Lagrangian of a Charged Particle in an Electromagnetic Field

Ask what a proton's momentum is, in the LHC, and you can get two answers that differ by more than the momentum itself — and both are correct. The mechanical momentum $\gamma m\mathbf{v}$ is what bends in the dipole. The canonical momentum $\gamma m\mathbf{v} + q\mathbf{A}$ is what Hamilton's equations and quantum mechanics are talking about. They are not the same object, and the gap between them is not an accounting error: the gap is the electromagnetic interaction. That is minimal coupling. Take the free-particle action, $-mc^2\!\int\!\mathrm{d}\tau$ , and add the only Lorentz scalar a charge and a four-potential can make between them: $-q\!\int\!A_\mu\,\mathrm{d}x^\mu$ . One term, no new constants, no freedom of choice. Out of it fall the Lorentz force, the field-strength tensor, and the statement that a magnetic field can never change a particle's energy. And it comes with a redundancy. Shift $A_\mu \to A_\mu + \partial_\mu\Lambda$ and the action changes by a pure endpoint term, which the variation cannot see. The physics is untouched; the canonical momentum is not. The one quantity minimal coupling introduces is the one quantity that is not gauge invariant — and understanding why that is a feature is what this article is for.

Key takeaways

Signature $(+,-,-,-)$ : $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , $u\!\cdot\!u = +c^2$ , $p\!\cdot\!p = +m^2c^2$ . SI units throughout, not Gaussian. Both locks re-derived below, never transcribed.
Minimal coupling: $S = -mc^2\!\int\!\mathrm{d}\tau \;-\; q\!\int\! A_\mu\,\mathrm{d}x^\mu$ , with $A^\mu = (\phi/c,\ \mathbf{A})$ so $A_\mu = (\phi/c,\ -\mathbf{A})$ . There is no $1/c$ — that factor is Gaussian, and this base carries it in three places while carrying SI in three others.
Canonical $\ne$ mechanical. $\;\mathbf{p} = \partial L/\partial\mathbf{v} = \gamma m\mathbf{v} + q\mathbf{A}$ , while $\boldsymbol{\pi} = \gamma m\mathbf{v} = \mathbf{p} - q\mathbf{A}$ . Only $\boldsymbol{\pi}$ appears in $E^2 = (\pi c)^2 + (mc^2)^2$ ; only $\mathbf{p}$ is conjugate to position and gets quantised.
$\boxed{H = \sqrt{(\mathbf{p} - q\mathbf{A})^2c^2 + m^2c^4} + q\phi}$ — the $c^2$ multiplies the whole bracket. The base prints $\sqrt{(c\mathbf{p} - q\mathbf{A})^2 + m^2c^4}$ , which subtracts a momentum from an energy.
Euler–Lagrange gives $\;\mathrm{d}p^\mu/\mathrm{d}\tau = q\,F^{\mu\nu}u_\nu\;$ with $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ — no $1/c$ — whose spatial part is the Lorentz force and whose time part is $\mathrm{d}E/\mathrm{d}t = q\,\mathbf{v}\!\cdot\!\mathbf{E}$ : no $\mathbf{B}$ , so a magnetic field can never change $\gamma$ .
Owned result. Under $A_\mu \to A_\mu + \partial_\mu\Lambda$ the action changes by exactly $-q[\Lambda]_{x_1}^{x_2}$ — an endpoint term, invisible to a variation with fixed endpoints. So $F_{\mu\nu}$ , the force and every trajectory are untouched, while $\mathbf{p} \to \mathbf{p} + q\nabla\Lambda$ moves. For a 1 T field and a 30 keV electron that shift is 1.9 times the electron's entire mechanical momentum.
Key idea: minimal coupling adds one scalar and buys the whole of electrodynamics — at the price of a canonical momentum that is not gauge invariant and is not a momentum. That price is where the Aharonov–Bohm phase, the covariant derivative, and every gauge theory in physics come from.

The core insight

There is exactly one way to couple a point charge to a four-potential that is (i) a Lorentz scalar, (ii) linear in $A_\mu$ , and (iii) built from the worldline alone: $-q\!\int\!A_\mu\,\mathrm{d}x^\mu$ . That is not a modelling choice, it is the only candidate — which is why it is called minimal. Everything else in electrodynamics is bookkeeping on that one term, and its gauge redundancy is the seed of every gauge theory that came after.

Where this lives today

Every accelerator on Earth is equation (6) in hardware. The dipoles bend, because $\mathbf{v}\times\mathbf{B}$ is perpendicular to $\mathbf{v}$ ; the RF cavities accelerate, because only $q\,\mathbf{v}\!\cdot\!\mathbf{E}$ can change the energy; and the radius is $r = \pi/(qB)$ , which is why the LHC's 27 km tunnel and its 8.33 T dipoles determine each other. The deeper descendant is the gauge principle. Promote $\mathbf{p} \to \mathbf{p} - q\mathbf{A}$ to the operator statement $\partial_\mu \to D_\mu = \partial_\mu + \tfrac{iq}{\hbar}A_\mu$ and you have the rule that generates all of quantum electrodynamics — and, with $A_\mu$ promoted from a number to a matrix, the whole Standard Model. The gauge-dependence of the canonical momentum, which looks like a defect in the classical theory, is the exact place where the rest of physics gets in. The Aharonov–Bohm effect is the experimental proof that this is not bookkeeping: an electron that never feels a force nevertheless remembers the flux it went around.

The physical picture

The free action of the previous articles is $-mc^2\!\int\!\mathrm{d}\tau$ — $-mc$ times the length of the worldline. To put a charge in a field you must add something, and relativity is extremely restrictive about what. The addition must be a Lorentz scalar, and it must be built out of the worldline and the field, and nothing else. The field is the four-potential $A_\mu$ ; the worldline supplies its tangent $\mathrm{d}x^\mu$ . There is exactly one scalar you can build from those two — their contraction. Integrate it along the curve, multiply by the one constant the particle carries, and

S_{\text{int}} \;=\; -\,q\!\int\! A_\mu\,\mathrm{d}x^\mu.

That is the whole electromagnetic interaction of a point charge, and there was no choice about it. Minimal means precisely that: not the simplest of several options, but the only option consistent with Lorentz invariance, linearity in the field, and locality on the worldline. (A magnetic-dipole coupling exists, but it needs an extra structure — a spin — that a point charge does not have.) Now the two consequences, and they pull in opposite directions.

One: everything works. Vary, and out comes the Lorentz force in covariant form, with $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ assembling itself out of the antisymmetric combination that survives the integration by parts. You do not put $F$ in. You get it out.

Two: the momentum breaks. The interaction $q\,\mathbf{v}\!\cdot\!\mathbf{A}$ depends on the velocity, so it contributes to $\partial L/\partial\mathbf{v}$ : the canonical momentum is no longer $\gamma m\mathbf{v}$ but $\gamma m\mathbf{v} + q\mathbf{A}$ . And $\mathbf{A}$ is not a physical field — a gauge transformation changes it without changing a single measurable thing. So the canonical momentum is gauge-dependent, and is not what any instrument measures. Not a small effect: the worked example below finds a gauge shift nearly twice the electron's own mechanical momentum. The first fact is why minimal coupling is used; the second is why it matters.

Figure

The field tensor is not put in — it falls out. LEFT: F_μν = ∂_μA_ν − ∂_νA_μ, built from OUR signature (+,−,−,−) with A^μ = (φ/c, A), so A_μ = (φ/c, −A) and ∂_μ = ((1/c)∂_t, ∇). Read off the entries: F_0i = E_i/c and F_ij = −ε_ijk B_k. Nobody chose this; the antisymmetric combination is simply what survives the integration by parts. RIGHT: contract with the four-velocity, dp^μ/dτ = q F^{μν} u_ν — with NO 1/c, because these are SI units — and the four-vector equation splits into TWO familiar laws. The SPATIAL part is the Lorentz force. The TIME part is the work–energy theorem, dE/dt = q v·E, and it carries the punchline: it contains no B at all. A magnetic field can never change a particle's energy, and therefore never changes γ. In the LHC's 8.33 T dipoles the 7 TeV protons are bent through 27 km without gaining one electron-volt from the magnets

The core equations

The four-potential is owned by `aq-sr-four-potential` and the field tensor by `aq-sr-electromagnetic-field-tensor`; both are recalled here, and both are re-derived from our metric rather than copied.

\boxed{\;S \;=\; -mc^2\!\int\!\mathrm{d}\tau \;-\; q\!\int\! A_\mu\,\mathrm{d}x^\mu\;}, \qquad A^\mu = \Big(\frac{\phi}{c},\ \mathbf{A}\Big), \qquad A_\mu = \Big(\frac{\phi}{c},\ -\mathbf{A}\Big) \qquad (1)

where: no $1/c$ . In SI, $A_\mu\,\mathrm{d}x^\mu = (\phi - \mathbf{v}\!\cdot\!\mathbf{A})\,\mathrm{d}t$ , so $-q\!\int\!A_\mu\mathrm{d}x^\mu = \int(q\,\mathbf{v}\!\cdot\!\mathbf{A} - q\phi)\,\mathrm{d}t$ exactly — which is the interaction that (2) needs. Put a $1/c$ there and the two disagree by a factor of $c$ . Note also that the free term is $-mc^2\!\int\!\mathrm{d}\tau$ : $mc^2$ against $\mathrm{d}\tau$ , but $mc$ against $\mathrm{d}s$ . This base writes $-mc\!\int\!\mathrm{d}\tau$ , units kg·m, not an action.

L \;=\; -\frac{mc^2}{\gamma} \;-\; q\phi \;+\; q\,\mathbf{v}\!\cdot\!\mathbf{A} \qquad (2)

\boxed{\;\mathbf{p} \;=\; \frac{\partial L}{\partial \mathbf{v}} \;=\; \gamma m\mathbf{v} + q\mathbf{A}\;}, \qquad\qquad \boldsymbol{\pi} \;\equiv\; \gamma m\mathbf{v} \;=\; \mathbf{p} - q\mathbf{A} \qquad (3)

where: $\mathbf{p}$ is the canonical momentum — conjugate to $\mathbf{x}$ , gauge-dependent, and the thing that becomes $-i\hbar\nabla$ . $\boldsymbol{\pi}$ is the mechanical momentum — gauge-invariant, and the only one satisfying $E_{\text{mech}}^2 = (\pi c)^2 + (mc^2)^2$ . Confusing them is the commonest error in this subject.

\boxed{\;H \;=\; \sqrt{\big(\mathbf{p} - q\mathbf{A}\big)^2c^2 + m^2c^4} \;+\; q\phi\;} \qquad (4)

where: the $c^2$ multiplies the whole bracket — every term under the root must be an energy $^2$ . The base's $\sqrt{(c\mathbf{p} - q\mathbf{A})^2 + m^2c^4}$ subtracts a momentum from an energy inside the bracket. (`aq-sr-relativistic-hamiltonian-formalism` gets it right; the two bases disagree.)

\boxed{\;\frac{\mathrm{d}p^\mu}{\mathrm{d}\tau} \;=\; q\,F^{\mu\nu}u_\nu\;}, \qquad F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \qquad (5)

where: again no $1/c$ . With $\partial_\mu = \big(\tfrac1c\partial_t,\ \nabla\big)$ and $A_\mu = (\phi/c,\,-\mathbf{A})$ , the entries come out as $F_{0i} = E_i/c$ and $F_{ij} = -\epsilon_{ijk}B_k$ , and (5) unpacks into

\frac{\mathrm{d}(\gamma m\mathbf{v})}{\mathrm{d}t} = q\big(\mathbf{E} + \mathbf{v}\times\mathbf{B}\big), \qquad\qquad \boxed{\;\frac{\mathrm{d}E}{\mathrm{d}t} = q\,\mathbf{v}\!\cdot\!\mathbf{E}\;} \qquad (6)

where: the second is the time component, and it is the one people forget. There is no $\mathbf{B}$ in it. A magnetic field does no work — ever — so it cannot change $\gamma$ , cannot change $|\mathbf{v}|$ , and cannot accelerate anything in the energy sense. It only bends.

A_\mu \;\to\; A_\mu + \partial_\mu\Lambda \qquad\Longrightarrow\qquad \delta S = -q\big[\Lambda\big]_{x_1}^{x_2}, \qquad F_{\mu\nu} \to F_{\mu\nu} \qquad (7)

where: a pure endpoint term — which a variation with fixed endpoints cannot see. Hence the equations of motion, and every trajectory, are gauge invariant. The canonical momentum is not: $\mathbf{p} \to \mathbf{p} + q\nabla\Lambda$ .

Working it out

Derivation AVary the coupling and the field tensor assembles itself

The goal is to get from the one-line interaction $-q\!\int\!A_\mu\mathrm{d}x^\mu$ to the covariant Lorentz force without assuming $F_{\mu\nu}$ exists. The field tensor is not an input to this calculation; it is the antisymmetric object that the integration by parts leaves behind, and watching it appear is the point of the exercise. We do the whole thing in SI, in our signature, and check the $1/c$ that the base carries.

Step 1. Vary the interaction, with $\lambda$ arbitrary. $S_{\text{int}} = -q\!\int\!A_\mu(x)\,\dot x^\mu\,\mathrm{d}\lambda$ , and $A_\mu$ depends on $\lambda$ only through $x(\lambda)$ , so

\delta S_{\text{int}} \;=\; -q\!\int\!\Big(\partial_\nu A_\mu\,\delta x^\nu\,\dot x^\mu \;+\; A_\mu\,\delta\dot x^\mu\Big)\mathrm{d}\lambda.

Why: two terms, because $A_\mu$ is a field — it changes when the worldline moves through it. That is the only difference from the free case, and it is enough.

Step 2. Integrate the second term by parts. Endpoints are fixed, so

-q\!\int\! A_\mu\,\frac{\mathrm{d}(\delta x^\mu)}{\mathrm{d}\lambda}\,\mathrm{d}\lambda \;=\; +\,q\!\int\! \partial_\nu A_\mu\,\dot x^\nu\,\delta x^\mu\,\mathrm{d}\lambda.

Why: the chain rule $\mathrm{d}A_\mu/\mathrm{d}\lambda = (\partial_\nu A_\mu)\dot x^\nu$ converts a total derivative into a field gradient. This is where the second index enters.

Step 3. Relabel and collect. Swap the dummy names $\mu\leftrightarrow\nu$ in the first term of Step 1 so that both terms multiply $\delta x^\nu$ :

\delta S_{\text{int}} \;=\; -\,q\!\int\!\Big(\partial_\nu A_\mu - \partial_\mu A_\nu\Big)\dot x^\mu\;\delta x^\nu\;\mathrm{d}\lambda \;\equiv\; -\,q\!\int\! F_{\nu\mu}\,\dot x^\mu\,\delta x^\nu\;\mathrm{d}\lambda.

Why: there it is. The antisymmetric combination $\partial_\nu A_\mu - \partial_\mu A_\nu$ was not postulated; it is the only thing that survives, because the symmetric part of $\partial_\nu A_\mu$ cancels between the two terms. $F_{\mu\nu}$ is a derived object. And it is the reason gauge invariance is automatic — $\partial_\mu\partial_\nu\Lambda$ is symmetric, so it dies in the antisymmetrisation.

Step 4. Add the free part and set $\delta S = 0$ . The free variation gives $-m\,\mathrm{d}u_\nu/\mathrm{d}\tau$ per unit $\delta x^\nu$ (`aq-sr-relativistic-action-proper-time`), so with $\lambda = \tau$ gauge-fixed after varying,

m\frac{\mathrm{d}u_\nu}{\mathrm{d}\tau} \;=\; q\,F_{\nu\mu}\,u^\mu \qquad\Longleftrightarrow\qquad \frac{\mathrm{d}p^\mu}{\mathrm{d}\tau} \;=\; q\,F^{\mu\nu}u_\nu.

Why: no $1/c$ . It appears only if the coupling is written $-\tfrac{q}{c}\!\int\!A_\mu\mathrm{d}x^\mu$ , which is Gaussian — and this base does exactly that in eq[0], eq[4] and eq[6] while its eq[1] and eq[2] are SI. A single article cannot be in two unit systems.

Step 5. Unpack. Build $F^{\mu\nu}$ from $\eta = \mathrm{diag}(+1,-1,-1,-1)$ : $F_{0i} = \partial_0 A_i - \partial_i A_0 = -\tfrac1c(\partial_t A_i + \partial_i\phi) = E_i/c$ , and $F_{ij} = -\epsilon_{ijk}B_k$ . Then contracting with $u_\nu = \gamma(c,-\mathbf{v})$ and using $\mathrm{d}/\mathrm{d}\tau = \gamma\,\mathrm{d}/\mathrm{d}t$ gives equations (6) — the Lorentz force and the work–energy theorem, both from one four-vector equation. Why: the $\mu = 0$ component is the free gift. $\mathrm{d}E/\mathrm{d}t = q\,\mathbf{v}\!\cdot\!\mathbf{E}$ contains no $\mathbf{B}$ , because the magnetic block of $F$ contracts an antisymmetric tensor with a symmetric velocity product and vanishes identically. A magnetic field cannot do work. That is not an extra fact you learn; it is the time component of the equation you already have. *Verified in verify_aq-sr-charged-particle-lagrangian-em-field.py: sympy builds $F_{\mu\nu}$ from $A_\mu = (\phi/c,-\mathbf{A})$ and $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , raises the indices, contracts with $u_\nu$ , and confirms exactly that $\mathrm{d}\mathbf{p}/\mathrm{d}t = q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$ and $\mathrm{d}E/\mathrm{d}t = q\,\mathbf{v}\!\cdot\!\mathbf{E}$ , with no $1/c$ ; and that inserting the base's $q/c$ makes both wrong by a factor of $c$ .*

Go deeper · Derivation B · 2 minThe owned result: gauge invariance is an ENDPOINT term, and the canonical momentum pays for it

Derivation BThe owned result: gauge invariance is an ENDPOINT term, and the canonical momentum pays for it

This is what the article exists to establish, and it has two halves that must be held together. (a) A gauge transformation changes the action, but only by a term evaluated at the endpoints — where the variation vanishes — so no trajectory moves. (b) The canonical momentum, the one object minimal coupling actually introduces, does move. Physics is gauge invariant; the formalism's central variable is not. Both facts, precisely.

Step 1. Transform. Under $A_\mu \to A_\mu + \partial_\mu\Lambda(x)$ for any smooth $\Lambda$ ,

\delta S_{\text{int}} \;=\; -\,q\!\int_{x_1}^{x_2}\!\big(\partial_\mu \Lambda\big)\,\mathrm{d}x^\mu \;=\; -\,q\!\int_{x_1}^{x_2}\!\mathrm{d}\Lambda \;=\; -\,q\Big[\Lambda(x_2) - \Lambda(x_1)\Big].

Why: $(\partial_\mu\Lambda)\,\mathrm{d}x^\mu$ is the total differential $\mathrm{d}\Lambda$ — that is what a gradient means. The line integral of an exact differential depends only on its endpoints, and on nothing about the path between them.

Step 2. Observe that the variation cannot see it. In the variational problem the endpoints $x_1, x_2$ are held fixed, so $\Lambda(x_2) - \Lambda(x_1)$ is the same number for every competing worldline. It shifts $S$ by a constant, and $\delta(\text{constant}) = 0$ . Why: this is the whole proof. The stationary point of $S$ and the stationary point of $S + \text{const}$ are the same worldline. Every trajectory, every force, every observable is gauge invariant — and it took one line, because the endpoint condition of the variational principle did all the work.

Step 3. Confirm it at the level of the fields: $F_{\mu\nu} \to F_{\mu\nu} + (\partial_\mu\partial_\nu - \partial_\nu\partial_\mu)\Lambda = F_{\mu\nu}$ . Why: second derivatives commute, so the extra piece is symmetric and dies in the antisymmetrisation. $\mathbf{E}$ , $\mathbf{B}$ and the Lorentz force are gauge invariant — the same conclusion as Step 2, reached from the other end.

Step 4. Now the price. The canonical momentum is $\mathbf{p} = \gamma m\mathbf{v} + q\mathbf{A}$ , and under the gauge transformation $\mathbf{A} \to \mathbf{A} + \nabla\Lambda$ (the spatial part of $\partial_\mu\Lambda$ , with a sign from the lowered index), so

\mathbf{p} \;\to\; \mathbf{p} + q\nabla\Lambda, \qquad\text{while}\qquad \boldsymbol{\pi} = \gamma m \mathbf{v} = \mathbf{p} - q\mathbf{A} \;\to\; \boldsymbol{\pi}.

Why: the canonical momentum is not gauge invariant, and is therefore not measurable. The mechanical momentum is, and is. Minimal coupling introduced exactly one new object into the mechanics — the canonical momentum — and that object is precisely the one that is not physical. The gap, $\mathbf{p} - \boldsymbol{\pi} = q\mathbf{A}$ , is the coupling. It is not an artefact of it.

Step 5. Do not conclude that the canonical momentum is useless. It is conjugate to $\mathbf{x}$ : it is what Hamilton's equations evolve and what becomes $-i\hbar\nabla$ . Every gauge theory in physics is the observation that a wavefunction's phase must rotate, $\psi \to e^{iq\Lambda/\hbar}\psi$ , exactly so as to keep $(\mathbf{p} - q\mathbf{A})\psi$ covariant — which is why $D_\mu = \partial_\mu + \tfrac{iq}{\hbar}A_\mu$ is the covariant derivative, and why $q\mathbf{A}$ is detectable as a phase even where $\mathbf{B} = 0$ (Aharonov & Bohm, 1959). Why: the gauge-dependence of $\mathbf{p}$ is not a bug to quarantine — it is the door to the rest of physics. *Verified in verify_aq-sr-charged-particle-lagrangian-em-field.py: sympy confirms $(\partial_\mu\Lambda)\mathrm{d}x^\mu = \mathrm{d}\Lambda$ symbolically and that $F_{\mu\nu}$ is invariant for a generic $\Lambda(x)$ ; the canonical and mechanical momenta are computed in two explicit gauges of a uniform field (Landau and symmetric), confirming that $\mathbf{p}$ shifts by $q\nabla\Lambda$ while $\boldsymbol{\pi}$ , $\mathbf{B}$ , and the force are identical — with the numbers printed below.*

Figure

PUNCHLINE — the canonical momentum is not a momentum. Same electron, same point, same velocity, same magnetic field B = 1.00 T ẑ. Only the GAUGE differs. LEFT, Landau gauge A = (−By, 0, 0); RIGHT, symmetric gauge A = ½(−By, Bx, 0). They differ by ∇Λ with Λ = Bxy/2 = 1.00×10⁻⁶ T·m². The blue arrow — the MECHANICAL momentum π = γmv — is IDENTICAL: 9.4942×10⁻²³ kg·m/s (a 30 keV electron, 177.65 keV/c). The orange arrow, qA, is not, and the red CANONICAL momentum p = π + qA swings wildly between the two panels. The shift |q∇Λ| = 1.7913×10⁻²² kg·m/s is 1.89 TIMES the electron's entire mechanical momentum. B is the same. The force is the same. The trajectory is the same. Nothing measurable moved — and the quantity that the Hamiltonian formalism and quantum mechanics are built on moved by more than the particle's own momentum

Limiting cases

Case	Result	Meaning
$q \to 0$	$S \to -mc^2\!\int\!\mathrm{d}\tau$	the free particle (`aq-sr-relativistic-lagrangian-free-particle`)
$v \ll c$	$L \to \tfrac12 mv^2 - q\phi + q\,\mathbf{v}\!\cdot\!\mathbf{A}$	the non-relativistic charged Lagrangian — the $\mathbf{v}\!\cdot\!\mathbf{A}$ term survives intact
$\phi = 0$	$\mathrm{d}(\gamma mc^2)/\mathrm{d}t = 0$	pure magnetic: $\gamma$ is constant. The magnets bend; they never accelerate

Common trapThis base is in TWO unit systems at once — and its Hamiltonian is dimensionally broken

Four defects, all in one base, and they interlock.

(1) SI vs Gaussian, inside one article. eq[1] ( $L = -mc^2/\gamma - q\phi + q\mathbf{v}\!\cdot\!\mathbf{A}$ ) and eq[2] ( $\mathbf{p} = \gamma m\mathbf{v} + q\mathbf{A}$ ) are SI, and the symbol table declares $A_\mu = (\phi/c, -\mathbf{A})$ , which is SI. But eq[0], eq[4] and eq[6] all carry a Gaussian $q/c$ . They cannot both be right, and the one-line check settles it: with $A^\mu = (\phi/c,\mathbf{A})$ , $\;-q\!\int\!A_\mu\mathrm{d}x^\mu = \int(q\mathbf{v}\!\cdot\!\mathbf{A} - q\phi)\mathrm{d}t$ , which is eq[1] exactly. There is no $1/c$ .

(2) $\;-mc\!\int\!\mathrm{d}\tau$ . eq[0]'s free term has units $\mathrm{kg\,m}$ . That is not an action. Against $\mathrm{d}\tau$ the prefactor is $mc^2$ ; only against $\mathrm{d}s$ is it $mc$ .

(3) $\;H = \sqrt{(c\mathbf{p} - q\mathbf{A})^2 + m^2c^4}$ . eq[3] subtracts a momentum from an energy inside the bracket. The correct form is $\sqrt{(\mathbf{p} - q\mathbf{A})^2c^2 + m^2c^4}$ — the $c^2$ multiplies the whole bracket. The base for `aq-sr-relativistic-hamiltonian-formalism` prints it correctly: the two bases contradict each other.

Worked examples

1 · Why the LHC needs 8.33 tesla, and why its magnets do no work

The LHC's design beam is 7 TeV protons; its dipoles reach $B = 8.33\ \mathrm{T}$ . What bending radius does that buy — and how much energy do the magnets give the beam?

Step 1 — identify. A pure magnetic field: $\phi = 0$ , so equation (6) gives $\mathrm{d}E/\mathrm{d}t = q\,\mathbf{v}\!\cdot\!\mathbf{E} = 0$ . $\gamma$ is constant. With $|\mathbf{v}|$ fixed, $q\,\mathbf{v}\times\mathbf{B}$ is purely centripetal: $\gamma m v^2/r = qvB$ , so $r = \pi/(qB)$ with $\pi$ the mechanical momentum.

Step 2 — the momentum. At 7 TeV the proton is ultra-relativistic: $\gamma = E/(m_pc^2) = \mathbf{7461}$ , so $\pi \simeq E/c$ to one part in $10^7$ :

\pi = \frac{7\times10^{12}\times 1.602\times10^{-19}}{2.998\times10^{8}} = \mathbf{3.7410\times10^{-15}\ kg\,m/s}.

Step 3 — the radius. $\;r = \pi/(qB) = 3.7410\times10^{-15}\big/\big(1.602\times10^{-19}\times 8.33\big) = \mathbf{2803\ m}$ .

Step 4 — the boundary number. Check it against the machine. $2\pi r = \mathbf{17{,}612\ m}$ , and the LHC's circumference is $26{,}659\ \mathrm{m}$ , so the dipoles must occupy $\mathbf{66.1\ \%}$ of the ring — the rest being straight sections for the experiments, the RF and the quadrupoles. That is the machine. The 8.33 T figure is not a specification handed down; it is what $r = \pi/(qB)$ demands once you decide to put 7 TeV protons in a 27 km tunnel. And the punchline is the time component. Over $10^{8}$ turns the magnets bend the beam through millions of kilometres of arc and transfer exactly zero energy to it, because $\mathrm{d}E/\mathrm{d}t = q\,\mathbf{v}\!\cdot\!\mathbf{E}$ has no $\mathbf{B}$ in it. All the acceleration is done by the RF cavities — an $\mathbf{E}$ field, along $\mathbf{v}$ . One four-vector equation tells you which piece of hardware does which job.

2 · A "momentum" that changes when you change nothing

A 30.0 keV electron sits at $(x,y) = (1.00, 2.00)\ \mathrm{mm}$ in $\mathbf{B} = 1.00\ \mathrm{T}\,\hat{\mathbf{z}}$ . Compute its canonical momentum in the Landau gauge and in the symmetric gauge.

Step 1 — identify. Both gauges give the same $\mathbf{B} = \nabla\times\mathbf{A}$ , so they describe the same physical field. Landau: $\mathbf{A}_L = (-By,\,0,\,0)$ . Symmetric: $\mathbf{A}_S = \tfrac12(-By,\,Bx,\,0)$ . They differ by a gradient, $\mathbf{A}_S - \mathbf{A}_L = \nabla\Lambda$ with $\Lambda = Bxy/2 = \mathbf{1.00\times10^{-6}\ T\,m^2}$ .

Step 2 — the mechanical momentum. $\gamma = 1 + 30.0/511.00 = 1.058709$ , $\beta = 0.328376$ , so $\pi = \gamma\beta m_ec = \mathbf{177.65\ keV}/c = \mathbf{9.4942\times10^{-23}\ kg\,m/s}$ . Identical in both gauges. It is what bends, and it is what appears in $E^2 = (\pi c)^2 + (m_ec^2)^2$ .

Step 3 — the vector-potential term. With $q = -e$ ,

q\mathbf{A}_L = \big(\mathbf{3.2044},\;0,\;0\big)\times10^{-22}\ \mathrm{kg\,m/s}, \qquad q\mathbf{A}_S = \big(\mathbf{1.6022},\;\mathbf{-0.8011},\;0\big)\times10^{-22}\ \mathrm{kg\,m/s}.

Step 4 — the boundary number. $\mathbf{p} = \boldsymbol{\pi} + q\mathbf{A}$ therefore differs between the gauges by

\big|q\nabla\Lambda\big| \;=\; \mathbf{1.7913\times10^{-22}\ kg\,m/s} \;=\; \mathbf{1.89}\times \big|\boldsymbol{\pi}\big|.

The "momentum" changed by nearly twice the electron's entire mechanical momentum, and nothing physical moved. Same $\mathbf{B}$ , same force, same trajectory, same energy. Worse: in the Landau gauge $|q\mathbf{A}_L| = 3.2044\times10^{-22} = 3.38\,|\boldsymbol{\pi}|$ — the canonical momentum is dominated by a quantity you can change by writing down a different $\Lambda$ . And yet $\mathbf{p}$ is the one conjugate to $\mathbf{x}$ , the one Hamilton's equations evolve, and the one that becomes $-i\hbar\nabla$ . Both statements are true, and holding them together is what it means to understand minimal coupling.

Physics of the portrait

Minimal coupling is a rule so short it looks like a convention — add $-q\!\int\!A_\mu\mathrm{d}x^\mu$ — and so deep that the Standard Model is its generalisation.

Origins · 2 minMinimal coupling and how this was discovered

Minimal coupling

1903–1959 · Schwarzschild casts electron theory as an action principle (1903) · Fock introduces the minimal substitution in the relativistic wave equation (1926) · London reads Weyl's scale factor as a quantum PHASE (1927) · Weyl states the gauge principle in its modern form (1929) · Aharonov and Bohm show that qA is physical where B is not (1959)

The action came first, and from an unexpected direction. In 1903 Karl Schwarzschild — a decade before the star that carries his name — published a three-part series Zur Elektrodynamik in the Göttingen Nachrichten, its first part titled Zwei Formen des Princips der Action in der Elektronentheorie. It cast the electron theory of the day in Lagrangian form, and the interaction term it needed was the contraction of the charge with the potential. Two years before special relativity existed, the coupling was already written down in the shape it still has. What no one had was a reason. $-q\!\int\!A_\mu\mathrm{d}x^\mu$ was the term that worked; nothing explained why it should be the term, or why the potential — manifestly not unique, since a gradient can be added at will — should appear in the fundamental description at all. Quantum mechanics supplied the missing half. In 1926 Vladimir Fock showed that a charged particle's relativistic wave equation is invariant under local phase rotations only if the ordinary derivative is replaced by $\partial_\mu + \tfrac{iq}{\hbar}A_\mu$ — the minimal substitution, which is classical minimal coupling read off at the level of the wave equation. Fritz London, in 1927, identified this as the quantum reading of an idea Hermann Weyl had floated in 1918 and abandoned: Weyl's local rescaling of length, which failed as physics, becomes a local rotation of phase, which succeeds completely. Weyl accepted the correction and, in Elektron und Gravitation I (Z. Phys. 56, 330, 1929), stated the modern gauge principle: demand invariance under $\psi \to e^{iq\Lambda/\hbar}\psi$ , and the electromagnetic potential — with its gauge freedom and its minimal coupling — is forced to exist. The redundancy stopped being an embarrassment and became the generating principle. The last word belongs to the experiment. If $A_\mu$ were a mere convenience, a region with $\mathbf{B} = 0$ would be electromagnetically inert. In 1959 Yakir Aharonov and David Bohm (Phys. Rev. 115, 485) pointed out that an electron beam split around a confined magnetic flux — passing only through regions where $\mathbf{B} = \mathbf{0}$ and the force is exactly zero — nonetheless acquires a measurable interference phase $\tfrac{q}{\hbar}\!\oint\!\mathbf{A}\!\cdot\!\mathrm{d}\boldsymbol{\ell}$ . The gauge-dependent object inside the canonical momentum leaves a gauge-invariant fingerprint. The potential is not a convenience. It is the field.

History

1903

— Schwarzschild, "Zur Elektrodynamik. I. Zwei Formen des Princips der Action in der Elektronentheorie" (Gott. Nachr.), casts electron theory as an action principle with the charge-potential coupling.

1926

— Fock shows that gauge invariance of the relativistic wave equation for a charged particle requires the minimal substitution, replacing the derivative by a gauge-covariant derivative.

1927

— London reinterprets Weyl's 1918 scale factor as a quantum phase, connecting gauge invariance to the phase of the wavefunction.

1929

— Weyl, "Elektron und Gravitation. I" (Z. Phys. 56, 330), states the gauge principle in its modern form: local phase invariance forces the electromagnetic coupling to exist.

1959

— Aharonov and Bohm, "Significance of Electromagnetic Potentials in the Quantum Theory" (Phys. Rev. 115, 485), show that a charged particle acquires a measurable phase from A in a region where E and B both vanish.

Going further

Texts. Jackson, Classical Electrodynamics, 3rd ed., ch. 12 (SI throughout — note that Jackson's earlier editions are Gaussian) · Landau & Lifshitz, The Classical Theory of Fields, §§16–17 (Gaussian; translate before copying) · Griffiths, Introduction to Electrodynamics, 4th ed., ch. 12 · Aharonov & Bohm, Phys. Rev. 115, 485 (1959) · Weyl, Z. Phys. 56, 330 (1929).

Provenance. Signature $(+,-,-,-)$ ; SI units; $A^\mu = (\phi/c,\mathbf{A})$ ; $u\!\cdot\!u = +c^2$ . $F_{\mu\nu}$ , its index-raised form and the covariant Lorentz force are all re-derived from that metric in verify_aq-sr-charged-particle-lagrangian-em-field.py, never transcribed. Every printed number is recomputed at 50-digit precision and greped from this file, with an artefact guard.

Base defects found and repaired — four. (i) eq[0] writes $-mc\!\int\!\mathrm{d}\tau$ : units kg·m, not an action. (ii) eq[0], eq[4] and eq[6] carry a Gaussian $q/c$ while eq[1], eq[2] and the symbol table are SI — two unit systems in one article — and the SI check settles it: no $1/c$ . (iii) eq[3] prints $H = \sqrt{(c\mathbf{p}-q\mathbf{A})^2 + m^2c^4}$ , subtracting a momentum from an energy. (iv) DERIV[1] closes with the same Gaussian factor. All four corrected above and asserted symbolically.

Scope. The four-potential and the field tensor as objects are `aq-sr-four-potential` and `aq-sr-electromagnetic-field-tensor` — recalled, not re-derived. The free action and its variation are `aq-sr-relativistic-lagrangian-free-particle` and `aq-sr-relativistic-action-proper-time`. The Legendre transform to $H$ and its quantisation are `aq-sr-relativistic-hamiltonian-formalism`. Spin transport in a field is `aq-sr-bmt-equation`. Note: `aq-sr-covariant-lorentz-force` is linked for its force treatment only; its signature declaration is defective and is not inherited here.

Builds on. `aq-sr-relativistic-lagrangian-free-particle` · `aq-sr-relativistic-action-proper-time` · `aq-sr-four-potential` · `aq-sr-electromagnetic-field-tensor` · `aq-sr-four-momentum`. Leads to. `aq-sr-relativistic-hamiltonian-formalism` · `aq-sr-bmt-equation` · `aq-sr-thomas-precession`.

Why is there no 1/c in the interaction term? Landau and Lifshitz have one.

Because Landau & Lifshitz work in Gaussian units and this corpus works in SI. In Gaussian units $A^\mu = (\phi, \mathbf{A})$ and the coupling is $-\tfrac{q}{c}\!\int\!A_\mu\mathrm{d}x^\mu$ ; in SI, $A^\mu = (\phi/c, \mathbf{A})$ — the $1/c$ absorbed into $A^0$ — and the coupling is $-q\!\int\!A_\mu\mathrm{d}x^\mu$ . Both are correct; taking one line from each is not. The check that settles it every time: expand the coupling and demand that it reduce to $\int(q\mathbf{v}\!\cdot\!\mathbf{A} - q\phi)\,\mathrm{d}t$ . With SI potentials and no $1/c$ it does, exactly. This article's base fails that check in three equations while passing it in two others.

Is the canonical momentum real, or just bookkeeping?

Both, and the tension is the physics. It is not measurable — the worked example changes it by nearly twice the particle's own momentum without touching anything physical. But it is not arbitrary either: it is the variable conjugate to position, so it is what Hamilton's equations evolve and what becomes $-i\hbar\nabla$ when you quantise. The resolution is that gauge-dependent is not the same as unphysical. Certain functionals of $\mathbf{p}$ are gauge invariant — most famously $\oint(\mathbf{p} - q\mathbf{A})\!\cdot\!\mathrm{d}\boldsymbol{\ell}$ around a closed loop, whose $q\mathbf{A}$ part is the Aharonov–Bohm phase $\tfrac{q}{\hbar}\Phi_B$ , and that is measured routinely. A gauge-dependent variable with gauge-invariant content is exactly what "gauge theory" means.

Why does a magnetic field do no work? I can feel a magnet pull a nail.

The nail is not a point charge. What pulls it is the force on the bound currents and dipoles inside the iron, whose energy comes from the field's own store — not from $q\mathbf{v}\times\mathbf{B}$ acting on a moving charge. For a genuine point charge the statement is exact, and it is one line: the time component of the covariant equation is $\mathrm{d}E/\mathrm{d}t = q\,\mathbf{v}\!\cdot\!\mathbf{E}$ , with no $\mathbf{B}$ anywhere, because $\mathbf{v}\times\mathbf{B} \perp \mathbf{v}$ so $\mathbf{F}\!\cdot\!\mathbf{v} = 0$ identically. Structurally it is because $F^{\mu\nu}$ is antisymmetric: $u_\mu F^{\mu\nu}u_\nu = 0$ for any $u$ — which is also the statement that the four-force never pushes the particle off its mass shell.

Why does the field tensor have to be antisymmetric?

It does not have to be — it comes out that way, and that is the more interesting statement. Vary the coupling and integrate by parts: the result is proportional to $\partial_\nu A_\mu - \partial_\mu A_\nu$ , because the symmetric part appears twice with opposite signs and cancels. Nothing was imposed. The consequences are immediate: antisymmetry means six independent components — exactly $\mathbf{E}$ and $\mathbf{B}$ ; it means gauge invariance, since $\partial_\mu\partial_\nu\Lambda$ is symmetric and dies; and it means $u_\mu F^{\mu\nu}u_\nu = 0$ , i.e. the four-force is orthogonal to the four-velocity and the mass shell is preserved. Three of the deepest facts about electromagnetism are one fact: $F$ was born as a commutator of derivatives.

Does the coupling change if the particle has spin?

Yes, and the extra piece is not minimal. Minimal coupling is defined by linearity in $A_\mu$ and construction from the worldline's position and tangent alone. A spin gives you a new object, and with it another Lorentz scalar — the magnetic-dipole coupling $\propto \mu\,F_{\mu\nu}S^{\mu\nu}$ . It is legitimate and it is not forced on you; that the electron's magnetic moment is very nearly the minimal Dirac value $g = 2$ , departing from it only through radiative corrections, is one of the deepest facts in physics. The relativistic dynamics of that spin in a field is `aq-sr-bmt-equation`. Minimal coupling is the floor, not the ceiling.

Can I read the Lorentz force off aq-sr-covariant-lorentz-force?

Read it for the force law, but do not inherit its conventions. That live article declares $(-,+,+,+)$ and prints $p^\mu p_\mu = -m^2c^2$ , contradicting the shipped SR corpus, which locks $(+,-,-,-)$ and $p\!\cdot\!p = +m^2c^2$ ; it is in the fix queue. Its force result is fine — the physical force does not care about your signature — but the signature line and the mass-shell sign are wrong for this corpus, and copying them propagates the defect. This article therefore re-derives $F_{\mu\nu}$ , $F^{\mu\nu}$ and $\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$ from $\eta = \mathrm{diag}(+1,-1,-1,-1)$ from scratch. The general rule, which has now caught defects in four separate bases: never transcribe a metric expression. Re-derive it.