The curvature of spacetime has been observed many times from the deflection of light around massive astronomical objects. But has it been observed around small objects in a lab?

In the Cavendish experiment, the gravitational attraction between two masses does sufficient effect for it to be measured on Earth. Thus, it raises the question whether light deflection from curved spacetime is also measurable in the lab.

If it has not been achieved, how far are we from it? How much precision would be needed?

  • 1
    $\begingroup$ Note that the Cavendish experiment does measure, in a sense, the curvature of spacetime. So the answer to the title is "yes", and your own example proves it. $\endgroup$ – AccidentalFourierTransform Sep 8 '17 at 16:07
  • $\begingroup$ @AccidentalFourierTransform, it does not make a difference between Newton's or Einstein's gravity though. Or does it ? $\endgroup$ – fffred Sep 8 '17 at 16:13
  • 1
    $\begingroup$ No, but so what? the fact that Newton explains the experiment as well does not mean that you are not measuring a GR effect after all. The effect is not unique to GR, but it is, however, a manifestation of curvature as we understand it. $\endgroup$ – AccidentalFourierTransform Sep 8 '17 at 16:14

Yes. A relatively direct measurement of curvature of spacetime has recently been performed using a 16 cm atom interferometer setup: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.118.183602 (paywalled).

The basic idea of this type of experiment is that a packet of atoms is prepared in a quantum superposition of locations, then recombined much like light on a beamsplitter, and the phase change is read out. The two spatially separated parts of the wavefunction experience difference phase shifts due to the small changes in the local gravitational field. Rather amazingly, the curvature that they measure is not due to the Earth's field- they need something they can move, so that they can show a phase change that depends on its position. So, they measure the curvature from an 84 kg test mass placed near the interferometer. As a result, this is a measurement of curvature on the human scale both in terms of size and in terms of the amount of mass generating the curvature.

Note that the authors make a distinction between this measurement, which equivalently can be seen as a measurement of the (non-local) gravitational tidal force, and the (local) measurement of acceleration at a point or gravitational redshift. This curvature measurement, unlike those two types of measurements, is coordinate-independent and in particular can not be replicated by going into an accelerated frame of reference. To do this, their device actually involves two separate interferometers, whose outputs are themselves interfered to effectively make a comparison between two spatially separated measurements of the gravitational shift.

See here for more details at a relatively accessible level (open access).


Yes it has, on 11th Feb last year - it did make many mainstream news reports. In one of the largest "labs" ever built, each arm is 2.5km long. I guess that fits the criteria "human scale" within certain limits.


You may also read about various other methods here


  • $\begingroup$ Thank you for this answer. However, my question is about human scale masses. $\endgroup$ – fffred Sep 10 '17 at 12:37
  • $\begingroup$ Well I know that LIGO suffers noise from local sources, rabbits and rodents in nearby fields, but I suspect that is a vibrational problem. $\endgroup$ – JMLCarter Dec 5 '17 at 0:19

If you are willing to make gravitational time dilation count and if about a mile of elevation counts as human scale, then this is the coolest of my bookmarks I can share: http://www.leapsecond.com/great2005/

The author is an expert on atomic clocks and he set up an experiment to show they ran differently 1340 meters higher, by driving up Mt Rainier, for the education of his kids. Scores high in the category "my dad is a geek!" But cool.

  • $\begingroup$ This effect seems to be due to the earth mass, not a human scale object $\endgroup$ – fffred Sep 10 '17 at 12:39
  • $\begingroup$ True enough, I misinterpreted your question. $\endgroup$ – user154997 Sep 10 '17 at 14:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.