\section{theory} \label{theory}

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 dopplar shifts corresponding to
the various orbits that intersect our line of sight (see figure
number). The hydrogen cloud with the greatest Doppler shift will
correspond to an orbit tangential to our line of sight.  
\par
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 (number), 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{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 dopplar 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.

