In previous posts I discussed two-body and three-body Sudakov decompositions (here, here, and here). By Sudakov decomposition, I mean the process of associating a null vector to each massive vector. In the two-body problem this decomposition is unique, but for the three-body problem there are least four distinct ones (one contiguous and three composite). Here in this post, I will consider four-body problems and the many kinds of Sudakov decompositions that can be done.
There are plenty of kinematic variables in a four-body problem. You have four on-shell constraints:
$$ m_{i}^{2} = -\left\Vert p_{i} \right\Vert^{2}; $$
six 2-Mandelstam invariants:
$$ s_{ij} \equiv -\left\Vert p_{i} + p_{j} \right\Vert^{2}; $$
six Regge-Mandelstam invariants:
$$ r_{ij} \equiv \frac{s_{ij} - m_{i}^{2} - m_{j}^{2}}{2 m_{i} m_{j}}; $$
six 2-Gram invariants:
$$ G_{ij} \equiv -\det{ \begin{pmatrix} \left\Vert p_{i} \right\Vert^{2} & p_{i} \cdot p_{j} \\ p_{j} \cdot p_{i} & \left\Vert p_{j} \right\Vert^{2} \end{pmatrix} }; $$
four 3-Mandelstam invariants:
$$ s_{ijk} \equiv -\left\Vert p_{i} + p_{j} + p_{k} \right\Vert^{2}; $$
four 3-Gram invariants:
$$ G_{ijk} \equiv -\det{ \begin{pmatrix} \left\Vert p_{i} \right\Vert^{2} & p_{i} \cdot p_{j} & p_{i} \cdot p_{k} \\ p_{j} \cdot p_{i} & \left\Vert p_{j} \right\Vert^{2} & p_{j} \cdot p_{k} \\ p_{k} \cdot p_{i} & p_{k} \cdot p_{i} & \left\Vert p_{k} \right\Vert^{2} \end{pmatrix} }; $$
one 4-Mandelstam invariant:
$$ s_{1234} \equiv -\left\Vert p_{1} + p_{2} + p_{3} + p_{4} \right\Vert^{2}; $$
and one 4-Gram invariant:
$$ G_{1234} \equiv -\det{ \begin{pmatrix} \left\Vert p_{1} \right\Vert^{2} & p_{1} \cdot p_{2} & p_{1} \cdot p_{3} & p_{1} \cdot p_{4} \\ p_{2} \cdot p_{1} & \left\Vert p_{2} \right\Vert^{2} & p_{2} \cdot p_{3} & p_{2} \cdot p_{4} \\ p_{3} \cdot p_{1} & p_{3} \cdot p_{2} & \left\Vert p_{3} \right\Vert^{2} & p_{3} \cdot p_{4} \\ p_{4} \cdot p_{1} & p_{4} \cdot p_{2} & p_{4} \cdot p_{3} & \left\Vert p_{4} \right\Vert^{2} \end{pmatrix} }. $$
You also have Regge-Gram invariants:
$$ g_{ij} \equiv \frac{G_{ij}}{m_{i}^{2} m_{j}^{2}}, $$
$$ g_{ijk} \equiv \frac{G_{ijk}}{m_{i}^{2} m_{j}^{2} m_{k}^{2}}, $$
$$ g_{1234} \equiv \frac{G_{1234}}{m_{1}^{2} m_{2}^{2} m_{3}^{2} m_{4}^{2}}. $$
All of these invariants should be enough to describe any four-body problem.
Contiguous
The four-body generalization of the contiguous three-body Sudakov decomposition is
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{12}} n_{2}, $$
$$ p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{23}} n_{3}, $$
$$ p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{34}} n_{4}, $$
$$ p_{4} = n_{4} + \frac{m_{4}^{2}}{S_{41}} n_{1}. $$
Here the \(n_{i}\) vectors are null, and I have introduced 2-Sudakov invariants:
$$ S_{ij} \equiv -\left\Vert n_{i} + n_{j} \right\Vert^{2} = -2 \left( n_{i} \cdot n_{j} \right). $$
There are two other 2-Sudakov invariants besides the four that appear explicitly in the decomposition. Like before, the goal is to write the 2-Sudakov invariants in terms of 2-Mandelstam invariants and masses. Introduce Regge-Sudakov invariants:
$$ R_{ij} \equiv \frac{S_{ij}}{m_{i} m_{j}}. $$
Using the six identities of the form
$$ r_{ij} = -\frac{p_{i} \cdot p_{j}}{m_{i} m_{j}}, $$
you find the relations:
$$ 2r_{12} = R_{12} + \frac{R_{13}}{R_{23}} + \frac{1}{R_{12}}, $$
$$ 2r_{23} = R_{23} + \frac{R_{24}}{R_{34}} + \frac{1}{R_{23}}, $$
$$ 2r_{34} = R_{34} + \frac{R_{13}}{R_{41}} + \frac{1}{R_{34}}, $$
$$ 2r_{41} = R_{41} + \frac{R_{24}}{R_{12}} + \frac{1}{R_{41}}; $$
$$ 2r_{13} = R_{13} + \frac{R_{41}}{R_{34}} + \frac{R_{23}}{R_{12}} + \frac{R_{24}}{R_{12} R_{34}}, $$
$$ 2r_{24} = R_{24} + \frac{R_{12}}{R_{41}} + \frac{R_{34}}{R_{23}} + \frac{R_{13}}{R_{23} R_{41}}. $$
This set of equations is harder to solve than the analogous set for the three-body case.
Actually, there are two other non-equivalent contiguous four-body Sudakov decompositions:
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{41}} n_{4}, $$
$$ p_{4} = n_{4} + \frac{m_{4}^{2}}{S_{24}} n_{2}, $$
$$ p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{23}} n_{3}, $$
$$ p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{13}} n_{1}; $$
and
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{12}} n_{2}, $$
$$ p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{24}} n_{4}, $$
$$ p_{4} = n_{4} + \frac{m_{4}^{2}}{S_{34}} n_{3}, $$
$$ p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{13}} n_{1}. $$
Non-Contiguous
With four bodies you can also perform a non-contiguous Sudakov decomposition where you combine two two-body decompositions. For example,
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{12}} n_{2}, $$
$$ p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{12}} n_{1}; $$
$$ p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{34}} n_{4}, $$
$$ p_{4} = n_{4} + \frac{m_{4}^{2}}{S_{34}} n_{3}. $$
Then
$$ 2 r_{12} = R_{12} + \frac{1}{R_{12}}, $$
$$ 2 r_{34} = R_{34} + \frac{1}{R_{34}}; $$
and you have a linear system for the remaining Regge-Sudakov invariants. In a previous post I discussed a particular case of this decomposition for a four-point process.
Just like contiguous decompositions, there are two other non-equivalent non-contiguous four-body Sudakov decompositions:
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{13}} n_{3}, $$
$$ p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{13}} n_{1}; $$
$$ p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{24}} n_{4}, $$
$$ p_{4} = n_{4} + \frac{m_{4}^{2}}{S_{24}} n_{2}; $$
and
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{41}} n_{4}, $$
$$ p_{4} = n_{4} + \frac{m_{4}^{2}}{S_{41}} n_{1}; $$
$$ p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{23}} n_{3}, $$
$$ p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{23}} n_{2}. $$
Composite
There are two kinds of composite decompositions. Both kinds begin with a two-body step involving a 3-composite vector and a simple vector. For example, with
$$p_{123} \equiv p_{1} + p_{2} + p_{3}; \qquad s_{1234} = -\left\Vert p_{123} + p_{4} \right\Vert^{2}. $$
The two-body decomposition is
$$ p_{123} = n_{123} + \frac{s_{123}}{S_{(123)4}} n_{4}, $$
$$ p_{4} = n_{4} + \frac{m_{4}^{2}}{S_{(123)4}} n_{123}; $$
where \(n_{123}\) is a null vector, and
$$ S_{(123)4} \equiv -\left\Vert n_{123} + n_{4} \right\Vert^{2}. $$
From
$$ \frac{p_{123} \cdot p_{4}}{\sqrt{s_{123}} m_{4}} = \frac{s_{1234} - s_{123} - m_{4}^{2}}{2 \sqrt{s_{123}} m_{4}}, $$
it follows that
$$ S_{(123)4} = \frac{1}{2} \left( s_{1234} - s_{123} - m_{4}^{2} + \sqrt{ \left( s_{1234} - s_{123} - m_{4}^{2} \right)^{2} - 4 s_{123} m_{4}^{2}} \right). $$
At this step you have a null vector for \(p_{4}\). What is left is to find null vectors for the rest of the massive momenta. Here is where the fork arises. You can either perform a contiguous three-body decomposition, or a composite three-body decomposition. The outcome in both cases will be very different from the contiguous four-body decomposition above.
