Why do gravitational waves circularize a binary?

Question

I understand that a binary orbiting around one another will circularize due to the emission of GWs due to Peters equations and that highly eccentric binaries evolve faster. But GW emission also removes energy and angular momentum (wouldn’t the latter increase the eccentricity from the relation between the eccentricity and the angular momentum?). What is the physical picture behind this?

Answer 1

The other answers have given good “rigorous mathematical” arguments for why this happens, but I'd like to add a simpler hand-waving one.

Gravitational waves are emitted when massive bodies accelerate. The acceleration is strongest at periapsis (i.e. when the bodies are closest). The GW emission removes energy. As a result, the orbiting body has less kinetic energy left to rise out of the periapsis, i.e. the next apoapsis will be lower.

By comparison, at the apoapsis there's not so much acceleration, thus the height of the periapsis doesn't change very much. But of course the next time you are at that periapsis, more energy is lost. Repeat, until eventually the apoapsis isn't really higher than the periapsis anymore: you have a circular orbit.

Answer 2

Note that the emission of gravitational waves does not necessarily makes an orbit more circular.

This happens to be the case in the weak field (as detailed in G. Smith's answer). However, for binaries with sufficiently different masses, emission of gravitational waves can actually increase the eccentricity in the strong field regime. This effect was discovered by Glampedakis and Kennefick in gr-qc/0203086, and has since been confirmed by many independent calculations.

Since whether gravitational wave increase or decrease eccentricity, apparently depends on the precise properties of the binary, we should not expect a simple qualitative argument explaining why its does so one way or the other.

Answer 3

The shape of a Keplerian orbit can be characterized geometrically by its semimajor axis $a$ and eccentricity $e$ or dynamically by its energy $E$ and angular momentum $L$.

The latter are expressible in terms of the former as

$$E=-\frac{Gm_1m_2}{2}\frac{1}{a}\tag{1}$$

and

$$L^2=\frac{Gm_1^2m_2^2}{m_1+m_2}a(1-e^2)\tag{2}.$$

The former are expressible in terms of the latter as

$$a=-\frac{Gm_1m_2}{2}\frac{1}{E}\tag{3}$$

and

$$e^2=1+2\frac{m_1+m_2}{G^2m_1^3m_2^3}EL^2\tag{4}.$$

Note that $E$ is negative for a bound orbit.

Gravitational radiation carries energy and angular momentum (and also linear momentum) off to infinity. $E$ decreases and becomes more negative, so its absolute value increases; $L^2$ decreases and becomes less positive, so its absolute value decreases. Whether the absolute value of the negative number $EL^2$ increases or decreases, and thus what happens to the eccentricity, is not obvious.

One must do the calculation! Here is what Peters did.

He first derives/rederives the formulas

$$\frac{dE^\text{rad}}{dt}=\frac{G}{c^5}\left(\frac15\dddot{Q}_{ij}\dddot{Q}_{ij}\right)\tag{5}$$

and

$$\frac{dL_i^\text{rad}}{dt}=\frac{G}{c^5}\left(\frac25\epsilon_{ijk}\ddot{Q}_{jl}\dddot{Q}_{kl})\right)\tag{6}$$

for the rate at which energy and angular momentum are carried to infinity by gravitational waves, in the leading order of a multipole expansion. Here

$$Q_{ij}=\sum_n m^{(n)}\left(x_i^{(n)}x_j^{(n)}-\frac13\delta_{ij}x_k^{(n)}x_k^{(n)}\right)\tag{7}$$

is the system's traceless mass quadrupole moment tensor when the system is considered as $n$ point masses.

He then applies this to a Keplerian binary and averages over one elliptical orbit. Using the conservation of energy and angular momentum, he finds that the binary's energy and momentum decrease at the average rate

$$\left\langle\frac{dE}{dt}\right\rangle=-\frac{32}{5}\frac{G^4}{c^5}\frac{m_1^2m_2^2(m_1+m_2)}{a^5}\frac{1+\frac{73}{24}e^2+\frac{37}{96}e^4}{(1-e^2)^{7/2}}\tag{8}$$

and

$$\left\langle\frac{dL}{dt}\right\rangle=-\frac{32}{5}\frac{G^{7/2}}{c^5}\frac{m_1^2m_2^2(m_1+m_2)^{1/2}}{a^{7/2}}\frac{1+\frac{7}{8}e^2}{(1-e^2)^2}\tag{9}.$$

Differentiating (3) gives

$$\frac{da}{dt}=\frac{Gm_1m_2}{2}\frac{1}{E^2}\frac{dE}{dt}\tag{10}$$

and differentiating (4) gives

$$e\frac{de}{dt}=\frac{m_1+m_2}{G^2m_1^3m_2^3}\left(L^2\frac{dE}{dt}+2EL\frac{dL}{dt}\right)\tag{11}.$$

Substituting (1), (2), (8), and (9) into (10) and (11) gives

$$\left\langle\frac{da}{dt}\right\rangle=-\frac{64}{5}\frac{G^3}{c^5}\frac{m_1m_2(m_1+m_2)}{a^3}\frac{1+\frac{73}{24}e^2+\frac{37}{96}e^4}{(1-e^2)^{7/2}}\tag{12}$$

and

$$\left\langle\frac{de}{dt}\right\rangle=-\frac{304}{15}\frac{G^3}{c^5}\frac{m_1m_2(m_1+m_2)}{a^4}\frac{e(1+\frac{121}{304}e^2)}{(1-e^2)^{5/2}}\tag{13}.$$

You can see that the rate of decrease of the eccentricity is very rapid for a highly eccentric orbit with $e$ near $1$, due to the $(1-e^2)^{5/2}$ in the denominator. In others words, the orbit rapidly circularizes.

From these equations, Peters proceeds to find $a$ as a function of $e$ (with two unusual exponents, $12/19$ and $870/2299$) and a differential equation for $e(t)$ from which the lifetime of the binary can be found.

Answer 4

A circular orbit is the one that minimises the energy (kinetic plus potential) for a given angular momentum.

If a process radiates energy away from the system without carrying angular momentum away, then the orbit will naturally relax to a circular configuration. This might be the approximate case for an object on an eccentric orbit interacting with a disc of gas and dust for example, where energy can be dissipated and radiated away.

A similar things happens for gravitational waves, except the complication here is that the gravitational waves also take away some angular momentum. It turns out - and the the answer of @G.Smith shows why - that the rate of angular momentum loss is not high enough, compared to the energy losses, to prevent circularisation.

The effect is quite extreme in short period eccentric binary systems because the luminosity in gravitational waves is dimensionally proportional to $L \sim M^2R^2 \omega^6$, where $\omega$ is the angular velocity and $R$ is the orbital separation. But if we set $v \sim R\omega$, then $L \sim M^2 R^{-4} v^6$. This strong dependence on orbital speed (and inverse dependence on orbital radius) is what makes this more efficient in eccentric binaries, because they spend parts of their orbits with a smaller orbital radius and faster orbital speed than a circular binary.