1
$\begingroup$

It's my understanding that many inverse-square laws can be explained as a central point emitting "interaction rays" in all directions equally. And that when another object with some area is "impacted" by those rays, it will then feel an effect proportional to the amount of rays that hit it. This amount can be found geometrically to be the inverse-square of the distance between the objects.

These laws are often used to predict the behavior of tiny particles like electrons, protons, etc. Some of these objects are sometimes conjectured to be point particles.

But then they would have no area. Which means that no matter how close or far they were to the central emitter, they would only ever be hit with a single ray. And the inverse-square law would not be observed. Interaction would be the same at all distances.

Does this mean that particles that follow inverse-square laws cannot be points? That they must have some non-zero area, no matter how 'elementary' they may be?

$\endgroup$
4
  • $\begingroup$ These interaction rays do not exist! for example in gravitation the same two masses, but one of 1/4 the area have the same force to attracting Them to the source go gravitation. So this picture of spreading rays is wrong . $\endgroup$ – trula Sep 19 '20 at 15:33
  • 2
    $\begingroup$ You are right. There is no way to reasonably explain a point particle. Even the comment above points out that The mass of a particle has nothing to do with the area it takes up. You always have particles that are smaller or larger than other ones. There’s no way to prove otherwise. $\endgroup$ – Bill Alsept Sep 19 '20 at 17:24
  • $\begingroup$ You a little bit misinterpret inverse-square law general description, which is, quoted from wikipedia "hence, the intensity of radiation passing through any unit area (directly facing the point source) is inversely proportional to the square of the distance from the point source" Thus if field strength drops as $1/r^2$ per unit area,- then it drops in each unit area $x,y$ coordinate where point particle may be located. So, imho, nothing contradictory here. $\endgroup$ – Agnius Vasiliauskas Sep 19 '20 at 18:09
  • $\begingroup$ Maybe the questioner is asking why it's not proportional to Area1*Area2/r^2 as it would be for the energy exchange between two black bodies? $\endgroup$ – Roger Wood Sep 19 '20 at 18:22
0
$\begingroup$

The language of "interaction rays" is wrong and conceptually misleading, but there is just enough validity in the question to be worth an answer, at least in so far as quantum electrodynamics is concerned (gravity is understood as a geometric effect, and the idea has no application there). The electromagnetic force is transmitted by photons (conceived as massless "point" particles, not rays). We can represent the transmission of a photon from a charge to a spherical surface centred centred on the charge in this diagram.

Since the photon may land randomly anywhere on the surface, the force transmitted by many photons to each part of the sphere is inversely proportional to the area of the sphere.

This is contrasted in qed, in which the force is transmitted by gluons and the gluon-gluon interaction is also taken into account.

enter image description here

The result is that the strong force does not decrease with distance, resulting in quark confinement because energies would rise to the point of particle creation when quarks are separated (e.g. by collisions in high energy scattering).

$\endgroup$

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