Show that
$$
s+t+u=m_A^2+m_B^2+m_C^2+m_D^2,
$$
where $m_i$ is the rest mass of particle $i$.
To display the kinematic (or physical) regions of processes related by crossing, we construct a two-dimensional plot which maintains the symmetry of $s, t, u$. The three axes $s, t, u=0$ are drawn (Fig. 4.7) to form an equilateral triangle of height $\sum m_i^2$. From any point inside and also outside (if attention is paid to the signs of $s, t, u)$ the triangle, the sum of the perpendicular distances to the axes is equal to the height of the triangle [see (4.44)].
It is easy to show that $s$ is the square of the total center-of-mass energy of the process $\mathrm{AB} \rightarrow \mathrm{CD}$ [see (3.12) or Exercise 4.6]. It is conventional to take the reaction under study to be the $s$ channel process. In our last example, this was $\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{e}^{-} \mathrm{e}^{+}$scattering. The crossed reactions $\mathrm{A} \overline{\mathrm{D}} \rightarrow \mathrm{C} \overline{\mathrm{B}}$ and $\overline{\mathrm{D}} \mathrm{B} \rightarrow \mathrm{C} \overline{\mathrm{A}}$ are called the $u$ and $t$ channels, respectively, since $u$ and $t$ are equal to the square of the total center-of-mass energy in the respective channels (see Exercise 4.7).