Wednesday, 4 March 2020

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

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.


Please explain!



Answer



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).


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