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

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