Blog 30: Galactic Rotation

Next week we will begin our second lab of the semester. We will use the millimeter-wave telescope, a radio telescope, in order to observe giant molecular clouds in the Milky Way. This is a lot of new information at once, so we are going to answer some questions before we actually head to lab.

Why? - We are observing GMCs in order to find their redshifts. From redshift, we can find the velocity of a GMC along our line of sight. With that velocity, we can find the circular velocity of a GMC rotating in the Milky Way using \[ V_{cir} = V_r(max) + V_{\odot} sin(l) \] where Vr is the radial velocity of the GMC moving away from us and the second term is the velocity of the sun in the direction of the cloud.

In previous blog posts  we explored why the rotation curve of the galaxy is important in astronomy. The basic idea is that we observe objects to be rotating too quickly, too far out in the galaxy just based on visible matter. This is our main indicator that there is another type of dark matter in the universe. Additionally, we can learn more about the Milky Way's visible matter from this experiment. For example, we can estimate the number of stars in the galaxy. 

How? - The telescope will detect radiation at the wavelength near 2.6 mm, or a frequency of 115.271 GHz. This wavelength corresponds with a strong emission line of Carbon Monoxide. We will obtain this spectrum from the telescope's spectrometer, one of many of the telescope's parts among an antenna for collecting and focusing the signal, a liquid helium cooler, and a computer for recording and interpreting data.

What is going to be the typical integration time per point? - What integration time would be needed to detect the peak of 12CO with SNR = 10 if I use a filter bank with 256 channels that are 0.5 MHz wide. TA for CO is about 2-3 K and Tsys = 500K?

In worksheet 9.2 we found \[ \tau = \frac{ SNR^2 T_{sys}^2 }{T_A^2 \Delta \nu } \] \[ \tau = \frac{ (10)^2 (500)^2 }{(2.5)^2 (0.5 \times 10^6) } \approx 8 \: seconds \]

Over what range of longitude do you plan to observe? - We are planning on observing from \(10^{\circ} \: \text{ to } 70^{\circ} \). 

How many positions do you plan to observe? - We plan on observing 4 GMCs, so we will most likely observe 4 positions.

At what LST are you going to start observing? At what EST? - I have been assigned to Monday's lab from 11:00 - 3:00 EST. Let's say that means we start observing at 11:30 EST. The LST on that date and time at our longitude (71.11 degrees west) would be 15:55.

Additionally, we can solve question number two on the worksheet to learn more about the experiment.  

Resolution of a single dish radio telescope: The spatial resolution of a telescope is θ = 1.2 λ/D , where D is the diameter of the dish, and λ is the observing wavelength. In radio astronomy, θ is also known as the half-power beam width (or full-width half-max of the beam). 

(a) Find an equation for the beam width, in arcminutes, of a single-dish radio in terms of frequency ν in GHz, and diameter D in meters. Use a calculator to determine this to 1 decimal place. Write an equation of the form \[ \theta = degrees \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] We can solve this by redefining theta: \[ \theta = \frac{c}{\nu D} = \frac{1.2 (3 \times 10^8) }{(1 \times 10^9)(1)} = 0.36 \: rad \left( \frac{180^{\circ}}{ \pi rad } \right) = 20.6^{\circ} \] \[ \theta = 20.6^{\circ} \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] 

(b) What is the beamwidth, in arcminutes, for the CfA 1.2 m telescope at the 12CO frequency?

Using the formula we just found, we know \[ \theta = 20.6^{\circ} \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] \[ \theta = 20.6^{\circ} \left( \frac{ 115.271 \: GHz}{GHz} \right)^{-1} \left( \frac{1.2 \: m}{m} \right)^{-1} = 0.149^{\circ} = 8.95 \: arcminutes \] 

(c) What linear dimension in pc does this correspond to at the Galactic center (8.5 kpc)?

We can call on our trusty parallax formula for this problem: \[ d = \frac{1}{ \theta} \] Where 8.95 arcminutes is 537 arcseconds, so \[ d = \frac{1}{ 537} = 1.9 \times 10^{-3} \: pc \] Now we should be all set to start lab.

Correction: The parallax is not a terribly useful number in this case. Rather the question is asking for resolvable size, which we could find with simple trigonometry. \[sin( \theta) = \frac{l}{8.5 \: kpc} \] With the small angle approximation, this is \[ l \approx 0.149 \cdot \frac{\pi}{180} \cdot 8,500 \approx 22 \: pc\]