Suppose an infinitesimal continuous transformation has variations $\delta\psi,\delta\bar\psi$ and changes the Lagrangian density by $\delta\mathcal L=\partial_\mu K^\mu$. Expanding by the chain rule and integrating derivatives by parts gives
$$
\begin{aligned}
\delta\mathcal L
&=\left[\frac{\partial\mathcal L}{\partial\psi}
-\partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\right]\delta\psi
+\left[\frac{\partial\mathcal L}{\partial\bar\psi}
-\partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\bar\psi)}\right]\delta\bar\psi\\
&\quad+\partial_\mu\left[
\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\delta\psi
+\delta\bar\psi\frac{\partial\mathcal L}{\partial(\partial_\mu\bar\psi)}
\right].
\end{aligned}
$$
On solutions of the <Euler-Lagrange equations>, the first two brackets vanish. Hence <Noether's theorem> gives the conserved current
$$
j^\mu
=\frac{\partial\mathcal L}{\partial(\partial_\mu\psi)}\delta\psi
+\delta\bar\psi\frac{\partial\mathcal L}{\partial(\partial_\mu\bar\psi)}-K^\mu,
\qquad \partial_\mu j^\mu=0.
$$
For spacetime transformations, the coordinate variation supplies the corresponding energy-momentum term.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 301 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-301/2/solution