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I have read that "thanks to conservation of momentum" there is no dipole gravitational radiation. I am confused about this, since I cannot see the difference with e.m. radiation. Is this due to the non-existence of positive and negative gravi-charges? Is this due to the non-linearity of Einstein equations? How the conservation of momentum enters here?

An example of my confusion below

Q: Why I cannot shake a single mass producing dipole gravi-radiation? A: You need another mass to shake it. Q: Isn't it the same with electromagnetism?

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The simple Newton-like explanation of dipole gravitational radiation unexistence is following.

The gravitational analog of electric dipole moment is $$ \mathbf d = \sum_{\text{particles}}m_{p}\mathbf r_{p} $$ The first time derivative $$ \dot{\mathbf d} =\sum_{\text{particles}}\mathbf p_{p}, $$ while the second one is $$ \ddot{\mathbf d} = \sum_{\text{particles}}\dot{\mathbf p}_{p} = 0, $$ indeed due to momentum conservation.

"Magnetic" dipole gravitational radiation is analogically impossible due to conservation law of angular momentum. Indeed, since by the definition it is the sum of cross products of position of point on corresponding current: $$ \mathbf{M} = \sum_{\text{particles}}\mathbf r_{p}\times m_{p}\mathbf{p}_{p} = \sum_{\text{particles}}\mathbf{J}_{p} \Rightarrow \dot{\mathbf M} = \ddot{\mathbf M} = 0 $$

What's about general relativity? As you know, the propagation of gravitational waves is described by linearized Einstein equations for perturbed metric $h_{\mu \nu}$, and in this limit they coincide with EOM for helicity 2 massless particles in the presence of stress-energy pseudotensor $\tau_{\mu \nu}$: $$ \square h_{\mu \nu} = -16 \pi \tau_{\mu \nu}, \quad \partial_{\mu}h^{\mu \nu} = 0, \quad \partial_{\mu}\tau^{\mu \nu} = 0 $$ Since $\tau^{\mu \nu}$ is conserved, this protects $h_{\mu \nu}$ from the contributions from monopole or dipole moments of sources as well as from additional helicities.

Formally the deep difference between gravitational and EM radiations is that we associate General relativity symmetry $g_{\mu \nu} \to g_{\mu \nu} + D_{(\mu}\epsilon_{\nu )}$ (it is infinitesimal version of $g_{\mu \nu}(x)$ transformation under $x \to x + \epsilon$ transformation) with covariant stress-energy tensor conservation (indeed, tensor current conservation, from which we can extract conservation of 4-momentum vector current), while EM gauge symmetry is associated with vector current conservation (from which we can extract the conservation of electrical charge scalar quantity). So that corresponding conservation laws affect on different quantities; the nature of radiation in EM and GR cases are different, and the first one rules primarily by Maxwell equations (and hence conservation of charge plays the huge role), while the second one rules by linearized Einstein equations (and hence the momentum conservation is genuine).

For example, heuristically speaking, due to conservation of EM charge EM monopole radiation is impossible (it is expressed therough the time derivative of charge), but nothing restricts dipole moment radiation. In GR due to conservation of momentum vector, which is related to metric (an so to gravitational waves, in the sense I've shown above), dipole moment radiation is impossible.

This, indeed as anna v said in comments, is connected with the fact that EM field represents helicity-1 particles, while linearized gravitational field coincides with the field which represents helicity-2 particles. As you see, such thinking doesn't require presence of plus minus masses.

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what about electromagnetism, what is the difference? – Arnaldo Maccarone Mar 1 at 17:18
Quantum mechanically it is the spin 2 of the graviton that makes the difference with the spin 1 of the photon. – anna v Mar 1 at 17:26
@ArnaldoMaccarone : the reason is that electrical dipoles radiation intensities are expressed through quantities which aren't conserved (but monopole radiation intensity, for example, is given through derivative of full electric charge, which is conserved in time). – Name YYY Mar 1 at 17:29
@ArnaldoMaccarone : also, look to the answer (I've added some part). – Name YYY Mar 1 at 17:41
You don't need to invoke negative charges to get dipole radiation in E&M. It is enough to use an electrically neutral body to absorb the reaction. – dmckee Mar 1 at 17:44

While there may be no dipole interaction in a vacuum, or monopole, for that matter, there is dipole and monopole gravitational radiation that is associated with underlying mass transfer. For example this paper on arXiv: "Monopole gravitational waves from relativistic fireballs driving gamma-ray bursts" (see link below)

Monopole Gravitational waves exist?

Basically these monopole waves don't exist if you use explanations like NameYYY's, but in the case cited in the paper, with a neutrino burst, the underlying driver for the monopole effect is virtually transparent - so to an observer without a neutrino detector these look like monopole gravitational waves.

Dipoles are simple to construct once you have a monopole wave source. One 'simply' shakes the monopole source up and down.

Thus the analogy with electromagnetism, where monopole waves are not allowed due to charge conservation is not the best way to think about your question. After all - monopole gravitational waves are created when energy in any form leaves a region, while electromagnetism only has one type of carrier which is a em charge. If the energy leaving (or arriving or both in a periodic fashion) is invisible to the thing/person detecting the gravitational effects is it a 'real' monopole or not?

Since there are many proofs about the non - existence of monopole and dipole gravitational radiation, the realization that experimentally there exists both can be surprising.

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