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If we detect a gravitational wave with a strain of, for example, $h=10^{-20}$, what is the flux of power carried by this wave, in SI units, $W/m^2$ ?

How can flux of power be calculated for a given strain, such as $h=10^{-50}$ or $h=10^{-100}$?

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  • $\begingroup$ voting to close because you are only allowed one question at a time. You could ask two separate questions. Also please look at out code of conduct and site tour. $\endgroup$ Apr 26, 2023 at 21:02
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    $\begingroup$ I think this is really one question. I've tried to edit it. The English was very bad but I think this is what the OP was asking. It doesn't really make sense. Perhaps if the OP asked for the power. After all a wave that lasts for one second will have less energy than a wave with the same strain lasting for 1000 years. Or perhaps it should be Watts/m^2 as what the OP really wants is a measure of "illuminance" for gravitational waves. $\endgroup$
    – James K
    Apr 26, 2023 at 21:15
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    $\begingroup$ @JamesK vote retracted. $\endgroup$ Apr 26, 2023 at 22:22
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    $\begingroup$ I have a source, but it gives power per unit area in terms of dh/dt tapir.caltech.edu/~teviet/Waves/gwave_details.html (towards the end) $\endgroup$
    – James K
    Apr 27, 2023 at 5:47
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    $\begingroup$ Not an expert on this, but if you are thinking about gravitational waves uniformly distributed in space (as opposed to signals originating from discrete events, as others are writing about), then you can directly associate an energy density. See this answer. $\endgroup$
    – Sten
    Apr 27, 2023 at 15:41

2 Answers 2

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Some inital considerations. The total energy would depend on the length of time that the wave was continuous for, and the total area through which the wave passes. That is there is no hope to find the "Energy, in Joules" but instead to find the "flux" in Watts per square metre.

Further thought: it is not the strain that will be proportional to the energy, but how fast the strain is changing. Consider two waves with the same strain, but one is at a much higher frequency. And recall Feynman's sticky bead thought experiment. The higher frequency wave would move the bead more quickly and so be of a higher power. At the opposite end of the spectrum, a wave with (nearly) infinite wavelength, would give a (nearly) constant strain, and so not transfer energy at all.

So the relationship will be "flux" (in SI units of Joules/second/square-metre) is a function of $\dot h$, the rate of change of strain wrt time. For convenience, it is normal to resolve $h$ and $\dot h$ into + and × components.

http://www.tapir.caltech.edu/~teviet/Waves/gwave_details.html gives a formula for the flux:

$$\def\d{\mathrm{d}}\text{flux}=\frac{\d E}{\d A\, \d t}=\frac{c^3}{16\pi G} (\dot h_+^2 + \dot h_×^2)$$

For sinusoidal waves, $\dot h$ is proportional to $f h$ where f is the frequency of the waves, so the flux would be in proportion to $f^2h^2$, and so a wave with a strain of $h=10^{-50}$, which is $10^{-30}$ times smaller would carry $10^{-60}$ times less energy (per second per square metre) than a wave with $h=10^{-20}$ if the frequencies were equal. (That's like the difference between a the sound energy released when a mote of dust lands and a supernova)

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    $\begingroup$ Which is the same as Albert's result, to a factor 2.5, for sinusoidal waves, for the reason I commented. $\endgroup$
    – ProfRob
    Apr 27, 2023 at 21:51
  • $\begingroup$ To be clear - I think this is accurate and Albert's answer is out by a factor 2.5. $\endgroup$
    – ProfRob
    Apr 28, 2023 at 6:16
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    $\begingroup$ Note that technically this result needs to be averaged over either a sufficiently long time scale or a sufficiently long length scale. (On shorter length/time scales the energy content of the gravitational wave is simply ill defined/coordinate dependent.) $\endgroup$
    – TimRias
    Apr 28, 2023 at 15:46
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    $\begingroup$ @Pyrania One formula is in terms of $\dot{h}^2$ and the other in terms of $h^2$. Did you differentiate a sine wave, square it, take its time-average and double it for the two polarisations? $\endgroup$
    – ProfRob
    Apr 5 at 14:33
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    $\begingroup$ @Pyrania 2.5 times more. $\endgroup$
    – ProfRob
    Apr 5 at 17:49
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As explained in the thread How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance & what a short derivation looks like?

If $m_1$ and $m_2$ are the respective masses of an orbitally bound binary body, and we define $M_c$ as the chirp mass: $$M_c=\dfrac{(m_1 m_2)^{3/5}}{(m_1+m_2)^{1/5}}$$ We call $f$ the frequency of the gravitational wave, which is twice the orbital frequency of the binary pair. $$f=2 f_{orb}$$ $c$ and $G$ are respectively the speed of light and the Universal Gravitational Constant and $D$ is the distance between the binary pair and us, then the strain $h$ of a gravitational wave is defined as the amount that distances are stretched or compressed $\Delta L$ by a passing gravitational wave, relative to the original length $L$. It is, of course, a dimensionless number. $$h=\dfrac{\Delta L}L$$ Then, the expression for the strain $h$ is: $$h=\dfrac{4GM_c}{c^2 \ D} \ \left ( \dfrac{\pi G}{c^3} \ f M_c \right )^{2/3}$$ Operating: $$h=\dfrac{4 \pi^{2/3} G^{5/3}}{c^4 D} \ \dfrac{m_1 m_2}{(m_1+m_2)^{1/3}} \ f^{2/3}\tag{1}$$ The power emitted by the binary pair, assumed circular orbits of radius $R$ $$P_e=\dfrac{32 G^4}{5 c^5} \ \dfrac{m_1^2 m_2^2(m_1+m_2)}{R^5}$$ Kepler's third law: $$R^3=\dfrac{G (m_1+m_2)}{(2 \pi f_{orb})^2}$$ At a certain distance $D$ from the binary pair, the detected power flux ($W/m^2$) will be: $$p=\dfrac{P_e}{4\pi D^2}$$ Combining the last 3 expressions: $$p=\dfrac{8 \pi^{7/3} G^{7/3}}{5 c^5 D^2} \ \dfrac{(m_1 m_2)^2}{(m_1+m_2)^{2/3}} \ f^{10/3} \tag{2}$$ Combining (1) and (2) we obtain, units $W/m^2$ : $$\boxed{p=\dfrac{\pi c^3}{10 G} \ f^2 \ h^2}$$ The power flux ($W/m_2$) is proportional to the square of the strain and the square of the frequency measured at the gravitational wave detector.

As an example, for the first gravitational wave detected by LIGO (GW150914) the peak frequency was $f=250 Hz$ and the strain was $h=10^{-21}$

$c=299792458 \ m/s$

$G=6.6743\cdot 10^{-11}$ I.S. units

We obtain that the order of magnitude of the peak flux of power is: $$p=0.008 \ \dfrac J{s\cdot m^2}=8 \ mW/m^2$$

The source I have used, are notes entitled "Relativistic Astrophysics - Lecture. Gravitational Waves" authored by J. Wheeler. I have the notes on paper; unfortunately, I have not found the PDF freely available on the Internet.

Best regards.

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    $\begingroup$ This is a reasonable dimensional analysis but the power emitted by a binary system isn't isotropic, with more power emitted in the direction of the orbital axis. In a similar vein, the strain amplitude depends on which direction you observe the binary from and is a mixture of two orthogonal polarisations. $\endgroup$
    – ProfRob
    Apr 27, 2023 at 16:04
  • $\begingroup$ thanks,but if we do not have a rotating binary source, and one Mass oscillating statice at its own place , how strain Energy of flux formula (P) modified with 1 mass and its frequency? $\endgroup$ May 3, 2023 at 23:03
  • $\begingroup$ To estimate it, use the same equation, where $h$ is the measured strain and $f$ is the measured frequency of the gravitational wave. $$p=\dfrac{\pi c^3}{10 G} \ f^2 \ h^2$$ $\endgroup$
    – Albert
    May 4, 2023 at 6:16
  • $\begingroup$ but in strain formula; h, we m1 & m2 , but in this example we have only one mass , we should ignore one of the mass in strain formula for Mc? $\endgroup$ May 4, 2023 at 6:25

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