 Solution

In the pion rest frame, conservation of momentum makes the muon and massless neutrino momenta equal and opposite, of magnitude $p$. The neutrino energy is $pc$, while the muon satisfies $E^2=p^2{c}^2+m^2{c}^4$. Energy conservation gives $E+pc=M{c}^2$. Squaring after substituting $E=M{c}^2-pc$ yieldsThe relativistic two-body decay speed is thusChoose $S^{\prime }$ moving with velocity $+v$ along $x$ relative to $S$, so the pion has velocity $-v$ in $S^{\prime }$ and speed $v$. With ${\Gamma }_v=(1-v^2/c^2{)}^{-1/2}$, the four-dimensional Lorentz transformation isThe muon's four-momentum is $(E/c,0,p,0)$ before the boost. The same matrix givesVelocity and momentum have the same spatial direction. The unsigned deflection angle from the positive $y$ axis therefore satisfiesChoosing the opposite boost makes the pion move along positive $x$ and reverses $p_x^{\prime }$, leaving this speed-based angle formula unchanged.