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When $\delta_{CP}= 0^o$ and Matter Effects are neglected then,
\begin{eqnarray}
	P_{\nu_\mu\rightarrow\nu_e} 
	& \simeq & 4 c_{13}^2 s_{13}^2 s_{23}^2 \sin^2 \Delta_{31} \\
	& +      & (8  c_{12} s_{12} c_{13}^2 s_{13} s_{23} c_{23} - 8 s_{12}^2 c_{13}^2 s_{13}^2 s_{23}^2)
		\cos \Delta_{23} \sin\Delta_{31} \sin\Delta_{21}  
\end{eqnarray}
As $\Delta_{23} = \Delta_{21} + \Delta_{13}$ then
$$
	\cos \Delta_{23} = \cos \Delta_{21} \cos \Delta_{13} - \sin \Delta_{21} \sin \Delta_{13}
$$
and in numerical application $\Delta_{21} = 1.27 \delta m^2_{21} L/E \approx O(10^{-2})$. So,
\begin{eqnarray}
	P_{\nu_\mu\rightarrow\nu_e} 
	& \simeq & 4 c_{13}^2 s_{13}^2 s_{23}^2 \sin^2 \Delta_{31} \\
	& +      & (8  c_{12} s_{12} c_{13}^2 s_{13} s_{23} c_{23} - 8 s_{12}^2 c_{13}^2 s_{13}^2 s_{23}^2)
		\Delta_{21}  \cos \Delta_{13} \sin\Delta_{31}  
\end{eqnarray}
If one uses $s_{12}^2 = 0.314$ and $s_{23}^2 = 0.44$ then one realizes that in the parenthesis the second term is of the order $s_{13}$ compared to the first term, so it may be neglected hereafter as we will focus on $s_{13} < 10^{-2}$. Then, it yields
\begin{eqnarray}
	P_{\nu_\mu\rightarrow\nu_e} 
	& \simeq & \alpha \cos\beta \sin^2 \Delta_{31} + \alpha \sin\beta \cos \Delta_{13} \sin\Delta_{31}  
\end{eqnarray}
with
\begin{eqnarray}
	\alpha \cos\beta & \equiv &  4 c_{13}^2 s_{13}^2 s_{23}^2 \\
	\alpha \sin\beta & \equiv & 8  c_{12} s_{12} c_{13}^2 s_{13} s_{23} c_{23}
\end{eqnarray}
It is remarkable that now the oscillation probability may be written as
\begin{eqnarray}
	P_{\nu_\mu\rightarrow\nu_e} 
	& \simeq & \frac{\alpha}{2}\left[
	 \cos\beta - \cos\left( 2\Delta_{13} + \beta\right) \right]  
\end{eqnarray}
The maximum of the probability is obtained at
\begin{eqnarray}
\Delta_{31} & = & \frac{\pi}{2} - \frac{\beta}{2} \\
\mathrm{with}\ \tan \beta & = & 2 \Delta_{21}\frac{c_{12} s_{12}c_{23} }{ s_{13}s_{23}}
\end{eqnarray}
The "usual" 2-famillies case is obtain with $\beta = 0$. In case of $E\sim 0.3$~GeV, $L\sim130$~km and $\sin^22\theta_{13} = 10^{-3}$ then
\begin{eqnarray}
\left(\delta m_{31}^2\right)_{max} & = & 2.9\, 10^{-3} \ \mathrm{if} \ \beta = 0 \\ 
															 & = & 1.8\, 10^{-3}
\end{eqnarray}



 


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