# Why is the strong nuclear force > electrostatic repulsion?

In a nucleus there is a gravitational force between the nucleons and also electrostatic repulsion between the protons, and since electrostatic repulson >> gravitational attraction, it follows that there must be an additional attractive force acting on the nucleons or else there is nothing stopping them from flying apart.

So if we let the gravitational and electrostatic forces be $g$ and $e$, respectively, and denote the additional attractive force by $x$, then we would need $g+x=e$ (because the attractive and repulsive forces must balance).

This gives $x=e-g<e$, implying that the additional attractive force must be less than electrostatic repulsion. Since the strong nuclear force must be part of $x$, we then have strong nuclear force $\le x<e$.

But in my revision guide, it says that the strong nuclear force is more than electrostatic repulsion, which seems counter-intuitive according to the above.

• "then we would need $g+x=e$ (because the attractive and repulsive forces must balance)" You are thinking of nucleons like little billiard balls and that model isn't really appropriate in this intrinsically quantum realm. That equality does not hold. – dmckee May 7 '15 at 18:57
• @dmckee: Thanks! But is there still an intuitive quantum explanation for why strong force > electrostatic repulsion? – user45220 May 7 '15 at 19:06

Consider the Earth-Moon system. They are subject to an attractive force (gravitation) and to no repulsive forces (neglecting solar tides, anyway), yet they stay at a nearly constant distance from one another because of their dynamics.

A a static analysis of this system would prompt us to postulate some repulsive force holding the bodies apart (and you can find it by using a non-inertial frame of reference: it is the centrifugal pseudoforce). The lesson is that static analysis will break when applied to dynamic systems.

You are trying to analyze the nucleus in terms of statics when it is a dynamic system (and moreover a dynamic quantum system). As nuclear particles are confined to a limited region in space they necessarily acquire a larger range of momenta as a consequence of the commuter between positions and momentum (we can wave our hands and say "Heisenberg Uncertainty Principle" if you want a shorter label for this effect).

• @dmckee in the Feynman lectures, Feynman said that the nuclear force can be approximated as $F=(1/r^2)exp(-r)$. It seems to me that although it falls off really fast, it has infinite range. So how come it has short range? – Omar Nagib Jul 13 '15 at 17:01