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

  • $\begingroup$ Do you include the weak force in "strong forces" (not a rhetorical question and not playing with words)? $\endgroup$ Commented Mar 20, 2018 at 8:14
  • $\begingroup$ No I don't, since weak force decrease with distance, it is as expected $\endgroup$
    – STAIN
    Commented Mar 20, 2018 at 9:08
  • $\begingroup$ Related question: physics.stackexchange.com/questions/396004/… $\endgroup$
    – rob
    Commented Mar 30, 2018 at 15:07

1 Answer 1


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.

  • 4
    $\begingroup$ More precisely, inverse-square forces follow from massless, chargeless exchange bosons. Half of the exchange bosons don't have this property. $\endgroup$ Commented Mar 19, 2018 at 17:54
  • 1
    $\begingroup$ Indeed, should I edit my answer? $\endgroup$ Commented Mar 19, 2018 at 17:57
  • 1
    $\begingroup$ Probably, as this is a better explanation than "just 'cuz." $\endgroup$ Commented Mar 19, 2018 at 18:19
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    $\begingroup$ 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.) $\endgroup$
    – David Z
    Commented Mar 19, 2018 at 20:56
  • 2
    $\begingroup$ 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. $\endgroup$
    – Džuris
    Commented Mar 19, 2018 at 22:27

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