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Zero, if I understand correctly. This system has no quadropole moment, so does not emit gravitational waves. You would need to spin the system (like the orbiting black holes recently detected) in order to get any emission.

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Okay, let us change the problem to a spinning problem too.Zero, if I understand correctly. This system has no quadropole moment, so does not emit gravitational waves. You would need to spin the system (like the orbitting black holes recently detected) in order to get any emission.

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could you also prove why it's zero. Mathematically.Zero, if I understand correctly. This system has no quadropole moment, so does not emit gravitational waves. You would need to spin the system (like the orbitting black holes recently detected) in order to get any emission.

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For two orbiting bodies the answer is given here: https://en.wikipedia.org/wiki/Gravitational_wave#Power_radiated_by_orbiting_bodies

I don't know what the answer for orbiting bodies attached by a spring is. Horribly complicated, I guess, because the stress-energy of the spring is non-negligible and possibly rapidly varying. I'll leave it to more knowledgeable heads to confirm that, however.

As to why the answer is zero in the backwards and forwards case, an answer is that the symmetry of the system won't allow it. You can have a mass monopole, but it can't be made to fluctuate so you can't get monopole radiation. You can't have a mass dipole because you can't have a negative mass and you can't attach a mass to a zero-stress-energy object and wave it around, so you can't have dipole radiation. And you described a one dimensional system which can't have quadropole or higher moments.

I'd recommend Ben Crowell's book on General Relativity, available for free download from www.lightandmatter.com for a relatively gentle approach to this, or Sean Carroll's lecture notes on GR if you want another free but more mathematical presentation. The above is a very short version of my understanding of @bcrowell's book's approach.

*Edit: It's probably worth noting that I don't think you will find a mathematical proof of anything to do with gravitational waves at the "B" level. Did you follow Pervect's response to your last thread? If not, you will need to work on your maths if you want mathematical answers.*

I don't know what the answer for orbiting bodies attached by a spring is. Horribly complicated, I guess, because the stress-energy of the spring is non-negligible and possibly rapidly varying. I'll leave it to more knowledgeable heads to confirm that, however.

As to why the answer is zero in the backwards and forwards case, an answer is that the symmetry of the system won't allow it. You can have a mass monopole, but it can't be made to fluctuate so you can't get monopole radiation. You can't have a mass dipole because you can't have a negative mass and you can't attach a mass to a zero-stress-energy object and wave it around, so you can't have dipole radiation. And you described a one dimensional system which can't have quadropole or higher moments.

I'd recommend Ben Crowell's book on General Relativity, available for free download from www.lightandmatter.com for a relatively gentle approach to this, or Sean Carroll's lecture notes on GR if you want another free but more mathematical presentation. The above is a very short version of my understanding of @bcrowell's book's approach.

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I don't think you will find a mathematical proof of anything to do with gravitational waves at the "B" level.

This is correct; in fact gravitational waves in general are not really a "B" level topic except for very basic questions, which the OP question is not. I have changed the level of this thread to "I" accordingly.

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Zero, if I understand correctly. This system has no quadropole moment, so does not emit gravitational waves. You would need to spin the system (like the orbitting black holes recently detected) in order to get any emission.

Hmm. I thought it had to do with spherical symmetry. If a distribution of mass is vibrating in such a way that it maintains spherical symmetry, then it will not radiate (gravitationally), but if it departs from spherical symmetry, it will in general have a quadrupole moment.

The mathematical definition of the quadrupole moment for a system of two point-masses is:

[itex]Q_{ij} = m_1 (3 r_1^i r_1^j - \delta{ij} (r_1)^2) + m_2 (3 r_2^i r_2^j - \delta{ij} (r_2)^2)[/itex]

where [itex]\vec{r_1}[/itex] is the position of the first mass, and [itex]m_1[/itex] is its mass, and [itex]\vec{r_2}[/itex] is the position of the second mass, and [itex]m_2[/itex] is its mass.

I think [itex]Q_{ij}[/itex] be nonzero for a linear system of two masses

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However, note that you are describing the quadrupole moment for two masses, not two masses connected by a spring whose stress-energy is, I think, relevant as a source term for gravity. I'm not sure what the implications of that are.

OP: If Steven is correct, then your system does radiate. It appears I may be out of my depth here, which I didn't think I was. Apologies for any confusion.

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Hmm. I thought it had to do with spherical symmetry. If a distribution of mass is vibrating in such a way that it maintains spherical symmetry, then it will not radiate (gravitationally), but if it departs from spherical symmetry, it will in general have a quadrupole moment.

The mathematical definition of the quadrupole moment for a system of two point-masses is:

[itex]Q_{ij} = m_1 (3 r_1^i r_1^j - \delta{ij} (r_1)^2) + m_2 (3 r_2^i r_2^j - \delta{ij} (r_2)^2)[/itex]

where [itex]\vec{r_1}[/itex] is the position of the first mass, and [itex]m_1[/itex] is its mass, and [itex]\vec{r_2}[/itex] is the position of the second mass, and [itex]m_2[/itex] is its mass.

I think [itex]Q_{ij}[/itex] be nonzero for a linear system of two masses

[wrong: Since the existence of gravitational waves cannot be coordinate dependent, you have to ask if change of coordinates can remove the change of quadrupole moment. If motion is confined to one dimension, and you use the COM frame, the quadrupole moment is constant. Therefore no GW.]

[oops, no, I Stevendaryl is correct; you can still have changing quadrupole moment, and the OP configuration should radiate].

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I think ##Q_{ij}## be nonzero for a linear system of two masses

It is, but it's worth noting that the power radiated as gravitational waves does not depend on the quadrupole moment itself, it depends on time derivatives of the quadrupole moment.

I thought I'd seen @PeterDonis saying that you couldn't get gravitational waves from 1d oscillation

It's possible that I speculated along those lines without checking the math; intuitively a system of two point masses oscillating in a line seems like a time-varying dipole moment, not a time-varying quadrupole moment. But stevendaryl's math looks correct to me, and as far as I can tell, it gives nonzero time derivatives of the quadrupole moment.

note that you are describing the quadrupole moment for two masses, not two masses connected by a spring whose stress-energy is, I think, relevant as a source term for gravity.

It is, and the presence of the spring will make the quadrupole moment of the system more complicated, since the stress-energy of the spring will be changing and will contribute to it. I would expect that the system would still radiate GWs, but the detailed description of them would look more complicated.

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It's possible that I speculated along those lines without checking the math; intuitively a system of two point masses oscillating in a line seems like a time-varying dipole moment, not a time-varying quadrupole moment.

Yeah, but remember that an electric dipole consists of

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If the two charges are the same sign, then you don't get a dipole moment. (I've read that there is no gravitational dipole, and I think that's the reason--because there is no negative mass.)

Yes, agreed; I didn't mean to imply that my intuition was justified.

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I learned that the mass dipole amounts to the center of mass, and its derivative is total momentum, which is conserved, thus no radiation can be attributed to it (else conservation would be violated).Yeah, but remember that an electric dipole consists ofoppositecharges connected in a straight line. If the two charges are the same sign, then you don't get a dipole moment. (I've read that there is no gravitational dipole, and I think that's the reason--because there is no negative mass.)

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Agreed - the dipole moment is identically zero in the center of mass frame so cannot be responsible for radiation. But I think Steven is saying, and Peter is agreeing, that this system has a non-zero and non-constant quadrupole moment, and radiation is attributable to that. And my re-read of bcrowell's GR text certainly implies that he believes that 1d motion can generate gravitational waves. Which is all a bit surprising to me, but I can't see anything wrong with Steven's maths...I learned that the mass dipole amounts to the center of mass, and its derivative is total momentum, which is conserved, thus no radiation can be attributed to it (else conservation would be violated).

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I already agreed Stevendaryl on quadrupole moment and GW for a linear system. I was noting that dipole moment is not zero in any frame other than a COM frame. It is constant in general, but not zero. The general statement is the derivative of the dipole moment vanishes, not that the dipole moment vanishes.Agreed - the dipole moment is identically zero in the center of mass frame so cannot be responsible for radiation. But I think Steven is saying, and Peter is agreeing, that this system has a non-zero and non-constant quadrupole moment, and radiation is attributable to that. And my re-read of bcrowell's GR text certainly implies that he believes that 1d motion can generate gravitational waves. Which is all a bit surprising to me, but I can't see anything wrong with Steven's maths...

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I learned that the mass dipole amounts to the center of mass, and its derivative is total momentum, which is conserved, thus no radiation can be attributed to it (else conservation would be violated).

Yeah. That's a better way to put it. Whether the dipole vanishes or not depends on exactly how you define things, but it's time derivative is zero in any case.

A little heuristic derivation:

For simplicity, let's just deal with one spatial dimension. Let [itex]x_1[/itex] the location of a point mass [itex]M_1[/itex], and let [itex]x_2[/itex] be the location of a point mass [itex]M_2[/itex], and let [itex]X[/itex] be the location of the center of mass. Now, compute the Newtonian gravitational potential at a point [itex]x[/itex]. (Assume [itex]x_1 < x_2 \ll x[/itex]).

[itex]U(x) = -G (\frac{M_1}{x-x_1} + \frac{M_2}{x-x_2})[/itex]

In terms of the center of mass [itex]X[/itex]:

[itex]U(x) = -G (\frac{M_1}{(x-X) - (x_1-X)} + \frac{M_2}{(x-X) - (x_2 - X)})[/itex]

Now, expand it in a power series, assuming [itex]x \gg x_1[/itex] and [itex]x \gg x_2[/itex] and [itex]x \gg X[/itex]:

[itex]U(x) = -G(\frac{M_1}{x-X} (1 + \frac{x_1 - X}{x-X} + (\frac{x_1 - X}{x-X})^2 + ...) + \frac{M_2}{x-X} (1 + \frac{x_2 - X}{x-X} + (\frac{x_2 - X}{x-X})^2 + ...))[/itex]

[itex] = -G(\frac{M_1 + M_2}{x-X} + \frac{M_1(x_1 - X) + M_2(x_2 - X)}{(x-X)^2} + \frac{M_1 (x_1 - X)^2 + M_2 (x_2 - X)^2}{(x-X)^3} + ...)[/itex]

So, without getting into spherical harmonics (which is the proper way to do multipole moments), we can tentatively identify:

- potential due to monopole moment = [itex]-G \frac{M_1 + M_2}{x-X}[/itex]
- potential due to dipole moment = [itex]-G \frac{M_1 (x_1 - X)+ M_2(x_2 - X)}{(x-X)^2}[/itex]
- potential due to quadrupole moment = [itex]-G \frac{M_1 (x_1 - X)^2+ M_2(x_2 - X)^2}{(x-X)^3}[/itex]

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The dipole contribution to potential is one thing; the question is what to define as the dipole moment. In the same sense as the quadrupole moment tensor is defined for cartesian coordinates, the dipole moment (with mass for charge) is just the center of mass times the total mass (i.e. just sum of mass times position vector, over the particles). Clearly, this does not generally vanish. Further, in general, its derivative does not vanish. Its derivative is just the total momentum, which is constant, not vanishing. Its second derivative vanishes. Only in a COM frame, does the dipole moment vanish (per this definition).Yeah. That's a better way to put it. Whether the dipole vanishes or not depends on exactly how you define things, but it's time derivative is zero in any case.

...

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the dipole moment (with mass for charge) is just the center of mass times the total mass (i.e. just sum of mass times position vector, over the particles).

Okay, I thought that the dipole moment was computed from the

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For a system with total charge = 0, the monopole moment vanishes, and this makes the dipole moment coordinate independent - pick any cartesian coordinates and you get the same magnitude vector for the dipole moment. For gravity, since the monopole (total mass) never vanishes, all the higher moments are coordinate dependent. You could insist on using the COM frame which makes total momentum vanish and the dipole moment vanish.Okay, I thought that the dipole moment was computed from therelativeposition vectors (that is, relative to the center of mass). So for two point-masses of the same mass and opposite charge, then the magnitude of the electric dipole moment would be [itex]Q D[/itex], where [itex]Q[/itex] is one of the charges, and [itex]D[/itex] is the distance between them.

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