Why bother with Mandelstam variables?#

When we calculate scattering cross sections in quantum field theory, we get amplitudes that depend on the four-momenta of the particles. But different observers (in different reference frames) measure different energies and momenta. If we express our cross section in terms of frame-dependent quantities like $E_\text{CM}$ and $\theta$, we’re tied to one specific frame.

Mandelstam variables fix this. They are Lorentz-invariant combinations of four-momenta. They have the same value in every inertial frame. This means:

• We can do the calculation in the most convenient frame (usually CM)
• Express the final result in terms of invariants
• Any experimenter in any frame can use our result

For a $2 \to 2$ scattering process $a + b \to c + d$, the three Mandelstam variables are:

Only two are independent (since $s + t + u = m_a^2 + m_b^2 + m_c^2 + m_d^2$), but keeping all three makes the symmetry of the problem manifest.

Setup: $e^+ e^- \to \mu^+ \mu^-$ in the CM frame#

Now the specific process $e^+ e^- \to \mu^+ \mu^-$. Label the four-momenta:

In the center-of-mass frame, the initial particles collide head-on along the $x$-axis. Without loss of generality, we can choose the final-state particles to scatter in the $xy$-plane:

where $\theta$ is the scattering angle, i.e. the angle between the incoming electron and the outgoing muon.

The CM frame is convenient because momentum conservation tells us $\vec{p}_1 + \vec{p}_2 = 0$ and $\vec{p}_3 + \vec{p}_4 = 0$. In the CM frame, the particles have equal and opposite momenta, which dramatically simplifies the algebra.

The massless case: $m_e = m_\mu = 0$#

First the simpler case, where all particles are massless. This is a good approximation for high-energy colliders where $E_\text{CM} \gg m_e, m_\mu$.

In the CM frame with massless particles:

where $E_\text{CM}$ is the total center-of-mass energy. The four-momenta become:

Each particle has energy $E_\text{CM}/2$, and the spatial components reflect the head-on collision geometry with the final-state particles scattering at angle $\theta$.

Calculating $s$: the total energy available for scattering#

This is the Mandelstam variable $s$, the total invariant mass squared of the initial state (which equals that of the final state):

So $s$ is just the total center-of-mass energy squared. This is why collider energies are quoted as $\sqrt{s}$: it is the invariant energy available for making particles.

Calculating $t$: the momentum transfer#

$t$ is the invariant momentum transfer squared. You can think of it as measuring how much the electron’s four-momentum changes when it turns into a muon:

$t$ is spacelike (negative) for physical scattering. It measures how “hard” the scattering is: small $|t|$ means glancing collisions (small $\theta$), large $|t|$ means hard scattering ($\theta \approx \pi$). In particle physics experiments, measuring the $t$-distribution tells us about the interaction’s strength at different momentum scales.

Calculating $u$: the crossing symmetry variable#

$u$ might look redundant at first, but it makes the crossing symmetry of the theory manifest:

Physically, the $u$-channel corresponds to crossing the final-state muon to an initial-state anti-muon. This might seem abstract now, but crossing symmetry is a strong constraint on scattering amplitudes: the same function that describes $e^+ e^- \to \mu^+ \mu^-$ also describes $e^+ \mu^- \to e^+ \mu^-$ after appropriate substitutions.

The relation $s + t + u = 0$#

Adding the three Mandelstam variables:

This is the massless limit of the general relation $s + t + u = \sum_i m_i^2$. For massless particles, the sum of the Mandelstam variables is zero. In practice this is very handy, both for simplifying expressions and for checking a calculation.

Rewriting the cross section in terms of invariants#

The differential cross section for this process in the CM frame is:

Using $E_\text{CM}^2 = s$ and the relation we just derived:

we get:

A caveat: this is still the angular distribution in the CM frame, because $d\Omega$ is a CM-frame solid angle, so $d\sigma/d\Omega$ is not itself Lorentz-invariant. The fully invariant object is $d\sigma/dt$. With massless kinematics $t = -\tfrac{s}{2}(1-\cos\theta)$, so $dt = \tfrac{s}{2}\,d\cos\theta$ and, after the trivial azimuthal integral, $d\Omega = 2\pi\,d\cos\theta = \tfrac{4\pi}{s}\,dt$:

The point is that $d\sigma/dt$ depends only on invariants. An experimenter at any collider, in any frame, can plug in their measured values of $s, t, u$ and compare with this prediction. We did the calculation in the CM frame, but the result is valid everywhere.

Adding masses back in: the general case#

For completeness, here is what happens when the electron and muon masses are included. Define:

The four-momenta in the CM frame become:

where $E_1 = E_2 = \sqrt{p^2 + m_1^2}$ and $E_3 = E_4 = \sqrt{p'^2 + m_2^2}$.

The general relation is:

The masses enter purely through the on-shell conditions $p_i^2 = m_i^2$. Expanding $s + t + u$ and using momentum conservation $p_1 + p_2 = p_3 + p_4$ leaves exactly $\sum_i p_i^2$. At high energies ($\sqrt{s} \gg m_1, m_2$), we recover $s + t + u \approx 0$.

Why the invariants matter#

So the Mandelstam variables are Lorentz-invariant, they make crossing symmetry manifest, and they let you write results without reference to a frame. We computed $\frac{d\sigma}{d\Omega}$ in the CM frame, but once the answer is in terms of $s, t, u$, anyone can use it, whether they are at LEP, the LHC or a future collider. The same machinery works for any $2 \to 2$ process.