TN-50 "Muon polarisation in the $\pi_{\mu 2}$ decay}": Pierre Depommier
(Wed, 3 Jan 2001)

Pierre Depommier

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\begin{document}

\title{TRIUMF Experiment E614 \\
       Technical Note No. 50 \\
       {~} \\
       Muon polarisation in the $\pi_{\mu 2}$ decay}

\author{ {~} \\
       Pierre Depommier \\
       {~} \\
       {\sl Department of Physics, University of Montreal}, \\
       {\sl P.O. Box 6128, Station ``Downtown'', Montreal, Quebec, Canada H3C 
       3J7} \\
       {\sl pom@lps.umontreal.ca}
       }
\date {January 3, 2001}
\maketitle

\section{Introduction}
Formulae are given for the kinematics and the muon polarization in the muonic
decays $\pi^+ \to \mu^+ \nu_\mu$, $\pi^- \to \mu^- \nu_\mu$ and the 
corresponding kaon decays, at rest or in flight. The same formulae can be used 
for the electronic decays. 

This work was undertaken in the framework of another experiment (KEK-E246
experiment, ``Search for a transverse polarization of the muon in the
$K^+ \to \pi^0 \mu^+ \nu$ decay''), following extensive discussions with
Peter Gumplinger. It will also be useful for E614 since this experiment makes 
use of muons from $\pi^+ \to \mu^+ \nu$ decays (either from rest or in flight).

As an application, the effect of thermal motion on the muon polarization is
evaluated.

\section{The kinematics}
Since the discussion of the polarization of the muon involves the understanding
of the kinematics, the latter will also be discussed in detail.

\subsection{Notations}
The notations used in this report are given in Table \ref{TAB1}. The notations 
for the angles $\theta_{\pi \mu}$ and $\theta_{\mu \nu}$ are
self-explanatory. 

\begin{table}
\begin{center}
\begin{tabular}{||c|c|c|c||} \hline \hline
Particle                &Pion           &Muon           &Neutrino       \\ 
\hline \hline
Mass                    &$m_\pi$        &$m_\mu$        &   0           \\ 
Energy-momentum 4-vector&$\tilde{p}_\pi$&$\tilde{p}_\mu$&$\tilde{p}_\nu$\\
Total energy            &$E_\pi$        &$E_\mu$        &$E_\nu$        \\
Momentum 3-vector       &$\vec p_\pi$   &$\vec p_\mu$   &$\vec p_\nu$   \\
Velocity                &$\vec v_\pi$   &$\vec v_\mu$   &$\vec v_\nu$   \\
Momentum (modulus)      &$p_\pi$        &$p_\mu$        &$p_\nu$        \\
Velocity (modulus)      &$v_\pi$        &$v_\mu$        &$c$            \\
Polarization 4-vector   & n.a.          &$\tilde{s}_\mu$& not used      \\
Polarization 3-vector   & n.a.          &$\vec \xi$     & not used      \\
\hline \hline
\end{tabular}
\end{center}
\caption{Notations used for the kinematics and polarization}
\label{TAB1}
\end{table}

\subsection{Angular distribution for the muon}
Energy-momentum conservation:
\begin{equation} 
\tilde{p}_\pi = \tilde{p}_\mu + \tilde{p}_\nu \end
{equation}
leads to:
\begin{equation} 
m_\pi^2 + m_\mu^2 - 2 E_\pi E_\mu + 2 \vec p_\pi \cdot \vec p_\mu = 0 
\end{equation}
and to the second degree equation $(a p_\mu^2 + 2 b^\prime p_\mu + c = 0)$:
\begin{equation} 4 p_\mu^2 [ E_\pi^2 - p_\pi^2 \cos^2 \theta_{\pi \mu} ]
- 4 p_\mu (m_\pi^2 + m_\mu^2) p_\pi \cos \theta_{\pi \mu}
+ 4 m_\mu^2 p_\pi^2 - (m_\pi^2 - m_\mu^2)^2 = 0 \end{equation}
with:
\begin{equation} 
a = 4 (E_\pi^2 - p_\pi^2 \cos^2 \theta_{\pi \mu}) 
\end{equation}
\begin{equation} 
b^\prime = - 4 m_\mu E_0 p_\pi \cos \theta_{\pi \mu} 
\end{equation}
\begin{equation} 
c = 4 m_\mu^2 p_\pi^2 - (m_\pi^2 - m_\mu^2)^2 = 4 m_\mu^2 (p_\pi^2 - p_0^2)  
\end{equation}
where:
\begin{equation} 
p_0 = \frac{m_\pi^2 - m_\mu^2}{2 m_\mu} = 39.35 \hbox{ MeV/c} 
\end{equation}
The determinant is:
\begin{equation} 
\Delta^\prime = 16 m_\mu^2 E_\pi^2 (p_0^2 - p_\pi^2 \sin^2 \theta_{\pi \mu}) 
\end{equation}
$\vec p_0$ is the transition momentum. The corresponding total energy is:
\begin{equation} 
E_0 = \frac{m_\pi^2 +m_\mu^2}{2 m_\mu} = 145.01 \hbox{ MeV} 
\end{equation}
and the corresponding kinetic energy:
\begin{equation} 
T_0 = \frac{(m_\pi - m_\mu)^2}{2 m_\mu} = 5.44 \hbox{ MeV} 
\end{equation}
If there are roots, their product has the sign of $c/a$, hence the sign of 
$p_\pi - p_0$, and their sum has the sign of $(- b^\prime/a)$, hence the sign 
of $\cos \theta_{\pi \mu}$ (note that $a$ is always positive).

There are three cases (note that $\sin \theta_{\pi \mu}$ is positive by 
definition):
\begin{itemize}
\item{$p_\pi < p_0$ 

There is no restriction on the angle $\theta_{\pi \mu}$ since:
\begin{equation} 
\sin \theta_{\pi \mu} \le 1 < \frac{p_0}{p_\pi} 
\end{equation}
There are two roots of opposite sign, one positive and one negative. Only the
positive root has physical significance. 
\begin{equation} 
p_\mu = m_\mu \frac{E_0 p_\pi \cos \theta_{\pi \mu} 
+ E_\pi \sqrt{p_0^2 - p_\pi^2 \sin^2 \theta_{\pi \mu}}}{E_\pi^2 -
p_\pi^2 \cos^2 \theta_{\pi \mu}} 
\end{equation}
For a given value of 
$\theta_{\pi \mu}$ there is only one value for the muon momentum. The muon can
be emitted backward as well as forward.}
\item{$p_\pi > p_0$ 

All muons are emitted forward ($\cos \theta_{\pi \mu}$ is always positive):
\begin{equation} 
2 E_\pi E_\mu - m_\pi^2 - m_\mu^2 > 
2 m_\pi \frac{(m_\pi^2 + m_\mu^2)}{2 m_\mu} - (m_\pi^2 + m_\mu^2) 
=(m_\pi^2 + m_\mu^2) \bigg( \frac{m_\pi}{m_\mu} -1 \bigg) > 0 
\end{equation}
There is a limitation on the angle $\theta_{\pi \mu}$:
\begin{equation} 
\sin \theta_{\pi \mu} < \frac{p_0}{p_\pi}\hspace{1.0cm}
\theta_{\pi \mu} < \theta_0 
\end{equation}
where:
\begin{equation} 
\sin \theta_0 = \frac{p_0}{p_\pi} 
\end{equation}
There are two positive roots. Both have physical significance. 
\begin{equation} 
p_\mu = m_\mu \frac{E_0 p_\pi \cos \theta_{\pi \mu} 
\pm E_\pi \sqrt{p_0^2 - p_\pi^2 \sin^2 \theta_{\pi \mu}}}{E_\pi^2 -
p_\pi^2 \cos^2 \theta_{\pi \mu}} 
\end{equation}
The muon is emitted in the forward direction and for a given value of 
$\theta_{\pi \mu}$ there are two values of the muon momentum.}
\item{$p_\pi = p_0$

The roots are given by:
\begin{equation} 
p_\mu = 0 
\end{equation}
and:
\begin{equation}
p_\mu = 2 m_\mu \frac{E_0 p_0 \cos \theta_{\pi \mu}} 
{E_0^2 - p_0^2 \cos^2 \theta_{\pi \mu}} 
= 2 m_\mu \frac{(m_\pi^2 + m_\mu^2) (m_\pi^2 - m_\mu^2)  
\cos \theta_{\pi \mu}}{(m_\pi^2 + m_\mu^2)^2 
- (m_\pi^2 - m_\mu^2)^2 \cos^2 \theta_{\pi \mu}} 
\end{equation}}
\end{itemize}

\subsection{Angular distribution for the neutrino}
Energy-momentum conservation gives:
\begin{equation} 
2(E_\pi E_\nu - \vec p_\pi \cdot \vec p_\nu) = m_\pi^2 - m_\mu^2 
\end{equation}
\begin{equation} 
E_\nu = \frac{m_\pi^2 - m_\mu^2} 
{2 (E_\pi - p_\pi \cos \theta_{\pi \nu})} 
\end{equation} 
with the limiting values:
\begin{equation} 
E_\nu = \frac{m_\pi^2 - m_\mu^2}{2 m_\pi^2} (E_\pi + p_\pi) 
\hspace{0.5cm}
\hbox{for}\hspace{0.5cm}\theta = 0 
\end{equation}
\begin{equation} 
E_\nu = \frac{m_\pi^2 - m_\mu^2}{2 m_\pi^2} (E_\pi - p_\pi) 
\hspace{0.5cm}
\hbox{for}\hspace{0.5cm}\theta = \pi 
\end{equation}
There is no restriction on the angle $\theta_{\pi \nu}$.

\subsection{Figures}
\begin{itemize}
\item{Figure \ref{FIG1}: muon momentum vs pion-muon angle for five values of
the pion momentum, below the critical momentum.}
\item{Figure \ref{FIG2}: muon kinetic energy vs pion-muon angle for five 
values of the pion momentum, below the critical momentum.}
\item{Figure \ref{FIG3}: muon momentum vs pion-muon angle for five values of
the pion momentum, above the critical momentum.}
\item{Figure \ref{FIG4}: muon kinetic energy vs pion-muon angle for five 
values of the pion momentum, above the critical momentum.}
\item{Figure \ref{FIG5}: muon momentum vs pion-muon angle. The pion momentum 
is equal to the critical momentum.}
\item{Figure \ref{FIG6}: muon kinetic energy vs pion-muon angle. The pion
momentum is equal to the critical momentum.}
\item{Figure \ref{FIG7}: muon momentum vs pion-muon angle. The transition
across the critical momentum.}
\item{Figure \ref{FIG8}: muon kinetic energy vs pion-muon angle. The transition
across the critical momentum.}
\end{itemize}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1a.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon momentum as a function of the pion-muon angle. The pion 
momentum is below the critical momentum. There is no limiting angle, 
the pion-muon angle ranging from 0 to 180 degrees. For every pion-muon angle 
there is only one muon momentum.}
\label{FIG1}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1b.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon kinetic energy as a function of the pion-muon angle. The 
pion momentum is below the critical momentum. There is no limiting angle, 
the pion-muon angle ranging from 0 to 180 degrees. For every pion-muon angle 
there is only one muon kinetic energy.}
\label{FIG2}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine2a.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon momentum as a function of the pion-muon angle. The 
pion momentum is above the critical momentum. There is a limiting angle, 
the pion-muon angle ranging from 0 to a certain maximum. For every pion-muon 
angle there are two muon momenta.}
\label{FIG3}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine2b.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon kinetic energy as a function of the pion-muon angle. The 
pion momentum is above the critical momentum. There is a limiting angle, 
the pion-muon angle ranging from 0 to a certain maximum. For every pion-muon 
angle there are two muon kinetic energies.}
\label{FIG4}
\end{figure}
\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kinela.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon momentum as a function of the pion-muon angle. The 
pion momentum is equal to the critical momentum. There is a limiting angle, 
the pion-muon angle ranging from 0 to 90 degrees. For every pion-muon 
angle there are two muon momenta; one of them is equal to zero: in this case
the pion-muon angle is undetermined (any value from 0 to 180 degrees.}
\label{FIG5}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kinelb.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon kinetic energy as a function of the pion-muon angle. The 
pion momentum is equal to the critical momentum. There is a limiting angle, 
the pion-muon angle ranging from 0 to 90 degrees. For every pion-muon 
angle there are two muon kinetic energies; one of them is equal to zero:
in this case the pion-muon angle is undetermined (any value from 0 to 180
degrees).}
\label{FIG6}

\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kinesupa.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon momentum as a function of the pion-muon angle. This figure
shows the transition across the critical momentum.}
\label{FIG7}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kinesupb.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The muon kinetic energy as a function of the pion-muon angle. This 
figure shows the transition across the critical momentum.}
\label{FIG8}
\end{figure}

\newpage

\subsection{The maximum {\boldmath $\pi$-$\mu$} angle vs. the pion momentum}
Figure \ref{Fa} shows the variation of the $\pi$-$\mu$ angle as a function of
the pion momentum.

\begin{figure}[h]
\begin{center}
\epsfig{file=max_angle.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The maximum $\pi$-$\mu$ angle vs. the pion momentum}
\label{Fa}
\end{figure}

\newpage

\section{Distribution of the {\boldmath $\pi$-$\mu$} laboratory angle}

An analytical calculation is possible, but tedious. It is simpler to use
a Monte-Carlo calculation.

First the cosine of the C.M. $\pi$-$\mu$ angle is chosen at random between
$-1$ and $+1$ (Figure \ref{F1}). Then the C.M. $\pi$-$\mu$ angle is obtained
(Figure \ref{F2}). The components of the C.M. muon momentum are:
\begin{eqnarray}
P_x^* & = & P^* \cos \theta^* \nonumber \\
P_y^* & = & P^* \sin \theta^*
\end{eqnarray}
where: 
\begin{equation}
P^* = \frac{M_\pi^2 - M_\mu^2}{2 M_\pi} = 29.8 \hbox{ MeV/c}
\end{equation}
and the corresponding total energy:
\begin{equation}
E^* = \frac{M_\pi^2 + M_\mu^2}{2 M_\pi} = 109.8 \hbox{ MeV/c}
\end{equation}
$P_x^*$ and $P_y^*$ are shown in Figures \ref{F3} and \ref{F4}.

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=pimu_ang1_a.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{$\pi$ C.M. Cosine of $\pi$-$\mu$ angle.}
\label{F1}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=pimu_ang1_b.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{$\pi$ C.M. $\pi$-$\mu$ angle (in degrees).}
\label{F2}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=pimu_ang1_c.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{$\pi$ C.M. $P_x$ component of muon momentum.}
\label{F3}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=pimu_ang1_d.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{$\pi$ C.M. $P_y$ component of muon momentum.}
\label{F4}
\end{figure}

\newpage

The components of the LAB. muon momentum are obtained by applying a Lorentz
boost:
\begin{eqnarray}
P_x & = &\gamma P_x^* + \beta E^* 
\nonumber \\
P_y & = & P_y^*  
\end{eqnarray}
As an example, let us consider the $K^+ \to \pi^+ \pi^0$ decay. The momentum
of the $\pi^+$ is 205.1 MeV/c, $\beta = 0.827$, $\gamma = 1.78$.
  
The components of the muon momentum are shown in Figures \ref{F5} and 
\ref{F6}. The LAB. angle is given by:
\begin{equation}
\theta = \tan^{-1}(P_y/P_x)
\end{equation}
and shown in Figure \ref{F7}.

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=pimu_ang1_e.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{LAB. $P_x$ component of muon momentum.}
\label{F5}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=pimu_ang1_f.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{LAB. $P_y$ component of muon momentum.}
\label{F6}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=pimu_ang1_g.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{LAB. $\pi$-$\mu$ angle (in degrees).}
\label{F7}
\end{figure}

\newpage

\section{The polarization vector of the muon}

In $V - A$ theory the matrix element is:
\begin{equation} 
M = <0 \vert J_V^\alpha + J_A^\alpha \vert \pi> 
[\overline{u}_\mu \gamma_\alpha (1 - \gamma_5) u_\nu]
= F(q^2) m_\mu [\overline{u}_\mu (1 - \gamma_5) u_\nu] 
\end{equation} 
\begin{equation} 
F(q^2) = F(m_\pi^2) = F_\pi 
\end{equation}
We have to calculate the trace:
\begin{equation} 
M^2 = F_\pi^2 m_\mu^2 ~Tr [ (1 - \gamma_5) u_\nu \overline{u}_\nu 
(1 + \gamma_5) u_\mu\overline{u}_\mu] 
\end{equation}
We use the projection operators:
\begin{equation} 
u_\nu \overline{u}_\nu = \not p_\nu \hspace{1.0cm}u_\mu \overline{u}_\mu 
= (- \eta m_\mu + \not p_\mu) (1 + \gamma_5 \not s_\mu) 
\end{equation}
with the convention:
\begin{equation} 
\eta = + 1 \hspace{0.5cm}\hbox{for $\mu^+$ decay} 
\end{equation}
\begin{equation} 
\eta = - 1 \hspace{0.5cm}\hbox{for $\mu^-$ decay} 
\end{equation}
\begin{equation} 
M^2 = 8 F_\pi ^2 m_\mu^2 [ (p_\nu \cdot p_\mu) 
- \eta m_\mu (p_\nu \cdot s_\mu) ] 
\end{equation} 
One has:
\begin{equation} 
(\tilde{p}_\nu \cdot \tilde{p}_\mu) = E_\nu (E_\mu - p_\mu 
\cos \theta_{\mu \nu}) = \frac{m_\pi^2 - m_\mu^2}{2} 
= m_\mu p_0 
\end{equation}
\begin{equation} 
m_\mu (\tilde{p}_\nu \cdot \tilde{s}_\mu) = E_\nu (\vec \xi \cdot \vec p_\mu)
- \vec p_\nu . \bigg[ m_\mu \vec \xi +  (\vec \xi \cdot \vec p_\mu) 
\frac{\vec p_\mu}{E_\mu + m_\mu} \bigg] 
\end{equation}
The polarization vector is given by:
\begin{equation} 
\vec P = \eta \frac{1}{m_\mu p_0} \bigg\{ - \bigg[ E_\nu -
\frac{\vec p_\mu \cdot \vec p_\nu}{E_\mu + m_\mu} \bigg] \vec p_\mu
+ m_\mu \vec p_\nu \bigg \} 
\end{equation}
It is easy to check that $(\vec P)^2 = 1$.

In the rest system of the pion:
\begin{equation} 
\vec p_\nu = - \vec p_\mu\hspace{0.5cm}\hbox{and}\hspace{0.5cm} 
p_\mu = \frac{m_\pi^2 - m_\mu^2}{2 m_\pi} 
\end{equation}
This gives:
\begin{equation} 
\vec P = - \eta \vec n_\mu 
\end{equation}
where $\vec n_\mu$ is a unit vector along the muon momentum.

The longitudinal polarisation is 100\%. Since it has the negative sign for
a $\mu^+$, or a positive sign for a $\mu^-$, the decay is helicity-suppressed.

In the general case the longitudinal polarisation is:
\begin{equation} 
P_L = \eta \vec P \cdot \frac{\vec p_\mu}{p_\mu} =
\frac{1}{m_\mu p_0} E_\nu ( E_\mu \cos \theta_{\mu \nu} - p_\mu) 
\end{equation}
where $P_L$ can be positive or negative.
\begin{equation} 
\vec P_L = \eta 
\frac{E_\nu}{m_mu p_0} \bigg[
\frac{E_\mu}{p_\mu} \cos \theta_{\mu \nu} - 1 \bigg] 
\vec p_\mu\hspace{2.0cm}
P_L^2 = \frac{E_\nu^2}{m_\mu^2 p_0^2} 
( E_\mu \cos \theta_{\mu \nu} - p_\mu )^2 
\end{equation}
The transverse polarization is:
\begin{equation} 
\vec P_T = \vec P - \vec P_L = \eta \frac{E_\nu}{p_0} \bigg[
- \frac{\vec p_\mu}{p_\mu} \cos \theta_{\mu \nu} +
\frac{\vec p_\nu}{p_\nu} \bigg] \hspace{1.0cm}
P_T^2 = \frac{E_\nu^2}{p_0^2} \sin^2 \theta_{\mu \nu} 
\end{equation}
The longitudinal polarization is equal to zero if:
\begin{equation} 
\cos \theta_{\mu \nu} = \frac{p_\mu}{E_\mu} = 
\frac{v_\mu}{c} 
\end{equation}
Then the transverse polarization is:
\begin{equation} 
P_T^2 = \frac{E_\nu^2}{p_0^2} \frac{m_\mu^2}{E_\mu^2} 
\end{equation}
Using the relation:
\begin{equation} 
(E_\mu + E_\nu)^2 - (\vec p_\mu + \vec p_\nu)^2 = m_\pi^2 
\end{equation}
one gets:
\begin{equation} 
\frac{E_\nu}{E_\mu} = \frac{p_0}{m_\mu}\hspace{1.0cm}
E_\mu = E_\pi \frac{m_\mu}{E_0} \hspace{1.0cm}
E_\nu = E_\pi \frac{p_0}{E_0} 
\end{equation}

This leads to:
\begin{equation} 
P_T^2 = 1 
\end{equation}
In this kinematical configuration the muon is transversely polarized.
It is interesting to calculate the angle $\theta_{\pi \mu}$ between the pion
and the muon:
\begin{equation} 
\cos \theta_{\pi \mu} 
= \frac{2 E_\pi E_\mu - m_\pi^2 - m_\mu^2}{2 p_\pi p_\mu}
= \sqrt{1 - \frac{p_0^2}{p_\pi^2}} 
\end{equation}
If the condition:
\begin{equation} 
p_\pi \sin \theta_0 = p_0 
\end{equation}
is satisfied, then:
\begin{equation} 
\cos \theta_{\pi \mu} = \cos \theta_0 
\end{equation}
This correspond to the maximum angle for $\theta_{\pi \mu}$.

Finally, we give simple expressions for the polarization vector:
\begin{equation} 
\vec P_L = \eta \frac{E_\mu \cos \theta_{\mu \nu} - p_\mu}
{E_\mu - p_\mu \cos \theta_{\mu \nu}}~\vec n_\mu \hspace{1.0cm}
\vec P_T = \eta \frac{m_\mu}
{E_\mu - p_\mu \cos \theta_{\mu \nu}}~( \vec n_\nu 
- \vec n_\mu \cos \theta_{\mu \nu}) 
\end{equation}
where $\vec n_\mu$ and $\vec n_\nu$ are unit vectors along the directions of 
the muon and neutrino, respectively.

For $m_\mu = 0$ we get:
\begin{equation} 
\vec P_L = - \eta \vec n_\mu \hspace{1.0cm} \vec P_T = 0 
\end{equation}
and the decay is helicity-forbidden.

It is also possible to express the polarization in terms of the angle between
pion and muon. Starting from:
\begin{equation} 
m_\mu p_0 \vec P = - \eta \bigg[ E_\nu - \frac{\vec p_\mu \cdot \vec p_\nu}
{E_\mu + m_\mu} \bigg] \vec p_\mu + m_\mu \vec p_\nu 
\end{equation}
with:
\begin{equation} 
E_\nu = E_\pi - E_\mu 
\end{equation}
\begin{equation} 
\vec p_\nu = \vec p_\pi - \vec p_\mu 
\end{equation}
one finds:
\begin{equation} 
\vec P_L = \eta \frac{1}{m_\mu p_0} 
\bigg[ E_\mu p_\pi \cos \theta_{\pi \mu} - E_\pi p_\mu \bigg] \vec n_\mu =
\eta \frac{1}{m_\mu p_0} \bigg[ E_\mu p_\pi (\vec n_\pi \cdot \vec n_\mu) 
- E_\pi p_\mu \bigg] \vec n_\mu
\end{equation}
The longitudinal polarization is equal to zero if:
\begin{equation} 
\cos \theta_{\pi \mu} = \frac{p_\mu}{E_\mu} \frac{E_\pi}{p_\pi} =
\frac{v_\mu}{v_\pi} 
\end{equation}
The transverse polarization is:
\begin{equation} 
\vec P_T = \eta \frac{p_\pi}{p_0} \bigg[ \vec n_\pi 
- (\vec n_\pi \cdot \vec n_\mu) \vec n_\mu \bigg] 
\end{equation}
We also have:
\begin{equation} 
P_T^2 = \frac{p_\pi^2}{p_0^2} \sin^2 \theta_{\pi \mu} 
\end{equation}
or:
\begin{equation} 
P_T = \frac{\sin \theta_{\pi \mu}}{\sin \theta_0} 
\end{equation}
The transverse polarization is linear in $\sin \theta_{\pi \mu}$ and reaches
its maximum value $P_T = 1$ when $\theta_{\pi \mu}$ is equal to the limiting
angle $\theta_0$.

It should be noted that the angular distribution of the muon is peaked
at the limiting angle, therefore most muons will be transversely polarized. 
If the beam has cylindrical symmetry, this transverse polarization will 
average to zero.

\subsection{Transversely polarized muons used in experiments}

Bardon et al.\cite{bar} have used transversely polarized muons in a Mott
scattering experiment. They produced a parallel beam of 28 $\pm$ 2.5 MeV 
negative $\pi$ mesons and used a selection on the angle to produce transversely
polarized muons with a polarization $P_y = 0.9$. The kinetic energy of the
muons was 9 MeV $< T_\mu <$ 33 MeV. 

Later on, the same technique was used by Beltrami et al.\cite{BEL,BUR}
in a $\mu$SR experiment to measure $P_\mu \xi =1.0027 \pm 0.0084$. The design
of the apparatus and associated Monte Carlo calculations are described in a
paper by B\"unzli\cite{BUEN87} (this work, done at PSI, was supervised by Peter
Gumplinger).
\subsection{A visual representation of the polarization}
For a pion decaying from rest the angular distribution has
spherical symmetry. The polarization vector of the muon is aligned on
the momentum (longitudinal polarization). Figure \ref{FIG10} shows the 
components $p_x$ et $p_y$ of the muon momentum. The circles represent the
polarization vector. Figure \ref{FIG11} shows the same but the pion momentum 
is 50 MeV (above the critical momentum). For the muons emitted at the limiting 
angle the polarization vector is orthogonal to the momentum (transverse 
polarization).

When the pion is Lorentz-boosted the sphere takes an elongated shape resembling
an ellipsoid. In Figure \ref{FIG12} the pion momentum is 20 MeV/c (below the
critical momentum). The long lines represent the muon momenta (the momentum 
of the pion is along the X axis). The short lines represent the polarization 
vector of the muon (in the rest system). Only muons emitted in the same 
direction as the pions (forward or backward) are longitudinally polarized 
(negative polarisation for the positive muons).

In Figure \ref{FIG13} the pion momentum is 50 MeV/c (above the critical 
momentum). The long lines represent the muon momenta (the momentum 
of the pion is along the X axis). The two groups of muons emitted forward
in the direction of the pion have opposite polarizations. In the case of 
positive muons the polarization is negative for the high-energy muons and
positive for the low-energy ones. For the muons emitted at the limiting angle
the polarization vector is orthogonal to the momentum (transverse 
polarization). See Figure \ref{FIG9}.

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=ellipse1c.eps,height=6.5cm}
\end{center}
\caption{Polarization vector in $\pi_{\mu 2}$ decay. The pion is at rest.}
\label{FIG10}

\vspace{1.0cm}

\begin{center}
\epsfig{file=ellipse1.eps,
height=6.5cm}
\end{center}
\caption{Polarization vector in $\pi_{\mu 2}$ decay. The momentum of the
pion is 50 MeV/c.}
\label{FIG11}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=polarvec1.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The momentum of the pion is 20 MeV/c, below the critical momentum.}
\label{FIG12}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=polarvec2.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The momentum of the pion is 50 MeV/c, above the critical momentum.}
\label{FIG13}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine2_150.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{The kinematics for a pion momentum = 150 MeV/c and the muon emitted 
at the maximum angle. The muon is transversely polarized. Note the direction
of the neutrino momentum.} 
\label{FIG9}
\end{figure}

\newpage

\subsection{Effect of pion thermal motion}

It has been suggested that one should consider the influence of the thermal 
motion of the pion on the polarization of the muon. The question is: is this
effect important for E614? A rough calculation is presented here.

Assuming a temperature of 500 K in the target, the thermal kinetic energy of
the pion is:
\begin{equation}
E = \frac{3}{2} k T 
= 8.617 \times 10^{-5} \hbox{ ev K}^{-1} \times 500 \hbox{ K} 
= 0.0646 \hbox{ eV}
\end{equation}
This correspond to a pion momentum of:
\begin{equation}
p = \sqrt{2 m E} = 3.70 \hbox{ KeV}
\end{equation}
A calculation with $p = 4$ KeV gives the following results:
\begin{itemize}
\item{Figure \ref{FIG14}: muon momentum vs pion-muon angle. The momentum 
spread is $\sim 2 \times 10^{-4}$.}
\item{Figure \ref{FIG15}: muon kinetic energy vs pion-muon angle. The energy 
spread is $\sim 4 \times 10^{-4}$.}
\item{Figure \ref{FIG16}: neutrino energy vs pion-muon angle. The energy 
spread is $\sim 5 \times 10^{-5}$.}
\item{Figure \ref{FIG17}: muon-neutrino angle vs pion-muon angle. What is 
plotted is the difference ($\times 10^8$) between the cosine of the angle 
($- 1$ for a pion decaying from rest) and $- 1$. The angular spread is 
$\sim 10^{-8}$.}
\item{Figure \ref{FIG18}: muon longitudinal polarisation vs pion-muon angle.
What is plotted is the difference ($\times 10^8$) between the longitudinal
polarisation ($- 1$ for a pion decaying from rest) and $- 1$. The deviation
from $-1$ is $\sim 5 \times 10^{-9}$.}
\item{Figure \ref{FIG19}: muon transverse polarization ($\times 10^4$).
This transverse polarization reaches a maximum $\sim 10^{-4}$.}
\item{Figure \ref{FIG20}: muon polarization component on the neutrino
momentum. What is plotted is the difference ($\times 10^8$) between this
component ($- 1$ for a pion decaying from rest) and $- 1$. The deviation from
$-1$ is very small, $\sim 5 \times 10^{-10}$. At such a low momentum the 
polarization of the muon tends to ``follow'' the polarization of the neutrino.}
\end{itemize}

The effects are of the order of $10^{-8}$ on the muon-neutrino angle and on the
longitudinal polarization of the muon. On the other hand, the transverse
polarization could be of the order of $10^{-4}$. 

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1sa.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{Muon momentum vs pion-muon angle.}
\label{FIG14}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1sb.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{Muon kinetic energy vs pion-muon angle.}
\label{FIG15}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1sc.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{Neutrino energy vs pion-muon angle.}
\label{FIG16}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1sd.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{Muon-neutrino angle vs pion-muon angle.}
\label{FIG17}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1se.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{Muon longitudinal polarization vs pion-muon angle.}
\label{FIG18}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1sf.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{Muon transverse polarization vs pion-muon angle.}
\label{FIG19}
\end{figure}

\newpage

\begin{figure}[h]
\begin{center}
\epsfig{file=kine1sg.eps,width=14.5cm,height=14.5cm}
\end{center}
\caption{Muon polarization component on the neutrino momentum vs 
pion-muon angle.}
\label{FIG20}
\end{figure}

\newpage

\section{Acknowledgments}
The author wishes to thank Peter Gumplinger for interesting discussions and
for bringing to his attention the paper by B\"unzli\cite{BUEN87}.

\begin{thebibliography}{1}
\bibitem{bar}{M. Bardon et al., {\em ``Helicity of $\mu^-$ mesons; Mott
Scattering of Polarized muons''}, Phys. Rev. Lett. {\bf 7}, 23, (1961).}
\bibitem{BEL}{I. Beltrami et al., {\em ``Muon Decay: Measurement of the 
Integral Asymmetry Parameter''}, Phys. Lett. {B194}, 326, (1987).}
\bibitem{BUR}{H. Burkard, {\em ``Messung der integralen Asymmetrie im 
$\mu$-Zerfall''}, Ph.D. Thesis, ETH, Z\"urich, Switzerland.}
\bibitem{BUEN87}{Daniel B\"unzli, {\em ``Analyse einer Apparatur zur Messung 
der Richtungsverteilung der Positronen im $\mu^+$-Zerfall''}, Diplomarbeit am
Institut f\"ur Mittelenergiephysik der ETH Z\"urich, Sommer 1987.}
\end{thebibliography}

\end{document}

\bibitem{SCHE78}{Florian Scheck, {\em ``Muon physics''}, Phys. Rep. 
{\bf 44}, 187 (1978).}







\end{document}

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