GR says that monopole gravitational radiation does not exist. I understand the reasons for this.

However there is this effect (which seems to me to have the hallmarks of a wave). Paper at arXiv:

Monopole gravitational waves from relativistic fireballs driving gamma-ray bursts (http://arxiv.org/abs/astro-ph/0309448)

I would like some feedback on that paper, my thoughts on it are below, which may or may not be relevant.

Basically what happens is that you have a mass that sheds energy. So if you are standing at a point outside the mass, the gravitational potential changes very quickly as the ejected mass passes by you. Since we detect the gravitational field as the slope of the potential, it seems there will be a boost applied to our motion as the potential changes.

This change in motion is why I think that the paper has the correct name - these are waves.

The author of the paper had no trouble calling this effect a monopole gravitational wave, but is it properly called a gravitational wave?

  • $\begingroup$ Your explanation sounds about right, that's what I am getting, too. There would be no problem with the usual notion of general relativity that free gravitational waves can only exist as quadrupoles or higher. As long as the wave is coupled to moving mass (or an electromagnetic field), the quadrupole restriction would not apply. I think this is better thought of as a kind of analog to plasmons, which are excitations of coupled systems. Since plasmons are usually characterized as quasi-particles, would "quasi-gravitational wave" make sense to distinguish it from the free-wave case? $\endgroup$
    – CuriousOne
    Dec 24, 2014 at 22:28
  • $\begingroup$ CuriousOne - like your point of associated mass or em wave, but since this is GR - where 'Spacetime tells matter how to move; matter tells spacetime how to curve.' (Wheeler) - that means that these monopole waves can also exist associated with GR quadropole emission, along with one would think these monopole waves themselves. $\endgroup$ Dec 24, 2014 at 23:43
  • $\begingroup$ I think the most interesting question is that of coupling. The monopole wave works because the propagation velocity of em waves and gravitational waves is the same, therefor they can couple tightly and move at the speed of light, just like the free wave would. The gravitational wave coupled to matter can only move at the speed of the matter, so there are two very distinct regimes of these waves. I am not sure I would call them physically equivalent or equivalent to a free gravitational wave, hence the quasi-wave suggestion. Thanks for brining it up, though, I learned something, today! $\endgroup$
    – CuriousOne
    Dec 24, 2014 at 23:50

2 Answers 2


These kind of effects apply to any kind of mass or energy. A massive particle moving at below the speed of light will carry a "subluminal gravitational wave". Dipolar radiation will carry a "dipolar gravitational wave" etc.

In principle, you can assign the label "gravitational wave" to any kind of gravitational field propagating along with some matter-energy object. But this is not usually done. A more purist viewpoint would be to call the free gravitational waves as the only gravitational waves. I.e., if it is a vacuum excitation of the space-time itself, it is a gravitational wave, if it is actually the gravity associated with some object, it is not. Such gravitational waves radiated from isolated sources can indeed be only quadrupole.

However, this line is harder to draw in the case of null (moving at speed of light) matter. In that case, you often do not know which part is the "free" wave and the one "carried along" because they would both move at the speed of light along the matter/radiation. And this is also a fundamental issue; there is no truly fundamental difference between the "space-time at rest" and the "vibrating space-time" through which the wave is passing. We are able to speak of the difference between a "wave" and a "non-wave" only thanks to a predefined meaning of a "still background". But when you are for example in the thin shell of gamma radiation travelling away from a source, the "correct background" is basically impossible to define.

Hence, people often talk about waves dragged with null matter simply as gravitational waves and do not make a distinction between the "free" and "dragged along" part (Griffiths & Podolský have a few chapters with examples). In the paper you link, the distinction could be made because one could show that there is no extra freedom in the polarization or strength of the wave. Simply put, the gravitational wave has no free properties and it is fully determined by the shell of gamma radiation (up to non-physical gauge transformations). So we could either call the mentioned excitation in the metric a "monopole wave" or a "dragged-along gravitational field", it is really just a question of a finer naming convention.

  • $\begingroup$ Thanks. So in the electromagnetic case, we have a similar 'dragged along' concept. In em waves a collapsing B field generates the E. This means that there are no dipole electric waves. We need the other half, B. Similarly in gravity, we can create waves with monopole and dipole configurations by having gravitational energy as the companion. $\endgroup$ Sep 29, 2016 at 13:11

I think it is important to emphasize that monopole gravitational radiation does not exist in a vacuum.

In the gamma ray fireball case, we experience something like a monopole gravitational wave because a chunk of the progenitor's mass energy is expelled in the form of (in the idealized case) a spherically expanding shell of radiation. While we are outside this shell of radiation, we see no change in the gravitational field, but as the shell of radiation passes by us, we see a corresponding reduction in the gravitational potential. Depending on how thin the shell is (how instantaneous the burst was) this reduction in potential can be accompanied by a sharp gradient.

Having said that, I would be reluctant to call it a gravitational wave. Unlike "true" gravitational waves, this "wave" does not exist on its own. Rather, it effectively "surfs" on the expanding shell of radiation. If somehow the radiation was stopped (e.g., if we surrounded the fireball by a concentric, spherically symmetric perfect mirror) this gravitational "wave" would be stopped, too. No real gravitational wave can be shielded this way.

I should also mention that this effect is by no means confined to relativistic fireballs. In principle, the same monopole "wave" can be experienced in any expanding or contracting spherically symmetric object, so long as the observer is inside the object.

So personally, I would refrain from calling it a gravitational wave. However, I can see why, from an observational perspective, it may be tempting to call it that nonetheless. As the radiation wavefront from a fireball passes us by, the monopole gravitational "wave" would be present, too, and may in principle be observed. However, given the typical distances to such fireballs, their unpredictability, and the timescales of the explosion, I suspect that the effect would be much harder to detect than quadrupole radiation from inspiraling binaries.

  • $\begingroup$ This is very confusing to me. As far as I know, GRBs are two relativistic jets pointing in opposite directions (cf. this Google image search), so what "idealized spherical shell" are you talking about here? $\endgroup$
    – Kyle Kanos
    Jul 18, 2015 at 14:48
  • $\begingroup$ @Kyle, I was trying to address a hypothetical idealized spherically symmetric case. If the fireball is anisotropic (and you are of course right, GRBs are supposedly rather anistropic on account of being relativistic jets) then this could be modeled as a spherically symmetric expansion with higher-order contributions on top, and my answer would still apply for the spherically symmetric component. The main point of my answer is that these "waves" are not free gravitational radiation in empty space but arise as a result of being inside the object's (radiation included) outermost radius. $\endgroup$ Jul 18, 2015 at 15:45
  • $\begingroup$ There's no if or supposedly here: GRB jets are well-established and not even close to spherically symmetric (except for maybe a few microseconds after collapse). It's probably not even zeroth-order correct to suggest it is spherical. $\endgroup$
    – Kyle Kanos
    Jul 18, 2015 at 16:06
  • $\begingroup$ I certainly didn't mean to suggest anything unconventional about GRBs. I was simply trying to provide an interpretation of the paper referenced by the OP. No matter how anisotropic the source is, to the extent that it produces monopole gravitational "waves", that would be due to the monopole component of a spherical harmonics expansion of the fireball. $\endgroup$ Jul 18, 2015 at 19:17
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    $\begingroup$ On the spherical shell thing: It really does not matter where or how non spherical the mass-energy ejection is, the monopole wave will be spherical, as it measures the mass left in the hole. Birkhoff's theorem and all that. $\endgroup$ Jul 30, 2015 at 22:38

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