$\ldots$ so say LIGO [Abbott *et al*, "Observation of gravitational waves from a binary black hole merger", (Hulse) Taylor and Weisberg, "The relativistic binary pulsar B1913+16: thirty years of observation and analysis", and Peters and Mathews, "Gravitational radiation from point masses in a Keplerian orbit".

In a tutorial Hilborn, in his tutorial on the derivation of gravity waves without general relativity, asks if the luminosity of gravitational waves,

$$L=\frac{2}{5}\frac{\mu}{c^5}a^4\omega^6$$

where $\mu=GM$, can be understood in simple terms. He asks " Come up with a better explanation of the physics contained in [the above] and share it with the author of this tutorial. He is always looking for better ways of understanding important results." How about writing it as

$$L=\frac{2}{5}\mathcal{T}\cdot v\left(\frac{v}{c}\right)^5?$$

The first term is the rate of energy loss: $GM^2/a^2\cdot v$, where $a$ is the separation between the binaries, and $v=a\omega$ is the angular speed, $\omega=2\pi/P$, with $P$ the period of the orbit. The last term is the aberrational factor---which is of fifth order! This says that gravitational waves aberrate just like electromagnetic waves---except weaker than because it is of fifth order.

Why they insist on the fifth order is because of strained analogy with the source of the radiation: a quadrupole. The quadrupole, $Q$, is proportional to the moment of inertia, $Mr^2$ and its third order derivative is $\ddot{Q}\sim Mr^2\omega^3$. Hence, the rate of energy loss is proportional to $|\ddot{Q}|^2$, and this is the connection with Einstein's calculation of a rotating dumbbell. Conservation of mass prohibits monopole radiation, conservation of momentum that of dipole radiation so the next best thing is quadrupole radiation. However, viewed as an aberrational effect there is nothing that cancels out the third order term, which is

$$L\sim\mathcal{T}\cdot v\left(\frac{v}{c}\right)^3=\frac{\mu}{a^2}a\omega\left(\frac{a\omega}{c}\right)^3.$$

By the same token, this term would be proportional to the square of the second time derivative of the dipole moment, $\mu a\omega^2$.

Setting this term equal to the surface luminosity would result in a temperature,

$$kT=\sqrt[4]{\frac{\mu}{a}(\hbar\omega)^3\left(\frac{a\omega}{c}\right)},$$

which would show a weak aberrational dependence. However, the analogy is flawed, and this has nothing to do with dipole radiation.

The luminosity is given by the thermal black-body formula

$$L=\sigma T^4 4\pi a^2,$$

where $\sigma=k^4/\hbar^3c^2$, leaving out inessential numerical factors. Equating the two expressions we derive an expression for the absolute temperature,

$$kT=\sqrt[4]{\frac{GM}{a}(\hbar\omega)^3\left(\frac{a\omega}{c}\right)^2}.$$

This is, indeed, surprising because it says that the temperature will undergo aberration, appearing as a squared term as if for a potential. First order aberration would also not work. But, for third-order we get

$$kT=\sqrt[4]{\frac{GM}{a}(\hbar\omega)^3},$$

which shows that temperature does not aberrate. It also tells you the composition of factor that make up the absolute temperature: one part gravitational and three parts electromagnetic. The latter can only be turned into the former by considering specific examples as we do below.

For black body cavity travelling at the speed $v$, the temperature increases by the factor $1/\sqrt{1-v^2/c^2}$, where $v$ is the translational speed of the cavity. If light is traveling at speed $c$ in a given direction, and the cavity at speed $v$ in another, then the decrease in velocity is proportional to the distance $c\sqrt{1-v^2/c^2}$, since $c$ is the hypotenuse of a right triangle with base $u$. This is not aberration.

Thermodynamically, the temperature can be increased by the motion because the heat absorbed increases by the same factor, but their ratio, the entropy, must remain invariant. Said plainly, disorder cannot be used to tell how fast the cavity is moving. But, as $v\rightarrow 0$, so the temperature returns to its stationary value. Not so if the temperature is affected by aberration.

Consider the Hertzprung--Russell diagram, where there is a known relation between luminosity and mass for binary stars. For the lower part of the curve, the mass $M$ increase from $0.1 M_{\circledcirc}$ to $M_{\circledcirc}$ as $L$ increases by a factor of $!0^3$. This means $L\sim M^3$ in this part of the curve. In contrast in the upper part of the curve $M$ changes by a factor of $2$ results in a change in $L$ by a factor of $20$. This means that $L\sim M^4$ for binary stars on this part of the curve.

Rather than considering the surface luminosity, $L$, we consider the total energy density (again omitting inessential numerical factors),

$$E=\frac{(kT)^4}{(\hbar c)^3}a^3,$$

and equate it with the energy

$$\frac{\mu}{a}\left(\frac{a\omega}{c}\right)^3,$$

which is the effect that aberration has due to a finite speed of propagation of electromagnetic waves

$$\frac{GM}{a}\psi,$$

where the angle

$$\psi=\theta-\tan^{-1}\theta=\theta-\theta+\frac{1}{3}\theta^3-\frac{1}{5}\theta^5+\cdots$$

and $\theta=a\omega/c$, since $a/c$ is the time it takes light to propagate a distance $a$. Recall that $\psi$ is the difference between where the planet is presently at orbiting a much larger mass, $M$, and where it would have been had it been linearly extrapolated from its past position. The first non-vanishing term is the cubic, not the fifth order as LIGO has assumed. But LIGO took their expression from Hulse and Taylor, and they took it from Peters and Mathew.

Equating the radiation energy to the gravitational one leads to a temperature of

$$kT=\sqrt[4]{\frac{\mu}{a}(\hbar\omega)^3},$$

without aberration, as it should be. The temperature is one quarter gravitational and three-quarters electromagnetic. For the binaries in the upper part of the HR curve, $\hbar\omega=GM/a$, while in the lower part of the curve, $\hbar\omega=(GM/a)^{2/3}$. This means that for all binaries on the main sequence the ratio of the mass to radius is constant and the core temperature is of the order $10^7 K$.

All the analyses use Keplerian dynamics, so why shouldn't we? Newton wrote the centrifugal force as

$$\mathcal{T}=\mathcal{F}\cos\vartheta=\mathcal{F}\sin\alpha=\frac{v^2}{\varrho},$$

where $\vartheta$ is the angle between the centripetal, $\mathcal{T}$, and centrally pointed force, $\mathcal{F}$, and $\alpha$ is the complementary angle. Since it is is given by

$$\sin\alpha=\frac{a\omega}{v_G},$$

Newton considered a first-order aberrational effect. $\varrho$ is the radius of curvature, and Newton found it to be $\varrho=a/\sin\alpha$. Hidden away is the fact that the radius of curvature depends upon the speed at which gravity propagates, $v_G$. For an infinite speed the curvature tends to zero.

If we use the above expression in conjunction with Kepler's II,

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

the conservation of angular momentum, which is impervious to the speed at which gravity propagates. Introducing the expression for $\sin\alpha$, converts Kepler's second into Kepler's third

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

In fact, Newton's law of inverse square,

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

implies $av^2=\mbox{const}$, which is just Kepler's III. Of course we can always multiply

$$\mathcal{F}\sin\alpha=\frac{\mu}{\varrho}$$

by any power of $\sin\alpha$ without disturbing the Keplerian nature of the problem. What it does is to change the expression for the centripetal force.

Consider the secular expression for the square of the semi-major axis,

$$a^2(t)=a_0^2+4\frac{\mu}{v_G}(t-t_0).$$

we can always arrange it that $a_0^2=4\mu t_0/v_G$, at the initial time. Taking the equation to the power $3/2$ and multiplying it by $\omega^2$, we get

$$\frac{v}{v_G}=\frac{1}{M},$$

where

$$M=\omega t=E-\varepsilon\sin E$$

is the mean anomaly, $E$ (not to be confused with total energy used above), the eccentric anomaly, and $\varepsilon$ the eccentricity. To verify the validity of the above expression for the mean anomaly, we differentiate the equation with respect to time we find

$$\frac{\dot{v}}{v_G}=-\frac{v}{a} v_G\left(\frac{v}{v_G}\right)^2=-\frac{\mu}{a^2}\frac{v}{v_G}=-\mathcal{F}\sin\alpha,$$

which is the first equality in Newton's relation, except for the minus sign, where we have used

$$\dot{E}=\sqrt{\frac{\mu}{a}}(1-e\cos E)^{-1}.$$

It is not hard to verify that Kepler's III is satisfied for **all** powers of the aberration, but the first non-vanishing term in the expansion of $\psi$.

That term has very simple and important properties with regard to luminosity. The secular equation is

$$a^3(t)=a_0^3+4\frac{\mu^2}{v_G^3}(t-t_0),$$

and again we cancel initial conditions. Taking the time derivative leads to

$$\dot{a}=\frac{4}{3v_G^3}\left(\frac{\mu}{a}\right)^2.$$

Using Kepler's III in the form

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

where $P$ is the period of the orbit, we come out with

$$\dot{P}=12\pi\psi,$$

where the angle $\psi$ is given by the cubic term in the above expansion. This says that the rate of change of the period is entirely due to the retarded action of the force which causes the motion and travels at a finite speed, $v_G$.

I hasten to add that, except for the connection with the luminosity expression and absolute temperature, the basic formulas and general conclusion that gravity does not aberrate and gravitational waves, if they exist, do not travel at speed $c$, can be found in van Flandern's paper, "The speed of gravity: What the experiments say". This paper was unduly criticized by Carlip and Kevin Brown. It will not due to replace the permittivity $\epsilon_0$ and permeability $\mu_0$ of free space by

$$\frac{1}{4\pi\epsilon_0}\rightarrow-G\hspace{20pt}\mbox{and}\hspace{20pt}\frac{\mu_0}{4\pi}\rightarrow-\frac{G}{c^2},$$

and use electromagnetism to derive the expression for the rate of decrease of energy for "gravitational waves". There is no conclusive experiment that warrants such a substitution. When comparing the observed and predicted periods and their rates of change of the binary pulsars PSR1913+16 and PSR1534+12, van Flandern wrote that "[a]t a glance, we see there is no possible match. The predicted period changes that would result if gravity propagated at the speed of light in a manner analogous to electromagnetic forces, are orders of magnitude larger than the observed period changes. For PSR1913+16, they have opposite signs as well. From PSR1534+12, we can set a lower limit to the speed of gravity as an electromagnetic-type propagating force $2800c$."

These words are as true as when they were uttered two decades ago. I wouldn't be concerned about the sign of the rate of change of the period for we obtain a sign change when considering Newton's relation from a different viewpoint. However, what we should be concerned about is that if we use the above expression for $\dot{P}$, the order of magnitude is twice that what has been observed, while if the fifth power

$$\dot{P}\sim\left(\frac{a\omega}{c}\right)^5,$$

is used by all the above authors, turns out to be seven powers smaller than what is observed,* if gravitational waves propagate at the speed of light.* There is nothing in between to fill the gap: aberration is always odd powers in the relative speeds. Even if the latter expression were correct, which it isn't, they would still have to explain why the cubic term is missing. That too satisfies Kepler's law, and even more important it is congruous with the black-body expression for luminosity, which the other expressions are not.

The connection between the third order aberration and the total radiation energy of black bodies is too good to be a mere coincidence. When dealing with rotational motion we note a peculiarity insofar as the index of refraction is proportional to the speed rather than inversely proportional as it it for the usual form of Snell's law. Here too, there is a peculiarity in that the mean anomaly, $M$, which would correspond to the angle $\theta$ in the above expansion of $\psi$ should be proportional to one another, and not inversely proportional to each other as it turns out. Recall that $\theta$ is the angle of the sector traced out by the planetary motion, from past to present positions.

As a final point, claims have been made that there is a miraculous cancellation of the first order aberration terms that was made by Carlip, "Aberration and the speed of gravity" and Ibison *et al, *"The speed of gravity revisited". They concluded that time-retardation only affects the magnitude of the force, but not the direction, of the field. However, there is one small point, they considered the time-retarded *radially directed *force, the electric field, analogous to $\mathcal{F}$, and, of course, this does not manifest aberration, as Newton well-knew.