27
$\begingroup$

As I understand, we usually talk about gravity at a macro scale, with "objects" and their "centre(s) of mass". However, since gravity is a property of mass generally (at least under the classical interpretation), it should therefore apply to individual mass-carrying particles as well.

Has this ever been shown experimentally? For example, isolating two particles in some manner and then observing an attraction between them not explained by other forces.

To pose the question another way, let's say I have a hypothesis that gravitation is only an emergent property of larger systems, and our known equations only apply to systems above some lower bound in size. Is there any experiment that reasonably disproves this hypothesis?

Improve this question
$\endgroup$
10
  • 2
    $\begingroup$ Isn't this basically asking about a theory for quantum gravity? A theory which we don't have yet? You know, the biggest unknown in physics right now? The unification of relativity and quantum mechanics? $\endgroup$
    – DKNguyen
    Commented May 21, 2022 at 0:30
  • 19
    $\begingroup$ @DKNguyen I do not study physics in any serious capacity so it may seem like a silly question. Still, I'm not asking for a theory, but rather if it has ever been observed. $\endgroup$
    – Hymns For Disco
    Commented May 21, 2022 at 1:31
  • 3
    $\begingroup$ It hasn't because we don't have the tech which is what makes it difficult. That's why physicists are only able to try and reconcile the existing math between the two theories and hoping that whatever they come up is unique and says something about quantum gravity. They don't have direct observations of quantum gravity which they could use to guide the way to reconcile the other two theories, or to come up with a separate theory. $\endgroup$
    – DKNguyen
    Commented May 21, 2022 at 1:48
  • 5
    $\begingroup$ At the very least we know photons of light are affected by gravity, as gravitational lensing is routinely observed and used as a tool in astronomy. Granted, I highly doubt this has ever been verified for an individual photon. $\endgroup$
    – RC_23
    Commented May 22, 2022 at 4:33
  • 1
    $\begingroup$ Isn't the fact that the Earth, or the Moon, or Sun, exist, proof that small particles attract? How can it be that only big collections of particles cause gravity? If the electric forces had gravity's strength, would only large collections of charges cause electricity? What's the rationale for thinking this in the first place? $\endgroup$
    – MatterGauge
    Commented May 23, 2022 at 8:16

5 Answers 5

Reset to default
48
$\begingroup$

For the interaction of one small (atom scale) mass and one large mass, measurements of the Earth's atmosphere that anyone could do with a homemade barometer and a nearby mountain constitute direct experimental confirmation. We find more gas molecules at low altitudes than at high altitudes. Only gravity acting on each gas molecule independently could be responsible for the observed behavior - they behave like a gas, but they don't just float away and uniformly distribute themselves across the cosmos, but instead assemble themselves into a pressure gradient pointing towards the center of the planet.

For the interaction of two small masses, rather than one small mass and Earth, or large distributions of small masses (e.g. nebula formation), the smallest I've read about is this one from last year, using ~90mg gold spheres. See arXiv: 2009.09546 [gr-qc].

Improve this answer
$\endgroup$
27
$\begingroup$

Here is an easy way to grasp how difficult it would be to make the direct measurement you propose.

Take two protons and place them one centimeter apart. They will exert a certain tiny amount of gravitational attraction, which we measure by some magic means, and a certain amount of electrostatic repulsion, which we will also measure.

Now, how far apart would we need to separate those two protons in order for the strength of the electrostatic force they experience to diminish to the point where it is as weak as the gravitational force they experience when they are one centimeter apart? Answer: 1.8 light years.

This means that when performing experiments where we have to account for electrostatic forces between individual subatomic particles, those forces will be stronger than the gravitational forces between them by a factor of (1.8 light years/1 centimeter).

And that means that we have no hope of ever, ever directly measuring the force of gravitational attraction between two protons in an experiment: the electrostatic force will utterly overwhelm that experiment.

Improve this answer
$\endgroup$
7
  • 7
    $\begingroup$ That is a very interesting and surprising (to me) statistic. What about 2 neutral charge particles? I suppose it will still be nearly impossible to compensate for external gravitational influences. $\endgroup$
    – Hymns For Disco
    Commented May 21, 2022 at 4:11
  • 1
    $\begingroup$ @HymnsForDisco, quite right- because you've got the earth right there... $\endgroup$
    – niels nielsen
    Commented May 21, 2022 at 6:30
  • 4
    $\begingroup$ If Earth was the only problem, we could have done it in space. Expensive, but doable. However, the mass of any space station is going to be by many magnitude orders higher than mass of neutron (or probably any other particle). Weighting a grain od sand added to a large heap seems to be much easier… $\endgroup$
    – v6ak
    Commented May 21, 2022 at 20:43
  • 7
    $\begingroup$ The problem isn't outside gravitational interference - with a judicious choice of reference frame, it can be worked around - the problem is the magnetic, residual strong and weak forces, all of which neutrons experience, and all of which (yes, even the non-binding weak force) are stronger than gravity by orders of magnitude at this scale. $\endgroup$
    – No Name
    Commented May 22, 2022 at 15:15
  • 2
    $\begingroup$ As long as your experiment is sensitive enough, you can distinguish the contributions from electromagnetic and gravitational force. For example by comparing to a second experiment involving a proton and an anti-proton. $\endgroup$
    – laolux
    Commented May 23, 2022 at 6:50
11
$\begingroup$

FWIW, small particles react to the big ones:Experiments have been done with neutrons in a gravity field. The phase of their wavefunction was shifted, as was shown by interference. If the neutron didn't have it's own gravity field, would it react? Would an electron without charge accelerate in an electric field? Food for thought.

Improve this answer
$\endgroup$
1
$\begingroup$

That's exactly what we are observing when we look at the night sky and see galaxies form out of nothing but dust.

Ever wondered why we are here to ask this question? Because gravity pulled together nothing really just dust of particles in the very early universe, forming all the galaxies and solar systems etc.

As a side note, if gravity would not exist at the particle level, this would make it very hard to explain the dark matter halos around the galaxies, since these particles only interact via gravity.

Since the dark matter does not dissipate as it only interacts gravitationally, it remains distributed outside the disk in what is known as the dark halo.

https://en.wikipedia.org/wiki/Galaxy_formation_and_evolution

So at the particle level, the existence of complex (gravitationally bound) systems is the very proof that gravity (understanding how it binds simple dust of particles) really works at the particle level.

Improve this answer
$\endgroup$
0
$\begingroup$

Mass Spectrometer

I don't recall the formal name for the instrument but I was able to tour a lab at Ohio State University that had mass spectrometers that lofted ionized individual molecules a few feet up a column and waited for them to pass a detector as they fell due to gravity. Organic molecules aren't quite point masses but they are about as close as you can reliably get measurements for. The instrument was considered to be best in class at the time and this was only a few years ago.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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