[BLANK_AUDIO] After World War II, radio astronomy surged to the forefront of the astronomical frontier. Vast numbers of radio sources were discovered in the sky, which added greatly to our understanding of the processes that go on in our universe. But one of the problems with early radio astronomy was that because of the wavelength of the observed radio-light was so long, the positions of the sources were exceedingly hard to pinpoint. So it was very difficult to correlate the newly discovered radio emissions with known optical counterparts in the sky. But by 1960, many positions were able to be refined and we began to see what these cosmic objects were doing in the optical as well as the radio regime. One of the techniques used to identify some of the sources involved looking at lunar occultations. If a radio source just happened by chance to be along the path of the moon's orbit, the moon would pass in front of the object, thereby shutting of temporarily the earth bound radiation. By precise timing of the disappearance and reappearance of the source, accurate positions could be obtained. Two such sources located were 3C48, and 3C273. The 3C stands for the Third Cambridge Catalogue of Radio Sources. Cambridge University in England was a pioneer in the radio astronomy field, and the numbers after 3C, were ordered by right ascension of the objects looked at. When Allan Sandage of Caltech saw the spectrum in visible light of 3C48, he said, and I quote, the thing was exceedingly weird, unquote. Indeed it was an object unlike any previously seen. It's optical appearance was extremely blue. And although it looked like a star, it's spectrum was very strange indeed. None of the known elements appeared to be there. The well studied fingerprints of hydrogen, calcium and other stellar constituents seem to be gone. Instead, other lines in the spectrum seemed to emerge at odd wavelengths corresponding to nothing we knew about in the laboratory. Then in 1963, the Dutch Astronomer Maarten Schmidt realized that the pattern of lines in the spectrum of 3C273 were identifiable. But they corresponded to wave lengths red shifted by an astounding amount. Never was a star like this seen before. Thus the objects, which now number in the thousands, were dubbed, quasi-stellar objects or QSO's or simple quasars for short. Let's look at the optical spectrum of 3C273. The three strong lines seen in the quasar spectrum are those of hydrogen, marked H delta, H gamma, and H beta. At rest on the Earth they correspond to the following wavelengths. H beta equals 486.1 nanometers. H gamma, 434 nanometers. H delta 410.2 nanometers. If you go back to the set of spectra we looked at when we studied stellar spectra, the A1 star shown here has for its most prominent features exactly these lines. You can also find a copy of this figure posted in the supplementary materials section of the course navigation bar. These lines and some others are identified in the comparison spectrum below the quasars. This comparison spectrum is taken in the observatory, at rest. And represents what a mixture of gases looks like when nothing is moving, with respect to the telescope. The nm here, stands for nanometers, and represents a unit of length equal to ten to the minus nine meters or ten to the minus seven centimeters. Here is another representation of the same spectra. Let's get the velocity and distance to 3C273. Print out the optical spectrum of the quasar plus comparison, which is available at the navigation bar under course supplementary materials. We will measure the relative positions of several lines in the comparison spectrum with an ordinary ruler. First we need to figure out how many nanometers of wavelength on the spectrum corresponds to one millimeter on the paper. This is called the dispersion or scale of the spectra. Measure several pairs of lines and take the average for a more accurate value. Note that the answers that we're talking and be given here, will be different from your results because of variations in printing of the page from computer to computer. For example, if the lines of H beta and H delta are separated by 36.5 millimeters in the comparison spectrum, the plate scale will be 2.08 nanometers per millimeter. So if we look at H beta, and H delta, and they are separated by a distance equal to 36.5 millimeters. We can calculate knowing what the wavelengths of H beta and H delta are that the scale will be equal to 2.08 nanometers on the spectrum per millimeter on your piece of paper. Now, we can choose one of the hydrogen lines in the Quasar spectrum, and see how far in millimeters it is from the corresponding rest wavelength. For example, if H delta line appears in the quasar spectrum at a distance of 33 millimeters to the right of the laboratory position, its wave length shift, delta lambda, will be equal to the plate scale 2.08 nanometers per millimeter times 33 millimeters or a displacement of about 68.7 nanometers. Now, we can derive the velocity. If you remember, for our Doppler shift, V over C, is about equal to delta lambda over lambda, and if our line is displaced 68.7 nanometers, while the denominator is 410 nanometeres. We find, that V is about equal to 0.17 times the velocity of light. Do this for the other hydrogen lines as well, and get an average value for increased accuracy. Remember, V is the velocity, C is the velocity of light, delta lambda is the amount in nanometers of the displacement of the line in the quasar spectrum, and lambda is the wavelength in nanometers of the line at rest. See how close you can get to the correct red shift of 0.158. Finally, now we can get the distance to the object. If V equals HR and H is equal to 70 kilometers per second per mega-parsec, we know what V is now and with the equal to about 5 times 10 to the 4 kilometers per second. R is equal to 700 million parsecs or about 2.2 billion light years. So when you see this object through a telescope, you are viewing it the way it appeared 2.2 billion years ago, when only bacteria and algae were present on the Earth. Mammals were still, two billion years away from existence, on our planet. Wow. Thus, these objects, though they look like stars, are so distant that they must be more luminous than the brightest galaxies. At over two billion light years and more, since 3C273 turns out to be the closest quasar, they are over one thousand times more distant than the Andromeda Galaxy, M31. Yet even at such gargantuan distances, many including 3C273 are bright enough to be seen in small telescopes. To find out just how bright they really are, let's go to DS9 and check out 3C273. We open DS9, go to Analysis, click on Virtual Observatory, do the usual, connect using web-proxy. Retguss primary MOOC x-ray analysis server. That brings up a window with all of the observations in it. We scroll down until we see 3C273 as the title of the observation, its obs ID number 1712. We click on the title and there it is 3C273. Note how different this source appears from KSA. It looks much much smaller and there's a little jet like protrusion coming out from the lower right-hand side of the object. Let's kind of zoom in to make this a little clearer. Okay, now we've got it. Also, the image shows a ring of emission that seems to be black in the center. This is not really the way 3C273 is in the sky. It is an artifact of the satellite and it occurs because the object is so bright in x-rays, that Chandra's counters get saturated. We call this phenomenon pile up, and it is similar to overexposure in a photographic image, but some x-rays are still there, they are just spread out along a column of the detector. See if you can adjust the contrast and brightness in DS9 by right clicking to see the line of radiation. We'll try that now. We'll kind of go this way, and there now we can see it okay. Let's change the color to make it a little bit, stand out a little bit more. Okay. Let's do that. Adjust it. There, you can kind of see that faint little line over there. If you can't do that on your own, you can get a good look at this at home by selecting the BB color scheme, go to the Color menu, click on Color Map Parameters and there you can adjust those parameters to get the contrast that you desire. A good one for you to try might be a contrast of about four, and a bias of 0.06 or so, there we go and that looks pretty good. So, let's find the luminosity of 3C273. Let's enclose the image of 3C273 and its jet, within a circular region. Here's our region. We'll make it a little bit bigger and we'll grab it and move it. Well, we'll make it even bigger, so it encloses 3C273, proper, and its jet. And you can see we will, we will be excluding, some of these pile up photons. But we're just interested in an order of magnitude estimate of the energy output from the object. So now, what we do is fit the spectrum to a model and the way we do that is by looking at the Chandra Ed Analysis Tools and go to CIAO/Sherpa Spectral Fit. We're going to fit a power lower model. The model really isn't all that important, for what we are trying to do here right now, which is just to find out how much stuff, what the flux is coming from the region around 3C273, so we can get an estimate of its luminosity. And we're going to display the Sherpa logs. We want to see the output of this fit. And then we hit OK. And there it is, this is what the energy spectrum of 3C273 looks like, fit with a particular model, so we have counts versus energy, and now we can look at the log and we can see that the flux for this data Is about 7.5 times 10 to the minus 12, ergs per square centimeter per second. We can round this off to about 10 to the minus 11 ergs per square centimeter per second. And what this means is that each second, every square centimeter of Chandra's detectors, received about 10 to the minus 11 ergs, from the region of the sky around 3C273. Since this is the only strong source in the field a few of the satellite, we can say that this represents roughly the x-ray energy received from 3C273 in the energy band that Chandra is sensitive to. But, think for a moment. 3C273 is pouring out these photons everywhere in the sky. Chandra only picks up a very tiny percentage of them. The rest keep streaming out into space, where no x-ray satellite is there to see them. In fact, we can imagine a huge ball centered on 3C273 whose radius is equal to the distance from the source to the Earth. Each square centimeter of the tiny satellite's area must be multiplied by the area of the ball which is 4 pi d squared, where d is the distance from 3C273 to the Earth, to get the amount of x radiation that 3C273 is giving off into space. So, if 10 to the minus 11 ergs per second of energy crosses each square centimeter of surface area at the distance of the Earth, to 3C273, what is the x-ray output of 3C273? Let's return to the blackboard and find out. Think for a moment, 3C273 is pouring out these photons everywhere in the sky. Chandra only picks up a very tiny percentage of them. The rest keep streaming out into space, where no x-ray satellite is there to see them. In fact, we can imagine a huge ball, centered at 3C273, whose radius is equal to the distance from the source to the Earth. Each square centimeter of our satellite's area must be multiplied by 4 pi d squared, where d is the distance from 3C273 to the Earth, to get the amount of x radiation that 3C273 is giving off. So if 10 to the minus 11 ergs per second of energy crosses each square centimeter of surface, at the distance of the Earth to 3C273, what is the x-ray output of 3C273? Well, it's very easy. The X-ray luminosity must be the flux that we measure in the sky time 4 pi d squared. If the flux the observed brightness is 10 to the minus 11 ergs per square centimeter per second and our sphere, is, a distance d of 2.2 times 10 to the 27 centimeters, we square that multiply by 4 pi and we end up with an x-ray luminosity of about 6 times 10 to the 44 ergs per second. This is close to one trillion times the entire energy output of our sun. And ten times the luminosity of our entire galaxy. Finding a mechanism to produce this much energy would be difficult under any circumstances. But the quasars present an even more difficult puzzle. These objects fluctuate in brightness and because of this they must be rather small. As in the side, note that this equation is nothing more than the inverse square law of a light, if instead of knowing the distance, we know the true brightness of the object along with its apparent brightness, here expressed as the flux, we can use this to solve for the distance. This is what we do for example with a Cepheid variable stars. [BLANK_AUDIO]
