On Boundary Variations in Noether's Conservation Laws

1. Introduction

In classical mechanics and field theory, a continuous set of transformations is termed a symmetry if applying the transformation leaves the action invariant up to first order. Standard derivations of Noether's theorem often rely on fixed boundaries, which creates confusion when analyzing spacetime symmetries. This note clarifies the process: deriving the correct conserved currents requires strictly accounting for how transformations shift those boundaries.

We begin our analysis with point particles, first examining the case where the Lagrangian is strictly invariant, and then generalizing to scenarios where the Lagrangian differs by a total time derivative (which preserves the equations of motion). Later, we extend this exact sequence to classical fields.

For a field defined over a spacetime domain \(\Omega\), the case of a strictly invariant Lagrangian is expressed as:

Following this, the corresponding field-theoretic case where the Lagrangian differs by a four-divergence is also discussed.

2. Particle Dynamics and Fixed Boundaries

Consider an action in classical mechanics:

with fixed boundary conditions \(q^i(t_1) = q^i_1\) and \(q^i(t_2) = q^i_2\). We apply a smooth variation to the coordinates:

\[ q'^i = q^i + \epsilon \, \theta^i(q^i, t) \]

such that \(\delta S = 0\) up to first order in \(\epsilon\). Because standard textbooks assume the boundary conditions at \(t_1\) and \(t_2\) remain unaltered, the surface terms are typically discarded. This leads to:

Integrating by parts yields:

For a trajectory that satisfies the classical equations of motion, the integral vanishes, leaving:

However, assuming the boundary term remains the same is not valid in general, especially when the time coordinate itself is subjected to variation.

3. Time Translation and Surface Terms

Let us rigorously apply a time translation symmetry, \(t \to t + \alpha\). The boundary shifts, giving \(dt' = dt\) (i.e., the Jacobian \(|J| = 1\)), and the varied action becomes:

where \(q'(t') = q'(t+\alpha) = q(t)\). Under this transformation, the physical path is simply shifted in time, meaning the new coordinates evaluated at the new time are identical to the old coordinates at the old time. Therefore:

\[ q'(t) = q(t - \alpha) \approx q(t) - \dot{q}(t)\alpha \]

To find the total variation \(\delta S = S' - S\), we must shift the domain of integration back to \([t_1, t_2]\). For a generic function \(f(t)\), shifting the integration bounds by a small parameter \(\alpha\) generates boundary terms via Taylor expansion:

Applying this property to our integral, and keeping only terms up to first order in \(\alpha\) for the boundary evaluation:

\[ S' = \int_{t_1}^{t_2} \left[ L(q,\dot{q},t) - \alpha \frac{\partial L}{\partial q^i} \dot{q}^i - \alpha \frac{\partial L}{\partial \dot{q}^i} \ddot{q}^i + \alpha \frac{\partial L}{\partial t} \right] dt + \alpha \, L(q,\dot{q},t)\bigg|_{t_1}^{t_2} \]

Subtracting the original action \(S = \int_{t_1}^{t_2} L(q, \dot{q}, t) \, dt\):

By the product rule, the integrand collapses into a total time derivative:

\[ -\alpha \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}^i} \dot{q}^i \right) \]

Grouping the evaluated boundary terms together yields the final result:

For the case where the Lagrangian has no explicit time dependence (\(\frac{\partial L}{\partial t} = 0\)), the symmetry demands:

4. Classical Field Theory: Form vs. Total Variation

Transitioning to classical field theory, consider a Lagrangian density \(\mathcal{L}(\phi, \partial_\mu \phi)\) integrated over a spacetime domain \(\Omega\):

4.1 Case I: Fixed Integration Domain

When the coordinates \((x,t)\) remain unchanged and only the fields vary (\(\phi(x) \to \phi(x) + \epsilon\,\theta(x)\)), the domain \(\Omega\) is fixed:

Using the Euler–Lagrange equations, the second term vanishes. Using the generalized Stokes' theorem (the divergence theorem in four-dimensional spacetime), we can rewrite the integral over the 3-dimensional boundary hypersurface \(\partial\Omega\) as an integral over the 4-dimensional spacetime volume \(\Omega\):

\[ \int_{\partial\Omega} \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \epsilon\,\theta(x) \cdot \hat{n} \, dS = \int_{\Omega} \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \epsilon\,\theta(x) \right) d^4x \]

For the action to remain invariant (\(\delta S = 0\)) over an entirely arbitrary spacetime volume \(\Omega\), the integrand itself must vanish. Factoring out the constant arbitrary variation \(\epsilon\), we identify the conserved Noether current \(j^\mu\) as:

4.2 Case II: Deformation of the Integration Domain

We now address the general case where \((x,t)\) coordinates are varied, directly altering the domain of integration \(\Omega \to \Omega'\). Under a spacetime symmetry \(x'^\mu = x^\mu + \epsilon\,\theta^\mu\), the field transforms as:

\[ \phi'(x'^\mu) = \phi(x'^\mu - \epsilon\,\theta^\mu) \approx \phi(x'^\mu) - \epsilon \, \partial_\mu \phi \cdot \theta^\mu \]

The functional variation at a fixed coordinate:

\[ \delta\phi \equiv \phi'(x) - \phi(x) = -\epsilon^\nu \partial_\nu \phi \]

The total variation in the action is:

\[ \delta S = \int_{\Omega'} \mathcal{L}(\phi', \partial_\mu \phi') \, d^4x' - \int_{\Omega} \mathcal{L}(\phi, \partial_\mu \phi) \, d^4x \]

This arises from two geometric sources:

\[ \delta S = \delta S^{\text{boundary}} + \delta S^{\text{internal}} \]

1. Boundary Contribution (Domain Shift): The translation shifts the 4-volume \(\Omega\). The change in the action evaluating the original Lagrangian over this new boundary \(\partial\Omega\) is converted to a volume integral via Stokes' theorem:

2. Internal Contribution (Functional Variation): Inside the fixed volume \(\Omega\), the field varies by \(\delta\phi\). Using the Euler–Lagrange equations, the internal variation of the Lagrangian collapses into a total divergence:

3. The Conserved Tensor: Combining both contributions and substituting \(\delta\phi = -\epsilon^\nu \partial_\nu \phi\):

Since \(\epsilon^\nu\) is arbitrary and the volume \(\Omega\) is arbitrary, the term inside the divergence must be a locally conserved current. Factoring out \(-\epsilon_\nu\) yields the Canonical Energy-Momentum tensor:

5. Quasi-Symmetries and Gauge Invariance

An action may not be strictly invariant under a transformation; rather, the Lagrangian may vary by a total divergence term, which leaves the equations of motion unaffected.

For a classical field, if \(\delta\mathcal{L} = \alpha \, \partial_\mu K^\mu\), and the field variation is \(\delta\phi = \alpha \, \Delta\phi\), then tracking the variation yields:

Equating this with the divergence term gives the generalized conservation law:

6. Illustrative Examples

Example 1: Rotational Symmetry (No Total Derivative)

Consider the Lagrangian:

\[ L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - \frac{kq}{\sqrt{x^2 + y^2}} \]

Under an infinitesimal rotation \(x' = x + \epsilon y\) and \(y' = y - \epsilon x\), the Lagrangian transforms as \(L(x', y') = L(x, y) + \mathcal{O}(\epsilon^2)\). The conserved quantity is the angular momentum:

Example 2: Gauge Invariance (Internal Symmetry)

Consider the complex scalar field:

\[ \mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^* \phi \]

Under a global phase transformation \(\phi \to e^{i\alpha}\phi \approx \phi + i\alpha\phi\), the Lagrangian is strictly invariant (\(\delta\mathcal{L} = 0\)). The conserved Noether current is:

This leads to the conserved charge (assuming spatial boundaries vanish):

\[ Q \propto \int d^3x \left( \phi^* \frac{\partial \phi}{\partial t} - \phi \frac{\partial \phi^*}{\partial t} \right) = \text{constant} \]

Example 3: Galilean Boost (Point Particle Mechanics)

Consider the free particle Lagrangian:

\[ L = \frac{1}{2}m\dot{x}^2 \]

Under an infinitesimal Galilean transformation \(x \to x + \epsilon t\) (where the boost velocity \(v = \epsilon\)), the variation of the coordinates is \(\delta x = \epsilon t\) and \(\delta \dot{x} = \epsilon\). The Lagrangian is not strictly invariant, but changes by a total time derivative:

\[ \delta L = m\dot{x}\,\delta\dot{x} = m\dot{x}\,\epsilon = \frac{d}{dt}(mx\epsilon) \]

Here, the total derivative term is \(F = mx\epsilon\). The conserved Noether charge incorporates this term:

Factoring out the arbitrary parameter \(\epsilon\), we obtain the conserved quantity:

Example 4: Spacetime Translation (Spacetime Symmetry)

Consider a real scalar field:

\[ \mathcal{L} = \frac{1}{2}(\partial_\mu \phi)(\partial^\mu \phi) - V(\phi) \]

This is the case covered in Section 4.2, which directly leads to the canonical energy-momentum tensor \(T^\mu_{\ \nu}\).

7. Differential Geometric Derivation (V.I. Arnold)

Case I: Lagrangian of particle is invariant under spatial transformation

Let the configuration space of the system be a smooth manifold \(M\). We consider a one-parameter group of diffeomorphisms (transformations) on this manifold:

\[ h_s : M \to M \]

parameterized by \(s \in \mathbb{R}\). For any coordinate \(q\), the transformed coordinate is \(q_s = h_s(q)\). This continuous transformation defines a vector field \(v(q)\) on the manifold, which represents the "velocity" of the coordinate under the transformation parameter \(s\):

A Lagrangian \(L(q, \dot{q}, t)\) is strictly invariant under the transformation \(h_s\) if, for any trajectory \(q(t)\) and for all \(s\), the value of the Lagrangian remains unchanged:

\[ L\!\left(h_s(q), \frac{d}{dt}h_s(q), t\right) = L(q, \dot{q}, t) \]

We proceed by differentiating the invariance identity with respect to the parameter \(s\) and evaluating it at \(s = 0\). Since the right side is independent of \(s\), its derivative is zero:

\[ \left.\frac{d}{ds} L\!\left(h_s(q), \frac{d}{dt}h_s(q), t\right)\right|_{s=0} = 0 \]

Applying the multidimensional chain rule:

Because the partial derivatives with respect to \(s\) and \(t\) commute (the transformation does not depend on the dynamic time \(t\)), we can swap the order of differentiation in the second term:

\[ \left.\frac{\partial}{\partial s} \frac{d}{dt} h_s(q)\right|_{s=0} = \frac{d}{dt}\left.\frac{\partial}{\partial s} h_s(q)\right|_{s=0} = \frac{d}{dt} v = \dot{v} \]

Substituting \(v\) and \(\dot{v}\) into the chain rule expansion yields a kinematic identity (true for any path, not just classical ones):

\[ \frac{\partial L}{\partial q} \cdot v + \frac{\partial L}{\partial \dot{q}} \cdot \dot{v} = 0 \]

To find the conserved quantity, we restrict our attention to a classical trajectory that satisfies the Euler–Lagrange equations:

\[ \frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \]

Substituting this dynamic relation into the first term of our kinematic identity gives:

\[ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \cdot v + \frac{\partial L}{\partial \dot{q}} \cdot \dot{v} = 0 \]

By recognizing the reverse of the product rule, we can rewrite the equation as a total time derivative:

Since \(\frac{\partial L}{\partial \dot{q}}\) is the generalized momentum vector \(p\), we arrive at the conserved quantity (the Noether charge) \(I\):