# Diaspora

## Chapter 4: Lizard Heart

### Mathematical Details

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.

 T2 = (4π2/(GM)) a3

where M = m1+m2 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 = (32G4M3μ2/(5c5)) / a5 (1)

where μ = (m1m2)/(m1+m2) 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 mi and its centrifugal acceleration ω2ai (where ai 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”, m1a1 = m2a2, 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 (a1+a2. So the amplitude of the combined wave a distant observer receives is:

 A ~ m1a1(a1+a2)ω3/r ~ μa2ω3/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, ω2 ~ M/a3, and integrating over a sphere of radius r (and hence surface area proportional to r2) yields:

 L ~ A2r2 ~ M3μ2 / a5 (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 ~ μa2ω2/r

Substituting again from Kepler’s Law, and inserting factors of G and c yields:

 dx/x ~ (G2Mμ/c4) / (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 + m1ω2a12/2 + m2ω2a22/2 = –GMμ/a + μω2a2/2 = –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/(2a2) da/dt

Substituting this, and Eqn (1), into dE/dt = –L gives:

 GMμ/(2a2) da/dt = –(32G4M3μ2/(5c5)) / a5 da/dt = –(64G3M2μ/(5c5)) / a3 dt/da = –(5c5/(64G3M2μ))  a3 t = –(5c5/(256G3M2μ))  a4 (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 discussion accompanying Figure 36.1.  Diaspora / Chapter 4: Lizard Heart (detailed) / created Saturday, 25 October 1997 / revised Friday, 12 June 1998