We want to figure out how energy depends on $x^\prime$ so that we can correct for energy dependence precession around delivery ring.
$$x^\prime = \frac{p_\perp}{p_0}$$ where $p_\perp$ is the momentum of the muon transverse to the boost direction (the pion momentum vector) and $p_0$ is the reference momentum. We will 
define $p_0=p_\mu^{max}$ that is the maximum momentum of a muon corresponding to the case where $p_\perp=0$. 
We suppose that the pion momentum is in the z-direction so that the boost matrix is
$$L(\beta) = \begin{pmatrix}\gamma & 0 & 0 & \gamma\beta\cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr \beta\gamma & 0 & 0 &\gamma\end{pmatrix}$$
where $\gamma =E_\pi/m_\pi$ and $\beta= p_\pi/E_\pi$.
In the pion rest frame the muon four vector is
$$p_r = \begin{pmatrix} \gamma_rm_\mu\cr p^r_x \cr 0 \cr p^r_z\end{pmatrix},$$ where the $p^r_x$ is the momentum transverse to the boost direction, and $p^r_z$ is the momentum in the boost direction. 
In the muon rest frame the polarization is ${\bf s}$. Transform to the pion rest frame by boosting in the -${\bf s}$ direction
using
\begin{eqnarray*}
{\bf S} &=& {\bf s}+\frac{\gamma_r^2}{\gamma_r+1}(\beta_r s){\bm \beta} = {\bf s}(1+\frac{\gamma_r^2}{\gamma_r+1}\beta^2)= {\bf s}(1+(\gamma_r-1))={\bf s}\gamma_r\\
 S_0&=&\gamma_r\beta_r s\\
S&=&\gamma_r(\beta_r s, {\bf s})\rightarrow \gamma_r(\beta_r s, s_x, 0, s_z) = \gamma_r\begin{pmatrix}\beta_r s\cr s_x \cr 0 \cr s_z\end{pmatrix}
\end{eqnarray*}
The muon is left handed and in the pion rest frame $${\bf s} =-\frac{{\bf p^r}}{\bf |p^r|}.$$

In the lab frame the muon four vector becomes
$$p_\mu = \begin{pmatrix}\gamma \gamma_rm_\mu+\beta p^r_z \cr p^r_x \cr 0 \cr \gamma(\beta\gamma_r m_\mu + p^r_z)\end{pmatrix}$$
The maximum energy and momentum corresponds to $p_x^r=0$ and $$p_z^r = p_r = \frac{m_\pi^2-m_\mu^2}{2 m_\pi}$$
and the spin is
$$ S_\pi = \gamma_r\begin{pmatrix}\gamma(\beta_rs+\beta s_z)\cr s_x \cr 0 \cr \gamma(\beta \beta_r s + s_z)\end{pmatrix}$$
And in the lab frame 
\begin{eqnarray*}
{\bf s^\prime\cdot p^\prime}&=& \gamma_r(s_x p_x^r +\gamma^2(\beta\beta_r +s_z)(\beta\gamma_r m_\mu +p_z^r))\\
\frac{\bf s^\prime\cdot p^\prime}{\bf |s^\prime||p^\prime|}&=&\frac{ \gamma_r(s_x p_x^r +\gamma^2(\beta\beta_r +s_z)(\beta\gamma_r m_\mu +p_z^r))}
{[({p_x^r}^2 + \gamma^2(\beta\gamma_r m_\mu + p_z^r)^2)(\gamma_r^2(s_x^2+\gamma^2(\beta\beta_r  + s_z)^2))]^{1/2}}\\
&=&\frac{ (s_x p_x^r +\gamma^2(\beta\beta_r +s_z)(\beta\gamma_r m_\mu +p_z^r))}
{[({p_x^r}^2 + \gamma^2(\beta\gamma_r m_\mu + p_z^r)^2)(s_x^2+\gamma^2(\beta\beta_r  + s_z)^2)]^{1/2}}\\
&\sim&\frac{ (s_x p_x^r +\gamma^2(\beta\beta_r +s_z)(\beta\gamma_r m_\mu +p_z^r))}
{\gamma(\beta\gamma_r m_\mu + p_z^r)(\gamma^2(\beta\beta_r  + s_z)^2)^{1/2}}\left(1 - \onehalf \frac{{p_x^r}^2}{\gamma^2(\beta\gamma_r m_\mu 
+p_z^r)^2}-\onehalf \frac{s_x^2}{\gamma^2(\beta\beta_r+s_z)^2}\right)\\
&\sim& 1+\frac{ s_x p_x^r}
{\gamma^2(\beta\gamma_r m_\mu + p_z^r)(\beta\beta_r  + s_z)} - \onehalf \frac{{p_x^r}^2}{\gamma^2(\beta\gamma_r m_\mu 
+p_z^r)^2}-\onehalf \frac{s_x^2}{\gamma^2(\beta\beta_r+s_z)^2}\\
\beta&\rightarrow& 1\\
&\sim& 1+\frac{ s_x p_x^r}
{\gamma^2(\gamma_r m_\mu + p_z^r)(\beta_r  + s_z)} - \onehalf \frac{{p_x^r}^2}{\gamma^2(\gamma_r m_\mu 
+p_z^r)^2}-\onehalf \frac{s_x^2}{\gamma^2(\beta_r+s_z)^2}\\
&\sim& 1+\frac{ -s^2_x \gamma_r\beta_rm_\mu}
{\gamma^2(\gamma_r m_\mu + p_z^r)(\beta_r  + s_z)} - \onehalf \frac{{s_x}^2(\beta_r\gamma_r m_\mu)^2}{\gamma^2(\gamma_r m_\mu 
+p_z^r)^2}-\onehalf \frac{s_x^2}{\gamma^2(\beta_r +s_z)^2}\\
&\sim& 1+\frac{ -s^2_x \gamma_r\beta_rm_\mu}
{\gamma^2(\gamma_r m_\mu + p_z^r)(\beta_r  + s_z)} - \onehalf \frac{{s_x}^2(\beta_r\gamma_r m_\mu)^2}{\gamma^2(\gamma_r m_\mu 
+p_z^r)^2}-\onehalf \frac{s_x^2}{\gamma^2(\beta_r +s_z)^2}\\
&\sim& 1-\frac{s_x^2}{\gamma^2}\left(\frac{\gamma_r\beta_rm_\mu}
{(\gamma_r m_\mu + p_z^r)(\beta_r  + s_z)} + \onehalf \frac{(\beta_r\gamma_r m_\mu)^2}{(\gamma_r m_\mu +p_z^r)^2}+\onehalf \frac{1}{(\beta_r +s_z)^2}\right)\\
s_z=p_z^r&\rightarrow 0\\
&\sim& 1-\frac{s_x^2}{\gamma^2}\left(\frac{\gamma_r\beta_rm_\mu}
{(\gamma_r m_\mu )(\beta_r)} + \onehalf \frac{(\beta_r\gamma_r m_\mu)^2}{(\gamma_r m_\mu)^2}+\onehalf \frac{1}{(\beta_r)^2}\right)\\
&\sim& 1-\frac{s_x^2}{\gamma^2}\left(1 + \onehalf \beta_r^2+\onehalf \frac{1}{(\beta_r)^2}\right)=\cos\phi \sim 1-\onehalf\phi^2\\
\rightarrow \phi \sim \frac{s_x}{\gamma\beta_r}
\end{eqnarray*}
If $p_z^r=s_z=0$ then
\begin{eqnarray*}
{\bf s^\prime\cdot p^\prime}&=& \gamma_r(s_x p_x^r +\gamma^2\beta^2\beta_r\gamma_r m_\mu)\\
&=& \gamma_r( p_x^r +\gamma^2\beta^2\beta_r\gamma_r m_\mu)\\
&=& \gamma_r p_x^r(1 +\gamma^2\beta^2)\\
&=& \gamma_r p_x^r \gamma^2\\
\frac{{\bf s^\prime\cdot p^\prime}}{{\bf |s^\prime||p^\prime|}}&=& \frac{\gamma_r p_x^r\gamma^2}{(1+(\gamma\beta\beta_r)^2)^{1/2}({p_x^r}^2 
+ (\gamma\beta\gamma_r m_\mu)^2)^{1/2}}= \frac{\gamma^2}{(1+(\gamma\beta\beta_r)^2)^{1/2}(1+\gamma^2)^{1/2}}
\end{eqnarray*}


Since $p_x$ is unchanged by the boost, 
\begin{equation}
x^\prime = \frac{p_x}{p_{max}} = \frac{p_x}{\beta\gamma_r m_\mu + p^r}\label{eq:xprime}
\end{equation}
We want $p=\sqrt{p_x^2+p_z^2}$ in terms of $x^\prime$. That will give us $\delta = (p-p_{max})/p_{max}$ in terms of $x^\prime$.
$$p_z = \gamma\beta\gamma_r m_\mu +\gamma p_z^r=\gamma\beta\gamma_r m_\mu \pm\gamma\sqrt{p_r^2-p_x^2}$$
With the approcximation that $\sqrt{p_r^2-p_x^2}\ll \beta\gamma_r m_\mu$
\begin{eqnarray*}
p^2 &=& p_z^2 + p_x^2 = (\gamma\beta\gamma_r m_\mu)^2\left(1\pm 2\frac{\sqrt{p_r^2-p_x^2}}{\beta\gamma_r m_\mu}\right)+p_x^2\\
&=&(\gamma\beta\gamma_r m_\mu)^2 \pm 2\sqrt{p_r^2-p_x^2}\gamma^2\beta\gamma_r m_\mu+p_x^2\\
\end{eqnarray*}
Can we assume that $p_x\ll p_r$? Suppose we only consider muons with momentum within 3\% of the maximum.
\begin{eqnarray*}
 \frac{p_{max} - p}{p_{max}} &=&\frac{p^r-p_z^r}{\beta\gamma_r m_\mu +p^r}=\frac{1-p_z^r/p^r}{\beta\gamma_r m_\mu/p^r +1}\\
&=&\frac{p^r-p_z^r}{\beta\gamma_r m_\mu +p^r}=\frac{1-p_z^r/p^r}{\beta\gamma_r m_\mu/p^r +1}\\
&=& \frac{1-p_z^r/p_r}{\beta\frac{m_\pi^2+m_\mu^2}{m_\pi^2-m_\mu^2}+1}\sim 0.2189(1-p_z^r/p^r)\\
\end{eqnarray*}
implies that for $\delta<0.03$, $p_z^r/p^r > 0.86$ and
$$p_x=p^r(1-(p_z^r/p_r)^2)^{1/2} < p_r(1-0.93) <0.07p_r$$
So, if the momentum of the muon in the lab frame is to be within 3\% of the maximum possible value then the rest frame momentum transverse 
to the boost direction is necessarily less that $7\%$ of the total rest frame momentum. 
Now we can go back to that equation meant to give us $p$ as a function of $p_x$ and assume that $p_x/p_r\ll 1$.
Then
\begin{eqnarray*}
p^2&\sim& (\gamma\beta\gamma_r m_\mu)^2+ 2p_r(1-\onehalf\frac{p_x^2}{p_r^2})\gamma^2\beta\gamma_r m_\mu + p_x^2\\
&\sim& (\gamma\beta\gamma_r m_\mu)^2+ 2p_r(\gamma^2\beta\gamma_r m_\mu)-\frac{p_x^2}{p_r}\gamma^2\beta\gamma_r m_\mu\\
\end{eqnarray*}
\begin{eqnarray}
p_{max}^2-p^2&=& (p_{max}+p)(p_{max}-p) = \frac{p_x^2}{p_r}\gamma^2\beta\gamma_r m_\mu\nonumber\\
\rightarrow \delta& =&\frac{ \frac{p_x^2}{p_r}\gamma^2\beta\gamma_r m_\mu}{ (\gamma\beta\gamma_r m_\mu + \gamma p_r)^2}\nonumber\\
& \sim&  \frac{p_x^2}{p_r (\beta\gamma_r)^2 m_\mu}\label{eq:deltapx}
\end{eqnarray} 
Now that we have $\delta(p_x)$ we write $p_x$ in terms of $x^\prime$ by substituting ~\ref{eq:xprime} into ~\ref{eq:deltapx}.
\begin{eqnarray*}
\delta \sim \frac{(\beta\gamma_r m_\mu+p^r)^2 {x^\prime}^2}{p_r(\gamma_r\beta)^2 m_\mu}\sim \frac{m_\mu}{p_r}{x^\prime}^2=
\frac{2m_\pi m_\mu}{m_\pi^2-m_\mu^2}{x^\prime}^2= 3.4 {x^\prime}^2
\end{eqnarray*}
Now 
$$\Delta \phi = \der{\phi}{\delta}\Delta \delta  = \der{\phi}{\delta} (3.4) {x^\prime}^2

