Neutron star collisions have long been proposed as "The Great Filter" explaining Fermi's paradox. I haven't looked for it, but I am almost surprised this has not been evoked in the aftermath of the news on my stream, yet -- except now, a bit indirectly, in this post by +Greg Egan -- his usual magistral self.
The chirp of death

Almost exactly twenty years ago, my novel Diaspora was published. The story begins about a thousand years in the future — just before Earth’s biosphere is ravaged by the gamma-ray burst created when a pair of neutron stars a mere 100 light-years away from us collide. This event doesn’t happen without warning: a gravitational wave observatory tracks the decay of the neutron stars’ orbit in the run-up to the collision.

As anyone who’s read the news today will know, twenty years later, we’ve actually observed gravitational waves from exactly this kind of collision — as well as the gamma rays, light and radio waves it produced. Needless to say, since it was about 130 million light years away, rather than 100 light years, we won’t be suffering any ill effects.

Back in 1997, I wrote a little Java applet to simulate the effects of the gravitational waves produced by this kind of source. Today, since more and more browsers refuse to let their users run Java, I rewrote the applet in JavaScript, which most browsers support. The applet is at:

One thing I learned when I was writing the applet back in 1997 was that, although the general-relativistic calculations describing the gravitational waves from a binary system are fairly complicated (even in the linearised version of GR, which is good enough for most purposes prior to the collision itself), you can get the correct form of some of the relevant equations from simple heuristic arguments and ordinary old Newtonian orbital mechanics. You only really need GR to get the constants of proportionality exactly right. [For anyone with Misner, Thorne and Wheeler’s book Gravitation, this is covered in Chapters 35 and 36.]

For example, suppose you want to know the rate at which a binary pair of neutron stars is radiating energy in the form of gravitational waves. Each star is experiencing centrifugal acceleration of a_i ω^2, where a_1 and a_2 are the distances of the two stars from the centre of mass of the system and ω is the angular frequency of the orbit. The amplitude, A_i, of the gravitational radiation produced by each star is proportional to the product of its acceleration and the mass of the star, m_i, and inversely proportional to the distance r at which the wave is measured

A_i = m_i a_i ω^2 / r

A free-falling body isn’t really accelerating in the general-relativistic view of things, but in linearised GR, where you pretend that special relativity applies and that space-time curvature is like an extra field, you can think of a free-falling body as accelerating, and just as the amplitude of electromagnetic radiation from an accelerating object is proportional to the product of that acceleration and the object’s charge, the amplitude of gravitational radiation from an accelerating object is proportional to the product of the acceleration and the object’s mass.

Now, the tricky thing is that the two stars will always be accelerating in opposite directions, so their gravitational radiation will be 180° out of phase. What’s more, by the definition of “centre of mass”, the distances a_1 and a_2 of the stars from the centre of mass obey the equation:

m_1 a_1 = m_2 a_2

This means the amplitudes of the gravitational radiation from both stars will also be more or less equal (assuming we are observing them from far enough away that r is essentially the same in both cases):

A_1 ≈ A_2

Two waves of equal amplitude, 180° out of phase, will cancel out!

But what prevents this cancellation is the fact that, in general, the two stars will not be equidistant from the observer, and though the difference in r won’t be significant, the time lag between the waves due to the different distances they travel will introduce a further phase difference. This will be proportional to the distance between the stars, a_1+a_2, and inversely proportional to the wavelength of the radiation ... and so proportional to the angular frequency, ω. So we have a total amplitude with the proportionality:

A ~ m_1 a_1 (a_1+a_2) ω^3 / r

If we define:

a = a_1+a_2


μ = m_1 m_2 / (m_1+m_2)

this simplifies to:

A ~ μ a^2 ω^3 / r

The power of radiation of amplitude A is proportional to A^2, and if we integrate over a sphere around the source, with surface area proportional to r^2, we end up with a total luminosity of:

L ~ μ^2 a^4 ω^6

But by Kepler’s Third Law, we have:

ω^2 ~ M/a^3

where M = m_1+m_2, and so:

L ~ M^3 μ^2 / a^5

The actual result from a careful calculation in linearised GR is:

L = (32/5) (G^4 / c^5) M^3 μ^2 / a^5

Now, a warning: don’t take this heuristic too literally! This argument makes it sound as if an observer positioned on the axis of the orbit, always equidistant from both stars, would measure no gravitational radiation, but that isn’t actually the case. We have ignored the fact that these waves are tensors, which introduces further subtleties, but nonetheless we did end up with the correct form for the gravitational luminosity.

If you ponder this result, you can see why there is an intense “chirp” of gravitational radiation as the two stars approach the moment of collision. As their separation, a, decreases, the rate at which energy is lost rises rapidly, which in turn drives the separation down ever faster, as well as increasing the frequency of the orbit and raising the “pitch” of the radiation.

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