Gravitational waves were discovered 35 years ago without fanfare in 1981/2 by Zumberge, R L Rinker and J E Faller, then completely ignored.

See: "A Portable Apparatus for Absolute Measurements of the Earth's Gravity", M A Zumberge, R L Rinker and J E Faller, Metrologia, Volume 18, Number 3, http://iopscience.iop.org/article/10.1088/0026-1394/18/3/006/meta

The variation in g over the course of a day due to the sun and moon was carefully measured from the Earth's surface in 1981 as shown below.

Strain measurements taken in 1981 of local planetary effects (sun, moon) on earth's gravitation clearly showing a gravitational wave.

This is the SAME gravitational wave effect measured by the LIGO researches recently (reported 11Feb2016).

LIGO actually detects, then filters out, this local gravitational wave in order to detect the remote ones producing the ultra weak gravitational waves from binary black holes. Although the sensitivity required to detect them is 3 orders of magnitude higher in both frequency and amplitude, the LIGO "gravitational" waves are otherwise exactly the same "gravitational" waves already discovered in 1981 in our own solar system.

The proof is in the fact that LIGO detects gravitational waves but can NOT detect "tidal" gravity waves. Thus categorization of Zumberge's waves as "tidal" gravity is incorrect as tidal waves are those between two surfaces as a consequence of gravity. These cannot be detected by an interferometer. Zumberge's measurements are of gravitational variations itself, hence gravitational.

Why has the 1981 work of Zumberge, Rinker and Faller been ignored?

(See also Leading-order cause of diurnal (not semidiurnal) variations in $g$?)

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Apr 22 '16 at 17:28

11 Answers 11


This represents a major misunderstanding of what a gravitational wave is. The effect presented is simply the semi-static gravitational field at earth due to the earth, moon and sun. It is predicted by Newtonian gravity. There is no 'wave' that propagated, it's the instant positions of the 3 bodies that change over 1 day (and over 1 year also).

It does not show that the change moved at the speed of light, which gravitational waves do. Nothing in Newton's equations talk about the speed of light. The GR equations for 3 bodies moving like the earth-sun-moon can only be solved approximately, and in this case it'd be through a post-Newtonian approximation. The pseudo-static term(s) would be the same but possibly some GR correction - and if it is (And I'm not sure if the strongest term correction might not be something like the term for the perihelion of mercury, or something else, in any case extremely small and not measurable in their g measurement). But that's not even a grav wave. The grav waves would be even smaller probably - you'd have to compute the rate of change of the quadrupole moment of the configuration, and do some other calculations. The simpler problem of just the grav radiation of the earth-sun rotation around each other gives a resultant power dissipated that translates in the orbit of the earth loosing altitude ('altitude' above the sun) of the size of 1 proton per day. That g change they measured in your graph is about 10 to the minus 7 g's. It isn't even dissipative, as the bodies keep doing the same thing over and over, in your approximation. If you don't see that dissipation you are not seeing the gravitational waves.

There is probably many other ways to see that what you're discussing, what the graphic measured, is not a gravitational wave, but rather a very slow change in a static gravity field, the one produced by the 3 bodies.

Grav waves produce something different than just a change in gravity in one direction, they do it in 2 directions at once, an asymmetrical squeezing of a circle first in one axis and then in the other, like squeezing a balloon in one direction, making it bulge in the other.

Like Nathaniel said, it's like comparing a (semi) static electric field (say produced by rubbing a couple rags together) and moving them around some, with light.

Note: yes, even changing static fields can not produce a change in what's observed at a distance faster than the speed of light, but that doesn't come in at all in your graphic, too small a differential effect for it to see it.

  • $\begingroup$ Please look at this recently published video (21Apr) by Nottingham University before making ill informed judgements as to the merits of my question. Especially the comments about gravitational wavelets in our solar system. --> youtu.be/hIgZG7A1fqc $\endgroup$ – Sandgroper Apr 27 '16 at 8:17
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    $\begingroup$ Did you even watch that video?! Ed Copeland says, "The tides don't exist because of the nature of gravitational waves." He explains that "they're different" phenomena, and says that the gravitational waves are "minuscule" compared to the tides. That's just what everyone here has been trying to tell you, including Bob Bee. $\endgroup$ – Mike Apr 27 '16 at 13:49

[Note: I work on gravitational waves, and am an author on several of the recent LIGO papers on GW150914 — though I am not a member of the LIGO collaboration. So if you're looking for conspiracy theories, that can be your reason to ignore me.]

Zumberge, Rinker, and Faller (ZRF) did not measure gravitational waves. To explain this, I'll start off with an analogy, then discuss gravitational waves directly, followed by the actual math from Einstein's equations that shows very explicitly what I'm talking about.

But first, I'll briefly point out the most obvious and (to me) persuasive point: ZRF found variations at about twice the frequency of the Earth's rotation — consistent with the usual tides, rather than gravitational waves. But Einstein's theory tells us that gravitational waves will only be produced strongly by the orbit of a pair of objects. The moon is orbiting earth on a much slower timescale, so the frequency of gravitational waves from its orbit will be roughly 27 times lower. The earth is orbiting on an even slower timescale, so the frequency of gravitational waves from its orbit will be roughly 365 times lower. Thus, ZRF's data are not measuring the dynamics spacetime effects that Einstein called gravitational waves.


The first point to understand is that variations are not the same things as waves. As I drive across hilly roads, my car's altitude varies up and down, but the hills are stationary; I haven't discovered seismic waves. You can argue that seismic waves would also show up in the data, just at a much smaller level and with different frequencies. That's true in principle, but my data are not precise enough to measure those quantities. My results are consistent with both the presence and absence of seismic waves. And more importantly, the point is that seismic waves are motion of the ground itself, whereas the plot of my car's altitude just shows how I moved across terrain that — for all I know — was perfectly stationary.

Gravitational waves

In exactly the same way, gravitational waves are disturbances of spacetime itself, rather than variations I measure as I move across a (nearly) static spacetime. The variations ZRF saw were consistent with a nearly static gravitational field — no waves. In particular, what they saw was consistent with both Newtonian gravity and Einstein's General Relativity, while gravitational waves are only consistent with General Relativity. Rather than the gravitational field changing in time at a particular place (which is what gravitational waves cause), ZRF's measuring device was moving through that field, because the device was moving along with the earth's surface through the field. This is why they saw variations in time, but those variations were not gravitational waves. [More technically, the OP seems to be confusing an advective derivative and a partial derivative.] The ZRF measurement looks like a very nice result, but it's just a different phenomenon.

To be more specific, gravitational waves are not just disturbances in spacetime itself; they are disturbances that propagate as waves at the speed of light. We know that the GW150914 signal moved — at least nearly — at the speed of light, because it was measured in one detector 7 milliseconds after the other. And since the detectors are located 10 milli-light-seconds apart, that means the signal traveled at least nearly as fast as light. [We believe that the signal traveled at the speed of light; we attribute the difference to the plane nature of these gravitational waves and the fact that the waves weren't going straight from one detector to the other. See here for an explanation.]

Math from Einstein's Equations

So now let's look at the key mathematics underpinning this definition. This is discussed in a section on wikipedia (which I think I wrote, actually). The basic point is that the gravitational field is described by the quantity $\bar{h}^{\alpha\beta}$, which obeys the equation \begin{equation} \tag{1} \frac{1}{c^2} \frac{\partial^2} {\partial t^2} \bar{h}^{\alpha\beta} = \nabla^2 \bar{h}^{\alpha\beta}. \end{equation} (At least this is true in empty space, and ignoring nonlinearities.) If you've taken a college-level physics course, you should recognize this equation as a basic wave equation. The left-hand side is measuring the rate of change in time at a particular place, while the right-hand side is measuring the rate of change as you move around in space at a particular time. As seen on that wiki page, this equation is derived directly from Einstein's equations. In particular, Einstein predicted equation (1), with $c$ as the speed of light. It turns out that Newton's theory of gravity assumes that the field propagates infinitely fast: $c \to \infty$, which is equivalent to $1/c^2 \to 0$. This means that in Newtonian gravity, you can ignore the left-hand side, which is the rate of change of the field at a particular place with respect to time. And indeed it turns out that the gravity of the solar system is pretty well described by that simpler Newtonian equation \begin{equation} \tag{2} \nabla^2 \bar{h}^{\alpha\beta} = 0. \end{equation} In fact, this is essentially the same as the well known Poisson's equation as it appears in Newtonian gravity. Now, $\bar{h}^{\alpha \beta}$ is still allowed to change in time, because the planets and moons (and sun) can move, but the time-dependence of $\bar{h}^{\alpha \beta}$ simply doesn't come into this equation.

So we can look at the field that ZRF were measuring. Their data are well modeled [as Floris showed beautifully here] by a fairly simple field that is pretty well specified as \begin{equation} \tag{3} \bar{h}^{00} = 2 \frac{G_\mathrm{N}} {c^2} \left( \frac{M_\mathrm{Earth}} {d_\mathrm{Earth}} + \frac{M_\mathrm{Moon}} {d_\mathrm{Moon}} + \frac{M_\mathrm{Sun}} {d_\mathrm{Sun}} \right), \end{equation} where $d_\mathrm{Earth}$ is the distance from the center of the Earth to the specified point, etc. [Floris's model is more sophisticated than this, with its Love numbers and whatnot, but this is enough to get the point across.] Now, this formula already satisfies equation (2) — Newton's theory — but it also very nearly satisfies equation (1), because the only time dependence involves how quickly the earth, moon, or sun are accelerating towards or away from the point in question — which are all very small numbers. Then, that's divided by $c^2$, which is a huge number, so the left-hand side of equation (1) is basically zero anyway. Plugging in some numbers I get at most \begin{equation} \frac{1}{c^2} \frac{\partial^2} {\partial t^2} \bar{h}^{00} \approx 10^{-30} \mathrm{m}^{-2}. \end{equation} So to satisfy Einstein's equation (1), you just have to add something to equation (3) to balance this out, which would be a really tiny quantity — well within the apparent errors in ZRF's data. So here's the key point: The ZRF data isn't accurate enough to tell us whether Einstein's equation (1) or Newton's equation (2) is a better description of reality. And Newton's version of gravity doesn't account for gravitational waves, which means that ZRF could not measure gravitational waves.

Gravitational waves are changes in the gravitational field that don't satisfy equation (2); they really require $c$ to be finite, and really require the full form of equation (1). The reason that GW150914 is important is because it is the first direct measurement of a gravitational field where equation (2) really is not enough; for GW150914, $c$ must be finite, and equation (2) is simply wrong. Even the gravitational field of a single black hole doesn't vary in time, so that $\partial^2/\partial t^2$ side is just $0$ [though there are things I left out of equation (1) that come in]. Seeing the dynamic (time-dependent) behavior of the gravitational field in this form is a big deal. ZRF did not do this; their data were consistent with both finite $c$ and with $c=\infty$.


Finally, if you still don't believe me, here's a sociological argument: Just look at the paper by Zumberge, Rinker, and Faller as cited by the OP. The only time the word "wave" is used by ZRF is when referring to the wavelength of their laser. Gravitational waves were predicted in 1916 and the topic of active research at the time ZRF published. So not only are they not being given credit for "discovering gravitational waves", they themselves did not claim to have done so — presumably because they knew that they had not. The effect that ZRF measured was varying in time primarily because the location of their measuring device was changing. Although the relative positions of the sun, earth, and moon were also changing, that was on a much slower time scale, such that what they saw was consistent with Newtonian gravity, and was not a detection of gravitational waves.

You can disagree about the definition of the phrase "gravitational waves" with ZRF, LIGO, the rest of the physics community, and everyone here on stackexchange all you want, but what matters ultimately is that LIGO has found something truly unique in the history of science.

  • $\begingroup$ @Mike thank you for this contribution, which is certainly detailed. However the assumption you (and Zumberge et al) make hinges on the Zumburge measurements being incapable of measuring the speed of propagation (Z waves varied in time), which I have already acknowledged. This does not mean the type of wave they were measuring was tidal (Newtonian) not gravitational (Einstein). To use your metaphor, the analogy I put to you is one of measuring continental drift versus seismic waves. As already stated, the physics of both systems are the same - planetary orbits versus binary black hole orbits. $\endgroup$ – Sandgroper Apr 23 '16 at 2:20
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    $\begingroup$ @Sandgroper The point is that ZRF's measurements were consistent with both Newton and Einstein. While we believe that the correct theory is Einstein's, the ZRF experiment provided no evidence to support that particular belief. You can analyze their experiment in either theory, and get the observed result. Furthermore, in neither Newton's nor Einstein's theory were the variations that they measured waves; there quite simply are no waves in ZRF — just variations. $\endgroup$ – Mike Apr 23 '16 at 2:59
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    $\begingroup$ Importantly - unless you measure at two distinct points, and demonstrate a phase shift, you cannot distinguish propagating waves from static fields. ZRF was a single point measurement. $\endgroup$ – Floris Apr 23 '16 at 3:00
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    $\begingroup$ @Mike the link you give in the comments to my answer is important. In the conclusion it says: "The differential part of the tidal deformation resembles a gravitational-wave signal and must be removed by actuating the mirrors ". Maybe you should incorporate it in your answer. $\endgroup$ – anna v Apr 23 '16 at 3:56
  • $\begingroup$ @mike thank you for the link and I have read the conclusion. Again, I appear not to be getting my point across. In a previous comment that was deleted, I noted that Pro David Blair confirmed "LIGO measures micron sized tidal variations but these are not gravitational waves." I disagree. LIGO filters out the local gravitational wave generated by our solar system with the tidal gravity wave due to the earths crust. The gravitational wave has peaks and troughs at odd places for various reasons but nevertheless propagates at the speed of light with a very long wavelength and days per cycle. $\endgroup$ – Sandgroper Apr 23 '16 at 13:56

This is the SAME gravitational wave effect measured by the LIGO researches recently

It is not. Bob already gave a nice answer, but I would like to add a couple of layman analogies.

Picture a pond. With simple instruments you can measure the water level over seasons and you will get some wavy trend like low in summer, high in winter, however this is just the water trend. Measuring water waves means being able to detect the tiny ripples on the surface when a drop falls into the pond, and this is a totally different story.

Considering air instead of water, the analogy consists in detecting the atmospheric pressure, easily done with a barometer, compared with the recording of sound, for which you need at least a phonograph. Note that between these two instruments there are more than two centuries of physics and engineering research!

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Apr 26 '16 at 12:11

I would like to add a description on the difference between LIGO and the experiment in the question by going into the quantum framework, accepting that the effective quantization used in cosmological problems is at work.

The comparison of gravity with electric and magnetic fields has been mentioned in the answers and the comments. The experiment in the question studies gravitostatics, analogous to electrostatics and magnetostatics.

The quantization framework of electrodynamics allows static electric and magnetic fields to be described by the exchanges of virtual photons. There is an interesting answer by Lubos Motl to the question "Virtual photon description of B and E fields" which one can extend analogously for the gravitational field as a confluence of virtual gravitons.

So the difference in the experiments is that LIGO is studying a beam of on shell gravitons from which the gravitational wave emerges, in the same way that optical experiments measure a beam of on shell photons from which the elecromagnetic field emerges.

The experiment in the question studies the equivalent of the electric (or magnetic) field, at best with virtual graviton interactions.

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    $\begingroup$ No, it is not. Both measure gravitational effects, but gravitational waves have on mass shell gravitons, i.e. elementary particles, the way electromagnetic waves have on mass shell photons. For example, it is the em waves that generate the photoelectric effect, not the electric field of a capacitor no matter how much it changes, because the photons between the plates of a capacitor are not on mass shell, not real . $\endgroup$ – anna v Apr 22 '16 at 12:11
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    $\begingroup$ to continue on this line: put an AC accross a capacitor. Will it make a good antenna radiating an EM wave? No , a very small part of the power may be radiated as em wav, but the changing electric field is just a changing electric field, i.e. a varying potential difference, with the frequency of the AC but no wave. In similar fashion, and much worse because the gravitational constant is so tiny, the asymmetrically changing mass distributions will radiate a tiny amount of gravitational waves away, but most of the power is in the gravitational field which is varying a potential difference. $\endgroup$ – anna v Apr 22 '16 at 12:24
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    $\begingroup$ Never heard of "tidal gravity waves", but LIGO does detect tides, as detailed in — just to take one example — this technical document from LIGO $\endgroup$ – Mike Apr 23 '16 at 3:13
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    $\begingroup$ Look, I think you completely confuse gravity_waves with gravitational waves. As I say, i believe that gravity will be quantized, and the difference between the two for me is clear as I describ above. Wiki for gravitational waves makes it clear "In certain circumstances, accelerating objects generate changes in this curvature, which propagate outwards at the speed of light in a wave-like manner. These propagating phenomena are known as gravitational waves." The experiment you claim measures gravitational waves measures only changes in gravitational potential energy. The assymmetric part $\endgroup$ – anna v Apr 23 '16 at 13:41
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    $\begingroup$ of the masses involved in the changes measured in the Z experiment, will generate very weak gravitational waves, but these are not what have been measured by Z. The equivalent for LIGO would be if one had detectors where the two holes were merging and was measuring the changes in the gravitational potential during the merge , they would be large and would not fit a newtonian model, whereas the Z is quite consistent with a newtonian model, so could not be a discovery. $\endgroup$ – anna v Apr 23 '16 at 13:47

A very easy way to see that this is not the same phenomenon as the gravitational waves predicted by GR is to consider the energy of the system.

To that end, consider a purely Newtonian Earth/Moon/Sun (or just Earth/Sun if you are put off by the three-body problem) system. Let each of the bodies be perfectly elastic, so no energy is lost through tidal effects (equivalently let each body be rigid, or a point particle: we just need to prevent the system being dissipative through friction).

Two things are now clear.

  1. This system does not radiate energy: at least in the two body case it will continue exist forever, as we can find an exact solution (for the rigid body / point mass case anyway) which shows that it will. In the three-body case it probably will as well: certainly it will not fail to do so because it has radiated away energy.

  2. A measurement of the local value of $g$ on the surface of the bodies will show variation, as above.

From these two things it is clear that energy is not being carried away from the system via this phenomena. In other words this phenomena is not a gravitational wave.

This is entirely different than the case in GR, where gravitational waves carry energy away from the system, causing an inspiral of the bodies and their eventual collision.

Another way of looking at this is to consider a two-body, Newtonian system, where both of the bodies are rigid spheres (this is an idealisation). The two bodies are orbiting each other, and we'll have one of them (or both, it does not matter) rotate as well, with a period much smaller than the period of the overall system but about the same axis (this is meant to be a rudimentary model of the Earth-Moon or Earth-Sun system). Now consider a device which measures $g$ on the rotating body: this device will clearly measure a variation in the value of $g$ which has angular frequency $\omega_r \pm \omega_o$ where $\omega_r$ is the angular frequency of the spinning planet and $\omega_o$ is the angular frequency of the whole system. And you can spin the planet as fast as you like, making this frequency as high as you like, but, quite clearly, this does not affect any measurement not made by a device attached to the spinning planet: the entire measurement is an artifact.

A final approach is to consider the power radiated from the Earth-Moon system. There's an approximation for this:

$$P = \frac{32 G^4}{5 c^5}\frac{(m_1 m_2)^2(m_1+m_2)}{r^5}$$

And you can plug in the various values for the masses and orbital radius, and you get a radiated power of about $7\times10^{-6}\mathrm{W}$. This is the power radiated by the whole system: it is simply not plausible that a detector could be built which was sensitive to respond to whatever tiny fraction of this it intercepts.

(The Earth/Sun system radiates about $200\mathrm{W}$.)

  • $\begingroup$ Sorry, @tfb, but the question that needs to be answered is how can LIGO detect the Zumburge waves if LIGO cannot detect tidal gravity waves? LIGO only detects gravitational waves and noise, and Zumburge waves are detected by LIGO. Therefore Zumberge waves are gravitational waves. $\endgroup$ – Sandgroper Apr 23 '16 at 1:12
  • $\begingroup$ This also reminds me of Fermat's last theorem. He claimed to have a simple solution but by the time he completed the simple case of n=4 it was patently obvious it was not going to be simple. Likewise, the above explanation is certainly not simple. Indeed it is very much a case of - ignoranus ignoratum - explaining the obscure with something even more obscure Please simply explain why LIGO is able to detect the Zumburge waves if LIGO cannot detect tidal gravity waves? LIGO only detects gravitational waves and noise, and Zumburge waves are detected by LIGO. $\endgroup$ – Sandgroper Apr 23 '16 at 1:44
  • $\begingroup$ @Sandgroper I have no idea whether LIGO is sensitive to variations in $g$, although it could well be. Pendulum clocks, which are sensitive to variations in $g$, are sensitive to temperature and (if they run in air) barometric pressure as well: this does not make them insensitive to $g$, it just means you need to compensate them. $\endgroup$ – tfb Apr 23 '16 at 9:03
  • $\begingroup$ wrote <quote>A very easy way to see that this is not the same phenomenon as the gravitational waves predicted by GR is to consider the energy of the system.</quote> Yes this is a very low energy wave, no doubt about that. So low that a LIGO detection method would never pick it up because its tuned to higher frequency ~300hz signals and removes anything less in the seconds per cycle bands. Thats the beauty of this. The Z technique IS capable of measuring the extremely long very low energy propagating lunar gravitational wave. $\endgroup$ – Sandgroper Apr 25 '16 at 14:42
  • $\begingroup$ ^"The Z technique IS capable of measuring the extremely long very low energy propagating lunar gravitational wave." Why are you saying this? What evidence do you have to support this statement? We need science, not just claims. $\endgroup$ – Mike Apr 25 '16 at 17:37

Detectable gravitational waves can have a frequency from 10^-7 Hx to 10^11 Hz. Frequency is largely irrelevant to whether they are considered waves.

I find it best in discussions to first be sure we all use the same term to describe the same thing. So I will at first avoid the term "wave", in order to best define it.

What matters is that they are propagating vibrations.

So let us consider, for simplicity, our earth-moon system, in the simple 2-dimensional "rubber sheet" model, with all other nearby bodies ignored. The planets are heavyish frictionless disks.

We consider them, initially, at rest, not rotating about each other. Each makes a dent or dimple in the sheet which reaches the other. In fact, since both are within eachother's dimple, we end up with the sum of the two being a sort of hourglass-shaped dimple.

Now we mark several points, equally spaced, on the perimeter of the "earth" disk. We place a "rubber stretch detector" at each marked point, and naturally, it detects a little more stretch in the direction of the "moon" disk, with more stretch the deeper that detector is within the dimple from the moon disk.

We start moving the two disks slowly about each other. The stretch-detectors at the points we marked will detect a "waveform" - which, for clarity I will call a "tide" - of increasing and then decreasing gravitational power.

The speed of the ups and downs in our marked points' measurement of this tide depend entirely on the speed we make the disks orbit around each other.

More importantly, the speed that the tide travels between the different detectors, also depends entirely on the speed with which we move the disks.

This is because the dimple from the "moon" plate stays the same, it's just that as the two plates orbit each other, the measuring point moves deeper into or farther out of the dimple caused by the moon plate. Because the orbits are circularish, the changes in measurements at each point happens to look like a waveform.

Next we drop a marble many feet away on the sheet. Ripples will spread out from it. They are traveling at the propagation-speed of the sheet ("light speed", if you will), and are not static to a dimple of the marble. They will eventually pass the detectors on our earth plate and be measured as a wave-shaped ripple. Another thing to note about these ripples is that they rise up out of the sheet as well as sinking into it: they are bidirectional.

Place a vibrator on the sheet some distance away. Again, our detectors measure ripples, not static dimples, moving at the propagation speed of the sheet.

It is hopefully intuitively obvious to us all that ripples are small, propagating bidirectional deformations, while tides are large, relatively static, unidirectional deformations.

The term "gravitational wave" refers only to the propagating ripples, not the static tide.

We can now address the specific claims made in the OP.

This is the SAME gravitational wave effect measured by the LIGO researches recently

This is not a gravitational wave, this is a measurement of tides caused by the movement of a single detector within the earth-moon dimple. If there had been a second detector, it would have been clear that these tides do not propagate across the earth at the speed of light, but at the speed of the movement of the moon.

(reported 11Feb2016).

Without a specific link to the claims you are debating, my response (and everyone else's) can only be guesswork. It's entirely possible that the operators of LIGO have gone bonkers and are reporting tidal effects as gravitational waves or something, but since we don't have access to this, we instead have to assume they did not.

LIGO actually detects, then filters out, this local gravitational wave

This uses the term "gravitational wave" incorrectly. It is more correct to say, "LIGO actually detects, then filters out, this local gravitational tidal noise."

in order to detect the remote ones producing the ultra weak gravitational waves from binary black holes. Although the sensitivity required to detect them is 3 orders of magnitude higher in both frequency and amplitude,

All the above appears correct.

the LIGO "gravitational" waves are otherwise exactly the same "gravitational" waves already discovered in 1981 in our own solar system.

The above is correct only in given the flawed definition of "gravitational wave" corrected already above.

It is more correct to say "The LIGO-filtered-out gravitational tidal noise is indeed the same gravitational tidal signal that was measured in 1981, and discovered in 1687 when Newton discovered that the moon was the cause of tides."

The proof is in the fact that LIGO detects gravitational waves but can NOT detect "tidal" gravity waves.

This is technically true: it can't detect them because it filters them out, as stated in the very first sentence I quoted above.

Thus categorization of Zumberge's waves as "tidal" gravity is incorrect

Nope, that's exactly what was being measured.

as tidal waves are those between two surfaces as a consequence of gravity.

More correctly, "tidal gravitation effects are those measured at a point as a mass moves about that point."

These cannot be detected by an interferometer.

Interferometers are explicitly designed to detect stretching of the "rubber sheet". The moon's gravity well stretches the rubber sheet. The movement of the moon about the earth moves its gravity well relative to the earth, causing tidal effects where the rubber sheet gets stretched more or less.

So the above is blatantly false. It would be more accurately written: "These must be very accurately filtered out of the interferometer measurements in order for the far smaller effects of gravitational waves to be detectable."

Zumberge's measurements are of gravitational variations itself, hence gravitational.

Yes. Changes in the strength of gravity as felt on the surface of the earth, due to the tidal forces of the moon moving around it.

Why has the 1981 work of Zumberge, Rinker and Faller been ignored?

Because in the paper you linked, they were just saying "Hey, we've developed this much smaller apparatus for measuring these tidal effects we've known about for thousands of years. It's portable, check it out." They are not claiming to have discovered any new gravitational effect; they are just showing the design of a better piece of equipment for measuring previously known effects.

Their graph was displayed without fanfare because everyone already knows about tides. The interest of their apparatus is that it lets one (after canceling out tidal effects) measure subtle variations in gravity due to geological effects.

So, corrected for errors in definition of "gravitational wave", and understanding of how interferometers work, the OP contains no unusual claims, but the question itself then disappears.

Some people invented a cool portable gizmo for measuring the strength of gravity at any point on earth. As many had before them, they then measured this over time, and drew a graph of the tidal effects over a period of a few days. That's all.

  • $\begingroup$ The earth LIGO is built on moves with tidal forces and LIGO removes this non propagating tidal gravity wave signal. In the process it also removes the propagating gravitational wave that gives rise to the tidal gravity wave. Zumberge waves also contain this non propagating wave component due to the rotation of the earth and also the more subtle propagating gravitational wave due to the lunar cycle. This gravitational wave can be detected tide free at the poles and modulates very slowly but still propagates at the speed of light. The Zumberge measurement is of a gravitational wave. $\endgroup$ – Sandgroper Apr 24 '16 at 0:58
  • $\begingroup$ @Sandgroper You found the link you didn't give, then? What is it? If not, from where are you pulling these claims? Because there's nothing in anything you've linked to which claims anything as outlandish as this. [edit: note I say "outlandish" rather than "nonsense". When relativity comes into play, things DO get outlandish. So you may be 100% right. But I can't say, unless I'm looking at the same sources that you are.] $\endgroup$ – Dewi Morgan Apr 24 '16 at 2:17
  • $\begingroup$ Thank you for your insights. I have answered my own question (see below). While we may not necessarily have communicated effectively I will however absorb your comments (and others) in ascertaining if it is possible to isolate the gravitational wave from Z apparatus measurements. I suspect it may not, but this will have more to do with the word "propagating" than the word "gravitational". So the complete answer may well be that Z waves are "gravitational" but not "propagating gravitational", but neither are they entirely "tidal". Clear definitions of each are vital. $\endgroup$ – Sandgroper Apr 24 '16 at 8:49
  • $\begingroup$ I had to think about it for a few minutes yesterday but yes of course it is possible to isolate the the gravitational wave from Z measurements - using harmonic (fourier) analysis. I've updated my (final) answer accordingly. $\endgroup$ – Sandgroper Apr 25 '16 at 14:47

Since "Sandgroper" doesn't give up :) (remarkable!), I'll also try a really short different answer :)

You are quite right, that the solar system probably also produces gravitational waves. But what is a wave?

The point about an electromagnetic wave is, that it falls with distance as $1/r$ and not as $1/r^2$ as the static field.

We can not measure the influence of gravity of a black hole. By e.g. seing some displacement of a spring dynamometer. It's just too far away and falls with $1/r^2$.

The influence of the relative position of a system of two black holes will fall even faster, like a quadrupole field, with $1/r^4$. No chance.

But they did measure something. And therefore conclude, that there is something transported as $1/r$. This nececcarily also means, that energy is radiated away. This is clear in the electromagnetic analogy.

  • $\begingroup$ wrote <quote>...the solar system probably also produces gravitational waves. But what is a wave?</quote> If I generate a radio wave with a 30 light-day wavelength, that may seem like a nonpropagating magnetic field but I can still send 1 bit of information every 30 days at the speed of light to anything capable of detecting it. That is a wave. $\endgroup$ – Sandgroper Apr 25 '16 at 16:13
  • $\begingroup$ If the moon-earth rotating oblong gravitational field system can be detected near the source (away from the centre of mass) with a reference point that is essentially fixed in space along the radius relative to the sun, then the detected gravitational wave will have propagated relative to the sun from the centre of mass of the system to the fixed observation point. A fixed observation point can be located on the earth if the measurement is taken at the same time each day. $\endgroup$ – Sandgroper Apr 25 '16 at 16:32
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    $\begingroup$ if you generate a radio wave, you do it by moving some charge to and fro. You will measure not only the wave, but also the changing static field of the charge. This are two distinct things. Newtonian mechanics predicts, that a moving mass will produce changing field; but no waves. $\endgroup$ – Ilja Apr 25 '16 at 16:46
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    $\begingroup$ so you are probably right, that the earth-moon system does produce waves, but this is a distinct phenomenon from pseudostatically changing field $\endgroup$ – Ilja Apr 25 '16 at 16:53

To demonstrate that gravity propagates like a wave, you need to measure both amplitude (phase) and velocity. This necessitates at least two spatially separated instruments. In LIGO this is exactly what was done: and it was only because the same pulse was observed at two stations _ at different times_ that we conclude that the gravitational field propagates in a wavelike manner.

The 1981 experiment did nothing of the sort.

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    $\begingroup$ +1 But I think a stronger argument you especially could make (and maybe the strongest possible argument) uses your truly excellent answer to the question the OP cites, in which you beautifully fit the plotted data to a model. That model is presumably purely Newtonian, and gravitational waves are simply not a prediction of Newtonian gravity. $\endgroup$ – Mike Apr 23 '16 at 14:40
  • $\begingroup$ Gravitational wave detection methods are neither "Newtonian" nor relativistic, and ascribing these terms to detection methods is both presumptuous and propagandist. $\endgroup$ – Sandgroper Apr 25 '16 at 14:58
  • $\begingroup$ @Sandgroper I believe you are addressing Mike's comment, not my post. Could you please refrain from name calling? It is OK to say "wrong", but "presumptuous" has connotations of a personal attack which we frown upon here. $\endgroup$ – Floris Apr 25 '16 at 15:02
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    $\begingroup$ Incidentally if LIGO measures distortion of space time how is that not a relativistic measurement? $\endgroup$ – Floris Apr 25 '16 at 15:26
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    $\begingroup$ @Sandgroper I didn't say the detection methods are Newtonian; I said that the model used to fit the data was Newtonian. It doesn't matter how they detected what they detected; the ZRF data are consistent with a model based on just Newtonian gravity, and that model does not allow for gravitational waves. $\endgroup$ – Mike Apr 25 '16 at 17:15

What Zumberge detected was the quasistatic, non-propagating variation of the Newton gravitational potential. In linearized gravity, the gravitational potential is a second rank tensor and the Newton potential is the component $V^{00}$. Gravitational waves, on the other hand, are propagating spatial components such as $V^{xy}$. Zumberge did not measure these components at all, not even their quasistatic variation, so his result has its merits but is absolutely not gravitational wave detection.

  • $\begingroup$ If you stare to much at the math you will go blind to my question being not of detail but of categorization. ZRF detected gravitational "tides" that belong on the gravitational spectrum. Earth's oceanic tides and waves have been historically categorically separated by the phenomena producing them, wind versus planetary rotation, however these are both waveforms propagating in water Likewise the gravitational spectrum does not separate out gravitational tides from gravitational waves. They are still waveforms propagating in the"fabric of space time". LIGO makes a distinction, not the science. $\endgroup$ – Sandgroper Oct 8 '20 at 9:45

In view of the information provided by the more professional responses I can restate my view with a little more clarity.

The earth LIGO is built on moves with tidal forces and LIGO adjusts for this non propagating tidal gravity wave signal. LIGO also ignores the propagating gravitational wave - that is a result of earth having a moon and that gives rise to the tidal gravity wave - because the science driving LIGO expects cycles per second frequencies from a far field source. LIGO is simply not looking for near field gravitational waves.

Zumberge measurements also contain a different non propagating wave component due to the rotation of the earth as well as the more subtle propagating gravitational wave due to the earth moon system. This "lunar" gravitational wave is observable without the tidal component at the poles due to its rotation around the observation point and modulates on the lunar cycle, but still propagates at the speed of light. However neither LIGO nor Zumberger apparatii can measure this propagating gravitational wave at the poles because Zumburge measurements only measure along the vertical axis, while LIGO depends on variations in the order of 1-1K Hertz.

However the propagating gravitational wave can be isolated using Zumberge apparatus with a digital sampling technique as follows:

  1. Set a time of day, say noon, at which a sample will be recorded.
  2. Locate the Zumberge apparatus as near as possible to the equator
  3. Sample daily for 30 days or more.

The sampling time means the sun is now the reference point for all samples, and the observation point on the earth is effectively stationary relative to the sun and the moon This will then quite nicely show the not insignificant amplitude 30 day cycle near field propagating earth-moon gravitational wave appearing as a relatively perfect sine wave.

Thus the Zumberge method is a method for detecting local gravitational waves in certain circumstances, and the Zumberge measurement in 1981/2 is effectively of a gravitational wave, not a tidal gravity wave, with the caveat regarding placement of the apparatus and filtering out of diurnal effects.

In Zumberge measurements the apparatus is ideally placed at the maximal G in the near field system - being the earths surface - but because it is also sufficiently away from the centre of mass/gravity at the point of measurement we can say that this is a wave and it propagates relative to the sun.

In response to criticism that this is not a wave but a static field effect, I disagree. The best analogy is a bar magnet and in this scenario it is the propagation of a magnetic wave that is of interest. If a wave can be defined as capable of propagating information, but a static field is not capable of propagating information, and by rotating a bar magnet I can propagate information across space that can be detected by a compass, then this is a magnetic wave that propagates at the speed of light.

Likewise, a gravitational wave is one that is capable of propagating information and the Zumberge apparatus can detect information (change in field strength) in the near field propagating from the monthly rotation of the earth-moon gravitational field around the relatively static observation point relative to the sun.

The near field gravitational wave is a consequence of the rotating oblong gravitational field of the earth-moon system relative to the sun. This rotating system produces a near field gravitational wave effect relative to the sun but is extremely weak and does not propagate beyond the near field.

One should not just assume that no propagating gravitational waves are generated by the earth-moon movement in our solar system simply because they do not propagate into the far field. The extent of the near field is inversely proportional to the frequency, so the near field of the earth moon system is possibly in the order of 30 to 60 light days from the earth.

Nevertheless, it is detectable because we are, in cosmic terms, infinitesimally close to the CoG/CoM, yet not at the CoG/CoM.

This pretty well wraps it up for me.

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    $\begingroup$ now the statement is probably right. One could measure. But you'd need a very high precision for this. Look at the graph in you question, how far some data points are away from the fit curve... $\endgroup$ – Ilja Apr 25 '16 at 17:01
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    $\begingroup$ You cannot detect a wave's propagation without at least some estimation of the propagation velocity. As the wavelength is inversely proportional with the frequency, you are looking at a VERY unfavorable setup (long wavelength = small slope, meaning that a phase difference is hard to measure). $\endgroup$ – Floris Apr 25 '16 at 18:32

If LIGO was on Mars it would detect the local planetary gravitational waves the same way it does now on earth. Except on Mars there will be hardly any gravitational wave to detect because it has no moon [errata: it has 2 moons] to push and pull the planet around in detectable ways. The influence of the earth and the moon would be also greatly diminished but not completely.

We are certainly too close to the sources of local solar system gravitational waves for LIGO or Zumberge measurements to determine their speed of propagation. That will be possible when LIGO can ascertain the distance of its remote sources of gravitational waves.

However the physical movement of both binary black holes and planetary orbits necessarily produce the same gravitational waves, albeit of different frequency. The significant difference is ONLY that of longer and shorter wavelengths (as amplitude difference is not significant) in spite of what others may wish to claim. They are simply and clearly mistaken in patently obvious ways.

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    $\begingroup$ Think of it this way. You're on a ship moving in a circle. You measure the distance of your ship from shore and you notice that if you graph it it forms a sine wave. So you declare: you've measured waves in the sea! Of course this is not the case. Waves in the sea is not the distance between you and the shore, it is the distance between the sea floor and the surface of the water. $\endgroup$ – slebetman Apr 22 '16 at 12:50
  • $\begingroup$ @slebetman This is so far away from the point that any specific comment from me would be seen as defensive. Please understand that I know the difference between tidal gravity waves and gravitational waves. I have been considering this issue for several years now - at least 7. The rhetorical question I put to you is the same one I posted before: How can LIGO detect the Zumburge waves if LIGO cannot detect tidal gravity waves? LIGO only detects gravitational waves and noise, and Zumburge waves are detected by LIGO. Therefore Zumberge waves are gravitational waves. A case of cosmic cringe! $\endgroup$ – Sandgroper Apr 23 '16 at 1:08
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    $\begingroup$ LIGO does not detect the magnitude of gravity. It detects the distortion of space. Basically what LIGO does is to see how much a very-very stable ruler stretches (which implies that the space between the atoms themselves have warped). The Zumburge measurement instead does not detect how deformed an ideal ruler becomes - it detects how heavy a given mass weighs. They are two very different measurements. Of course LIGO also detects vibrations from thermal expansion, human activity etc. so those sources of noise have to be ruled out. $\endgroup$ – slebetman Apr 23 '16 at 6:33
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    $\begingroup$ You are simply not understanding that LIGO is not measuring the same wave measured by Zumberge. Like I said. The Zumberge wave is analogous to the distance between the ship to shore. The LIGO waves is analogous to the waves on the surface of the water. LIGO DOES NOT MEASURE GRAVITY AT ALL. Zumberge measure gravity. Instead LIGO only measures wave-like distortion to space-time. Zumberge does not measure wave-like distortion to space-time at all but merely measure the weight of an object over time. One measures weight, the other measures distance. $\endgroup$ – slebetman Apr 23 '16 at 16:42
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    $\begingroup$ Granted, if the Zumberge measurement was sensitive enough it could measure gravity waves. But it would be plotted as ripples inside the bigger "wave". The Zumberge measurements were simply not sensitive enough. $\endgroup$ – slebetman Apr 23 '16 at 16:45

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