Are Newton's gravity waves detectable by a laser interferometer?

Newton's theory of gravity supports "gravity waves" in that moving objects cause changing gravitational fields. For example, two bodies rotating around their center of mass will have a stronger gravitational field when they are longitudinally oriented than when they are transverse oriented. Given two masses of mass $M$ orbiting on a circle of radius $r$, at a distance $d$ from an observer, the strength of the attraction is: $$\begin{matrix} F_{min} &=& \frac{2GM}{d^2},\\ F_{max} &=& \frac{2GM}{d^2}\frac{1+r^2/d^2}{(1-r^2/d^2)^2} \end{matrix}$$

This should be detectable at long distance.

My question is this: With that sort of gravity, would the laser interferometer based gravity wave detectors be able to detect the gravity wave? An example is the LIGO, Laster Interferometer Gravitational-wave Observatory.

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You're asking purely about a Newtonian treatment, not the full GR analysis, right? –  David Z Apr 7 '11 at 2:55
Dear Carl, there are no "waves" in Newtonian gravity. A "wave" is something that has shape - minima and maxima - that depends on space. There are no oscillations like that, as a function of space, in Newtonian gravity. The gravitational force is uniformly decreasing away from the source. This totally changes the situation. You apparently want to measure the time depedendence of the quadrupole moment - but it's very tiny, a hopeless power law. You can't measure "tides" from distant galaxies or binary stars. Gravity waves, which only exist in GR, have effects that drop slower with the distance. –  Luboš Motl Apr 7 '11 at 5:32
@Lubos; Instead of "waves" call it what you like. My question is whether it's detectable at LIGO. –  Carl Brannen Apr 8 '11 at 20:46

This is exactly the approach taken in Bernard Shutz's note "Gravitational waves on the back of an envelope" (Am. J. Phys. 50 vol 5 pp 412). The abstract reads:

Using only Newtonian gravity and a little special relativity we calculate most of the important effects of gravitational radiation, with results very close to the predictions of full general relativity theory. Used with care, this approach gives helpful back‐of‐the‐envelope derivations of important equations and estimates, and it can help to teach gravitational wave phenomena to undergraduates and others not expert in general relativity. We use it to derive the following: the quadrupole approximation for the amplitude h of gravitational waves; a simple upper bound on h in terms of the Newtonian gravitational field of the source; the energy flux in the waves, the luminosity of the source (called the ‘‘quadrupole formula’’), and the radiation reaction in the source; order‐of‐magnitude estimates for radiation from supernovae and binary star systems; and the rate of change of the orbital period of the binary pulsar system. Where our simple results differ from those of general relativity we quote the relativistic ones as well. We finish with a derivation of the principles of detecting gravitational waves, and we discuss the principal types of detectors under construction and the major limitations on their sensitivity.

(If you don't have access to Am J Phys, this talk seems to recapitulate the details.)

A major difference in this Newtonian scalar theory from the real GR theory of gravitational waves is in the effect of the waves on an inertial test particle. This Newtonian theory predicts that the waves would appear as an oscillating force along the direction between the source and the test particle. By contrast, GR predicts an oscillating differential tidal effect in the plane perpendicular to the line connecting the source and the test mass.

As a result, while I think LIGO would still detect the "Newtonian" form of gravitational waves, the antenna pattern of the detector would be different. The L-shaped LIGO detector has optimal sensitivity to a source located directly overhead in the GR case (allowing the gravitational wave to stretch one arm while it is compressing the other). There would be no sensitivity to a "Newtonian" source directly overhead. However, you could detect it if the "Newtonian" source were aligned with either arm.

By the way, "Newtonian noise" (the near-field action of Newtonian gravity arising from density waves in the material near the detector) is a real concern for terrestrial gravitational wave detectors!

P.S. To be pedantic, it is best to avoid the term "gravity wave" (as opposed to "gravitational wave"), since a "gravity wave" ("Newtonian gravity wave" even!) is something completely different.

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Thanks for posting something referenced! –  Sklivvz Apr 7 '11 at 18:10

I don't think so. To first order, the maximum variation in your Newton's gravity force is $$F_{min}-F_{max}=\frac{6GMmr^2}{d^4}$$

If we assume that the effects of gravity propagate at light speed, then in order for one mirror to experience $F_{max}$ while the other experiences $F_{min}$, the objects must rotate 90 degrees in the time $l/c$ where $l$ is the path length of the interferometer (4 kilometers for LIGO).

If we choose the speed of light as an upper bound for the speed of the rotating objects, this means that they can be separated by no more than about 5 kilometers ($r=2.5$).

A lower bound on $d$ can be 4 lightyears, the distance to the nearest star.

Let's suppose the mass of a mirror is $m = 100 kg$, and the mass of the rotating objects is 20 billion solar masses, the largest known mass in the universe.

Punching this into google calculator gives a force difference of about $10^{-28}$N, which is 4 orders of magnitude smaller than any force yet measured - and of course most of these assumptions are unphysically generous.

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It's newtonian gravity, so there is no propagation speed. The effect is instantaneous. That and the falling off as $1/d^4$ like you mention, makes this not very "gravitational radiation" like in my opinion. –  John Apr 7 '11 at 3:06
yeah, I wasn't sure about that.. I was thinking you could maybe treat it in a semi-relativistic manner by assuming a propagation speed of c, kind of like basic relativistic E&M and retarded potentials. There's probably lots of problems with this approach, but I think it's less wrong than assuming an instantaneous propagation which clearly violates special relativity. –  user2963 Apr 7 '11 at 3:24
The main problem is that there's no wave equation for the waves to satisfy. You can do something ad-hoc like applying retarded potentials, but that makes sense in Maxwell theory because retarded and advanced potentials solve Maxwell's equations. What is the "Newtonian" wave equation? –  Jerry Schirmer Apr 7 '11 at 4:59
The operation of LIGO is actually more easily understood in the long wavelength approximation, in which the gravitational wave strain changes very slowly compared to the time needed for light to circulate in the interferometer arms. It is not necessary to arrange for one mirror (of the two in an arm) to see a maximum of the wave while the other sees a minimum, since we are looking for a tidal effect. –  nibot Apr 7 '11 at 16:17

This does not imply gravity waves. This can be seen by the following calculation. The mass configuration is that of two masses in a circular orbit around each other. This is a dipole term, which defines ${\vec p}~=~m{\vec d}$, for $\vec d$ the distance between the two masses. So this dipole moment projected along the field determines a dipole potential $$\phi_d(r)~=~{\vec p}\cdot\nabla\frac{GMm}{r}$$ If this generates power in waves that escape the system, then a time derivative is dependent on $md{\vec d}/dt$. However, momentum conservation tells us this must be zero in the initial frame of the two masses.

So in order for gravity waves to exist one must have quadrupole moments and higher. However, this contributes to gravity waves in the post Newtonian expansion to second order. The

A gravity wave is a perturbation on a background metric $\eta_{ab}$ with the total metric $$g_{ab}~=~\eta_{ab}~+~h_{ab}.$$ The flat background metric has zero Ricci curvature so that to first order in the perturbation expansion $R_{ab}~=~\delta R_{ab}$, which enters into the Einstein field equation $R_{ab}~-~\frac{1}{2}Rg_{ab}~=~\kappa T_{ab}$, where $\kappa~=~8\pi G/c^4$ is the very small coupling constant between the momentum-energy source and the spacetime configuration or field. The Ricci curvature to first order is then $$R_{ab}~=~{1\over 2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\partial_b{h^c}_a~-~\partial_a\partial_bh)~-~\partial_c\partial^ch_{ab}\Big).$$ The harmonic gauge $g^{bc}\Gamma^a_{bc}~=~0$, to first order as $\partial_c{h^c}_a~=~\frac{1}{2}\partial_\mu h$ the Einstein field equation gives $$\partial^c\partial_ch_{ab}~-~\frac{1}{2}\eta_{ab}\partial^c\partial_ch~=~{{16\pi G}\over {c^4}}T_{ab},$$ which is well defined for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~\frac{1}{2}\eta_{ab}h$ with the simple wave equation $$\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over {c^4}}T_{ab}.$$ This is then a basic wave equation for a tensor field with two independent components $h_{\times\times}$ and $h_{++}$ for polarization directions.

What is of interest is the source of the gravity waves $T_{ab}$. The metric perturbation far from the source $r~=~|{\bf x}|~>>~|{\bf y}|$ is approximated by, $${\bar h}_{ab}~=~{{4G}\over{rc^4}}\int_V d^3yT_{ab}(t~-~r/c,~{\bf y}).$$ The covariant constancy of the momentum-energy tensor $\nabla\cdot {\bf T}~=~0$ results in a number of relationships. The energy $T_{tt}~=~\rho U_0U_0$ component of with these relationships gives that for the quadrupole moment $I_{ij}~=~\int_Vx_ix_j\rho(t,~{\bf x})$ the metric wave is given by $${\bar h}_{ij}~=~{{2G}\over {rc^4}}{{d^2}\over{dt^2}}I_{ij}(t,~{\bf x}).$$ This term is different from a naïve idea of gravity waves suggested here. The coupling constant is $~\sim~G/c^4$ which comes from the Einstein field equation and is extremely small. The post Newtonian terms to $c^{-4}$ are second order, and this is a departure from a Newtonian gravity.

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Isn't that one has to content with quadrupole moment, because mass dipoles don't exist? (Whether antimatter/matter might make up such a mass dipole I dont know.) Black holes circling each other are surmised to emit gravity waves. What is the difference to Carls masses? –  Georg Apr 7 '11 at 13:56