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From what I know, gravitational waves are produced when accelerated massive bodies move through space-time and create ripples in the gravitational fields throughout the space-time. This is significant on a cosmological scale. What about in small scale experiment? Is it possible to produce gravitational waves using some amount of a gas in a very small volume of space?

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    $\begingroup$ Is it possible to change the orbit of the Moon by moving a heavy rock to the top of my house, thus changing the gravitational pull on the Moon? $\endgroup$ – Aaron Stevens Jan 26 at 5:49
  • $\begingroup$ Even if the gas atoms had a significant mass (enough to produce g-waves) g-waves wouldn't be emitted. They collide (i.e. in an ideal gas) elastically. $\endgroup$ – descheleschilder Jan 26 at 7:02
  • $\begingroup$ According to GR, collisions of atoms produce gravitational waves that we cannot detect. Thus there is no perfectly elastic collision. They lose a tiny amount of energy in gravitational radiation. $\endgroup$ – G. Smith Jan 26 at 7:03
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    $\begingroup$ Mark, I wonder why you particularly mention gas - gas seems totally unrelated to the issue? If your question was "can we produce gravitational waves by moving two baseballs" that would make sense, great question. I don't really follow how "gas" comes in to the issue? $\endgroup$ – Fattie Jan 26 at 17:18
  • $\begingroup$ @Fattie , though we know gravity and gravitation are the same thing, 'gravity waves' and 'gravitational waves' are different, the former is discussed in fluid dynamics while the latter is discussed in general relativity. $\endgroup$ – M. A. Bromuela Jan 31 at 9:29
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Yes, even waving your hand creates gravitational waves. But the waves produced by non-astronomical sources are far too weak to detect.

The relevant formula for calculating the metric perturbation is the quadrupole formula in case you want to try out a particular scenario. For example, suppose we suddenly move 1 gram of gas by 1 centimeter in 1 microsecond. The metric perturbation 1 meter away is on the order of $10^{-39}$.

A device like LIGO can detect metric perturbations as small as about $10^{-21}$, so it would need to be made 1,000,000,000,000,000,000 times more sensitive and shrunk down considerably.

Physicists spent a billion dollars of taxpayers’ money building LIGO because they knew, based on this formula derivable from Einstein’s field equations, that a tabletop experiment would not work. And now we have a way of observing some of the universe’s most amazing events, like the merger of two black holes, rather than just observing something boring like a bit of gas moving in a lab. A new era of astronomy has begun. We can now study gravity under extreme conditions, and perhaps eventually find that General Relativity needs corrections.

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    $\begingroup$ Isn't it the derivative of acceleration that causes the ripples in spacetime? $\endgroup$ – descheleschilder Jan 26 at 6:57
  • $\begingroup$ No. As the formula shows, it is the second time derivative of the mass quadrupole moment. $\endgroup$ – G. Smith Jan 26 at 7:05
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    $\begingroup$ The dimensions of the second time derivative of a mass quadrupole moment are $ML^2T^{-2}$. The dimensions of the first time derivative of acceleration are $LT^{-3}$. Not the same thing! $\endgroup$ – G. Smith Jan 26 at 7:18
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    $\begingroup$ I don’t see the connection between that and spin 2. Gravitons have spin 2 because they are the quanta of a tensor field. Photons have spin 1 because they are the quanta of a vector field. Higgs bosons have spin 0 because they are the quanta of a scalar field. Explaining these connections requires the mathematics of group representation theory. $\endgroup$ – G. Smith Jan 26 at 7:24
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    $\begingroup$ The sensitivity of LIGO is a function of its size. If you want something better and smaller you need a completely new technology. $\endgroup$ – OrangeDog Jan 27 at 11:01

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