Past exam of the general relativity course of the University of Oxford / 2021 / 2 / a

No. A motionless isolated point mass produces a static gravitational field. All its mass multipole moments are time independent, so there is no outgoing gravitational radiation. In the quadrupole formula, $h_{ij}^{\mathrm{TT}}\propto d^2{Q}_{ij}^{\mathrm{TT}}/d{t}^2=0$.

No, for steady rotation of the uniform sphere. Its mass quadrupole moment is zero by spherical symmetry; its rotating mass current is stationary when the rotation axis and rate are fixed. A stationary source does not emit gravitational waves. Rotation by itself is insufficient: a time-dependent nonaxisymmetric distribution, or another changing radiative multipole moment, is required. The conclusion concerns a sphere with the stated constant density and steady motion, rather than a deformed or time-varying rotator.

No, for a uniform or otherwise axisymmetric disc in steady rotation. Although individual material elements accelerate, the mass density and mass current at every spatial point remain constant, and the mass quadrupole moment is time independent. There is no gravitational radiation. Axisymmetry of the material distribution is necessary for this conclusion: the source does not explicitly specify the disc's density. A counterexample to an unconditional answer is a disc with two equal dense spots at opposite radii; their rotating mass quadrupole moment produces gravitational waves at twice the rotation angular frequency.

Yes, for genuine precession of a nonspherical disc. Let its unit normal be $\mathbf{n}=(\sin \alpha \cos \omega t,\sin \alpha \sin \omega t,\cos \alpha )$. Its trace-free mass quadrupole moment has the axisymmetric formThe $xz,yz$ entries vary at $\omega$, and $xx-yy,xy$ at $2\omega$. Their nonzero second and third time derivatives give an oscillating far-field metric perturbation and positive average radiated power. For $\alpha =\pi /2$ the $\omega$ terms vanish but the $2\omega$ terms remain. The precise condition is $\omega \ne 0$ and $\sin \alpha \ne 0$: the literal inequality $\alpha \ne 0$ also allows $\alpha =\pi$, where the normal is fixed and no such gravitational radiation occurs. For a uniform thin disc of mass $M_d$ and radius $b$, $Q_0=-M_d{b}^2/4$, confirming that it is not zero.