Locate a collision with invariant variables

Throughout this module use $\hbar=c=1$ and metric $(+,-,-,-)$. Write a physical positive-energy momentum as $p=(E,\mathbf p)$, so $p^2=E^2-|\mathbf p|^2=m^2$. Energies, masses, and momenta can all be quoted in GeV in this convention; squared invariants have units of GeV squared. For $a+b\to c+d$, label the actual incoming momenta $p_a,p_b$ and outgoing momenta $p_c,p_d$. Do not silently switch to a convention in which every momentum points into a diagram.

Momentum conservation gives $p_a+p_b=p_c+p_d$. Squaring the total incoming momentum defines s, while differences between a chosen incoming and outgoing momentum define t and u. Expanding these expressions and using each on-shell condition gives their sum as the sum of the four squared masses. Only two invariants are independent for fixed external masses, but that identity alone does not establish that arbitrary proposed values correspond to a physical collision with positive energies and real scattering angles.

In the center-of-momentum frame, the incoming three-momenta are opposite, as are the outgoing ones. Let $\theta$ be the angle from particle a to particle c in that frame. The transfer t depends linearly on $\cos\theta$, so its physical interval follows from $-1\le\cos\theta\le1$. The incoming and outgoing momentum magnitudes are fixed by s and the masses. Threshold requires enough invariant energy for both final masses; below threshold there is no positive-energy two-daughter solution.

For massless two-to-two scattering, the incoming and outgoing energies are each half the center-of-momentum energy. Forward scattering has zero t, while backward scattering has the most negative t. The companion invariant u covers the complementary transfer. This familiar negative-transfer region belongs to the specified physical channel. Crossing particles into a different initial state changes which invariant plays the role of total energy; an analytic amplitude evaluated in a crossed region is not automatically a physical event in the original beam configuration.

For a stationary target b, the invariant is $s=m_a^2+m_b^2+2m_bE_{a,\mathrm{lab}}$. The projectile energy here is total energy, including rest mass. The laboratory total energy is larger than $\sqrt s$ whenever the overall momentum is nonzero. The boost to the center-of-momentum frame has velocity equal to total momentum divided by total energy. Equal-energy head-on beams can make that total momentum vanish, which explains the different mass reach of collider and fixed-target arrangements.

An invariant calculation is a powerful consistency check on a reconstruction. Compute the mass from the summed four-vector in one frame, boost the complete set of momenta, and verify that the result agrees. Three-momentum magnitudes and angles need not agree between frames. A detector's beam energy alone also does not determine the invariant energy of a partonic collision inside two hadrons, because the colliding constituents can carry different unknown fractions. Keep the level of description—whole beams or selected constituents—explicit.