Telescopes in the night, but actually used in the day

We are bounded in a nutshell of Infinite space: Blog Post #30, Worksheet # 9.2: Telescopes in the night, but actually used in the day

1. Noise properties of a single dish radio telescope

Radio telescopes (or radiometers) work like a standard radio antenna or satellite dish, converting an incoming radio signal into a pattern of oscillating voltages through the interaction of the wave with the electrons in the antenna. A radiometer receives a large number of signals, but filters allow only a certain chunk of the spectrum through. Our telescope is what has called a “heterodyne” receiver, which means that it takes the incoming signal at $\nu_0$, and mixes it down with a local oscillator of frequency $\nu_{LO}$ to produce a signal centered at a lower new frequency $\nu_{IF} = \nu_0 - \nu_{LO}$. The system is designed so that νif is a constant. So, when we “tune” the telescope, we are really setting $\nu_{LO}$. This signal is then fed into the filter bank and split into $N_{chan}$  channels of width $\Delta\nu_{RF}$. The power in each channel is plotted, and this is called a spectrum. Note that $T_{sys}$ is called the “system temperature” and has units of Kelvin; however, you should think of $T_{sys}$ as a measurement of noise and not a physical temperature.

When doing this, the signal has very particular noise properties. Recall that the root-mean-square (rms) error decreases as a function of the number of samples $\sigma \sim 1/\sqrt{N}$. For spectral line observations like ours, in a radiometer, each sample has noise $T_{sys}$, and the number of samples in a channel goes up with increasing channel width and integration time. The noise is characterized by the ideal radiometer equation.

Precisely, if I have a receiver with a system temperature $T_{sys}$, channel width $\Delta\nu$, and integration time $\tau$ , the radiometer equation is $$\sigma = \frac{T_{sys}}{\sqrt{\tau ~\Delta \nu}}.$$

(a) What is the expression for the signal to noise ratio (SNR)? The signal has a brightness temperature $T_{A}$.

(b) Find an equation for the velocity resolution $\Delta\nu$ km/s that corresponds to a channel width of $\Delta\nu$ MHz at the frequency of $^{12}CO, \nu =$115.271 GHz (2.6 mm). Write this in the form $$\Delta\nu = \_\_\_\_ km/s \left(\frac{\Delta\nu}{1~MHz}\right)$$  

(c) What integration time would be needed to detect the peak of $^{12}CO$ with SNR = 10 if I use a filter bank with 256 channels that are 0.5 MHz wide. $T_{A}$ for CO is about 2-3 K and $T_{sys} = 500 K$. (Hint: The peak appears in 1 channel. Do the algebra first to get the final expression and then put in the numbers.)

2. Resolution of a single dish radio telescope

The spatial resolution of a telescope is $\theta = 1.2\frac{\lambda}{D}$ , where D is the diameter of the dish, and $\lambda$ is the observing wavelength. In radio astronomy, $\theta$ 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 $\nu$ in GHz, and diameter D in meters. Use a calculator to determine this to 1 decimal place. Write an equation of the form $$\theta_{HPBW} = \_\_\_\_ degree \left(\frac{\nu}{GHz}\right)^{-1} \left(\frac{D}{m}\right)^{-1}$$

(b) What is the beamwidth, in arcminutes, for the CfA 1.2 m telescope at the $^{12}CO$ frequency. (to 1 decimal place)

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

3. Geometry of the experiment

Below is a schematic of the experiment. A molecular cloud a distance R from the galactic center has a velocity $V_{cir}$. When we observe this cloud, we detect a different velocity, $V_{obs}$, which is the combination of our velocity towards it and its velocity away. Note: We are using R to denote the distances because the sources are some radius from the Galactic center.

(a) For our experiment we will take a spectra which will have velocity components from multiple clouds along the line of sight. Looking at the figure above, which cloud will have the highest velocity and why?

(b) Plot, very roughly, the velocity of the clouds versus their distance along the line sight, what would the shape be?

(c) The velocity which we measure using the $^{12}CO$ line is the combination of the cloud’s velocity around the galactic center and the projection of the Sun’s velocity along our line of sight. What is the equation for the velocity we measure $V_{max}$?

(d) Write the equations for the rotational velocity, $V_{cir}$and $R_{tan}$in terms of the quantities we know - $V_{\odot}$, $V_{\odot}$, l, and $V_{max}$ .

(e) What is the equation for the mass profile of the galaxy? The ultimate goals of this lab are to estimate the total gravitational mass of the Galaxy within the Suns orbit and to infer something about how that mass is distributed in Galactic radius. To do this we will determine the so-called Galactic rotation curve by measuring the velocities of molecular clouds orbiting in the Galactic potential.

1. (a) For this post, we will be exploring the nature of a radio telescope and how these can be used and their slight limits corrected by knowing exactly what to look for and what effects to mitigate. First of all, we’ll find the ratio of the signal to the noise. So if the noise has already been defined as $$\sigma = \frac{T_{sys}}{\sqrt{\tau ~\Delta \nu}},$$ and the signal is defined as $T_A$, then it is simply a matter of dividing one by the other. So the Signal to Noise Ratio is simply: $$\frac{T_A}{\sigma}$$ $$\frac{T_A}{T_{sys}} \sqrt{\tau~\Delta\nu}.$$

(b) Next, we are asked to find the relationship between velocity and frequency, something that sounds very familiar to red shift. In fact, from the red shift equation, with a bit changed, we have: $$\frac{\nu_{obs} - \nu_{em} }{\nu_{em}} = \frac{v}{c} ,$$ which can be interpreted as: $$\Delta v = c\frac{\Delta\nu}{\nu}.$$ So taking this equation and plugging in the necessary values as seen above, we have: $$\Delta v = 3 \times 10^5 km/s \cdot \frac{\Delta\nu}{115.271 \times 10^9 Hz} ,$$ which yields: $$\Delta v = 2.6 \times 10^{-6} \Delta\nu \frac{km/s}{Hz} \times \frac{10^6 Hz}{MHz},$$ adding in the conversion factor to achieve the correct units as asked, we finally know that: $$\Delta v = 2.6 \frac{km/s}{MHz} \Delta\nu .$$

(c) Now, knowing how the velocity resolution equation works and how the SNR operates, we can use them to find the next factor, the time of integration for $^{12} CO$, $\tau$. Using the SNR equation, we have:    $$SNR = \frac{T_A}{T_{sys}} \sqrt{\tau~\Delta\nu},$$ and solving for the integration time, $$\tau =\frac{SNR^2\frac{T_{sys}^2}{T_A^2}}{\Delta \nu} .$$ By plugging in the values given to us by the question, we have: $$\tau = \frac{10^2\frac{500K^2}{2.5K^2}}{0.5 \times 10^6 Hz},$$ which simplifies down to: $$\tau = 8 s.$$

2.(a) Here, we are given a basic piece of information that essentially tells us what to do in order to find the final angle. Taking into consideration the $\theta = 1.2\frac{\lambda}{D}$, it can be turned into $\theta = 1.2\frac{\frac{c}{\nu}}{D}$, so by plugging this into the equation given, we can make it in the correct dimensions by multiplying by the correct orders of magnitude. So: $$\theta_{HPBW} = \_\_\_\_ degree \left(\frac{\nu}{GHz}\right)^{-1} \left(\frac{D}{m}\right)^{-1} ,$$ $$\theta_{HPBW} = 1.2\frac{\frac{c}{\nu}}{D} \left(\frac{\nu}{GHz}\right)^{-1} \left(\frac{D}{m}\right)^{-1} ,$$   $$\theta_{HPBW} = \frac{1.2 \cdot 3\times 10^8 m/s }{ \nu D} \times \frac{1~GHz}{10^{-9} Hz}  \times \frac{180~degrees}{\pi ~radians} ,$$ which it all simplifies to: $$\theta_{HPBW} = \frac{20.36~degrees~ m\cdot GHz}{\nu~D} .$$

(b) Now we take the equation we just found and plug in the values of D and $\nu$ they give. So the equation turns into: $$\theta_{HPBW} = \frac{20.36~degrees}{\nu~D} ,$$ $$\theta_{HPBW} = \frac{20.36~degrees~ m\cdot GHz}{1.2 m \cdot 115.271 GHz},$$ $$\theta_{HPBW} = 0.149 ~degrees = 8.9 ~arcminutes.$$

(c) Now that we know the angle with which we are observing, we can use it in the basic angle to distance relation that is parallax. From the information given by the question, we know that the distance from the object is 8.5 kpc, and if we have the distance and the angle, then we can know the width of our field of view. From earlier problems, we found how $$angle_{(in~ arcseconds)} =\frac{1 AU}{D_{in ~pc}},$$ but the relationship can be rewritten as: $$angle_{(in~ arcseconds)} \cdot D_{(in~pc)} = width ~of ~view ~in~ AU ,$$ and we know the values that need to be plugged in and the conversion factors to be used: $$(8.9 ~arcminutes \times 60) \cdot (8.5 \times 10^3 pc) = width ,$$ $$4.539 \times 10^6 AU  \ times \frac{1 ~pc }{2.05 \times 10^5 AU}= width,$$ $$Linear~Dimension_{(Width)} = 22.14 pc.$$

3. (a) After looking at the diagram for a bit, it becomes obvious that the object with the highest speed away from us, the object with the greatest Red Shift, is undoubtedly object B, since all the velocity we observe is pointed tangentially to its orbit path, thus making it the fastest among all the objects in our line of sight.

(b)

(c) Taking into consideration all the factors that are changing how we are perceiving the objects on the other side of the galaxy,  the best equation that relates the max speed of any object is: $$V_{Max} = V_{cir} - V_\odot \sin(l) ,$$ which recognizes how our velocity impacts the way we perceive the velocity of the observed object.

(d) By simply rewriting the above equation, we can solve for the circular velocity, which is simply: $$V_{cir} = V_{Max} + V_\odot \sin(l) ,$$ and the tangential radius is simply the component of the radius from the galactic core to the sun, such that: $$R_{tan} = R_\odot \sin(l) .$$

(e) And finally, to understand the mass profile for the galaxy, we must go back to a post from a couple of months ago ( http://ay17-rcordova.blogspot.com/2015/09/the-milky-way-is-now-sphere-trust-me.html ) where the velocity profile is described as: $$v(r) = \left(\frac{G M_{enc}}{r}\right)^{1/2} ,$$ and to find the mass profile, a bit of rearranging is all that is necessary: $$M_{enc} = \frac{V_{cir}^2 R_{tan}}{G}.$$