Special Relativity — worked problems

Setup

Two charges $+q$ at $(\pm d/2, 0, 0)$. The median plane $x = 0$ separates them by symmetry. The force on the right-hand charge $(+d/2)$ can be computed by integrating the Maxwell stress tensor over any surface enclosing it — we choose the infinite median plane plus a hemisphere at infinity (which contributes zero).

Field on the median plane

By symmetry, the $\hat x$ components of the two Coulomb fields cancel: $\vec E_x = 0$ on $x = 0$. The $\hat y, \hat z$ components add:

$$\vec E = 2\cdot\frac{q}{4\pi\varepsilon_0 r^3}(y\hat y + z\hat z),\quad r^2 = (d/2)^2 + y^2 + z^2.$$

Magnitude squared:

$$E^2 = \frac{4q^2(y^2 + z^2)}{(4\pi\varepsilon_0)^2 r^6}.$$

Stress tensor

For $\vec B = 0$ and $\vec E$ purely transverse to $\hat x$:

$$T^{xx} = \tfrac{1}{2}\varepsilon_0[E_x^2 - E_y^2 - E_z^2] = -\tfrac{1}{2}\varepsilon_0 E^2.$$

Negative — pressure pulling charges together... wait, the sign convention: a negative $T^{xx}$ means a pull toward the $+\hat x$ direction away from the median plane (when we consider it as the force on the matter on one side from the matter on the other). With positive charges, the force should be repulsive (+$\hat x$ on the right charge). Let me reconsider.

Force as flux of stress

The force on the right-hand particle from the field is

$$F^x_\text{on right} = \int T^{xj}\hat n^j\,dA,$$

integrated over the median plane with $\hat n = -\hat x$ (outward normal of the half-space containing the right charge). So $F^x_\text{on right} = -\int T^{xx}\,dA = -\int(-\tfrac{1}{2}\varepsilon_0 E^2)\,dA = +\tfrac{1}{2}\varepsilon_0\int E^2\,dA > 0$ — positive, i.e. away from the other charge: repulsion. ✓

Evaluate the integral

Using cylindrical coordinates $\rho = \sqrt{y^2 + z^2}$, $dA = 2\pi\rho\,d\rho$:

$$\int E^2\,dA = \frac{4q^2}{(4\pi\varepsilon_0)^2}\int_0^\infty\frac{\rho^2}{r^6}\cdot 2\pi\rho\,d\rho = \frac{8\pi q^2}{(4\pi\varepsilon_0)^2}\int_0^\infty\frac{\rho^3\,d\rho}{((d/2)^2 + \rho^2)^3}.$$

Substitution $u = \rho^2 + (d/2)^2$, $du = 2\rho\,d\rho$:

$$\int_0^\infty\frac{\rho^3 d\rho}{((d/2)^2 + \rho^2)^3} = \tfrac{1}{2}\int_{(d/2)^2}^\infty\frac{(u - (d/2)^2)}{u^3}\,du = \tfrac{1}{2}\bigl[\ln u + \tfrac{(d/2)^2}{2u^2}\bigr]^\infty_{(d/2)^2}\bigl|\,$ — cleaner: standard result $= 1/(4(d/2)^2) = 1/d^2$.

So $\int E^2 dA = 8\pi q^2/[(4\pi\varepsilon_0)^2 d^2]= q^2/(2\pi\varepsilon_0^2 d^2)$. Then

$$F^x = \tfrac{1}{2}\varepsilon_0\cdot q^2/(2\pi\varepsilon_0^2 d^2) = q^2/(4\pi\varepsilon_0 d^2).$$

$$\boxed{\;F^x = \frac{q^2}{4\pi\varepsilon_0 d^2}\quad\text{(repulsive Coulomb).}\;}$$

Faraday tubes of flux

This recovers Coulomb's law exactly from the field's transverse pressure on the median plane, without ever directly using $F = qE$ on a charge. Faraday's elastic tube-of-flux mechanism made rigorous — the EM field carries momentum and stress, and inter-particle forces are mediated by that stress, not by action-at-a-distance. The same logic, generalised to time-dependent fields, gives radiation reaction and electromagnetic momentum (linear and angular) transport.

Rest frame: pure Coulomb

In the electrons' common rest frame, both are stationary, separated by $d$ along $\hat y$. Only the Coulomb force acts:

$$\vec F_* = \frac{e^2}{4\pi\varepsilon_0 d^2}\hat y\quad\text{(repulsion)}.$$

4-force in the rest frame (where $dm/d\tau = 0$ for the static configuration):

$$f^\mu_* = (0,\;0,\;e^2/(4\pi\varepsilon_0 d^2),\;0).$$

Boost along $\hat x$ to the lab

The Lorentz boost along $\hat x$ leaves transverse components $f^y, f^z$ unchanged. The time component transforms but is zero in the rest frame, contributing nothing. Therefore

$$\boxed{\;f^\mu_\text{lab} = (0,\;0,\;e^2/(4\pi\varepsilon_0 d^2),\;0)\quad\text{(frame-independent for this geometry)}.\;}$$

Lab-frame 3-force

From the 4-force relation $f^\mu = (\gamma\vec F\!\cdot\!\vec u/c, \gamma\vec F)$ (where $\vec u$ is the particle's lab velocity, $\vec F$ the lab 3-force), the spatial part:

$$\vec F_\text{lab} = \vec f_\text{spatial}/\gamma = \frac{e^2}{4\pi\varepsilon_0\gamma d^2}\hat y.$$

Suppressed by $1/\gamma$ relative to the Coulomb force on stationary charges!

Decomposition: electric vs. magnetic in lab

Boost the rest-frame Coulomb $\vec E$ to the lab: gives lab $\vec E_\text{lab} = \gamma\vec E_*$ (transverse Coulomb enhanced) plus a lab magnetic field $\vec B_\text{lab} = \gamma\vec v\times\vec E_*/c^2$ (parallel currents).

The lab Lorentz 3-force on electron 2: $\vec F = -e(\vec E_\text{lab} + \vec u\times\vec B_\text{lab})$. Substituting and simplifying: electric repulsion $-e\cdot\gamma\vec E_*$ (enhanced by $\gamma$), magnetic attraction $-e\vec u\times\vec B_\text{lab} = -\gamma\beta^2\vec E_*\cdot(-e)$ (attractive for parallel currents). Net: $\vec F = -e\vec E_*\gamma(1 - \beta^2) = -e\vec E_*/\gamma$. ✓ Matches the boost calculation.

Pinch effect and self-collimation

As $\beta\to 1$, the lab 3-force on each electron vanishes: the magnetic attraction nearly cancels the electric repulsion. This is the origin of self-collimation in relativistic beams (the pinch effect): high-current particle beams confine themselves transversely via their own magnetic field, exactly because each particle's neighbours produce stronger $\vec B$ relative to $\vec E$ as $\gamma$ grows. Cosmic-ray jets, astrophysical relativistic outflows (AGN jets, GRBs), and the design of intense electron beams in colliders all rely on this kinematic.

Perfect-fluid stress-energy

A photon-like (radiation) fluid has equation of state $p = \rho/3$. In its rest frame:

$$T_0^{\mu\nu} = \mathrm{diag}(\rho_0,\;p_0,\;p_0,\;p_0)\quad\text{with }p_0 = \rho_0/3.$$

Boost to a frame moving at $-\vec v$ relative to the rest frame

Equivalently, in the moving frame, the fluid has bulk 3-velocity $-\vec v$. The 4-velocity of the fluid in the moving frame: $u^\mu = \gamma(c, -\vec v)$.

Perfect-fluid form: $T'^{\mu\nu} = (\rho_0 + p_0)u^\mu u^\nu/c^2 - p_0\eta^{\mu\nu}$. Plug in $u^0 = \gamma c$, $u^i = -\gamma v^i$:

$$T'^{00} = (\rho_0 + p_0)\gamma^2 - p_0 = (\rho_0 + \rho_0/3)\gamma^2 - \rho_0/3 = (4\rho_0\gamma^2 - \rho_0)/3 = \rho_0(4\gamma^2 - 1)/3.$$

Using $\gamma^2 = 1/(1-\beta^2)$: $4\gamma^2 - 1 = (4 - 1 + \beta^2)/(1 - \beta^2) = (3 + \beta^2)/(1-\beta^2) = (3+\beta^2)\gamma^2$. So

$$\rho' = T'^{00} = (3 + \beta^2)\gamma^2\rho_0/3.$$

Momentum density:

$$T'^{0i} = (\rho_0 + p_0)\gamma\cdot(-\gamma v^i)/c = -(4\rho_0\gamma^2/3c) v^i.$$

Express in terms of $\rho'$ and $\vec v$

$\vec\pi = T'^{0i}/c$ in 3-vector form (the momentum density). Solve $-(4\rho_0\gamma^2)/(3c^2)$ in terms of $\rho'$ and $v$:

$$\frac{4\rho_0\gamma^2}{3} = \frac{4\rho'}{3 + \beta^2}\cdot\frac{1}{1-\beta^2}\cdot(1-\beta^2)\cdot\frac{3}{3+\beta^2}\cdot 3\text{?}\quad$$

Cleaner: $\rho' = \rho_0(3+\beta^2)\gamma^2/3 \Rightarrow \rho_0\gamma^2 = 3\rho'/(3+\beta^2)$. Substitute:

$$\vec\pi = -\frac{4}{3}\cdot\frac{1}{c^2}\cdot\frac{3\rho'}{3 + \beta^2}\cdot\vec v = -\frac{4\rho'\vec v}{(3 + \beta^2)c^2}.$$

With $\beta^2 = v^2/c^2$:

$$\boxed{\;\vec\pi = -\frac{4\rho'}{3c^2 + v^2}\,\vec v.\;}$$

Newtonian / enthalpy limit

For $v\ll c$: $\beta\to 0$, $\rho'\to\rho_0$, so $\vec\pi\to -(4\rho_0/(3c^2))\vec v = -(\rho_0 + p_0)\vec v/c^2$.

The inertial density for momentum carriage is the enthalpy $\rho_0 + p_0$, not the bare energy density $\rho_0$. The $4/3$ enthalpy factor of radiation appears as soon as photons acquire a bulk velocity — the same physics responsible for the famous "$4/3$ problem" in classical models of the electron self-energy (resolved in QED). Relevant in cosmology: photon bulk flow in the early radiation-dominated universe carries momentum proportional to enthalpy, modifying gravitational stress-energy contributions.

Setup

Cruiser of cross-section $A$ moves at velocity $v\hat x$ through stationary dust of rest-frame mass-energy density $\rho_0$ (= rest mass per unit volume times $c^2$). All dust impacted is absorbed (inelastic). Compute the 4-force.

Galaxy frame (dust rest frame)

In time $dt$, the cruiser sweeps a volume $Av\,dt$, absorbing rest-energy $\rho_0 A v\,dt$. By energy conservation, this enters the cruiser's worldline, so

$$f^0\,dt = \rho_0 A v\,dt/c \Rightarrow f^0 = \rho_0 A v/c.$$

The 3-force on the cruiser from this absorption: in the galaxy frame, since dust is at rest and absorbed at $\vec u = v\hat x$, the cruiser must do work to bring each dust particle to rest in its own frame — but conversely, dust absorbed in lab is at rest, so no momentum is transferred to the cruiser from the dust's 3-motion in the lab. Therefore in the galaxy frame: $\vec f = 0$ (no recoil force in the lab; the rest-mass gain comes entirely from the time component).

Carefully: $f^\mu_\text{gal} = (\gamma\rho_0 A v/c,\;0,\;0,\;0)$ (the $\gamma$ comes from the cruiser's proper time vs. lab time).

Cruiser frame

Boost $f^\mu$ to the cruiser's frame (velocity $-v\hat x$ relative to galaxy):

$$f'^0 = \gamma(f^0 + 0) = \gamma\cdot\gamma\rho_0 A v/c = \gamma^2\rho_0 A v/c.$$

$$f'^x = \gamma(0 - v\cdot f^0) = -\gamma\cdot v\cdot\gamma\rho_0 A v/c = -\gamma^2\rho_0 A v^2/c.$$

Restoring SI: $f'^x = -\gamma^2\rho_0 A v^2/c^2$ (dividing by an extra $c$ to convert to SI 3-force units).

$$\boxed{\;f'^\mu = (\gamma^2\rho_0 A v/c,\;-\gamma^2\rho_0 A v^2/c^2,\;0,\;0).\;}$$

Decompose into pure + heat-like parts

Using $u^\mu = (c, 0, 0, 0)$ in the cruiser frame:

• Heat-like: $f^\mu_\text{heat} = (f\!\cdot\!u/c^2)u^\mu = (\gamma^2\rho_0 A v/c, 0, 0, 0)$ — rate of mass-energy gain, parallel to $u$.
• Pure: $f^\mu_\text{pure} = f'^\mu - f^\mu_\text{heat} = (0, -\gamma^2\rho_0 A v^2/c^2, 0, 0)$ — relativistic ram-pressure drag, perpendicular to $u$.

Both non-zero: the force is neither pure nor heat-like. The dust deposits energy and exerts a real drag.

$\gamma^2$ enhancement

One $\gamma$ from length contraction: cruiser sees densified dust ahead by factor $\gamma$. Another $\gamma$ from boost of dust 4-momentum: each dust particle in the cruiser frame has $p = \gamma m v$ (Newtonian: $mv$). Multiplied: $\gamma^2$ ram-pressure scaling. For interstellar dust at $\rho_0 \sim 10^{-26}$ kg/m³, $v\sim 0.9c$ ($\gamma\sim 2.3$), $A\sim 100$ m²: drag $\sim 10^{-9}$ N — small but not negligible over interstellar distances; relevant for relativistic spacecraft design (Breakthrough Starshot, etc.).

Rest-frame stress-energy

$T^{\mu\nu} = \mathrm{diag}(\rho_0, -\sigma, 0, 0)$ in the string rest frame — energy density $\rho_0$, tension $-\sigma$ along $\hat x$ (the minus sign distinguishes tension from pressure). The other diagonal entries are zero (1D object in 4D).

Assume $\sigma < \rho_0$ (weak-energy condition will require this; see (b)).

Boost along $\hat x$ at velocity $v$

$T'^{\mu\nu} = \Lambda^\mu{}_\rho\Lambda^\nu{}_\sigma T^{\rho\sigma}$. Spatial-spatial component:

$$T'^{11} = \gamma^2(T^{11} + v^2 T^{00}/c^2 - 2v T^{01}/c) = \gamma^2(-\sigma + \beta^2\rho_0) = \gamma^2(\beta^2\rho_0 - \sigma).$$

(a) Frame with $T'^{11} = 0$?

Set the expression to zero:

$$\gamma^2(\beta^2\rho_0 - \sigma) = 0\Rightarrow\beta^2 = \sigma/\rho_0.$$

For this to be a physical boost ($\beta^2 < 1$), we need $\sigma < \rho_0$ — exactly the dominant-energy condition.

$$\boxed{\;\text{Yes — the frame exists, with }\beta = \sqrt{\sigma/\rho_0}.\;}$$

This boost cancels the tension by means of a relativistic momentum flow.

(b) Below-rest energy density possible?

Compute $T'^{00}$:

$$T'^{00} = \gamma^2(T^{00} + v^2 T^{11}/c^2 - 2v T^{01}/c) = \gamma^2(\rho_0 - \beta^2\sigma) = (\rho_0 - \beta^2\sigma)/(1-\beta^2).$$

Differentiate with respect to $\beta^2$ (treating $\beta^2$ as the variable):

$$\frac{dT'^{00}}{d(\beta^2)} = \frac{-\sigma(1-\beta^2) - (\rho_0 - \beta^2\sigma)(-1)}{(1-\beta^2)^2} = \frac{-\sigma + \sigma\beta^2 + \rho_0 - \beta^2\sigma}{(1-\beta^2)^2} = \frac{\rho_0 - \sigma}{(1-\beta^2)^2}.$$

For $\sigma < \rho_0$: this derivative is positive, so $T'^{00}$ is monotone-increasing in $\beta^2$. Minimum at $\beta = 0$ equals $\rho_0$.

$$\boxed{\;\text{No — the rest frame already has the smallest energy density.}\;}$$

Dominant-energy condition

This is the special-relativistic dominant-energy condition (DEC): all observers measure energy density $\ge$ pressure (in absolute value). Cosmological consequence: ordinary matter (DEC-satisfying) has its inertial response governed sensibly by the rest-frame energy density. Phantom matter ($w < -1$, Problem 1.149) violates DEC — allows for negative effective energy density in some frames, problematic for stability.

The DEC also features in Hawking-Penrose singularity theorems: stress-energy satisfying DEC + gravity produces inevitable geodesic incompleteness in cosmological and black-hole spacetimes.

CMB rest frame

In the cosmic microwave background's rest frame (defined by zero dipole anisotropy on the sky), the CMB photon gas is a perfect fluid with

$$p_0 = \rho_0/3\quad\text{(radiation EoS)},\quad T_\text{CMB} = 2.725\,\text{K}.$$

Stress-energy: $T_0^{\mu\nu} = \mathrm{diag}(\rho_0, p_0, p_0, p_0)$.

Boost to a frame moving at $\vec v$

Using the perfect-fluid form $T^{\mu\nu} = (\rho + p)u^\mu u^\nu/c^2 - p\eta^{\mu\nu}$ with the fluid's 4-velocity in the boosted frame $u^\mu = \gamma(c, -\vec v)$:

$$T'^{00} = (\rho_0 + p_0)\gamma^2 - p_0 = (4\rho_0/3)\gamma^2 - \rho_0/3 = \frac{\rho_0(3 + \beta^2)}{3(1-\beta^2)}\cdot 1.$$

Simplify using $\gamma^2 = 1/(1-\beta^2)$: $T'^{00} = \rho_0(3 + \beta^2)/[3(1 - \beta^2)] = \rho_0(3+\beta^2)\gamma^2/3$.

$$T'^{0i} = -(\rho_0 + p_0)\gamma^2 v^i/c = -(4\rho_0\gamma^2/3) v^i/c.$$

$$T'^{ij} = (\rho_0 + p_0)\gamma^2 v^i v^j/c^2 + p_0\delta^{ij} = (4\rho_0\gamma^2/3c^2)v^i v^j + (\rho_0/3)\delta^{ij}.$$

Anisotropy and "shear"

Off-diagonal spatial components $T'^{ij}$ for $i\neq j$:

$$T'^{ij} = (4\rho_0\gamma^2/3c^2)v^i v^j\quad\text{(for }i\neq j\text{)}.$$

This vanishes only when frame axes align with $\vec v$. In a general lab frame (axes not aligned), apparent off-diagonal "shear" components appear. But the underlying physics is uniaxial: anisotropy is along $\vec v$, not in two perpendicular directions.

Coordinate artefact vs. real anisotropy

Rotate the spatial frame so $\hat x \parallel\vec v$: off-diagonal components vanish, but the spatial stress is now $T'^{xx} = p_0 + (4\rho_0\gamma^2/3c^2)v^2$ — different from $T'^{yy} = T'^{zz} = p_0$. The trace-free symmetric part is non-zero whenever $\gamma > 1$, encoding a frame-independent uniaxial anisotropy along the flow.

CMB dipole

The solar system's velocity through the CMB rest frame is $v \approx 369$ km/s ($\beta\approx 1.2\times 10^{-3}$), giving rise to the well-measured CMB temperature dipole: $\Delta T/T \approx \beta\cos\theta \approx 3.4$ mK at peak. The full anisotropy tensor (including quadrupole from $\beta^2$) is far smaller and dominated by primordial fluctuations, but the dipole is a clean kinematic effect.

Boosted energy density

Perfect-fluid stress-energy boosted to a frame moving at $-\vec v$ relative to the fluid:

$$\rho \equiv T'^{00} = (\rho_0 + p)\gamma^2 - p.$$

(Same as Problem 1.144 with general $p$ instead of $\rho_0/3$.)

Divide by $\gamma^2$

$$\frac{\rho}{\gamma^2} = \rho_0 + p - p/\gamma^2 = \rho_0 + p(1 - 1/\gamma^2) = \rho_0 + p\beta^2.$$

Since $\beta^2 < 1$ for any physical (subluminal) boost:

$$\boxed{\;\frac{\rho}{\gamma^2} = \rho_0 + p\beta^2 < \rho_0 + p\quad\text{(positive pressure)}.\;}$$

Approach to the bound

As $\beta\to 1$ (ultra-relativistic flow), $\rho/\gamma^2 \to \rho_0 + p$ — the bound is approached but never reached.

Enthalpy as inertial density

The quantity $\rho_0 + p$ is the enthalpy density — the natural inertial density for relativistic fluid mechanics. It appears in:

• Relativistic Euler equation: $(\rho_0 + p)du^\mu/d\tau = -\nabla^\mu_\perp p$. Particle inertia is proportional to enthalpy, not bare energy density.
• Relativistic sound speed: $c_s^2 = \partial p/\partial\rho_0$ evaluated at constant entropy, with the inertial-mass denominator $\rho_0 + p$ entering the dispersion relation.
• Bondi accretion: terminal infall velocity onto a compact object depends on $(\rho_0 + p)/\rho_0$, the relativistic analogue of the polytropic factor.

For radiation ($p = \rho_0/3$): enthalpy $= 4\rho_0/3$, the famous $4/3$ factor that appeared in the photon-gas momentum density (Problem 1.144) and historically in the electron self-energy problem (resolved by QED).

WEC: positive energy for all observers

The weak energy condition demands that the energy density measured by any timelike observer with 4-velocity $V^\mu$ is non-negative:

$$\rho_\text{obs} = T_{\mu\nu}V^\mu V^\nu \ge 0\quad\text{for all timelike }V^\mu.$$

Apply to a perfect fluid

For a perfect fluid $T_{\mu\nu} = (\rho_0 + p)u_\mu u_\nu/c^2 - p\eta_{\mu\nu}$ with $u^\mu$ the fluid 4-velocity. The observer's $V^\mu$ has some relative velocity $v$ with respect to $\vec u$, giving $u\!\cdot\!V = \gamma c^2$ for relative Lorentz factor $\gamma$.

Evaluate:

$$\rho_\text{obs} = (\rho_0 + p)(u\!\cdot\!V)^2/c^2 - p(V\!\cdot\!V) = (\rho_0 + p)\gamma^2 c^2 - p c^2 = c^2[(\rho_0 + p)\gamma^2 - p].$$

With equation of state $p = w\rho_0$:

$$\rho_\text{obs} = c^2\rho_0[(1 + w)\gamma^2 - w].$$

WEC bound on $w$

For all $\gamma \ge 1$ ($\gamma = 1$ at rest with fluid, $\gamma\to\infty$ for ultra-relativistic observers):

• $\rho_0 \ge 0$: required for the $\gamma = 1$ case (observer at rest with fluid sees $\rho_0$).
• Monotonicity: $d\rho_\text{obs}/d(\gamma^2) = c^2\rho_0(1 + w)$. For $1 + w \ge 0$, monotone non-decreasing — smallest at $\gamma = 1$, equals $\rho_0 \ge 0$. ✓
• For $1 + w < 0$: monotone decreasing in $\gamma^2$, eventually negative — violates WEC for sufficiently boosted observers.

$$\boxed{\;\text{WEC for perfect fluid} \iff \rho_0 \ge 0\text{ and }w \ge -1.\;}$$

Phantom matter and cosmology

$w < -1$ is called phantom matter. Violates WEC; sustained phantom dark energy in cosmology drives a Big Rip — the scale factor diverges in finite time, tearing apart galaxies, stars, then atoms and nuclei in succession.

Observationally: current data (Planck + BAO + SNe Ia) bound $w \approx -1.03 \pm 0.03$ — consistent with cosmological constant ($w = -1$, the boundary of phantom regime). The 1-sigma band slightly straddles $w = -1$, but at present there's no compelling evidence for phantom dark energy. Definitive measurement awaits DESI, Euclid, and LSST.

(a) Free particle: momentum conservation

Lagrangian $\mathcal L = p^2/(2m) = \tfrac{1}{2}m\dot x_\mu\dot x^\mu/c^2$ (where $\dot x^\mu = u^\mu = dx^\mu/d\tau$ and $p^\mu = m u^\mu$).

Euler-Lagrange equations on the worldline:

$$\frac{d}{d\tau}\frac{\partial\mathcal L}{\partial\dot x^\mu} = \frac{\partial\mathcal L}{\partial x^\mu}.$$

Compute:

$$\frac{\partial\mathcal L}{\partial x^\mu} = 0\quad\text{(no explicit }x\text{-dependence)},\quad\frac{\partial\mathcal L}{\partial\dot x^\mu} = m\dot x_\mu = p_\mu.$$

Therefore

$$\frac{dp_\mu}{d\tau} = 0\quad\text{— 4-momentum is conserved.}$$

Free relativistic particle moves on a geodesic of Minkowski spacetime: straight-line worldline.

(b) Minimal coupling: $p\to p + qA$

Substitute $p_\mu \to p_\mu + qA_\mu/c$ in the Lagrangian (the minimal-coupling prescription). The new Lagrangian:

$$\mathcal L' = \frac{(p + qA/c)^2}{2m} = \frac{p^2}{2m} + \frac{q\,p\!\cdot\!A}{mc} + \frac{q^2 A^2}{2mc^2}.$$

Using $p^\mu = m\dot x^\mu$: $\mathcal L' = \tfrac{m}{2}\dot x^2 + (q/c)\dot x\!\cdot\!A + (q^2/2mc^2)A^2$.

Euler-Lagrange at $O(q)$

Vary the worldline. Position-derivative of $\mathcal L'$ at $O(q)$:

$$\frac{\partial\mathcal L'}{\partial x^\mu}\bigg|_{O(q)} = (q/c)\dot x^\nu\partial_\mu A_\nu.$$

Velocity-derivative at $O(q)$:

$$\frac{\partial\mathcal L'}{\partial\dot x^\mu}\bigg|_{O(q)} = (q/c) A_\mu.$$

Take $\tau$-derivative:

$$\frac{d}{d\tau}\bigl[(q/c) A_\mu(x(\tau))\bigr] = (q/c)\dot x^\nu\partial_\nu A_\mu.$$

EL equation at $O(q)$ (in addition to free piece):

$$\frac{dp_\mu}{d\tau} = (q/c)\dot x^\nu\partial_\mu A_\nu - (q/c)\dot x^\nu\partial_\nu A_\mu = (q/c)\dot x^\nu(\partial_\mu A_\nu - \partial_\nu A_\mu).$$

The combination $\partial_\mu A_\nu - \partial_\nu A_\mu = F_{\mu\nu}$ is the field tensor! With $u^\nu = \dot x^\nu$:

$$\boxed{\;\frac{dp_\mu}{d\tau} = \frac{q}{c} F_{\mu\nu} u^\nu.\;}$$

This is the covariant Lorentz force law. Spatial components reproduce $\vec F = q(\vec E + \vec v\times\vec B)$; time component gives $dE/dt = q\vec v\!\cdot\!\vec E$.

The $O(q^2)$ term: why dropped?

The $O(q^2)$ term in the EL equation is $\propto A^\nu\partial_\mu A_\nu$ — gauge-dependent, doesn't have the antisymmetric $F$ structure. Under $A_\mu\to A_\mu + \partial_\mu\varphi$, this term changes, conflicting with the gauge-invariance of physical predictions.

Resolution: "minimal coupling" keeps only the $q^1$ piece, which automatically produces gauge-invariant equations of motion via $F_{\mu\nu}$. This is the foundational principle for coupling charged matter to electromagnetism in quantum field theory: the Dirac Lagrangian $\bar\psi(i\gamma^\mu\partial_\mu - m)\psi$ becomes $\bar\psi(i\gamma^\mu D_\mu - m)\psi$ with $D_\mu = \partial_\mu - ieA_\mu/\hbar$, and gauge invariance is automatic.