<p>As explained in the thread <a href="https://astronomy.stackexchange.com/a/53471/31201"><strong>How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance &amp; what a short derivation looks like?</strong></a></p>
<p>If <span class="math-container">$m_1$</span> and <span class="math-container">$m_2$</span> are the respective masses of an orbitally bound binary body, and we define <span class="math-container">$M_c$</span> as the <em>chirp mass</em>:
<span class="math-container">$$M_c=\dfrac{(m_1 m_2)^{3/5}}{(m_1+m_2)^{1/5}}$$</span>
We call <span class="math-container">$f$</span> the frequency of the gravitational wave, which is twice the orbital frequency of the binary pair.
<span class="math-container">$$f=2 f_{orb}$$</span>
<span class="math-container">$c$</span> and <span class="math-container">$G$</span> are respectively the speed of light and the Universal Gravitational Constant and <span class="math-container">$D$</span> is the distance between the binary pair and us, then the strain <span class="math-container">$h$</span> of a gravitational wave is defined as the amount that distances are stretched or compressed <span class="math-container">$\Delta L$</span> by a passing gravitational wave, relative to the original length <span class="math-container">$L$</span>. It is, of course, a dimensionless number.
<span class="math-container">$$h=\dfrac{\Delta L}L$$</span>
Then, the expression for the strain <span class="math-container">$h$</span> is:
<span class="math-container">$$h=\dfrac{4GM_c}{c^2 \ D} \ \left ( \dfrac{\pi G}{c^3} \ f M_c \right )^{2/3}$$</span>
Operating:
<span class="math-container">$$h=\dfrac{4 \pi^{2/3} G^{5/3}}{c^4 D} \ \dfrac{m_1 m_2}{(m_1+m_2)^{1/3}} \ f^{2/3}\tag{1}$$</span>
The power emitted by the binary pair, assumed circular orbits of radius <span class="math-container">$R$</span>
<span class="math-container">$$P_e=\dfrac{32 G^4}{5 c^5} \ \dfrac{m_1^2 m_2^2(m_1+m_2)}{R^5}$$</span>
Kepler's third law:
<span class="math-container">$$R^3=\dfrac{G (m_1+m_2)}{(2 \pi f_{orb})^2}$$</span>
At a certain distance <span class="math-container">$D$</span> from the binary pair, the detected power flux (<span class="math-container">$W/m^2$</span>) will be:
<span class="math-container">$$p=\dfrac{P_e}{4\pi D^2}$$</span>
Combining the last 3 expressions:
<span class="math-container">$$p=\dfrac{8 \pi^{7/3} G^{7/3}}{5 c^5 D^2} \ \dfrac{(m_1 m_2)^2}{(m_1+m_2)^{2/3}} \ f^{10/3} \tag{2}$$</span>
Combining (1) and (2) we obtain, units <span class="math-container">$W/m^2$</span> :
<span class="math-container">$$\boxed{p=\dfrac{\pi c^3}{10 G} \ f^2 \ h^2}$$</span>
The power flux (<span class="math-container">$W/m_2$</span>) is proportional to the square of the <strong>strain</strong> and the square of the <strong>frequency</strong> measured at the gravitational wave detector.</p>
<p>As an example, for the first gravitational wave detected by LIGO (GW150914) the peak frequency was <span class="math-container">$f=250 Hz$</span> and the strain was <span class="math-container">$h=10^{-21}$</span></p>
<p><span class="math-container">$c=299792458 \ m/s$</span></p>
<p><span class="math-container">$G=6.6743\cdot 10^{-11}$</span> I.S. units</p>
<p>We obtain that the order of magnitude of the peak flux of power is:
<span class="math-container">$$p=0.008 \ \dfrac J{s\cdot m^2}=8 \ mW/m^2$$</span></p>
<p>The source I have used, are notes entitled &quot;<em>Relativistic Astrophysics - Lecture. Gravitational Waves</em>&quot; authored by J. Wheeler. I have the notes on paper; unfortunately, I have not found the PDF freely available on the Internet.</p>
<p>Best regards.</p>

CC BY-SA 4.0
2023-04-27T16:04:03.883Z
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.
ProfRob
CC BY-SA 4.0
2023-05-03T23:03:27.207Z
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?
Hamidreza Abdollahi
CC BY-SA 4.0
2023-05-04T06:16:58.020Z
To estimate it, use the same equation, where $h$ is the measured strain and $f$ is the measured frequency of the gravitational wave. $$p=\dfrac{\pi c^3}{10 G} \ f^2 \ h^2$$
Albert
CC BY-SA 4.0
2023-05-04T06:25:31.923Z
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?
Hamidreza Abdollahi
CC BY-SA 4.0
2023-04-27T11:14:16.057Z
2023-04-28T06:50:33.473Z
2023-04-28T06:50:33.473Z
Albert
<p>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 &quot;Energy, in Joules&quot; but instead to find the &quot;flux&quot; in Watts per square metre.</p>
<p>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.</p>
<p>So the relationship will be &quot;flux&quot; (in SI units of Joules/second/square-metre) is a function of <span class="math-container">$\dot h$</span>, the rate of change of strain wrt time. For convenience, it is normal to resolve <span class="math-container">$h$</span> and <span class="math-container">$\dot h$</span> into + and × components.</p>
<p><a href="http://www.tapir.caltech.edu/%7Eteviet/Waves/gwave_details.html" rel="nofollow noreferrer">http://www.tapir.caltech.edu/~teviet/Waves/gwave_details.html</a> gives a formula for the flux:</p>
<p><span class="math-container">$$\def\d{\mathrm{d}}\text{flux}=\frac{\d E}{\d A\, \d t}=\frac{c^3}{16\pi G} (\dot h_+^2 + \dot h_×^2)$$</span></p>
<p>For sinusoidal waves, <span class="math-container">$\dot h$</span> is proportional to <span class="math-container">$f h$</span> where f is the frequency of the waves, so the flux would be in proportion to <span class="math-container">$f^2h^2$</span>,  and so a wave with a strain of <span class="math-container">$h=10^{-50}$</span>, which is <span class="math-container">$10^{-30}$</span> times smaller would carry <span class="math-container">$10^{-60}$</span> times less energy (per second per square metre) than a wave with <span class="math-container">$h=10^{-20}$</span> 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)</p>

CC BY-SA 4.0
2023-04-27T21:51:00.360Z
Which is the same as Albert's result, to a factor 2.5, for sinusoidal waves, for the reason I commented.
ProfRob
CC BY-SA 4.0
2023-04-28T06:16:44.377Z
To be clear - I think this is accurate and Albert's answer is out by a factor 2.5.
ProfRob
CC BY-SA 4.0
2023-04-28T15:46:24.123Z
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.)
TimRias
CC BY-SA 4.0
2024-04-05T14:05:52.400Z
@ProfRob I know I'm late to this, but looking at the formula of James's answer and the formula in Albert's answer, the only difference I see can be canceled out to just 1/(16pi) = pi/10, which gives me a difference between the two 15.7. I was just wondering how you got a difference of 2.5.
Pyrania
CC BY-SA 4.0
2024-04-05T14:33:43.137Z
@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?
ProfRob
CC BY-SA 4.0
2024-04-05T17:40:19.263Z
@ProfRob Ah, okay. That's what happened. Since my knowledge of mathematics is limited to basic calculus, would you mind telling me if it's 2.5 times more or less than Albert's answer? Thank you.
Pyrania
CC BY-SA 4.0
2024-04-05T17:49:20.633Z
@Pyrania 2.5 times more.
ProfRob
CC BY-SA 4.0
2023-04-27T20:14:32.910Z
2023-04-28T06:29:14.753Z
2023-04-28T06:29:14.753Z
James K
<p>If we detect a gravitational wave with a strain of, for example, <span class="math-container">$h=10^{-20}$</span>, what is the flux of power carried by this wave, in SI units, <span class="math-container">$W/m^2$</span> ?</p>
<p>How can flux of power be calculated for a given strain, such as  <span class="math-container">$h=10^{-50}$</span> or <span class="math-container">$h=10^{-100}$</span>?</p>

CC BY-SA 4.0
2023-04-26T21:02:54.783Z
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.
Deko Revinio
CC BY-SA 4.0
2023-04-26T21:15:51.137Z
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.
James K
CC BY-SA 4.0
2023-04-26T22:22:13.847Z
@JamesK vote retracted.
Deko Revinio
CC BY-SA 4.0
2023-04-26T23:46:22.513Z
Great question! Slightly related in Physics SE: [Do gravitational waves impart linear momentum to objects? (e.g. Quasar 3C 186)](https://physics.stackexchange.com/q/321166/83380) and especially [Transfer of energy from gravity back to other "more familiar" forms of energy?](https://physics.stackexchange.com/q/262926/83380) The total energy that goes into making a gravitational wave event is expressed as $mc^2$ and it's the $m$ in units of solar masses that's usually quoted.
uhoh
CC BY-SA 4.0
2023-04-26T23:52:34.080Z
See for example [Fraction of initial mass lost (radiated) by neutron star mergers compared to black hole mergers?](https://astronomy.stackexchange.com/q/38733/7982) and [GW190521 black hole merger total mass calculation and missing mass, how does this happen?](https://astronomy.stackexchange.com/q/38726/7982) You may find these answers interesting: [How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance & what a short derivation looks like?](https://astronomy.stackexchange.com/q/53465/7982)
uhoh
CC BY-SA 4.0
2023-04-26T23:57:02.630Z
...but the total energy of the event still needs to be calculated. Noe that strain falls of as $1/r$ and not $1/r^2$
uhoh
CC BY-SA 4.0
2023-04-27T05:47:08.087Z
I have a source, but it gives power per unit area in terms of dh/dt  http://www.tapir.caltech.edu/~teviet/Waves/gwave_details.html  (towards the end)
James K
CC BY-SA 4.0
2023-04-27T15:41:00.070Z
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](https://physics.stackexchange.com/a/661064/180843).
Sten
<gravity><general-relativity><gravitational-waves><gravitation>
Gravitational-Wave Strain and Power (watt/square metre)
Hamidreza Abdollahi
# Gravitational-Wave Strain and Power (watt/square metre)

(Asked by: username_0 on 2023-04-26)

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 $h=10^{-100}$?

**Tags:** gravity, general-relativity, gravitational-waves, gravitation

> **Comments:**
> **username_1**: 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.  
> **username_2**: 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.  
> **username_1**: @JamesK vote retracted.  
> **username_3**: Great question! Slightly related in Physics SE: [Do gravitational waves impart linear momentum to objects? (e.g. Quasar 3C 186)](https://physics.stackexchange.com/q/321166/83380) and especially [Transfer of energy from gravity back to other "more familiar" forms of energy?](https://physics.stackexchange.com/q/262926/83380) The total energy that goes into making a gravitational wave event is expressed as $mc^2$ and it's the $m$ in units of solar masses that's usually quoted.  
> **username_3**: See for example [Fraction of initial mass lost (radiated) by neutron star mergers compared to black hole mergers?](https://astronomy.stackexchange.com/q/38733/7982) and [GW190521 black hole merger total mass calculation and missing mass, how does this happen?](https://astronomy.stackexchange.com/q/38726/7982) You may find these answers interesting: [How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance & what a short derivation looks like?](https://astronomy.stackexchange.com/q/53465/7982)  
> **username_3**: ...but the total energy of the event still needs to be calculated. Noe that strain falls of as $1/r$ and not $1/r^2$  
> **username_2**: I have a source, but it gives power per unit area in terms of dh/dt  http://www.tapir.caltech.edu/~teviet/Waves/gwave_details.html  (towards the end)  
> **username_4**: 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](https://physics.stackexchange.com/a/661064/180843).  

---

## 2 answers

### Answer by username_2 on 2023-04-27. Score: 4 | Accepted answer

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](http://www.tapir.caltech.edu/%7Eteviet/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)

> **Comments:**
> **username_6**: Which is the same as username_5's result, to a factor 2.5, for sinusoidal waves, for the reason I commented.  
> **username_6**: To be clear - I think this is accurate and username_5's answer is out by a factor 2.5.  
> **username_7**: 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.)  
> **username_8**: @username_6 I know I'm late to this, but looking at the formula of James's answer and the formula in username_5's answer, the only difference I see can be canceled out to just 1/(16pi) = pi/10, which gives me a difference between the two 15.7. I was just wondering how you got a difference of 2.5.  
> **username_6**: @username_8  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?  
> **username_8**: @username_6 Ah, okay. That's what happened. Since my knowledge of mathematics is limited to basic calculus, would you mind telling me if it's 2.5 times more or less than username_5's answer? Thank you.  
> **username_6**: @username_8 2.5 times more.  

---

### Answer by username_5 on 2023-04-27. Score: 3

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?**](https://astronomy.stackexchange.com/a/53471/31201)

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.

> **Comments:**
> **username_6**: 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.  
> **username_0**: 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?  
> **username_5**: To estimate it, use the same equation, where $h$ is the measured strain and $f$ is the measured frequency of the gravitational wave. $$p=\dfrac{\pi c^3}{10 G} \ f^2 \ h^2$$  
> **username_0**: 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?  

---

