The Lagrangian Density · Elementary Particle Physics

A force stops being something that pushes and becomes something that makes an arbitrary local choice unobservable. The rest of the chapter is arithmetic performed inside that idea.

🎯 Why this matters

Once the interaction follows from the symmetry there is no adjustable version of it. Nobody can write down a slightly different electromagnetism and check which one nature picked — the structure is fixed and only the strength is measured.

Four chapters have catalogued particles. This one is about dynamics, and it opens with a claim that sounds like a slogan until you watch it work:

A local symmetry, imposed on a theory of free particles, generates the interaction between them.

Not constrains it. Not decorates it. Generates it — the interaction term is forced into the Lagrangian by demanding a symmetry that costs nothing to impose. QED is the worked example because its symmetry group is the simplest one that exists, the phases of a complex number.

📐 Physics you need first — what a Lagrangian is, and why anyone bothers

You have almost certainly met F = ma. The Lagrangian formulation is the same physics arranged so that symmetries are visible, and that is the only reason it takes over from here.

For one particle. Write $L = T - V$ , kinetic minus potential energy, as a function of position and velocity. Of all the paths a particle could take between two fixed endpoints, the one it actually takes makes the action $S = \int L\,\mathrm{d}t$ stationary. Demanding that gives the Euler–Lagrange equation

\htmlClass{t-q}{\frac{\partial L}{\partial q}} - \htmlClass{t-p}{\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q}} = 0

The Euler–Lagrange equation — the condition for the action to be stationary. It is F = ma rearranged so that symmetries are visible, and that rearrangement is the only reason this formalism takes over from here.

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and for $L = \tfrac12 m\dot q^2 - V(q)$ it returns $m\ddot q = -\partial V/\partial q$ , which is F = ma. Nothing new has been said yet.

What is new is that a symmetry of $L$ now has immediate consequences. If $L$ does not depend on $q$ , the equation says $\partial L/\partial\dot q$ is constant — momentum is conserved because space is uniform. That is Noether’s theorem, and it is why this formalism is worth the trouble.

For a field. A field has a value at every point, so instead of one $L$ we write a Lagrangian density 𝓛 and integrate it over space and time: $S = \int \mathcal{L}\,\mathrm{d}^4x$ . The variable is no longer $q(t)$ but $A^\mu(x)$ or $\psi(x)$ , and the Euler–Lagrange equation acquires a four-divergence in place of a time derivative. That is Eq. (5.8) below, and it is the only piece of machinery this section actually uses.

For an engineer: the action is a cost functional and the field equations are its stationarity condition — a variational formulation exactly like the ones behind finite-element methods or optimal control. You are not solving the dynamics directly; you are writing down what is being extremised and letting the Euler–Lagrange equation hand you the dynamics.

Electromagnetism, in four-vector form

⚠️ Erratum — the metric signature in Eq. (5.1)

The section opens by fixing the metric as $g_{00} = -1$ , $g_{11} = g_{22} = g_{33} = +1$ . That is the opposite of the convention this book uses everywhere else, and adopting it would flip the sign of every invariant in the previous four chapters.

Check it against the book’s own statements. Eq. (1.28) is $E^2 = p^2 + m^2$ , i.e. $p\cdot p = +m^2$ , which needs $g_{00} = +1$ . Section 5.6, four pages from here, states $s \geq 0$ and $t \leq 0$ — also only true with $(+,-,-,-)$ . And the $F^{\mu\nu}$ of Eq. (5.4) is the standard one, written for that same signature.

This site uses $g = \mathrm{diag}(+1, -1, -1, -1)$ throughout, as §1.4 established. It is one sign in one equation, but it is the sign that decides whether mass comes out real or imaginary.

Which signature reproduces the book's own m² = E² − p²?

import numpy as np

# a real 4-momentum: E = 5 GeV, p = (1, 2, 2) GeV, so |p| = 3 and m = 4
p = np.array([5.0, 1.0, 2.0, 2.0])

for name, g in [("(+,-,-,-)   used by ch01 and by Eq. (1.28)", np.diag([1., -1, -1, -1])),
                ("(-,+,+,+)   as Eq. (5.1) prints it",         np.diag([-1., 1, 1, 1]))]:
    print(f"  {name}:  p.p = {p @ g @ p:+7.3f} GeV^2")

print(f"\n  the book's own m^2 = E^2 - p^2 = {p[0]**2 - p[1:] @ p[1:]:+.3f} GeV^2, so m = 4 GeV")
print("  only the first signature reproduces it -- and only the first gives")
print("  s >= 0 and t <= 0, which this chapter relies on from Sec. 5.6 onwards")

prints

  (+,-,-,-)   used by ch01 and by Eq. (1.28):  p.p = +16.000 GeV^2
(-,+,+,+)   as Eq. (5.1) prints it:  p.p = -16.000 GeV^2

the book's own m^2 = E^2 - p^2 = +16.000 GeV^2, so m = 4 GeV
only the first signature reproduces it -- and only the first gives
s >= 0 and t <= 0, which this chapter relies on from Sec. 5.6 onwards

With that settled, the whole of classical electromagnetism collapses into two objects.

\htmlClass{t-A}{A_\mu} = \left(\phi/c,\ \mathbf{A}\right), \qquad \htmlClass{t-F}{F_{\mu\nu}} = \partial_\mu \htmlClass{t-A2}{A_\nu} - \partial_\nu A_\mu, \qquad \htmlClass{t-J}{J_\mu} = \left(\rho c,\ \mathbf{j}\right)

(5.2–5.5)

Bettini p. 185. The potential, the field tensor built from it, and the source.

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Transcribed from Bettini Eq. (5.4). The arrangement is the content: what a Lorentz boost does to the electromagnetic field is a similarity transformation on this matrix, and it moves entries between the shaded regions. A field that is purely magnetic in one frame has an electric part in another.

The Lagrangian, and Maxwell falling out of it

\mathcal{L} = -\frac{1}{4\mu_0}\,\htmlClass{t-FF}{F_{\mu\nu}F^{\mu\nu}} \;-\; \htmlClass{t-JA}{J^\mu A_\mu}

(5.6)

Bettini p. 185, with the sign corrected — see the erratum below. Two terms: the field on its own, and the field talking to charge.

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Erratum — the sign of the field term in Eqs. (5.6) and (5.18)

Both equations print $+\tfrac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu}$ . It should be minus, and the book says so itself two lines later: the expansion immediately below Eq. (5.6) opens with

\mathcal{L} = -\frac{1}{4\mu_0}\left(\partial_\mu A_\nu - \partial_\nu A_\mu\right)\left(\partial^\mu A^\nu - \partial^\nu A^\mu\right) - J^\mu A_\mu

and Eq. (5.7), the tidied result of that expansion, carries the minus too. So the sign in (5.6) disagrees with the book’s own algebra on the same page. Confirmed on the page render.

It is not a cosmetic point: with the wrong sign the field energy is negative and the Euler–Lagrange equation returns Maxwell with the source reversed.

🪜 From the Lagrangian (5.6) to the Maxwell equations (5.9)

Step 1 of 6 — expand F in terms of A

\mathcal{L} = -\frac{1}{4\mu_0}\left(\partial_\mu A_\nu - \partial_\nu A_\mu\right)\left(\partial^\mu A^\nu - \partial^\nu A^\mu\right) - J^\mu A_\mu

Why you may do this: F is not an independent object — it is defined as a derivative of A, and A is the field we vary. Substituting the definition is the only way to get an Euler–Lagrange equation for anything.

The whole derivation the book gives on p. 185, one algebraic move per step. Nothing here is deeper than the product rule — the content is in what you started from.

🔢 Worked example — half of Maxwell is free, and gauge invariance is exact

Two claims from this section can be checked numerically on any potential you like, and neither needs a formula. Take an arbitrary smooth $A_\mu(x)$ and an arbitrary gauge function $\chi(x)$ , build $F$ by finite differences, and look.

Reproduce it

import numpy as np
h = 1e-5
E = np.eye(4)

def A(x):                       # an arbitrary smooth four-potential A_mu(t,x,y,z)
    t, X, Y, Z = x
    return np.array([np.sin(X)*np.exp(-Z**2), X*Y + t, np.cos(Z) + t*X, Y**2 - t*Z])

def chi(x):                     # an arbitrary gauge function -- any smooth scalar will do
    t, X, Y, Z = x
    return np.sin(t*X) + Y*Z**2 + 0.3*t**2

def grad(f, x):                 # central differences, d_mu f
    return np.array([(f(x + h*E[m]) - f(x - h*E[m])) / (2*h) for m in range(4)])

def Ftensor(Afun, x):
    J = np.array([grad(lambda y: Afun(y)[n], x) for n in range(4)])   # J[n][m] = d_m A_n
    return J - J.T                                                   # F_mn = d_m A_n - d_n A_m

x = np.array([0.7, -0.4, 1.1, 0.5])
F0 = Ftensor(A, x)
F1 = Ftensor(lambda y: A(y) + grad(chi, y), x)          # A_mu -> A_mu + d_mu chi

print("1. gauge invariance of the field tensor")
print(f"   max |F(A) - F(A + d chi)| = {np.abs(F0 - F1).max():.1e}   (finite-difference floor)")
print(f"   F is antisymmetric:  max |F + F^T| = {np.abs(F0 + F0.T).max():.1e}")

print("\n2. the homogeneous Maxwell equations, for free")
def dF(l, m, n):
    return (Ftensor(A, x + h*E[l])[m, n] - Ftensor(A, x - h*E[l])[m, n]) / (2*h)
worst = 0.0
for (l, m, n) in [(0,1,2), (0,1,3), (0,2,3), (1,2,3)]:
    s = dF(l,m,n) + dF(m,n,l) + dF(n,l,m)
    worst = max(worst, abs(s))
    print(f"   d{l}F{m}{n} + d{m}F{n}{l} + d{n}F{l}{m} = {s:+.1e}")
print(f"   worst residual {worst:.0e} -- identically zero for ANY A, which is why")
print("   the homogeneous pair needs no derivation at all")

prints

1. gauge invariance of the field tensor
 max |F(A) - F(A + d chi)| = 4.4e-16   (finite-difference floor)
 F is antisymmetric:  max |F + F^T| = 0.0e+00

2. the homogeneous Maxwell equations, for free
 d0F12 + d1F20 + d2F01 = +0.0e+00
 d0F13 + d1F30 + d3F01 = +0.0e+00
 d0F23 + d2F30 + d3F02 = +0.0e+00
 d1F23 + d2F31 + d3F12 = +0.0e+00
 worst residual 0e+00 -- identically zero for ANY A, which is why
 the homogeneous pair needs no derivation at all

The second block is the identity $\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0$ , and it is exactly $\nabla\cdot\mathbf{B} = 0$ together with Faraday’s law. It came out zero not because the potential was well chosen but because F was built as an antisymmetrised derivative, and the second derivative of anything is symmetric in its two indices. Half of Maxwell is a consequence of writing E and B in terms of a potential at all.

Charge conservation, and the best limit anyone has

Charge conservation is not an extra postulate here. Take the four-divergence of Eq. (5.9): the left side is $\partial^\nu\partial^\mu F_{\mu\nu}$ , a symmetric operator contracted with an antisymmetric tensor, which is identically zero. So $\partial_\mu J^\mu = 0$ — the continuity equation (5.11), (5.12) — follows from the Maxwell equations rather than standing beside them.

📏 How do you test a conservation law that nothing violates?

You look for the one decay it would allow. The electron is the lightest charged particle, so it has nothing to decay into without violating charge conservation — and that makes its lifetime a direct measurement of how good the law is. BOREXINO searched for the photon from $e^- \to \nu_e + \gamma$ and found nothing:

\tau_e > 6.6 \times 10^{28}\ \text{yr}

That is 5 × 10¹⁸ times the age of the universe. The number is achievable only because the signature is clean and the detector is enormous — the same exposure arithmetic as §3.6’s proton-decay limit, and the ExposureLab there applies unchanged.

The redundancy that turns out to matter

\mathbf{A} \Rightarrow \mathbf{A} + \nabla\htmlClass{t-chi}{\chi}, \quad \phi \Rightarrow \phi - \frac{\partial\chi}{\partial t} \qquad\Longleftrightarrow\qquad \htmlClass{t-cov}{A_\mu \to A_\mu + \partial_\mu\chi}

(5.13–5.14)

Bettini p. 186. The same statement twice: in three-vector form and covariantly.

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gauge invariance · the same magnetic field, four different potentials

arrows: A · circles: B₀ ẑ, out of the page

probe x = 1.20probe y = 0.80

A at the probe	(-0.800, 0.000) B₀
∇χ at the probe	(-0.400, -0.600) B₀
B_z = ∂A_y/∂x − ∂A_x/∂y	1.000 B₀

Change gauge and watch every arrow move while the bottom row does not. ∇×(∇χ) = 0 for any χ whatsoever, which is Eq. (5.14) in three lines of vector calculus.

A = B₀(−y, 0, 0). All of the potential now points along x and depends only on y. The picture is completely different from the symmetric gauge — and describes the identical magnetic field. This is the standard choice when a problem has translational symmetry along one axis.

but the Lagrangian is not invariant — until ψ rotates too

term of 𝓛	what the gauge shift adds	value at the probe
−q_e ψ̄γ^μψ A_μ	−q_e ψ̄γ^μψ ∂_μχ (5.19)	0.400
i ψ̄γ^μ∂_μψ	nothing — ψ was left alone	0.000
total change in 𝓛	NOT invariant ✗	0.400

Values are per unit of q_e ψ̄γ^μψ, with the current taken along x so a single number suffices. The point is not the magnitude, it is that the second row exists only when ψ is rotated — and that when it does, the two entries are equal and opposite for every χ, at every point. That exact cancellation is what forces the interaction term −q_e ψ̄γ^μψ A_μ to be in the Lagrangian at all.

Picking one particular $\chi$ — and so one particular $\mathbf{A}$ out of the infinitely many that give the same $\mathbf{B}$ — is called gauge fixing . It is a convenience, never a physical statement: the Landau and symmetric gauges above describe the identical field, and any quantity that came out different in the two would be a bug in the calculation rather than a prediction.

⚙️ Engineer’s bridge — a local redundancy in the representation

You accept this move constantly and call it something else. Circuit potentials are defined only up to a choice of ground: shift every node by 10 V and no measurement changes, because only differences are physical. Phase in a communications system is defined only relative to a reference. A quaternion and its negation describe the same rotation.

Gauge freedom is that idea, with one upgrade that changes everything: the offset can be chosen independently at every point in space and time. A global redundancy is bookkeeping. A local redundancy is a constraint on what interactions are allowed to exist — because for the physics to be unaffected, something must compensate the mismatch between neighbouring points, and that something is the field $A_\mu$ .

Where the analogy breaks: shifting your ground reference genuinely changes nothing, so it teaches you nothing. Here, insisting on the freedom predicts that a massless vector field must exist and fixes the form of its coupling. A redundancy in the description turns out to be the strongest statement in the theory — which is a genuinely strange thing, and the section says so.

The fine-structure constant

\htmlClass{t-a}{\alpha} = \frac{1}{4\pi\varepsilon_0}\frac{\htmlClass{t-q}{q_e^2}}{\hbar c} \simeq \frac{1}{137}

(5.16)

Bettini p. 186. The dimensionless strength of the electromagnetic interaction.

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Is it really dimensionless, and is it really 1/137?

import numpy as np

qe   = 1.602176634e-19      # C      (exact, SI 2019)
eps0 = 8.8541878128e-12     # F/m
hbar = 1.054571817e-34      # J s
c    = 299792458.0          # m/s    (exact)

alpha = qe**2 / (4*np.pi*eps0*hbar*c)
print(f"  alpha    = {alpha:.10e}")
print(f"  1/alpha  = {1/alpha:.6f}")

# check it really is dimensionless: C^2 / (F/m * J s * m/s) = C^2 / (C^2/J * J) = 1
print("\n  units: C^2 / [(C^2 J^-1 m^-1)(J s)(m s^-1)] = C^2 / C^2 = 1  -> pure number")
print("  that is the whole reason a perturbative expansion in it converges")

prints

  alpha    = 7.2973525737e-03
1/alpha  = 137.035999

units: C^2 / [(C^2 J^-1 m^-1)(J s)(m s^-1)] = C^2 / C^2 = 1  -> pure number
that is the whole reason a perturbative expansion in it converges

Putting the electron in: the QED Lagrangian

Add the Dirac Lagrangian of §2.9, and let the electron field be the source of the vector current :

J^\mu = -q_e\,\bar\psi\gamma^\mu\psi, \qquad \mathcal{L} = \underbrace{\htmlClass{t-dirac}{i\bar\psi\gamma^\mu\partial_\mu\psi - m\bar\psi\psi}}_{\text{free electron}} \;\underbrace{-\ \tfrac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu}}_{\text{free photons}} \;\underbrace{-\ \htmlClass{t-int}{q_e\bar\psi\gamma^\mu\psi A_\mu}}_{\text{interaction}}

(5.17–5.18)

Bettini p. 187, with the field-term sign corrected. Three groups: the electron alone, the field alone, and the one term that couples them.

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Now test it. Under Eq. (5.14) the interaction term picks up a piece that has no business being there:

\htmlClass{t-was}{-q_e\bar\psi\gamma^\mu\psi A_\mu} \;\to\; \htmlClass{t-was}{-q_e\bar\psi\gamma^\mu\psi A_\mu} \;\htmlClass{t-junk}{-\; q_e\bar\psi\gamma^\mu\psi\,\partial_\mu\chi}

What the gauge transformation A_μ → A_μ + ∂_μχ does to the interaction term. The original survives; an extra piece appears that has no business being there.

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The second term breaks the invariance. Fock’s 1926 observation is that it can be cancelled — but only by transforming the matter field at the same time:

\htmlClass{t-rot}{\psi \to e^{-iq_e\chi}\psi}, \qquad \bar\psi \to e^{+iq_e\chi}\bar\psi

(5.20)

Bettini p. 187. The phases of modulus 1 form the group U(1) — the gauge symmetry of QED.

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💡 What this really says — the interaction is not optional

Read the logic backwards, which is how it was actually discovered.

Start with a free electron. Its Lagrangian is invariant under a global phase rotation — multiply ψ by a constant phase everywhere and nothing changes. Perfectly ordinary, and it buys you charge conservation by Noether.

Now demand more: let the phase be chosen independently at every point. The free Lagrangian immediately fails, because the derivative $\partial_\mu\psi$ compares ψ at neighbouring points and now those points have unrelated phases. The failure is a specific, calculable term, $q_e\bar\psi\gamma^\mu\psi\, \partial_\mu\chi$ .

To repair it you need a new field that transforms in exactly the compensating way — and $A_\mu \to A_\mu + \partial_\mu\chi$ is precisely that. The photon is what a local phase symmetry costs. Its coupling is not adjustable, its form is not adjustable, and it must be massless (a term $m^2A_\mu A^\mu$ is not gauge invariant and is therefore forbidden).

Chapter 6 runs the same argument with SU(3) instead of U(1) and gets eight gluons; Chapter 9 runs it with SU(2) ⊗ U(1) and gets the W and Z. Every force in this book except gravity is a different answer to the same question.

Aside — which conservation laws are fundamental, and which are only observed

The section closes with a remark that is easy to skim and worth stopping on. Electric charge is conserved because of a gauge symmetry: the conservation law and the interaction come from the same source.

Baryon number and lepton number have no such backing. They are conserved in everything anyone has measured, but no gauge symmetry demands it, so from a theoretical standpoint their conservation is not on the same footing at all. That is not an idle distinction — it is why the searches for proton decay (§3.6) and neutrinoless double beta decay (§10.7) are worth running: those laws are exactly the ones with no principle behind them.

The three gauge groups of the Standard Model, all built the same way

group	interaction	mediators	what makes it different
U(1)	electromagnetic	1	The group is abelian — two phase rotations commute — so the photon carries no charge of its own and photons ignore each other. Range infinite.
SU(3)	strong (Chapter 6)	8	Non-abelian: the gluons carry colour and therefore couple to each other. The mediator is still massless, but the range is short — a mechanism with no analogue in the row above.
SU(2) ⊗ U(1)	electroweak (Chapter 9)	4	Non-abelian and, uniquely, spontaneously broken — which is how mediators that gauge invariance forbids to have mass end up weighing 80 and 90 GeV. The Higgs field is what breaks it.

Chapters 3 and 4 used SU(2) and SU(3) as <em>approximate</em> symmetries, to sort hadrons into multiplets — accidents of the light quark masses (§4.11). The same groups reappear here as <strong>exact local</strong> symmetries, and their role is completely different: they no longer classify the particles, they generate the forces.

🔑 If you remember only three things

α is the only thing left to measure. Every structural feature follows from the symmetry, so the whole theory reduces to one number an experiment has to supply.
A gauge choice moves every potential and no observable. That is the definition of a redundancy, and here the redundancy is what generates the force.
Half of Maxwell costs nothing at all. Two of the four equations are identities of the potential, true before any physics is assumed.

Where this goes next

§5.2–5.3 is the experiment that made all this necessary: the Lamb shift, which the Dirac equation says is exactly zero.
§5.5 turns the interaction term above into a picture — the vertex of every Feynman diagram is $q_e\bar\psi\gamma^\mu\psi A_\mu$ and nothing else.
§5.8 is what happens to α once the vacuum is allowed to polarize: the “constant” of Eq. (5.16) turns out to depend on how closely you look.
Chapter 6 repeats this section’s argument with SU(3), and Chapter 9 with SU(2) ⊗ U(1). If you understand this page you have the shape of both.

✅ Check yourself — the Lagrangian density and gauge invariance

0/6 answered · 0 correct

1.Why do the homogeneous Maxwell equations ( $\nabla\cdot\mathbf{B} = 0$ and Faraday's law) never appear in the derivation from the Lagrangian?
2.In the GaugeLab, switching from the symmetric to a Landau gauge changes every arrow in the picture. What does not change, and why?
3.Turn off 'also rotate the electron field' in the ledger. What breaks, and what does that tell you?
4.Eq. (5.1) prints the metric as $g_{00} = -1$ , $g_{11} = g_{22} = g_{33} = +1$ . Why does this page use the opposite?
5.Why does it matter so much that α is dimensionless?
6.The section ends by noting that baryon and lepton number correspond to no gauge symmetry. Why is that worth saying?