If the quarks that make up the proton are point particles, and the forces binding them together is the strong force which is $137$ times stronger than the electromagnetic force (which makes the quarks repel), why don't the protons collapse to a point because the quarks don't have a finite radius?
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2The strong force is repulsive over small domains of separation – Nov 25 '17 at 23:09
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Thanks. You should put that as an answer though. – Nov 25 '17 at 23:12
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1Also, angular momentum is a thing. For example, in a gravitational system of two point masses, they only collapse to a point if angular momentum is zero. – probably_someone Nov 25 '17 at 23:13
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2There's also the Heisenberg uncertainty relation, which means that any particle confined to a tiny region must have an enormous amount of kinetic energy. – tparker Nov 25 '17 at 23:15
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1Downvoters care to comment? – Nov 25 '17 at 23:55
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Ultimately, for the same reason that electrons don't crash into the nuclei they “orbit”: because all massive particles obey the Heisenberg Uncertainty Principle, of the form $$ \Delta x\,\Delta p\gtrsim\frac12\hbar, $$ so that if the quarks' motion collapsed to a point, having $\Delta x$ zero (or very small) would require having infinite (or extremely large) $\Delta p$, and since, as a rough approximation, $$ \Delta p^2 = ⟨p^2⟩-⟨p⟩^2 = ⟨p^2⟩=T, $$ the kinetic energy, having large $\Delta p$ requires a lot of energy.
(In addition to that, there's the fact that the strong nuclear force is repulsive at low distances, but that is irrelevant - the uncertainty principle would preclude the collapse even if the strong nuclear force was attractive all the way through, like the Coulomb force on electrons is.)
Emilio Pisanty
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2I know that the residual strong force is repulsive over short distances but i am actually talking about the fundamental strong force, which isn't repulsive. – Nov 26 '17 at 23:12