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I recently got confused by an article that claimed that gravitational waves were interesting because that their amplitude of a gravitational wave falls as $1/r$ in the far field limit. Suggesting this was due to their quadrupole behaviour.

How do I determine how the amplitude of certain multipole expansion falls off in electromagnetism? I guess, I could verify the electric field directly. For example for a electric dipole antenna it goes like $$|E|\approx \frac{\mu_0 \ddot{p}}{4\pi}\frac{1}{r}$$ where $p$ is the dipole moment. So it is the case that the amplitude in general falls as $1/r$ (gravity or electromagnetically)?

If this is not what I should be looking for, and somehow the amplitude falls as $1/r^2$ for the dipole, and $1/r$ for the quadrupole, how do I obtain that? How would a sextupole fall off like?

Alternatively, if electromagnetic waves fall the same independent of the multipole configuration, but gravitational waves fall off differently to EM waves, where is the difference?

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The intensity of electromagnetic and gravitational waves falls off as r$^{-2}$. Since the intensity is the absolute square of the amplitude, the amplitude falls off as 1/r. There is no connection to the multipolar nature of the source, which however does determine the angular distribution of the radiated intensity.

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The answer by @my2cts is absolutely correct, but the article might also have been getting at an interesting point (although as stated it does seem wrong): while the amplitude and intensity of GW and EM radiation fall off in the same way, there is a crucial difference in how we can detect them on Earth.

The detection of photons is typically in the form of a certain area - a telescope dish, for example - configured to detect light that falls upon it. The relevant quantity to consider if we want to compute the "visibility" of an EM source, then, is the intensity of the light it emits multiplied by the area of our detector; this decays as $r^{-2}$ if we keep the area of the detector the same.

On the other hand, in the case of GW current strategies for detection are based (loosely speaking) on the measurement of the change in length of the arms of an interferometer, which scales like the amplitude of the GW. This means that, unlike the EM case, the "visibility" of a GW source scales with $r^{-1}$.

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  • $\begingroup$ Thanks that makes sense. But is there any physical constraint that makes it impossible to make an EM wave detector based on amplitude detection? $\endgroup$
    – Mauricio
    Commented Mar 27, 2021 at 8:38
  • $\begingroup$ That's an interesting question. I believe there is a good way to answer it, but I cannot seem to think of it right now. $\endgroup$ Commented Mar 28, 2021 at 9:50
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    $\begingroup$ I will make a new question then, thanks! $\endgroup$
    – Mauricio
    Commented Mar 28, 2021 at 11:12

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