<p>I don't think there is any short derivation and dimensional analysis is underdetermined. You can start from the basic equation for gravitational wave strain
<span class="math-container">$$ h = \frac{2G}{c^4r} \ddot{Q}\,$$</span>
where <span class="math-container">$h$</span> is the strain and <span class="math-container">$\ddot{Q}$</span> is the second time derivative of a mass quadrupole moment. (Note that these are <span class="math-container">$3\times 3$</span> matrices, but we can ignore that for the sake of proportionalities). The reciprocal <span class="math-container">$r$</span> dependence, where <span class="math-container">$r$</span> is the distance to the source, is easily understood because <span class="math-container">$h$</span> is a wave amplitude and so the square of <span class="math-container">$h$</span> integrated over a sphere is constant (i.e. <span class="math-container">$h^2$</span> obeys the inverse square law).</p>
<p>The derivation of the &quot;quadrupole formula&quot; would no doubt feature in text books on gravitational waves. It isn't a short derivation. Its general form can be argued on physical grounds and dimensional analysis, but that isn't short either.</p>
<p>From there we can do a rough analysis by writing the quadrupole mass moments as some moment of inertia, which will be some numerical fraction multiplied by <span class="math-container">$M_c a^2$</span>, where <span class="math-container">$M_c$</span> is the chirp mass and <span class="math-container">$a$</span> is the binary separation. However we are interested in the time-dependent component of the traceless quadrupole moment, which as seen from Earth varies as <span class="math-container">$\sin 2\omega t$</span>, where <span class="math-container">$\omega$</span> is the angular velocity of the binary. i.e. <span class="math-container">$Q \sim M_c a^2 \sin 2\omega t$</span>. (See footnote).</p>
<p>To use the first equation, we differentiate twice to get
<span class="math-container">$$\ddot{Q} \sim -4M_c a^2 \omega^2 \sin 2\omega t\ .$$</span>
Then use Kepler's third law, such that
<span class="math-container">$$GM_c \sim \omega^2 a^3\ ,$$</span>
to some small numerical factor. Thus writing <span class="math-container">$a\sim (GM_c/\omega^2)^{1/3}$</span>,
<span class="math-container">$$\ddot{Q}\sim -4M_c^{5/3}\omega^{2/3}\sin2\omega t$$</span>
and since <span class="math-container">$P = 2\pi/\omega$</span>, this is where the proportionality for the strain amplitude
<span class="math-container">$$ h \propto M_c^{5/3} P^{-2/3} r^{-1}$$</span>
comes from.</p>
<p><em>Footnote: The traceless quadrupole moment actually changes TWICE per orbit, so in fact the time dependence is <span class="math-container">$\sin 2\omega t$</span> and gravitational wave frequencies are TWICE the orbital frequency. This doesn't affect the proportionality arguments for the strain amplitude.</em></p>

CC BY-SA 4.0
2023-04-20T06:19:25.860Z
2023-04-20T07:18:34.773Z
2023-04-20T07:18:34.773Z
ProfRob
<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>
Note that <span class="math-container">$M_c$</span> has dimensions of mass. 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 luminosity distance between the binary pair and us.</p>
<p>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">$$\boxed{h=\dfrac{4GM_c}{c^2 \ D} \ \left ( \dfrac{\pi G}{c^3} \ f M_c \right )^{2/3}}$$</span>
From this:
<span class="math-container">$$h \propto \dfrac{M_c^{5/3} \ f^{2/3}}{D}$$</span>
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
2023-04-20T18:05:28.573Z
It would be good to mention that this is a weak field approximation.
TimRias
CC BY-SA 4.0
2023-11-15T16:34:21.703Z
Do you also know a solution for strong fields in the vicinity of a (double) star? Where is this described?
9herbert9
CC BY-SA 4.0
2023-04-20T07:10:24.747Z
2023-04-20T14:27:31.593Z
2023-04-20T14:27:31.593Z
Albert
<p>As a side note, the scaling of the strain can also be written as:</p>
<p><span class="math-container">$$ h \propto M \nu (Mf)^{2/3}/D, $$</span></p>
<p>where <span class="math-container">$\nu=\frac{m1m2}{(m1+m2)^2}$</span> is the symmetric mass ratio. This form is useful because <span class="math-container">$GM f/c^3$</span> is dimensionless. Note that this expression is only valid as a weak field/slow-motion approximation. However the general expression still takes the form (for non-spinning black hole binaries)</p>
<p><span class="math-container">$$ h = \frac{\nu M}{D} F(\nu, GM f/c^3), $$</span> where <span class="math-container">$F$</span> is a dimensionless analytic function of its two dimensionless arguments. In particular the peak strain in a BHB merger is given by</p>
<p><span class="math-container">$$h_{\textrm{peak}} = \frac{\nu M}{D} F_{\textrm{peak}}(\nu).$$</span></p>
<p>The peak strain of merger waveform is therefore simply proportional to the total mass. Moreover the frequency at which this peak is achieved is inversely proportional to the total mass.</p>

CC BY-SA 4.0
2023-04-20T18:44:42.237Z
2023-04-20T18:44:42.237Z
TimRias
<p>Our most familiar experience with wave propagation either from firsthand experience or in school comes from the phenomena of sound and light and radio electromagnetic propagation.</p>
<p>In air, we know that the strength of these waves as measured in sound pressure level and intensity (as measured by a photodetector or eye) drop off as <span class="math-container">$1/r^2$</span>. However if you measure the electric field of a radio wave rather than its intensity, that drops as <span class="math-container">$1/r$</span>.</p>
<p>For gravitational wave observations, we measure the strength and frequency of a strain wave and wonder where it is (direction and how far), the mass of the rotating system.</p>
<p>(Direction comes from having multiple GW detectors aligned in different directions.)</p>
<p>Thus it is helpful to know the scaling law between the magnitude of the strain wave and the mass of the system, the rotational speed, and the distance.</p>
<p>Thus I'd like to ask:</p>
<p><strong>Question:</strong> How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance and what would a short derivation look like?</p>

CC BY-SA 4.0
2023-04-19T22:23:36.720Z
[this answer](https://astronomy.stackexchange.com/a/53452/7982) seems to be an excellent fit to this question though it also helps there. I've now asked this new question to ensure that more future readers seeking information on GW scaling* will see it.
uhoh
CC BY-SA 4.0
2023-04-19T22:30:48.780Z
in rob's answer he has "Of course gravitational wave strain from a binary goes as something like $M^{5/3}P^{−2/3}d^{−1}$"  
,
James K
CC BY-SA 4.0
2023-04-20T18:04:17.480Z
To me "a rotating binary" (in the way used in the title) is a pleonasm. A binary by definition is rotating. This actually makes it confusing since, someone might infer that you are trying to say that the components of the binary are rotating.
TimRias
CC BY-SA 4.0
2023-04-20T21:15:28.377Z
@TimRias I don't think it's part of a definition so much as it is a *consequence* of gravity for large bodies. I could have a contact binary asteroid that wasn't rotating for example.
uhoh
<observational-astronomy><astrophysics><rotation><gravitational-waves>
How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance & what a short derivation looks like?
uhoh
# How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance & what a short derivation looks like?

(Asked by: username_0 on 2023-04-19)

Our most familiar experience with wave propagation either from firsthand experience or in school comes from the phenomena of sound and light and radio electromagnetic propagation.

In air, we know that the strength of these waves as measured in sound pressure level and intensity (as measured by a photodetector or eye) drop off as $1/r^2$. However if you measure the electric field of a radio wave rather than its intensity, that drops as $1/r$.

For gravitational wave observations, we measure the strength and frequency of a strain wave and wonder where it is (direction and how far), the mass of the rotating system.

(Direction comes from having multiple GW detectors aligned in different directions.)

Thus it is helpful to know the scaling law between the magnitude of the strain wave and the mass of the system, the rotational speed, and the distance.

Thus I'd like to ask:

**Question:** How does the gravitational wave strain from a rotating binary depend on the chirp mass, frequency and distance and what would a short derivation look like?

**Tags:** observational-astronomy, astrophysics, rotation, gravitational-waves
> **Comments:**
> **username_0**: [this answer](https://astronomy.stackexchange.com/a/53452/7982) seems to be an excellent fit to this question though it also helps there. I've now asked this new question to ensure that more future readers seeking information on GW scaling* will see it.  
> **username_1**: in rob's answer he has "Of course gravitational wave strain from a binary goes as something like $M^{5/3}P^{−2/3}d^{−1}$"   ,  
> **username_2**: To me "a rotating binary" (in the way used in the title) is a pleonasm. A binary by definition is rotating. This actually makes it confusing since, someone might infer that you are trying to say that the components of the binary are rotating.  
> **username_0**: @TimRias I don't think it's part of a definition so much as it is a *consequence* of gravity for large bodies. I could have a contact binary asteroid that wasn't rotating for example.  

---

## 3 answers

### Answer by username_3 on 2023-04-20. Score: 2

I don't think there is any short derivation and dimensional analysis is underdetermined. You can start from the basic equation for gravitational wave strain
$$ h = \frac{2G}{c^4r} \ddot{Q}\,$$
where $h$ is the strain and $\ddot{Q}$ is the second time derivative of a mass quadrupole moment. (Note that these are $3\times 3$ matrices, but we can ignore that for the sake of proportionalities). The reciprocal $r$ dependence, where $r$ is the distance to the source, is easily understood because $h$ is a wave amplitude and so the square of $h$ integrated over a sphere is constant (i.e. $h^2$ obeys the inverse square law).

The derivation of the "quadrupole formula" would no doubt feature in text books on gravitational waves. It isn't a short derivation. Its general form can be argued on physical grounds and dimensional analysis, but that isn't short either.

From there we can do a rough analysis by writing the quadrupole mass moments as some moment of inertia, which will be some numerical fraction multiplied by $M\_c a^2$, where $M\_c$ is the chirp mass and $a$ is the binary separation. However we are interested in the time-dependent component of the traceless quadrupole moment, which as seen from Earth varies as $\sin 2\omega t$, where $\omega$ is the angular velocity of the binary. i.e. $Q \sim M\_c a^2 \sin 2\omega t$. (See footnote).

To use the first equation, we differentiate twice to get
$$\ddot{Q} \sim -4M\_c a^2 \omega^2 \sin 2\omega t\ .$$
Then use Kepler's third law, such that
$$GM\_c \sim \omega^2 a^3\ ,$$
to some small numerical factor. Thus writing $a\sim (GM\_c/\omega^2)^{1/3}$,
$$\ddot{Q}\sim -4M\_c^{5/3}\omega^{2/3}\sin2\omega t$$
and since $P = 2\pi/\omega$, this is where the proportionality for the strain amplitude
$$ h \propto M\_c^{5/3} P^{-2/3} r^{-1}$$
comes from.

*Footnote: The traceless quadrupole moment actually changes TWICE per orbit, so in fact the time dependence is $\sin 2\omega t$ and gravitational wave frequencies are TWICE the orbital frequency. This doesn't affect the proportionality arguments for the strain amplitude.*


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### Answer by username_4 on 2023-04-20. Score: 2

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*:
$$M\_c=\dfrac{(m\_1 m\_2)^{3/5}}{(m\_1+m\_2)^{1/5}}$$
Note that $M\_c$ has dimensions of mass. 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 luminosity distance between the binary pair and us.

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:
$$\boxed{h=\dfrac{4GM\_c}{c^2 \ D} \ \left ( \dfrac{\pi G}{c^3} \ f M\_c \right )^{2/3}}$$
From this:
$$h \propto \dfrac{M\_c^{5/3} \ f^{2/3}}{D}$$
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_2**: It would be good to mention that this is a weak field approximation.  
> **username_5**: Do you also know a solution for strong fields in the vicinity of a (double) star? Where is this described?  

---

### Answer by username_2 on 2023-04-20. Score: 1

As a side note, the scaling of the strain can also be written as:

$$ h \propto M \nu (Mf)^{2/3}/D, $$

where $\nu=\frac{m1m2}{(m1+m2)^2}$ is the symmetric mass ratio. This form is useful because $GM f/c^3$ is dimensionless. Note that this expression is only valid as a weak field/slow-motion approximation. However the general expression still takes the form (for non-spinning black hole binaries)

$$ h = \frac{\nu M}{D} F(\nu, GM f/c^3), $$ where $F$ is a dimensionless analytic function of its two dimensionless arguments. In particular the peak strain in a BHB merger is given by

$$h\_{\textrm{peak}} = \frac{\nu M}{D} F\_{\textrm{peak}}(\nu).$$

The peak strain of merger waveform is therefore simply proportional to the total mass. Moreover the frequency at which this peak is achieved is inversely proportional to the total mass.


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