# Helicity angle of electron in $B^0 \rightarrow K^{*0} \gamma$, with $\gamma \rightarrow e^+ e^-$?

I haven't been able to find the exact definition of a helicity angle, and the ones that I found don't apply to this particular case, because they'd require to boost to the electrons' 4-momentum to the photon's rest frame. Which doesn't make sense.

My decay is $B^0 \rightarrow K^{*0} \gamma$, with $\gamma \rightarrow e^+ e^-$.

I need the helicity angle of the $e^+$ (or the $e^-$)... anyone know how to do it?

For each particle the four momentum in the laboratory frame was stored in the experiment. The analysis proceeds using only these four momenta as input. Due to the helicity amplitudes formalism every momentum has to be boosted back from the laboratory (lab.) frame to the mother rest frame (r. f.). You see the X(3872) decay to $J/\psi$ and $\rho$ below.
In your case, $J/\psi$ is $K^{*0}$ , $\rho$ is $\gamma$. So you need to boost $e^-$ or $e^+$ momenta first to the $B^0$ rest frame, then boost it to the $\gamma$ r.f with the $\gamma$ momentum in the $B^0$ r.f as it's shown in below.
$$P^{lab}_{e^-} \xrightarrow{-P_{B^0}} \xrightarrow{-P_{\gamma}^{B^0 r.f}}P_{e^-}^{\gamma r.f}$$