\subsection{Unraveling a fixed pitch as a Lorentz boost \label{ss:lorentzboost}}
Consider again the muon circulating in a uniform magnetic field $\bfB=B_z\bfhz$. The spiral trajectory of the muon, traveling in a clockwise direction 
has velocity components parallel and perpendicular to the magnetic field
 ${\bm\beta} = {\bm\beta}_{xy}(\cos\omega t)\ \bfhx -\sin(\omega t)\ \bfhy) +  \beta_z\bfhz$ (see ~\ref{fig:spiral}). We translate by a Lorentz boost $\beta_z$ in the z-direction 
into the frame where $\beta_z=0$. The momentum transverse to the boost, and the magnetic field (parallel) to the boost are invariant.
In the boosted frame the 
velocity is
\begin{equation}
{\bm\beta} =|{\bm\beta}|(\cos\omega^b_c t \bfhx - \sin\omega^b_c t \bfhy)
\end{equation}
with
cyclotron frequency is $$\omega_c^b = \frac{eB_z}{mc\gamma^\prime}.$$
According to the BMT equation (Equation~\ref{eq:j170}), the polarization of
the muon in the boosted frame will 
precess about the $z^\prime$-axis with frequency 
$$\omega^b_s =\frac{eB_z}{mc}(a_\mu+\frac{1}{\gamma^\prime})$$
such that
$$\bfs = |\bfs|(\cos\omega^b_s t \bfhx - \sin \omega^b_s t \bfhy)$$
where the superscript $b$ indicates the quantity measured by an observer in the boosted frame.

Then, the quantity of interest, 
$${\bm\beta}\cdot\bfs = -|{\bm\beta}||\bfs|(\cos\omega^b_c t \cos\omega^b_s t + \sin\omega^b_c t \sin\omega^b_s t) 
= -|{\bm\beta}||\bfs|\cos(\omega^b_c - \omega^b_s)t$$
and $\omega^b_a = \omega^b_s - \omega^b_c$. 

The observer in the boosted frame measures the frequencies $\omega^b_a,\omega^b_s,\omega^b_c$  with her local clock.
Now let's transform back to the lab frame.  
From the point of view of the observer in the lab frame, the time measured by the clock in the boosted frame is dilated. According to the
clock in the lab frame everything is running slower in the boosted frame, including the frequencies of revolution and precession by a factor $1/\gamma$ where
$\gamma =1/\sqrt{1-\beta_z^2}$. As measured in the lab frame
\begin{eqnarray}
\omega_a({\rm lab})&=& \omega^b_a\sqrt{1-\beta_z^2}\label{eq:lorentz}\\
&=& \omega^b_a(1-\psi^2\beta_{xy}^2)^{1/2}\nonumber\\
&\sim& \omega^b_a(1-\onehalf\psi^2)\nonumber
\end{eqnarray}
where $\psi$ is the pitch angle and in the limit where $\psi\ll 1$ and $\gamma \gg 1$.

\begin{comment}
The Lorentz transformation corresponding to the boost in the z-direction is
\begin{equation}
A(\beta_z)=\begin{pmatrix}\gamma & 0 & 0 & -\gamma\beta_z\cr
                          0 & 1 & 0 & 0\cr
                          0 & 0 & 1 & 0 \cr
                          -\gamma\beta_z & 0 & 0 & \gamma\end{pmatrix}
\end{equation} where $\gamma = (1-\beta_z^2)^{-1/2}$
The polarization 4-vector of a particle with velocity ${\bm\beta^\prime}$ is
\begin{equation}
S^\prime = \begin{pmatrix}{\bm\beta^\prime}\cdot\bfs^\prime\cr s^\prime_x \cr s^\prime_y \cr s^\prime_z\end{pmatrix}
\end{equation}
Note that the zeroth component of the polarization 4-vector, namely ${\bm\beta^\prime}\cdot\bfs^\prime$, is the quantity of interest. 
\begin{equation}
S =A^{-1} S^\prime = \begin{pmatrix} \gamma{\bm\beta^\prime}\cdot\bfs^\prime\cr s^\prime_x\cr s^\prime_y \cr -\gamma\beta_z({\bm\beta^\prime}\cdot\bfs^\prime)\end{pmatrix}=
\begin{pmatrix}{\bm\beta}\cdot\bfs \cr s_x \cr s_y \cr s_z\end{pmatrix}
\end{equation}
Therefore since
$$S^\prime = |\bfs^\prime|\begin{pmatrix}|{\bm\beta^\prime}|\cos(\omega^\prime_a t^\prime)\cr \cos\omega^\prime_s t^\prime \cr \sin\omega^\prime_s t^\prime 
\cr 0\end{pmatrix}=\begin{pmatrix}{\bm\beta^\prime}\cdot\bfs^\prime\cr s^\prime_x\cr s^\prime_y \cr s^\prime_z\end{pmatrix},$$
in the lab frame with finite $\beta_z$,  
\begin{equation}
S = |\bfs^\prime|\begin{pmatrix}|\gamma{\bm\beta^\prime}|\cos(\omega^\prime_a t^\prime)\cr \cos\omega^\prime_s t^\prime \cr \sin\omega^\prime_s t^\prime \cr -\gamma\beta_z
|\bm\beta^\prime|\cos(\omega_a^\prime t^\prime)\end{pmatrix}
=\begin{pmatrix}{\bm\beta}\cdot\bfs\cr s_x\cr s_y\cr s_z\end{pmatrix},\label{eq:slab}
\end{equation}
where again, $\gamma = (1-\beta_z^2)^{-1/2}$.


And while we're at it
\begin{eqnarray}
\gamma^0\beta^0 &=& A^{-1}\gamma^\prime\begin{pmatrix}1\cr|\bm\beta^\prime|\cos\omega_c^\prime t^\prime \cr |\bm\beta^\prime|\sin\omega_c^\prime t^\prime \cr 0 \end{pmatrix}
=\gamma^\prime\begin{pmatrix}\gamma \cr |\bm\beta^\prime|\cos\omega_c^\prime t^\prime \cr |\bm\beta^\prime|\sin\omega_c^\prime t^\prime \cr -\gamma\beta_z \end{pmatrix}\nonumber\\
\beta_0&=&\frac{\gamma^\prime}{\gamma^0}\begin{pmatrix}\gamma \cr |\bm\beta^\prime|\cos\omega_c^\prime t^\prime \cr |\bm\beta^\prime|\sin\omega_c^\prime t^\prime \cr -\gamma\beta_z \end{pmatrix}=\begin{pmatrix}1 \cr \frac{|\bm\beta^\prime|}{\gamma}\cos\omega_c^\prime t^\prime \cr \frac{|\bm\beta^\prime|}{\gamma}\sin\omega_c^\prime t^\prime \cr -\beta_z\end{pmatrix}\label{eq:blab}
\end{eqnarray}
where we use $\gamma\gamma^\prime = \gamma^0$ (see ~\ref{ss:velocityaddition}).


We measure the time dependence of $\bm\beta\cdot\bfs$. We find that $\bm\beta\cdot\bfs = |\bfs^\prime||\bm\beta^\prime|\gamma\cos\omega_a^\prime t^\prime$. Temporal intervals
on the clock in the lab frame $\Delta t$ are related to intervals in the boosted frame $\Delta t^\prime$ according to $\Delta t^\prime = (1/\gamma) \Delta t$. 
Then $\bm\beta\cdot\bfs 
= |\bfs^\prime||\bm\beta^\prime|\gamma\cos\omega_a^\prime t^\prime =|\bfs^\prime||\bm\beta^\prime|\gamma\cos\omega_a^\prime \gamma t\rightarrow \omega_a = (1/\gamma) \omega_a^\prime $. 
From Equation ~\ref{eq:slab} we see that $s_z$ is in phase with $\bm\beta\cdot\bfs$. The spin seems to rotate about in a plane with axis 
$\bfB_\perp$ tilted out of the x-y plane and about the x-axis. See \ref{ss:pitchanglebetaz} for derivation of relationship between pitch angle and $\beta_z$. 
\end{comment}


\begin{comment}
Next let's substitute our solution into the BMT equation (\ref{eq:j170_mm}). and see what happens.
First compute the cross products using $\bfB=B\bfhz$ and $\bm\beta = \beta_x\bfhx + \beta_y\bfhy + \beta_z\bfhz$.
\begin{eqnarray*}
\bfs\times\bfB &=& s_y B_z\bfhx -s_x B_z\bfhy = |\bfs^\prime|(\sin\omega_s^\prime t^\prime\bfhx- \cos\omega_s^\prime t^\prime\bfhy)B\\
\bfs\times\bm\beta &=& (s_y\beta_z-\beta_y s_z)\bfhx +(s_z\beta_x-s_x\beta_z)\bfhy + (s_x\beta_y-s_y\beta_x)\bfhz\\
&=& |\bfs^\prime|(-\sin\omega_s^\prime t^\prime\beta_z + \frac{|\beta^\prime|}{\gamma}\sin\omega_c^\prime t^\prime\gamma\beta_z|\bm\beta^\prime|\cos\omega_a^\prime t^\prime)\bfhx
+|\bfs^\prime|(-\cos\omega_s^\prime t^\prime\beta_z + \frac{|\beta^\prime|}{\gamma}\cos\omega_c^\prime t^\prime\gamma\beta_z|\bm\beta^\prime|\cos\omega_a^\prime t^\prime)\bfhy
 +\frac{|\beta^\prime||\bfs^\prime|}{\gamma}\sin\omega_a^\prime t^\prime\bfhz\\
&=& |\bfs^\prime|\beta_z(-\sin\omega_s^\prime t^\prime + {\beta^\prime}^2\sin\omega_c^\prime t^\prime\cos\omega_a^\prime t^\prime)\bfhx
+|\bfs^\prime|\beta_z(-\cos\omega_s^\prime t^\prime + {\beta^\prime}^2\cos\omega_c^\prime t^\prime\cos\omega_a^\prime t^\prime)\bfhy
 +\frac{|\beta^\prime||\bfs^\prime|}{\gamma}\sin\omega_a^\prime t^\prime\bfhz\\
\end{eqnarray*}
The BMT equation in terms of components
\begin{eqnarray*}
\der{}{t}\begin{pmatrix}s_x\cr\cr s_y\cr\cr s_z\end{pmatrix}&=& \frac{eB}{mc}\begin{pmatrix}(a_\mu + \frac{1}{\gamma_0})s_y  -a_\mu\frac{\gamma_0}{\gamma_0+1}\beta_z(s_y\beta_z-\beta_y s_z)\cr\cr
                                           (a_\mu + \frac{1}{\gamma_0})s_x - a_\mu\frac{\gamma_0}{\gamma_0+1}\beta_z(s_z\beta_x-s_x\beta_z) \cr\cr
                                            - a_\mu\frac{\gamma_0}{\gamma_0+1}\beta_z(s_x\beta_y-s_y\beta_x)\end{pmatrix}
\end{eqnarray*}
Substitution into the above
\begin{eqnarray*}
|\bfs^\prime|\der{}{t}\begin{pmatrix}\cos\omega_s^\prime t^\prime\cr\cr \sin\omega_s^\prime t^\prime\cr\cr -\gamma\beta_z|\bm\beta^\prime|\cos\omega_a^\prime t^\prime\end{pmatrix}
&=& \frac{eB}{mc}\begin{pmatrix}(a_\mu + \frac{1}{\gamma_0})|\bfs^\prime|\sin\omega_s^\prime t^\prime  -a_\mu\frac{\gamma_0}{\gamma_0+1}\beta_z |\bfs^\prime|\beta_z(-\sin\omega_s^\prime t^\prime + {\beta^\prime}^2\sin\omega_c^\prime t^\prime\cos\omega_a^\prime t^\prime)  \cr\cr
                                           (a_\mu + \frac{1}{\gamma_0})|\bfs^\prime|\cos\omega_s^\prime t^\prime - a_\mu\frac{\gamma_0}{\gamma_0+1}\beta_z |\bfs^\prime|\beta_z(-\cos\omega_s^\prime t^\prime + {\beta^\prime}^2\cos\omega_c^\prime t^\prime\cos\omega_a^\prime t^\prime)\cr\cr
                                            - a_\mu\frac{\gamma_0}{\gamma_0+1}\beta_z( \frac{|\beta^\prime||\bfs^\prime|}{\gamma}\sin\omega_a^\prime t^\prime)\end{pmatrix}
\end{eqnarray*}
Here are the expressions for spin and velocity that are (very nearly) consistent with BMT (integration)
\begin{eqnarray*}
s_x &=& -\cos(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2}\\
s_y &=& \sin(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2}\\
s_z &=&\beta_z\beta^\prime\sin(\omega_at)
\end{eqnarray*}
\begin{eqnarray*}
\beta_x &=& \beta^\prime/\gamma_z  \sin(\omega_c t)\\
\beta_y &=& \beta^\prime/\gamma_z  \cos(\omega_c t)\\
\beta_z &=& \beta_z
\end{eqnarray*}
Then
\begin{eqnarray*}
(\bfs\times\bm\beta)_x&=& \sin(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2}\beta_z - \beta_z\beta^\prime\sin(\omega_at)\beta^\prime/\gamma_z  \cos(\omega_c t)\\
(\bfs\times\bm\beta)_y&=& \beta_z\beta^\prime\sin(\omega_at)\beta^\prime/\gamma_z  \sin(\omega_c t) +\beta_z\cos(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2}\\ 
(\bfs\times\bm\beta)_z&=& -\cos(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2}\beta^\prime/\gamma_z  \cos(\omega_c t)-\sin(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2}\beta^\prime/\gamma_z  \sin(\omega_c t)\\
&=&(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2}\left(\beta^\prime/\gamma_z\cos\omega_a t\right)
\end{eqnarray*}
and
\begin{eqnarray*}
(\bfs\times \bfB)_x &=& \sin(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2} B\\
(\bfs\times \bfB)_y &=& \cos(\omega_s t)(1-(\beta_z\beta^\prime\sin(\omega_at))^2)^{1/2} B
\end{eqnarray*}

That looks pretty complicated. Maybe we should start with something simpler, namely substitution into Equation ~\ref{eq:j}.
First
\begin{eqnarray*}
\bmbetahat\cdot\bfs &=& |\bfs^\prime|\beta^\prime(\frac{1}{\gamma}+\beta_z^2\gamma)\cos\omega_a^\prime t^\prime\\
&=& |\bfs^\prime|\beta^\prime\gamma\cos\omega_a^\prime t^\prime\\
\end{eqnarray*}
In an inertial frame the polarization vector rotates but according to BMT it's length does not change. The normalization that preserves length is
$$N=(1+(\gamma\beta_z\beta^\prime\cos\omega_a^\prime t^\prime)^2)^{-1/2}$$ so that
$$\bfs = N(\cos\omega_s^\prime t^\prime \bfhx+\sin\omega_s^\prime t^\prime \bfhy -\gamma\beta_z\beta^\prime\cos\omega_a^\prime t^\prime \bfhz)$$
Then 
\begin{eqnarray*}
\der{}{t}\bmbetahat\cdot\bfs &=& -N \beta^\prime\gamma\omega^\prime_a\sin\omega_a^\prime t^\prime\left(1 -N^2(\gamma\beta_z\beta^\prime)^2\cos^2\omega_a^\prime t^\prime\right)\\
 &=& -N^3 \beta^\prime\gamma\omega^\prime_a\sin\omega_a^\prime t^\prime
\end{eqnarray*}
\begin{eqnarray*}
\der{}{t}\bmbetahat\cdot\bfs   &=& -\frac{e}{mc}a_\mu  \bfs_\perp \cdot(\bmbetahat\times \bfB + )\\ 
\der{}{t}\frac{\bm\beta\cdot\bfs}{|\bm\beta|}   &=& -\frac{e}{mc}a_\mu\bfs_\perp\cdot(\frac{\bm\beta}{|\bm\beta|}\times \bfB)\\
\der{}{t}\left(|\bfs^\prime||\gamma\bm\beta^\prime|\cos(\omega_a t)\left(\frac{p_\perp^2+p_z^2+m^2}{p_\perp^2+p_z^2}\right)^{1/2}\right)   
&=& -\frac{e}{mc}a_\mu|\bfs^\prime|(\cos\omega_st\bfhx+\sin\omega_s t\bfhy + s_z \bfhz)\cdot\frac{(p_\perp(\cos\omega_c t \bfhx + \sin\omega_c t\bfhy )-m\gamma \beta_z\bfhz)}
{p_\perp^2+ p_z^2}\times \bfB)\\
&=& -\frac{eB}{mc}a_\mu|\bfs^\prime|(\sin\omega_at)\frac{p_\perp}{(p_\perp^2+ p_z^2)^{1/2}}\\
\der{}{t}\left(|\bfs^\prime||\gamma\bm\beta^\prime|\cos(\omega_a t)\right)   
&=& -\frac{eB}{mc}a_\mu|\bfs^\prime|(\sin\omega_at)\frac{p_\perp}{(p_\perp^2+ p_z^2+m^2)^{1/2}}\\
\der{}{t}\left(|\bfs^\prime|\gamma\frac{p_\perp}{(p_\perp^2+m^2)^{1/2}}|\cos(\omega_a t)\right)   
&=& -\frac{eB}{mc}a_\mu|\bfs^\prime|(\sin\omega_at)\frac{p_\perp}{(p_\perp^2+ p_z^2+m^2)^{1/2}}\\
\der{}{t}\left(|\bfs^\prime|\gamma\cos(\omega_a t)\right)   
&=& -\frac{eB}{mc}a_\mu|\bfs^\prime|(\sin\omega_at)\frac{(p^2_\perp+m^2)^{1/2}}{(p_\perp^2+ p_z^2+m^2)^{1/2}}\\
\der{}{t}\left(\gamma\cos(\omega_a t)\right)   
&=& -\frac{eB}{mc}a_\mu(\sin\omega_at)\frac{(p^2_\perp+m^2)^{1/2}}{(p_\perp^2+ p_z^2+m^2)^{1/2}}\\
\der{}{t}\left(\gamma\cos(\omega_a t)\right)   
&=& -\frac{eB}{mc}a_\mu(\sin\omega_at)\frac{\cos\psi(p^2_\perp+m^2)^{1/2}}{(p_\perp^2+m^2\cos^2\psi)^{1/2}}\\
\der{}{t}\left(\frac{\cos(\omega_a t)}{\cos\psi}\right)   
&=& -\frac{eB}{mc}a_\mu(\sin\omega_at)\cos\psi
\end{eqnarray*}
\end{comment}
