Under an infinitesimal <diffeomorphism> generated by $\xi^a$,
$$
\delta g_{ab}=2\nabla_{(a}\xi_{b)},
\qquad
\delta\omega_a=\mathcal L_\xi\omega_a
=\xi^c\nabla_c\omega_a+\omega_c\nabla_a\xi^c.
$$
Insert these variations into
$$
\delta S_{\rm matter}=\int d^4x\sqrt{-g}
\left(-\frac12T^{ab}\delta g_{ab}+E^a\delta\omega_a\right)
$$
and integrate derivatives of $\xi^a$ by parts. Diffeomorphism invariance and arbitrariness of $\xi^b$ give the <Noether identity>
$$
\boxed{\nabla_aT^a{}_b=E^a\nabla_b\omega_a-\nabla_a(E^a\omega_b)}.
$$
On the covector equation of motion $E^a=0$, this reduces to <stress-energy conservation>, $\nabla_aT^a{}_b=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 2 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-309/2/a/ii/solution