Suppose we have a cylinder of mass $m$, radius $R$ and height $h$ in rotation with speed $\omega$ around its symmetry axis with no friction (ideal situation).

I'd expect this cylinder to emit gravitational waves, is this the case?

If the cylinder is emitting waves (thus energy) should it stop rotating after some time?

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    $\begingroup$ #GrishaKirilin has a great response, but to add a little qualitatively... Gravitational waves are produced when an observer can see (certain) changes in the grav field. This requires `quadrupole' acceleration---acceleration in the second order term. The monopole (zeroth order term) is the total mass, and must be conserved. The dipole (first-order) term is the center of mass and can't accelerate from cons. of momentum. Thus the quadrupole is the lowest order term which can generate grav waves. $\endgroup$ – DilithiumMatrix Jul 30 '12 at 19:44

There is no gravitational waves for a uniformly rotating axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fields, §88 The constant gravitational field:

However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the field of an axially symmetric body rotating uniformly about its axis still also be constant. However in this case the time directions are no longer equivalent by any means -- if the sign of time is changed, the sign of the angular velocity is changed. Therefore in such constant gravitational fields (we shell call them stationary fields) the components $g_{0i}$ of the metric tensor are in general different from zero.

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 $r'=r$, $z' = z$, $\phi' = \phi + \Omega t$ coordinate transformation ($r$, $\phi$, $z$ are cylindrical coordinates). Therefore all derivatives $\partial x_{\alpha}/\partial x_{\beta}$ do not depend on time. Hence the metric of a uniformly rotating axially symmetric body is time-independent.

For example, the component $g_{0i}$ outside of the slow rotating body ($M\ll c m r_{g}$) has the form:

$$ g_{0i}=-\frac{2G}{c^{3}}\,M_{ij} \frac{n_{j}}{r^{2}}, $$

where $M_{ij}$ is the total angular momentum antisymmetric tensor.

UPD. Concerning the answer of David Bar Moshe. The conclusions presented above are valid only for axial symmetric body. The total power radiated by an non axial symmetric body is proportional to the third power of difference of inertia tensor eigen values for axes transverse to the rotation axis, i.e., $(I_{1}-I_{2})^{3}$.

  • $\begingroup$ Ok, so bodies with constant mass density and rotating around one of their principal inertia axis do not emit gravitational waves, because of non-timedependent metric but when the rotation is around another axis they do, right? $\endgroup$ – linello Jul 30 '12 at 13:05
  • $\begingroup$ The question was about cylinder. Thus my answer was about an axial symmetric body such that mass density doesn't depend on $\phi$, i.e., $I_{1}=I_{2}$ (eigen values of inertia tensor). If $I_{1}\neq I_{2}$ then the rotating body emits gravitational waves according to the expression given in the review which David Bar Moshe mentioned in his answer. $\endgroup$ – Grisha Kirilin Jul 30 '12 at 13:19
  • $\begingroup$ Does a galaxy emit gravitational waves and how much ? And also when considering their discontinuous (stars) structure and dark matter. $\endgroup$ – gox Oct 22 '18 at 10:56
  • $\begingroup$ @GrishaKirilin here $\alpha$, $\beta $ from $\frac{\partial x_{\alpha}}{\partial x_{\beta}}$ are running over the spatial dimensions, right? I.e. the coordinate transformation to the non-rotating frame is time independent. Could you give an example with the Schwarzschild solution? Will this turn out the Kerr solution? $\endgroup$ – Alexander Cska Dec 21 '18 at 20:25

The first-order approximation to the radiation power from the quadrupole term is given by

$$ P = - \frac{128}{5 c^5} G M^2 R^4 \Omega^6 $$

where $\Omega$ is the angular speed and $M$ and $R$ are the mass and radial separation of a binary system of masses. See this summary for a detailed derivation

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    $\begingroup$ Since the question explicitly asks for the case of a uniform cylinder, I'm not sure that formula is very relevant. $\endgroup$ – genneth Jul 30 '12 at 18:30
  • $\begingroup$ @genneth: Physically this is the only relevant formula. There are no other systems that generate detectable gravitational waves. $\endgroup$ – CuriousOne Apr 20 '16 at 23:00

Rotating bodies emit gravitational waves. The emission is of quadrupole radiation. The radiation power (which is proportional to the fifth power of the angular speed) causes gradual reduction of the angular speed. Please see the following review by: Alessandra Buonanno equations 7.14, 7.17.

  • $\begingroup$ Extremely incomplete $\endgroup$ – DilithiumMatrix Jul 30 '12 at 19:40

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