ASTR 160: Frontiers and Controversies in Astrophysics

Lecture 7

 - Direct Imaging of Exoplanets


Class begins with a problem on transits and learning what information astronomers obtain through observing them. For example, radii of stars can be estimated. Furthermore, applying the Doppler shift method, one can find the mass of a star. Finally, a star’s density can be calculated. A second method for identifying planets around stars is introduced: the astrometry method. The method allows for an extremely accurate assessment of a star’s precise position in the sky. Special features of the astrometry method are discussed and a number of problems are solved. A short summary is given on the three methods astronomers use to identify exoplanets. Class ends with an overview of upcoming space missions and the hope of detecting the presence of biological activity on other planet.

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Frontiers and Controversies in Astrophysics

ASTR 160 - Lecture 7 - Direct Imaging of Exoplanets

Chapter 1. Calculating Planetary Density from Transits and Doppler Shifts [00:00:00]

Professor Charles Bailyn: So, we’ve been talking about transits and I want to say a couple things more just to round out that discussion. You’ll recall how this works. Transits–you get light blocked–light from the star blocked by the planet. And in order for this to occur, the orbit has to be exactly edge-on; otherwise, the planet doesn’t actually get in the way between you and the star. So, must be edge-on. And so, what you see, basically, although you’re too far away to see it in this kind of detail, looks something like this. Here’s a star, it’s got this big, bright, shining disc, and then crossing that star is a darker disc from the planet. So, it looks like you have a little spot there. And it’s clear, just from the geometry of this situation, how much light is obscured. It’s basically the ratio of the cross-section of the planet to that of the star. So, the depth of the transit, and by that I mean, you know, the fraction of the light that is obscured, is equal to–that will be, the cross-section of the planet divided by the cross-section of the star. And the cross-section is just the projected area. And so, what that means in practice is, here’s the radius of the star, and there’s an equivalent radius of the planet. And it’s just the ratio of the projected areas.

Now, the area of a circle, projected area of a sphere is a circle. The area of a circle, you probably remember from geometry, is πr2. And som what this is going to be is πr of the planet squared, divided by πr of the star squared and theπs cancel. So, it’s the ratio of the squares of the radii.

Okay, so to take an example–supposing you were an astronomer in some distant place and you’re looking at the–at our own Solar System. And you’re fortunate enough to be in a place where you can see our Solar System edge-on. What would you see when the Earth transits the Sun?

So the Earth’s–so here’s an example: Earth transiting the Sun. The radius of the Earth, it turns out, is something like 7 x 106 meters. The radius of the Sun is about 100 times that, 7 x 108 meters. And so, the depth of such a transit is the square of the ratios of these radii.

(7 x 106) / (7 x 108)2

That’s (10 -2)2 = 10-4 , or 0.01%.

So, that’s a lot less than what you found in section yesterday for a Jupiter-like planet. And the key is in this r2 factor, until the Earth is about 1/10–has about 1/10 the radius, is about 1/10 the size of Jupiter. But that tenth gets squared, so that’s 100. And so, instead of getting a 1% dip in the light, or approximately 1%, you get a 1/100 of 1%.

1/100 x 1% = 10-4.

So, as the planet gets smaller you–it becomes harder and harder to see these transits. And, if you think back on the pictures of these light curves, the graphs that were made, in the observations from the ground, you really needed to have about a 1% dip in order to be able to see it. From space you can do a lot better. Things are much more stable, and you might well be able to see something like this, but only from space and not necessarily from the ground.

Now, having done this little calculation, of course, in real life, you do it backwards. What I did here was I assumed that I knew what the radius is, and I calculated what the transit would be. That isn’t how it works with extra solar planets because, of course, you don’t know what the radius is. And instead, what happens is, you see the transit. And you see the transit, and you work backwards to figure out what the radius is. You have to make some assumption about what the radius of the star is likely to be. And so, basically, observing a transit is a way to get the radius of the star.

So transits give you the radius–and this is nice, because the radial velocity method, the Doppler shift method–Doppler measurements ‒ also give you something about the star. But it’s the mass not the radius. And then, if you have both of these things, then you get to put them together and figure out what the density is. And that’s why observing both of these things in the same system is so useful.

So let’s do this again for the Earth. The mass of the Earth is 6 x 1024 kilograms. The radius of the Earth we just figured out, 7 x 106 meters. And, our hypothetical astronomers out there in the rest of the galaxy somewhere would have been able to determine this if they saw the transit. So now, what is the density? Density is this Greek letter ρ, symbolically, and that is equal to mass over volume, which is equal to mass over 43 πr3.

π = 3. So that cancels, which is helpful. And let’s see, so the mass on top, 6 x 1024 divided by 4, times this quantity cubed. Now, what do we do about that? 7 x 7 = 50, right? So, this is 7 x 50, which is 350. So let’s see, this is equal to (6 x 1024) / (4 x 350 x (106)3), which is 1018, 6 x 3 = 18. And so what’s this? Oh! I don’t know; this is 1,400. And so this, then, in turn, is equal to (6 x 1024) / (1.4 x 1021) because this is 1.4 x 103 x 1018.

6 / 1.4 = 4, give or take, times 103. And that’s the answer, that’s the density.

In what units? Let’s see, this had better be kilograms per meter cubed. It’s an MKS calculation. Why? Because that’s what I started with, kilograms and meters. 4 x 103 kilograms per meter cubed, what does that tell you? What does that tell you? Four times more dense than water, excellent. So, this is 4 times water. Water is defined to be 103 kilograms per meter cubed–density of water–or rather, a kilogram is defined so that water has that particular density. It’s 4 times water. That makes it a rocky planet, because the ices and the gases that make up a typical gas giant planet don’t get–don’t achieve that kind of density.

And so, our hypothetical astronomers observing this transit would have determined not only that the Earth is there, but also that it’s got to be made out of rocks. And that is an additional piece of information that’s really useful in trying to figure out what’s going on in these kinds of systems. Okay.

Chapter 2. Astrometry to Assess Planetary Position [00:09:14]

Now, it turns out that there are some other methods for identifying planets around other stars. And I want to talk about one of them. And the reason I haven’t talked about this one more is because it hasn’t yet discovered any planets around other stars. There have been lots of planets discovered by the radial velocity method. There have been lots of planets discovered now by transits, now with the latest Hubble Space Telescope results, but those are not the first methods that were tried. For a long time, people have been trying something else, and it hasn’t really worked. And this is called astrometry, so here’s another method.

Astrometry means “measuring stars” in Greek. And basically what this is, is it’s determining the positions of stars to extremely high accuracy. Remember, the whole game here is that the star moves, too. And we’ve been using Doppler shift measurements to measure the velocity of the star, the radial velocity of the star, as it goes round and round. But you could, in principle, also try and figure out that the actual position of the star in space moves a little bit too.

So, observing change in position of the star–now that change is quite small. And you’ll recall that we started out a few weeks ago saying, you know, you don’t see the star all that accurately because of atmospheric distortion and optical problems in your telescope. You just see this big blob of light. And the key point here is that it is actually not–if you have enough light, you can determine the center of a big blob of light really quite accurately. Determine the center of a blob of light to much greater accuracy than the extent of the blob. So, you know, I take a picture of a star and I see something that kind of looks like this. And that’s big and ugly and hard to deal with, except, you know, the accuracy of the position of that blob isn’t limited by the extent. You know that the center is somewhere around here. And if that blob were to move slightly so that the center took a little–went around in circles like this, you’d actually be able to see that as this big blob sort of moves slowly around. And that’s the trick with astrometry. So, you do these repeated measurements of the precise position of the star, and you ought to be able to see it going around in circles.

So, we have, in fact, all the equations we need to deal with this already. And so, let’s just go straight to a particular exercise, a calculation. How big a change in the Sun’s position due to the planet Jupiter, because that’s the most massive and will make the most difference, would you see if you’re on Alpha Centauri?–would an Alpha Centauri astronomer observe? So here’s the situation: you’re sitting on Alpha Centauri. You’re looking back–which is the nearest star, you may recall. You’re looking back at the Solar System, and you’re hoping to see the change in movement of the Sun due to the presence of the planet Jupiter.

So, the first thing to say about this is, what do we mean by change in position? Change in position. What we’re talking about is an angle. We talked about this in a different context before. Here’s the observer. Here’s the star. And the star’s moving, it’s going to move up to here. And what we want is to draw this triangle, and we want an angle. And then D2 over here is the amount by which the star moves, D1 over here is the distance to the star. And we want the angular change in position, because that’s what we mean when we make a position measurement in space. Okay, so that’s the first thing is to understand what the answer ought to look like. It’s got to be some kind of angle.

And then the second thing is, I’ve done my usual trick here of not giving you a whole bunch of information that you’re going to need. And so, the first thing you have to do in order to line up–try and set up the answer is you have to figure out some actual answer. So, the mass of the Sun is obviously equal to the mass of the Sun; that’s easy. The mass of Jupiter is equal to 2 x 1027 kilograms, that’s 1/1000 of the mass of the Sun. The distance from Alpha Cen to us–Sun–from Alpha Cen to the Sun is about 1 parsec. I think I mentioned that before, because this is the closest star that there is. And the semi-major axis of Jupiter’s orbit is equal to 5 Astronomical Units. I wrote that down on the first day of class. So you’re going to need all this information.

Now, here’s the trick; here’s Jupiter doing its 5 Astronomical Unit thing, and then the Sun in the middle. Here’s the center of mass, center of mass. And the Sun is doing a little circuit in the other direction, always making sure that it’s on the other side of the center of mass from the planet. That’s what a center of mass means, right? And then we can take–at one point here’s the–let’s say, here’s the Sun and here’s Jupiter. And we take this little line here, and the whole of the distance here is atotal. And the distance from the planet to the center of mass is aplanet, and this tiny little distance here isastar. And what we need to look for is astar in this case, because we’re watching the star go round in circles, not the planet.

And then, the usual equations apply where Mstarastar is equal to Mplanet aplanet. We’ve seen this equation in dealing with velocities, but it’s the same thing in distances, because, the time that it takes to go around these circles is one orbital period–is the same in both cases. So, if the velocity is slower, then you cover less ground. Or, to put it a different way, if you’re going in a smaller circle, you’re not moving as fast. So, it’s the same equation.

And so, let’s do this. astar = (Mplanet / Mstar) x aplanet. That’s 5 A.U. in the case of–in this particular case. And I’ve already said that the mass of the planet is equal to 10-3 times the mass of the star here. So, this is 10-3. So, this is 5 x 10-3Astronomical Units. That’s how far the–that’s the radius of the Sun’s orbit around the center of mass with Jupiter, if you want to think of it that way. But that’s not the answer we want. We want an angle. And this in A.U., and this, here, in parsecs, will give us angle in arc seconds.

And this is a particularly easy one. We already have 5 x 10-3 A.U., and the distance is 1 parsec. And so, this is 5 x 10-3arc seconds. Five milli-arc seconds. Now, that’s much smaller than the size of the blob of light from Alpha Centauri. But, as I said, you can see the motion of the center of a blob of light to fairly high accuracy, and this is pretty readily observable. So, folks sitting on Alpha Centauri would know that the–who have our kind–our level of technology would already know that Jupiter exists from this method.

Interestingly, we can look at Alpha Centauri, right? You can do this backwards and you get exactly the same answer, because Alpha Centauri is a Sun-like star, has exactly the same mass, and we don’t observe any such motion in Alpha Cen. And so, there is no Jupiter-like planet around Alpha Cen. We know that because if there had been, we would have seen it through this–through this method.

Now, you can do the same thing, let me do it quickly, for the Earth. Supposing you want to detect the Earth in this way. So aplanet–so this is for Earth. a of the planet is now 1 A.U. instead of 5 A.U.s. Mass of the planet is 6 x 1024.

So mass of the planet over mass of the star is (6 x 1024) / (2 x 1030). That’s 3 x 10-6. And so the a of the Sun due to the Earth now, not due to Jupiter, is equal to 1 A.U. That’s the size of the planet times the ratio of the masses, which is 3 x 10-6. So it’s 3 x 10-6 A.U., much less than for Jupiter, which was 5 x 10-3. And then, Alpha equals D2 / D1.

(3 x 10-6) / 1.

Three micro-arc seconds. This is not observable, at least not with current technology. So, our friends on Alpha Centauri, who know that there’s Jupiter, don’t yet know that Earth exists, because this is so much smaller than the effect due to Jupiter.

And it’s smaller for two different reasons. One reason is that the mass of Earth is small, so this ratio is small. The other is that the Earth’s orbit is shorter, so it’s 1 A.U. instead of 5 A.U. So, you’re multiplying two numbers together. Instead of multiplying 5 by 10-3, you’re multiplying 1 by 10-6. And so, both of those numbers are smaller. And so, the effect is that much smaller. So we can’t yet–our friends on Alpha Centauri can’t tell whether we have Earths, and likewise we can’t tell whether Alpha Centauri has an Earth.

So, one of the interesting things about the astrometry method, and this is why I bring the whole thing up, is that it favors big orbits. The bigger the orbit, the bigger the motion of both the star and the planet. This is exactly opposite to what happens with the radial velocity method. Radial velocity method favors small orbits, because in small orbits, things go faster, and you’re looking at the velocity. So, this favors big orbits, and that’s the opposite of radial velocity measurements. And it also favors massive planets: the more massive the planet, the better off you’ll be, which is the same as the radial velocity method.

And so, this is useful, one of the–if you’re trying to figure out what kinds of things are out there, it’s helpful to be looking with several different methods, which select for different kinds of objects that you can look for. And so, you can call these selection effects. And, you know, if you’re looking at radial velocities, you’re going to miss all the large orbit stars. And so, it would be nice to also look with astrometry, so that you can see other kinds of stars–other kinds of planets.

Now, let’s see. The astrometry method has one other feature that it needs–star needs to be nearby. This is a–this is a question that doesn’t arise in radial velocity, and it’s because of this term. It’s because of this term here. You’re looking for alpha. So, even once you figure it out, how much the star moves, you’ve got to divide by the distance of the star. So, for example, if you’re looking at something 10 parsecs away instead of 1 parsec away then you’ve got only 1/10 the effect. And so, while people sitting on Alpha Centauri can see the effect of Jupiter, people ten times further away, on some other star, which is still quite close in galactic terms, would not be able to see that exact same effect even though, of course, the Sun’s moving the same way regardless of who’s looking at it. The further away you get, the smaller the effect becomes. That’s why nobody’s–largely, I think, that’s why nobody has successfully used the astrometry effect, because it only works out well on very nearby stars, and there’s just a limited number of those to look at.

Chapter 3. Summary of Planetary Identification Methods [00:23:55]

And so, let me summarize all these different methods now–oh good. By just making a little table of the so-called selection effects. These are what kinds of things are easy to see and what kinds of things are not so easy to see for a given method. And we’ve got three different methods, so let’s make a little table here. Method; what kind of orbits get picked out; what kind of planets get picked for a given orbit; and other considerations. And here we go.

So, the first one we talked about was radial velocities. This is the Doppler shift, and this picks out short or small orbits. Short if you’re thinking about the period, small if you’re thinking about the semi-major axis, because the smaller the orbit, the faster the planet has to go, and therefore, the faster the star has to respond. It picks out massive planets, because the more massive the planet, the greater the reflex motion of the star. And you’ll recall that it does make a difference how the thing is oriented. If it’s less than edge-on you see only part of the motion, so it helps a little to be close to edge-on.

Now, for the next method, which are transits, where the star gets in the way, this criteria that you have to be edge-on is much, much more important. Because, in the case of the radial velocity, you know, if you’re not edge-on, it sort of gradually becomes a smaller, and smaller, and smaller effect. But gradually, there’s no big moment where you can’t see it at all. But with transits, if you’re not edge-on, you don’t see anything. It’s not that the effect gets smaller, it’s that the effect disappears altogether if you’re not sufficiently edge-on so that the planet gets away–gets in the way of your line of sight to the star. This also favors short or small orbits. But–and that’s because it has to be edge-on. How close to edge-on you have to be depends on how far the star is from the planet.

Here’s what I mean. Here’s the star and here’s the–here’s you. And, supposing there’s a planet here, then that will just get in the way. But if there’s a planet here a little further away on that same–in the same kind of orbit, only more distant, then it doesn’t get in the way. And so more distant–let me phrase that differently. Larger semi-major axis makes it harder to have a transit. You have to–closer to edge-on. If you were out here, you’d have to–see here, anything within this little angle here will cause a transit. But if you’re out here, let’s see, it would have to be anything within this smaller angle. Does that make sense? Yeah, it’s just, the further away you hold your thumb from your eye, the less you have to move your arm to cover a given thing.

All right. And also, in this case, for transits, let’s see, the planet has to be large, physically, in terms of its size. And so, it’s not so much the mass, but the size of the thing that matters. So, if you have a big, but light, planet, a big fluffy thing, that’ll work well for transits, but not so well for radial velocities.

Then, finally, this astrometry method I just described favors large orbits in contrast to the other effects. It is irrelevant whether it’s edge-on, because if it’s face-on, then the star goes around in a circle, like this, and, you can see it. If it’s edge-on, the star goes around in a circle like this, and you see it moving back and forth as if it was in a line, so you can do it either way. So, that’s irrelevant, but has to be nearby–must be nearby. And, like the radial velocity method, you favor massive planets. So, these are three of the methods that are commonly being used now. Yes, question?

Student: So, does the orbit have to be larger between methods so that the center of mass is sufficiently far away?

Professor Charles Bailyn: Yes. So the–yes, that’s exactly right. The orbit has to be large so that the center of mass has to be sufficiently far away from the star, so that it travels a big enough distance to be able to see. That’s exactly right.

There’s one more method that’s commonly in use called “microlensing,” which is really cool, but I’m not going to talk about it, because it’s a relativistic method and so we haven’t talked about relativity yet. We’ll cycle back and talk about this later on in the course. And this has actually worked, and I will restrain myself from trying to explain it in the remaining fifteen minutes of this particular segment of the course. We’ll come back to it. But this sort of set of selection effects is a good thing to know, because then, if you do all of these things and you compare them to each other, and each method has a different selection effect, then you have a fighting chance of figuring out what’s actually out there rather than what you are simply able to see because of the particular method you’re using. S,o it’s a good idea to be trying to do all of these things at once.

Okay, let me end this segment by talking about some space missions–yes, question yes?

Student: So essentially there’s–if there’s a similar tech–people of technology somewhere out there, they can’t see this?

Professor Charles Bailyn: We can–yes, so there–if there’s people elsewhere with our technology, they can’t detect the Earth. That is the same as saying we can’t detect Earths around other stars. That is the case right now. However, technology marches on, and, as I’ll show you, in ten, fifteen years, that may no longer be true. So, if the aliens out there are even fifteen years further advanced than we are in technology, they might know that we’re here. Other questions, yes?

Student: Even supposing that we could detect the Earth around another star, would the presence of a more massive planet sort of mess that up so that we can’t?

Professor Charles Bailyn: Would the presence of a more massive planet disrupt our potential detections of an Earth-like planet? Not necessarily. Remember, I showed you that–the radial velocity curves where they had detected several planets in the same system? What you do is you take the effect of the massive planet, you figure out what the mass and orientation of the massive planet is, and you remove that from your data and then you look for the little wiggles left over. And you could do that effectively with any of these methods. The example I showed was the radial velocity thing, where there were superposed sine curves, but you could also do it–there could be three different transits, and so you’d get a big lump for the big star, and then a little lump for the little star. And if the little star’s closer, you’d get them more often. And, similarly, in astrometry, you could get it moving back and forth due to a massive planet, and then a motion on top of that due to some other planet. So it–for–in all of these cases the effect of one massive planet could be taken out so that you could see the result.

Now, what’s perhaps more important is, the presence of the massive planet could screw up the orbit of the Earth-like planet. This whole migration idea–this Jupiter comes cruising into the Inner Solar System, and then us earthlings are in trouble because our orbit gets disrupted and we fall into the Sun, or something like that. So, the bigger problem is not detecting it, but whether it permits the planet to be there at all. Other questions? Yes?

Student: Will we get our problem sets back?

Professor Charles Bailyn: Will you get your problem set back? Yes, that’s an easy question, end of the class. Other questions? Yeah.

Student: Will you be posting the slides before the test?

Professor Charles Bailyn: We’ll post these slides before the test, yes, yeah. Other things, yes?

Student: Why is it that when you calculate a it’s not actually–this was the method you were just doing, the astrometry method–why is it not 2a because it’s going from one side of the center of mass to the other?

Professor Charles Bailyn: Oh, the–a is defined as the semi-major axis. It’s just a definition thing.

Student: But isn’t the star going [inaudible]

Professor Charles Bailyn: Yes, so the star travels a circle with radius a, so you’re right. It goes from one side to the other, so the distance the star goes is actually 2a. And so that’s just a definition question of–

Student: But is it–can’t you observe the star at one point or another?

Professor Charles Bailyn: Yes, you’re right. So that the total angular change is–you’re right, is actually 2a. And what I was calculating there is the radius of the little circle that it goes around. But you’re absolutely right the–in fact the distance–it’s like for the radial velocity method, the actual change in radial velocity is from the peak to the trough, but the number we quote is the amplitude of the sine curve, so you’re right about that. Other questions?

Chapter 4. Kepler, SIM, and TPF [00:34:12]

Okay, let me show you some things in–work in progress, and in particular space missions that are being planned to explore this further. So, the first of these is the so-called Kepler Mission, which is going to look for transits. I showed you the Hubble Space Telescope, transit curve, which is much, much better than anything you can do from the ground. So, obviously the place to look for transits is in space. From space you could see transits the size of Earths–of planets the size of Earth, whereas from the ground, you wouldn’t be able to see that.

And also–and so what they want to do is send up a mission that’s going to do nothing but make high precision measurements of stars in order to look for transits. So, the problem with the space telescope is everybody wants to use it for all kinds of things. There’s relatively little–limited amounts of time for any one particular project. It’s also not perfectly designed for making these kinds of transits observations, so this is going to be a mission which is set up for the express purpose of making very high precision measurements of the brightness of stars over, and over, and over again.

And the problem with this is just the general problem with transits: that you only see the ones that are edge-on. So, you’ve got to look at lots and lots of stars. Even if lots of them have planets, you’ll only see a few transits. But this is designed to carry out that mission, so they’re going to be lots of transits observed, including transits from Earth-like–from Earth-sized stars. So, this will be the first moment when this thing gets launched that we’ll really have a good handle on being able to detect an Earth-sized planet. So, if the aliens have launched their equivalent of Kepler, they’re looking at us right now, provided our Solar System happens to be exactly edge-on from their perspective. This is due to launch in a year or two. It’s under construction. It’s ready to roll, and this thing will be up in space taking data by the time–within the next couple of years. So, that’s an exciting thing to look forward to.

A little bit further into the future is something called SIM, this is the Space Interferometry Mission, and this does astrometry. And it’s going to improve the accuracy of positions of stars from what we now have, which is in the milli-arc second range, 1/1000 of an arc second, to the micro-arc second range, 1/1,000,000 of an arc second. And again, as we just did in that little calculation, that will, in principle, enable us to look for Earth-like systems.

Now, the limitation on this is that–is this business that for astrometry, you have to look at nearby stars, and the nearer the star, the better. And if you get far out into these–into the galaxy, then the method becomes less sensitive. It’s also true that, you know, this method favors long orbits, and space missions tend not to last for forty years. And so, if you’re looking for Neptune, which has, you know, a many-decade-long orbit, it’s not clear that this is going to be around for long enough to actually make those measurements.

So, this astrometry thing is kind of squeezed between the requirement of the method to have long orbits and the requirement of human beings to get results in a relatively short amount of time. Nevertheless, this will be a factor of a thousand improvement over what we can do from the ground. This mission is authorized. It’s kind of under construction, but it’s one of those things where it’s going to launch in five to seven years. That was also true five to seven years ago. And it will probably–on our pessimistic days we think it will still be true five to seven years from now. It’s always five years in the future, both for technical reasons, and for funding and political reasons. But there’s some feeling that one of these days they’ll actually get around to doing it. But it’s certainly further in the future than the Kepler Mission is.

And then, the great granddaddy of all planet missions is something called the Terrestrial Planet Finder, the TPF. This is a huge NASA commitment, and they want to directly image the planet. So, this is fundamentally different from everything else. This is–we’re going to actually take a picture of the light from the planet. We’re going to give up this silly business of observing the star to infer the presence of the planet. We’re actually going to look at the planet itself. In order to do that, you have to get rid of the light from the star. You have to blank out the light of the star, because otherwise, as we saw, it’ll drown out the planet.

And there are a couple of different ways you can do it. One is basically just masking. You just put a big opaque thing somewhere in the optical path of the telescope, and it just blocks out the light of the star, but it’s carefully sized so that the light of stuff around it can still get in. There are also more sophisticated optical methods–something called interferometry, which I won’t explain right now, which can have the same effect of nulling out the light from the star. And so, we then observe the planetary light directly.

The best way to do this is not in optical light but in infrared. Optical light is where stars give out most of their light. Planets give out most of their light in the infrared. You may be aware, we all glow in the dark, but we glow in the dark in the infrared, that’s why the military uses those interesting infrared goggles, which turn everything green. What those do is they take infrared photons and they ingest infrared photons. And they emit green light so that you can see infrared, and that’s how they do these night vision goggles. And that’s because at the temperatures on ordinary planets, we glow in the infrared, which is why you don’t see each other glowing in the dark, because your eyes don’t detect infrared. But if you’re going to be looking for planets, you’re much better off looking in the infrared than in ordinary light.

And then, once you can see the light from a planet, you get to do a really cool thing, which is, you get to examine the spectrum of that planet to find out what the atmosphere is made of. And we talked a little bit about spectral features, which come from different atoms or molecules. And once you actually see the light from the planet as opposed to the light from the star, you can actually study in detail what that star’s–what that planet’s atmosphere is made of.

This, by the way, is how one goes about detecting life on planet Earth. There was a very cute experiment a little while ago. They were sending a mission to Jupiter, the Galileo Mission, which is now at Jupiter doing science. But the way they did it was funny. They sent it not outward toward Jupiter, but inward toward Venus. It took a swing around Venus. It turns out, if you do it right, and you swing around a planet, you can pick up extra speed. And so, to get this thing out to Jupiter, they had it swing around Venus, then come back to Earth and swing around Earth. And each time it did one of these swings, it picked up speed, so it got to Jupiter actually faster that way–a very clever bit of orbital dynamics.

But the consequence of it was that this mission, one of our prime missions to look at other planets with all the various instruments that we like to look at other planets with, swung by near to Earth. And so, somebody had the very clever idea, let’s use our standard suite of exploration instruments and see if we can detect the existence of life on Earth using one of the satellites that we used to explore the rest of the Solar System. And people had all kinds of thoughts about what would be detectable–you know, cities at night, you know, you would be able to see lights on the dark side, the Great Wall of China, you would be able to detect–no. Turns out that the strongest sign of life on Earth is the fact that the atmosphere contains methane.

Now, the atmosphere, of course, has a lot of oxygen. Methane, you may recall, is CH4, and if you put a little bit of methane into an oxygen atmosphere it immediately dissociates and you get water and carbon dioxide because of all the free oxygen. So there shouldn’t be any methane in the atmosphere, but there is, and you can detect it by looking at the infrared spectrum of the atmosphere.

Now, what does it mean that there’s methane in our atmosphere? It means there’s a continuous source of methane pumping into the atmosphere all the time. What is the source of that methane? It is, as Carl Sagan, delicately put it, bovine flatulence. Cow farts. And just digestion, in general, has an output of methane, which is continually piping into the atmosphere. And if that weren’t happening, that happens from biological activity, and if that weren’t happening, there wouldn’t be any methane observable in the atmosphere.

So, what you do is, if you find some other planet, if you look at the atmospheric chemistry, and you see if there’s something in there that’s out of equilibrium that could only be continually replenished by biological activity. And having both oxygen and methane at the same time is one such indicator. So, the hope would be that we could not only see the light from the planet, but we’d be able to detect the presence of biological activity as well. So how do you go about doing this? Bovine flatulence equals life–words to live by.

Let’s see, so here’s how you do it. You need to have very high resolution; you need to be able to separate the light from the star, from the light from the planet, to the greatest extent possible. Turns out, the way you do this is you have space telescopes spaced pretty widely apart. That gives you the best handle on the resolution. But they have to know where each other is to a precision very substantially greater than the wavelength of the light that they’re observing. And so, they need to be able to tell the distance apart of these things to tiny fractions of a meter–“picometer metrology” is the technical word. A picometer is 10-12 meters and metrology is just measuring how far apart you are.

So, they need to be able to figure out–to formation fly, and know how far apart they are to about the width of an atom or so. You want to image in the infrared. Mid-infrared imaging wants you to–pushes you beyond Jupiter. The problem is that inside the orbit of Jupiter, there’s all this warm dust in our Solar System. And so, it’s like the sky is bright all the time from all the infrared emission, from the dust. So, you’ve got to go out into the colder parts of the Solar System to do this.

So we’re–what we’re going to do, the plan is, you’re going to build a half dozen or so things, each of which is about ten times the size of the space telescope. We’re going to move them out beyond Jupiter and formation fly them in this–to this extremely high precision. This is not something we’re capable of doing right now. But there is a twenty-five year-old; twenty-five year so-called roadmap, a series of missions along the way, which is going to simultaneously be good missions in their own right, and develop the technology.

SIM is one of these missions. It’s going to develop the high precision spatial imaging that you need for this. JWST is the acronym for the next-generation space telescope, which is due to be launched in 2012. That’s basically one such telescope, but orbiting the Earth. That’s a milestone. And there are various other milestones along this roadmap. And the hope is that in twenty-five years, with, you know, 25 to 50 billion dollars, we could actually do this thing.

And so, here’s the little artist’s conception of these guys flying in formation outside of the orbit of Jupiter. Then, I guess this represents the Sun down in the middle of the Solar System. They’re looking out this way, finding life on other planets. And the big problem with this whole mission is that it’s been cancelled, because a decision has been made that it’s–this is a very big, expensive mission, and the decision has been made that if you’re going to do something big and expensive, you ought to go to Mars, instead. Put people on Mars, instead. However, this is a twenty-five year time scale. Things get cancelled and un-cancelled every three or four years, so, who knows exactly what’s going to happen. And I think this may reappear in the NASA plan at some point in the future. And that’s sort of the ultimate that people can think of now in terms of find–studying planets around other stars. Okay, that’s what I got to say about this, and a test on Thursday.

[end of transcript]

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