Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 / 1

The translation changes the Lagrangian density by $\delta {\mathcal{L}}_0={ϵ}^{\nu }{\partial }_{\nu }{\mathcal{L}}_0={\partial }_{\mu }({ϵ}^{\mu }{\mathcal{L}}_0)$. The Noether current is thereforeThis is the canonical stress-energy tensor. Directly,up to the analogous adjoint-field term, so the Euler-Lagrange field equations imply ${\partial }_{\mu }{T}^{\mu \nu }=0$ on shell. The Dirac equation also gives ${\mathcal{L}}_0=0$ on shell.

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After using the Clifford algebra, integrations by parts and the Dirac equation, every allowed Lorentz-covariant symmetric rank-two expression with one derivative reduces, modulo terms vanishing on shell and identically conserved improvements, to an overall multiple ofThus the general nontrivial tensor in the stated class isIt is manifestly symmetric. Differentiating it, commuting partial derivatives and usingmakes the terms cancel pairwise, proving ${\partial }_{\mu }{\hat{T}}^{\mu \nu }=0$ on shell. The normalization $C$ remains free at this stage.

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Choose $C=i/4$. Expandingwith the Clifford algebra, applying the product rule and then using the Dirac equation and its adjoint givesThis is the Belinfante-Rosenfeld stress-energy tensor written as an improvement of the canonical tensor.

The translation charges areFor spatial $\nu$, their difference is the integral of ${\partial }_i{S}^{i0\nu }-{\partial }^{\nu }(\bar{\psi }{\gamma }^0\psi )$. For $\nu =0$, use conservation of the Dirac current to replace ${\partial }^0(\bar{\psi }{\gamma }^0\psi )$ by a spatial divergence; also $S^{00\nu }=0$. Hence ${\hat{P}}^{\nu }-P^{\nu }$ is always a spatial boundary integral. Under the usual decay boundary condition it vanishes, so both currents generate the same four-momentum.

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