- Tue 01 December 2015
- Physics
- #sudakov, #two-body, #kinematics
In a previous post I discussed the two-body Sudakov decomposition. I found that the Regge-Mandelstam invariant \(r_{12}\) and the Regge-Sudakov invariant \(R_{12}\) satisfy the equation
$$ 2 r_{12} = R_{12} + \frac{1}{R_{12}}. $$
You can introduce a rapidity \(\rho_{12}\) via
$$ R_{12} = \exp{ \left( \rho_{12} \right) } \quad \Longrightarrow \quad \rho_{12} = \log{ \left( R_{12} \right) }. $$
Thus,
$$ r_{12} = \operatorname{cosh}{ \left( \rho_{12} \right) } \quad \Longrightarrow \quad \rho_{12} = \operatorname{acosh}{ \left( r_{12} \right) }. $$
You also have the Regge-Gram invariant:
$$ g_{12} \equiv \frac{G_{12}}{m_{1}^{2} m_{2}^{2}} = r_{12}^{2} - 1 = \operatorname{sinh}^{2}{ \left( \rho_{12} \right) } \quad \Longrightarrow \quad \rho_{12} = \operatorname{asinh}{ \left( \sqrt{ g_{12} } \right) }. $$
In terms of the Mandelstam invariant and the two masses, you have
$$ \rho_{12} = \log{ \left( \frac{s_{12} - m_{1}^{2} - m_{2}^{2} + \sqrt{ \left( s_{12} - m_{1}^{2} - m_{2}^{2} \right)^{2} - 4 m_{1}^{2} m_{2}^{2} }}{2 m_{1} m_{2}} \right) }. $$
The double and half argument formulas are also relevant:
$$ \operatorname{cosh}{ \left( 2 \rho_{12} \right) } = \operatorname{cosh}^{2}{ \left( \rho_{12} \right) } + \operatorname{sinh}^{2}{ \left( \rho_{12} \right) } = 2 r_{12}^{2} - 1, $$
$$ \operatorname{sinh}{ \left( 2 \rho_{12} \right) } = 2\operatorname{cosh}{ \left( \rho_{12} \right) } \operatorname{sinh}{ \left( \rho_{12} \right) } = 2 r_{12} \sqrt{r_{12}^{2} - 1} ; $$
$$ \operatorname{cosh}^{2}{ \left( \frac{\rho_{12}}{2} \right) } = \frac{r_{12} + 1}{2} = \frac{s_{12} - \left( m_{1} - m_{2} \right)^{2}}{4 m_{1} m_{2}}, $$
$$ \operatorname{sinh}^{2}{ \left( \frac{\rho_{12}}{2} \right) } = \frac{r_{12} - 1}{2} = \frac{s_{12} - \left( m_{1} + m_{2} \right)^{2}}{4 m_{1} m_{2}}. $$
These provide an argument for restricting only to the range \(s_{12} \geq \left( m_{1} + m_{2} \right)^{2}\).