# Does a pendulum necessarily emit gravitational waves?

A question about the behaviour of a pendulum in a frictionless vacuum recently made it back to the front page, and a few comments below John Rennie's excellent answer set me thinking about one specific aspect: dissipation in the form of emission of gravitational waves. It is well known that the lowest allowed multipolarity of gravitational radiation is quadrupolar, which means that to avoid dissipation via this method, the only oscillation would have to be on the dipole moment.

I can think of one way to achieve this. If you displace a monopolar distribution infinitesimally, you get a dipolar contribution. (Thus, for example, $e^{-r^2}$ is monopolar, and $-\frac{1}{2}\frac\partial{\partial z}e^{-r^2}=ze^{-r^2}$ is dipolar.) This means that if you have a spherical bob whose amplitude of oscillation is much smaller than its radius, any quadrupolar contribution is vanishingly small, and the bob will not radiate gravitationally.

However, I'm far from convinced that this is the only possible way. In particular, the quadrupolar polarizations of gravitational waves - $+$ and $\times$ - seem to me rather at odds with the geometry of an oscillating sphere. More specifically, if you put the $z$ axis horizontally, along the bob's motion, then the relevant quadrupole moment would have $m=0$, and it's not clear to me whether such a quadrupole would radiate.

With this in mind, then:

Consider a uniform sphere of radius $R$ and mass $m$ performing simple harmonic oscillations of amplitude $A$ along a horizontal $z$ axis, so the sphere's centre is at $z(t)=A\sin(\omega t)$.

1. What are the system's multipolar moments?

2. Does this system emit gravitational radiation?

• This article (arxiv.org/abs/gr-qc/9611051) shows how to construct non-spherically-symmetric extended bodies of uniform density behaving exactly as pointlike masses thus being effectively "perfect" gravitational monopoles! Such fascinating objects have been envisioned for high-precision Equivalence Principle experiments. It could be interesting to ask (to add to the question may be?) if one can optimize the building of a pendulum emitting as less gravitational waves as possible using a pair of such devices... Nov 18, 2013 at 15:35
• @laboussoleestmonpays: Even a point mass would emit gravitational waves if moving periodically with finite amplitude: there would be non-zero time derivatives of quadrupole moments. Nov 18, 2013 at 15:49
• @user23660 sure I agree and(but) I think your remark addresses the specific question of Emilio (more than my comment where I was trying awkwardly I confess to deal with a pair of devices to modelize the whole system : pendulum object + gravitationnal field source). So I guess the answer is a pendulum necessary emits gravitational waves... Isn't it ? Nov 18, 2013 at 16:48
• @laboussoleestmonpays: Yes, but I also think that your idea has merit: specifically designed system of oscillating masses could have suppressed quadrupole (and even higher -pole) moments. But this won't be equivalents of couple point masses. Nov 18, 2013 at 17:10
• Related, and inspired by @laboussoleestmonpays's link: How do aspherical gravitational monopoles look like? Nov 19, 2013 at 15:27

In the nonrelativistic limit the energy lost by the system due to gravitational radiation is defined by the third time derivative of quadrupole moment: $$- \frac{d E}{dt} = \frac{G}{45 c^5}\dddot{D}^2_{ij}.$$ Where indices $i$, $j$ correspond to (flat) 3D space, and dot denotes time derivative. This equation is taken from Landau & Lifshitz' 'Classical theory of fields', but the result goes back to Einstein (1918).

The quadrupole moment tensor is defined according to: $$D_{ij}= \int \mu(\mathbf{r})\cdot(3\,x_i x_j-r^2 \delta_{ij}) d^3\mathbf{r},$$ where $\mu$ is mass distribution.

For spherically symmetrical mass distribution multipole expansion with a center coinciding with the center of symmetry, produces only monopole term, while all higher multipole moments are zero. So all multipole terms come from the expansion of $\frac{1}{|\mathbf{r}-\mathbf{r}_m(t)|}$ with respect to powers of $\frac 1r$ and spherically symmetrical mass could be replaced by a point mass in calculations.

For a point mass moving along the $z$-axis the mass distribution is $\mu(\mathbf{r})=\delta(\mathbf{r}-z(t)\mathbf{e}_z)$. This gives us the quadrupole tensor of $$D_{ij} = m z^2(t)\,\mathrm{diag}(-1,-1,2).$$ Note, that this tensor oscillates around its average (nonzero) value at double the frequency of the pendulum.

After substituting the explicit dependence for $z(t)$ and averaging over time we get the following radiation power: $$- \left\langle \frac{d E}{dt} \right\rangle = \frac{192 G m^2 \omega^6 A^4}{45 c^5}.$$ For reasonable laboratory values of parameters the power lost for such gravitational radiation is extremely small. Taking values $m=1000\, \text{kg}$, $A = 10 \,\text{m}$ and $\omega = 1\, \text{s}^{-1}$ (the acceleration will have amplitude of ~ 1 g) radiated power will be $\sim 10^{-42}\, \text{W}$.

By using a system of several (at least three) masses we could suppress the quadrupole radiation, then the next term would require varying octupole moments and would contain even higher degree of $c$ in the denominator. To suppress both quadrupole and octupole radiation a system of at least 4 point masses could be used.

Polarization and angular distribution. Gravitational wave intensity of a polarization given by the polarization tensor $e_{ij}$ in the solid angle $d\Omega$ is: $$dI = \frac G {72 \pi c^5} (\dddot{D}_{ij}e_{ij})^2 d \Omega.$$ For example, the '$\times$' polarization in the direction $\mathbf{n}$ has tensor $e^\times_{ij}=\frac{1}{\sqrt 2}(e^\theta_i e^\phi_j+e^\theta_j e^\phi_i)$ ($\mathbf{e}^\theta$ and $\mathbf{e}^{\phi}$ are unit vectors on a sphere). It is easy to show, that for pendulum oscillating along the $z$-axis, $D_{ij} e^\times_{ij}=0$ for all directions at all times, so the gravitational radiation for this system has only the '$+$' polarization, and the intensity has angular distribution $\sim \sin^4 \theta$, so that the pendulum does not radiate along the $z$-direction.

• Could you comment on the polarization state of this radiation? Looking at the pendulum from the $+x$ axis, the $\times$ radiation on the $y,z$ plane is definitely off, which leaves the $+$ mode on that plane. However, that means that a half-period delay in time flips the motion of the source but maintains the same effect in the radiated field. Why is this not a contradiction? Nov 18, 2013 at 21:01
• @EmilioPisanty: Tensor of quadrupole moments oscillates at double frequency $2\omega$ around averaged value. So I see no contradiction. Will expand on polarization. Nov 19, 2013 at 4:22