Using publicly available data from NASA of actual satellite observations of astronomical x-ray sources, we explore some of the mysteries of the cosmos, including neutron stars, black holes, quasars and supernovae. We will analyze energy spectra and time series data to understand how these incredible objects work. We utilize an imaging tool called DS9 to explore the amazing diversity of astronomical observations that have made x-ray astronomy one of the most active and exciting fields of scientific investigation in the past 50 years. Each week we will explore a different facet of x-ray astronomy. Beginning with an introduction to the nature of image formation, we then move on to examples of how our imaging program, DS9, can aid our understanding of real satellite data. You will using the actual data that scientists use when doing their work. Nothing is "canned". You will be able to appreciate the excitement that astronomers felt when they made their important discoveries concerning periodic binary x-ray sources, supernovae and their remnants, and extragalactic sources that have shaped our understanding of cosmology.
Lecture 2 The Time Machine, Part 2
Loading...
來自 Rutgers the State University of New Jersey 的課程
Analyzing the Universe
69 個評分
Explore Related Videos
At Coursera, you will find the best lectures in the world. Here are some of our personalized recommendations for you
從本節課中
To the Ends of the Universe; Quasars, 3C273, and beyond
This week we wrap things up with trips to galaxies and exotic objects, seen long ago and far away. The mysterious quasars provide clues about the way our Universe is evolving in time. They are incredible objects (actually, come to think of it, what isn't incredible in the x-ray sky?) discovered almost exactly a half century ago, quite by accident. We will explore the astonishingly prodigious x-ray output of 3C 273, one of the nearest ones, at a mere 2.5 billion light years away.
與講師見面
Dr. Terry A. MatilskyProfessor
Physics and Astronomy
[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]
