Classical Fields and Their Actions

An action principle produces two pieces at once: a differential equation in the interior and a term on the boundary. The interior equation is universal for a chosen Lagrangian density; the boundary term tells us which variations define the physical problem. An action therefore organizes the dynamics, but it does not silently choose initial values, wall data, or asymptotic behavior for us.

Required background. Special Relativity supplies Minkowski tensors and invariant contractions; Calculus and Linear Algebra supplies multivariable differentiation and integration by parts. Helpful background. Ordinary Differential Equations supplies initial- and boundary-value reasoning; Differential Forms and Integration supplies orientations and Stokes’s theorem.

From a local density to field equations

Let $\phi^a(x)$ denote one or more real fields on a fixed spacetime region $\Omega$. The label $a$ distinguishes field components; it is not a spacetime index. For a first-derivative local theory,

$$S[\phi] =\int_\Omega d^4x\, \mathcal L\!\left(\phi^a,\partial_\mu\phi^a,x\right).$$

An allowed variation is a one-parameter family $\phi^a_s=\phi^a+s\,\delta\phi^a$ that obeys the chosen boundary restrictions. Differentiating at $s=0$ gives

$$\delta S =\int_\Omega d^4x\left[ \frac{\partial\mathcal L}{\partial\phi^a}\,\delta\phi^a +\frac{\partial\mathcal L}{\partial(\partial_\mu\phi^a)} \partial_\mu\delta\phi^a \right].$$

Define the field momentum density

$$\pi_a{}^\mu \equiv \frac{\partial\mathcal L}{\partial(\partial_\mu\phi^a)}.$$

The product rule isolates the derivative of the variation:

$$\pi_a{}^\mu\partial_\mu\delta\phi^a =\partial_\mu\!\left(\pi_a{}^\mu\delta\phi^a\right) -\left(\partial_\mu\pi_a{}^\mu\right)\delta\phi^a.$$

Applying the divergence theorem only now—not before the surface term is visible—yields

$$\boxed{ \delta S =\int_\Omega d^4x\, \mathcal E_a\,\delta\phi^a +\int_{\partial\Omega}d\Sigma_\mu\, \pi_a{}^\mu\delta\phi^a,}$$

where

$$\mathcal E_a \equiv \frac{\partial\mathcal L}{\partial\phi^a} -\partial_\mu\pi_a{}^\mu.$$

Choose variations with compact support inside $\Omega$. They vanish near the boundary, so stationarity for every such variation requires

$$\boxed{ \partial_\mu\left[ \frac{\partial\mathcal L}{\partial(\partial_\mu\phi^a)} \right] -\frac{\partial\mathcal L}{\partial\phi^a}=0.}$$

These are the Euler–Lagrange field equations. The same integration-by-parts calculation, with boundary variations set to zero from the outset, appears in Tong 2006–2007, §1.1, printed p. 8, PDF. Retaining the surface term makes the boundary choice explicit.

What the boundary term asks us to choose

On a smooth non-null boundary, let $n^\mu$ be the outward-directed unit normal and define $\varepsilon=n^\mu n_\mu=\pm1$. With outward-normal-first boundary orientation,

$$d\Sigma_\mu=\varepsilon n_\mu\,d\Sigma.$$

The pointwise boundary contribution is therefore

$$\varepsilon n_\mu\pi_a{}^\mu\,\delta\phi^a.$$

Three common variational problems are distinct:

Allowed boundary variation	What makes the variation stationary?	Usual name
Fix $\phi^a$ on the boundary, so $\delta\phi^a=0$	No extra condition on $\varepsilon n_\mu\pi_a{}^\mu$ follows	Dirichlet
Allow $\delta\phi^a$ freely and add no boundary action	Require $\varepsilon n_\mu\pi_a{}^\mu=0$	Natural or homogeneous Neumann
Allow $\delta\phi^a$ and add a boundary functional	Combine its variation with $\varepsilon n_\mu\pi_a{}^\mu\delta\phi^a$	Mixed or Robin

For example, adding a quadratic boundary functional can turn a zero-normal-momentum condition into a linear relation between the field and its normal derivative. The relation is derived from the total action; it is not an extra equation appended afterward.

On a timelike wall $\varepsilon=+1$; on a spacelike cap $\varepsilon=-1$. For example, the future cap with future-directed unit normal has $d\Sigma_0=+d^3x$, as required by the divergence theorem. The table does not say that every geometric boundary should be treated alike. Initial and final spacelike slices encode a hyperbolic initial-value problem; a timelike wall controls what enters or leaves the region. A null boundary has a directed surface element but no unit normal. The causal character of each boundary component and the well-posedness of the differential equation must be checked separately.

On a timelike wall, or in an elliptic boundary-value problem, freely prescribing both $\phi$ and its normal derivative on the same boundary component generally overdetermines the solution. A spacelike Cauchy surface is different: for a hyperbolic equation, the field and its normal derivative form initial data, subject to any constraints.

Scalar-field sign and boundary check

For a real scalar field, take

$$\mathcal L =-\frac12\partial_\mu\phi\,\partial^\mu\phi -\frac12m^2\phi^2.$$

The inherited metric convention turns the kinetic term into

$$-\frac12\partial_\mu\phi\,\partial^\mu\phi =\frac12(\partial_t\phi)^2-\frac12\lvert\boldsymbol\nabla\phi\rvert^2,$$

so the time derivative has the sign needed for positive canonical energy. Here

$$\pi^\mu=-\partial^\mu\phi, \qquad \frac{\partial\mathcal L}{\partial\phi}=-m^2\phi.$$

Substitution into the general variation gives

$$\delta S =\int_\Omega d^4x\,(\Box\phi-m^2\phi)\delta\phi -\int_{\partial\Omega}d\Sigma_\mu\, \partial^\mu\phi\,\delta\phi,$$

with $\Box=\partial_\mu\partial^\mu=-\partial_t^2+\nabla^2$. Thus

$$(\Box-m^2)\phi=0$$

is equivalent to

$$\partial_t^2\phi-\nabla^2\phi+m^2\phi=0.$$

This component form is a useful sign check: a spatially uniform mode obeys $\ddot\phi+m^2\phi=0$, not an exponentially unstable equation. On a timelike spatial wall, free boundary variations and no boundary action give $n^\mu\partial_\mu\phi=0$; fixed boundary values instead give no such Neumann condition.

Electromagnetism follows the same variational logic, with an important complication: $A_0$ has no independent time-derivative kinetic term. Its equation is a constraint, and gauge-related potential variations do not label distinct physical configurations.

What an action does—and does not—determine

The action packages bulk dynamics, boundary response, symmetries, conserved currents, and metric response. It does not by itself select a unique solution. A solution also needs compatible initial, boundary, or asymptotic data.

Adding a total divergence leaves the interior Euler–Lagrange equation unchanged, but it can change the surface term and the canonical variables. “The Lagrangians differ by a total derivative” is therefore enough for the same local bulk equation, not automatically for the same boundary-value problem.

An equality obtained after imposing $\mathcal E_a=0$ is on shell. The first-variation identity above is off shell: it holds for arbitrary sufficiently regular field configurations and variations. Keeping that distinction prevents equations of motion from being used circularly inside their own derivation.

Common pitfalls

Dropping the boundary term too early. Integration by parts does not erase a term; it moves it to $\partial\Omega$. State which variations or boundary functional make it vanish.

Varying $\partial_\mu\phi$ independently of $\phi$. For an ordinary field variation, $\delta(\partial_\mu\phi)=\partial_\mu\delta\phi$. Treating the two as unrelated changes the problem.

Confusing a boundary with initial data. A timelike wall, an initial spacelike slice, and a null hypersurface have different causal roles. A compact table of “Dirichlet versus Neumann” does not replace a well-posedness analysis.

Exercises

A free endpoint

For a static scalar field on $0\leq x\leq L$, let

$$S[\phi]=\frac12\int_0^L(\phi')^2\,dx.$$

Fix $\phi(0)=0$ but leave $\phi(L)$ free. Find the stationary field and its natural boundary condition.

Solution

Variation gives

$$\delta S =-\int_0^L\phi''\delta\phi\,dx +\left[\phi'\delta\phi\right]_0^L.$$

Because $\delta\phi(0)=0$ and $\delta\phi(L)$ is arbitrary, stationarity requires

$$\phi''=0, \qquad \phi'(L)=0.$$

The bulk solution is $\phi=ax+b$. The fixed value gives $b=0$, and the natural condition gives $a=0$, so $\phi=0$. If $\phi(L)$ were fixed instead, $\delta\phi(L)$ would vanish and no Neumann condition would follow.

A boundary action produces Robin data

Keep $\phi(0)=0$ and let

$$S[\phi] =\frac12\int_0^L(\phi')^2\,dx +\frac{\kappa}{2}\phi(L)^2-h\phi(L),$$

with $\phi(L)$ free. Derive the boundary condition and solve the stationary problem when $1+\kappa L\ne0$.

Solution

The bulk variation is unchanged, while the endpoint coefficient becomes

$$\left[\phi'(L)+\kappa\phi(L)-h\right]\delta\phi(L).$$

Therefore

$$\phi''=0, \qquad \phi'(L)+\kappa\phi(L)=h.$$

Writing $\phi=ax$ and applying the Robin condition gives

$$a(1+\kappa L)=h, \qquad \boxed{\phi(x)=\frac{h}{1+\kappa L}\,x.}$$

The extra boundary functional changed the natural data without changing the bulk equation.

Next, Maxwell’s Equations in Covariant Form applies this structure to a gauge potential and separates evolution equations from constraints.

References

• David Tong, Quantum Field Theory (PDF), §1.1 “The Dynamics of Fields,” printed p. 8, University of Cambridge Part III Mathematical Tripos lecture notes, Michaelmas Terms 2006 and 2007.