FR05-A · How three diagrams become one expression

Gluon Scattering Amplitudes

Why can changing a polarization reference move every diagram and leave their sum fixed? A polarization reference changes how each diagram describes the same transverse gluon. The exchange diagrams and contact term cancel the reference dependence. Their sum agrees with the compact helicity expression.

Drag an outgoing beam and compare three Feynman contributions with one helicity formula. Follow a fifth gluon toward soft or collinear emission and see factorization emerge.

Amplitude magnitude $1.49$

Comparison residual $0$

Four-point Ward residual $0$

Numerical condition estimate $8.37$

Change the geometry; follow the cancellation

A short answer to a long calculation

Drag the gold outgoing handle. Compare the graph contributions below, then follow a soft or collinear emission.

CalculationHard outgoing direction

\theta

70^\circ

Polarization reference

Polarization reference direction

\varphi_q

30^\circ

This auxiliary direction changes each graph, so compare the phasors below. It is not an extra particle in the scene.

Conventions and keyboard controls

$g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1),\quad E_{\rm beam}=1$

Incoming legs are 1 and 2. Negative helicities are assigned to those all-outgoing labels. Positive outgoing legs follow the displayed color order. Beam arrows show momentum directions, not particle tracks.

Focus the scene and use the left or right arrow key to turn the hard outgoing direction.

Model assumptions

The model evaluates coupling-stripped, color-ordered, massless tree amplitudes with two negative helicities. The momentum arrows show directions, with lengths rescaled for visibility. Soft and collinear comparisons use an explicitly mapped on-shell four-point state. Loop corrections, color sums, parton showers and detector events are outside this calculation.

Analytic tree model

The three arrows add to the same answer

First exchange

s_{12}

5.83+0.875i

Second exchange

s_{23}

-5.98+1.09i

Contact graph

\text{contact}

0.147-0.478i

First exchangeSecond exchangeContact graphSum / helicity answer

Complex arrows are placed head to tail. The white endpoint is the compact helicity result. Changing the reference moves the intermediate endpoints.

Numerical details

\sum_i k_i^\mu=0,\qquad k_i^2=0

The four-point condition is the sum of the magnitudes of the three graph contributions divided by the magnitude of their sum. In factorization modes it is the inverse opening-angle scale of the adjacent spinor bracket. Both are local estimates, not rigorous forward-error bounds.

Dixon, A brief introduction to modern amplitude methods (2013), sections 3–5

Start with the essentials

Focus question: Why can changing a polarization reference move every diagram and leave their sum fixed?
One-sentence intuition: The simple helicity answer contains the cancellation that the separate Feynman graphs perform explicitly.

Core mathematical model

Four-point amplitude with two negative helicities

\begin{aligned}A_4(1^-,2^-,3^+,4^+)&\\&=i\frac{\langle12\rangle^4}{\langle12\rangle\langle23\rangle\langle34\rangle\langle41\rangle}\end{aligned}

Angle brackets are antisymmetric products of two-component momentum spinors. Their product with the reversed square bracket gives the pair invariant. All momenta are outgoing; incoming energies are negative. Couplings and the color trace are stripped.

Replacing a polarization by its momentum

A_4(\ldots,\varepsilon_j\mapsto k_j,\ldots)=0

The three diagram contributions are contracted again after the replacement, for each of the four legs. The displayed maximum is normalized by the larger of one and the compact amplitude magnitude.

Adding a soft positive-helicity gluon

\begin{aligned}A_5(1,2,3,5^+,4)&\\\underset{E_5\to0}{\longrightarrow}\;&\frac{\langle34\rangle}{\langle35\rangle\langle54\rangle}A_4(1,2,3,4)\end{aligned}

The soft factor depends on the two neighbors in the color order. The finite-energy comparison maps the hard pair to back-to-back unit-energy momenta, so its relative residual tends to zero as the emission softens.

Common difficulties

A graph is one part of the calculation

Typical misconceptionEach graph should already be independent of the polarization reference.

Better mental modelThe reference cancels in the sum required by gauge symmetry. Dropping the contact graph generally spoils that cancellation.

A large amplitude needs careful arithmetic

Typical misconceptionA small factorization residual guarantees numerical stability at the pole.

Better mental modelThe residual tests the approximation at the selected point. Small spinor brackets or cancellations can amplify roundoff separately; the condition readout helps reveal this.

Run the experiment

01
Change the reference
Choose Three graphs, one answer. Open Polarization reference and move its direction through a full turn. Watch the three arrows in the data row.
What to observe: The separate arrows move; their sum stays on the compact result. The Ward residual remains near floating-point roundoff.
02
Make one gluon soft
Choose A gluon loses its energy. Predict the trend, then reduce the emitted energy by a factor of ten at a time.
What to observe: The amplitude grows, yet the relative difference from the soft approximation falls. The log plot shows this convergence across the full selected scan.
03
Bring two directions together
Choose Two outgoing beams merge. Keep the emitted energy fixed and reduce the opening angle. Save a snapshot with the full scan.
What to observe: The adjacent pair invariant tends to zero. The collinear splitting approximation improves, while the spinor-bracket condition estimate grows.

Gluon Scattering Amplitudes

The three arrows add to the same answer