Two-Dimensional Four-Point Kinematics (5)

Thu 17 November 2016
Physics
#kinematics, #two-dimensional

You have two incoming null momenta ( $P_{1}$ and $P_{2}$ ) and two outgoing null momenta ( $P_{3}$ and $P_{4}$ ). This problem does not have a non-relativistic limit.

Energy-Momentum Vectors

In a more traditional approach, I would have chosen the four external null momenta as

$P_{i} = \begin{pmatrix} k_{i} & k_{i} \end{pmatrix}.$

But then any sum or difference of momenta will be null too. This leads to all invariants being zero.

The four external null momenta are

$P_{1} = \begin{pmatrix} k_{1} & k_{1} \end{pmatrix};$

$P_{2} = \begin{pmatrix} k_{2} & -k_{2} \end{pmatrix};$

$P_{3} = \begin{pmatrix} k_{3} & k_{3} \end{pmatrix};$

$P_{4} = \begin{pmatrix} k_{4} & -k_{4} \end{pmatrix}.$

Here $k_{1} > 0$ , $k_{2} > 0$ , $k_{3} > 0$ , and $k_{4} > 0$ .

Conservation

The conservation of energy leads to

$k_{1} + k_{2} = k_{3} + k_{4}.$

Similarly, the conservation of momentum leads to

$k_{1} - k_{2} = k_{3} - k_{4}.$

These two expressions can be combined to yield two equivalent expressions:

$k_{1} = k_{3}; \qquad k_{2} = k_{4}.$

That is, this particular set of external momenta corresponds to the identity permutation.

2-Mandelstam Invariants

You can introduce three 2-Mandelstam invariants:

$s = -(P_{1} + P_{2})^{2} = -(P_{3} + P_{4})^{2};$

$t = -(P_{1} - P_{3})^{2} = -(P_{2} - P_{4})^{2};$

$u = -(P_{1} - P_{4})^{2} = -(P_{2} - P_{3})^{2}.$

Due to the conservation laws, you have

$s + t + u = 0.$

Regge-Mandelstam Invariants

You can introduce six Regge-Mandelstam invariants:

$r_{12} \equiv -\frac{P_{1} \cdot P_{2}}{k_{1} k_{2}} = \frac{s}{2 k_{1} k_{2}} = 2;$

$r_{34} \equiv -\frac{P_{3} \cdot P_{4}}{k_{3} k_{4}} = \frac{s}{2 k_{1} k_{2}} = 2;$

$r_{13} \equiv -\frac{P_{1} \cdot P_{3}}{k_{1} k_{3}} = -\frac{t}{2 k_{1}^{2}} = 0;$

$r_{24} \equiv -\frac{P_{2} \cdot P_{4}}{k_{2} k_{4}} = -\frac{t}{2 k_{2}^{2}} = 0;$

$r_{14} \equiv -\frac{P_{1} \cdot P_{4}}{k_{1} k_{4}} = -\frac{u}{2 k_{1} k_{2}} = 2;$

$r_{23} \equiv -\frac{P_{2} \cdot P_{3}}{k_{2} k_{3}} = -\frac{u}{2 k_{1} k_{2}} = 2.$

It follows that

$s \geq 0, \qquad t = 0, \qquad u \leq 0.$

These conditions define the range of values allowed for the 2-Mandelstam invariants. The first one says that $s$ only takes positive values. Thus, the vector $P_{s} = P_{1} + P_{2} = P_{3} + P_{4}$ is slow and you can write

$P_{s} = \begin{pmatrix} \sqrt{s} \cosh{(\xi_{s})} & \sqrt{s} \sinh{(\xi_{s})} \end{pmatrix}.$

The second condition says that $t$ is zero. But since $k_{3} = k_{1}$ , it follows that $P_{3} = P_{1}$ and thus the vector $P_{t} = P_{1} - P_{3}$ is identically zero. The third condition says that $u$ is negative. This means that the vector $P_{u} = P_{1} - P_{4} = P_{3} - P_{2}$ is fast and you can write

$P_{u} = \begin{pmatrix} \sqrt{-u} \sinh{(\rho_{u})} & \sqrt{-u} \cosh{(\rho_{u})} \end{pmatrix}.$

However, since $P_{s} \cdot P_{u} = 0$ , you have $\rho_{u} = \xi_{s}$ .

You can write $k_{2}$ , $k_{3}$ , and $k_{4}$ in terms of $k_{1}$ and $s$ :

$k_{2} = \frac{s}{4 k_{1}}, \qquad k_{3} = k_{1}, \qquad k_{4} = \frac{s}{4 k_{1}}.$

Composite Rapidities

The slow rapidity $\xi_{s}$ associated to $P_{s}$ can be writen as

$\xi_{s} = \log{\left( \frac{2 k_{1}}{\sqrt{s}} \right)}.$