<p>Acceleration isn't a property of spacetime per se. Two test particles at the same location can have different instantaneous accelerations, so just knowing that there is a passing gravitational wave doesn't tell you what an accelerometer will measure.</p>
<p>If your accelerometer is infinitesimal in size, in a perfect vacuum, and not propelled by any nongravitational means, then it will always register zero no matter what happens, essentially by definition.</p>
<p>If it's infinitesimal but affixed to the earth (as your question implies) then it will register something, since the passing gravitational wave will physically distort the earth and excite seismic waves. Those waves propagate rather slowly, but they are excited everywhere, so in principle there should be a ridiculously tiny but nonzero acceleration at the accelerometer's location as soon as the gravitational wave hits, unless the gravitational wave is perfectly symmetrical near that location.</p>
<p>If the accelerometer is a physical object with a nonzero size, then the gravitational wave will act on it directly. This is similar to the previous paragraph with the accelerometer itself replacing the earth. An extended body in curved spacetime doesn't have a single well-defined acceleration, so the reading is somewhat ambiguous in this case, but there's no reason why it should be precisely zero unless, again, there is perfect symmetry.</p>

CC BY-SA 4.0
2023-11-18T08:45:59.130Z
2023-11-18T08:45:59.130Z
benrg
<p>A gravimeter gauges the Earth's gravitational force by measuring the resistance it imposes on free fall. Technically, it is not measuring acceleration due to gravity, it measures the upward force exerted by the Earth that holds it still.  In a state of free fall, a gravimeter registers a measurement of zero. This aligns with the Equivalence Principle, which asserts that it is impossible to distinguish whether a spaceship is in free fall due to gravity or floating in space far from any masses, based on internal measurements.  This would include gravity waves since they are just changes in the warping of space, while the spaceship still follows a (more complex) geodesic.  Hence, if the Earth were to fully flex with the passing wave, gravimeters would be in free fall wrt the wave and measure nothing.  But the Earth does not fully flex, instead the surface and interfaces between density changes are pinged and seismic waves form.  These waves are smaller than we can presently detect.</p>
<p>To put it differently, gravity in GR is not a force; instead, it warps spacetime, causing the apparent curvature of paths without inducing detectable acceleration locally.  Therefore, a small gravimeter would not detect a gravity wave while traveling in a spaceship.  The casing, weight, and spring would move as a unit at the zero reading as the spaceship passed planets or stars or as gravity waves passed</p>
<p>When a gravity wave traverses a gravimeter positioned on the Earth's surface, the device, if (ridiculously) exceptionally sensitive and had amazingly fast readout speeds, would discern the solid Earth's resistance to the wave. This manifests as an oscillation in the local gravitational field (g), providing some information about one directional component of the wave, over an Earth radius, and its frequency.</p>
<p>One could obtain information about the other directional components by placing more gravimeters at other locations around the globe.</p>
<p>There are practical problems to using a gravimeter for the stellar mass black hole mergers that LIGO observes, aside from the fact that they are not yet sensitive enough or rapid enough, that may not be solvable.  LIGO is exquisitely isolated from vibrations of the Earth which would otherwise limit the noise floor, but a gravimeter could not be isolated for this purpose since the ground is the reference point.  Also, if one uses a spring based gravimeter, then there is complexity of deciphering it’s response to a changing amplitude and frequency forcing.</p>
<p>It may be that the merger of supermassive black holes or intermediate mass black holes could be observed by gravimeters because they can produce much greater strains.  When Andromeda mergers with the Milky Way, their BHs will merge and the strain here will be about <span class="math-container">$10^{-10}$</span> and that would produce accelerations above most of the seismic noise.  See    <a href="https://physics.stackexchange.com/questions/584239/andromeda-milky-way-merger-gravitational-waves">https://physics.stackexchange.com/questions/584239/andromeda-milky-way-merger-gravitational-waves</a></p>

CC BY-SA 4.0
2023-11-21T14:55:44.067Z
The properties of gravimeters have been well known for decades. The comparison of gravimeters of different designs (spring, magnetic field, free fall fountain; absolute and relative gravimeters) has led to a very high quality standard, including in the evaluation.
9herbert9
CC BY-SA 4.0
2023-11-22T09:38:11.600Z
To give a hint of the vibration damping needed: The arguably small GW detector near Hanover needs to isolate the mirror fixations against the vibrations induced by the waves Northern Sea clashing against the coast around 200km away...
planetmaker
CC BY-SA 4.0
2023-11-22T18:17:24.740Z
Plus, with a gravimeter, if you eliminate the vibration you will eliminate this signal.
eshaya
CC BY-SA 4.0
2023-11-25T01:10:31.717Z
"Equivalence Principle, which asserts that it is impossible to distinguish whether a spaceship is in free fall due to gravity or floating in space far from any masses" is only true for a uniform field, is it not? Like the assumption of an infinite flat Earth. In the case of free fall with no windows near the Earth, you could detect the gradient by watching the distances between some bowling balls that are falling with you. This implies the gravimeter will be null, but an instrument like Forward's rotating graviometer will register, due to a torque in any cruciform shape as the wave passes.
C. Towne Springer
CC BY-SA 4.0
2023-11-25T01:50:56.677Z
The gravity waves that LIGO detects have wavelengths of order c/100 hz = 3000 km and are extremely weak.  Maybe if you had a 100 km sized Forward gradiometer, there would be a chance.  Anyway, I interpret the question as asking if gravity waves can be measured by a local (point) measurement and thus a question about the equivalence principle.  Forward gradiometers work by measuring at 4 points which is, by definition, non-local.
eshaya
CC BY-SA 4.0
2023-11-19T02:41:51.187Z
2023-11-26T02:03:33.470Z
2023-11-26T02:03:33.470Z
eshaya
<p>In principle yes, in practice, probably not with current technology.</p>
<p>If a gravitational wave passes through the Earth then the Earth will respond by expanding and contracting. As a result, there will be a small acceleration measured by something attached to the surface of the Earth. This was first suggested by <a href="https://journals.aps.org/pr/abstract/10.1103/PhysRev.117.306" rel="nofollow noreferrer">Weber (1960)</a>, and <a href="https://ui.adsabs.harvard.edu/abs/1969ApJ...156..529D/abstract" rel="nofollow noreferrer">Dyson (1969)</a> investigated this by treating the Earth as a sphere of elastic solid. He concluded that Earth would flex by about <span class="math-container">$2\times 10^{-15}$</span> m as a result of a <span class="math-container">$h \sim 10^{-22}$</span> gravitational wave at 1 Hz (which is basically just the radius of the Earth multiplied by the strain).</p>
<p>Obviously that would produce a stupidly small deviation in the local gravity (1 part in <span class="math-container">$10^{44}$</span>, and this is NOT how a gravitational wave might be measured) but a gravimeter would also experience a local acceleration modulated at the frequency of the gravitational wave (assuming that the exciting source was continuous). If the displacement is modelled as <span class="math-container">$x = a \sin(2\pi f t)$</span>, where <span class="math-container">$f$</span> is the frequency and <span class="math-container">$a$</span> the displacement amplitude, then the acceleration magnitude is <span class="math-container">$4\pi^2 f^2 a$</span>. For the waves considered by Dyson, this would be <span class="math-container">$\sim 10^{-13}$</span> m/s<span class="math-container">$^2$</span>
modulated at 1 Hz<span class="math-container">${\dagger}$</span>.</p>
<p><a href="https://www.eas.slu.edu/GGP/fullpapers/shiomi_gravphys_sgs08.pdf" rel="nofollow noreferrer">Shiomi (2008)</a> discusses the use of gravimeters to detect gravitational waves. They say that the precision of superconducting gravimeters can reach sensitivites <span class="math-container">$10^{-11}$</span> m/s<span class="math-container">$^2$</span> for measurements taken over a year (not over 1 second!) and have stabilities of <span class="math-container">$\sim 10^{-8}$</span> m/s<span class="math-container">$^2$</span> on those timescales. Although this then sounds hopeless, the Earth does have resonant &quot;normal modes&quot; where the response to a gravitational wave might be boosted by factors of <span class="math-container">$\sim 100$</span> over a narrow range of frequencies (<span class="math-container">$\Delta f/f \sim 100$</span>). A network of superconducting gravimeters would probably have the best chance of &quot;seeing&quot; a wave response in the mHz band where there are some of the normal modes and a local miminum in the noise spectrum due to seismic activity. For example, <a href="https://arxiv.org/abs/1406.1147" rel="nofollow noreferrer">Coughlin &amp; Harm (2014)</a> use 10 years of the Global Geodynamics Project data from a network of superconducting gravimeters to put constraints (not detections) on the gravitational wave background at frequencies of around a mHz.</p>
<p>Lower frequencies are less favourable. Lower frequency gravitational wave sources are <em>a priori</em> weaker gravitational wave emitters unless very close, the strain is probably smaller than <span class="math-container">$10^{-22}$</span>. More importantly, the acceleration signal declines as <span class="math-container">$f^2$</span> for a given displacement amplitude, whilst the seismic noise actually increases towards lower frequencies. Higher (LIGO-like) frequencies are not favoured because the wavelength is then smaller than the size of the Earth and the response is muted.</p>
<p><span class="math-container">$\dagger$</span> Note that <em>of course</em> Dyson considers the scenario where you look at a time series of measurements taken over 10 years so that you can reduce the bandwidth of the noise to <span class="math-container">$\sim 10^{-8}$</span> Hz around the predicted frequency of the continuous gravitational wave source. He concluded that with a network of 100 of the best detectors available at the time (it has improved), that this level of gravitational wave would fall short of being detectable in 10 years by at least 5 orders of magnitude.</p>

CC BY-SA 4.0
2023-11-21T11:04:07.063Z
Everything is correct, only the assumption in the first line is too pessimistic. The reason: Everyone is always looking for noticeable spectral lines (SNR>9?) in broadband spectra and does not take any measures to reduce the noise level. Incomprehensible because $f_{GW}$ of binary stars is known very well. If you suppress the noise with filters ($BW\approx1$ nHz), the GW are unmistakable. I managed to do this without any problems with 20 close binary systems, which cannot be called a coincidence.
9herbert9
CC BY-SA 4.0
2023-11-21T13:08:46.573Z
@9herbert9 your assumption that you are cleverer and have thought of something that thousands of scientists who have lived and breathed gravitational waves for decades is astounding. You continue to promote your own (rejected) work through a social media website rather than through peer-review.
ProfRob
CC BY-SA 4.0
2023-11-21T13:10:16.250Z
@9herbert9 The seismic noise levels for close binary systems (frequencies of $10^{-5}-10^{-4}$ Hz become larger and the gravitational wave strain much smaller than $10^{-22}$. I am sceptical that you have even got within 5 orders of magnitude of detecting an actual gravitational wave signal. However, you can prove me wrong by getting your work published in a peer-reviewed journal or referring to other peer-reviewed work that has made a similar detection.
ProfRob
CC BY-SA 4.0
2023-11-21T14:37:16.883Z
Why so irritable? The disaster with the data connection between the Huygens probe and Cassini orbiter shows how clever some scientists are. Back to GW: This question could be answered very easily if someone were to repeat the experiments.
9herbert9
CC BY-SA 4.0
2023-11-25T22:10:33.777Z
@ProfRob I like to think of Stack Exchange as not exactly social media, and at least a little bit like peer review (that's roughly what's happeneing here), but yes, if there are only two baskets and not three, I guess social media it is.
uhoh
CC BY-SA 4.0
2023-11-21T09:08:47.347Z
2023-11-21T15:10:45.717Z
2023-11-21T15:10:45.717Z
ProfRob
<p>This depends entirely on the signal-to-noise ratio (SNR &gt; 10?). The safest way is to try it out, because a repeatable experimental result is better than any philosophical discussion. The stress <span class="math-container">$h_{GW}\approx 10^{-25}$</span> can be estimated using formulas, but so far there is no experimental confirmation that the formulas are correct. The inherent noise of a gravimeter can be determined fairly precisely because different designs can be compared.</p>
<p>Regarding your last sentence: There are probably more than a thousand gravimeters on earth that can simultaneously detect variations in gravity. The records from the last decades are available free of charge. Gravimeters measure acceleration and cannot distinguish between gravity and gravitational waves.</p>

CC BY-SA 4.0
2023-11-17T16:40:43.620Z
“Gravimeters measure acceleration and cannot distinguish between gravity and gravitational waves.”    So that means that “normal” gravity and gravitational waves are both detectable by a gravimeter. Right?
Peter
CC BY-SA 4.0
2023-11-17T16:48:54.807Z
I am convinced of that.
9herbert9
CC BY-SA 4.0
2023-11-17T11:49:31.200Z
2023-11-17T15:34:23.273Z
2023-11-17T15:34:23.273Z
9herbert9
<p>All events that are currently detected by the Virgo interferometer are associated with binary systems.</p>
<p>At the inspiral phase, it is obvious that super-sensitive gravimeter will detect a change in the magnitude and vector of the gravitational potential of the binary system. Therefore, the answer is positive.</p>
<p>Detailed estimates were first obtained by: Peters, P. Gravitational Radiation and the Motion of Two Point Masses // Physical Review: journal. — 1964. — Vol. 136, No. 4B. — P. 1224-1232. — doi:10.1103/PhysRev.136.B1224. — Bibcode: 1964PhRv..136.1224P</p>
<p>P.S.</p>
<p>At the merge and ringdown phases, super-sensitive gravimeter will detect the defect mass (radiated GW energy).</p>
<p>P.P.S.
In the general case. The thought experiment about sticky balls, originally proposed by Richard Feynman, and popularized by Herman Bondi, completely refers to an super-sensitive gravimeter.</p>

CC BY-SA 4.0
2023-11-21T15:11:29.650Z
2023-11-22T06:47:14.153Z
2023-11-22T06:47:14.153Z
Serge3leo
<p>A gravimeter does not measure resistance, it works <a href="https://www.sciencedirect.com/topics/earth-and-planetary-sciences/gravimeter" rel="nofollow noreferrer">like this</a>: A housing is firmly connected to the earth's surface. There is a ball in the housing that is held in position by springs or magnetic fields. Due to the inertia of the mass, the ball follows fast movements of the housing with a delay. The <em>vertical</em> distance between the ball and the housing is the measurement result (1D-gravimeter); it is measured without contact. Since the Earth rotates on its own axis, a 3D gravimeter makes evaluation more difficult without providing additional information.</p>
<p>A gravimeter reacts depending on frequency:</p>
<p>a) Constant or extremely slow changes in <span class="math-container">$G$</span> cause a shift in the ball's rest position. This is used to search for underground deposits of iron ore or salt.</p>
<p>b) When <span class="math-container">$G$</span> changes quickly, the ball oscillates around a center position. The ball-spring-housing system is damped so that the vibrations stop quickly when the excitation stops.</p>
<p>Behavior at GW (in the absence of the moon!):</p>
<ol>
<li><p>Viewed from a great distance, the sphere and the center of the earth describe circular orbits around the sun. The sphere is so small (<span class="math-container">$d_K\approx0.03$</span> m) that the GW cannot measurably change the diameter. GW <em>does not</em> affect the position of the sphere's center.</p>
</li>
<li><p>The Earth is so large (<span class="math-container">$d_E\approx 10^7$</span> m) that the diameter changes periodically in the rhythm of the GW. Therefore, the distance between the casing and the sphere of the gravimeter oscillates at the same frequency. The gravimeter housing <em>must</em> be connected to the earth's surface. A mechanical resonance like with Weber bars is not necessary.</p>
</li>
</ol>
<p>Earthquakes and BH mergers move the housing briefly. This can be integrated away through long-term observation. Continuous GWs change the housing-sphere distance periodically at a constant frequency. The signal can be easily detected using long-term observation (<span class="math-container">$T&gt;10$</span> years) and extremely narrow-band filters (<span class="math-container">$BW&lt;1$</span> nHz). Try it!</p>

CC BY-SA 4.0
2023-11-19T12:56:31.743Z
*Welcome to Stack Exchange!* I don't see why you need to use two separate answer posts, you can just edit your original post and add this to it.
uhoh
CC BY-SA 4.0
2023-11-19T15:12:24.253Z
"The signal can be easily detected...". No it can't - or refer to a *peer-reviewed* publication that has done so.
ProfRob
CC BY-SA 4.0
2023-11-19T15:54:08.700Z
What came first? The chicken or the egg? My attempts to publish measurement methods and results in Nature and AAS were rejected on the grounds that "we don't believe that". Is there no one who can build a receiver for 60 Hz? Apparently many people confuse experimental physics with theology.
9herbert9
CC BY-SA 4.0
2023-11-22T13:00:23.150Z
@9herbert9 so you have a technical idea and no-one wants to spend time and money to (dis)prove your concept? Build it yourself and demonstrate its value.
planetmaker
CC BY-SA 4.0
2023-11-19T11:01:43.193Z
2023-11-19T11:10:47.860Z
2023-11-19T11:10:47.860Z
9herbert9
<p>Imagine you have a super sensitive 3D accelerometer/gravimeter alongside the Virgo interferometer.
A gravitational wave passes by and Virgo detects a variation in the length of the 3000m long arms.
Will the gravimeter be able to, theoretically, simultaneously detect variations in gravity?</p>

CC BY-SA 4.0
2023-11-16T21:08:22.777Z
Questions about detection of gravitational waves need to recall just how mind boggling weak they are.  strain is only about 1 part in a billion-trillion. Now, what is the accuracy of the best gravimeters?
James K
CC BY-SA 4.0
2023-11-17T10:09:32.990Z
@JamesK you've missed the point of the question. The OP's "super-sensitive" and "theoretically" are their way to express that this is a gedankenexperiment, not a quesiton about instrumental sensitivity. I've made a small edit to make that more explicit.
uhoh
CC BY-SA 4.0
2023-11-17T21:14:55.800Z
I rolled back your edit.  It supposed a context that wasn't in the question, and seemed to be an attempt to hijack the question to some extent.  My comment is just referring to the practicalities, and so isn't an attempt to answer the question. The final paragraph indicates that the OP is interested in the practicalities of measurement
James K
CC BY-SA 4.0
2023-11-17T23:10:40.637Z
@Peter were you asking about whether existing equipment would be sensitive enough, or asking if an arbitrarily supersensitive device would measure something or there is in fact no acceleration to measure at all? In other words, are you asking primarily 1) about the device's sensitivity itself, or 2) about the existence or non-existence of acceleration associated with a GW? l've assumed it's 2) but JamesK would like it to be 1). If it's not 2), then maybe that can be asked separately because it's a great question!
uhoh
CC BY-SA 4.0
2023-11-18T08:55:41.400Z
Uhoh, you are puttin words into my mouth!  I wouldn't "like it to be 1)"  Peter asked a question, initially framed in practical terms (I was going to write an answer about the practicalities, but I decided not to - I commented on the practicalities.  So Peter edited to focus on the theoretical aspects. That is the purpose of comments! Peter's edits were useful and made an answerable question. I can't answer the theoretical question, so I didn't.  I think there is a good answer now.  The practical fact remains that you can't use a gravimeter to detect GW, because their effects are too small.
James K
CC BY-SA 4.0
2023-11-18T10:42:08.760Z
@JamesK "...you can't use a gravimeter..." that you've happened to come across. You'd also tell Einstein that trains can't go close to the speed of light so he shouldn't think about it? [Look at all the scientists that were perfectly fine with thought experiments](https://en.wikipedia.org/wiki/Thought_experiment)!
uhoh
CC BY-SA 4.0
2023-11-18T10:49:03.737Z
I'm fine with them - so back off please.  Peter asked a question that mixed practicalities and theory I pointed out the difficulty of the practicalities. He edited to improve his question.  Everyone's happy except you - for some reason.
James K
CC BY-SA 4.0
2023-11-18T19:58:37.197Z
-The Earth is deformed by a passing GW. Virgo can measure this deformation.
-A gravimeter on Earth measures the gravity where it is placed. I assume here that this is an extremely sensitive superconductor gravimeter with a mass of Niobium hanging freely in a magnetic field. 

My question was whether this gravimeter, which measures "ordinary gravity", also can detect a GW.
If yes. How and in which direction does the Niobium sphere moves, in relation to the rest of the gravimeter itself?
Will it follow one of the red dots shown in this video? https://youtu.be/AXgljRvI_Tg?si=sGv_mI8soy9RFHsf
Peter
CC BY-SA 4.0
2023-11-18T19:59:31.023Z
Sorry if I phrased myself imprecisely and poorly. I have no education in the subject and my English is not good either.
Bear with me.
Peter
CC BY-SA 4.0
2023-11-20T16:12:57.753Z
@Peter: Gravimeters measure the vertical movement of the ball once per second. The average value of a minute is usually published. Since the earth rotates, the direction of the vertical rotates 360 degrees every day. If you know the direction to the source, it makes sense to only evaluate the measured values if there is a right angle (GW are transverse waves).
9herbert9
<gravity><gravitational-waves><newtonian-gravity>
Gravitational waves vs. "normal gravity"
Peter
# Gravitational waves vs. "normal gravity"

(Asked by: username_0 on 2023-11-16)

Imagine you have a super sensitive 3D accelerometer/gravimeter alongside the Virgo interferometer.
A gravitational wave passes by and Virgo detects a variation in the length of the 3000m long arms.
Will the gravimeter be able to, theoretically, simultaneously detect variations in gravity?

**Tags:** gravity, gravitational-waves, newtonian-gravity

> **Comments:**
> **username_1**: Questions about detection of gravitational waves need to recall just how mind boggling weak they are.  strain is only about 1 part in a billion-trillion. Now, what is the accuracy of the best gravimeters?  
> **username_2**: @JamesK you've missed the point of the question. The OP's "super-sensitive" and "theoretically" are their way to express that this is a gedankenexperiment, not a quesiton about instrumental sensitivity. I've made a small edit to make that more explicit.  
> **username_1**: I rolled back your edit.  It supposed a context that wasn't in the question, and seemed to be an attempt to hijack the question to some extent.  My comment is just referring to the practicalities, and so isn't an attempt to answer the question. The final paragraph indicates that the OP is interested in the practicalities of measurement  
> **username_2**: @username_0 were you asking about whether existing equipment would be sensitive enough, or asking if an arbitrarily supersensitive device would measure something or there is in fact no acceleration to measure at all? In other words, are you asking primarily 1) about the device's sensitivity itself, or 2) about the existence or non-existence of acceleration associated with a GW? l've assumed it's 2) but JamesK would like it to be 1). If it's not 2), then maybe that can be asked separately because it's a great question!  
> **username_1**: Uhoh, you are puttin words into my mouth!  I wouldn't "like it to be 1)"  username_0 asked a question, initially framed in practical terms (I was going to write an answer about the practicalities, but I decided not to - I commented on the practicalities.  So username_0 edited to focus on the theoretical aspects. That is the purpose of comments! username_0's edits were useful and made an answerable question. I can't answer the theoretical question, so I didn't.  I think there is a good answer now.  The practical fact remains that you can't use a gravimeter to detect GW, because their effects are too small.  
> **username_2**: @JamesK "...you can't use a gravimeter..." that you've happened to come across. You'd also tell Einstein that trains can't go close to the speed of light so he shouldn't think about it? [Look at all the scientists that were perfectly fine with thought experiments](https://en.wikipedia.org/wiki/Thought_experiment)!  
> **username_1**: I'm fine with them - so back off please.  username_0 asked a question that mixed practicalities and theory I pointed out the difficulty of the practicalities. He edited to improve his question.  Everyone's happy except you - for some reason.  
> **username_0**: -The Earth is deformed by a passing GW. Virgo can measure this deformation. -A gravimeter on Earth measures the gravity where it is placed. I assume here that this is an extremely sensitive superconductor gravimeter with a mass of Niobium hanging freely in a magnetic field.   My question was whether this gravimeter, which measures "ordinary gravity", also can detect a GW. If yes. How and in which direction does the Niobium sphere moves, in relation to the rest of the gravimeter itself? Will it follow one of the red dots shown in this video? https://youtu.be/AXgljRvI_Tg?si=sGv_mI8soy9RFHsf  
> **username_0**: Sorry if I phrased myself imprecisely and poorly. I have no education in the subject and my English is not good either. Bear with me.  
> **username_3**: @username_0: Gravimeters measure the vertical movement of the ball once per second. The average value of a minute is usually published. Since the earth rotates, the direction of the vertical rotates 360 degrees every day. If you know the direction to the source, it makes sense to only evaluate the measured values if there is a right angle (GW are transverse waves).  

---

## 6 answers

### Answer by username_4 on 2023-11-18. Score: 5

Acceleration isn't a property of spacetime per se. Two test particles at the same location can have different instantaneous accelerations, so just knowing that there is a passing gravitational wave doesn't tell you what an accelerometer will measure.

If your accelerometer is infinitesimal in size, in a perfect vacuum, and not propelled by any nongravitational means, then it will always register zero no matter what happens, essentially by definition.

If it's infinitesimal but affixed to the earth (as your question implies) then it will register something, since the passing gravitational wave will physically distort the earth and excite seismic waves. Those waves propagate rather slowly, but they are excited everywhere, so in principle there should be a ridiculously tiny but nonzero acceleration at the accelerometer's location as soon as the gravitational wave hits, unless the gravitational wave is perfectly symmetrical near that location.

If the accelerometer is a physical object with a nonzero size, then the gravitational wave will act on it directly. This is similar to the previous paragraph with the accelerometer itself replacing the earth. An extended body in curved spacetime doesn't have a single well-defined acceleration, so the reading is somewhat ambiguous in this case, but there's no reason why it should be precisely zero unless, again, there is perfect symmetry.


---

### Answer by username_5 on 2023-11-19. Score: 4

A gravimeter gauges the Earth's gravitational force by measuring the resistance it imposes on free fall. Technically, it is not measuring acceleration due to gravity, it measures the upward force exerted by the Earth that holds it still. In a state of free fall, a gravimeter registers a measurement of zero. This aligns with the Equivalence Principle, which asserts that it is impossible to distinguish whether a spaceship is in free fall due to gravity or floating in space far from any masses, based on internal measurements. This would include gravity waves since they are just changes in the warping of space, while the spaceship still follows a (more complex) geodesic. Hence, if the Earth were to fully flex with the passing wave, gravimeters would be in free fall wrt the wave and measure nothing. But the Earth does not fully flex, instead the surface and interfaces between density changes are pinged and seismic waves form. These waves are smaller than we can presently detect.

To put it differently, gravity in GR is not a force; instead, it warps spacetime, causing the apparent curvature of paths without inducing detectable acceleration locally. Therefore, a small gravimeter would not detect a gravity wave while traveling in a spaceship. The casing, weight, and spring would move as a unit at the zero reading as the spaceship passed planets or stars or as gravity waves passed

When a gravity wave traverses a gravimeter positioned on the Earth's surface, the device, if (ridiculously) exceptionally sensitive and had amazingly fast readout speeds, would discern the solid Earth's resistance to the wave. This manifests as an oscillation in the local gravitational field (g), providing some information about one directional component of the wave, over an Earth radius, and its frequency.

One could obtain information about the other directional components by placing more gravimeters at other locations around the globe.

There are practical problems to using a gravimeter for the stellar mass black hole mergers that LIGO observes, aside from the fact that they are not yet sensitive enough or rapid enough, that may not be solvable. LIGO is exquisitely isolated from vibrations of the Earth which would otherwise limit the noise floor, but a gravimeter could not be isolated for this purpose since the ground is the reference point. Also, if one uses a spring based gravimeter, then there is complexity of deciphering it’s response to a changing amplitude and frequency forcing.

It may be that the merger of supermassive black holes or intermediate mass black holes could be observed by gravimeters because they can produce much greater strains. When Andromeda mergers with the Milky Way, their BHs will merge and the strain here will be about $10^{-10}$ and that would produce accelerations above most of the seismic noise. See <https://physics.stackexchange.com/questions/584239/andromeda-milky-way-merger-gravitational-waves>

> **Comments:**
> **username_3**: The properties of gravimeters have been well known for decades. The comparison of gravimeters of different designs (spring, magnetic field, free fall fountain; absolute and relative gravimeters) has led to a very high quality standard, including in the evaluation.  
> **username_6**: To give a hint of the vibration damping needed: The arguably small GW detector near Hanover needs to isolate the mirror fixations against the vibrations induced by the waves Northern Sea clashing against the coast around 200km away...  
> **username_5**: Plus, with a gravimeter, if you eliminate the vibration you will eliminate this signal.  
> **username_7**: "Equivalence Principle, which asserts that it is impossible to distinguish whether a spaceship is in free fall due to gravity or floating in space far from any masses" is only true for a uniform field, is it not? Like the assumption of an infinite flat Earth. In the case of free fall with no windows near the Earth, you could detect the gradient by watching the distances between some bowling balls that are falling with you. This implies the gravimeter will be null, but an instrument like Forward's rotating graviometer will register, due to a torque in any cruciform shape as the wave passes.  
> **username_5**: The gravity waves that LIGO detects have wavelengths of order c/100 hz = 3000 km and are extremely weak.  Maybe if you had a 100 km sized Forward gradiometer, there would be a chance.  Anyway, I interpret the question as asking if gravity waves can be measured by a local (point) measurement and thus a question about the equivalence principle.  Forward gradiometers work by measuring at 4 points which is, by definition, non-local.  

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### Answer by username_8 on 2023-11-21. Score: 3

In principle yes, in practice, probably not with current technology.

If a gravitational wave passes through the Earth then the Earth will respond by expanding and contracting. As a result, there will be a small acceleration measured by something attached to the surface of the Earth. This was first suggested by [Weber (1960)](https://journals.aps.org/pr/abstract/10.1103/PhysRev.117.306), and [Dyson (1969)](https://ui.adsabs.harvard.edu/abs/1969ApJ...156..529D/abstract) investigated this by treating the Earth as a sphere of elastic solid. He concluded that Earth would flex by about $2\times 10^{-15}$ m as a result of a $h \sim 10^{-22}$ gravitational wave at 1 Hz (which is basically just the radius of the Earth multiplied by the strain).

Obviously that would produce a stupidly small deviation in the local gravity (1 part in $10^{44}$, and this is NOT how a gravitational wave might be measured) but a gravimeter would also experience a local acceleration modulated at the frequency of the gravitational wave (assuming that the exciting source was continuous). If the displacement is modelled as $x = a \sin(2\pi f t)$, where $f$ is the frequency and $a$ the displacement amplitude, then the acceleration magnitude is $4\pi^2 f^2 a$. For the waves considered by Dyson, this would be $\sim 10^{-13}$ m/s$^2$
modulated at 1 Hz${\dagger}$.

[Shiomi (2008)](https://www.eas.slu.edu/GGP/fullpapers/shiomi_gravphys_sgs08.pdf) discusses the use of gravimeters to detect gravitational waves. They say that the precision of superconducting gravimeters can reach sensitivites $10^{-11}$ m/s$^2$ for measurements taken over a year (not over 1 second!) and have stabilities of $\sim 10^{-8}$ m/s$^2$ on those timescales. Although this then sounds hopeless, the Earth does have resonant "normal modes" where the response to a gravitational wave might be boosted by factors of $\sim 100$ over a narrow range of frequencies ($\Delta f/f \sim 100$). A network of superconducting gravimeters would probably have the best chance of "seeing" a wave response in the mHz band where there are some of the normal modes and a local miminum in the noise spectrum due to seismic activity. For example, [Coughlin & Harm (2014)](https://arxiv.org/abs/1406.1147) use 10 years of the Global Geodynamics Project data from a network of superconducting gravimeters to put constraints (not detections) on the gravitational wave background at frequencies of around a mHz.

Lower frequencies are less favourable. Lower frequency gravitational wave sources are *a priori* weaker gravitational wave emitters unless very close, the strain is probably smaller than $10^{-22}$. More importantly, the acceleration signal declines as $f^2$ for a given displacement amplitude, whilst the seismic noise actually increases towards lower frequencies. Higher (LIGO-like) frequencies are not favoured because the wavelength is then smaller than the size of the Earth and the response is muted.

$\dagger$ Note that *of course* Dyson considers the scenario where you look at a time series of measurements taken over 10 years so that you can reduce the bandwidth of the noise to $\sim 10^{-8}$ Hz around the predicted frequency of the continuous gravitational wave source. He concluded that with a network of 100 of the best detectors available at the time (it has improved), that this level of gravitational wave would fall short of being detectable in 10 years by at least 5 orders of magnitude.

> **Comments:**
> **username_3**: Everything is correct, only the assumption in the first line is too pessimistic. The reason: Everyone is always looking for noticeable spectral lines (SNR>9?) in broadband spectra and does not take any measures to reduce the noise level. Incomprehensible because $f_{GW}$ of binary stars is known very well. If you suppress the noise with filters ($BW\approx1$ nHz), the GW are unmistakable. I managed to do this without any problems with 20 close binary systems, which cannot be called a coincidence.  
> **username_8**: @username_3 your assumption that you are cleverer and have thought of something that thousands of scientists who have lived and breathed gravitational waves for decades is astounding. You continue to promote your own (rejected) work through a social media website rather than through peer-review.  
> **username_8**: @username_3 The seismic noise levels for close binary systems (frequencies of $10^{-5}-10^{-4}$ Hz become larger and the gravitational wave strain much smaller than $10^{-22}$. I am sceptical that you have even got within 5 orders of magnitude of detecting an actual gravitational wave signal. However, you can prove me wrong by getting your work published in a peer-reviewed journal or referring to other peer-reviewed work that has made a similar detection.  
> **username_3**: Why so irritable? The disaster with the data connection between the Huygens probe and Cassini orbiter shows how clever some scientists are. Back to GW: This question could be answered very easily if someone were to repeat the experiments.  
> **username_2**: @username_8 I like to think of Stack Exchange as not exactly social media, and at least a little bit like peer review (that's roughly what's happeneing here), but yes, if there are only two baskets and not three, I guess social media it is.  

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### Answer by username_3 on 2023-11-17. Score: 2

This depends entirely on the signal-to-noise ratio (SNR > 10?). The safest way is to try it out, because a repeatable experimental result is better than any philosophical discussion. The stress $h\_{GW}\approx 10^{-25}$ can be estimated using formulas, but so far there is no experimental confirmation that the formulas are correct. The inherent noise of a gravimeter can be determined fairly precisely because different designs can be compared.

Regarding your last sentence: There are probably more than a thousand gravimeters on earth that can simultaneously detect variations in gravity. The records from the last decades are available free of charge. Gravimeters measure acceleration and cannot distinguish between gravity and gravitational waves.

> **Comments:**
> **username_0**: “Gravimeters measure acceleration and cannot distinguish between gravity and gravitational waves.”    So that means that “normal” gravity and gravitational waves are both detectable by a gravimeter. Right?  
> **username_3**: I am convinced of that.  

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### Answer by username_9 on 2023-11-21. Score: 0

All events that are currently detected by the Virgo interferometer are associated with binary systems.

At the inspiral phase, it is obvious that super-sensitive gravimeter will detect a change in the magnitude and vector of the gravitational potential of the binary system. Therefore, the answer is positive.

Detailed estimates were first obtained by: username_0s, P. Gravitational Radiation and the Motion of Two Point Masses // Physical Review: journal. — 1964. — Vol. 136, No. 4B. — P. 1224-1232. — doi:10.1103/PhysRev.136.B1224. — Bibcode: 1964PhRv..136.1224P

P.S.

At the merge and ringdown phases, super-sensitive gravimeter will detect the defect mass (radiated GW energy).

P.P.S.
In the general case. The thought experiment about sticky balls, originally proposed by Richard Feynman, and popularized by Herman Bondi, completely refers to an super-sensitive gravimeter.


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### Answer by username_3 on 2023-11-19. Score: -1

A gravimeter does not measure resistance, it works [like this](https://www.sciencedirect.com/topics/earth-and-planetary-sciences/gravimeter): A housing is firmly connected to the earth's surface. There is a ball in the housing that is held in position by springs or magnetic fields. Due to the inertia of the mass, the ball follows fast movements of the housing with a delay. The *vertical* distance between the ball and the housing is the measurement result (1D-gravimeter); it is measured without contact. Since the Earth rotates on its own axis, a 3D gravimeter makes evaluation more difficult without providing additional information.

A gravimeter reacts depending on frequency:

a) Constant or extremely slow changes in $G$ cause a shift in the ball's rest position. This is used to search for underground deposits of iron ore or salt.

b) When $G$ changes quickly, the ball oscillates around a center position. The ball-spring-housing system is damped so that the vibrations stop quickly when the excitation stops.

Behavior at GW (in the absence of the moon!):

1. Viewed from a great distance, the sphere and the center of the earth describe circular orbits around the sun. The sphere is so small ($d\_K\approx0.03$ m) that the GW cannot measurably change the diameter. GW *does not* affect the position of the sphere's center.
2. The Earth is so large ($d\_E\approx 10^7$ m) that the diameter changes periodically in the rhythm of the GW. Therefore, the distance between the casing and the sphere of the gravimeter oscillates at the same frequency. The gravimeter housing *must* be connected to the earth's surface. A mechanical resonance like with Weber bars is not necessary.

Earthquakes and BH mergers move the housing briefly. This can be integrated away through long-term observation. Continuous GWs change the housing-sphere distance periodically at a constant frequency. The signal can be easily detected using long-term observation ($T>10$ years) and extremely narrow-band filters ($BW<1$ nHz). Try it!

> **Comments:**
> **username_2**: *Welcome to Stack Exchange!* I don't see why you need to use two separate answer posts, you can just edit your original post and add this to it.  
> **username_8**: "The signal can be easily detected...". No it can't - or refer to a *peer-reviewed* publication that has done so.  
> **username_3**: What came first? The chicken or the egg? My attempts to publish measurement methods and results in Nature and AAS were rejected on the grounds that "we don't believe that". Is there no one who can build a receiver for 60 Hz? Apparently many people confuse experimental physics with theology.  
> **username_6**: @username_3 so you have a technical idea and no-one wants to spend time and money to (dis)prove your concept? Build it yourself and demonstrate its value.  

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