Eccentricities in Gravitational Wave Binaries
“The most exciting phrase to hear in science, the one that heralds new discoveries, is not ‘Eureka’ but ‘That’s funny...’”
— Isaac Asimov
Recently, I have noticed increased discussion on the importance of eccentricity in the gravitational wave (GW) compact object coalescence literature. However, I lack intuition for why eccentricity is considered significant, for the following reasons:
- The orbits circularize ($e \to 0$) as the compact binary approaches coalescence due to gravitational wave emission. So why include eccentricity at all stages?
- The eccentricities of binary neutron star (BNS) systems formed through isolated binary evolution channels are expected to be small ($e < 0.3$), as shown in \cite{Ihm:2005tw}. Dynamically formed binaries contribute only a small fraction of the total BNS population.
- The orbital period does not depend on eccentricity. To explore this, I begin by revisiting the basics of orbital dynamics.
The Lagrangian for a two-body system in the center-of-mass frame is:
$$ \mathcal{L} = \frac{1}{2}\mu\dot{r}^2 + \frac{1}{2}\mu r^2 \dot{\theta}^2 + \frac{Gm_1m_2}{r} $$
Here, $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass. Since $\theta$ is a cyclic coordinate, we have:
$$ \mu r^2 \dot{\theta} = \ell $$
And the radial equation of motion becomes:
$$ \mu \ddot{r} = \frac{\ell^2}{\mu r^3} - \frac{Gm_1m_2}{r^2} $$
Using the substitution $u = \frac{1}{r}$, the solution is a conic:
$$ r(\theta) = \frac{\ell^2}{\mu^2 GM (1 + e \cos\theta)} $$
Here $M = m_1 + m_2$ and $e$ is the orbital eccentricity. The orbital radius $r$ depends on $e$, but the period $T$ is independent of it for bound orbits. By Kepler's second law:
$$ \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} = \frac{\ell}{2\mu} = \text{constant} $$
Integrating over a full orbit:
$$ A = \frac{\ell T}{2\mu} $$
And for an ellipse:
$$ A = \pi a b = \pi a^2 \sqrt{1 - e^2} $$
From the conic solution:
$$ a(1 - e^2) = \frac{\ell^2}{\mu^2 GM} $$
Combining these results:
$$ T^2 = \frac{4\pi^2}{GM} a^3 $$
This is Kepler’s third law, which confirms that the period $T$ is independent of eccentricity.
My skepticism about the emphasis on eccentricity is rooted in these points. However, the inclusion of gravitational radiation may alter this picture. I will explore this further in future work.