Gravitational-Wave Strain and Power (watt/square metre)

Question

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 $h=10^{-100}$?

Answer 1

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)

Answer 2

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.