Gravitational quadrupole moment?!?!?

So I was reading this paper in Nature on cosmic strings, and I understood very little of it, but mention was made of a gravitational quadrupole moment.

Could someone explain to me what that is?

(FWIW, the paper was about cosmic strings vibrating and ending up rocketing out of the centers of galaxies into deep space, probably with “Die Walkure” playing in the background. No, wait, there’s no sound in space- I probably misread the last part.)


I just finished reading a book about string theory, and I have no idea (of course I didn’t understand much of the book, but I FEEL smarter for having read it!:))…

Just a bump, 'cause I want to see an answer to this, too…

wait… this might help, if you can wade through it… over MY head…:wink:

Or do a google search on “Gravitational quadrupole moment”… you’ll get a bunch of hits which are over my head… but might lead you to an answer…

Alright, it’s been ages since I had any General Rel, so I’ll probably make a fool of myself. But here goes.

If you’re looking for Gravity Waves the lowest order component you can look for is the one associated with the Gravitational Quadruople Moment.

Gravity waves are analogous to electromagnetic waves (radio, light, you know – that stuff), but a lot weaker, and still not successfully detected, despite much ingenuity on the part of the searchers. One reason is that gravity is a much weaker force than electromagnetism. (Actually, I’ve always hated that statement – you’re comparing apples and oranges, but let it pass for now.) Another reason is that in electromagnetism you can detect the Dipole Moment radiation, which dies off less rapidly with distance.

You make gravity waves the same way you make EM waves – you wiggle something up and down, or otherwise accelerate it. You can break down the resulting radiation into components that die away at different rates with distance. If you jiggle an atom up and down the lowest order term would be the Monopole Moment term – but it’s zero, because the effect ofjiggling the positive nucleus exactly cancels the effect of jiggling the negative electrons. But the Dipole Moment term is nonzero, because the electrons and protons move in opposite directions and have opposite sign charges. The dipole radiation drops off with the square of the distance – the infamous “inverse square” law. There is a higher order term, the quadrupole moment term, that is also nonzero (so are all the higher-order terms), but this one drops off with the inverse fourth power of the distance, so it’s a lot smaller.

In the case of gravity waves, the dipole moment term cancels out – jiggle two masses against each other (or, more likely, have one rotate around the other, like the earth and the sun) and the two masses move in opposite directions, but they have the same “sign” – all mass is positive, and there ain’t no negative mass (I know all your objections and special cases – can 'em. They don’t count)-- so the dipole term is zero. That means that the lowest order contribution is the quadrupole term, which is not only intrisically smaller in magnitude than the dipole term, but dies off much more rapidly with distance.

People doing experiments to try to validate string theory would love to be able to detect quadrupiole radiation to see the gravity waves, but the detectors have to be built to see quadfrupole forms of radiation and have to be able to see a VERY weak signal. The biggest problem is noise and interfering effects due to non-gravity wave sources (like trucks rumbling by on a nearby highway).

In an otherwise great post:

You sure about that Cal? I’m likely equally as far from my physics classes as you are, but I recall that monopole radiation is inverse square, and dipole is inverse cubed. I know that is the case with static fields, a magnet (always a dipole at most) follows the inverse cube law. Am I wrong about radiation?

I’m perhaps a little less separated from my classes, but more importantly, I’m a lot less separated from my copy of Jackson. :slight_smile:

Radiation fields are complicated, so they get broken down into different zones. In the near zone (where the wavelength is long compared to the distance from the source), the fields look static, and a dipole field in the near zone is an inverse cube. In the radiation zone (at distances long compared to the wavelength), the fields go as 1/r. So at long distances, where we can actually see radiation, dipole fields and quadrupole fields both decay as 1/r.

The difference is in the dependence of the total power radiated on frequency; dipole radiation goes as frequency to the 4th, and quadrupole radiation goes as frequency to the 6th. This has been used as indirect evidence of gravitational radiation, as I recall (don’t quote me on this; I’ll see if I can dig up a reference). In rotating binary systems, you have an oscillating quadrupole. The effects of the power lost in gravitational radiation can be seen in a slowing down of the rotations and it agrees with the predictions made by theory. So even though we can’t see the radiation from the system (for precisely the reasons Cal mentioned), we CAN see the effects of the radiation.

You’re all making this too complicated. Basically, dipole, quadrapole, octapole and hexadecipole moments (and all the rest, but I’ve run out of Greek) just tell you how far from spherical something is. The monopole moment is just the total charge (in electromagnetism) or mass (in gravity). The other moments will depend on the origin of your coordinate system, but you usually use the “center” of your distribution as the origin. Since mass is never negative, it works out that the gravitational dipole moment is always zero, in that coordinate system. In general, then, any mass distribution (such as the Earth) that’s not spherically symmetric has a quadrapole moment. It’s actually possible to come up with a contrived special case that has no quadrapole moment, but an octapole or higher moment, but that’s hardly ever seen anywhere other than physics homework problems. In general, most real distributions have all of their moments nonzero, but if you get far enough away, it’s only the lowest nonzero one that matters. In electromagnetism, with positive and negative charges available, it’s easy to make something without a monopole moment, or without a dipole, etc., but mass always has a monopole moment, so the higher ones are only relevant if you’re close in.

Of course, if you’re talking about cosmic strings (which have nothing to do with string theory, by the way), then you’re going to have to work with General Relativity for the gravity, but you can talk about the quadrapole moment of a mass distribution even without any forces at all.

If you have a single stationary charge (monopole) it can’t radiate because of conservation of charge.

However a single oscillating charge will radiate but it won’t be monopole radiation it will be dipole radiation (plus higher orders).

By far the strongest radiation is produced by the dipole moment, however this requires the center of charge to accelerate with respect to its center of mass. For example this could happen when the charge on a spinning bar is positive at one end and negative at the other.

An oscillating mass will also have a time varying dipole moment but the only way it can oscillate is via the oscillation of another mass with equal but opposite momentum (conservation of momentum). The time varying dipole moment of this second mass will
exactly offset the dipole moment of the first mass and ergo no radiation.

In other words the “center of gravitational charge” is, obviously, identical to the center of mass, so by definition it cannot have a dipole moment. Quadrupole radiation is similar to having the same charge at both ends of the spinning bar. Much less radiation is emitted because the only thing that changes is the extent and the shape of the charge distribution in space.