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Most of the forces induced by a point particle follows the $1/r^2$ rule. Then why does the strong force not obey it?

Peter Mortensen
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STAIN
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1 Answers1

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Most of the forces induced by a point particle follows the 1/r^2 rule

No, it's the forces mediated by point particles with no mass and charge that follow the the 1/r^2 rule.

then why does strong force don't obey it?

The inverse square law is a consequence of the particles having no mass and/or charge. Such particles have long/infinite lifetimes and can travel to long distances so that they have time and space to "spread out" and cause their force to fall off with distance.

The weak force's W's and Z's have mass. Thus they have very short lifetimes, so they don't travel very far. As a consequence, they act over very short distances and then basically disappear. The weak force looks inverse square at very short distances, but disappears at longer ones.

The strong force's gluons are massless, so at first glance they could follow the inverse square. However, they also have color charge (as well as electric), which has entirely different physics. This gives rise to the creation of new particle pairs (mesons) that carry this residual strong force that holds nuclei together. These mesons are massive, so we're back to the first case again.

So two of the basic forces are inverse square. Two are not. And that's because of the particles that mediate them.

Maury Markowitz
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    More precisely, inverse-square forces follow from massless, chargeless exchange bosons. Half of the exchange bosons don't have this property. – probably_someone Mar 19 '18 at 17:54
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    Indeed, should I edit my answer? – Maury Markowitz Mar 19 '18 at 17:57
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    Probably, as this is a better explanation than "just 'cuz." – probably_someone Mar 19 '18 at 18:19
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    It might be worth more strongly emphasizing that it's the mass (and other properties) of the mediator particle that matters. I think it would be a little too easy for someone to glance at that answer, notice the bold phrase "point particles with no mass and charge", and come away with the impression that you're talking about the particles subject to the force. (Good answer otherwise, though.) – David Z Mar 19 '18 at 20:56
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    Double on what David asks. I thought you are explaining that $1/r^2$ forces are between massless and chargeless point particles and had to reread to understand the point. – Džuris Mar 19 '18 at 22:27
  • If only point particle with no mass follows the 1/r^2 rule then why gravity work on it – STAIN Mar 20 '18 at 01:14
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    @STAIN You've misunderstood what Maury said. He said that the mediator (the photon for E&M, the graviton for gravity if you wish to formulate it as a quantum field theory) has no mass. – dmckee --- ex-moderator kitten Mar 20 '18 at 01:47
  • Is inverse square law about intensity distribution https://en.m.wikipedia.org/wiki/Inverse-square_law – STAIN Mar 20 '18 at 02:45
  • Doesn't the strength of the color-charge coupling constant matter nearly as much as the existence of color charge on the gluon? – rob Mar 20 '18 at 03:35
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    Why is it that if a mediator has mass, it "disappears" at large distance? – Ruslan Mar 20 '18 at 07:29
  • @STAIN - the "particle of gravity" in QM is the graviton. It is chargeless and massless. It also might not exist - gravitation remains best explained by GR, where it is not mediated by particles at all. In this case, the 1/r^2 "falls out" of the math from a different source, but in the end the same geometry of "spreading out" applies and you get the same effect. – Maury Markowitz Mar 20 '18 at 13:17
  • @MauryMarkowitz does the existence of graviton theoretically proven, if then there is nothing in this universe that has mass, which is like everything is made up of strings and strings has no mass. If then how does mass is defined in QM – STAIN Mar 20 '18 at 13:24
  • @STAIN - strings have mass. E=mc^2, and they definitely have E. What they may or may not have is rest mass, which is different. In QM, rest mass is the sum of the particle's interactions with the fields it sits in, including the "background" Higgs field. Particles may or may not have interactions with these fields, which is why they have different masses. – Maury Markowitz Mar 20 '18 at 13:34
  • @Ruslan - the lifetime of a virtual particle is the inverse of its mass due to the uncertainty principle. As a result, massive virtual particles like a W have very short lifetimes. Their range is simply a function of how far they can go in that time, limited by the speed of light. So that's the range of their force. – Maury Markowitz Mar 20 '18 at 13:37
  • @MauryMarkowitz so basically particles are group of groups of strings interacting with each other and hence they have mass. Is it how it works – STAIN Mar 20 '18 at 13:38
  • @STAIN - yes - if you believe in string theory anyway :-) QM has its own solutions to these problems. You might want to read "The Great Design: Particles, Fields, and Creation" - far from perfect, but covers most of this in an easy-to-read fashion. – Maury Markowitz Mar 20 '18 at 13:41
  • @MauryMarkowitz ok I'll refer it, tnx for the suggestion. Do u believe in strings, and how much backed up is string in theoretical physics. Is there any one made any models of universe based on string theory? – STAIN Mar 20 '18 at 13:44
  • I am not sure about the 'no charge condition' here. If it means mediating particles do not interact with themselves (probably at tree level), then gravitons do not obey this condition, if we are to believe quantised GR as an EQFT. rob's comment seems quite crucial. – Dexter Kim Mar 21 '18 at 01:27