With this software, you can analyze fits images, found online or stored locally on your computer. For this lab, I used the file found on Professor Johnson’s website with a binary system. Loaded into the Aladin viewer, a FITS file is represented visually as a field of stars:
From this specific FITS file, we can determine the location of the targets: from Aladin, I found the following for each source:
| Source | RA | DEC | RA (degrees) | DEC (degrees) |
| A | 18:05:27.73 | +2:29:48.0 | 270.5455 | 2.4967 |
| B | 18:05:27.18 | +2:29:44.4 | 270.5363 | 2.4956 |
We have found the angular separation of the binaries to be 0.009 degrees. This distance was found using the distance tool on Aladin (just click on your two sources, and the program computes the exact distance between them!). Also, it can be found manually using the source positions.
Given that the angular resolution is based on the ‘airy’ pattern – it would be useful to know, what is an airy pattern? The airy pattern refers to the lighter and darker regions, where different strengths of intensities of light are visible.
[I start to go down the wrong road here… skip ahead if you want!]
The sinc function, which is the Fourier transform of the box function, is defined by:
http://www.codecogs.com/eqnedit.php?latex=\bg_black\textup{sinc}(x)=\textup{sin}(πx)/(πx) \textup{ if } x ≠0, \textup{and sinc}(0) = 1" target="_blank">
You can load the Simbad database into Aladin as well. Also, you can select individual sources and see an image of each source. If you do this with this data file, you get what looks like a giant region of exposure.
What I wanted to do next was to find the sinc function of the light from these images. Theoretically, you could do this by extracting a spectrum of the region containing the point source, and plotting how intensity oscillates as you move further away from the center (maximum). I played around with Aladin for a ridiculously long amount of time until I found a feature that gives you the full width at half maximum in each direction. To find this, I used the “Phot”[ometry] command to select a circular region containing the point source, then examined the table of values to find the full width values. This was harder for me than it should have been, because I wasn’t sure exactly what I was looking for. I only vaguely remembered that “full width at half maximum” can refer to a sinc function.
The values I found for full width at half maximum for each source were:
| Source | FWHM (appx) |
| A | 13 |
| B | 11 |
We want to see how the full width at half maximum (FWHM) relates to other qualities of the telescope.
WAIT! At this point, I asked Professor Johnson if I was going about this project the right way, and he suggested a much easier way to do it. Instead of using the FWHM (which I am still unclear if this would actually work), he suggested I change the display range and measure the airy rings.
I was having some trouble finding the way to change the display range (sad, I know) until I found that there is an undergraduate mode for the Aladin software! It looks like the undergraduate mode if just the regular version with a bunch of features locked out, but it’s a little less overwhelming to start with.
If you zoom in, you can see what look like airy rings. These are the distances at which there appears to be a dark spot. We can see the cause of these if we look at the function of light reflected on a screen. The airy ring, therefore, occurs at this first node. On the image, we can measure how far it is between the center (max) and this first mode (min). The image gives a value of approximately 0.1 degrees.
How can we use this to find the angular resolution of the image? It is given that the image is in K-band. This means that the frequency of light is between 20 and 40 GHz. This corresponds to 7.5*10^-1 to 1.5 cm for the wavelength. Using the average-ish value, we can say that lamba equals 1.15 cm. Using the formula:
Substituting theta with the value found by the airy ring method. Note that these values are in radians and nanometers. Also, the small angle approximation is used here.
So,
This implies that the diameter of the telescope is roughly 10 meters. Which telescopes do we know of that have an aperture of 10 meters? The first one that springs to mind for me is Keck. However, how could this be? The K-band is infrared. Now, there are two ways to figure out from which telescope this file was observed. One way is to look at the declination of the pair.
The source is at about +2:30. This means that the object is just slightly north of the celestial equator. Also, I found the observation date in the image – 4/27/2010, at 11:28 – so we can check if the pair would have been visible from Keck at that time. I will assume the time given is PM, rather than military time (though it is possible there is some mix-up with times zones)? 11:30 pm is probably an acceptable approximation, however). The right ascension of the image is around 18 hours. At the vernal equinox, the RA is 0 hours at noon. So, we must calculate how much that changes in a month.
Each sidereal day is four minutes shorter than a solar day. So, the sidereal time will be at about 3:00 at noon on April 27 (this is a rough estimation). At 18:00, sidereal time, it will be roughly be 3:00 AM April 28. This is the time that the source would be directly overhead. It would totally be possible to see this source from Keck at 11:28 PM, April 27.
If you look at this paper: http://www.astro.cornell.edu/academics/courses/astro233/symp06/symp06.pdf, you will see that there is actually a way for Keck to observe in infrared! The NIRC is a near-infrared camera that was on Keck at one point. NIRC2 would have been operational on April 27, 2010, when this data was taken.
Great detective work!
ReplyDeleteI think you made a couple of errors that managed to result in a reasonable answer. This happens even to the best astronomers.
First, those double-tick marks ('') in your airy pattern zoom-in stand for arcseconds, not degrees. Recall that the typical seeing is of order 1 arcsecond, which is 1/(60*60) = 0.00027 degrees, and that AO makes this much smaller.
Also, the "Theta" term in your lambda/D equation refers to the distance between nulls on either side of the peak. Multiply your value by 2 and make sure it's in radians, since its in the argument of a trig function.
Finally, the Ks-band in the FITS header info refers to the JHKs photometric system. Do a Google search to learn about these near-IR bands, which are much smaller wavelengths than cm.
Try again. You may end up with the same answer or you may find something else. I'm interested to see what you get!
AH!! I knew the arcsecond thing, too :( I did it that way at first, but I kept getting a 40 meter aperture, so I thought I was doing something wrong. But compounded with my other errors, that makes sense. I will fix that by Wednesday!! (I have a paper to finish now...)
ReplyDeleteThis is fantastic, Juliette! It's really good that you're going through this thought process now because I didn't start to learn all these things (what is K band? there are TWO of them?!) until I'd already been at grad school for a while, and boy was it confusing!
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