Fairly straightforward. I just thought it wouldn't be too hard to produce a ripple in spacetime several times smaller than a proton radius in a particle accelerator or something. It seems like it should be going on all the time.

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    $\begingroup$ Well, we do - they are just too small to measure. $\endgroup$
    – Jon Custer
    Commented May 16, 2017 at 20:44
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    $\begingroup$ How many spare black holes do you have lying around? $\endgroup$
    – Jon Custer
    Commented May 16, 2017 at 20:50
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    $\begingroup$ The time is irrelevant. Yes, a long way away, but a really huge (HUGE!) amount of gravitational energy was released. $\endgroup$
    – Jon Custer
    Commented May 16, 2017 at 20:52
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    $\begingroup$ You might check out Raymond Chiao's work on detecting gravity waves. en.wikipedia.org/wiki/Raymond_Chiao or blog.physicsworld.com/2009/04/29/can-gravitional-waves-be-detec $\endgroup$
    – David Elm
    Commented May 16, 2017 at 22:52
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    $\begingroup$ Why can't you hear the sounds my keyboard made when I typed this? $\endgroup$ Commented May 18, 2017 at 20:47

6 Answers 6


In 1973, Grishchuk and Sazhin proposed in their paper, "Emission of gravitational waves by an electromagnetic cavity", a method by which to generate gravitational waves for experiments, using the argument that while the generation would be very weak, it would also not suffer from the decay in $r^{-2}$.

The idea was to generate a rapid quadrupole moment change in the electrons of a metal cavity by applying very high-frequency EM radiations to it. The average of the energy flux was found to be of the order $G c^{-3} R^{-2} \lambda^2 r_0^2 \varepsilon^2$, with $\lambda$ the frequency of the gravitational radiation, $r_0$ the characteristic dimension of the cavity, and $\varepsilon$ the energy density of the EM wave.

From the appearance of $G c^{-3}$, you can tell that it's going to be fairly small. I don't know if this idea was ever discussed again (this paper was almost never referenced), or if it would be more plausible today, but it is certainly a rather complex setup for very small results.


While it's true that LIGO is incredibly sensitive, such that it can measure ripples in spacetime up to a few $1000$ times smaller than the width of a proton, it's also equally important to consider, for example, GW150914 event, thereby highlighting few other factors involved in producing and detecting gravitational waves:


  • The length of LIGO's arms is $4 km$ each physically, and $1120 km$ virtually – see official source and Fabry–Pérot interferometer.

  • The pair of black holes in this event was thought to be of masses between $30M_{\odot}$ and $35M_{\odot}$ each, orbiting sufficiently close to each other, at about $0.3c$ to $0.6c$ (where $c = 3 \times 10^{8}ms^{-1}$), eventually resulting in a merger in $0.2$ seconds. The merger radiated away the mass-energy of about $3M_{\odot}$ as gravitational waves.

  • The black holes pair was estimated to be at a distance of about $1.4\pm0.6$ billion light years from earth (additional note: radius of the Observable Universe is $46$billion light years). So at cosmological scales, that distance is almost in our neighborhood.

  • When these black holes merged, they emitted the peak energy of about $3.6\times10^{49} Watts$ (approx. $5.3\times10^{47} Joules$). Wiki states that this is more than the combined power of light from all the stars in the observable universe. Just to add a bit of perspective, the most powerful particle accelerator LHC has had a maximum power of $13TeV$, i.e. $2\times10^{-6} Joules$.

Despite the above mentioned numbers of masses of objects, their speeds and the energy emitted due to their collision being so overwhelmingly enormous, plus the black hole merger event taking place almost in our neighbourhood, the decayed ripples that reached Earth barely had enough strength to be detected by LIGO at a frequency of $35 Hz$ to $250 Hz$ and create the distortion measurable at the scale of a thousandth of a width of a proton!

Hence, with these factors in mind, one can imagine the likelihood of creating a lab that could accommodate such masses, let alone accelerate them to relativistic speeds and bring about a collision between them, and yet carry this all out in a controlled environment to be able to stand a chance of producing / detecting gravitational waves.

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    $\begingroup$ Do you know, how much was the actualy energy that hit LIGO? $\endgroup$ Commented May 17, 2017 at 6:37
  • $\begingroup$ @BjarkeFreund-Hansen An indirect measurement of the energy from GWs is made via (1) Strain, which was found to be approx. $10^{-21}$ for GW150914. The other method is using (2) Amplitude Spectral Density (which is the square root of Power Spectral Density: $Watts/Hz$) - please see a related Q&A and a web app for more. I'm not entirely sure (and can't really figure out) if or how these values can be converted into Joules($10^{-xx}$?), but that's indeed something quite interesting to think about! $\endgroup$ Commented May 18, 2017 at 0:39
  • $\begingroup$ No criticism about the answer, but even if my city was 92 km wide, I would not call something 1.4 (or 2) km away as my neighbourhood ;P $\endgroup$ Commented May 18, 2017 at 16:25
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    $\begingroup$ @another'Homosapien' I have to disagree - I have a 1.2 mile walk to the bus stop for the commute into my city. For me, that bus stop is local compared to the 15 miles on the bus. London is 50 km across and at that scale I'd fully say that 1km is local. $\endgroup$
    – Tim
    Commented May 18, 2017 at 23:55
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    $\begingroup$ @BjarkeFreund-Hansen I've been thinking about the comment you made. In fact, found that there's a very closely related question here on Physics SE, unanswered for a year: How much energy does the Earth absorb when a gravitational wave passes through it?. Because we have specific information about Energy Density and Frequency (in effect, also the Wavelength) of the GW here, it should ideally be possible to integrate it over the distance to compute total absorbed energy by Earth (using its diameter), but I'm not sure exactly how. $\endgroup$ Commented May 20, 2017 at 21:14

Short answer

You can, but to make them detectable, you need a lot of mass, and you can't (realistically) compensate for this just by getting closer.

Long answer

In the case of two orbiting masses (e.g., a binary system or a pair of protons), the measured quantity – the strain amplitude – of the emitted gravitational waves scales as:

$$|h|\propto \frac{v^2M}{R}$$

where $M$ is the mass of the system, $R$ is the distance between the detector and the source, and $v$ is the orbital speed. Note that this isn't an inverse-square dependence on $R$, so distance doesn't matter as much as for EM telescopes, which measure energy flux across an imaging sensor, proportional to $R^{-2}$.

Given that the GW150914 event was close to the threshold of detectability, with $|h| \approx 10^{-21}$, we can work out roughly how the amplitude would scale if we were to conduct an experiment with a pair of accelerated protons observed from, say, $1\,\mathrm{m}$.

$$\frac{M_\text{proton}}{M_\text{GW150914}} \approx \frac{1.7\times10^{-27}\,\mathrm{kg}}{3\times10^{31}\,\mathrm{kg}} \approx 2.8\times10^{-58}$$

$$\frac{R_{GW150914}}{1\,\mathrm{m}} \approx \frac{1.4\times10^{25}\,\mathrm{m}}{1\,\mathrm{m}} \approx 1.4\times10^{25}$$

So already we need to overcome a drop in amplitude of a factor of $3.9\times10^{33}$. Even with the LHC, where $v>0.9c$, there's no way you can recover 33 orders of magnitude! Also, bear in mind that you can only observe waves in the far-field limit $R \gg r$ (where $r$ is the orbital radius).

There's a discussion of this in Fundamentals of Interferometric Gravitational Wave Detectors.

  • $\begingroup$ The amplitude of an electromagnetic fields also dies off as 1/r, doesn't it? That's the energy flux that dies off as 1/r^2, isn't it? Apart from that, your answer is the best of the bunch imho: concise and precise! $\endgroup$
    – user154997
    Commented May 17, 2017 at 9:31
  • $\begingroup$ @LucJ.Bourhis Yes, you're right :) However, interferometric GW detectors measure the amplitude (of strain) directly, so the $R^{-1}$ scaling is the relevant one. I've clarified the answer. $\endgroup$ Commented May 17, 2017 at 10:15
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    $\begingroup$ @peterh Bear in mind that for GW150914, at peak power, $r$ was about 350 km and $f$ was about 100 Hz. Making an experiment that rotates two 5000-tonne lead balls in this way will be challenging :) $\endgroup$ Commented May 17, 2017 at 11:09
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    $\begingroup$ @peterh Really? The lead balls would have to be travelling at about half the speed of light. $\endgroup$ Commented May 17, 2017 at 11:39
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    $\begingroup$ @peterh, let's apply some ridiculously optimistic numbers (following Will's formula from above): 3.3×10^8kg (mass of Empire State Building) accelerated to go around a circle with r=1km a hundred times per second (more than 6 times faster than the fastest ever man-made object) placed 1km above a LIGO detector (ignoring that this isn't exactly far-field) and we're still 6 orders of magnitude short of detecting this! $\endgroup$
    – Emil
    Commented May 17, 2017 at 13:26

Simply because the gravity is too weak.

You need literally astronomical amount of matter to produce significant gravitational fields. And to produce significant gravitational radiation, "system requirements" are even higher: you need astronomical amount of matter moving around violently.

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    $\begingroup$ You're sort of right, but this answer would be much better if you could provide quantities. $\endgroup$ Commented May 17, 2017 at 7:28

You're missing one very important thing. Spacetime is already distorted on Earth, much more so than the waves we've been detecting - and spacetime distortions aren't the only thing the LIGO detectors pick up. The detectors are extremely sensitive, but to be useful, they must be able to tell the difference between a gravitational wave and a truck passing by.

One solution to this is that there are two separate LIGO observatories. When a distortion appears on one but not the other, we know it's from an environmental source, like thermal noise, ground vibrations, possibly even the shifting of Earth's own gravity (I'm not sure if the changes are large enough to be detectable by LIGO). So to detect a gravitational wave, the "wavefront" must be large enough to affect both observatories at precise time intervals (related to the source). We certainly don't have the energy or equipment to produce a gravitational wave massive enough for that.

Could LIGO detect a man-made gravitational wave? Possibly - as you noted, it's very sensitive. Could it isolate it from the noise? No, not by far. The Livingston observatory is actually inside a worked forest, and it can hear the trees falling; that is, it's sensitive enough to detect the micro-seismic vibrations of the tree hitting the ground, as well as the logging trucks carrying the logs. I'm pretty sure there isn't a place on Earth where you could put a similar observatory without having environmental noise far in excess of what would be expected of man-made gravitational waves, and at these scales, the difference in speed of propagation of gravitational waves and accompanying seismic waves from the source might be less than our ability to detect them - not to mention that even with just one observatory, you still need a wavefront that looks "flat" to the detector itself, with its 4kmx4km dimensions; that's quite a bit of a problem given that the waves spread out in three dimensions.

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    $\begingroup$ If you could modulate the generated wave, you should in principle be able to see it over the noise - it would just be a matter of collecting data over a sufficiently long timeframe. $\endgroup$ Commented May 18, 2017 at 0:42

We can produce gravitational waves in the lab! For instance, we just have to move our body and we will already produce these waves. However, the waves will be so weak that with today's measurement techniques we will not be able to measure them. At the moment the sensitivity of the LIGO experiment is several orders of magnitude too low to measure gravitational waves produced by humans.

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    $\begingroup$ The real question is probably why are not able to measure them with today's measurement techniques. $\endgroup$ Commented May 18, 2017 at 6:34

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