0:12

After World War II, radio astronomy surged

Â to the forefront of the astronomical frontier.

Â Vast numbers of radio sources were discovered in the sky, which added

Â greatly to our understanding of the processes that go on in our universe.

Â But one of the problems with early radio

Â astronomy was that because of the wavelength of

Â the observed radio-light was so long, the

Â positions of the sources were exceedingly hard to pinpoint.

Â So it was very difficult to correlate the newly discovered radio

Â emissions with known optical counterparts in the sky.

Â But by 1960, many positions were able to be refined and we began

Â to see what these cosmic objects were doing in the optical as

Â well as the radio regime. One of the techniques used to

Â identify some of the sources involved looking at lunar

Â occultations. If a radio source just happened by chance

Â to be along the path of the moon's orbit, the moon

Â would pass in front of the object, thereby shutting of temporarily

Â the earth bound radiation. By precise timing of the disappearance

Â and reappearance of the source, accurate positions could be obtained.

Â Two such sources located were 3C48, and 3C273.

Â The 3C stands for the Third Cambridge Catalogue

Â of Radio Sources. Cambridge University in England was a

Â pioneer in the radio astronomy field, and the numbers after

Â 3C, were ordered by right ascension of the objects looked at.

Â When Allan Sandage of Caltech saw the spectrum in visible light of

Â 3C48, he said, and I quote, the thing was exceedingly weird, unquote.

Â Indeed it was an object unlike any previously seen.

Â It's optical appearance was extremely blue.

Â And although it looked like a star, it's spectrum was very strange indeed.

Â None of the known elements appeared to be there.

Â The well studied fingerprints of hydrogen, calcium

Â and other stellar constituents seem to be gone.

Â Instead, other lines in the spectrum seemed to emerge at

Â odd wavelengths corresponding to nothing we knew about in the laboratory.

Â Then in 1963,

Â the Dutch Astronomer Maarten Schmidt realized that

Â the pattern of lines in the spectrum of 3C273

Â were identifiable. But they corresponded to wave lengths

Â red shifted by an astounding amount. Never was a

Â star like this seen before. Thus the objects,

Â which now number in the thousands, were dubbed,

Â quasi-stellar objects or QSO's or simple quasars for short.

Â Let's look at the optical spectrum of 3C273.

Â The three strong lines seen in the quasar spectrum are those of hydrogen,

Â marked H delta, H gamma, and H beta. At rest on

Â the Earth they correspond to the following wavelengths.

Â H beta equals 486.1 nanometers.

Â H gamma, 434 nanometers. H delta

Â 410.2 nanometers. If you go back to the set of spectra

Â we looked at when we studied stellar spectra, the A1 star

Â shown here has for its most prominent features exactly these lines.

Â You can also find a copy of this figure posted

Â in the supplementary materials section of the course navigation bar.

Â These lines and some others are identified

Â in the comparison spectrum below the quasars.

Â This comparison spectrum is taken in the observatory, at

Â rest. And represents what a mixture of gases

Â looks like when nothing is moving, with respect to the telescope.

Â The nm here, stands for nanometers, and represents a unit of length

Â equal to ten to the minus nine meters or ten to the minus seven centimeters.

Â 5:25

Let's get the velocity and distance to 3C273.

Â Print out the optical spectrum of the quasar plus comparison,

Â which is available at the navigation bar under course supplementary materials.

Â We will measure the relative positions of several

Â lines in the comparison spectrum with an ordinary ruler.

Â First we need to figure out how many nanometers of wavelength on

Â the spectrum corresponds to one millimeter on the paper.

Â This is called the dispersion or scale of the spectra.

Â Measure several pairs of lines and take the average for a more accurate value.

Â Note that the answers that we're talking and be given

Â here, will be different from your results because of variations

Â in printing of the page from computer to computer.

Â For example, if the lines of H beta and H delta are

Â separated by 36.5 millimeters in the comparison

Â spectrum, the plate scale will be 2.08

Â nanometers per millimeter. So if we look at

Â H beta, and H delta,

Â 6:54

and they are separated by a distance

Â equal to 36.5

Â millimeters. We can

Â calculate knowing what the wavelengths of H

Â beta and H delta are

Â that the scale will be equal to

Â 2.08 nanometers

Â on the spectrum per millimeter on your piece

Â of paper.

Â Now, we can choose one of the hydrogen lines in the Quasar spectrum, and see how

Â far in millimeters it is from the corresponding rest wavelength.

Â For example, if H delta line appears in the quasar spectrum

Â at a distance of 33 millimeters to the right of the laboratory position,

Â its wave length shift, delta lambda, will

Â be equal to the plate scale 2.08

Â nanometers per millimeter times

Â 33 millimeters or a displacement

Â of about 68.7 nanometers.

Â Now, we can derive the velocity. If you remember,

Â for our Doppler shift, V over C,

Â is about equal to delta lambda

Â over lambda, and if our line is

Â displaced 68.7 nanometers,

Â while the denominator is 410

Â nanometeres. We find, that V is

Â about equal to 0.17 times the

Â velocity of light. Do this for the other hydrogen

Â lines as well, and get an average value for increased accuracy.

Â Remember, V is the velocity, C is the

Â velocity of light, delta lambda is the amount

Â in nanometers of the displacement of the line in the quasar spectrum,

Â and lambda is the wavelength in nanometers of the line at

Â rest. See how close you can get to

Â the correct red shift of 0.158.

Â Finally, now we can get the distance to the object.

Â If V equals HR and H is equal to 70

Â kilometers per second per mega-parsec,

Â we know what V is now and with the equal

Â to about 5 times 10 to the 4 kilometers

Â per second. R is equal

Â 11:27

Yet even at such gargantuan distances, many including

Â 3C273 are bright enough to be seen in small telescopes.

Â To find out just how bright they really are,

Â let's go to DS9 and check out 3C273.

Â We open DS9, go to Analysis, click on Virtual

Â Observatory, do the usual, connect using web-proxy.

Â Retguss primary MOOC x-ray analysis server.

Â That brings up a window with all of the observations in it.

Â We scroll down until

Â we see 3C273 as the title of the

Â observation, its obs ID number 1712.

Â We click on the title and there it is

Â 3C273. Note how different this

Â source appears from KSA. It looks much much smaller and

Â there's a little jet like protrusion coming out

Â from the lower right-hand side of the object.

Â Let's kind of zoom in to make this a little clearer.

Â 12:55

Okay, now we've got it.

Â Also, the image shows a ring of emission that seems to be black in the center.

Â This is

Â not really the way 3C273 is in the sky.

Â It is an artifact of the satellite and it occurs because

Â the object is so bright in x-rays, that Chandra's counters get saturated.

Â We call this phenomenon pile up, and it is similar to

Â overexposure in a photographic image, but some x-rays are still there,

Â they are just spread out along a column of the detector.

Â See if you can adjust the contrast and brightness in DS9 by right

Â clicking to see the line of radiation. We'll try that now.

Â We'll kind of go this way, and there

Â now we can see it okay. Let's change the color to make it a little

Â bit, stand out a little bit more. Okay.

Â Let's do that.

Â Adjust it. There, you can kind of see that faint

Â little line over there. If you can't do that on your own,

Â 15:05

So, let's find the luminosity of

Â 3C273. Let's enclose the image of 3C273 and

Â its jet, within a circular region. Here's our region.

Â We'll make it a little bit bigger and we'll grab it and move it.

Â Well, we'll make it even

Â bigger, so it encloses 3C273,

Â proper, and its jet. And you can see we will, we

Â will be excluding, some of these pile up photons.

Â But we're just interested in an order of

Â magnitude estimate of the energy output from the object.

Â So now, what we do

Â is fit the spectrum to a model and

Â the way we do that is by looking at the Chandra Ed

Â Analysis Tools and go to CIAO/Sherpa

Â Spectral Fit. We're going to fit a power lower model.

Â The model really isn't all that important, for what we are trying to do here right

Â now, which is just to find out how much stuff, what the flux

Â is coming from the region around 3C273,

Â so we can get an estimate of its luminosity.

Â And we're going to display the Sherpa logs.

Â We want to see the output of this fit. And then we hit OK.

Â 16:47

And there it is, this is what the energy spectrum

Â of 3C273 looks like, fit with a particular

Â model, so we have counts versus energy,

Â and now we can look at the log and we can see that the

Â flux for this data Is about 7.5

Â times 10 to the minus 12, ergs per

Â square centimeter per second. We can round this off to about

Â 10 to the minus 11 ergs per square centimeter per second.

Â And what this means is that each

Â second, every square centimeter of Chandra's detectors,

Â received about 10 to the minus 11 ergs, from the region of the sky around 3C273.

Â 18:09

3C273 is pouring out these photons everywhere

Â in the sky.

Â Chandra only picks up a very tiny percentage of them.

Â The rest keep streaming out into space, where no x-ray satellite is there

Â to see them. In fact, we can imagine a huge ball

Â centered on 3C273 whose radius is equal to the distance from the source to the Earth.

Â Each square centimeter of the tiny satellite's area must

Â be multiplied by the area of the ball which is 4 pi

Â d squared, where d is the distance from 3C273 to the

Â Earth, to get the amount of x radiation that

Â 3C273 is giving off into space. So, if 10 to

Â the minus 11 ergs per second of energy crosses each square centimeter

Â of surface area at the distance of the Earth, to 3C273,

Â what is the x-ray output of 3C273?

Â Let's return to the blackboard and find out.

Â 19:28

3C273 is pouring out these photons everywhere in the sky.

Â Chandra only picks up a very tiny percentage of them.

Â The rest keep streaming out into space, where no x-ray satellite

Â is there to see them. In fact, we can imagine a huge ball,

Â centered at 3C273, whose

Â radius is equal to the distance from the source to the Earth.

Â Each square centimeter of our satellite's area

Â must be multiplied by 4 pi d squared, where d

Â is the distance from 3C273 to the Earth,

Â to get the amount of x radiation that

Â 3C273 is giving off. So if 10 to the

Â minus 11 ergs per second of energy crosses

Â each square centimeter of surface, at the distance of the Earth

Â to 3C273, what is the x-ray output

Â of 3C273? Well, it's very easy.

Â The X-ray luminosity must be the flux that we

Â measure in the sky time 4 pi d

Â squared. If the flux the observed brightness

Â is 10 to the minus 11 ergs per square

Â centimeter per second and our sphere,

Â is, a distance d of

Â 2.2 times 10 to the 27

Â centimeters, we square that

Â multiply by 4 pi and

Â we end up with an x-ray luminosity of

Â about 6 times 10 to the 44 ergs

Â per second. This

Â is close to one trillion times the entire

Â energy output of our sun. And ten times the luminosity

Â of our entire galaxy.

Â Finding a mechanism to produce this much

Â energy would be difficult under any circumstances.

Â But the quasars present an even more difficult puzzle.

Â These objects fluctuate in brightness and because of this they must be rather small.

Â