Gravitational waves from orbiting binaries without general relativity: a tutorial on not how to do physics

The best way to show how ridiculous is the theory of gravitational waves is to present the arguments used by Hilbron in a tutorial on gravitational waves that does not use general relativity. Although his expressions for the  polarization, angular distributions, and overall power results differ from those of GR, waveforms that are very similar to the pre-binary-merger portions of the signals observed by the Laser Interferometer Gravitational-Wave Observatory (LIGO-VIRGO) collaboration.

The power radiated by gravitational waves without keeping track of numerical factors is

$$\dot{E}=\mathcal{T}v$$

where the transverse force per unit mass is

$$\mathcal{T}=\frac{\mu}{a^2}\sin^5\alpha,$$

with

$$\sin\alpha=\frac{v}{c},$$

is the aberration factor and $v=a\omega$ is the transverse velocity. 

Everyone claims that this is compatible with Keplerian motion. This is inaccurate. Newton wrote the transverse force per unit mass as

$$\mathcal{T}=\mathcal{F}\sin\alpha=\frac{v^2}{\rho},$$

where $\mathcal{F}$, is the force per unit mass, directed to the center of the source, $\rho$ is the radius of curvature. Introducing Kepler's II,

$$av\sin\alpha=K,$$

into the above results in

$$\rho\sin^3\alpha=\frac{K^2}{\mathcal{F}a^2}.$$

Newton calculated the radius of curvature using his theory of fluxons  and found that the left-hand side is constant. Hence, he established his inverse square law,

$$\mathcal{F}=\frac{\mu}{a^2}.$$

QED

The transverse force is therefore not given by the above expression but rather by

$$\mathcal{T}=\mathcal{F}\sin\alpha,$$

as Newton reasoned. 

If we plug in $\sin\alpha=a\omega/c$ in the expression for the conservation of angular momentum, we get

$$av^2=Kc,$$

which amalgamates $Kepler's II with his third, 

$$a^3\omega^2=GM.$$

The secular variation of the semi-major axis of the ellipse is

$$a^2=a^2_0+4\frac{\mu}{c}(t-t_0).$$

If we agree that $a_0^2=4\mu t_0/c$ at the initial instant, then on the strength of Kepler's II or III, which are now equivalent,

$$\omega^{-4/3}=\omega_0^{-4/3}+\left(\frac{\mu}{cK}\right)^{1/3}(t-t_0).$$

Hibron finds the differential equation for the semi-major axis

$$\dot{a}=-2N\eta\frac{(GM)^3}{c^5r^3},$$

where $N\eta$ are numerical factors. He "repackages" it to read

$$\dot{a}=-\frac{\eta Nc}{4}\left(\frac{r_S}{a}\right)^3,$$

by introducing the Schwarzschild radius, $r_S=2\mu/c^2$. This he integrates to

$$a^4=a_0^4-N\eta r_S^3c(t-t_0),$$

which definitely does not correspond to the Keplerian result. 

We can see this more clearly by considering the equation for the angular speed, $\omega$, which he finds as

$$\dot{\omega}=3\eta N\mu^{5/3}\omega^{11/3}/c^5$$.

Rather than integrating the last equation, he expresses the so-called chirp mass, $\eta^{5/3}M$ in terms of it, viz.,

$$(\eta N)^3 M=\frac{c^3}{3^{3/5}G}\left(\dot{\omega}/\omega^{11/3}\right)^{3/5},$$

which is void of any meaning. Rather, we choose to integrate it, and come out with

$$-\omega^{-8/3}=-\omega_0^{-8/3}+8N\eta\mu^{8/3}(t-t_0).$$

Notice the incorrect negative sign. It is also apparent that when the time is eliminated by the equation for $a^4$, Kepler's III does not result.