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The Nuclear Force

Friday, April 1, 2022

Characteristics of the Nuclear Force

Nuclei have no nuclear excited states (bringing a proton and neutron together does not emit several photons). Instead, scattering neutrons from protons and doing experiments with heaver nuclei have resulted in the following properties being learned about the nuclear force (sometimes called the strong force):

  1. The nuclear force is the strongest of the known forces. For adjacent protons, the nuclear interaction is 1-2 orders of magnitude stronger than the electromagnetic interaction.
  2. The strong force has very short range (∼10−15 m\sim10^{-15}~\text{m}), which explains constant nuclear density. Nucleons only experience a bonding force with their nearest neighbors (unlike gravity or the electromagnetic interaction, which have infinite range).
  3. The nuclear force between any two nucleons does not depend on the type: the n-p nuclear force is the same as the n-n nuclear force, which is the same as the nuclear part of the p-p force.

The strong force can be modeled as an exchange force, where a neutron emits a particle that a proton feels a strong attraction to, so much so that it absorbs it. Then the proton emits a particle that the neutron feels a strong attraction to, and so on. This appears as an attractive force between the neutron and proton itself.

This appears to violate conservation of energy, but that is not a problem due to the uncertainty principle ΔtΔE∼ℏ\Delta t\Delta E\sim \hbar: if the energy ΔE\Delta E is used for a time at most Δt=ℏ/ΔE\Delta t=\hbar/\Delta E, conservation of energy has not been violated. By extension of the rest energy being mc2mc^2, we say

Δt=ℏmc2\Delta t=\frac{\hbar}{mc^2}

Using the fact that the particle can't move faster than the speed of light (meaning the farthest it can move is x=cΔtx=c\Delta t), we see that the rest energy can be described as

mc2=ℏcxmc^2=\frac{\hbar c}{x}

Using the range of the strong force, 1 fm1~\text{fm}, we see the rest energy of the exchanged particle:

mc2=ℏcx=200 MeV⋅fm1 fm=200 MeVmc^2=\frac{\hbar c}{x}=\frac{200~\text{MeV}\cdot\text{fm}}{1~\text{fm}}=200~\text{MeV}

Note: the particle cannot be observed in laboratory experiments without violating conservation of energy. However, by providing energy to the proton in the form of a photon, the particle can be observed. This particle is found to be pi mesons (pions), with rest energies of 140 MeV140~\text{MeV}.