Sunday, November 8, 2015

30 | Radio Astronomy Lab, Galactic Rotation

In this lab, we will be using the millimeter-wave telescope at the Harvard-Smithsonian Center for Astrophysics to ultimately determine the mass of the Milky Way interior to the Sun's orbital radius. We will accomplish this by measuring the Doppler shifts of the 115 GHz spectral line of CO. This line corresponds to a strong transition in the rotational energy levels of carbon monoxide. At this frequency, we observe photons emitted by the transition from the first energy level back down to the ground state.

 Here are some considerations for observing with the radio telescope:

What is going to be the typical integration time per point?

To calculate a reasonable integration time, we need to know a but about noise, as we hope to have an integration time long enough that our signal-to-noise ratio (SNR) is quite high. The root-mean-square error \(\sigma\) is given by:
\[\sigma = \frac{T_{\text{sys}}}{\sqrt{\tau \Delta \nu}}\]
Where \( T_{\text{sys}}\) is the noise of each sample, \(\tau\) is the integration time, and \(\Delta \nu\) is the channel width.
We will define the signal strength to be \(T_A\). Therefore, the SNR is defined as:
\[\text{SNR} = \frac{T_A}{\sigma}\]
If we solve this for \(\sigma\) and plug it into the equation we found above, then solve for \(\tau\), we get:
\[\tau = \frac{T_{\text{sys}}^2 (\text{SNR})^2}{T_A^2 \Delta \nu}\]
We are told that \(T_{\text{sys}}\) is about 500 K and that the average signal peak will be about 2 K. Furthermore, we will consider using the 0.5 MHz wide channel. As for the SNR, let's calculate the integration time that corresponds to an average error of 1%, meaning SNR=100.
\[\tau = \frac{(500 \text{ K})^2 (100)^2}{(2\text{ K})^2 (0.5\times 10^6 \text{ s}^{-1})} = 1250 \text{ s} \]
This means that a good integration time would be about 21 minutes.

Over what range of longitude do you plan to observe?

According to the lab manual, we will be observing at Galactic longitudes of 10° to 70°.

How many positions do you plan to observe?

The manual seems to suggest that we will be observing four giant molecular clouds, so we will be observing at four positions.

At what LST are you going to start observing? At what EST?

We will probably begin our observations 1:00pm EST on (for example) November 10. This corresponds to about 15:34 LST.

Are you going to position switch or frequency switch to "flat field" the spectrum? 

We want the flat-field correction to correspond to the sensitivity of the CCD at the specific frequency at which we are observing. As a result, we do not want to switch the frequency to flat field our images. Instead, we can perform a "position switch" to do this. 

Above, we started Problem 1 on Worksheet 9.2, so let's finish it:

(b) Find an equation for the velocity resolution \(\Delta v\) km/s that corresponds to a channel width of \(\Delta \nu\) MHz at the frequency of \(^{12} CO\), \(\nu = 115.271 GHz\). Write this in the form:
\[\Delta v = \text{____} \text{km/s} \left(\frac{\Delta \nu}{1 \text{ MHz}}\right)\]

To do this, we can use the Doppler equation, which tells us:
\[\Delta v = c \frac{\Delta \nu}{\nu}\]
To find the equation in the desired form, we simply need to multiply the speed of light \(c\) (\(3\times 10^5 \text{ km/s}\)) by \(10^6\), which corresponds to the MHz term we put in the denominator. This gives us:
\[\boxed{\Delta v = 3.8\times 10^{11} \text{km/s}\left(\frac{\Delta \nu}{1 \text{ MHz}}\right)}\]

I've shown solutions very similar to (a) and (c) above.

1 comment:

  1. (b) is way off. It should just be ~2.6 km/s. You calculate \( \Delta v \) for a MHz and then scale it by \( \nu \) in MHz.