Frontiers and Controversies in Astrophysics

ASTR 160 - Lecture 13 - Stellar Mass Black Holes (cont.)

Chapter 1. Defining Black Holes in Terms of the Schwarzschild Radius

Professor Charles Bailyn: Last time, I wrote down the Schwarzschild metric, which was the sort of culmination of the theoretical part of the--of this section of the course. And I realized, as I was talking to some of you afterwards about some of the implications of this, that the way I--one of the ways in which I wrote it down might have been a little confusing. So, let me write it down again and I'll say a couple things.

And what I'm going to do is I'm going to write it down with a little r instead of a capital R and I'll tell you why in just a minute. Obviously, it doesn't matter. I've just changed the notation slightly. And the reason I want to do this is because I want to emphasize that this r, the little r, what I've now written down as a little r in this equation, is not the radius of the object; r is just a coordinate. It can be anything. It's not the radius of the object for which the mass is giving you the Schwarzschild radius. The Schwarzschild radius is a very specific radius.

rs = 2GM / c2 so that's a specific number. The little r, here, is just a coordinate and so is φ and θ, which combine into this thing which I've called Omega [Ω] and T.

And so if you have an object of mass M and radius--I'll give--I'll call that R0 just to make sure that you understand that this is the radius of the object. The radius of the object doesn't come into this metric. The mass does, because the mass turns into the Schwarzschild radius and that's part of the equation. But the radius of the object doesn't matter.

And so, if you then ask, well, what is a black hole? A black hole is something in which the radius of the object is less than the Schwarzschild radius. Because if that's true, then there is some coordinate r, some little r where little r = RS. And then, if you go up into these equations here very--these very strange things happen when little r is equal to RS. Yes, question?

Student: Yesterday, the big group had [Inaudible]

Professor Charles Bailyn: Oh you're right, you're right, it has to be the other way around, I'm sorry.

Student: [Inaudible]

Professor Charles Bailyn: This has to be squared. You're right on both counts, good. RS / r. Yes, it better be that way because RS / r in the normal part of space is less than 1. So, thank you very much, sorry about that. Okay, I think we're all right. Are we all right? Yeah, good.

Okay, so there is some r where r is equal to RS, and then these very bizarre things start to happen. If R0, the radius of the actual object, is bigger than its Schwarzschild radius, then that isn't true. There's no r where r is equal to RS--then no requal to RS. Now, you might think that somewhere inside the object, there would be a little r, which is equal to the Schwarzschild radius. But that isn't--that turns out not to be right either, because the relevant M, the relevant mass-

Student: Can you lower the top slide?

Professor Charles Bailyn: Oh sure.

Student: The whole--the slide not the -

Professor Charles Bailyn: Oh yeah, okay. So, yeah. Let me put this back. The relevant mass is the mass inside r. So if you're--if you've got some mass M, like this, and you're outside, if you're here, then the whole of the mass counts. But if you're sitting inside this thing, then the only mass that counts is this mass here. And so, for r--for coordinates for positions inside the object, then the relevant mass that you put into the equation is less than the mass of the whole object.

And the consequence of that is that, as you go further and further inside the thing, the mass becomes less and less, and the Schwarzschild radius becomes less and less. And so, if the total size of the object is greater than the Schwarzschild radius, then there is no position inside the thing which is less than the Schwarzschild radius, because the relevant mass decreases, and so does the Schwarzschild radius as you go in toward the center of the object.

So, I just wanted to make sure you understood that the key to using this metric is that the--this r term in it tells you where you are. It has nothing to do with the size of the object. And you become a black hole when the size of the object is smaller than the Schwarzschild radius. Because under those circumstances there's some position, or set of positions, where you can be inside the Schwarzschild radius, and therefore all of these very exotic effects start to happen. Questions on this? Or have I just baffled you further? Well, we'll see.

Student: Is that the radius of the Sun?

Professor Charles Bailyn: Sorry.

Student: Is that the radius of the Sun?

Professor Charles Bailyn: Oh no, this is the radius of the object. Sun is--Sun has a dot in the middle--not that you could tell with my handwriting. Okay, I don't know if--whether that was helpful or not, but I felt the need to say it. Yes, please.

Student: Could you please explain the diagram that you [Inaudible]

Professor Charles Bailyn: This right here? Yeah, so the situation is that--what is the relevant M to put into this equation? This is true, by the way, in Newtonian physics also. If you're inside an object then the mass outside of your position, this stuff, doesn't count toward the M. And you only want to count the mass inside of where you are.

Student: Okay, so that whole circle is an object?

Professor Charles Bailyn: Yeah, so this whole circle is an object--object with mass and some--with some mass and some radius. But now you're asking the question, "What happens if you're somewhere inside this thing?" Yes?

Student: Does that mean that the dotted line is RS?

Professor Charles Bailyn: No, the dot--well, the dotted line could be anything. You're at some position inside and this--the dotted line is the radius at which your particular position is.

Student: [Inaudible]

Professor Charles Bailyn: No, it's just wherever you are inside the thing. Less mass counts than you think it's going to because it's only the stuff inside your own radius that counts toward the metric, or toward the gravitational force in the Newtonian thing. You can be at--at whatever point you choose to be inside here, you can draw this same thing. And the point of this is that if you are now down in here, where you thought the Schwarzschild radius was going to be, for the whole of this mass--in fact, the Schwarzschild radius is still going to be inside your position because much less mass will count toward it. This isn't helping, I can tell by the look on your face. Okay, let me try again. Let me try again.

Okay, here's an object. And this object has a radius bigger than its Schwarzschild radius, and it's got some mass. Okay, so, now I'm going to draw, in dotted lines, its Schwarzschild radius. That's inside here, okay? Now, supposing you're inside this object. Supposing you're living here. Then, at that point, you--I draw this sort of interior radius, here. And if I'm at this point, then I ask, what is the Schwarzschild radius that goes into the metric for something that's sitting inside for person X, here. And the answer to that is it's actually smaller than the Schwarzschild radius of the entire object, because less mass counts. Only this mass counts towards the Schwarzschild radius. So, there's some part of the object whose mass no longer counts, because it's outside your position. And so, as I move in here to some interior point, the Schwarzschild radius that is appropriate for me to use becomes smaller. Does that make more sense?

Student: Can you draw that if it were a black hole?

Professor Charles Bailyn: Ah, so if I was black--if it was a black hole, right. Okay, so now, let's do the opposite case. Good question--is less than the Schwarzschild radius. So, now here's my object, and here's the Schwarzschild radius. That's the difference, right? And therefore--and the consequence of this, by the way, is that this whole object sinks down to a point. Because, remember, things inside the Schwarzschild radius have to contract down to a point. So, I can't be part way through it, no matter what I do. And, you know, if I'm outside this tiny little thing, if I'm here, I have the same Schwarzschild radius, because all of the mass is interior to my position. So, the key thing is, is the object smaller or bigger than its own Schwarzschild radius?

Student: So, in that case, all the mass is going to be closer than--

Professor Charles Bailyn: All the--yeah, all--well, all the mass--

Student: [Inaudible] is going to [Inaudible]

Professor Charles Bailyn: If the radius of the object is smaller than its Schwarzschild radius, then all the mass is inside the Schwarzschild radius. And any point in our Universe where, you know--and so, all r greater than RS is out here, somewhere, and it doesn't matter where you are. All that mass counts because it's all inside the Schwarzschild radius anyway. And then, when you get to this point, you have this magic moment where r = RS and everything blows up. Right? And that can't happen here. Because as you move in, in the middle of the object, the Schwarzschild radius shrinks down away from you as you get in, because less and less mass counts towards it, therefore it gets smaller. Yes?

Student: Would it be possible to have an object [inaudible] density--like such that there would be a point at which you would [inaudible]

Professor Charles Bailyn: Ah, okay. So, could you have some incredible density gradient where you're actually gaining more--you're actually catching up to the Schwarzschild radius as you go in. In principle, yes. But if you work out what that density gradient would have to be, it would be incredibly steep. Remember, there was a problem set at some point where we discovered that the less mass there is in a black hole, the denser it has to be in order to be inside its Schwarzschild radius. So, as you're losing mass as you get down in here, the amount by which it has to be denser on the inside goes up really sharply.

But yes, such a thing can be. Imagine, for example, a black hole with an atmosphere. Imagine there's some--or some, just, gas floating around out here. If you think of that as one object, then when you're outside the gas, if you think of the outer edge of the gas as the edge of the object, then that's not a black hole. And if you come on--if you go down--sink down into the atmosphere, eventually you'll find it. So, that's an incredibly steep density change because you've gone from an atmosphere to some object. So, in principle, that's possible. In any kind of real object that's holding itself up, probably not, because the density gradient would have to be so severe. Yes?

Student: How would this work for non-spherical objects?

Professor Charles Bailyn: How does it work for non-spherical objects? Excellent. Okay. So, let's see. What happens is, if you have a non-spherical--first of all, the gravitational forces are sufficiently strong near a--if you're anything close to a black hole, that you're likely to end up spherical.

The only way you don't is if you're spinning. Because then, you know, you bulge out at the center. Spinning black holes exist. They have a different metric. And the angular term becomes important because you're spinning around and it's a different metric. And the shape of the event horizon where r is equal to RS or the equivalent in this other metric, is non-spherical. And so, you can have non-spherical event horizons if the thing is spinning, because that has the effect of making the mass distribution different. And this is called Kerr black hole, it has a Kerr metric, K-E-R-R. And that's a different metric, and it's got a whole other set of terms that have to do with the rotation of the angular component.

But again, the same thing happens. When you get to some place, all the terms blow up but it's a non-spherical surface where that occurs. And you have to do it in cylindrical coordinates. It becomes a mess--not that it isn't a mess already. Other questions?

Chapter 2. Perihelions and Deflection of Light [00:15:44]

All right, let me go back to the effects of general relativity. We've already done one of these. So, what I'm going to talk about is the post-Newtonian gravitational effects. So, these are things where it just barely starts to deviate away from what the predictions of Newton's laws would be.

So, just to go back, if you take that metric and you kind of expand it in small quantities where the Schwarzschild radius is much, much smaller than wherever you are. And you take the first term of the expansion. Mathematically, it turns into the same things as Newton's laws. And so, when you're out there in more or less flat space, or close to flat space, then Newton's laws still apply, because the expansion of the metric gives you the exact same result.

As the gravitational forces become larger, you start to get post-Newtonian terms, and things start to be a little bit different from Newton. And we've already had one of these--so, post-Newtonian effect number one is the precession of the perihelion, where, unlike the prediction for Newton, the orbit doesn't exactly repeat. It sort of moves on a little bit each time. And you get one of these nice, sort of, Spirograph patterns, as an elliptical orbit gradually precesses, or gradually moves around.

So, that was the first one. And that was the first one that was seen. That was seen in Mercury, which is the planet you would expect to see such a thing, because it's closest to the Sun and therefore, RS / r is the largest number for Mercury--for any planet in the Solar System. So, Mercury. And, that actually turns out to have been known before Einstein actually worked out the general relativity.

So, now, post-Newtonian effect number two is the deflection of light. The whole idea, remember, is that there's the--is that gravity works because it curves space and objects moving along this curved space appear to have curved trajectories. That's true of light also. Light is also moving across this curved space. So, there's a prediction in general relativity that if you shine light near a massive object, it's the path that that light will take. It's curved. It's as if the light's being gravitationally attracted by this object,

So, let's see here. Imagine the following situation. Here are you, you're observing something. Here's a star. And the path that light takes to this star, from the star to your eye, is some kind of straight line. And so, that's where you perceive the star to be.

Now, supposing that along--that between you and this star, there's some object. So, here's an object with mass M, some massive object. Now, because the path is going to curve, the path that the light takes from the star to your eye is going to now look like something like this. And it's deflected by the presence of the mass, which curves the space that it's in.

That means, you think, because you're interpreting this as if the light is coming straight towards you, you think the star is here. Or it appears to be there. Because, you say, all right, what is the angle at which the light is coming into my eye? It's coming from there. You don't realize that the star is actually over there and the light has gone up toward the ceiling and bent back down to your eye. So, it's as if I'm looking up there and seeing the exit sign. Because the light from the exit sign has gone up to the ceiling, been bent around by some black hole or something, up at the top of the classroom, and come back down to my eye. Yeah?

Student: Well, I mean, wouldn't the light on the other side of the planet showing the massive thing also be [Inaudible]

Professor Charles Bailyn: Ah, it is possible, yes. It is possible. In this particular case, probably not, because there wouldn't be enough bend in order to get it in here. So, light going this way, for example--let me take another color before this gets too complicated. Light coming this way would bend like that and it would miss your eye. However, there are situations in which you see two images of the same object, one from each side. These are called gravitational lenses.

Student: Can you see many, many images?

Professor Charles Bailyn: In principle, if it's lined up exactly right. Imagine that it's perfectly aligned. Then you see this way, you see that way, and you see out of the paper and you can see a whole ring. This has actually been observed. Not the whole ring, but large parts of the ring, because it's never perfectly aligned. But you can see these things. These are called Einstein's rings, and you can actually see them. You can see these kinds of multiple images in cases where this is a very distant galaxy or quasar. Quasars are very bright, very small, extragalactic objects. And this is an intervening galaxy or galaxy cluster. And you can see these multiple images. I'll bring some pictures in next time. They're very--they're amazing-looking things.

It's often true that you can see four images. That comes about when the intervening mass is not spherical. And so, you get weird deflections and you--they're multiple directions. You see the same object--a very spectacular thing. These were discovered only recently but it turns out that you can observe this effect even in our own Solar System. Because what you can do--I'll bring in some pictures of gravitational lenses, those are amazing things. And what also happens is, if you're looking at something big, like a galaxy, that has a shape, it distorts the shape. It stretches the whole thing out, and you get these amazing pictures. All right.

Chapter 3. Hunting Eclipses [00:22:44]

In any case, in our own Solar System, you see this in the following way. Here's the Sun, and here's a star right behind the Sun. And the light from the star comes like this, and so you think the star is here. So, if you look at a star when it's right next to the Sun, right at the edge of the Sun, it seems to be in a different position relative to stars further--relative to other stars up here. Here's another star. Then it would--if you looked at it at night when the Sun wasn't in the way. Yes?

Student: Then how do you do astronomy during the daytime?

Professor Charles Bailyn: Excellent question, I'm getting to that. Okay, so--but the principle is--supposing the Sun were dark, okay. Look at star behind the Sun, or right next to the Sun, really. You can't see through the Sun. I'm not that crazy. Star behind Sun. And it appears to be in a different place, position, than when observed at night, without the Sun in the way.

Okay, so how do you do astronomy during the daytime? That's a good question.

Student: Eclipses.

Professor Charles Bailyn: Eclipses, yes. You wait until there's a solar eclipse and then you do your observation. Do this by observing during an eclipse. And if you do this, the position, the change in position of the star, if it's right at the edge of the Sun--and you've now blocked the Sun, because here's the Moon, right? And the light from the Sun doesn't come through. This change in position, that's an angle, is around 1 arc second. And that's a positional change you can observe. Remember the size of--the apparent size of a star, when you look at it, is about 1 arc second. So, it seems like it shifted a significant amount.

And so, you go, you take a picture of the sky, of the stars behind the Sun, during a total eclipse of the Sun. Now, the problem with this is a practical problem. Total eclipses of the Sun happen only for very short amounts of time, and only occasionally, and only in a small--over a small portion of the Earth at any one time. How many people have seen an eclipse of the Sun? Pretty spectacular thing, huh? But, you know, you have to travel to get there. It doesn't come to you.

So, here's what happened. In 1917, Einstein publishes his theory of general relativity. And then, in 1919, there's going to be an eclipse in Brazil, I think it was. Eclipse in Brazil. And it seems like a good idea that you should actually go and test this theory, which has made a very specific prediction about what ought to be going on with these stars. So Eddington, we've heard of him before, he's the guy who trashed his student Chandrasekhar some years later. But at the time he was young, up-and-coming British scientist, mounts an expedition to Brazil to test this theory.

And it works. Goes down there, he takes these pictures. It does just what Einstein said. And this is the event that makes Einstein into a great international figure. This is front-page headlines around the world. And part of the reason of why it was is, look at these dates. Einstein is a German scientist. This is the middle of World War I. And so, in the middle of World War I a German scientist--he's later thought of as a Jewish scientist, but at the time, Hitler was not yet on the scene. And so, Einstein was thought of as a German scientist. So, this German scientist publishes a grand, bizarre, new theory of the Universe right after the war. The Brits mount an operation to go and check and see if this is right, and it turns out to be correct.

So, this is now right after World War I, in the era when everybody thought, well, we're never going to have a war again. There's going to be peace and justice and brotherhood throughout the world for the rest of time. And as part of this, the British mount an expedition to confirm a German theory of science. And this is regarded as a huge sign of the great new future that we're moving into of international cooperation. And a consequence of this--this is front-page news all over the world, and it makes Einstein famous. This is what makes Einstein famous.

So, the fable, here, is the 1919 eclipse expedition. And the moral that was drawn at the time is, the great international, is how science is this wonderful international--maybe even universal--quest. And, in retrospect, what it is, is, it's a great demonstration of, you know, how science is supposed to work. New theory makes a prediction. You go out and measure the effect, and it works. So, this is, science works as advertised.

Except, you know, maybe it doesn't. Because subsequently, in subsequent years, people looked back at the apparatus that Eddington and his team had put together and said, you know, you can't actually make measurements that accurate with this stuff. So, how were they able to make this measurement? And, oddly enough, for about fifty years, the photographic plates ‒ what they had done was taken photographs of the sky ‒ had disappeared. And so, nobody could go back and check the original data. At some point, subsequently, they kind of reappeared inside--just recently, actually, inside the personal effects of a British astronomer who had gone to live in Chile. And nobody quite knows how he got a hold of them. And, I actually have to say, I'm not sure what the outcome of this is--whether people have actually gone back and tried to measure this again.

So--and at the time, you know, Eddington, who, in his subsequent life, demonstrated a real propensity for believing his own theories, if you know what I mean--which served him poorly when he was dealing with Chandrasekhar, who was thinking of other things. Eddington had said before he went on the expedition that he was absolutely certain this was going to work because he had studied Einstein's theory and had come to believe it, because it was so beautiful and elegant. And so, there's a famous story where somebody said, well, Mr. Eddington what's--or said to somebody else on the expedition, what will Eddington do if it comes out wrong? And they said, well, Eddington will just go insane.

And so, there's a little bit of suspicion attached to this particular experiment. However, subsequent to that this has been measured repeatedly, many times, during many eclipses, and it's very clear that this works. And as I said, there's now evidence from other astronomical objects for the deflection of light, all of which confirms Einstein's theory. Yes?

Student: What does the second line say?

Professor Charles Bailyn: What does the second line say? Let's see, eclipse expedition, scientists and international, universal quest, science works as advertised, maybe. That's what I was trying to write down. So, that's the second post-Newtonian test. And the fact that it works in 1919 is what makes people really believe relativity, and what makes people believe that Einstein's this intense genius.

Chapter 4. Gravitational Redshift [00:31:31]

Okay, there are a couple of other post-Newtonian effects. So here's, now, post-Newtonian effect number three. And this is something called the gravitational redshift. You've heard of a redshift. Redshifts and blueshifts. They come about because of the Doppler shift. Something moves towards you, its light gets shifted to the blue. Something moves away from you, its light gets shifted to the red. Turns out, the wavelengths of light also change due to gravity. Or due to being emitted in a curved space-time, if you want to use the relativistic formulation.

Basically, here's how it works. You have some object with radius R, mass M, and you have a light source on the surface of that object. And you look at it from a distance. You're out here at infinity watching this light. And the light, if you emit green light here, by the time it reaches you, it'll have shifted into the red. And there's an equation. Let me write down the equation. So, this is delta lambda [Δλ]--oops, that one's not working anymore. Δλ over λ0. That's the shift in light.

And remember that the Doppler shift has some equation, here. Well, here's the equation for this shift, due to the gravitation of an object. RS / r, where this is the distance to the given mass. Or if you're emitting light from the surface of an object, the distance to the mass is the distance to the center of the object. So, in this particular case it would be the radius of the object, minus 1. So, you can see that if there's no mass, then the Schwarzschild radius is zero, and this is 1 - 1, and there's no shift.

Okay, now why does this happen? This has to happen. Imagine that you're standing on the surface of an object and you throw an object up into the air. I have an object here. I'm going to do this. What will happen? As I throw it up into the air, it's going to slow down. If I throw it faster than the escape velocity and I don't have a ceiling and stuff, then it'll go--it'll keep going away forever, but it will, nevertheless, slow down. If I throw it up slower than the escape velocity, then it's going to rise to some height, stop, turn around and fall back. Right? So, that's what happens with an object.

Why is this happening? The reason it happens is because as it moves--as it tries to move away from the Earth, it has to lose energy, because it's pushing its way out of the gravitational field. And so, it gradually slows down, because its kinetic energy decreases, as it goes further away, to balance the change in the potential energy, for those of you who remember high school physics. It balances the change in potential energy, so the total energy is zero. And so, the kinetic energy has to slow down. And whether it slows down enough to stop it or not depends on whether it's moving faster than its escape velocity or not.

Supposing I do this with a flashlight. And instead of throwing a keychain up in the air, I throw a photon up in the air. The photon starts out at the speed of light. Now, that, too, has to lose energy as it moves away from the Earth. But light, as we know, always travels at the speed of light. It doesn't slow down. It can't slow down. So, what it has to do is, it has to lose energy in some other way.

And the trick is that the energy of a photon is equal to a couple of constants, h times c, over the wavelength. So, as the energy gets less, the wavelength gets longer--has to get bigger. And so, what is happening here is, as you shine light off a gravitating object, as it gets further and further away from the object, it loses energy. But it loses energy not by slowing down, as an ordinary mass of objects would, but by changing its wavelength.

And now, go back and look at this equation again. Supposing you're shining your flashlight from the event horizon of a black hole--from something where the distance to the object is equal to the Schwarzschild radius. Then, you get Δλ equals infinity. And you've redshifted your way out to an infinitely large wavelength. The consequence of that is that the energy of the light is equal to zero.

And so, this is how it happens that light can't escape from a black hole. Because, you know, if you have the Newtonian vision of the escape velocity in mind--the thing goes up, stops, turns around and comes back--and you imagine shining a flashlight from inside the event horizon of a black hole, you imagine that the light would go up, stop, turn around and fall back. But light doesn't do that, because light always goes at the speed of light. So, what happens instead is, it redshifts itself out of existence altogether. And so, that's the way light is prevented from leaving a black hole. Because if you shine from either inside, or at, the event horizon, it'll redshift itself to infinity. Yeah?

Student: Everything you just said still makes perfect sense in the context of the Newtonian concept, why do we need to [Inaudible]

Professor Charles Bailyn: Okay, so there is a Newtonian version of this. You can imagine how you expand this thing. But this particular equation is relativistic. Oh, the other thing is that, of course, in Newtonian physics, there's no reason for light not to slow down. This business that light always goes at the speed of light, regardless of the observer, is a relativistic effect. So, you know, if you take a photon and throw it in the air, in a Newtonian concept, there's no reason it can't slow down and turn around and come back. Yes? Yeah?

Student: [Inaudible]

Professor Charles Bailyn: Δλ over λ. Δλ over λ.

Student: Sorry, next page.

Professor Charles Bailyn: This?

Student: [Inaudible]

Professor Charles Bailyn: Okay, so this is speed of light -

Student: Yeah.

Professor Charles Bailyn: And this is another famous constant called Planck's Constant. I don't think we're going to use that equation for anything in particular. But it's a couple of constants. So, basically, the energy is proportional to 1 / λ.

Student: Okay.

Professor Charles Bailyn: Yeah. Other questions? Okay. If you turn it around--if you sit on the surface of, I don't know, a neutron star or something, and look at distant starlight, that light will pick up energy as it comes towards you. So, if you sit on a gravitating object and look at a distant light, that light will actually be blueshifted, because it will gain energy falling down toward you. And so, while this is always referred to as the gravitational redshift, because we're always imagining that we're some distant observer in a flat space looking at some gravitating object, in principle, if you could live on a neutron star, all the stars would look bluer than they do to us because of this effect. So, it works both ways.

This, by the way, has now been tested in the laboratory. Here's how you do it. You set up--you sit on top of a ladder and you look at light coming from the surface and coming up toward you. And then, you have somebody--you have some confederate, or graduate student, or something, who sits on the ground and looks at light coming sideways. And you determine that the wavelength detected here is slightly larger than the wavelength detected here. That's a really small effect, because the Earth's gravitational field--you know, if you do this little calculation and you try and figure out RS / r for the surface of the Earth, it's a really, really small difference from 1. Nevertheless, because you're in the laboratory and we know how to measure the wavelengths of light with incredible accuracy, this effect can actually be measured in the lab, and has been. Yes?

Student: So, if you're on a black hole or whatever, shining a light upward and instead of doing the arc thing just--see through the--stretches out and winks out of existence at the top, then winks back into existence, falls back?

Professor Charles Bailyn: Well, it can't fall back, right. It can't fall back because it can't stop, turn around and come back.

Student: So, it just winks out of existence up here?

Professor Charles Bailyn: Well, notice what happens. Supposing you go from the event horizon only to slightly above the event horizon. The difference in the wavelength you see is this quantity at the event horizon, minus this quantity a little bit above the event horizon, because it's still going to be redshifted. But this is infinite at the event horizon. So, by the time it moves even a tiny bit off the event horizon, it's already gone.

Student: Okay.

Professor Charles Bailyn: Yeah, it's because this goes--this term becomes infinity large. Oh, but I should say, this brings up a good point. Supposing you're here and it's not a black hole. So, it's not going to--it's not going to be disastrous. But you're observing it, not at infinity, but close. So here's--this is at some R, and this is at R1 a little bit further away. So, what you would observe, here, is Δλ / λ = Δλ at R1 minus Δ--let me write this so that there's some hope of reading it.

So, this gives you the Δλ you observe at R1, over λ0, is equal to Δλ at R1, calculated by that formula, minus Δλ at R. Because, it redshifts a little bit from here to here. It would redshift more from here out to infinity. And so, the way you calculate that is you calculate redshift from here to infinity, minus redshift from here to infinity. And since--for something that is at the event horizon, the redshift from there to infinity is infinite, even if you're observing it just a little bit higher, it's already gone.

Student: Okay.

Professor Charles Bailyn: Does that make sense? How are we doing? Other questions? Yes?

Student: [Inaudible]

Professor Charles Bailyn: Oh, the photons, remember--there's a relationship between the whole way wave particle duality works for photons. There's a relationship between the energy of the photon and the wavelength. That was this E= hc / λ thing. So, as the wavelength becomes infinite, the energy in that photon goes to zero. So, it's still a photon. But if you've got a photon with zero energy, you're not going to notice it. Yes?

Student: Just as we, like, you know, use the gravitational field of, like, Mercury, Jupiter, or whatever to kind of slingshot probes into outer space, could light, like, pick up the energy by, like, you know, going around massive objects?

Professor Charles Bailyn: Well, light picks up energy as it falls towards a massive object. That's the inverse gravitational redshift. You sit on the gravitating object. You watch light coming at you from a distance. So, it does that. That slingshot effect only works if you add energy to the thing and kick yourself back out of orbit at just the right moment. You can't do that with light, because you can't give it extra--you can't give it an extra velocity kick, at any moment, to push it back out. But it is true, that if you sit--you can do this experiment backwards, right? You have your light source up here, and you observe it from down here, and then you'll see it blueshifted.

Chapter 5. Gravitational Waves [00:45:14]

Other questions? How are we doing? Okay. Let me mention one last post-Newtonian effect. This is post-Newtonian effect number four. This is the existence of gravitational waves. And the way this works is--let me put these over here. Remember the bed sheet and the basketball, over here, from a few days ago? All right, now, imagine that that basketball, which is some massive object, is moving back and forth on the--it's orbiting some other thing on that bed sheet. What will happen? Well, the bed sheet will ripple in different ways as the thing moves around. And this creates, essentially, a wave of distortions of space-time that propagates outward, as it turns out, at the speed of light.

So, the way this works is, as a mass moves back and forth--perhaps because it's in orbit, or for some other reason--it generates ripples in space-time. These propagate outward at the speed of light. And we'll talk later, maybe, about how one--what it means to have a ripple in space-time, and how one might detect it. But there's a more immediate effect, which is that the energy in these so-called gravitational waves--they carry energy with them. And the energy comes from--is extracted from the orbit of the object that's moving back and forth.

So, the orbit gradually loses energy. And that means that the orbit gets smaller, gradually. So, the orbit gets gradually smaller. And so, in the Newtonian approximation, things stay in orbit and they're perfectly stable forever and ever, and they just go orbiting around, and around, and around. But, in general relativity, there's a gradual, slow loss of energy, and objects will gradually spiral in.

Now, this happens really slowly. It has not been detected in the Solar System. That is to say, even the orbit of Mercury, we can't measure its orbital period accurately enough to see this slow down. Certainly all the planets in the Solar System have been happily going in their orbits for four-and-a-half billion years with no sign of this effect. But you can see this in certain kinds of binary stars. Stars which happen to have very short orbits. And in particular, there's something called the binary pulsar. Pulsar is a kind of neutron star. And these gravitational wave effects, the gradual decrease in the orbit, has been observed in binary pulsars. And we'll talk about that, I guess, next time.

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