0:12

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.

5:25

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,

6:54

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

11:27

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.

12:55

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,

15:05

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.

16:47

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.

18:09

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.

19:28

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.