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?

  • $\begingroup$ this answer 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. $\endgroup$
    – uhoh
    Apr 19 at 22:23
  • $\begingroup$ in rob's answer he has "Of course gravitational wave strain from a binary goes as something like $M^{5/3}P^{−2/3}d^{−1}$" , $\endgroup$
    – James K
    Apr 19 at 22:30
  • $\begingroup$ 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. $\endgroup$
    – TimRias
    Apr 20 at 18:04
  • $\begingroup$ @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. $\endgroup$
    – uhoh
    Apr 20 at 21:15

3 Answers 3


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.


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.

  • 1
    $\begingroup$ It would be good to mention that this is a weak field approximation. $\endgroup$
    – TimRias
    Apr 20 at 18:05
  • $\begingroup$ Do you also know a solution for strong fields in the vicinity of a (double) star? Where is this described? $\endgroup$
    – 9herbert9
    Nov 15 at 16:34

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .