The Running of the Quark Masses; The Origin of the Hadron Mass

A quark mass is not a property of a quark. It is a parameter carrying a scheme and a scale, and quoting one without both of them says nothing at all.

🎯 Why this matters

Two published quark masses therefore cannot be compared until you check they were quoted in the same scheme at the same scale. The number on its own does not contain enough information to be either right or wrong.

Two short sections that belong together, because they are both about the same uncomfortable fact: a confined particle has no rest frame in which to be weighed. §6.6 asks what a quark’s mass even means; §6.7 asks where the proton’s mass comes from, given that it is not the quarks’.

§6.6 A mass that depends on the scale

To measure a mass you need a free particle. Quarks are never free, so “the mass of the b quark” is not, on its face, a well-defined quantity. QCD rescues it by extending the definition — at the price of two dependencies that a mass is not normally supposed to have.

📐 Physics you need first — why a quark mass carries two labels

The mass parameter in the Lagrangian is not observable, and renormalization — the same procedure as §5.8’s — makes it finite by subtracting infinities. Two choices enter and neither is physical:

The scheme. How you subtract. QCD almost always uses modified minimal subtraction , written $\overline{\text{MS}}$ . A different scheme gives a different number for the same quark.
The scale μ. Renormalization introduces a reference energy, exactly as it did for the coupling. The mass then runs with it, like α_s does.

So a quoted quark mass is meaningless without both labels. The book’s own Tables 4.1 and 4.5 follow the convention: light quarks (u, d, s) at μ = 2 GeV; the heavy c and b at μ equal to the mass itself, so you see $m_b(m_b) = 4.18$ GeV — a mass quoted at its own value, which looks circular and is merely a convention.

The split at $\Lambda_{\text{QCD}}$ matters: u, d, s are lighter than Λ and their masses cannot be got perturbatively at all — they come from the lattice (§6.10) or from chiral symmetry (§6.8). Only c and b are heavy enough for perturbation theory.

\htmlClass{t-m}{m_b(\mu^2)} = m_b(4.2^2)\left[\frac{\htmlClass{t-a}{\alpha_s(\mu^2)}}{\alpha_s(4.2^2)}\right]^{\htmlClass{t-e}{12/23}}

(6.59)

Bettini p. 255. The mass runs because the coupling runs — and it falls, because α_s falls.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

QCD prediction, Eq. (6.59)
m_b(m_b) = 4.18 GeV — the anchor
LEP: ALEPH, DELPHI, OPAL at μ = M_Z
LHC: ATLAS, CMS from H → bb̄ at μ = M_H

Fig. 6.28, redrawn. The b quark weighs 4.18 GeV at its own scale and about 2.8 GeV at the Z — a 33 % change in a quantity most people think of as fixed. The LEP and LHC points are displaced horizontally for legibility, as they are in the book; all the LEP measurements are at μ = M_Z and the LHC ones at μ = M_H. Error bars are large because the effect is small: it scales as m_b²/Q², which at the Z is 0.002.

🔢 Worked example — measuring a mass by counting three-jet events

The trick is that a heavier quark radiates gluons less readily. So compare the rate of three-jet events for b quarks against light quarks — Eq. (6.60):

R_3^{b,udsc} = \frac{\htmlClass{t-b}{\Gamma^b_{3\text{jet}}/\Gamma^b}} {\htmlClass{t-l}{\Gamma^{udsc}_{3\text{jet}}/\Gamma^{udsc}}}

(6.60)

A double ratio, built so that everything except the quark mass cancels. This is how a quantity that is not directly observable — the mass of a confined particle — gets measured to a few per cent.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

If the b were massless this would be exactly 1. ALEPH, DELPHI and OPAL all found about 0.97 — a 3 % suppression, and from it $m_b$ at $\mu = M_Z$ .

Two features worth noticing.

The effect is tiny on purpose. Mass effects go as $m^2/Q^2$ , and at the Z that is $m_b^2/M_Z^2 = 0.002$ . You want $Q \gg \Lambda_{\text{QCD}}$ so the QCD calculation is trustworthy — and the price of a large $Q$ is a small signal. There is no scale that gives you both.

Identifying which jet is the b needs a vertex detector: a B hadron flies a few hundred microns before decaying, and that displaced vertex is the tag. It is the same technique that found the top quark (§9.11).

At the LHC the same measurement is done differently — the partial width of $H \to b\bar b$ gives $m_b$ at $\mu = m_H \approx 125$ GeV. Different machine, different process, same curve.

the b mass at three scales

import numpy as np
L5, mb, MZ, MH = 0.215, 4.18, 91.1876, 125.25
als = lambda Q: 12*np.pi / (23 * np.log(Q**2 / L5**2))      # one loop, n_f = 5

print("Eq. (6.59)'s exponent: 4/beta0 with beta0 = 11 - 2 n_f/3")
print(f"  n_f = 5: beta0 = {11 - 2*5/3:.4f} = 23/3, so 4/beta0 = {4/(11-2*5/3):.4f} = 12/23")
print()
for mu, lab in [(mb, "mu = m_b   "), (MZ, "mu = M_Z   "), (MH, "mu = M_H   ")]:
    m = mb * (als(mu) / als(mb))**(12/23)
    print(f"  {lab} ({mu:6.2f} GeV):  alpha_s = {als(mu):.4f},  m_b = {m:.3f} GeV")
print()
print("  the accepted m_b(M_Z) is about 2.83 GeV, and the LEP points sit there")
print(f"  the b quark loses {(1 - 4.18*(als(MZ)/als(mb))**(12/23)/4.18)*100:.0f}% of its mass")
print("  between its own scale and the Z -- in a quantity usually thought fixed")
print()
print(f"  why the measurement is hard: m_b^2/M_Z^2 = {mb**2/MZ**2:.4f}")
print("  mass effects are that small, and you cannot go lower in Q without")
print("  losing control of the QCD calculation.")

prints

Eq. (6.59)'s exponent: 4/beta0 with beta0 = 11 - 2 n_f/3
n_f = 5: beta0 = 7.6667 = 23/3, so 4/beta0 = 0.5217 = 12/23

mu = m_b    (  4.18 GeV):  alpha_s = 0.2762,  m_b = 4.180 GeV
mu = M_Z    ( 91.19 GeV):  alpha_s = 0.1355,  m_b = 2.882 GeV
mu = M_H    (125.25 GeV):  alpha_s = 0.1287,  m_b = 2.807 GeV

the accepted m_b(M_Z) is about 2.83 GeV, and the LEP points sit there
the b quark loses 31% of its mass
between its own scale and the Z -- in a quantity usually thought fixed

why the measurement is hard: m_b^2/M_Z^2 = 0.0021
mass effects are that small, and you cannot go lower in Q without
losing control of the QCD calculation.

§6.7 Where the proton’s mass comes from

A proton weighs 938 MeV. Its three valence quarks weigh about 9 MeV between them. Ninety-nine per cent of the mass of ordinary matter is not the mass of anything — it is the energy of the colour field.

💡 What this really says — the hydrogen atom, and then the opposite of it

Start with what is familiar. The hydrogen atom’s mass, Eq. (6.61):

m_H = \htmlClass{t-c}{m_p + m_e} \htmlClass{t-b}{- \;13.6\ \text{eV}}

(6.61)

The hydrogen atom's mass — the familiar case, written down so the proton's can be compared against it clause by clause. Every term here reverses in the next equation.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

The mass of the constituents, minus the work you must do to pull them into a configuration where they no longer interact — which for an atom means pulling them apart, and the work is negative and utterly negligible: 13.6 eV against a GeV.

Now the proton, and every clause flips.

Where do the constituents stop interacting? Not far apart — QCD gets stronger with distance. They stop interacting when they are on top of each other, because there the three antiscreening clouds cancel exactly (they are in a colour singlet). Asymptotic freedom, §6.5.
What sign is the work? Positive. Letting the quarks come together releases energy — the spring contracts — so assembling the proton at its actual size costs energy relative to the non-interacting configuration.
How big is it? Not negligible. It is essentially all of it.

The atom is light constituents plus a tiny negative correction. The proton is negligible constituents plus a correction that is the entire answer.

🔢 Worked example — the proton’s mass and radius, from two lines

The estimate for the mass. The energy scale separating “strongly interacting” from “not interacting” is $\Lambda_{\text{QCD}}$ , and there are three quarks:

m_p \approx 3\,\Lambda_{\text{QCD}} \approx 3 \times 340\ \text{MeV} \approx 1\ \text{GeV}

Now check the radius, which is a genuine test because nothing above used it. Take the colour-field energy to grow linearly with separation, $kx$ , as §6.5’s flux tube requires. The quarks are ultra-relativistic, so their kinetic energy is their momentum, and the uncertainty principle gives $p \approx 1/x$ . For three quarks, Eq. (6.62):

E(x) = \htmlClass{t-k}{\frac{3}{x}} + \htmlClass{t-p}{kx}

(6.62)

The proton's energy as a function of its size, with two terms pulling in opposite directions. Everything in this estimate — the mass, the radius, and the relation between them — falls out of minimising this one line.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Minimise, Eq. (6.63): $\mathrm{d}E/\mathrm{d}x = -3/x^2 + k = 0$ , so $x_p = \sqrt{3/k}$ . Substituting back, Eq. (6.64):

m_p = E(x_p) = \htmlClass{t-a}{\frac{3}{x_p}} + \htmlClass{t-b}{k x_p} = \htmlClass{t-r}{\frac{6}{x_p}}

(6.64)

Evaluated at the minimum, the two competing terms are equal — so the answer is twice either one, and the string tension k has vanished from it.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

That is the payoff: the minimum energy is exactly twice the kinetic term, so $x_p = 6/m_p$ with no unknown left. Putting $m_p = 0.94$ GeV gives $x_p = 1.26$ fm — the mean inter-quark distance. Three quarks on an equilateral triangle sit at radius $x_p/\sqrt3$ , Eq. (6.65):

r_p = \frac{\htmlClass{t-x}{x_p}}{\htmlClass{t-g}{\sqrt3}} \approx \htmlClass{t-r}{0.7\ \text{fm}}

(6.65)

From the inter-quark distance to a radius — pure geometry, and the only step in the estimate that uses no physics at all.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

The measured charge radius is 0.84 fm. The book calls this “even too good for the very rough calculation we made”, and it is right to — though 0.7 against 0.84 is a 17 % underestimate, not an exact hit.

kinetic, 3/x — localisation costs energy
colour field, kx — the flux tube
total E(x) = 3/x + kx
the minimum: x_p = 1.26 fm, E = 0.94 GeV

Eq. (6.62) plotted. Squeeze the quarks and the kinetic term explodes; separate them and the flux tube charges you linearly. The proton sits at the bottom of the trade — and the value at the bottom is its mass. This is the same construction as §5.2's Bohr radius from the uncertainty principle, with one crucial change: there the potential fell as −1/r, here it rises as +kx, and that single sign flip turns a bound atom into a confined hadron.

the proton from an uncertainty-principle argument

import numpy as np
hbarc = 0.1973269804                    # GeV fm
mp = 0.9382720882

# E(x) = 3/x + kx, minimised at x_p = sqrt(3/k), where E = 6/x_p.
# So the mass fixes everything with no free parameter left.
xp_nat = 6 / mp                         # GeV^-1
xp = xp_nat * hbarc                     # fm
k = 3 / xp_nat**2                       # GeV^2
print(f"m_p = {mp:.3f} GeV  ->  x_p = 6/m_p = {xp:.2f} fm     (book: 1.2 fm)")
print(f"                        r_p = x_p/sqrt(3) = {xp/np.sqrt(3):.2f} fm  (book: 0.7 fm)")
print(f"                        k   = 3/x_p^2     = {k/hbarc:.2f} GeV/fm")
print(f"  the measured proton charge radius is 0.84 fm, so this underestimates by "
      f"{(1 - xp/np.sqrt(3)/0.84)*100:.0f}%")
print(f"  and the lattice string tension is ~0.9 GeV/fm, so k is low by ~2x --")
print("  about what a two-line estimate has any right to")
print()
print("the independent estimate, from the only scale QCD has:")
print(f"  m_p ~ 3 Lambda_QCD(3) = 3 x 0.340 GeV = {3*0.340:.2f} GeV")
print()
mq = 2*0.00216 + 0.00467                # two u and one d, current masses
print(f"the three valence quarks weigh {mq*1000:.1f} MeV between them,")
print(f"which is {mq/mp*100:.2f}% of the proton.  The other {100-mq/mp*100:.1f}% is")
print("the energy of the colour field.  Almost all the mass of ordinary matter")
print("is not the mass of anything.")

prints

m_p = 0.938 GeV  ->  x_p = 6/m_p = 1.26 fm     (book: 1.2 fm)
                      r_p = x_p/sqrt(3) = 0.73 fm  (book: 0.7 fm)
                      k   = 3/x_p^2     = 0.37 GeV/fm
the measured proton charge radius is 0.84 fm, so this underestimates by 13%
and the lattice string tension is ~0.9 GeV/fm, so k is low by ~2x --
about what a two-line estimate has any right to

the independent estimate, from the only scale QCD has:
m_p ~ 3 Lambda_QCD(3) = 3 x 0.340 GeV = 1.02 GeV

the three valence quarks weigh 9.0 MeV between them,
which is 0.96% of the proton.  The other 99.0% is
the energy of the colour field.  Almost all the mass of ordinary matter
is not the mass of anything.

⚙️ Engineer’s bridge — where the mass of your desk comes from

Two consequences worth carrying out of this section, because they are the kind of thing that sounds like rhetoric and is not.

Mass is mostly not stuff. You, your desk and the Earth are about 99 % proton and neutron mass by weight, and 99 % of that is colour-field energy rather than quark mass. $E = mc^2$ is usually taught as “mass can be converted to energy”. The proton says the stronger thing: most mass simply is energy, already, with no conversion pending. If the Higgs mechanism were switched off and every quark became exactly massless, the proton would lose about 1 % of its weight.

And a modelling lesson. The whole of §6.7 is one variational estimate: write the energy as a sum of two competing terms, minimise, and read off the answer. Localisation costs $1/x$ ; the field costs $kx$ ; the system sits at the minimum. It is the same one-line method as the Bohr radius in §5.2 and problem 5.2, and the same method an engineer uses for any equilibrium of two opposing costs — buffer size against latency, cache line against false sharing. What makes it physics rather than arithmetic is the observation that at the minimum $E = 2 \times$ (kinetic term), so the answer carries no free parameter at all.

Where it breaks: the estimate has no free parameter because the trial form was chosen to leave none. Write the kinetic term as $3/x$ and the virial relation gives $m_p = 6/x_p$ with everything cancelling; write a different but equally defensible ansatz and the coefficient moves. So the calculation is right to a factor and is not a determination of the proton mass — the real number needs the lattice of §6.10, at a cost of $(L/a)^4$ . What survives the caveat is the structural claim, which is what the section is for: the mass is field energy, and it comes out of order Λ_QCD without any quark mass in the calculation.

quantity	this estimate	measured / accepted
proton mass, from 3Λ_QCD	1.02 GeV	0.938 GeV
inter-quark distance x_p = 6/m_p	1.26 fm	— (not directly observable)
proton radius x_p/√3	0.73 fm	0.84 fm (charge radius)
string tension k = 3/x_p²	0.37 GeV/fm	≈0.9 GeV/fm (lattice)
fraction of m_p that is quark mass	—	1.0 %
m_b at its own scale vs at the Z	4.18 → 2.88 GeV	4.18 → ≈2.83 GeV

Two rows deserve honesty. The radius is 17 % low, not exact — the book's “too good for the calculation” is about the order of magnitude, and it is fair. The string tension is a factor of two low, which is worse and is the price of treating three quarks as a single one-dimensional spring. Neither undermines the conclusion, which is about the <em>origin</em> of the mass rather than its precise value.

🔑 If you remember only three things

The proton’s mass comes from Λ, not from its quarks. Three light quarks and a scale generated by the running give the number; the constituents contribute a percent or two.
A mass nobody can weigh has to be inferred from a rate. Heavier quarks radiate less readily, so counting three-jet events measures something no balance could.
The word changes meaning between chapters. In Chapter 1 mass was the norm of a four-vector; here it is a renormalized parameter, and they are not the same kind of object.

Where this goes next

§6.8–6.9 is the vacuum this energy lives in, and why the pion is so much lighter than everything else.
§6.10 computes the whole light-hadron spectrum properly, on a lattice, from three inputs — and gets the proton right to a per cent instead of a factor.
§6.5’s flux tube is where the linear term $kx$ comes from; the widget there uses the lattice tension rather than this section’s estimate.
§9.16 is the Higgs, which gives the quarks the 1 % — and is often described as giving matter its mass. It gives matter 1 % of its mass.

✅ Check yourself — running masses and the origin of the proton's mass

0/5 answered · 0 correct

1.Why does a quoted quark mass need both a scheme and a scale attached, when an electron's mass needs neither?
2.The b-quark mass is measured at LEP from a 3 % suppression of three-jet events. Why not use a lower energy, where the effect would be bigger?
3.For a hydrogen atom the binding work is negative and negligible; for a proton it is positive and dominant. What flips?
4.Minimising E(x) = 3/x + kx gives m_p = 6/x_p. Why is that relation the useful part?
5.What is the strongest correct statement about the Higgs and the mass of everyday matter?