There's some recent buzz about a new indirect measurement of gravitational waves in Rapid Orbital Decay in the 12.75-minute WD+WD Binary J0651+2844. It presents a measurement the period decrease in a binary system of two white dwarfs which orbit each other every 13 minutes:
Summary figure from phys.org: on the top is the accumulated shift in eclipse time, and on the bottom are some light curves of eclipse measurements.
In the comments on the associated Phys.org news post there was some confusion on how exactly the six second time shift over 380 days translated into the two objects merging in two million years. And I have seen (and experienced) similar confusions when looking at the canonical Hulse-Taylor plots for example the one on the Cardiff Gravitational Physics Tutorial on the Hulse-Taylor pulsar , which show cumulative shift in the periastron time of the orbits:
Figure (from Weisberg and Taylor (2004) via the Cardiff site) showing the accumulating shift in periastron time over years of measurements.
If we want to estimate the time taken by some number of orbits $n$, if the time per orbit is a slowly varying function, we can approximate by Taylor expanding:
$$ t_n = t_0 + \frac{dt}{dn} (n- n_0) + \frac{1}{2} \frac{d^2t}{dn^2} (n - n_0)^2 +... $$
The period $P$ of the orbit is then $\frac{dt}{dn}$ in this notation, and
$$ \begin{aligned}
\frac{d^2t}{dn^2} &= \frac{d}{dn}\left( \frac{dt}{dn} \right) \\
&= \frac{dt}{dn} \frac{d}{dt} \left(\frac{dt}{dn}\right) \\
&= P \dot{P}
\end{aligned} $$
where $\dot{P} = dP/dt$ and we throw infinitesimals around like physicists. This means that we have after $n$ orbits:
$$ t_n = t_0 + P (n- n_0) + \frac{1}{2} P \dot{P} (n - n_0)^2 $$
compared to the constant period estimate,
$$ t_n^{\text{no-GW}} = t_0 + P (n- n_0) $$
so the accumulated time shift $T$ over $N$ orbits is quadratic in $N$:
$$ T = \frac{1}{2} P \dot{P} N^2 $$
In the case of the WD-WD system, we have $T \simeq - 6$ seconds, $ P \simeq 13$ minutes, and
$$ N \simeq \frac{380 \text{ days}}{13 \text{ minutes} / \text{ orbit}} = 42000 \text{ orbits} $$
therefore
$$\begin{aligned}
\dot{P} &= \frac{2 T}{P} \frac{1}{N^2} = \frac{ 2 T} { (P N) N }\\
&\simeq 2 (- 6 \text{ seconds}) / (380 \text{ days}) / (42000 \text{ orbits}) \\
&= - 8.7 \times 10^{-12} \text{ s / s} = - 0.27 \text{ ms / year} \\
\end{aligned}$$
with some Google unit conversions, which is almost the same as the result in the paper abstract. If we just keep extrapolating this with a constant $\dot{P}$, we will have a period of $0$ minutes in
$$\frac{0 - 13 \text{ minutes} }{- 0.27 \text{ ms / year}} = 3 \text{ million years} $$
which is pretty soon, astronomically speaking.
This white dwarf system would be a powerful "verification binary," detectable in a future space-based gravitational-wave telescope like LISA or eLISA; Antoine Petiteau on the LISA Facebook group estimated signal-to-noise ratios of 40-70 in LISA and 20 in eLISA based on their current designs. The frequency is too low for the ground-based detectors to see, since Earth's seismic noise and gravity-gradient noise get in the way.
