Action for particles and electromagnetic fields

Use signature $(+,-,-,-)$, $x^0=ct$, $A^\mu=(\phi/c,\mathbf A)$, and $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$, preserving $F^{0i}=-E_i/c$. A massive particle's free action is minus its rest energy times elapsed proper time. Writing it with an arbitrary monotonically increasing parameter lambda makes the path dependence explicit through the square root of the tangent norm. Fixed endpoints are assumed in the variation. Reparameterizing the same timelike path does not change the geometric integral.

Do not first set lambda equal to proper time and then regard the integrand as an immutable constant during arbitrary path variations. Each varied path has its own proper time. Vary the square-root action before making the proper-time choice, or use a constrained formulation that retains the normalization condition. The electromagnetic coupling is the line integral of the potential covector. In laboratory time its Lagrangian contribution is $q\mathbf A\cdot\mathbf v-q\phi$, so the familiar electric-potential sign and the covariant tensor-force sign agree.

The vacuum field action is a spacetime integral of $-F^{\mu\nu}F_{\mu\nu}/(4\mu_0c)$. The coordinate measure is $d^4x=c\,dt\,d^3x$, so the explicit inverse c is needed to obtain joule-seconds. Expanding the invariant gives the laboratory Lagrangian density $\epsilon_0E^2/2-B^2/(2\mu_0)$. This difference is not the positive sum defining energy density. A plane wave can have zero field Lagrangian density at an event while carrying nonzero energy, making action and energy particularly easy to distinguish.

For a prescribed smooth current, add $-c^{-1}\int J^\mu A_\mu d^4x$. Varying the potential, integrating by parts, and taking its boundary variation to vanish gives $\partial_\mu F^{\mu\nu}=\mu_0J^\nu$. The homogeneous Maxwell equations already follow from defining F in terms of A. For dynamical point charges, their worldline coupling supplies the source instead; do not add the same current coupling again. Singular point self-fields require extra care, so the simple prescribed-field particle examples continue to neglect radiation reaction.

With our signs choose $A^\mu\mapsto A^\mu+\partial^\mu\chi$, equivalently $\phi\mapsto\phi+\partial_t\chi$ and $\mathbf A\mapsto\mathbf A-\nabla\chi$. The particle action changes by $-q(\chi_f-\chi_i)$. For fixed endpoints this does not alter its stationary paths. The field tensor and its action stay unchanged. For a prescribed current, integration by parts relates the bulk gauge variation to $\partial_\mu J^\mu$; charge conservation removes that bulk term when the boundary contribution is controlled.

Canonical spatial momentum is $\mathbf P=\partial L/\partial\mathbf v=\gamma m\mathbf v+q\mathbf A$. It changes under a gauge transformation, whereas mechanical momentum $\mathbf p=\gamma m\mathbf v$ and the measured force do not. A uniform static vector potential in a simply connected region can have zero fields but nonzero canonical contribution. This is not a hidden classical acceleration. Translating a canonical conservation statement into a physical momentum claim requires the chosen gauge and the symmetry of the complete Lagrangian, including its possible boundary terms.