Questions: Classical Field Theory and Lagrangian Density

Question 1

In particle mechanics, the Lagrangian depends on generalized coordinates q_i(t) and their time derivatives. When transitioning to field theory, a student writes L = L(phi, dphi/dt). What critical modification is missing?

• A. The Lagrangian must also depend on the spatial derivatives of the field, because a field's dynamics depend on its spatial variation
• B. The Lagrangian must be replaced by a Hamiltonian for fields
• C. The field phi must be complex-valued for the formalism to work
• D. The Lagrangian must include an explicit dependence on spacetime coordinates x and t

Answer: (A)

A field phi(x,t) varies in both time and space, so its dynamics depend on spatial gradients as well as time derivatives. The Lagrangian density L(phi, partial_mu phi) depends on the field and all its spacetime partial derivatives. In particle mechanics, only time derivatives appear because the generalized coordinates are functions of time alone. The spatial derivative dependence is what makes field equations partial differential equations rather than ordinary differential equations.

Question 2

The Euler-Lagrange equation for a field, partial_mu (partial L / partial (partial_mu phi)) - partial L / partial phi = 0, reduces to the Klein-Gordon equation when L = (1/2)(partial_mu phi)(partial^mu phi) - (1/2)m^2 phi^2.

• T. True
• F. False

Answer: True

Applying the Euler-Lagrange equation to this Lagrangian density: partial L / partial phi = -m^2 phi, and partial L / partial (partial_mu phi) = partial^mu phi. Taking partial_mu of the latter gives the d'Alembertian of phi. The resulting equation is (partial_mu partial^mu + m^2)phi = 0, which is exactly the Klein-Gordon equation for a free scalar field of mass m. This demonstrates how the Lagrangian density encodes the field equation.

Question 3

A Lagrangian density that depends explicitly on the spacetime coordinates (not just through the fields) still yields valid Euler-Lagrange equations, but it breaks Poincare invariance.

• T. True
• F. False

Answer: True

The Euler-Lagrange derivation works for any L, regardless of explicit coordinate dependence. However, explicit dependence on spacetime coordinates means the physics is different at different points in spacetime, which violates translational invariance (a component of Poincare invariance). Fundamental Lagrangian densities in particle physics do not have such explicit dependence, which is precisely what guarantees conservation of energy-momentum via Noether's theorem.

Question 4

Explain why the action S = integral L d^4x must be a Lorentz scalar, and what this requirement imposes on the Lagrangian density L.

Model answer: The action must be Lorentz-invariant because the equations of motion derived from it (via the principle of least action) must be the same in all inertial frames. Since d^4x transforms as a Lorentz scalar (the Jacobian of a Lorentz transformation is 1), L itself must also be a Lorentz scalar. This means L must be built from Lorentz-invariant combinations of the fields and their derivatives: scalar products of four-vectors, traces of tensor products, and contractions with the metric tensor. This constraint dramatically restricts the allowed terms in L.

This is why Lorentz invariance is such a powerful organizing principle. Rather than guessing field equations directly, you construct the most general Lorentz-scalar L from your fields and their derivatives (subject to additional constraints like gauge invariance and renormalizability), and the field equations follow automatically. The physical content is encoded in the symmetry requirements on L.