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ASTR 160: Frontiers and Controversies in Astrophysics
Lecture 19
 Omega and the End of the Universe
Overview
Class begins with a review of the issues previously addressed about the origin and fate of the universe. The role of gravity in the expansion of the universe is discussed and given as the reason why the rate of expansion cannot remain constant and will eventually slow down. The actual density of the universe is calculated using various methods. Finally, the unsolved problem of dark matter is addressed and two explanatory hypotheses are proposed. One is that the universe is comprised of WIMPs (Weakly Interactive Massive Particles) that fulfill two requirements: they have mass and do not interact with light. The second hypothesis is that dark matter is made of MACHOs (Massive Astrophysical Compact Halo Objects), which scientists have attempted to identify through gravitational lenses.
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htmlFrontiers and Controversies in AstrophysicsASTR 160  Lecture 19  Omega and the End of the UniverseChapter 1. Review of Issues in Cosmology [00:00:00]Professor Charles Bailyn: Okay, we’re talking about the origin and fate of the Universe. And let me remind you of the story so far. There are basically two sets of observations that are important here. One is the existence of the Hubble Diagram and Hubble’s Law, which is the observational relationship between distance and velocity for galaxies. And this leads you to the idea of a universal expansion. And the other is what we discussed last time: that if you look back into the past, if you observe at a large distance–that is to say, a large lookback time–what you discover is that things were different in the past. That the Universe, as a whole, looked somewhat different and, in particular, was significantly denser, which is exactly what you would predict if the Universe was expanding. And these two things–these two observational facts put together are really what lead to the idea of a Universe with a Big Bang cosmology. And this is great because you can then use this assumption that everything is governed by the scale factor of the Universe. And the scale factor starts either at zero, or very close to zero, and gets bigger with time. And you can use that concept to do all sorts of wonderful things. You can describe the past. And in particular, one of the things we did last time was to calculate the age of the Universe from the observations of the Hubble Constant. And you can predict the future. And the future depends on how the expansion of the scale factor changes. If the scale factor just continues to expand at its current rate, the Universe will continue to expand and gradually get sparser and sparser, and colder and colder, and more and more boring. But it’s not expected that the expansion rate stays the same. It’s expected that the expansion rate will change. And, in particular, it’s expected that the expansion rate will slow down. Why? Because there’s matter in the Universe, and matter exerts gravity, and gravity tends to pull things back together again. And so, this is where we ended up last time. If you assume that gravity is the dominant force–that is to say that any changes in the expansion rate of the Universe will be due to gravity, then, you can derive this critical density, which we did last time, which is a quantity equal to 3H^{2} / 8 π G. H, you measure. The other things are just constants, and you can calculate what this quantity is. Now, at this point, let me write down a piece of astronomical jargon, which I didn’t do last time. The actual density of the Universe, divided by this critical density, is given a letter of its own. This is written down as a capital Omega. So Ω is the true–the actual density of the Universe, whatever that turns out to be, divided by the critical density. And then, you can describe the future of the Universe, depending on what Ω is. If Ω is greater than 1, that means that the density’s greater than the critical density. And this leads to recollapse and the “Big Crunch”–whereas, if Ω is less than 1, the Universe expands forever. Somebody asked, what happens if Ω is exactly equal to 1? In that case, there is no Big Crunch. The Universe expands forever, but the expansion rate asymptotically approaches zero. But, of course, in real life, it’s very hard to get something that’s exactly some–any physical quantity to be precisely equal to any theoretical value. And so, with this in mind, it then becomes very important to actually go out and measure the average density of the Universe because, then, you could divide it by this critical density. We’ve already measured H, so we know what this quantity is. And then, you could figure out what’s going to happen. So, the goal here is to determine the density of the Universe. And conceptually, this isn’t such a hard thing to do. You go out and measure the mass of everything you can see. You try and do it over a large volume, because what you want to avoid–the mistake you want to avoid is to measure the density of a piece of the Universe that doesn’t represent the overall average. If we measured the density of material in this room, it would be something like 27 orders of magnitude bigger than the critical density. And if we assume that the Universe were just like this room, obviously, it would recollapse. In fact, it would have recollapsed long ago. But, we don’t do that because, of course, most of the Universe is not like this room. Most of the Universe is empty. So, you say, well, we better include a lot of stars and the empty spaces between them. But even that’s a mistake, because you’re measuring stars in our galaxy. So, you say, well, we better include lots of galaxies and the empty spaces between them. That still doesn’t work for a while, because there are clusters of galaxies. There are clusters of clusters of galaxies. And so, you have to go really, quite far out, before you have a fair sample of what the average conditions in the Universe are like. But, in principle, that’s certainly possible to do. You just keep measuring things further and further and further away, until you get to a point where, if you increase the distance–where, as you increase the distance, that density doesn’t change anymore. So, you’re out to the part where you’ve really achieved the average. How do you know you’ve achieved the average? Well, you look out twice as far and you get the same answer. And so, in principle, the way you do this is, you add up all the mass in some sizeable chunk of the Universe–in a sufficiently large chunk of the Universe, where sufficiently large is sufficiently large to average over any local perturbations. So, you add up all the mass and you divide by the volume. You divide by the volume that that mass occupies. And so, obviously, you have to identify all the different kinds of mass. And you have to make sure that whatever volume you’ve taken, you’ve found all the mass in it. You add it all up. You divide by volume. You determine–that gives you a value for density. You divide by the critical density and you know what’s going to happen to the Universe. Chapter 2. Determining Mass [00:08:28]Okay. Now, how do you find the mass of things? Determining mass. Well, one way you can do it is you can just go out and measure how bright ‒ yes, go ahead. Student: Can you put the other slide up? Professor Charles Bailyn: Oh, put this back for a second. Top part? Bottom part? What do you Student: [Inaudible] if you don’t mind putting it on. Professor Charles Bailyn: Yeah, yeah. So, you’ve determined the density of the Universe by adding up the mass. Divide it by volume. And then, the question becomes, “How do you determine the mass?” And one way you can do it is, you look at how bright things are. Add up the light you see. And then, you assume some value for the amount of mass it takes to create a certain amount of light. So, that’s assuming what’s called a masstolight ratio. And so, you can do that, you know. If it’s the Sun, then one solar mass produces one solar luminosity. If all stars–if all objects are exactly like the Sun, then everything would be like that. It turns out that isn’t the case, but you can take local samples of stars and figure out what the average masstolight ratio is. And if you have some value that you’re happy with, of masstolight ratio, then you multiply the amount of light by the masstolight ratio, and this gives you a mass. Student: Do you need to adjust for distance? Professor Charles Bailyn: Sorry. Student: Do you need to adjust for distance? Professor Charles Bailyn: Well, what you mean by light is the–do you need to adjust for distance? What you mean by light is the intrinsic light. You mean the equivalent of the absolute magnitude, which takes the distance into account. So, what you need to ask is not how bright it looks, but its intrinsic brightness in this particular case. Yes. So, you do need to account for the distance, and so, you need to be thinking about absolute magnitude rather than apparent magnitude, yes. And that’s one of the problems. That’s hard to do. The other problem, of course, is this awkward word, here [“assume”], which is the kind of thing that makes people nervous, because you could get that wrong. If you’re looking at one kind of star and it’s actually some other kind of star, which happens to be much more massive but dimmer, like white dwarfs or something like that, then you’re going to make a mess of this. So, there’s an alternative method, which you may already have considered, because we’ve done it in both of the previous parts of this class, which is, you measure orbits. And you do the same thing we did with–in part one and part two of the class. You find some star in the distant portion of the galaxy, orbiting around the galaxy. You figure out how fast the thing is going. You figure out how far the thing is going. You use Kepler’s Laws. And you determine the mass from orbital theory, from Kepler’s Laws, basically. And, in particular, you know, V^{2} = GM/a. And so, you can measure this from the Doppler shift. You can determine this; basically, in the case of galaxies, galaxies are big objects. You can physically measure the angular separation on the sky. Use the small angle formula, if you know the distance, to determine this. So, this can also be measured, and therefore, this can be calculated. And so, you go and do that for a whole bunch of galaxies. And this has been done. And let me give you some examples, here. Let me actually write down some numbers and do some calculations. Supposing you have a galaxy at a distance of 20 megaparsecs [Mpc]. And supposing it has an apparent magnitude of, something like, 14. These are kind of typical numbers for nearby galaxy clusters. There’s a particular–the nearest big galaxy cluster to us is a cluster in the constellation of Virgo, known as the Virgo cluster. If you want to know about the Virgo cluster, ask Hugh Crowll [ a graduate teaching assistant for the course] who is devoting his life to the study of this object and the galaxies within it. But these are sort of quasitypical numbers, adjusted slightly because it’s actually 17 Mpc, which is kind of a pain. All right. So, what do you know about the mass? What can you determine about the mass of such a galaxy? Well–oh, and let me warn you before we even begin that, of course, astronomers have played you a dirty trick–namely, that the symbol we use for magnitude is M. The symbol we use for mass is also M. So, you’ve got to keep those clear in your mind. All right. So, what do we know about this? We know the relationship between apparent and absolute magnitude. And, as I said just a minute ago, it’s the absolute magnitude that we need to know in order to actually determine anything. m  M = 5 log (D / 10 parsecs). So, let’s figure out the righthand side first. That’s 5 log (2 x 10^{7}). That’s 20 Mpc. 1 Mpc is 10^{6}. Over 10. That’s 5 log (2 x 10^{6}). Now, what do I do about that? Let’s see. That’s 5 times log of 10^{6}, that’s pretty straightforward, plus the log of 2. Because, if you add logs, then you multiply the thing inside the parentheses. So, log (2) + log (10^{6}) = log (2 x 10^{6}). log (10^{6}) = 6. log (2) = .3. It’s just a useful number to know. The log of 2 is around .3. The log of 3 is around .5. The log of 5 is around .7. You could look it up. And so, this is equal to 5 x 6.3. 5 x 6 = 30. 5 x .3 = 1.5. So, this is 31.5. Let me caution you at this point. So, let me give you a little side note, here. Do not approximate magnitudes. Why not? I mean, we approximate everything else in this course. Magnitudes are a logarithmic quantity, right? And so, you don’t approximate magnitudes for the same reason that you don’t approximate the exponents. You can’t say, 10^{7} is equal to 10^{6}. You can say 7 equals 6, but you can’t say 10^{7} is equal to 10^{6}, because that’s a factor of 10 difference, whereas the difference between 7 and 6 is just a little more than 10%. Similarly, this .3. You would have been tempted to get rid of it, right? Because who cares about the difference between 6 and 6.3? But, in fact, it comes out of this log of 2. And so, .3 in the log is actually a factor of 2. And so, you got to not approximate the exponents. This is important. Yes? Student: Does this mean we should also try to be more precise when we’re dealing with magnitudes? Professor Charles Bailyn: Well, yes. That’s saying–I guess that’s saying the same thing. You should be more precise. That means you shouldn’t approximate. Yeah, so, I guess. But, it’s for the same reason that you don’t approximate the exponents. And it’s also true that the numbers are easier to work with, because it turns out that you add them rather than multiplying them most of the time, so, it’s not such a bad thing. Anyway, here we are at 31.5, so what have we got? We’vem  M = 31.5. This M was stated in the problem to be 14. So, 14  31.5 = M. So, M = 17.5. Okay. That’s not such a bad number. We can work with that. So, now we know the absolute magnitude. We know how bright the thing is. So, now we can figure out how many times brighter than the Sun it is. Why is that a useful thing? Because if you then make the assumption that the masstolight ratio is the same as the Sun, that this galaxy consists entirely of Sunlike stars, then you can determine how massive it is. So, let’s do that. How many Suns–and this is the other magnitude equation. This is, you know, M_{1} ‒ M_{2} is equal to–for two different objects, is equal to ^{5}⁄_{2} log of the brightness of 1 over the brightness of the other. But I think I want it in the other form. I think I want it in the form of 10^{0.4}, or 10^{2/5 (M1M2)} = b_{1} / b_{2}. This is the exact same equation, as you’ll recall, just having been–getting rid of the log, taking everything, putting it into 10 to the something power. The reason I want it in this form is that this is the answer I want. I want b_{1} / b_{2}. I want one to be the galaxy. I want two to be the Sun. So, then, I’ve got 10^{2/5}, and then, the galaxy is 17.5, that’s the absolute magnitude. The Sun is 5, has an absolute magnitude of 5. And that’s going to give me the brightness of the galaxy over the brightness of the Sun. That’s 10^{2/5 (22.5)} . Let’s see. The minuses cancel out, so that’s a plus, actually. ^{2}⁄_{5} x 22.5 ‒ well, let’s see. 2 x 22.5 = 45. A fifth of 45 is 9. So, this is equal to 10^{9}. So, this galaxy is a billion times brighter than the Sun, 10^{9} times brighter than the Sun. So, if it were made out of Sunlike stars, it would have a mass of a billion solar masses. So, mass would equal 10^{9} times the mass of the Sun, if all Sunlike stars. But, it turns out that galaxies tend to be somewhat dimmer than the Sun, per unit mass. Most stars are a little bit less massive than the Sun, but a lot less bright. This is just the way stars turn out to be. And so, typical masstolight ratios of populations of stars tend to be on the order of 10, or something like that, times the Sun. So, probably it needs to be more massive, because typical stars are fainter than the Sun. Typically, stars are fainter. So, you could guess and say, mass, maybe, should be, I don’t know, 10 times greater than that, 10^{10} solar masses. And you can see why this particular line of reasoning starts to get pretty dubious, because I picked this number completely out of the air. There’s actually some modest basis for it, but you could pick other numbers. You could argue about this endlessly and you wouldn’t get very far. Why should it be 10 times the Sun? Maybe it’s 100. Maybe it’s 1,000. Maybe it’s less than the Sun. How would you really know? And so, let’s go back and do the other approach–namely, figure out its mass from orbits of things around it. So, let’s look at–supposing it’s an edgeon galaxy. Here’s the center of the galaxy, or–and, actually, let’s look at it from the top. So, here’s a nice spiral galaxy of some kind. Here’s the center of the spiral galaxy. Here’s some star way out on the edge. That star is moving around the center of the galaxy. It has to be, or it’s going to fall in. So, it’s orbiting around the center of the galaxy, presumably in some circular orbit. You’re down here, looking at this thing. And, of course, you can measure the velocity of that star by the Doppler shift, because it’s moving away from you. And so, one can measure this velocity. You can measure this distance. That would be the equivalent of a in our formulas, because it’s the distance between the orbiting object and the center. Stars are much less massive than galaxies so we don’t have to worry about the motion of the galaxy. And you can use a familiar equation–namely, V^{2} = GM / a. So, now, let’s give this some numbers. Typical velocities of things orbiting around the galaxy turn out to be something like 200 kilometers a second, or 2 x 10^{5} meters per second. And the size of a typical galaxy, you know, out to where it stops being easy to see stars is, oh, I don’t know, what number did I take here? Yeah. Let’s call it 20 kiloparsecs, which is 2 x 10^{4} parsecs. And a parsec is 3 x 10^{16} meters. So, this is 6 x 10^{20} meters. So, now, let’s calculate M. M = V^{2} a / G. [(2 x 10^{5})^{2}(6 x 10^{20})] / (7 x 10^{11}). Get rid of those–let’s see, that’s (4 x 10^{30}) / 10^{11}. 4 x 10^{41}, this is in kilograms. One solar mass, you recall, is 2 x 10^{30}. So, this mass, in units of the Sun, (4 x 10^{41}) / (2 x 10^{30}), which is something like 2 x 10^{11} solar masses. And now, we have a problem, right? You probably don’t remember what the answer to the previous version of this problem was, where we did it with light. That came out to a magnitude of–the brightness was about 10^{9} times the Sun. Maybe the mass is 10^{10} times the Sun. But now we’ve just calculated it in this other, more reliable way, and it’s 2 x 10^{11}. It’s 20 times more massive than you thought it was going to be, given how bright the light from this thing was. Yes, question? Student: [Inaudible] mass of the galaxy? Professor Charles Bailyn: This is the mass of the galaxy, yes. Now, before I go on let me just point out–those of you who have taken a look at the problem set–what I’ve just done here, this calculation I’ve just done, is problem one of the problem set, except done backwards. On the problem set, what I did is, I told you what the density was, what the critical density was, and then, you had to derive characteristics of the galaxies from that. Here I’ve told you what the galaxies are like. We figured out how big–how massive they are. If we divide by the volume, we’ll get a density. So, we’re doing the same problem backwards. I should say, the numbers I’ve chosen here are different, so, you can’t know the answer to the problem set by looking at the premises of these particular things. But, what I’m doing is the exact same set of calculations, only done backwards. So, that may or may not be helpful. Chapter 3. Dark Matter: WIMPs? [00:26:39]But let’s pause here for a moment, because this is now–we’re now up to–we’re making progress. We’re now up to Frontiers and Controversies circa 1985. You’ll remember, in 1920, they were worried about whether the spiral nebulae were actually galaxies. In 1950 they were worried about, maybe the “steady state” was the correct response. And by the time 1985 rolls around, the big issue is mass is determined by orbital rotation. So, what you might call dynamical masses–that is to say, determined by orbits of things around galaxies. Orbits around galaxies. And also, I should say, galaxy clusters. You can have galaxies orbiting around each other and galaxies orbiting around whole clusters of galaxies, and the same thing is true. And so, around galaxies and galaxy clusters–are much bigger than you expect from the light they give off. And therefore–by about a factor of 10. By approximately a factor of 10. So, there’s 10 times more mass than you can account for by adding up all the stars. Now, there’s mass in other forms than stars. There’s also dust. There’s also gas. These are things you can detect in other ways. You add them all up and you’re still off by about a factor of 10. So, there’s 10 times more mass than you have any way of accounting for. This is the socalled dark matter problem. So, this is Frontiers and Controversies in 1985. There’s all this dark matter. Most of the matter in galaxies is in some form that we can’t detect. It’s dark matter, and what is it? Now, unlike Frontiers and Controversies in 1920 and 1950, this is one that we haven’t solved yet, so I don’t know the answer. For a quarter of a century, people have been busy trying to figure this out. There’s still no good answer. And ten years ago, when I taught this course, this question of what is the dark matter was a big focus of this part of the course. Now, I’m going to talk about it only in this class, only in one lecture, because we got way bigger problems, even, than this. That’s saying a lot. I’ve just told you that we don’t know what 90% of the mass in the Universe is, and then, we’ve got bigger problems than that. So, things are getting a little murky, here, and not just because the matter is dark. Okay. But, let me pause a little bit on dark matter, because it’s an interesting problem. And, as I say, we have no idea what this stuff is. What are the possibilities? So, here’s a hypothesis. Hypothesis #1 is that what this stuff is, is some kind of unknown subatomic particle. And it has to have two characteristics, this subatomic particle, for it to work out. It has to have mass. That’s pretty basic. If you’re using it to explain mass, you can’t have photons, right? Photons don’t carry any mass. It has to have mass, but it has to not interact with light. No interaction with light. If it absorbs light, it would be opaque, and we would know it was there, because galaxies behind this stuff would look dim. Alternatively, if it gives off light, then we’d see it. And so, it has to not interact with light, or interact with light only very weakly. And so, these are given the name, generically, Weakly Interactive Massive Particles, or WIMPs. So, here’s the hypothesis: the Universe is 90% WIMPs. This is not such a crazy idea as it might, at first, seem. There are known subatomic particles that have these properties. There’s something called the neutrino. There are trillions of them going through this room every second. They have mass and they don’t interact very much with anything. They’re known to exist from particle accelerator experiments, and they have been detected from celestial sources. Now, we know that–for various reasons, that the dark matter doesn’t consist of neutrinos. But, there could be many other kinds of particles with these kinds of characteristics, and indeed, some are predicted by current particle theories. As I say, WIMPs have been detected–I’m sorry, WIMPs have not been detected, but neutrinos have been detected. Here’s how they do it. It’s kind of an amazing experiment. They took a mineshaft in South Dakota and filled it with cleaning fluid. And the reason they did that was that every so often–neutrinos don’t interact with light, but they do interact, occasionally, with chlorine atoms. And the effect of a neutrino banging into a chlorine atom is to turn it into argon. And so, this happens–there are–as I say, trillions of neutrinos flow this mine every second. Once a day or so, one of them will hit a chlorine atom just right, create an argon atom. So, here’s what you do. You fill your mineshaft with cleaning fluid, a large fraction of which is chlorine, and you count the argon atoms that bubble off the top. And this has been successful. They detected neutrinos emitted from the Sun. The Sun is–all stars that have nuclear reactions going on in them, emit neutrinos as part of the output of these nuclear reactions. And then, they had a problem, because they had predicted how many neutrinos you ought to see from the Sun in an experiment of this kind, and they didn’t see enough of them. They only saw a third of them. And it turns out–and then, there was a big debate for a long time. This is Frontiers and Controversies circa about 1975. There was a big debate for a while. Where are all the solar neutrinos? Is it possible that we don’t understand nuclear reactions in the Sun? Is it possible that we don’t understand the chemistry of chlorine or argon? After all, you’re counting individual argon atoms, so that’s kind of a difficult task. No, it turned out that what was going on was, we didn’t understand neutrinos. And it turns out there are three kinds of neutrinos. And neutrinos switch back and forth between these different kinds, and you could only detect one kind by the chlorine. And so, they were all emitted from the Sun as if they were in the form that you would have been able to detect them. But as they traveled from the Sun to us, some fraction of them flipped back and forth between all these other kinds, and you ended up only with about a third of them. So, it was a big piece of particle physics that was discovered. We have also detected, by now, neutrinos coming from supernova explosions. So, there are–11 of them, I think, were detected, all at once. And if you’re detecting things, sort of, once per day, and then you suddenly detect 11 of them over the course of a few minutes, you’ve seen something exciting occur. And that is now known to be this supernova explosion that occurred in a neighboring galaxy. And so, there are a bunch of–so, by analogy with that, people are looking for the WIMPs that make up the dark matter. If all this dark matter is in WIMPs, there are lots, and lots, and lots of these things, and they’re going through us every second. So, there are a whole bunch of experiments with the same basic characteristics. You have a huge vat of something, and something is supposed to happen, occasionally, when one of these WIMPs hits whatever’s in the vat. So, the Japanese have, sort of, a cubic mile of distilled water, and they’re looking for little light flashes when the neutrino runs into the water molecule. They busted all their detectors recently, and they had a sort of earthquake, and it was bad for the little light detectors they had put on the inside of these things. But there are a lot of such experiments. Dan McKinsey, here in the Physics Department, is a big player in one of them. And the hope is that you will see the interaction between one of these WIMPs, of which there must be an incredibly large number, with something. This has, so far, failed. So, there is no direct evidence from WIMPs. The other hope, I should say, is that every time you build a bigger collider, you make new kinds of subatomic particles, and that they’ll eventually make something that looks like it could be a WIMP. And that hasn’t been–happened either. So, no detections yet. No direct detections. With considerable effort, you know, this is going to turn out to be 90% of the mass in the Universe. So, you would like to detect it because if you do, they’ll give you a Nobel Prize. All right, that’s one hypothesis. There’s another hypothesis. So, here’s Hypothesis #2. It’s just, you know, dark chunks of something that doesn’t glow. Ordinary matter–chunks. Student: Do these hypotheses exist today or [inaudible] Professor Charles Bailyn: Yes, yes, yes, all of the–we don’t know what it is, and so, nothing has yet been ruled out. What happens is that they–you know, they continue to conduct these experiments, so, you can rule out WIMPs with certain kinds of characteristics, because you would have detected them. Similarly, you can rule out some of these other things with certain characteristics, because you would have noticed they were there. But both of these hypotheses are still more or less viable. Chapter 4. Dark Matter: MACHOs? [00:37:30]Chunks of ordinary matter that just don’t glow, that don’t emit light. Now, there’s some limitations. These chunks can’t be too small, because if what you’ve got are tiny, you know, micronsized particles, we call that dust. And, basically, that’s what it is. It would just be dust. The problem with dust is, dust in large quantities is opaque, and you can’t see through it. And therefore, you would know it was there, because it obscures the light of things behind it. And, indeed, we see cosmic dust this way all the time. It’s just, there isn’t nearly enough of it to account for any substantial fraction of the dark matter. So, dust would be observed because it–by obscuring light. And it also tends to glow in the infrared. And so, we know that dusts exists but we can count how much of it there is, because it obscures light and it makes its presence known in other ways. It’s also true that these chunks of ordinary matter can’t be too big. They can’t be the size of whole galaxies, or even a substantial fraction of a galaxy. You can’t take all your dark matter and put it into one lump per galaxy, or even 100 lumps per galaxy because if they were very large masses, you’d see it, because it would disrupt the orbits of stars around the galaxy. So, if there was some huge unknown mass, you’d see things orbiting around it. And, in fact, we do. We see these supermassive black holes in the centers of galaxies and we know they’re there, because we see stars orbiting around them, just like the problem on the last Midterm. And so, it can’t be too small. It can’t be too big. But, you could, perhaps, have, sort of, a bunch of star massed; so, you could have sort of a bunch of star massed, or planet massed dark things in–it would have to be, for various technical reasons that I won’t go into, it has to be in the outer parts of galaxies, in the halos of galaxies. So, that, in principle is possible. We wouldn’t have any direct way of detecting them. These things are called Massive Astrophysical Compact Halo Objects. [Laughter] Some people get it. Massive, because they have to carry mass. Astrophysical, because they’re not particles. Compact, because if they were big you’d–you know, they’d block light and you’d see them. Halo, because that’s the part of the galaxy they’re in. These are MACHOs, right? And so, the alternative to WIMPs is MACHOs. And so, the alternative explanation is that 90% of the Universe is MACHOs. There’s been a very clever experiment carried out to try and find these things. Here’s how you do it. You do it with gravitational lensing. Lensing MACHO searches; remember gravitational lensing? This is this business that mass bends light. So, here you are. You’re looking at some star. And, in between you and the star is a MACHO of some kind. So, here’s the MACHO. You can’t see the MACHO, but the presence of the MACHO changes the direction of the light. So, it comes into you like this, and it basically acts like a lens. And, in particular, the way it acts like a lens, in the case of MACHOs lensing stars, is it makes it brighter–makes the star brighter. Now, in order for this to work, the alignment has to be essentially perfect. All of these objects are moving around. They’re orbiting the galaxy and stuff. So, the alignment holds for a few weeks, typically. So, what you’ll see is, you’ll see this star become much brighter. And it can really become much brighter–we’re talking tens to hundreds of times brighter than it ordinarily was. This lasts for a few weeks, and then, it goes away. These have been observed. These lensing events have been observed. Lensing events observed. But there are too few of them to explain the dark matter. Now, there are still ways out. Let’s see. If you have particularly low mass MACHOs, so–supposing the whole Universe is filled with things about the mass of Earth, those cause lensing events that might be too small to see. Alternatively, supposing you have things that are many thousands of times the mass of a star, but not big enough to totally disrupt galactic orbits, then, there are many fewer of them for a given amount of mass, and there aren’t enough MACHO events that you would have expected to see any substantial number of them. So, there’s still a way around the result of these experiments, if you want to believe in MACHOs. But it’s getting very tough. So, no WIMPs detected so far. No MACHOs. You could still postulate kinds of WIMPs and kinds of MACHOs that might explain the dark matter, but it’s getting kind of tough. Most people, I think, believe in WIMPs. Most people tend to believe in this. But, and as far as I can tell, that’s because the particle physicists keep coming up with new candidate WIMPs that might exist, but that we haven’t quite been able to see, so far. And so, there’s a theoretical basis for the existence of these things, whereas, with these MACHOs, if you ask the astronomers ‒ well, fine. So, you want to have 90% of the Universe be in little Earthlike things just floating around with no star, how did they–how did that happen? How did these come into being? We really have no answer at all for that. So, there’s no theoretical basis for any of the stillallowed categories of MACHOs. And so, at the moment, people tend to believe WIMPs over MACHOs, although there’s no direct evidence for either. Yes? Student: If 90% of the matter of the Universe is made of little Earthlike objects, then wouldn’t that be 90% of the Universe is made of metal? Professor Charles Bailyn: Oh, Earth. Earthmass objects is what I meant. I don’t care what it’s made out of. Yeah, maybe there are little Earthsized balls of hydrogen. That would be fine too. Except how do you get them? We know something about how balls of hydrogen form and what they become. They turn into stars. This is well known. And one of the popular kinds of MACHOs was just very, very dim stars. And this is one of the things that the space telescope helped to rule out, because it can see really faint objects, and they weren’t there. And so, no WIMPs. No MACHOs. And so, we don’t know what’s going on. That was a digression. And what I digressed from was the fact that this galaxy that we had measured the mass of turned out to be 2 x 10^{11} solar masses, or around 4 x 10^{41} kilograms. If you have these things, one such galaxy every–I don’t know, 2 Mpc, or so, what’s the density of the Universe? Remember, that’s where we started–of the Universe. So, now, let’s finish this calculation. Let’s see the density is equal to M/V. 4 x 10^{41}, from observing these orbits. And the volume, down here, is going to 2 Mpc cubed. That’s 2 x 10^{6}, times 2–sorry, times 3 x 10^{16}. That’s 1 parsec. So, this is 6 x 10^{22}. I want to cube it. 6^{3}. 6 x 6 = 36, times another 6, is 200. So, that’s 200 x 10^{66}, or 2 x 10^{68}. So, then, the density of the Universe. (4 x 10^{41}) / (2 x 10^{68}), that’s equal to 2 x 10^{27} kilograms per meter cubed. And real critical can be calculated–turns out to be, as you’ll discover on the problem set, 6 x 10^{27} in these units. ρ over ρ_{critical} is equal to about ⅓. So, if you buy that, the Universe is going to keep expanding, because Ω, the ratio of the density to the critical density is only about ⅓. But the problem is, we’ve got all this dark matter around and what we’re doing is, we’re adding up galaxies. How do you know that there isn’t a whole bunch of dark matter where there aren’t galaxies? And where there’s nothing to see orbiting around, you have no idea what this stuff is. And, indeed, most of the WIMP kinds of ideas, sort of, postulate some kind of dark matter that, kind of, pervades the Universe. And so, you’d expect there to be somewhat more of it than you can see in any given galaxy. Well, somewhat more than 1/3 gets you into dangerous territory; namely near one, which is the thing we’re trying to distinguish–whether this number is greater than 1 or not. And so, you need a new approach. This isn’t going to get you the answer. And so, there is a different approach. And that’s what we’ll talk about next time. And that will finally bring us up to Frontiers and Controversies in the twentyfirst century. [end of transcript] Back to Top 
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