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In the first reply to this question it is written that:

There is no gravitational waves for a uniformly rotating axially symmetric body, because the metric doesn't depend on time.[..]

The reason is very simple. For an axially symmetric body, the distribution of mass in the lab frame coincides with that in the rotating system, thus the solution of Einstein equation can be found in the rotating system where the body and metric are static and then in the lab frame by means of r′=r, z′=z, ϕ′=ϕ+Ωt coordinate transformation (r, ϕ, z are cylindrical coordinates). Therefore all derivatives ∂xα/∂xβ do not depend on time. Hence the metric of a uniformly rotating axially symmetric body is time-independent.

So what about a non-uniformly rotating axially symmetric body?

Consider, for example, a disk rotating around its axis with a certain angular acceleration $\dot\Omega$, which could be caused by either a torque or a changing moment of inertia

In that case the metric would be time-dependent, right?

I've doubts about this because I've read about non-uniformly rotating McLaurin disks, and those seem to emit gravitational waves due to their periodic expansion/collapse cycles, not because of their non-uniform rotation.

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Povel
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The lowest multipole moment that needed to change to produce gravitational radiation is a quadrupole moment: $$ Q_{ij}=\int \,\rho ({\mathbf {r}})(3r_{i}r_{j}-\|{\vec {r}}\|^{2}\delta _{ij})\,d^{3}{\mathbf {r}}. $$ Energy carried away by gravitational radiation in this case is proportional to the square of third derivative of quadrupole moment: $$ -\frac{d\mathcal{E}}{dt}=\frac{G}{45 c^5} \left(\dddot{Q}_{ij}\right)^2. $$

So if the angular acceleration of a disk is caused by its changing moment of inertia tensor ($I_{ij}$) in such a way that quadrupole moment is also changing then yes, there would be gravitational radiation. Note, that not all deformations of a body would lead to a change in quadrupole moment. For example varying the radius of a massive sphere would change its moment of inertia (and if it was rotating it would change angular velocity), however quadrupole moment would remain zero and so there would be no gravitational radiation (in that order of approximation).

If however the quadrupole moment remains constant, gravitational radiation could still be generated by a changing higher multipole moment or by gravitomagnetic effects, that is changes in mass-currents while mass densities remain the same. For example if disk has inner and outer parts that could spin independently, periodically varying angular velocities of those two parts (while total angular momentum remains constant) would produce changing gravitomagnetic quadrupole moment. The power of radiation would then be proportional to the square of third derivative of gravitomagnetic quadrupole moment.

Note that in most situations contributions from gravitomagnetic multipole moments are very small.

A.V.S.
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  • 1/2 Thank you, now it's more clear. I've read a bit about current quadrupoles, and I found out that what is required specifically for radiation to be produced is a time-varying dipole moment of the angular momentum.

    I'd have two further questions:

    (1)Since mass momenta and their constancy are irrelevant in this case, would a system composed of two counter-rotating spheres, or a system composed of a rotating sphere inside a counter-rotating spherical shell, produce this type of radiation too?

     – Povel May 22 '18 at 12:20
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    (2) In the case of a disk with separate inner and outer part, the inner (or outer) part could vary its angular velocity by changing its radius, and so its moment of inertia. In this way the total angular momentum would be unchanged, and there would be emission due to both change in mass quadrupole and in mass currents, right?

     – Povel May 22 '18 at 12:21
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    @Povel: rotating sphere at a first glance would not have a gravitomagnetic quadrupole (not until it becomes deformed or relativistic effects become relevant). Two counterrotating spheres with zero total angular momentum would not even have a dipole gravitomagnetic field. So the radiation would be zero even if angular velocities are varying. – A.V.S. May 22 '18 at 16:55
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    @Povel (to 2nd comment), yes, but mass current (aka gravitomagnetic) term contributions would be much smaller (by about $v^2/c^2$), so in most cases if mass quadrupole derivatives are nonzero then this would be the leading term. – A.V.S. May 22 '18 at 17:26
  • Hmm.. what is exactly the difference between the case of two counter-rotating disks and the case of two counter-rotating spheres? To be clear,I'm thinking to a situation where both disks/spheres couples share a common rotational axis. Both have constant mass quadrupole (one is different and the other is equal to zero, sure), and each disk/sphere has non-vanishing angular momentum, even if each couple has zero total angular momentum. So why having two counter-rotating spheres doesn't result in a dipole moment, but two counter-rotating disks work? – Povel May 22 '18 at 18:24
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    @Povel: Disk has quadrupole moment, rotating disk has gravimagnetic quadrupole moment. Counterrotating disks of different sizes would have zero gravitomagnetic dipole moment (which is simply proportional to total angular momentum) but nonzero gravitomagnetic quadrupole moment. – A.V.S. May 22 '18 at 18:53
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    The reason for that is dipoles and quadrupoles scale differently with size. If we match geometrically similar objects of different sizes to have equal dipole moment (by appropriately changing angular velocity) there would be discrepancy in quadrupole moment. – A.V.S. May 22 '18 at 19:19
  • I see, you are referring to the "dipole moment" as the second term of the multipole expansion, associated with total angular momentum. The book I looked into after your first answer ("Gravitation: From the Hubble Length to the Planck Length" by Ciufolini) talks about the source of the current quadrupole moment as the "dipole (or first) moment of the angular momentum distribution", and make the example of two non uniformly counter-rotating rings that share the rotation axis, similar to your own example with disks. Thank you again for patience. – Povel May 23 '18 at 17:37