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We know that gravitational waves are emitted (at least in GR) when the system has a time-varying quadrupole (or higher) moment. My question is

Is it possible to easily tell (e.g. just by looking) if a system has such moments?

Is there some kind of symmetry to look for that lets you know that the system will emit such radiation?

user12345
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2 Answers2

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I don't think it possible to easily tell, but here is how you might approach this. The multipole moments $T_{\ell, m}(t)$ of the energy density $T^{00}(t, \mathbf x)$ of a source can be written on surfaces of constant time in terms of spherical harmonics as follows: $$ T_{\ell m}(t) = \int d^3x \, Y^*_{\ell, m}(\theta, \phi)r^\ell\,T^{00}(t, \mathbf x) $$ Notice that if we write the energy density in terms of spherical harmonics; $$ T^{00}(t,\mathbf x) = \sum_{\ell, m}c_{\ell, m}(t,r)Y_{\ell, m}(\theta, \phi) $$ then performing the integration in spherical coordinates and using orthogonality of spherical harmonics (I can give details if you want), we find $$ T_{\ell, m}(t) = \int dr\, r^{\ell+2} c_{\ell, m}(t,r) $$ Therefore, we see that if certain of the coefficients $c_{\ell, m}$ in the expansion of the density in terms of spherical harmonics vanish, then the corresponding multipole moment vanishes. The monopole moment corresponds to $\ell = 0$, the dipole moment to $\ell = 1$, and the quadrupole moment to $\ell = 2$.

So say, for example, that the energy density has spherical symmetry, namely $$ T^{00}(t,\mathbf x) = f(t,r), \qquad r = |\mathbf x| $$ Then its expansion contains no spherical harmonics with $\ell>0$, so all multipole moments other than the monopole moment vanish. In particular, there is no quadrupole.

So the best advice I can give is to look at the pictures of spherical harmonics, and attempt to get some intuition for when certain energy distributions will contain them in their expansions.

I hope that helps! Let me know of any mistakes and/or typos.

Cheers!

joshphysics
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  • I like the idea of using the pictures* of spherical harmonics but I'm a little confused on one point. Take a binary system for example, to me that looks like a dipole, am I thinking of this in the wrong way?

    *http://en.wikipedia.org/wiki/File:Harmoniki.png

     – user12345 Feb 07 '13 at 11:23
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    Yeah I don't have very good intuition for this myself; have you tried calculating the $c_{\ell,m}$ for an example binary system for some small values of $\ell,m$ to see what you get? That's what I would do (and I might if I have time). – joshphysics Feb 07 '13 at 18:33
  • I haven't calculated any spherical harmonics, no. Although I think I could, I don't see how it would help me with developing an intuition, if that's even possible. Thanks for interest and suggestions though! – user12345 Feb 07 '13 at 19:05
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    My pleasure. My thought process is that computing them for a few simple systems might begin to reveal qualitative features of distributions with each kind of moment. Then, one could make some sort of a hypothesis in general about what which multipole moments do and don't vanish and under which conditions this is the case. If I end up doing this; I'll let you know what I learn. – joshphysics Feb 07 '13 at 19:22
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The simplest way to construct a quadrupole is to have two identical dipole displaced by some distance opposed to each other, there is a quadrupole. In some sense, a dark mode dipole (with dipole canceling each other in far field, but not canceling in near field), gives you a quadrupole. An octupole is constructed by placing quadrupoles in a similar fashion.

Or actually, as long as the source is smeared in a finite volume and oscillating with time, you will have all multipole modes up to arbitrary order (just increasingly vanishing). This is the result of multipole expansion.

Bo Zeng
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  • Taking the case of a binary system of stars, it almost sounds like what you describe as "to have two identical dipole displaced by some distance opposed to each other, there is a quadrupole" except I can't see why a star would be a dipole, or am I thinking of this in the wrong way? – user12345 Feb 07 '13 at 19:08
  • That part I also don't know since gravity has only one kind of pole (not like charge or magnet). But is mass is not concentrated into a singular point, then Multipole Expansion is useful to get dipole, quadrupole etc. even though there is one type of "mass". The dipole component should look like $$ mr/R^2 $$ while quadrupole looks like $$ mr^2/R^3 $$ etc. – Bo Zeng Feb 07 '13 at 21:29