The relationship between the orbital period, T, and the separation between the neutron stars, a, is given by Kepler’s Law: the period squared is proportional to the separation cubed.
T^{2} | = | (4π^{2}/GM) a^{3} |
where M = m_{1}+m_{2} is the combined mass of the two stars, and G is the universal gravitational constant.
The total power being radiated in gravitational waves is inversely proportional to the fifth power of the separation.
L | = | (32G^{4}M^{3}μ^{2}/5c^{5}) / a^{5} | (1) |
where μ = (m_{1}m_{2})/(m_{1}+m_{2}) is the “reduced mass” of the system. A rough justification for this equation is that the amplitude of gravitational radiation from each star is proportional to its mass m_{i} and its centrifugal acceleration ω^{2}a_{i} (where a_{i} is the distance from the centre of mass, and ω = 2π/T), and inversely proportional to its distance from the observer, r. But by the definition of “centre of mass”, m_{1}a_{1} = m_{2}a_{2}, so the two stars generate gravitational radiation of precisely equal amplitude, 180° out of phase. The only thing that stops the two waves from cancelling each other out is the time lag across the orbit, which introduces a phase difference proportional to (a_{1}+a_{2})ω. So the amplitude of the combined wave a distant observer receives is:
A | ~ | m_{1}a_{1}(a_{1}+a_{2})ω^{3}/r | |
~ | μa^{2}ω^{3}/r | (2a) |
The power of the radiation is proportional to the square of its amplitude. Substituting for ω with the value given by Kepler’s Law, ω^{2} ~ M/a^{3}, and integrating over a sphere of radius r (and hence surface area proportional to r^{2}) yields:
L | ~ | A^{2}r^{2} | |
~ | M^{3}μ^{2} / a^{5} | (2b) |
The correct factors of G and c can be found easily from dimensional arguments, but the 32/5 in Eqn (1) only comes from a proper analysis using General Relativity. A full analysis is also needed to derive the detailed motion of the ring of test particles, but multiplying Eqn (2a) by T/2 = π/ω — to allow for the cumulative effect over half the period of each successive wave as it pushes the particles away from their initial position — gives an estimate for the distortion of the ring:
dx/x | ~ | μa^{2}ω^{2}/r |
Substituting again from Kepler’s Law, and inserting factors of G and c yields:
dx/x | ~ | (G^{2}Mμ/c^{4}) / (a r) | (3) |
To calculate the time until the stars collide, note that the total energy (potential plus kinetic) of the binary system is:
E | = | –GMμ/a + m_{1}ω^{2}a_{1}^{2}/2 + m_{2}ω^{2}a_{2}^{2}/2 | |
= | –GMμ/a + μω^{2}a^{2}/2 | ||
= | –GMμ/a + GMμ/(2a) | ||
= | –GMμ/(2a) |
The rate of change of this with respect to time (which must match the power being lost in gravitational radiation) is:
dE/dt | = | GMμ/(2a^{2}) da/dt |
Substituting this, and Eqn (1), into dE/dt = –L gives:
GMμ/(2a^{2}) da/dt | = | –(32G^{4}M^{3}μ^{2}/5c^{5}) / a^{5} | |
da/dt | = | –(64G^{3}M^{2}μ/5c^{5}) / a^{3} | |
dt/da | = | –(5c^{5}/64G^{3}M^{2}μ) a^{3} | |
t | = | –(5c^{5}/256G^{3}M^{2}μ) a^{4} | (4) |
Eqn (4) assumes that the stars collide at t=0; it gives the time prior to collision, for a given separation a.
Reference: Gravitation by Misner, Thorne and Wheeler; Chapters 35 & 36. Apart from factors to convert from geometrical units, Eqn (1) is essentially MTW’s Eqn (36.16a), Eqn (4) is MTW’s Eqn (36.17b). The argument leading to Eqn (2a) & (2b) is a paraphrase of MTW’s for Eqn (36.1).