The most general \(2 \rightarrow 2\) process has the form
$$A(p_{1}) + B(p_{2}) \longrightarrow Z(p_{3}) + Y(p_{4}).$$
Note that this process has no permutation symmetry; it is inelastic. The four energy-momentum vectors \(p_{I}\) satisfy on-shell constraints,
$$p_{I}^{2} = - m_{I}^{2} \qquad I = 1, 2, 3, 4;$$
and a conservation constraint,
$$p_{1} + p_{2} = p_{3} + p_{4}.$$
I will work with the assumption that quanta \(1\) and \(2\) are incoming, and quanta \(3\) and \(4\) are outgoing.
Momentum Invariants
Due to the conservation constraint, there are only three 2-Mandelstam invariants:
$$s \equiv -(p_{1} + p_{2})^{2} = -(p_{3} + p_{4})^{2} \qquad t \equiv -(p_{1} - p_{3})^{2} = -(p_{2} - p_{4})^{2} \qquad u \equiv -(p_{1} - p_{4})^{2} = -(p_{2} - p_{3})^{2}. $$
These satisfy a linear relation involving the masses:
$$s + t + u = m_{1}^{2} + m_{2}^{2} + m_{3}^{2} + m_{4}^{2}.$$
In terms of the 2-Mandelstam invariants and the masses, the 2-Gram invariants are:
$$G_{IJ}(s) = \frac{1}{4} [s - (m_{I} - m_{J})^{2}] [s - (m_{I} + m_{J})^{2}] \qquad (I, J) = (1, 2), (3, 4);$$
$$G_{IJ}(t) = \frac{1}{4} [t - (m_{I} - m_{J})^{2}] [t - (m_{I} + m_{J})^{2}] \qquad (I, J) = (1, 3), (2, 4);$$
$$G_{IJ}(u) = \frac{1}{4} [u - (m_{I} - m_{J})^{2}] [u - (m_{I} + m_{J})^{2}] \qquad (I, J) = (1, 4), (2, 3).$$
There is also a nontrivial 3-Gram invariant, but I will not bother with it.
Center-of-Momentum Frame
In the center-of-momentum reference frame you can write the energy-momentum vectors in the form
$$p_{1} = (E_{1}, \mathbf{p}_{1}) \qquad p_{2} = (E_{2}, -\mathbf{p}_{1}) \qquad p_{3} = (E_{3}, \mathbf{p}_{3}) \qquad p_{4} = (E_{4}, -\mathbf{p}_{3}).$$
This way, part of the conservation constraint is explicitly satisfied. The remaining part is the conservation of energy
$$E_{1} + E_{2} = E_{3} + E_{4}.$$
While in the center-of-momentum frame, you can write most kinematic quantities in terms of the 2-Mandelstam invariants and the masses.
Energy and Momentum
Using the on-shell constraints, you can write the energies in terms of the spatial momenta:
$$E_{1} = \sqrt{m_{1}^{2} + \mathbf{p}_{1}^{2}} \qquad E_{2} = \sqrt{m_{2}^{2} + \mathbf{p}_{1}^{2}} \qquad E_{3} = \sqrt{m_{3}^{2} + \mathbf{p}_{3}^{2}} \qquad E_{4} = \sqrt{m_{4}^{2} + \mathbf{p}_{3}^{2}}.$$
The four energies are manifestly positive. Using \(s = -(p_{1} + p_{2})^{2}\) you can find that
$$s = (E_{1} + E_{2})^{2} = (E_{3} + E_{4})^{2},$$
which tells you that \(s\) has to be positive. With these relations you can solve for some of the energies:
$$E_{2} = \sqrt{s} - E_{1} \qquad E_{4} = \sqrt{s} - E_{3}.$$
Using these you can find expressions for the magnitude of the spatial momenta in terms of \(s\) and the masses:
$$\vert \mathbf{p}_{1} \vert = \frac{\sqrt{\Lambda_{12}(s)}}{2 \sqrt{s}} \qquad \vert \mathbf{p}_{3} \vert = \frac{\sqrt{\Lambda_{34}(s)}}{2 \sqrt{s}} \qquad \Lambda_{IJ}(s) \equiv [s - (m_{I} - m_{J})^{2}] [s - (m_{I} + m_{J})^{2}].$$
With these magnitudes, you can find expressions for the energies in terms of \(s\) and the masses:
$$E_{1} = \frac{s + (m_{1} - m_{2})(m_{1} + m_{2})}{2 \sqrt{s}} \qquad E_{2} = \frac{s - (m_{1} - m_{2})(m_{1} + m_{2})}{2 \sqrt{s}},$$
and
$$E_{3} = \frac{s + (m_{3} - m_{4})(m_{3} + m_{4})}{2 \sqrt{s}} \qquad E_{4} = \frac{s - (m_{3} - m_{4})(m_{3} + m_{4})}{2 \sqrt{s}}.$$
Note that the requirement that the energy be positive leads to a restriction on the values that \(s\) can take.
Scattering Angle
With the results for the energies, you can obtain
$$E_{1} - E_{3} = \frac{(m_{1} - m_{2})(m_{1} + m_{2}) - (m_{3} - m_{4})(m_{3} + m_{4})}{2 \sqrt{s}},$$
and
$$E_{1} - E_{4} = \frac{(m_{1} - m_{2})(m_{1} + m_{2}) + (m_{3} - m_{4})(m_{3} + m_{4})}{2 \sqrt{s}}.$$
These particular combinations are important, because you also have
$$\vert \mathbf{p}_{1} - \mathbf{p}_{3} \vert^{2} = (E_{1} - E_{3})^{2} - t \qquad \vert \mathbf{p}_{1} + \mathbf{p}_{3} \vert^{2} = (E_{1} - E_{4})^{2} - u,$$
and thus
$$4 (\mathbf{p}_{1} \cdot \mathbf{p}_{3}) = \vert \mathbf{p}_{1} + \mathbf{p}_{3} \vert^{2} - \vert \mathbf{p}_{1} - \mathbf{p}_{3} \vert^{2} = (E_{1} - E_{4})^{2} - (E_{1} - E_{3})^{2} - u + t,$$
giving
$$\mathbf{p}_{1} \cdot \mathbf{p}_{3} = \frac{(m_{1} - m_{2})(m_{3} - m_{4})(m_{1} + m_{2})(m_{3} + m_{4}) - s (u - t)}{4 s}.$$
This result allows you to find an expression for the cosine of the angle between \(\mathbf{p}_{1}\) and \(\mathbf{p}_{3}\):
$$z_{s} \equiv \cos{(\theta_{s})} \equiv \frac{\mathbf{p}_{1} \cdot \mathbf{p}_{3}}{\vert \mathbf{p}_{1} \vert \vert \mathbf{p}_{3} \vert} = \frac{(m_{1} - m_{2})(m_{3} - m_{4})(m_{1} + m_{2})(m_{3} + m_{4}) - s (u - t)}{\sqrt{\Lambda_{12}(s)} \sqrt{\Lambda_{34}(s)}}.$$
Here I have labeled the angle with an \(s\) since this is the scattering angle along the \(s\)-channel.
Using
$$\vert \mathbf{p}_{1} \vert^{2} + \vert \mathbf{p}_{3} \vert^{2} = \frac{(m_{1} - m_{2})^{2}(m_{1} + m_{2})^{2} + (m_{3} - m_{4})^{2}(m_{3} + m_{4})^{2} - 2s (u + t)}{4 s},$$
and
$$\vert \mathbf{p}_{1} \vert^{2} + \vert \mathbf{p}_{3} \vert^{2} = \vert \mathbf{p}_{1} \vert \vert \mathbf{p}_{3} \vert \left( \frac{\vert \mathbf{p}_{1} \vert}{\vert \mathbf{p}_{3} \vert} + \frac{\vert \mathbf{p}_{3} \vert}{\vert \mathbf{p}_{1} \vert} \right),$$
leads to another expression for \(z_{s}\):
$$z_{s} = \left[ \frac{\Lambda_{12}(s) + \Lambda_{34}(s)}{\sqrt{\Lambda_{12}(s) \Lambda_{34}(s)}} \right] \left[ \frac{(m_{1} - m_{2})(m_{3} - m_{4})(m_{1} + m_{2})(m_{3} + m_{4}) - s (u - t)}{(m_{1} - m_{2})^{2}(m_{1} + m_{2})^{2} + (m_{3} - m_{4})^{2}(m_{3} + m_{4})^{2} - 2s (u + t)} \right].$$
This result is useful when making some of the masses identical.
Speed and Rapidity
A quantum with mass \(m \geq 0\) and energy \(E \geq m\) has a speed \(0 \leq \vert \mathbf{v} \vert < 1\) (we work with units where \(c = 1\)) given by
$$\vert \mathbf{v} \vert = \frac{\sqrt{E^{2} - m^{2}}}{E}$$
In the center-of-momentum frame, you can write the individual speed of the external quanta in terms of \(s\) and the masses:
$$\vert \mathbf{v}_{1} \vert = \frac{\sqrt{\Lambda_{12}(s)}}{s + (m_{1} - m_{2})(m_{1} + m_{2})} \qquad \vert \mathbf{v}_{2} \vert = \frac{\sqrt{\Lambda_{12}(s)}}{s - (m_{1} - m_{2})(m_{1} + m_{2})},$$
and
$$\vert \mathbf{v}_{3} \vert = \frac{\sqrt{\Lambda_{34}(s)}}{s + (m_{3} - m_{4})(m_{3} + m_{4})} \qquad \vert \mathbf{v}_{4} \vert = \frac{\sqrt{\Lambda_{34}(s)}}{s - (m_{3} - m_{4})(m_{3} + m_{4})}.$$
With the speed \(\vert \mathbf{v} \vert\) you can find the rapidity \(\varphi\) via
$$\varphi \equiv \operatorname{artanh}{(\vert \mathbf{v} \vert)}.$$
Note that since \(0 \leq \vert \mathbf{v} \vert < 1\), then it follows that \(0 \leq \varphi < \infty\). In the center-of-momentum frame, the individual rapidity of the external quanta are
$$\varphi_{1} = \frac{1}{2} \log{\left[ \frac{(m_{1} - m_{2})(m_{1} + m_{2}) + s + \sqrt{\Lambda_{12}(s)}}{(m_{1} - m_{2})(m_{1} + m_{2}) + s - \sqrt{\Lambda_{12}(s)}} \right]} \qquad \varphi_{2} = \frac{1}{2} \log{\left[ \frac{(m_{1} - m_{2})(m_{1} + m_{2}) - s - \sqrt{\Lambda_{12}(s)}}{(m_{1} - m_{2})(m_{1} + m_{2}) - s + \sqrt{\Lambda_{12}(s)}} \right]},$$
and
$$\varphi_{3} = \frac{1}{2} \log{\left[ \frac{(m_{3} - m_{4})(m_{3} + m_{4}) + s + \sqrt{\Lambda_{34}(s)}}{(m_{3} - m_{4})(m_{3} + m_{4}) + s - \sqrt{\Lambda_{34}(s)}} \right]} \qquad \varphi_{4} = \frac{1}{2} \log{\left[ \frac{(m_{3} - m_{4})(m_{3} + m_{4}) - s - \sqrt{\Lambda_{34}(s)}}{(m_{3} - m_{4})(m_{3} + m_{4}) - s + \sqrt{\Lambda_{34}(s)}} \right]}.$$
Other results include the sum of the two incoming rapidities,
$$\varphi_{1} + \varphi_{2} = \frac{1}{2} \log{\left[ \frac{m_{1}^{2} + m_{2}^{2} - s - \sqrt{\Lambda_{12}(s)}}{m_{1}^{2} + m_{2}^{2} - s + \sqrt{\Lambda_{12}(s)}} \right]} = \log{\left[ \frac{(m_{1} + m_{2})^{2} - s + \sqrt{\Lambda_{12}(s)}}{(m_{1} + m_{2})^{2} - s - \sqrt{\Lambda_{12}(s)}} \right]},$$
and the sum of the two outgoing rapidities,
$$\varphi_{3} + \varphi_{4} = \frac{1}{2} \log{\left[ \frac{m_{3}^{2} + m_{4}^{2} - s - \sqrt{\Lambda_{34}(s)}}{m_{3}^{2} + m_{4}^{2} - s + \sqrt{\Lambda_{34}(s)}} \right]} = \log{\left[ \frac{(m_{3} + m_{4})^{2} - s + \sqrt{\Lambda_{34}(s)}}{(m_{3} + m_{4})^{2} - s - \sqrt{\Lambda_{34}(s)}} \right]}.$$
These two expressions appear in loop amplitudes.
Physical Scattering Region
Quantities like the energies \(E_{I}\) or the spatial momenta \(\mathbf{p}_{I}\) have a clear physical interpretation. For a scattering process you expect the energies and the magnitude of the spatial momenta to be positive. Requiring \(\vert \mathbf{p}_{1} \vert > 0\) and \(\vert \mathbf{p}_{3} \vert > 0\) leads to
$$s > \operatorname{max}{\lbrace (m_{1} + m_{2})^{2}, (m_{3} + m_{4})^{2} \rbrace}.$$
In other words, \(s\) must be above the \(s\)-channel thresholds. This means that inside the physical scattering region you have \(G_{12} > 0\) and \(G_{34} > 0\).
Similarly, requiring \(\vert \mathbf{p}_{1} - \mathbf{p}_{3} \vert^{2} \geq 0\) and \( \vert \mathbf{p}_{1} + \mathbf{p}_{3} \vert^{2} \geq 0\) leads to
$$4 s t \leq [(m_{1} - m_{2})(m_{1} + m_{2}) - (m_{3} - m_{4})(m_{3} + m_{4})]^{2},$$
and
$$4 s u \leq [(m_{1} - m_{2})(m_{1} + m_{2}) + (m_{3} - m_{4})(m_{3} + m_{4})]^{2}.$$
These inequalities can be reformulated as bounds for the values that \(t\) and \(u\) can take as functions of \(s\) and the masses.
Forward Scattering
When \(z_{s} \rightarrow 1\) (i.e. \(\theta_{s} \rightarrow 0\)), you are in the regime of forward scattering. Using
$$u_{f} = m_{1}^{2} + m_{2}^{2} + m_{3}^{2} + m_{4}^{2} - s - t_{f},$$
and setting \(z_{s} = 1\) leads to
$$t_{f} = \frac{[(m_{1} - m_{2})(m_{1} + m_{2}) - (m_{3} - m_{4})(m_{3} + m_{4})]^{2} - [\sqrt{\Lambda_{12}(s)} - \sqrt{\Lambda_{34}(s)} ]^{2}}{4s},$$
which in turn leads to
$$u_{f} = \frac{[(m_{1} - m_{2})(m_{1} + m_{2}) + (m_{3} - m_{4})(m_{3} + m_{4})]^{2} - [\sqrt{\Lambda_{12}(s)} + \sqrt{\Lambda_{34}(s)} ]^{2}}{4s}.$$
Backward Scattering
When \(z_{s} \rightarrow -1\) (i.e. \(\theta_{s} \rightarrow \pi\)), you are in the regime of backward scattering. Using
$$u_{b} = m_{1}^{2} + m_{2}^{2} + m_{3}^{2} + m_{4}^{2} - s - t_{b},$$
and setting \(z_{s} = -1\) leads to
$$t_{b} = \frac{[(m_{1} - m_{2})(m_{1} + m_{2}) - (m_{3} - m_{4})(m_{3} + m_{4})]^{2} - [\sqrt{\Lambda_{12}(s)} + \sqrt{\Lambda_{34}(s)} ]^{2}}{4s},$$
which in turn leads to
$$u_{b} = \frac{[(m_{1} - m_{2})(m_{1} + m_{2}) + (m_{3} - m_{4})(m_{3} + m_{4})]^{2} - [\sqrt{\Lambda_{12}(s)} - \sqrt{\Lambda_{34}(s)} ]^{2}}{4s}.$$
Orthogonal Scattering
Forward (backward) scattering corresponds to the special case when the incoming spatial momentum \(\mathbf{p}_{1}\) and the outgoing spatial momentum \(\mathbf{p}_{3}\) are collinear (anticollinear). When \(z_{s} \rightarrow 0\) (i.e. \(\theta_{s} \rightarrow \pi / 2\)), you are in regime of orthogonal scattering (i.e. \(\mathbf{p}_{1}\) and \(\mathbf{p}_{3}\) are orthogonal). Using
$$u_{o} = m_{1}^{2} + m_{2}^{2} + m_{3}^{2} + m_{4}^{2} - s - t_{o}$$
and setting \(z_{s} = 0\) leads to
$$t_{o} = \frac{[(m_{1} - m_{2})(m_{1} + m_{2}) - (m_{3} - m_{4})(m_{3} + m_{4})]^{2} - \Lambda_{12}(s) - \Lambda_{34}(s)}{4s},$$
which in turn leads to
$$u_{o} = \frac{[(m_{1} - m_{2})(m_{1} + m_{2}) + (m_{3} - m_{4})(m_{3} + m_{4})]^{2} - \Lambda_{12}(s) - \Lambda_{34}(s)}{4s}.$$
Note that you can interpret \(t_{o}\) as the midpoint of the interval \(t_{b} < t < t_{f}\). Similarly, you can interpret \(u_{o}\) as the midpoint of the interval \(u_{f} < u < u_{b}\).
Cross Processes
You can find two other \(2 \rightarrow 2\) processes by crossing an incoming quantum with an outgoing quantum. If you cross \(B\) with \(Z\), you find the process
$$A(p_{1}) + \bar{Z}(\bar{p}_{2}) \longrightarrow \bar{B}(\bar{p}_{3}) + Y(p_{4}).$$
You can use the same kinematic variables above if you let \(\bar{p}_{2} = -p_{3}\) and \(\bar{p}_{3} = -p_{2}\).
Similarly, if you cross \(B\) with \(Y\), you find the process
$$A(p_{1}) + \bar{Y}(\bar{p}_{2}) \longrightarrow Z(p_{3}) + \bar{B}(\bar{p}_{4}).$$
Again, you can use the same kinematic variables above if you let \(\bar{p}_{2} = -p_{4}\) and \(\bar{p}_{4} = -p_{2}\). Note that all of these processes are inelastic.
