In a previous post I discussed the kinematics of three-body problems. The contiguous three-body Sudakov decomposition is given by
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{12}} n_{2}, $$
$$ p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{23}} n_{3}, $$
$$ p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{31}} n_{1}. $$
Here \(n_{1}\), \(n_{2}\), and \(n_{3}\) are (null) Sudakov vectors, and I have introduced three 2-Sudakov invariants:
$$ S_{ij} \equiv -\left\Vert n_{i} + n_{j} \right\Vert^{2} = -2 \left( n_{i} \cdot n_{j} \right).$$
The labeling of the Sudakov vectors is such that in a massless limit \(m_{i} \rightarrow 0\) the momentum \(p_{i}\) becomes the Sudakov vector \(n_{i}\). From the two-body decomposition you expect each Sudakov vector to carry the content of many individual quanta. The above decomposition can be written in matrix form:
$$ \begin{pmatrix} p_{1} \\ p_{2} \\ p_{3} \end{pmatrix} = \mathcal{S}_{3} \cdot \begin{pmatrix} n_{1} \\ n_{2} \\ n_{3} \end{pmatrix}, \qquad \mathcal{S}_{3} \equiv \begin{pmatrix} 1 & \dfrac{m_{1}^{2}}{S_{12}} & 0 \\ 0 & 1 & \dfrac{m_{2}^{2}}{S_{23}} \\ \dfrac{m_{3}^{2}}{S_{31}} & 0 & 1 \end{pmatrix} . $$
The determinant of this matrix is
$$ \det{\left( \mathcal{S}_{3} \right)} = 1 + \frac{m_{1}^{2} m_{2}^{2} m_{3}^{2}}{S_{12} S_{23} S_{31}} = \frac{S_{12} S_{23} S_{31} + m_{1}^{2} m_{2}^{2} m_{3}^{2}}{S_{12} S_{23} S_{31}} . $$
The inverse of this matrix is
$$ \operatorname{inv}{\left( \mathcal{S}_{3} \right)} = \frac{1}{\det{\left( \mathcal{S}_{3} \right)}} \begin{pmatrix} 1 & - \dfrac{m_{1}^{2}}{S_{12}} & \dfrac{m_{1}^{2} m_{2}^{2}}{S_{12} S_{23}} \\ \dfrac{m_{2}^{2} m_{3}^{2}}{S_{23} S_{31}} & 1 & - \dfrac{m_{2}^{2}}{S_{23}} \\ - \dfrac{m_{3}^{2}}{S_{31}} & \dfrac{m_{3}^{2} m_{1}^{2}}{S_{31} S_{12}} & 1 \end{pmatrix}. $$
Thus, each Sudakov vector can be written as a linear combination of the three momenta. In other words, each massive body contributes to each null vector.
My goal is to write the 2-Sudakov invariants in terms of the Mandelstam invariants and the masses. In order to do this, it is helpful to also introduce Regge-Sudakov invariants:
$$ R_{ij} \equiv \frac{S_{ij}}{m_{i} m_{j}}. $$
Using the identities
$$ r_{ij} = -\frac{p_{i} \cdot p_{j}}{m_{i} m_{j}}, $$
you find the relations
$$ 2r_{12} = R_{12} + \frac{R_{31}}{R_{23}} + \frac{1}{R_{12}}, $$
$$ 2r_{23} = R_{23} + \frac{R_{12}}{R_{31}} + \frac{1}{R_{23}}, $$
$$ 2r_{31} = R_{31} + \frac{R_{23}}{R_{12}} + \frac{1}{R_{31}}.$$
Note that you can obtain the second and third equations from the first one by cyclically shifting the integer labels. The solution should also exhibit this property. Solving this system of equations gives
$$ R_{12} = \frac{r_{12} \left[ g_{123} - 1 + \sqrt{ g_{123} \left( g_{123} - 1 \right)} \right] + r_{23} r_{31}}{g_{123} + g_{31}}, $$
$$ R_{23} = \frac{r_{23} \left[ g_{123} - 1 + \sqrt{ g_{123} \left( g_{123} - 1 \right)} \right] + r_{31} r_{12}}{g_{123} + g_{12}}, $$
$$ R_{31} = \frac{r_{31} \left[ g_{123} - 1 + \sqrt{ g_{123} \left( g_{123} - 1 \right)} \right] + r_{12} r_{23}}{g_{123} + g_{23}}. $$
Thus,
$$ S_{12} = \frac{1}{4} \left[ \frac{ 2 \left( s_{12} - m_{1}^{2} - m_{2}^{2} \right) \left[ G_{123} - m_{1}^{2} m_{2}^{2} m_{3}^{2} + \sqrt{ G_{123} \left( G_{123} - m_{1}^{2} m_{2}^{2} m_{3}^{2} \right) } \right] + m_{1}^{2} m_{2}^{2} \left( s_{23} - m_{2}^{2} - m_{3}^{2} \right) \left( s_{31} - m_{3}^{2} - m_{1}^{2} \right) }{G_{123} + m_{2}^{2} G_{31}} \right], $$
$$ S_{23} = \frac{1}{4} \left[ \frac{ 2 \left( s_{23} - m_{2}^{2} - m_{3}^{2} \right) \left[ G_{123} - m_{1}^{2} m_{2}^{2} m_{3}^{2} + \sqrt{ G_{123} \left( G_{123} - m_{1}^{2} m_{2}^{2} m_{3}^{2} \right) } \right] + m_{2}^{2} m_{3}^{2} \left( s_{31} - m_{3}^{2} - m_{1}^{2} \right) \left( s_{12} - m_{1}^{2} - m_{2}^{2} \right) }{G_{123} + m_{3}^{2} G_{12}} \right], $$
$$ S_{31} = \frac{1}{4} \left[ \frac{ 2 \left( s_{31} - m_{3}^{2} - m_{1}^{2} \right) \left[ G_{123} - m_{1}^{2} m_{2}^{2} m_{3}^{2} + \sqrt{ G_{123} \left( G_{123} - m_{1}^{2} m_{2}^{2} m_{3}^{2} \right) } \right] + m_{3}^{2} m_{1}^{2} \left( s_{12} - m_{1}^{2} - m_{2}^{2} \right) \left( s_{23} - m_{2}^{2} - m_{3}^{2} \right) }{G_{123} + m_{1}^{2} G_{23}} \right]. $$
I have implicitly assumed that all Regge-Mandelstam invariants, as well as all Regge-Gram invariants, are positive.
The massless limits are useful. For \(S_{12}\) you have:
$$ \lim_{m_{1} \rightarrow 0} S_{12} = \lim_{m_{1} \rightarrow 0} \frac{G_{123} \left( s_{12} - m_{2}^{2} \right)}{G_{123} + m_{2}^{2} G_{31}}, $$
$$ \lim_{m_{2} \rightarrow 0} S_{12} = s_{12} - m_{1}^{2}, $$
$$ \lim_{m_{3} \rightarrow 0} S_{12} = \lim_{m_{3} \rightarrow 0} \frac{1}{4} \left[ \frac{ 4 \left(s_{12} - m_{1}^{2} - m_{2}^{2} \right) G_{123} + m_{1}^{2} m_{2}^{2} \left( s_{23} - m_{2}^{2} \right) \left( s_{31} - m_{1}^{2} \right) }{G_{123} + m_{2}^{2} G_{31}} \right]; $$
For \(S_{23}\) you have:
$$ \lim_{m_{2} \rightarrow 0} S_{23} = \lim_{m_{2} \rightarrow 0} \frac{G_{123} \left( s_{23} - m_{3}^{2} \right)}{G_{123} + m_{3}^{2} G_{12}}, $$
$$ \lim_{m_{3} \rightarrow 0} S_{23} = s_{23} - m_{2}^{2}, $$
$$ \lim_{m_{1} \rightarrow 0} S_{23} = \lim_{m_{1} \rightarrow 0} \frac{1}{4} \left[ \frac{ 4 \left(s_{23} - m_{2}^{2} - m_{3}^{2} \right) G_{123} + m_{2}^{2} m_{3}^{2} \left( s_{31} - m_{3}^{2} \right) \left( s_{12} - m_{2}^{2} \right) }{G_{123} + m_{3}^{2} G_{12}} \right]; $$
For \(S_{31}\) you have:
$$ \lim_{m_{3} \rightarrow 0} S_{31} = \lim_{m_{3} \rightarrow 0} \frac{G_{123} \left( s_{31} - m_{1}^{2} \right)}{G_{123} + m_{1}^{2} G_{23}}, $$
$$ \lim_{m_{1} \rightarrow 0} S_{31} = s_{31} - m_{3}^{2}, $$
$$ \lim_{m_{2} \rightarrow 0} S_{31} = \lim_{m_{2} \rightarrow 0} \frac{1}{4} \left[ \frac{ 4 \left(s_{31} - m_{3}^{2} - m_{1}^{2} \right) G_{123} + m_{3}^{2} m_{1}^{2} \left( s_{12} - m_{1}^{2} \right) \left( s_{23} - m_{3}^{2} \right) }{G_{123} + m_{1}^{2} G_{12}} \right]. $$
Thus, when all three masses vanish the Sudakov vectors become the momenta of the quanta, and the 2-Mandelstam invariants become the 2-Sudakov invariants.
