolution is determined by two forces. One is gravity, which has the tendency to hold the star together. And now, the thing about gravity is, it ought to work. But stars have no solid surface. So, if you can imagine gravity pulling on an atom on the surface of the star, why shouldn’t that atom just fall all the way down to the bottom of the star? Because there’s no solid surface to prevent it from doing so. And the answer is that gravity isn’t the only force operating, there’s also pressure.

So, there’s gravity, which pulls stuff in, and pressure, which has the tendency to push out. And these things balance in most stars. In the Sun, for example, these two forces are in balance at all points in the Sun, and this balance goes by the technical name hydrostatic equilibrium. Hydro, because it’s a fluid, it’s not a solid surface. Static–nothing’s moving, and equilibrium is just balance.

And, to be a little more precise, the way this works–here’s the surface of some star. Here’s some point within the star, and there are two kinds of forces acting on this point. There’s gravity, which is pulling the thing toward the center of the star, and then there’s pressure forces in two different ways. The outer regions of the star exert a pressure inward. So, there’s an inward pressure. And the inner regions of the star exert a pressure outward. And the outward pressure has to be greater than the inward pressure by exactly the right amount to counteract gravity.

So, it basically looks like P_out - P_in = gravity. And this holds true at all points. Now, in order–it’s the other way around right? Thank you. P_in - P_out–yeah, right. P_in minus P_out has to equal gravity. And that requires that the pressure on the inside has to be bigger than the pressure on the outside, because you don’t have negative gravity; at least, not until the third part of this course. But, at the moment, you don’t have any negative gravity, and so the pressure on the inside has to be greater than the pressure on the outside.

Okay, so what is pressure? Cast your mind back to high school chemistry. Remember high school chemistry? It doesn’t matter if you don’t. I’ll tell you everything you need to know. There’s something called an ideal gas, and there’s something that the ideal gas does, which is to exert gas pressure. Your high school chemistry teacher probably wrote down something that looks like this: PV = nRT. And here’s the thing about this. V is volume in this case; P is the pressure;n is the number of particles per volume.

And so, the key thing here is that n divided by V is equal to the density, basically, times a constant. Because, remember, density is mass per volume. If you take the number of particles and you multiply by the mass of each particle, that’ll give you the total amount of mass in a given region. You divide by the volume, that equals the density. And so, this also is a constant, this R thing. And so, what you get is P is equal to a constant, times the density, times the temperature. So, this is how physicists think of the ideal gas law, because we prefer to work in terms of the density.

Okay, so here’s the pressure. And the pressure on the inside had better be bigger than the pressure on the outside, or this isn’t going to balance, which means, either the density or the temperature, or both, had better be larger in the middle of the star than it is in the outer parts of the star. So, inside of the star, the T and/or the has to be bigger than it is on the outside.

Now, it turns out that if you just keep the temperature constant all the way through the star, you never achieve this balance. So, if only the density varies, then inner regions do have higher pressure but the increase in density also increases the force of gravity, because gravity is dependent on how much mass there is. And, if you increase the density, you also increase the amount of mass. So, you have higher pressure, but you also have higher gravity. And it turns out that you can prove, mathematically speaking, that no balance is possible, because it always ends up being the case, for gas pressure, that the amount you have to increase the density by, if you’re only increasing the density, will also increase the gravity, and you’ll never get a balance.

So, the consequence of that is that the inner parts of a star must be hotter than the outer parts. Otherwise, the star wouldn’t exist; it would collapse. And this is true. The inside of the Sun turns out to be something like 10⁷ degrees. The surface of the Sun turns out to be something like 6 x 10³. So yes, indeed, the inside very much hotter than the Sun, that is very much hotter than the outside, in the Sun, and that’s what keeps the Sun in balance.

And there’s a problem with this. And the problem is that there is something called thermodynamics. And one of the laws of thermodynamics is that heat tends to flow from places where it’s hot to places where it’s cool. This is evident in everyday life. If you take a little piece of, I don’t know, molten lead or something, and you drop it in a bucket of water, the heat from the lead will spread into the water. The water will increase in temperature very slightly. The heat will come out of that piece of lead. The lead will solidify, and everything will come into a kind of temperature balance. Similarly, if you put a snowball in some hot place, it’ll melt. Why? Because the heat from around it will go into the snowball. The temperatures will try and equalize each other and they’ll–it’ll come out even. So, this law of thermodynamics is why the snowball has no chance in hell.

And so, this happens in stars too, right? So, the heat in the center of the star flows out. And when it gets to the surface, it’s radiated. At the surface, it radiates, and that’s the energy that we see coming from the star. It’s this heat that was in the center. It’s gotten to the surface. It’s now radiating away out into the cold depths of space, and that’s what we see.

But, that means that the temperature in the center of the star, which is holding the star up, decreases, and then the star wouldn’t hold itself up. So you require–in order for the star to hold itself out–an energy source at the center of the star. And this does two things. It replaces all that lost heat, and it preserves the equilibrium of the star so it doesn’t collapse.

Chapter 5. From Electron Degeneracy Pressure to the Chandrasekhar Limit [00:28:06]

Okay, so this is all pretty abstract. And it was known that this had to be true before they figured out what the energy source was. It was known for people just sort of thinking in very general terms about how the Sun could exist–understood that there had to be some kind of large source of energy down in the middle of the star. And, notice it has to be in the middle of the star. It does no good for the energy to be created all the way through the star, because if it’s created all the way through the star, then the temperature is distributed throughout the star and you don’t get a situation where it’s much hotter in the middle than it is on the outside.

So, everybody knew for quite a while that there was energy being created in the center of the star. They just didn’t understand how that was done. And then when people invented nuclear physics in the 1930s and ’40s and ’50s, it was understood that this comes from nuclear reactions–nuclear fusion, in particular. In the case of the Sun, it’s the fusion of hydrogen, atoms together to make helium, that does this. And that releases energy in the same way that a nuclear bomb does.

The problem with this is that eventually you run out of hydrogen, or whatever your nuclear fuel is, because you’ve got only a limited amount of it in any one star. So, eventually, the nuclear fuel runs out. And then the star has many adventures before it settles down. And for these, I will have to refer you either to a textbook or to some other course, because I’m not going to take you through the whole exciting life of a star once its nuclear fuel is exhausted. Suffice it to say that you know in advance what the outcome has to be, because there’s no way it can hold itself up, in the long run, because it doesn’t have an energy source down at the center of the star. So, the consequence of this has to be that the star collapses.

Now, it doesn’t necessarily collapse all the way down to being a black hole, because there are other kinds of pressure besides the pressure exerted by an ideal gas. So, at very high densities, you get other kinds of pressure. In particular, there’s something called “electron degeneracy pressure.” This is sometimes called Fermi pressure, after the guy who thought it up.

Degeneracy is another one of those words that means something different to physicists than they do to ordinary, normal people. I remember the first time I taught Astro 110, right after I came here, about fifteen years ago. I started talking in the middle of a class about degenerate white dwarfs. And you could feel this sort of wave of something between confusion and anxiety permeate through the class, and I had no idea what was going on. Some teaching assistant had to pull me aside afterward and remind me that these words mean different things. So, I apologize if this sounds like a sort of adult version of a Grimm’s fairy tale, you know, with degenerate white dwarfs wandering around and stuff, but such are the words we have to work with.

Okay. So you have this electron degeneracy pressure–that’s a different kind of pressure ‒ and that can stabilize the star. So it stabilizes a star–stabilizes the star at–around the radius of the Earth. So, that’s very high density material. We can calculate the density. You’ll remember our equation for density: mass over volume. And so let’s see, how does this work? This will be the mass of the Sun, let’s say, divided by the volume of the Earth. That’s ⁴⁄₃ πr³. Radius of the Earth is (7 x 10⁶)³. So, let’s see, (2 x 10³⁰) / 4. 7³–

7² = 57, cubed is 350. Times 10¹⁸. That’s 6 x 3–yeah, okay, we’re doing all right. And let’s say that that’s 2,000 x 10²⁷over, I don’t know, 1,400 x 10¹⁸. Let’s cancel those.

10⁹ kilograms per meters squared. That’s about a million times denser than water. Water, you’ll remember, is 10³kilograms per meters squared. And water is defined to have a density of 1 gram for a cubic centimeter. So, if you were to pick up one gram of this white dwarf–one cubic centimeter of this white dwarf, it would be a million times more massive than a gram. That’s about a ton. So, a thimbleful of this stuff weighs about a ton, very dense. And the Sun will end its life as a white dwarf with this electron pressure balancing the gravity. These–such stars are called white dwarfs, and there are many of them known. White dwarfs–this is the end point of the Sun.

And then, in the 1930s, what happened to make black holes inevitable, was that one of the great theoretical astrophysicists of the twentieth century, a man named Subramanyan Chandrasekhar, discovered that electron degeneracy pressure doesn’t always do the job. So, in the 1930s, Chandrasekhar discovers–proves that if the mass of an object is greater than 1.4 times the mass of the Sun, electron–this kind of electron pressure is insufficient. And the star continues to collapse.

Now, Chandra was a graduate student in England at the time he figured this out, and he presented this rather dramatic result at a big meeting of the Royal Astronomical Society in London. And then, every graduate student’s worst nightmare took place. Chandra’s thesis advisor was a very famous man named Arthur Eddington, one of the great physicists of the early twentieth century. And after Chandra had presented his results to all these assembled scientific dignitaries, Eddington got up and denounced his own student, and said, “This can’t possibly be true.” And Eddington made the famous remark, “There ought to be a law of nature to prevent stars from behaving in this foolish manner.”

And the consequence of that is that a lot of people didn’t follow up on Chandra’s idea. Chandra got miffed, as you might understand. He got on the boat to the United States. He spent the rest of his career at the University of Chicago. On the boat, he wrote a great textbook still in use on the structure of stars, in which he laid out in detail all of the arguments that the Chandrasekhar limit must actually exist. And fifty years later he got the Nobel Prize for it. But that–there was this kind of time lag there, and it is, I think, important for those of us, say, on the faculty to recall that if Eddington had listened to his student instead of to his intuition, the study of black holes would probably be forty years more advanced than it is. So, another fable for our times.

Chandra was always very gracious about this, actually. He would praise Eddington to the skies as a wonderful advisor, and then with this little asterisk. So, fable: Chandrasekhar’s limit is the title, and the moral here is, “believe your student, not your intuition.”

And actually, the story of Eddington was actually kind of interesting. As Eddington got older, he became more and more convinced that, you know, he could guess the right answer. And so, it wasn’t just the Chandrasekhar limit, it was other things. He got a little weird toward the end of his life, and he started believing his intuition. Einstein did this too, right? Einstein famously discovers all this great stuff, and then the second half of his life is completely useless, scientifically, because he becomes convinced that his gut is telling him that quantum mechanics is wrong. This famous remark, “God does not play dice with the universe”–but it turns out, that isn’t true. And so, there is this probabilistic nature of reality that quantum mechanics shows. And so, Einstein spent the second half of life trying to prove that his intuition was correct, that quantum mechanics couldn’t really be true and thus, did no physics worth doing for about thirty or forty years. So, you have to watch out for this. If you’re too smart and start believing yourself, you can get into trouble.

Chapter 6. Neutron Stars [00:37:59]

All right. Now, having said all that, Eddington was partly right. There is, sort of, a law of nature that prevents stars from behaving in this foolish manner. So, to understand that, what happens when white dwarfs–when the white dwarf collapses? It’s got to get rid of its electrons, because the problem is, the way this electron degeneracy pressure works–you can’t squeeze electrons any closer together than they are in a white dwarf. So now, you have to get rid of electrons. And so what do you do? You combine the electrons and the protons and you turn them into neutrons plus neutrinos. These are neutrinos. They stream out. And so, you end up with something that’s made entirely out of neutrons. So, the whole star turns into neutrons.

A chemist would think of this as essentially turning the whole star into one atom, into one atomic nucleus. An atomic nucleus with no–an atom with no protons, no electrons, and 10⁵⁷ neutrons. And you could imagine putting that somewhere on the Periodic Table.

Astronomers call these things neutron stars, and they exist. They were discovered in the 1960s. And a typical neutron star, a couple times the mass of the Sun has–mass equals 2 times the mass of the Sun. Radius of about 10 kilometers. And you can work out the density for that. Density is a billion times greater–I’ll leave this as an exercise–greater than for white dwarfs. So, instead of a cubic centimeter of the stuff weighing a ton, it now weighs a billion tons. And you’re having a tough time moving it around.

But 10 kilometers ‒ that’s getting close to the Schwarzschild radius. Remember, we calculated the Schwarzschild radius of the Sun. It was about 3 kilometers. And, in fact, if you calculate the Schwarzschild radius of a star in terms of the Schwarzschild radius of the Sun–let’s see, you get 2GM / c², where M is the mass of the star, divided by 2G mass of the Sun, over c². So the Gs and the cs all cancel, here. And you get M_star / M_sun.

So, if the Chandrasekhar mass of the Sun is equal to 3 kilometers, as we calculated, the Schwarzschild radius of a star with–whose mass happens to equal 3 times the mass of the Sun, is going to be 3 x three kilometers, or 10 kilometers. That’s equal to the radius of a neutron star. So, a neutron star with mass greater than 3 times the mass of the Sun has a radius less than its Schwarzschild radius. And that’s a black hole. Remember?

And the key thing here is that there are lots of stars with mass more than 3 times the mass of the Sun. We don’t see them as black holes because they’re still in hydrostatic equilibrium. But eventually, they’re going to run out of nuclear fuel, and they’re going to collapse. Now, in fact, during the course of the star’s life, one of the things I glossed over is stars tend to lose mass as they live. And so, they don’t end up with the same mass they started with. But stars with initial masses, at the beginning of their lifetime, greater than–I don’t know, something like thirty times the mass of the Sun, will end up with masses greater than three times to the mass of the Sun. And then, there’s nothing to stop their collapse. What happens is, they turn into neutron stars, but they turn into neutron stars whose Schwarzschild radius–whose radii are smaller than their Schwarzschild radius, and that is a black hole. So, they collapse down into black holes.

Chapter 7. Conclusion [00:42:38]

And so, you expect a large number of black holes to actually exist. This is what happens to massive stars at the end of their life. And so, we expect there to be–that there are many black holes. And so, the question we’ll be exploring in the rest of this segment of the class is, how can you find these things? What are the properties of these things from a theoretical point of view? What does Einstein’s Theory of Relativity suggest that these things are going to behave like? And then, the big question is, once you’ve found some, and you have a theory for what they behave like–then you can ask the question, does the actual behavior of these objects conform to the theoretical expectations? Another way of saying that is, was Einstein right? Is general relativity the correct theory to describe these very exotic objects? And so, that’s what we’ll be talking about in the rest of this section of the course. Now, let me turn back to the previous section of the course, which was–which culminated last time in this little test. And I think we’re ready to hand these back. Is that true?

[end of transcript]

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