{c}
{wiki=Belinfante–Rosenfeld_stress–energy_tensor}

The Belinfante-Rosenfeld construction adds the divergence of a spin-current superpotential to the <canonical stress-energy tensor>, producing a symmetric conserved tensor with the same integrated four-momentum under suitable boundary conditions.

Belinfante-Rosenfeld stress-energy tensor

The translation changes the <Lagrangian density> by $\delta\mathcal L_0=\epsilon^\nu\partial_\nu\mathcal L_0=\partial_\mu(\epsilon^\mu\mathcal L_0)$. The <Noether current> is therefore
$$
j^\mu=\frac{\partial\mathcal L_0}{\partial(\partial_\mu\psi)}\delta\psi-\epsilon^\mu\mathcal L_0
=\epsilon_\nu T^{\mu\nu},
\qquad
T^{\mu\nu}=i\bar\psi\gamma^\mu\partial^\nu\psi-\eta^{\mu\nu}\mathcal L_0.
$$
This is the <canonical stress-energy tensor>. Directly,
$$
\partial_\mu T^{\mu\nu}
=\left(\frac{\partial\mathcal L_0}{\partial\psi}
-\partial_\mu\frac{\partial\mathcal L_0}{\partial(\partial_\mu\psi)}\right)\partial^\nu\psi
$$
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.

Solved by gpt-5.6-sol high.

Solution

Paper 301 / 1 / a / Solution

{title2=$T^{\mu\nu}_{\mathrm{can}}$}
{wiki=Stress–energy_tensor\#Canonical_stress–energy_tensor}

Spacetime translations give the canonical stress-energy tensor
$$
T^\mu{}_{\nu}=\sum_a
\frac{\partial\mathcal L}{\partial(\partial_\mu\Phi_a)}
\partial_\nu\Phi_a-\delta^\mu_\nu\mathcal L.
$$


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