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A field configuration or identity is on shell when the <Euler-Lagrange field equations>[equations of motion] hold.

On shell

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

Expanding $F_{\mu\nu}$, using antisymmetry of $\epsilon^{\mu\nu\rho\sigma}$, and cyclically rearranging the <matrix trace> gives
$$
\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu})
=\partial_\mu K^\mu,
$$
where the <Chern-Simons current> is
$$
K^\mu=2\epsilon^{\mu\nu\rho\sigma}
\operatorname{Tr}\left(
A_\nu\partial_\rho A_\sigma-\frac{2ig}{3}A_\nu A_\rho A_\sigma
\right).
$$
The theta contribution to the action is therefore a boundary term. For variations that vanish at the spacetime boundary, its variation vanishes, so it contributes nothing to the classical <Euler-Lagrange field equations>. Its integral can still distinguish topologically nontrivial gauge configurations in the quantum theory.

Solved by gpt-5.6-sol high.

Paper 305 / 2 / iv / Solution

For a field $\Phi$,
$$
\frac{\partial\mathcal L}{\partial\Phi}
-\partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\Phi)}=0.
$$


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 Incoming links: Euler-Lagrange field equation