The <Dirac action> is
$$
S_D=\int d^4x\,\overline\psi(i\gamma^\mu\partial_\mu-m)\psi.
$$
To first order,
$$
\Lambda^\mu{}_\nu
=\delta^\mu{}_\nu+\Omega^\mu{}_\nu,\qquad
S[\Lambda]=1+\frac12\Omega_{\rho\sigma}S^{\rho\sigma}.
$$
The commutator found in part i gives
$$
S[\Lambda]^{-1}\gamma^\mu S[\Lambda]
(\Lambda^{-1})^\nu{}_\mu=\gamma^\nu+O(\Omega^2).
$$
Together with $\det\Lambda=1$, the transformations in part ii therefore leave both the kinetic and mass terms invariant through first order. Thus $S_D'=S_D+O(\Omega^2)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 301 / 3 / iii / Solution

{c}

The free Dirac action is
$$
S_D=\int d^4x\,\bar\psi(i\gamma^\mu\partial_\mu-m)\psi.
$$
Its Euler-Lagrange equations are the Dirac equation and its adjoint.


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