Gauge independence of the gravitational-wave frequency and gauge dependence of the binary separation?

Question

In the seminal paper by Cutler & Flanagan (1994), which uses multi-timescale analysis to derive waveforms in the post-Newtonian approximation without spin effects, they state that,

"In Eq. (3.9) and below, we use $(\pi Mf)^{1/3}$ as our post-Newtonian expansion parameter, instead of $(M/r)^{1/2}$. We note that $(\pi Mf)^{1/3}$ equals $(M/r)^{1/2}$ up to but not including terms of order $(M/r)^{3/2}$. This change of variables is advantageous because the frequency of the wave is a directly measurable, gauge-independent quantity (unlike the radius of the orbit)."

Why is the gravitational wave frequency gauge-independent, and why is the binary separation not gauge-independent? Is this related to the notion that general relativity is a scale-free theory?

Answer 1

The radius of the orbit is a coordinate dependent quantity. A different choice of radial coordinate, will lead to a different orbital radius for the same orbit. This is not a consequence of the scale invariance of GR, as much as it is a simple consequence of the coordinate covariance of GR.

The invariance of the gravitational wave frequency derives from invariance of the gravitational wave strain. This is literally a quantity that you can measure with a device like LIGO. It therefore cannot depend on the arbitrary coordinates you use to model the physics. Since the strain as a whole is gauge invariant, the frequency of the strain is as well.

To make this last point more explicit, lets look at how the gravitation wave frequency is obtain from the strain. Suppose you are given a complex strain as a function of time

$$ h(t) = h_{+}(t) - i h_{\times}(t)$$,

then the gravitational wave phase $\phi$ is simply (minus) the argument of the strain

$$ \phi(t) = -\arg h{t} = -\operatorname{im}\log h(t)$$.

The gravitational wave frequency $\omega$ (or $f$ in text quoted in the OP) is the time derivative of the phase

$$ \omega(t) = \dot{\phi}(t) = -\operatorname{im} \frac{\dot{h}(t)}{h(t)} $$.