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Is there any equation which shows how far a train of gravitational waves will travel from a source until it loses its ripples and amplitude?

The wave train will lose energy as it travels. It has a large strain in the beginning at its source and then when it reaches our Earth It has lost its ripple power. Then after passing through our Earth, it will continue losing energy until it has no energy left for rippling in space-time.

For example, for a strain of $h=10^{-20}$, how far away can these waves travel from their radiation source until they vanish in space-time?

Albert
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Hamidreza Abdollahi
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    Why do you suppose it would vanish? It never vanishes..... Same as light. Gravitational waves will continue to propagate indefinitely. – James K Jun 01 '23 at 21:44
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    i do not think so, it is loose its energy, as you can see in reference we have large strain in the beginning at source and then when it is reach to our earth loose its ripples power then after pass through our earth continue loosing untill no energy for rippling in space-time. – Hamidreza Abdollahi Jun 01 '23 at 22:17
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    but i am not sure, if you have a source for your comment , pop it up , thanks. you meant it is infinitly tarvel in space??? – Hamidreza Abdollahi Jun 01 '23 at 22:18
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    @HamidrezaAbdollahi it's a good question. As the strain wave expands, it drops as 1/r in strain the same way the electric field amplitude of an expanding light wave drops as 1/r. But 1/r never goes to zero. I think you are asking about additional losses associated with mass distributions in space. See my answer to Does a gravitational wave loses energy over distance? and answers to Transfer of energy from gravity back to other "more familiar" forms of energy? in Physics SE. – uhoh Jun 02 '23 at 04:05
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    Your question is not a duplicate of the first one because you're asking for "how far". It's a good question! Perhaps there is a theoretical estimate for the 1/e attenuation length for GW over very long distances! – uhoh Jun 02 '23 at 04:07
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    It may be better to ask this question as "until we are unable to detect them" The three processes: "spreading out", "transfer of energy to the medium" and "metric expansion of space" cause the strain to decrease, but only asymptotically: none of these can actually reduce the strain to zero. Moreover the latter two are very weak/slow - space is mostly empty and expansion is very slow. On the other hand the sensitivity of the current generation of detectors is known, and it would make a good question to ask how far it would travel until we can't detect it. – James K Jun 02 '23 at 08:15
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    I've found some resources about GW absorbsion https://arxiv.org/pdf/1906.04853.pdf which cites https://articles.adsabs.harvard.edu/pdf/1966ApJ...145..544H (By Stephen Hawking, 1966 - but only deals with the case when the frequency is very low, or the collision frequency is very high) or https://journals.aps.org/prd/abstract/10.1103/PhysRevD.97.123506 Which considers the effect on cold dark matter, but concludes "it's too small to detect". – James K Jun 02 '23 at 20:41
  • This question needs to be rephrased or closed. Even an attenuated wave continues on at a non-zero level. If we ask James' question of "until we can't detect them" then it is the trivial question of how far away can Ligo see, which anyone can look up. They have already traveled cosmologically significant distances and there is no sign of dissipation otherwise the Ligo results would be goofy. – eshaya Jun 06 '23 at 21:39
  • @eshaya please read this excellent answer and the two papers linked at the bottom. This demonstrates that the question is a good one and should not be closed. Remember the reason to close a question is to prevent answers, that would obviously be a bad thing in this case. Also, in Stack Exchange we do not rephrase questions significantly once others have taken time to write answers. *When a wave decays exponentially it is commonly said to "vanish" or become "vanishingly small" over a finite distance. "Zero" is never mentioned in the question!! – uhoh Jun 06 '23 at 22:30
  • @eshaya please first read this answer and the two papers linked at the bottom. This demonstrates that the question is a good one and should not be closed. Remember the reason to close a question is to prevent answers, that would obviously be a bad thing in this case. Also, in Stack Exchange we do not rephrase questions significantly once others have taken time to write answers. When a wave decays exponentially it is commonly said to "vanish" or become "vanishingly small" over a finite distance. "Zero" is never mentioned in the question! – uhoh Jun 06 '23 at 22:36
  • Vanishingly small has no meaning in the context of GR waves that displace objects across the planet by less than the size of an atom. With one exception; its dissipation as it travels through essentially empty space. That is vanishingly small. – eshaya Jun 07 '23 at 15:56
  • Both papers conclude dissipation is negligible for Ligo frequencies. The dust paper gives a distance of a million Mpc for 10% drop if the whole universe were at MW interstellar densities and all of that were in the form of dust and the dust responded in a maximal way; which it would not. – eshaya Jun 07 '23 at 16:31
  • @eshaya "vanishingly small" is a perfectly reasonable expression throughout physics. Waves are waves, each type has its peculiarities but they share commonalities; they propagate and decay, sometimes quickly, sometimes very slowly. The question is fine. That it is difficult to answer does not make it a bad question - on the contrary, that's what makes it such a good question! GW exist whether LIGO does or not - LIGO is not mentioned in the question; there's no restrictions on frequency at all. – uhoh Jun 07 '23 at 21:25

2 Answers2

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Till infinity

Your question is based on the false premise that gravitational waves require energy to propagate through space and after a certain distance, the waves lose energy and disappear, which is FALSE.

Gravitational waves, are waves in spacetime that are generated by the accelerated masses of an orbital binary system that propagate as waves outward from their source at the speed of light. As @uhoh pointed out in the comments, As the strain wave expands, it drops as $\dfrac{1}{r}$ in strain the same way the electric field amplitude of an expanding light wave drops as $\dfrac{1}{r}$. Here $r$ is the radius or distance from the centre of the propagation. Which means that no matter how far you are from the origin of the gravitational waves, they will still be non-zero. They might be extremely miniscule in the strain the produce (GW150914 produced a distortion that was about $\frac{1}{2000}$ the width of a proton) after a very long distance, but still, they will never be zero.

Alastor
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    While true in the idealized setting of vacuum GR, our universe is not vacuum. Every time the gravitational wave passes through a planet it will deposit a tiny amount of energy in seismic waves, similarly for stars. So technically we should expect a (very very tiny) exponential attenuation on top of the 1/r decay. (There is also the cosmological redshift). – TimRias Jun 02 '23 at 06:53
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    No. The question is not based on that false premise. You've mis-construed the OP's question in order to write a sensationally-introduced answer. You also missed my comment about a long-range attenuation due to Feynman's "beads on a string" mechanism. "they will never be zero" is so far scientifically unsupported. -1 as written, add some supporting sources or qualify your assertions and I can change it. – uhoh Jun 02 '23 at 07:05
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    Not sure about your interpretation uhoh. The question asks "how far can [they] travel in space" and not "how far can they travel in a medium. And "a vacuum" is a pretty good model for space! CMB photos have happily travelled for 13.7 billion years in space. and they are attenuated by the medium far more than gravitational waves. There might be an interesting physics question on the penetration of GW through a solid (I guess this would be negative exponential wrt intensity or strain^2) Though that again is only asymptotic to zero. So "Infinity" seems likey to be the correct answer. – James K Jun 02 '23 at 08:28
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    @JamesK Oh, my interpretation of space is that it is the "space" that science books and TV shows talk about, not empty space which doesn't exist. Space has baryonic matter (plasma, stars, dust) photons (CMB) dark matter, dark energy and even quantum fluctuations. If it said "space except for everything in it" that would be different. – uhoh Jun 02 '23 at 11:04
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    If you add dissipation, then the falloff is $1/r^{1+\epsilon}$ which also reaches zero only at infinity. – eshaya Jun 02 '23 at 18:35
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    @eshaya dissipation should yield something of the form $$\frac{\text{e}^{-r/a}}{r}.$$ – uhoh Jun 02 '23 at 21:37
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    That also is never zero. – eshaya Jun 03 '23 at 02:41
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    @eshaya it's close enough – uhoh Jun 03 '23 at 04:52
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    Perhaps quantum gravity will say that the strain becomes zero at some point in time. If there were enough Tiny quadrupole masses in space they could rotate and absorb the gravitational energy. –  Jun 03 '23 at 08:11
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    The accelerating expansion of the universe ensures that the small amount of dissipation will get smaller. The net result is that after traveling for a Hubble Time, the most important term in the time dependence of a gravity wave's intensity is the redshift term from the expansion. Of course, by then we will not be able to see the most distance gravity events anymore. – eshaya Jun 03 '23 at 16:43
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    Strain is dimensionless and can't be expressed as a fraction of a proton diameter. – ProfRob Jun 10 '23 at 19:34
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In addition to the decrease in the vacuum due exclusively to the distance from the source of the gravitational wave, the gravitational wave undergoes an additional damping when it passes through matter (stars, planets, dust, gas, etc)

According to the book Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity, Steven Weinberg, (Wiley, NY, 1972), chapter "10 General-relativistic theory of small fluctuations", page 585, equation 15.10.42, there it says:

...the energy density $\tau_g^{00}$ of these gravitational waves decreases as

$$\tau_g^{00} \sim R^{-4} \cdot \boldsymbol{e}^{\boldsymbol{-}\dfrac{16\pi G}{c^2} \displaystyle \int \eta \ dt}$$

(NOTE: I have added $c^2$ to the original formula in the book, which is expressed in units c=1)

...The factor $R^{-4}$ is just what we should expect for the free expansion of any wave representing a massless particle. [Compare Eq. (15.1.23)] The extra factor in (15.10.41) tells us that gravitational waves in a viscous medium are absorbed at a rate

$$\Gamma_g=16\pi \dfrac G{c^2} \eta$$

$\eta=$ coefficient of shear viscosity of the traversed matter

The $R^{-4}$ term is identical to the term for photons (massless particles) in an expanding universe. The interesting term is the one that expresses the damping when passing through matter $\Gamma_g$

I have performed a dimensional analysis:

$[G]=M^{-1}L^3 T^{-2}$

$[c]=L T^{-1}$

$[\eta]=M L^{-1} T^{-1}$

Therefore:

$[\Gamma_g]=T^{-1}$

As can be seen, everything is dimensionally coherent.

We see that the theoretical mathematical damping is a negative exponential and that therefore, theoretically, the energy density is never zero, although we know that physically we can consider that it has extinguished to zero when its value is below the limit in which the approximations of the model carried out to obtain the expressions, cease to be fulfilled.

Arxiv articles on this topic:

Damping of gravitational waves by matter

The damping of gravitational waves in dust

Best regards.

Albert
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