13
$\begingroup$

I‘ve been searching for it on many different websites and they all basically say the same thing. The subatomic particles’ mass is so small that other forces (as the weak force, strong force and electromagnetic force) act on them much more effectively, in such a way that particle physicists may completely ignore gravity when attempting to explain a physical phenomenon involving them.

Ok, my question is: Although the gravitational pull may be extremely weak, it still exists. Shouldn’t it account for something, and thus be considered, in some special situations? For example: Let‘s say we could isolate one subatomic particle (vacuum) inside of a high container in such a way that all the other interactions (weak, strong and electromagnetic) could be mitigated to a value lesser than the (tiny) gravitational pull acting upon this subatomic particle. Question: In this situation, would we observe a tendency of this particle to be dragged to the bottom of the container (gravitational drag towards the center of the earth), after a sufficient amount of time?

$\endgroup$
  • 1
    $\begingroup$ Yes? It just gets dragged to the bottom much slower than most things particle physicists like to talk about. $\endgroup$ – user253751 Aug 20 at 15:44
  • $\begingroup$ You would observe the charge falling in the gravitational field. Its movement won't be a simple descent due to quantum effects though. Interestingly, the distortion gravity causes to the charge's electric field is what makes the charge accelerate downwards, so technically the electromagnetic force is still relevant. $\endgroup$ – Laff70 Aug 20 at 16:43
  • 3
    $\begingroup$ The short answer? Yes. The long answer? Read the rest of the page. :) $\endgroup$ – Mark Morales II Aug 21 at 6:34
  • 3
    $\begingroup$ A “sufficient amount of time” won’t be very long at all by everyday standards. Subatomic particles near the surface of the Earth accelerate towards its centre at 10m/s/s, just like anything else. $\endgroup$ – Mike Scott Aug 21 at 17:33
20
$\begingroup$

Ultracold neutrons have velocities of a few to tens of meters per second and can be transported from source to experiment on ballistic trajectories (governed by gravity). (See the PF2 experiment at Institut Laue-Langevin )

enter image description here

In terms of high energy experiments, say a positron beam scattering from atomic electrons, the gravitational effect would not only be unmeasurable, but they would also be theoretically uninterpretable.

The experimentalists measure a probability for a beam of positrons to scattering at some angle with some energy. Theorists cannot calculate exactly what happens; instead, they do successive approximations using Feynman diagrams.

The simplest diagrams are easy to calculate (even the experimentalist can do it), and it dominates the process:

enter image description here

In higher-order diagrams, each pair of vertices includes a factor of

$$ \alpha = \frac{e^2}{\hbar c} \approx \frac 1 {137}$$

Second-order diagrams look like:

enter image description here

That figure only shows two diagrams. The next level includes all connected diagrams that you can draw.

Enter gravity. Gravity is $10^{36}$ times weaker than the electromagnetic force so that the leading order gravity term is the same strength as the 17-th (plus or minus) order QED term. Note that you need to compute every diagram of that and up to that order, of which there must be $10^{\rm a lot}$.

The most precise calculation ever done, the electron g-factor, involved up to at least fourth-order diagrams, of which there are thousands, and the computations took many researchers years.

So, while one could compute the gravitational contribution to your experiment, it will always be much smaller than electromagnetic terms that you could compute in theory, but simply do not have the resources to do so. In practice, as with the muon magnetic moment, there are also hadronic contributions with uncertainties that are many orders of magnitude larger than the gravity term, so that you cannot even compute them well enough in theory.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Best answer! The only thing missing is the outcome of the cold neutron experiments in a gravitational field which showed a change in the wavefunction's phase. I didn't read the links though. $\endgroup$ – Deschele Schilder Aug 20 at 20:38
  • $\begingroup$ On the subject of ultra-cold neutron experiments, the UCN$\tau$ Collaboration is making a neutron lifetime measurement by putting ultra-cold neutrons in an open-topped bowl made of magnets. Gravity keeps them from spilling over, and the neutron magnetic dipole moment keeps them from falling through. $\endgroup$ – probably_someone Aug 21 at 18:29
  • $\begingroup$ This answer makes me think that somebody already has a Theory of Everything, except that they usually ran out of time and of grant money whenever they tried to use it. What am I missing? Perhaps the fact that Theory of Everything isn't needed with the simplest and thus strongest pure gravity diagrams, but only a bit later in the enumeration where QED and gravitons start to appear in the very same diagram? $\endgroup$ – Jirka Hanika Aug 22 at 16:27
  • 1
    $\begingroup$ @JirkaHanika that effective quantization of gravity works in first order, but when the fool toolkit of QFT is applied there are divergences of course not seen in real life.(as classical gravity works ). That is why a lot of theoretical physicists put their money andtheir work on string theories, which have quantization of gravity, but unfortunately no unique string theory has been found. Future high energy experiments might find supersummetry ( necessary for strings) and further constraints to lead to a theory of everything (strings embed the standard model which has the other three interac) $\endgroup$ – anna v Aug 23 at 5:23
  • $\begingroup$ @annav - Thank you! $\endgroup$ – Jirka Hanika Aug 26 at 16:22
6
$\begingroup$

The electron, an elementary particle, is a subatomic particle. The LEP accelerator beams were beams of electrons and positrons necessary to be controlled with great accuracy in order to collide and the collisions studied. In this conference report, the necessary corrections to the beams due to earth tide effects that induced a change in gravity is described. Note that the electron/positron beams move in a very good vacuum, similar to your requirement.

Fluctuations in the Energy Calibration data [2] were correlated to gravity variations in the Geneva area related to tidal forces.

Thus the answer is that there exists experimental evidence that subatomic particles are affected by changes in the gravitational field.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Very interesting, but it appears that the corrections are necessary not because of gravity affecting the beam trajectory but because the Earth deforms under the tidal forces of the Moon and the Sun: "[The tidal forces] move the Earth surface up and down by as much as -0.25 m which represents a relative local change of the Earth radius of 0.04 ppm. This motion has also lateral components resulting in a change of the LEP circumference...". The position of the quadrupoles changes and thus the geometry of the LEP "optics" is altered, leading to beam deflections. $\endgroup$ – Peter - Reinstate Monica Aug 21 at 7:24
  • 4
    $\begingroup$ The direct deflection due to gravity, let alone gravity differentials (aka tides), is much smaller than that: The particles move close to light speed, covering the circumference of 26.7km in less than 0.1ms. If I didn't screw up on the back of my envelope, 0.1ms of free fall in Earth's gravity should translate to a 5E-8m deviation. The tides move the Earth's surface by 0.25m. $\endgroup$ – Peter - Reinstate Monica Aug 21 at 7:35
  • $\begingroup$ @Peter-ReinstateMonica it is a composite effect they are studying, trying to improve the knowledge of the energy of the beam , they call them "tide-induced local gravity changes." $\endgroup$ – anna v Aug 21 at 10:34
  • $\begingroup$ @Peter-ReinstateMonica it is probably a collective effect, that said, I have not gone through their calculations. Each bunch contained some 250 billion particles. physicsmasterclasses.org/exercises/keyhole/it/accelerators/… $\endgroup$ – anna v Aug 21 at 10:42
5
$\begingroup$

The gravitational field has been measured to induce a measurable phase difference in neutron interferometry. See, for example, the COW experiment, originally published by Colella, Overhauser, and Werner in Phys.Rev.Lett. 34 (1975).

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

If we were able to create a large container, of perfect vacuum, somehow isolated from all external electromagnetic and nuclear forces, and then release a particle into it at the top, then from a classical perspective, the motion of the particle is extremly simple.

It would fall downwards with an acceleration of 9.81 $ms^2$, just as every other mass would. It wouldn't fall any slower, per the definition of the experiment and our understanding of classical mechanics. It would simply accelerate downwards.

Naturally, we can't really do this experiment in real life, at least not in this form, as it is very difficult to prepare perfect vacuums completely isolated from all external forces (and measurement of individual atoms/particles is not well explained by classical mechanics). However, some experiments have come close enough to confirm that atoms fall as expected [1], and it would be weird to think that gravity mysteriously turns off when you break an atom apart into subatomic particles.

But you don't even need to do this experiment to know that atoms are affected by gravity. You can tell by the fact that you are breathing! The earth's atmosphere is made up of a large number of atoms. If they did not respond to gravity, then they would simply fly in straight lines, bounce off the earth, and eventually escape the earth's sphere of influence, never to be seen again. Imagine you threw a large number of perfectly elastic bouncy balls down into a deep pit. If they were unaffected by gravity, they would simply bounce out, and keep going upwards. If they are affected by gravity, then they will bounce upwards, but eventually fall back down again, reaching a height determined by the initial kinetic energy they were given, and the strength of the gravitational field. This is a massive oversimplification, but it is broadly similar to the earth's atmosphere. The fact that molecules in our atmosphere reach heights of hundreds of kilometres before falling back is due to the fact that the force of gravity on an individual atom is very small compared to the forces of collisions with the other molecules. Never the less, they go up, and come back down, same as anything else does


  1. https://news.stanford.edu/pr/99/atomgravity990825.html (Someone please let me know if they can find a reference to the original work.)
| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Question: In this situation, would we observe a tendency of this particle to be dragged to the bottom of the container (gravitational drag towards the center of the earth), after a sufficient amount of time?

Absolutely. It would fall just as fast as any other object. There are three factors that distinguish this from quantum calculations:

  1. There is an external field. While electromagnetic force has polarity, allowing positive an negative charges to cancel out, all particles have nonnegative gravitational mass. Thus, while when considering the forces between just two charged particles, the electromagnetic force dominates gravitational forces, the gravitational force from the earth dominates the electromagnetic force of the earth. The external gravitational force is relevant to your question as to whether the particle will drift down, but the gravitational force between an electron and the nucleus of an atom is not a significant factor in its orbital.

  2. You are looking from the perspective of an outside frame of reference. According to the Equivalency Principle, gravitational fields are locally equivalent to a (generally noninertial) frame of reference without any gravitational force. Since we generally consider particle interactions in the frame of reference of their center of mass, gravity can be disregarded. For instance, when an electron orbits the nucleus, gravity will pull the electron down, but it will also pull the nucleus down. Since everything's moving together, it's not affecting what's happening to the electron relative to the nucleus. If scientists are measuring the spectral lines of an atom, the question of whether they're in a spaceship in outer space or hurtling towards a planet in freefall would be relevant to the scientists themselves, but would not be relevant to the experiment.

  3. Quantum mechanics generally takes place over very short time frames. Particles therefore don't have much time to accelerate due to gravity.

So when physicists say that gravity can be disregarded for quantum mechanics, what they mean is that the gravitational force within a system is too small to matter (unless you're going for an extremely high level of precision), and outside gravitational forces generally don't have time to act, and even if they do, they just move the center of mass of the system, rather than having a significant effect of what's going on inside the system.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The answer is yes, the particle would be dragged towards the Earth, the point is it would be very slow. For almost all physical calculations on the atomic scale the effects of gravity are treated as if they weren't there because gravity is so much weaker than the other 3 forces, for a comparison see here.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But do you know any experiment actually done to prove it? Or it is just an assumption, since our gravity theories say that every object with mass (no mattering how tiny it is) is affected by gravitaty? $\endgroup$ – DiBeos Aug 20 at 16:06
  • $\begingroup$ @DiBeos Maybe try this link? $\endgroup$ – Charlie Aug 20 at 16:15
  • $\begingroup$ @DiBeos physicsworld.com/a/neutrons-reveal-quantum-effects-of-gravity $\endgroup$ – G. Smith Aug 20 at 16:26
0
$\begingroup$

What you're describing is partly impossible. You can't shield from the weak and strong forces, you'd be able to do some weird stuff if you could, like preventing radioactive decay and blowing protons up at will.

But, let's assume we have a setup where all effects other than gravity are close to zero (very, very tricky). What happens? Nobody knows. The most "sensible" answer is that you would see the particle fall just like any other object, or possibly jitter around but on average follow a normal "falling due to gravity" path.

But the simple answer is nobody knows, this is one of the areas where the mismatch between relativity and quantum mechanics become obvious. We know that in some way a group of particles have to fall like a rock does...a rock is just a group of particles after all, but what an individual particle does is less clear.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ An experiment with cold neutrons has been performed which showed the effect of the neutron's wavefunction: the phase of the wavefunction shifted. $\endgroup$ – Deschele Schilder Aug 20 at 20:41
  • $\begingroup$ What's an "individual particle"? Does a neutron count? Because if so, there are ultra-cold neutron traps that are just bowls made of magnets, that rely on gravity to keep neutrons from spilling over the sides. $\endgroup$ – probably_someone Aug 21 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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