# Reduced mass vs. Total mass in gravitational wave estimations

When people do back of the envelope calculations about GW physics, they always use a very abstract mass scale $$M$$ and I want to figure out the identity of said scale for different relevant magnitudes in GW processes.

GW Intensity $$h$$

The intensity of the gravitational waves emitted by a gravitational system with masses $$m$$ and $$M$$ are proportional to its quadrupole moment

$$$$h\propto \frac{G}{c^4}\frac{\ddot{Q}}{r}$$$$

We want highly nonspherical systems to get a big quadrupole moment so we can assume

\begin{aligned} Q&\propto \mu L^2 \\ \ddot{Q}&\propto \mu v^2 \\ \ddot{Q} &\propto E_{kin} \end{aligned} where $$\mu=\frac{m\cdot M}{m+M}$$ is the reduced mass of the system. Using the virial theorem, the kinetic energy is proportional to the gravitational potential so we get

\begin{aligned} E_{kin}&\propto E_{pot}\\ &\propto \frac{G\mu M_{tot}}{r} \end{aligned} We are interested in the gravitational waves at minimum distance (since they will be the strongest). These will happen at distances equal to the sum of the Schwarzschild radius of each object

$$$$L_{smallest}\approx \frac{2Gm}{c^2}+\frac{2GM}{r}=\frac{2GM_{tot}}{c^2}$$$$ So we get a kinetic energy proportional to reduced mass

\begin{aligned} E_{kin}\propto \mu c^2 \end{aligned}

Cool. So we found that the intensity of a gravitational wave goes like the reduced mass

$$$$h\propto \frac{G}{c^2}\frac{\mu}{r}$$$$

Frequency of GW $$f$$

A similar analysis can be made for the frequency of the gravitational waves. Using Kepler's third law, we get that the orbital frequency has to be

$$$$f_{orb}=\big[\frac{GM_{tot}}{L^3}\big]^{1/2}$$$$

which, again, for the smallest separation of $$L\propto M_{tot}$$ gives

$$$$f_{orb}\propto \frac{1}{M_{tot}}$$$$

Gravitational waves are proportional to two times the orbital frequency (Because they are proportional to the quadrupole moment) so we get that the frequency of GW is

$$$$f_{detection}\propto \frac{1}{M_{tot}}$$$$

Conclusion

So the conclusion is that the intensity of gravitational waves is proportional to the reduced mass while the frequency is proportional to the total mass.

Observations

• Does this mean that the frequency of a binary system of supermassive black holes with masses $$M\approx 10^6M_{\odot}$$ and the frequency of a binary system with one supermassive black hole and one stellar-mass black hole with mass $$M\approx M_{odot}$$ will be the same? Since it's given by the inverse of the total mass, two supermassive black holes or one supermassive black hole and one stellar one should have roughly the same order of mass. And that's why LIGO can't detect systems with supermassive black holes (either one or two): Because it works on the high frequency regime between 1HZ and $$10^4$$ Hz so it needs to deal with binary systems where both objects are stellar mass.

• Similarly, does this mean that a system of two stellar mass black holes and one with one stellar and one supermassive blackholes would have roughly the same intensity? Since the intensity is proportional to the reduced mass it would be $$m/2$$ for two stellar black holes and $$m$$ for a stellar and a supermassive black holes.

• Thanks for the answer! So frequency is indeed proportional to the inverse total mass. But in the PN expansion the intensity of fthe waves actually depends on both masses separately and the multipolar simplified argument breaks down. Is that right? Commented Sep 27, 2023 at 3:09

These statements are true (only) as a leading order approximation. More generally, the gravitational wave strain will look something like

$$h = \mu \left(A(f) + B(f)\nu + O(\nu^2)\right),$$

for non-spinning binaries, and $$\nu=\frac{m M}{M_{\rm tot}^2}$$ the symmetric mass-ratio. (For spinning systems, there will be additional dependence on the spins, and the anti-symmetric combination of the masses).

When talking about the frequency, one needs to be a bit more careful about exactly what frequency one means. When talking about the frequency at fixed separation $$r$$, the dependence is actually (as can be easily inferred from Kepler's third law)

$$f = \sqrt{\frac{GM_{\rm tot}}{r^3}} + O(\nu,r^{-5/2}),$$

i.e. to leading order $$f \propto \sqrt{M_{\rm tot}}$$.

However, the frequency of the peak of a merger waveform, $$f_{peak}$$ will have the form

$$f_{peak} = \frac{1}{M_{\rm tot}}\left(a + b \nu + O(\nu^2)\right).$$

Somewhat interesting the peak gravitational wave luminosity $$\mathcal{L}_{\rm peak}$$, is to leading order independent of the total mass,

$$\mathcal{L}_{\rm peak} = \alpha(\nu).$$

The peak luminousity of a stellar mass (non-spinning) binary merger of equal mass is the same as the peak luminousity of a supermassive black hole binary merger!