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Nuclear, particle & quantum-information sequence · Interactive lesson

Interactions, mediators and gravity's exclusion

Before any cross-section, you should be able to say how many force carriers the Standard Model has and why, which of them were made heavy and by what, and what a mediator's mass does to the reach of its force. This lesson makes those answers routine, then counts mass dimensions to show why gravity's coupling cannot join the list.

UPV-18.02 · Lesson 2 of 9 in this unit. Position is not a completion score.
Starting model: Interactions, mediators and gravity's exclusionLog–log exchange amplitude against momentum transfer q. Dashed: the photon, e²/q² with α = 1/137, unbounded as q → 0 — infinite range. Solid: a mediator of mass M = 80.4 GeV and coupling g = 0.653, g²/(q² + M²), parallel to the photon above q = M and flat at g²/M² = 6.597 × 10⁻⁵ GeV⁻² below it. Markers sit at the probe q = 1.00 GeV.g²/(q² + M²) in GeV⁻², log axesdashed: photon, α = 1/137solid: M = 80 GeV, g = 0.65q = Mplateau g²/M²0.1101000 GeV10⁻⁸1
Start with a picture. Change the model in the lab below.

Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.

1

Build the model

Generators count the mediators, mass sets their reach, and gravity fails on dimensions.

The key idea

A mediator is the vector field a gauge symmetry demands for each of its generators; the mediator's mass M sets the range of its force through λ = ℏ/Mc and its low-energy strength through g²/M².

Example. For the W, λ = ℏc/(MW c²) = 197.3 MeV fm ÷ 80 400 MeV = 2.45 × 10⁻³ fm, and √2 g²/(8MW²) with g = 0.653 gives 1.17 × 10⁻⁵ GeV⁻², which is Fermi's constant.

The complete physical model and assumptions

A gauge theory does not let you choose how many force carriers it has. Name the group and the covariant derivative Dμ = ∂μ − i gₛ Gaμ Ta − i g Wiμ τi/2 − i g′ Bμ Y/2 must carry one vector field per generator to absorb the derivative of a local rotation. SU(3) has eight generators, SU(2) three and U(1) one: twelve mediators, each in the adjoint representation — gluons carry colour, the Wi weak isospin, the B nothing — and gauge invariance forbids all twelve a mass term. Which of them end up heavy is the Higgs doublet's decision, made when its vacuum value v = 246 GeV breaks three of the four electroweak generators and leaves the photon's combination untouched. Mass then sets the force's reach: the propagator 1/(q² − M²) becomes the Yukawa tail e(−r/λ)/r with λ = ℏ/Mc, so the photon reaches to infinity while the W and Z stop near 2 × 10⁻³ fm, and far below M the W collapses into Fermi's constant GF = √2 g²/8MW². The weak force is weak because its mediators are heavy, not because g is small. Gravity has no place here: it gauges spacetime rather than an internal phase, its quantum would be spin 2, and its coupling GN has mass dimension −2 — the dimension that made Fermi's theory fail — so amplitudes grow as (E/Mₚₗ)² and no finite set of counterterms tames them. The price of the model is honesty: g, g′, gₛ, v and every mass are fitted, and gravity is left outside.

1Count the generators and you have counted the mediators

Demand that the Lagrangian survive a rotation U(x) = exp(−i αa(x) Ta) that differs from point to point, and ∂μ ψ picks up an unwanted (∂μ U) ψ. The only cure is one vector field per generator, Aaμ, transforming so that Dμ = ∂μ − i g Aaμ Ta rotates with ψ. The count is therefore the dimension of the Lie algebra: N² − 1 for SU(N) and 1 for U(1), so SU(3)c × SU(2)L × U(1)Y supplies 8 + 3 + 1 = 12 gauge fields — eight gluons Gaμ, three Wiμ and one Bμ. Their quantum numbers are fixed by the same algebra, because they sit in the adjoint representation: gluons carry colour–anticolour (an octet, not nine states, since the singlet combination is absent), the Wi form a weak-isospin triplet, and B is a singlet with zero hypercharge. The structure constants decide who talks to whom: Faμν = ∂μAaν − ∂νAaμ + g fabc Abμ Acν contains cubic and quartic self-couplings, so gluons scatter off gluons and W's off W's, while the abelian B has f = 0 and no self-coupling at all. And M² Aμ Aμ is not gauge invariant, so at this stage every one of the twelve is massless.

2Breaking decides which mediators weigh anything

Symmetry breaking, not the group, sorts the twelve into massless and massive. The Higgs doublet's vacuum value v = 246 GeV is invariant under the combination Q = T₃ + Y/2 and under all of SU(3)c, so the photon and the eight gluons keep zero mass, while the three broken generators hand mass to W⁺, W⁻ = (W¹ ∓ iW²)/√2 and to Z = cos θW W³ − sin θW B. The kinetic term |Dμ φ|² gives MW = gv/2 = 0.653 × 246.2/2 = 80.4 GeV and MZ = MW/cos θW; with sin²θW = 0.231 that is 80.4/0.877 = 91.7 GeV against the measured 91.2 GeV, the half-percent gap being loop corrections. Read the bookkeeping honestly: v comes from GF, g from MW and GF, θW from neutral-current data — the relations are predictions, the numbers are fits. The gluons are a warning in the other direction. They are exactly massless, yet the strong force between nucleons reaches only about 1 fm, because confinement lets only colour singlets propagate and the residual force is carried by the lightest of them, the pion, with range ℏ/mπ c = 197.3/139.6 = 1.41 fm. A massless mediator gives infinite range only when it can travel freely.

3Mass sets range: the Yukawa tail from the propagator

Exchange of a mediator of mass M between two static sources gives the momentum-space amplitude g²/(|q|² + M²) — the Feynman propagator 1/(q² − M²) with the energy transfer set to zero, so that q² = −|q|². Its Fourier transform is the potential: ∫ d³q/(2π)³ e(iq⋅r)/(|q|² + M²) = e(−Mr)/(4πr), a contour closed around the pole at |q| = iM, so |V(r)| = (g²ℏc/4πr) e(−r/λ) with λ = ℏ/Mc once the constants are restored. M = 0 is Coulomb: the pole sits at |q| = 0 and the 1/r tail never ends. For the W, λ = ℏc/MW c² = 197.3 MeV fm ÷ 80 400 MeV = 2.45 × 10⁻³ fm; for the Z, 197.3/91 200 = 2.16 × 10⁻³ fm. A proton's charge radius is 0.84 fm, so the weak force reaches under a three-hundredth of it, and at r = 0.01 fm the exponential has already cut the W's potential to e(−4.1) = 0.017 of what a massless carrier would deliver. The range depends on M alone; the coupling g only scales the height of a curve whose shape M has already fixed.

4Below M the propagator collapses into a contact constant

For |q²| ≪ M² the W propagator loses its q-dependence: g²/(q² − MW²) → −g²/MW². Two currents joined by a W then look like four fermions meeting at a point, and matching the charged-current amplitude to Fermi's Leff = −(GF/√2) Jμ Jμ† gives GF/√2 = g²/(8MW²). Run it backwards with the measured GF = 1.166 × 10⁻⁵ GeV⁻² and MW = 80.4 GeV: g² = 8 × 6464 × 1.166 × 10⁻⁵/1.414 = 0.426, g = 0.653, and αW = g²/4π = 0.034 = 1/29.5 — four and a half times the electromagnetic α = 1/137. The weak interaction is weak because its mediator is heavy, not because g is small. Compare the two amplitudes at a momentum transfer of 1 GeV: e²/|q²| = 0.092 GeV⁻² against g²/(|q²| + MW²) = 6.6 × 10⁻⁵ GeV⁻², a ratio of 1.4 × 10³. Push q above MW and the denominators become alike, the ratio tends to e²/g² = 0.22, and the weak amplitude is the larger. Fermi's constant also carries the warning that opens the last section: with mass dimension −2 it makes cross-sections grow as GF² s, which cannot continue past √s of a few hundred GeV — and the W propagator is exactly what stops it.

5Rank the four by a dimensionless number; watch gravity move

A fair comparison of interactions uses dimensionless couplings, because a dimensionful one changes size with the units. Three of the four oblige: αₛ(MZ) = 0.118, αW = 1/29.5, α = 1/137 — strong, weak, electromagnetic, in that order, which upsets the folklore only for the weak one. Gravity does not oblige. GN = 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻² has to be multiplied by a squared mass before it becomes a pure number, αG = GN m²/ℏc, so its strength depends on who is being asked. For two protons αG = 6.674 × 10⁻¹¹ × (1.673 × 10⁻²⁷)²/(3.162 × 10⁻²⁶) = 5.9 × 10⁻³⁹; for two W bosons it is 4.3 × 10⁻³⁵; at the Planck mass Mₚₗ = (ℏc/GN)¹⁄² = 1.22 × 10¹⁹ GeV it reaches 1. In natural units this is the statement [GN] = −2: αG = (E/Mₚₗ)² grows quadratically with energy, while α, αW and αₛ move only logarithmically. That growth is what the twelve gauge couplings never do, and it is the technical reason gravity cannot be quantised the way they are.

6Why gravity cannot be a thirteenth mediator

Look for gravity's generator in SU(3)c × SU(2)L × U(1)Y and there is none: each rotates an internal label, while gravity gauges spacetime itself, and its source is the energy-momentum tensor Tμν rather than the Noether charge of a phase. A rank-two source needs a rank-two carrier, so the graviton would be spin 2, not spin 1, and its exchange is universally attractive: no anticharge, no screening. Then the dimensions bite. In ℏ = c = 1 the Lagrangian has dimension 4 and the twelve gauge couplings dimension 0, so each loop adds only logarithms and a finite list of counterterms absorbs every divergence: the theory is renormalisable. The gravitational vertex carries √GN with dimension −1, so the n-loop term grows as (E/Mₚₗ)(2n) relative to tree level and every order demands new counterterms — Fermi's disease, which a W at 80 GeV cured. Nobody knows gravity's W. Below Mₚₗ the model is silent about it and gravity works as an effective theory; at 10¹⁹ GeV both fail together. Gravity is not left out for being weak but because its coupling has the wrong dimension for the machinery the other three share.

2

Key equations

Read the units, assumptions and limits before substituting.

Covariant derivative: one field per generator

Dμ = ∂μ − i gₛ Gaμ Ta − i g Wiμ τi/2 − i g′ Bμ Y/2

Eight gluons, three W and one B: twelve mediators, fixed by SU(3) × SU(2) × U(1) before any experiment is done.

a = 1…8, i = 1…3; Ta = λa/2 the Gell-Mann generators, τi the Pauli matrices, Y the hypercharge; gₛ, g, g′ dimensionless

Yukawa potential and range

|V(r)| = (g²/4π) ℏc e(−r/λ)/r, λ = ℏ/Mc = ℏc/Mc²

M = 0 returns Coulomb's 1/r and infinite reach; MW = 80.4 GeV gives λ = 2.45 × 10⁻³ fm.

M the mediator mass, r the separation, g dimensionless; ℏc = 197.3 MeV fm, so λ in fm is 197.3 ÷ (Mc² in MeV); like charges repel under vector exchange

Propagator to contact constant

g²/(q² − M²) → −g²/M² for |q²| ≪ M²GF/√2 = g²/(8MW²)

Solving for g gives 0.653, larger than e = 0.303. The weakness lives in the denominator, not at the vertex.

ℏ = c = 1; q the four-momentum transfer; GF = 1.166 × 10⁻⁵ GeV⁻²

Electroweak masses from the vacuum value

MW = gv/2, MZ = MW/cos θW, Mγ = 0, v = (√2 GF)(−1/2) = 246 GeV

The group fixes the count; only the breaking fixes which mediators weigh anything, and every input here is fitted.

g, g′ dimensionless; cos θW = g/√(g² + g′²); tree level, corrected at the percent level

Dimensionless strengths of the four

αₛ(MZ) = 0.118, αW = g²/4π = 1/29.5, α = 1/137, αG = GN mₚ²/ℏc = 5.9 × 10⁻³⁹

By coupling alone the weak force outranks electromagnetism; gravity sits thirty-six orders below it.

αG is quoted for two protons; unlike the other three it grows as E², because GN is not dimensionless

Mass dimension and renormalisability

[gₛ] = [g] = [g′] = 0[GF] = [GN] = −2Mₚₗ = (ℏc/GN)¹⁄² = 1.22 × 10¹⁹ GeV

Fermi's −2 was cured by resolving GF into W exchange at 80 GeV; the model offers no such cure for GN.

ℏ = c = 1, so every quantity is a power of mass; a coupling of dimension −2 makes amplitudes grow as (E/M)²
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

80.4 GeV
1 to 300 GeV; steps of 0.1.
0.653
0.1 to 1.5; steps of 0.001.
0.0
−1 to 3; steps of 0.1.

Try this. Drag M from 80 GeV down to 1 GeV and the knee at q = M slides left until the solid curve runs parallel to the photon's line, just above it — the weak force is weak only because M is large — then, with M back at 80.4 GeV, push the probe up past 40 GeV and watch the photon ÷ X readout fall below 1, because g = 0.65 exceeds e = 0.30.

Live results

RANGE ℏc/M
2.45 ×10⁻³ fm
CONTACT √2 g²/8M²
1.166 ×10⁻⁵ GeV⁻²
αX = g²/4π
0.0339
PHOTON ÷ X AT PROBE q
1390.34

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics V modelLog–log exchange amplitude against momentum transfer q. Dashed: the photon, e²/q² with α = 1/137, unbounded as q → 0 — infinite range. Solid: a mediator of mass M = 80.4 GeV and coupling g = 0.653, g²/(q² + M²), parallel to the photon above q = M and flat at g²/M² = 6.597 × 10⁻⁵ GeV⁻² below it. Markers sit at the probe q = 1.00 GeV.g²/(q² + M²) in GeV⁻², log axesdashed: photon, α = 1/137solid: M = 80 GeV, g = 0.65q = Mplateau g²/M²0.1101000 GeV10⁻⁸1
Log–log exchange amplitude against momentum transfer q. Dashed: the photon, e²/q² with α = 1/137, unbounded as q → 0 — infinite range. Solid: a mediator of mass M = 80.4 GeV and coupling g = 0.653, g²/(q² + M²), parallel to the photon above q = M and flat at g²/M² = 6.597 × 10⁻⁵ GeV⁻² below it. Markers sit at the probe q = 1.00 GeV.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Mediator mass M: 80.4 GeV; Coupling g at the vertex: 0.653; Probe log₁₀(q / GeV): 0.0

Baseline 80.4; choose 1 to 300, in steps of 0.1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The weak interaction is not weak because its coupling is small. Solve GF/√2 = g²/8MW² for g and you get 0.653, more than twice the electron's e = 0.303; as a fine-structure constant, αW = 1/29.5 beats α = 1/137 by a factor of 4.6. What is small at ordinary energies is g²/MW² — the propagator's denominator, not its vertex — and it is small only while |q²| ≪ MW². Above the W mass the two amplitudes share a 1/q² and the weak one is the larger; once collision energies reach MW and MZ the suppression is gone, which is why LEP and the LHC produce W and Z bosons in enormous numbers. The same slip appears in the range: students tie a short range to a weak coupling, but λ = ℏ/Mc contains no coupling at all. Mass, and only mass, sets how far a force reaches; the coupling scales how hard it pulls within that reach.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

The Z boson has mass 91.2 GeV. Find the range of the force it carries, and compare it with the proton's charge radius of 0.84 fm and with the 1.41 fm range of pion exchange.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

Use GF = 1.166 × 10⁻⁵ GeV⁻² and MW = 80.4 GeV to extract the SU(2) coupling g. Then find αW = g²/4π, compare it with α = 1/137, and recover the Higgs vacuum value v by two routes.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

Show that Newton's constant has mass dimension −2 in units ℏ = c = 1, compute the Planck mass, then find the energy at which gravity's dimensionless strength αG = GN E²/(ℏc⁵) would match αW = 0.034. Contrast with the scale 1/√GF at which Fermi's theory failed.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A hypothetical W′ carries the W's coupling g but a mass of 804 GeV, ten times MW. For |q²| ≪ M², how does its four-fermion constant G′ compare with GF?
7

Review & continue

A mediator is the vector field a gauge symmetry demands for each of its generators; the mediator's mass M sets the range of its force through λ = ℏ/Mc and its low-energy strength through g²/M².

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.