How Kepler's Laws Destroy the LIGO Analysis of the Interferometer Strain Signal

We will show that Kepler's III destroys the independence of the frequency $\omega$ and its rate of change in time, $\dot{\omega}$ so that they cannot be varied independently to determine the 'chirp' mass $\eta^{3/5}M$ from

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

where $N$ is a numerical factor, $32/5$ for linear GR, and $N=2/5$ for the EM-like model discussed by Hilborn. For equal masses $\eta^{3/5}M=M/\sqrt[5]{2}$ so that Kepler's third law

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

will always apply, where $a$ is the distance of separation between the two masses. However, Kepler's III is used when convenient, and, at other times, forgotten.

Kepler's III implies that the right-hand side of the equation be constant, 

$$\frac{\dot{\omega}}{\omega^{11/3}}=\kappa=\mbox{const.}$$

It can be immediately integrated to yield the period, $P=2\pi/\omega$

$$P(t)=\left(P_0^{8/3}-\frac{8}{3}\kappa(t-t_0)\right)^{3/8},$$

where $P_0$ is the period of the orbit at time $t=t_0$. The constant $\kappa$, is a constant times $(a\omega/c)^5$ that we discussed in  previous blogs.

From the LIGO-VIRGO data for GW950914, and using the linear GR value of $N$ they found a chirp mass of approximately $30 M_{\circledcirc}$. According to the collaboration, different values of $M$ will lead to different results for the mass separation, $a(t)$, for a given orbital frequency. This is incorrect because of the Keplerian constraint,

$$3\frac{\dot{a}}{a}=-2\frac{\dot{\omega}}{\omega}; $$

that is a change in $a(t)$ imposes a change in $\omega(t)$. But let's not worry about this for the moment.

The waveform plots "strain" versus time and looks like the one in the diagram that LIGO obtained for the event GW150914 (magnified $10^{21}$).

strain vs time

The question, among many others, is how they got an explicit time dependency for the wave strain? Not being able to follow the logic behind numerical relativity, I will limit myself to the EM-like model discussed by Hilborn, which captures the salient features except for the angular dependencies.

The EM-like model uses retarded times and positions to extract explicit time dependencies. Let $R$ be the distance between the center of mass and the point of observation. The gravitational analogue of the vector potential is claimed to be

$$\vec{A}=\frac{G\eta M\omega^2}{c^3 R}\vec{r}(\hat{R}\cdot\vec{r}).$$

Retarded and present times are related by

$$c(t-t_R)=R.$$

And the field generated at $\vec{x}_0$ is related to it at time $t$ by

$$\vec{r}(t)=\vec{x}_0-\vec{v}t.$$

The position and velocity of one of the masses is

$$\vec{r}=r\left[\hat{x}\cos(\omega t_R)+\hat{y}\cos(\omega t_R)\right],$$

and

$$\vec{v}=\omega r\left[-\hat{x}\sin(\omega t_R)+\hat{y}\cos(\omega t_R)\right],$$

where $\hat{x}$ and $\hat{y}$ are unit normals in the $x$- and $y$- directions, respectively. It is apparent that $r$ and $\omega$ have nothing whatsoever to do with $a$ and $\omega$ of the previous discussion.

Is the assumption that $\vec{r}$ and $\vec{v}$ vary periodically? True the latter is the time derivative of the former, but they must also satisfy

$$r\left[\hat{x}\cos(\omega t_R)+\hat{y}\sin(\omega t_R)\right]=\vec{x}_0-\omega r\left[-\hat{x}\sin(\omega t_R)+\hat{y}\cos(\omega t_R)\right]t.$$

For simplicity, we place the source at the origin, and compare components. Along the $x$-axis, we have

$$\cos(\omega t_R)=\omega t\sin(\omega t_R)t.$$

From this, and the relation between past and present times we get

$$\cot(\omega t_R)-\omega t_R=\omega R/c,$$

which cannot be valid for all times, and values of $\omega$ since $R>0$, by definition. Yet, this is the origin of the explicit time dependencies in the EM-like model which is claimed to be consistent with linear GR.

Hilborn then claims that on the strength of the periodicities in $\vec{r}$ and $\vec{v}$ that

$$\frac{d\vec{v}}{dt}=-\omega^2\vec{r},$$

apparently not realizing that this is the condition for the vanishing of the radial force,

$$\vec{F}_{rad}=\ddot{\vec{r}}-\omega^2\vec{r}=\vec{0}.$$

But, this is, indeed, a consequence of the assumption that position and velocity are periodic in time.

It is, therefore, no wonder that the gravitational vector potential is written as

$$\vec{A}=\frac{G\eta M\omega^2r^2}{2c^3R}\sin\theta\left[\cos\theta(1+\cos(2\omega t_R)\hat{\theta}+\sin(2\omega t_R)\hat{\phi}\right],$$

where $\hat{\theta}$ and $\hat{\phi}$ are the polarization unit vectors. The fundamental expression for the gravitational acceleration, $\vec{g}$, is the negative time derivative. Since we intend to project along the $x$-axis, where $\hat{x}\cdot\hat{\phi}=0$, the expression for the gravitational acceleration vector is

$$\vec{g}=\frac{\partial\vec{A}}{\partial t_R}=-\frac{G\eta M\omega^3r^2}{3c^3R}\sin(2\theta)\sin(2\omega t_R)\hat{\theta}.$$

This means that $r$ and $\omega$ have  absolutely nothing to do with $a$ and $\omega$ that we introduced previously, and claimed satisfies Kepler's laws.

Assuming the mass is free to move in the $x$-direction, projecting $\vec{g}$ along $x$, $\hat{x}\cdot\hat{\theta}=\sin\beta$, Hilborn get the harmonic oscillator equation

$$\ddot{x}=\vec{g}\cdot\hat{x}=\frac{G\eta M\omega^3r^2}{3c^2R}\sin\beta\sin(2\theta)\sin(2\omega t),$$

conveniently forgetting to indicate retarded time of which all the variables are function of. Now, of all things, we are told to integrate this equation to get

$$x(t)=-\frac{\vec{g}\cdot\hat{x}}{(2\omega)^2},$$

where the $2$ comes from the fact that the wave frequency is twice the orbital frequency. We are told that "the minus sign  simply tells us that the displacement and forcing term are π radians out of phase for the sinusoidal driving of an object at a frequency far above its natural oscillation frequency." But the radial force is zero!

The above integration in time is valid provided

$$ \omega^3a^2=\mbox{const}.$$

This is not Kepler's III, $a^3\omega^2=\mbox{const}.$ We can stop here, but let us continue.

By a series of unscrupulous procedures, the strain equation finally emerges,

$$\Delta L=\Delta x_2(t)-\Delta x_1(t)=\frac{G\eta M\omega^2r^2}{4c^3R}\Delta t\sin\beta\sin(2\theta)\cos(2\omega t).$$

Anyone who has persevered   this far will be rewarded: Introducing $\Delta t=\Delta R/c=L\cos\beta/c$, where $L$ is the length of the arm of the interferometer, and using "Kepler's law to replace $r$ in favor of $\omega$"  result in

$$\Delta L=\frac{(GM)^{5/3}\eta\omega^{2/3}}{8c^3R}L\sin(2\beta)\sin(2\theta)\cos(2\omega t).$$

Regarding the first substitution, $\Delta t=t-t_R=R/c$, and no one has varied the past position, while regard the second only  confirms our suspicions that $r$ is $a(t)$ and $\omega$ is $\omega(t)$. Holding them constant in above integration is in violation of Kepler's III.

Now the procedure used to calculate the gravitational waveform is

1. Select an initial value of the separation of the masses that is a few times the combined Schwarzschild radii of the binary objects.  (The strong in-spiral of the binary objects begins when the separation of the masses is a few times the combined Schwarzschild radii.)  

2.  Use

$$\dot{r}=r_s\left[\left(\frac{r_0}{r_s}\right)^4-\frac{N\eta c}{r_s}(t-t_0)\right]^{1/4},$$

which we discussed in the previous blog, to find $r(t)$ for a sequence of time values.

3.  From $r(t)$, use Kepler’s III to find $\omega(t)$ .

4.  Use that value of $\omega(t)$ in the above expression for $\Delta L$, and repeat steps 3 and 4 for the next time value.

5.  Plot the results.

It's step 4 that ruins the procedure for it has been derived using $\omega^3a^2=\mbox{const}$, and not Kepler's III, $a\omega^2=GM/a^2$.

The conclusion that "the relative displacement oscillates at twice the orbital frequency of the source and the amplitude depends on the combination $\eta M^{5/3}$ multiplied by $\omega^{2/3}$, exactly the dependence predicted by linear GR" is inaccurate. The explicit periodic time dependencies comes from a misuse of retarded time and position, i.e. $\vec{r}$ and $\vec{v}$ cannot be periodic functions of time because past will become present and present will become past on a regular basis. Moreover, Kepler's law cannot be used when it proves convenient: either it applies, or doesn't apply. The expression for the chirp mass explicitly depends on Kepler's III, but the analysis leading up to the final result that relates the strain $\Delta L$ does not--nor does the analysis of the LIGO-VIRGO data.