\section{Theoretical foundation}

The actual experiment we performed consisted of two parts. In the
first part, we calibrated our equipment by finding the effective
temperature of the Sun in the frequency region near 21~cm. The second
part was the actual measurement of the 21 centimeter hydrogen line
spectrum of the galaxy.

A radio telescope dish antenna functions as a circular aperture.  Its
power response is therefore a Bessel function.  The full width at half
power of this Bessel function is termed the {\em beam angle} of the
telescope, $ \theta_{b} $, and is a measure of the linear angle
subtended by the telescope. Our telescope selects frequencies around
21 centimeters, so we can determine the effective temperature of the
Sun at 21 centimeters.  We need only measure the antenna power of the
telescope when it is aimed at the sun, multiply this power by a
conversion factor.


In the second part of the experiment, we measure the radio line at 21
centimeters, produced by spin-flip transitions of monatomic hydrogen.
The ground energy state of monatomic hydrogen is affected by the
interaction between the electron and proton spins.  The energy of the
state in which the proton and the electron have anti-parallel spins is
a little lower than that of the parallel spin state.

Collisions between atoms, can exite an atom of hydrogen to the
parallel spin state.  Since this state has a slightly higher
energy than the other one, the atom will try to change spin alignment
and emitting a photon with energy equal to the difference between the
two energy levels.  (This corresponds to a frequency of 1420~MHz or
wavelength of 21.12~cm.)  This transition is commonly known as the
{\em spin-flip transition}.

Because of the nature of the coupling involved, though, the spin-flip
transition has extremely low probability.  In even the best laboratory
conditions available on Earth, an atom in the parallel-spin state will
leave that state through a collision long before the spin-flip
transition can occur.  However, the density of interstellar medium is
sufficiently low so that the collisions are fairly rare.  Since the
amount of hydrogen in the universe is enormous, there will be some spin-flip
transitions despite their low probability. Therefore, we will be able to
observe photons with wavelengths close to 21~cm.

Our ultimate goal is to derive the velocity curve of the galaxy
interior to the sun's orbit. We accomplish this by observing the
Doppler shift of the 21 centimeter line. Due to the rotation of the
galaxy, we expect to see a spread of Doppler shifts corresponding to
the various orbits that intersect our line of sight (see figure
\ref{rot}). The hydrogen cloud with the greatest Doppler shift will
correspond to an orbit tangential to our line of sight.        

We must determine the frequency at which we observe the 21 centimeter
line at its greatest Doppler shift.  This frequency corresponds to the
hydrogen cloud source at point P on figure \ref{rot}, the cloud that
is moving away from us with the greatest velocity.  From simple
geometry, we know that this cloud will rotate around the center of the
galaxy with an orbital radius of

\begin{figure}[ht]
\begin{center}
\leavevmode
\epsfxsize=0.5\hsize
\epsfbox{/mit/bert/Class/JuniorLab/paper/rotation.eps}
\end{center}
\caption{Determining the radial dependence of the velocity of galactic disk}
\label{rot}
\end{figure}


\begin{equation} 
R = R_\sun \sin\lambda \label{radius}
\end{equation}

In this equation, $R_\sun$ is the distance from the sun to the center
of the galaxy, and $\lambda$ is the angle of our line of sight in
galactic coordinates.  Using the well-known formula for Doppler shift,

\begin{equation}
\nu' = \frac{1 -\left(\frac{u}{v} \right)} {\sqrt{1 - \left( \frac{u}{c} \right)^{2} }}
\label{dopplar}
\end{equation}
 
we can find the velocity at which the hydrogen cloud is receeding from
us, in terms of the unshifted and shifted frequencies: 

\begin{equation}
u = \frac{\nu^{2} - \nu'^{2}}{\nu^{2} + \nu'^{2}} \label{velocity}
\end{equation}

If we call the rotational velocity of the sun around the center of the
galaxy $\omega_\sun$, and the rotational velocity of the hydrogen
cloud $\omega$, then the relative rotational velocity of the hydrogen
cloud is $(\omega-\omega_\sun)$. The tangential velocity is simply the
radius times the rotational velocity, so we can combine this equation
with equation \ref{radius} to get:

\begin{equation}
\Delta V_{\rm max} = (\omega-\omega_\sun) R_\sun \sin\lambda.
\label{deltav}
\end{equation}

Finally, we substitute our expression inovlving the Doppler shifted
frequencies for $\Delta V_{max}:$

\begin{equation}
\frac{\nu^{2} - \nu'^{2}}{\nu^{2} + \nu'^{2}} = (\omega - \omega_\sun) R_\sun \sin\lambda 
\label{combined}
\end{equation}

Rewriting equation \ref{combined}, we get:

\begin{equation}
\omega=\frac{1}{R_\sun \sin\lambda}\;\frac{\nu^{2}-\nu'^{2}}{\nu^{2}+\nu'^{2}}
\label{final}
\end{equation}

Thus, from the value of the maximum Doppler shift, we can determine
the angular velocity of a hydrogen cloud at radius 
$R = R_\sun\sin\lambda.$
\par
In this derivation, we have not taken into account the Earth's
rotation around the the sun, or the sun's extra-rotational velocity.
Appendix \ref{corrections} describes our corrections to the raw data
to account for these effects.






