SIZE   DOES   MATTER

For I dipped into the Future
Far as the human eye could see;
Saw the vision of the world
And all the wonder that would be.

                                                                                                       Alfred, Lord Tennyson, 1842

Part I: GALAXIES

	When Edwin Hubble uncovered the key to the sizes of galaxies in the 1920’s, we were awestruck by the sheer enormity of these “island universes”. It is hard to fathom an object that is so large that light, which is capable of traveling over 7 times around the world in one second, takes over 100,000 years to get from one end of a typical spiral galaxy to the other.  Could there possibly be anything larger than this in the universe?
	The answer was not long in coming; within 10 years of Hubble’s epochal discovery, it was clear that galaxies were not distributed randomly about the sky. Instead, many vast clusters of galaxies exist, sometimes containing thousands of individual galaxies, each of which, in turn, contain up to 100 billion stars similar to our own Sun.  (Interestingly, Sir William Herschel recognized the existence of the Virgo cluster, but at that time in the 18th century, it was not known whether or not these objects were part of our own galaxy, the Milky Way. Incidentally, many of the larger cluster are named after  the constellation in which they are located.)
	To get some idea of these enormous structures, let us look at our own neighborhood.  The Milky Way is in a small group, with the only other large galaxy being M31 (the Andromeda spiral).  However, if we were located in the center of the Coma cluster, we would see 100 large galaxies in the same volume of space contained in a cubical box enclosing us and M31.  The sky would be littered with fuzzy patches of light every 20 degrees, which corresponds approximately to the distance between the top and bottom of the constellation of Orion.

Activity 1:  Let’s look at M31 in x-rays and visible light

A)  Open DS9.  Load the M31 Chandra image.  Go to: Color bb,  Scale  log.   What you now see is a bunch of point-like  x-ray sources.  These are most likely binary star systems emitting x-rays from the  nucleus of M31.  (The dark “crosses” you see are part of the satellite hardware, and are spaces between the x-ray sensitive chips that you can clearly see in the image.)  Now, let’s see exactly where in the sky these sources are.  Go to:
Frame  tiles frames
Analysis  DSS server  retrieve.  This will allow you to place, side-by-side with your x-ray image, the visible light seen through a large telescope in that region of the sky.  Set: 
Scale  sqrt for the best image detail.  

Here is the nucleus of M31, our closest spiral galaxy neighbor, a mere 2 million light years away.  Imagine, light that you see coming from M31 on a clear  autumn evening anywhere in the northern hemisphere (yes, you can see it with your naked eye if you know where to look), left the galaxy long before modern humans (homo sapiens) ever walked the Earth.   See the activities for 3C273 to determine the size of M31.  [clickable link here!]

Now we can do something really neat.  Go to:  Frames  match frames  WCS and Frame  lock crosshairs  WCS.  Then Edit  crosshairs.

Now, crosshairs will appear on both images, and you can “grab” either crosshair by left clicking and holding down the mouse button on the intersection of the two lines;  now you can roam around the image and the two crosshairs will track together through space.  So you can see exactly where in the galaxy each of the x-ray sources are coming from.  Notice that almost all of them seem to be quite faint and nothing special in visible light.  But let’s see how much x-ray energy is coming out of these bright, “point” sources.

B) Using an analysis similar to that described in our discussion of 3C273  [ADD LINK HERE TO ACTIVITY 3], verify that the luminosity of the source near physical pixel (4185, 4081) is about 1037 ergs/sec.   So these innocuous appearing objects are radiating about 10,000 times the total light output of our Sun, and about 10 billion times more than our Sun’s x-ray production!

C) Can you think of another way to verify the luminosity of this source?  (Hint:  use the pixel table in DS9, plus the total length of the observation, which can be obtained by going to File  Display FITS header.)

Activity 2:  When is a pixel not a pixel?  Answer: when it’s 4, 9, 16 or more pixels!

There are many ways to look at astronomical data.   Sometimes you want to see as wide a field-of-view as possible.  Sometimes, you want to zoom in to a small region of the sky.  So displaying the same data can take many forms.  To see this, open DS9, and go to Frame  tile frames.  Then load the image that reads center of M31.
Select Scale  log.  Then go to Frame  New frame, and load the regular “M31” image.

What you are now looking at is the same data file, but displayed in two different ways.  Notice that the image on the left (the “center image”) looks magnified.  Although it occupies the same amount of screen space as the other, you are only seeing a portion of the sky that the “full” image (on the right) displays.  How is this done?  The center image is displaying the native 0.5 arc-second resolution of the Chandra satellite.  I.e. each pixel edge you see displayed corresponds to 0.5 arc-second on the sky.  It is a portion of the data that was obtained in observation #303 in 1999.  The image on the right takes the same data, and sums a 4x4 set of 16 pixels together and displays them as a single pixel on the screen.  Therefore, you can see 16 times as much sky area for a given number of displayed screen pixels.  To see this graphically, zoom in the right image.  Either select zoom  4 in the drop down menu, or click the “zoom” button on the menu bar just above the image displays.   Then click “in” twice to get the same result as from the drop down menu.  (Many functions in DS9 can be accessed in different ways; when you use the program frequently you will see the utility for so doing).

Now you see the two images with the same magnification, but the one on the right seems very “grainy”.  That’s because 16 pixels of the left hand image went into each (now larger displayed) pixel on the right.  In order to see more of the sky on our screen, we sacrificed resolution, since each pixel edge now represents  2 arc-sec on the sky.  To see this clearly, go to Physical Pixel (4068, 4136) on each image.   Notice that it “sees” the same part of the sky, namely, a region where there are two point sources very close to each other. Notice also the size of the magnified pixel in the “magnifier” at the upper right of the DS9 display, as you cruise across from one image to the other. (It is vital to use Physical Pixels here, because they represent the “true” sky position of the object in the data file, regardless of where they appear on the image display).  But see how you can barely distinguish the sources on the right hand image.  So many pixels (16) were summed for each part of that image , that the separation of the two objects was almost lost, and they virtually appear as one “blob”.  

Now you can begin to appreciate what kind of advance the Chandra satellite had over previous missions.  Because of its incredibly high resolution capability, it could distinguish sources separated from each other by tiny angular distances on the sky.

You may wonder why one would ever use anything but the “best” resolution that Chandra has?  Just imagine a very faint source in the field of an observation.  It may be virtually indistinguishable from the “background”.  But now, if you sum adjacent pixels (each one just a bit brighter than the background surrounding the source), you may be able to see the source more clearly.  Thus, depending on what you want to do with your data, you may elect to display it in different ways.

Part II:  CLUSTERS OF GALAXIES

	Perhaps it is not surprising that these huge galaxies with their enormous gravitational fields would cluster together.  But it would still appear that the individual galaxies are the largest structures known in the universe.  Indeed, some of the giant radio lobes associated with radio wavelength emission of certain peculiar galaxies are known to be over 5 million light years in extent.  That means that the sizes of these objects are greater than the entire distance to our nearby galaxies.
	Yet, this view was shattered in the 1970’s, when the Uhuru and HEAO satellites detected X-rays pouring out from between the galaxies in rich clusters.  Spectral analysis of X-rays at different wavelengths revealed that this emission was due to gas heated to 10-100 million (degrees) Kelvin. 

	Activity 3:  A look at the Coma cluster of galaxies in X-ray and visible light

A)	Load the Coma image; select “sqrt” for the scale and “b” for the color.  This will give you a good look at the x-ray emission as seen from the Chandra satellite.  Notice that there is a diffuse elliptically shaped emission region as well as some point-like sources in the field of view.  What you are seeing here is only the central (bright) portion of the cluster.

B)  To compare this to the visible light, we do the same thing we did for M31 above.
Go to: Frame  tiles frames; Analysis  DSS server  retrieve.  What you  see in the new, right hand frame are hundreds of galaxies.  There are some stars (from our own Milky Way) as well, but all those slightly elongated elliptical blobs are galaxies, each with tens of billions of stars similar to our own Sun.  To match the two images, select the x-ray image (the first one you loaded) by clicking with your mouse anywhere in that left hand image window and then do:  Frame  match frames  WCS and Frame  lock crosshairs  WCS.  Then Edit  crosshairs.
Now you can cruise around and see that the “point” sources in the x-ray light really correspond to entire galaxies in the visible picture!

Activity 4:  The size and temperature of the x-ray gas

A)  In the same way as we did for 3C273, we can find the distance to Coma.  Since the   cluster is far closer, the change in the wavelengths of the spectral lines is much smaller, however.  In this case, Dl/l = 0.02, instead of  0.13.  Show that this implies a distance to the cluster of about 120 Mpc.

B)  Now, estimate the size (in physical pixels) of the “long” dimension of the diffuse emission, and show that the extent of the x-ray gas is about 500 kpc in size.  (Note that your estimates for this will differ because of different estimates for the diameter of the region on the image).  This is roughly the same order of magnitude as the entire distance to M31 from our own galaxy.  Also remember that this is merely the central, brightest portion of the cluster that extends for a far greater distance into space.  This central region is called the core radius of the cluster.

C)	To find out the temperature of the gas, we can fit the emission with a spectral 
model (thermal bremsstrahlung).  This model attempts to quantify the way rarified, hot gas would behave in an environment similar to that of the Coma cluster.  First, select an elliptical region by going to Editpointer, then Regionshapeellipse.  Then, estimate where the “center” of the elliptical (red appearing) x-ray emission is, and click once to place the ellipse on the cluster.  You can then click again to select the region, and drag the corners to the proper shape and size of the cluster.  You may want to try this several times to get a better fitting region from which to do the analysis.  (Remember, if you want to delete a region and start over, just select the region by clicking once within it, and with the cursor anywhere inside the region, hit the “delete” key on your keyboard).  Now, if we were really trying to get some accurate numbers, we would select a background region to subtract from our source region estimates, but since the Coma cluster is so bright, and we are just interested in reasonable estimates, we will omit this step.  So, after your region is selected, go to:  AnalysisCIAO/Sherpa Spectral Fit.  Then, when the dialog box pops up, select “bremsstrahlung” for the Model type, and hit OK.  After a few moments you will see the data displayed, as well as the model fit to the data.  From the text file display, you will find a line that gives you the temperature for the final fitted model.  This should have a value of about 7 to 9 keV.   This corresponds to a temperature of about 100 million degrees (on the absolute scale)!

Activity 5:  How bright is the gas?

A)  In the text box you generated above, you will find another very useful bit of information.  Look for the line that gives you the “flux”.  This should be a number approximately equal to 10-10 ergs/sec/cm2.  What this means is that through each cm2 of area (about the size of your little finger’s nail) at the satellite, about 10-10 ergs pass each second from the direction of the Coma cluster.  This is a tiny amount of energy, but since the cluster emits x-rays in all directions, if we take this number and multiply it by the number of square centimeters in all space at the distance of the Earth from the cluster, we will get the entire energy output of the gas.  To visualize this more clearly, imagine a basketball, with the cluster gas at the center.  The Chandra satellite (and the Earth) is on the surface of this ball, and through each little 1 cm2 box on this surface about 10-10 ergs pass each second.  Therefore, if we multiply this flux by 4*pi*r2 (where r is the ball’s radius), we will get the total energy passing each second through the ball.  This must be the total output of the gas.  Do this!  You should get an answer of about 1044 ergs/sec. This is about the same as the output of 100 billion Suns!  (Since we are dealing with only the central part of the cluster, the actually result will be greater still).

B)  Now, just for fun, suppose we call up our friends on another galaxy that is twice as far away from Coma as we are.  They, too, have just launched a duplicate Chandra satellite, and are measuring the energy from the Coma cluster also.  What will they get for the flux? What about the total energy output? 

     	Further research showed that the mass of this x-ray emitting gas was roughly comparable to the total mass of the (optically visible) member galaxies.  Where could this gas come from?
	At first, it was thought that the clusters captured this gas from intergalactic space, but soon, it was found that a spectral line of highly ionized iron appears in all the x-ray spectra at an energy of about 6 keV.  (Note that in the spectrum that you examined in Activity 4, this feature was not prominent.  This is because bombardment by cosmic rays early in the Chandra mission degraded the instrument somewhat and spreads out the emission over a greater range of energy so that it is not obvious here). Iron can only be formed within the interiors of stars, and later released in supernova explosions.  Thus, the gas must be the product of the galaxies themselves.  But how does it get distributed between the galaxies?

	Several mechanisms have been proposed.  For example, if two galaxies collide, the stars, being widely separated, will pass by each other like ships in the night, virtually unaware of each other’s existence.  But the gas will interact, and be left behind.  In this picture, as the galaxies recede from each other, the gas will remain between the galaxies.  However, further calculations seemed to indicate that there wouldn’t be enough collisions between galaxies to account for all the X-ray emitting gas that is observed.

	More likely is a process we call  “ram pressure stripping”.  In this scenario, the gas is lost by having the galaxies just move through intergalactic space. This process is similar to the wind knocking the hat off a rapidly pedaling cyclist  (ALWAYS wear a helmet!).  Calculations show that gas already expelled from stars via supernova explosions would have sufficient energy to blow enough interstellar gas out of even large spiral galaxies to account for the quantity of observed x-ray gas.

	So the gas eventually fills an enormous region from one end of the cluster to the other. Because this gas responds and orients itself (as does all mass in the Universe) to the gravitational fields present in the cluster of which it is a part, it acts like a “map” of this field and the material therein.  Thus, by examining the distribution of X-ray gas, we can deduce the amount of mass in the entire cluster.  Let’s see how this is done.

	Activity 6:  get a radial profile of the surface brightness of the Coma cluster

Just like we did for Cas-A in another set of pages, do the following: 

A) Select an annulus for your region shape, and center the cursor over the brightest part of the cluster. Click once to create the region, and again to select it. Grab a corner of the region and drag it outward until you are covering most the x-ray cluster gas with the outer annulus ring. 

 B)  Go to the Region Get info... dialog box and enter an inner radius of
 "0" , and 10 for the number of annuli. (Note: you MUST select the region by clicking once anywhere inside it before you go to the "get info" dialog box). 

 	C)  Click on "generate" and then "apply". You will notice that 10 concentric
 rings appear like a bulls-eye surrounding the cluster. What we will find
 out is how bright the gas is as a function of how far you are away from the center. 
D) Under analysis, click on "radial profile". 

 	Your graph shows you that the intensity of the cluster steadily decreases from the center.  (How is this different from the Cas-A result?)  This is indicative of a mass of x-ray gas filling a region with the brightest area at the gravitational center of the cluster.  Using this brightness profile, coupled with the result for the temperature you saw earlier, scientists can model the cluster to determine how much mass must be present in order to have the x-ray gas “float” in the gravitational field set up by all the galaxies. 

	What we find is astonishing.  In order for the gas to be at the brightness and temperature observed, we need about 100 times more mass in the cluster than that which is seen in the star-filled galaxies.  This is the famous “missing-mass” problem, which has been dogging astronomers for well over 50 years.  What this is apparently telling us is that there is unseen matter in the universe, amounting to 100 times what we can see with all our telescopes.  This result is consistent no matter which cluster we look at.  All clusters appear this way.

	This “missing matter” is inferred by observations of individual galaxies as well.  Here, the stars in these galaxies appear to be moving in such a way as to require, again,  about 100 times more mass in any galaxy than we can see in gas and stars.

	More recently, though, this extraordinary result is being called into question.  For if we just assume that this missing matter exists, we are forced to predict other consequences in the appearances and motions of stars and galaxies that don’t really seem to fit the observed data.  This has led some scientists to question our assumptions about the nature of the law of gravity itself.  Our understanding of this law is definitely incomplete (since, for example, we have no quantum theory of gravity at all, and therefore have no understanding about the way masses behave gravitationally on the scale of atoms), so a new form of this law may allow us to explain this puzzle without the  need for “missing” mass.  In any event, we are left with an amazing legacy: either 99% of the universe is in the form of some exotic type of unseen matter, or else the known forces in the Universe are going to be radically changed.  Stay tuned for the latest in this intensively studied area of research.