# Why is frequency different for gravitational waves coming from SMBHs vs small compact binaries?

Why do gravitational waves emitted by supermassive black holes binaries have low-frequency range (~10$^{-9}$ - 10$^{-2}$ Hz) than those emitted by stellar mass compact binaries (~10$^{-4}$ - 10$^2$ Hz)?

Below, a NASA graphic showing the wavelength ranges of various phenomenon found here. Click for full size.

• You should include a link or a reference to where you are quoting those numbers from.
– uhoh
Mar 6, 2018 at 5:19
• As a note, the frequency range for supermassive black hole binaries should be narrower, maybe $10^{-10}$ to $10^{-7}$ Hertz. $10^{-2}$ Hertz is much larger than you would see in that sort of system, even if it's small for the comparatively low-mass systems LIGO and VIRGO observe. Mar 6, 2018 at 5:44
• Source of the above frequency ranges: imagine.gsfc.nasa.gov/features/satellites/archive/images/… Mar 6, 2018 at 6:30
• @AwaisMirza thanks! I'll add that to your question. "You should include a..." means include within the question itself. Comments are considered temporary, so we try to put all the important things back into the original question when possible.
– uhoh
Mar 6, 2018 at 7:45

The reason for the upper limit is that gravitational force scales as the product of masses over separation squared and that the closest approach of black holes is limited by their event horizons. The lower limit is because the amplitude of the waves gets smaller and becomes undetectable at lower frequencies.

In more detail:

The frequency of gravitational waves is twice the orbital frequency.

Kepler's third law tells us that the orbital period $P$ is related to the total mass $M$ and orbital separation $a$ as $$P^2 \propto M^{-1} a^3$$

The orbital frequency $\omega =2\pi/P$ so $$\omega \propto M^{1/2} a^{-3/2} \tag{(1)}$$

The frequency of gravitational waves therefore depends on both the mass and separation of the binary system.

However, the maximum frequency occurs at closest approach, which for black holes is when their event horizons come into contact. The Schwarzschild radius is $2GfM/c^2$, where $f$ is the fraction of the system mass of each black hole. Thus the closest separation $a_{\rm min}$ is the sum of the Schwarzschild radii and is simply $2GM/c^2$, since $f_1+f_2=1$.

Using $a_{\rm min}$ for $a$ in equation (1) we see that the maximum frequency $\omega^{\rm max} \propto M^{1/2}a_{\rm min}^{-3/2} \propto M^{-1}$ Thus the maximum frequency gets bigger as the mass gets smaller, and as supermassive black holes are a million to a billion times more massive than stellar black holes, the maximum frequencies are correspondingly smaller.

The lower limits to frequency are not really limits at all. The frequencies can be anything smaller than the maximum. However the amplitude of the gravitational waves depends on the acceleration of the masses, which in turn decreases as they become more widely separated. This effectively gives a window between where the waves become detectable and the maximum frequency.

In general, the frequency of gravitational waves emitted by a binary system is $f=2f_{\text{orb}}$, where $f_{\text{orb}}$ is the orbital frequency of the system. In the late evolutionary phases during which gravitational waves from supermassive black hole binaries are detectable, the orbital periods are still comparatively large, measured in the months or years.

Near coalescence, we still have an upper frequency bound, because the maximum frequency $f_{\text{max}}$ is proportional to $\sqrt{G\rho}$, where $\rho$ is some sort of density for the system. In the case of a black hole, $\rho\propto M/R^3$, and since the Schwarzschild radius $R_s\propto M$, we find that $f_{\text{max}}\propto\sqrt{M/M^3}=M^{-1}$. Therefore, the more massive the black hole, the lower the maximum orbital frequency. For supermassive black holes, this turns out to be in the nanohertz regime, give or take an order of magnitude.

• What can be the physical reason for the dependence of maximum frequency on black hole masses? Mathematically, you answered the question correctly. Mar 6, 2018 at 6:25
• @AwaisMirza because the radius of the horizon of the black hole limits the orbital radius, and the orbital radius along with the masses determines the orbital period. Mar 6, 2018 at 18:13