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Does the weak nuclear force play a role (positive or negative) in nuclear binding?

Normally you only see discussions about weak decay and flavour changing physics, but is there a contribution to nuclear binding when a proton and neutron exchange a $W^\pm$ and thus exchange places? Or do $Z$ exchanges / neutral currents contribute?

Is the strength of this interaction so small that it's completely ignorable when compared to the nuclear binding due to residual strong interactions?

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Well, intuitively, even repulsive EM force between protons is weak compared to the attractive nuclear force that holds nucleus together. So it should not be surprising that weak force has even much lesser say. Of course there will be some contribution but probably completely negligible one. –  Marek Jan 3 '11 at 11:20
This is a GREAT question. Its answer is not obvious to me and it could shed light on beta decay - which is mediated by the weak force. As for the relative strengths of the various forces, keep in mind that when in a strong-field regime, as inside a nucleus, such distinctions can break down. At high-energies the strong force becomes weaker and the weak force becomes stronger :) –  user346 Jan 3 '11 at 19:57

2 Answers 2

up vote 5 down vote accepted

A back of the envelope calculation suffices, meaning all factors of $2,3, \pi$ etc. have been ignored.

The residual strong force is mainly due to pion exchange and can be modeled by this plus a short range repulsion due to exchange of $\omega$ mesons. The Yukawa potential from pion exchange is $e^{-m_\pi r}/r$. The weak interactions arise from $W^\pm$ and $Z$ exchange and will give rise to a similar potential with $m_\pi$ replaced by $m_Z$ (in the spirit of the calculation I take $m_W \sim m_Z$). At typical nuclear densities of $.16/(fermi)^3$ nuclei are separated by distances which are within a factor of two of $1/m_\pi$. Thus the ratio of the Yukawa potential due to weak exchange to the Yukawa potential due to pion exchange at these densities is roughly $e^{-m_W/m_\pi} \sim e^{-640}$. I've ignored the fact that there are different coupling constants in front of the two potentials, but this difference is irrelevant compared to the factor of $e^{-640}$. So yeah, you can ignore the weak contribution to the binding energy.

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Nice! You've made it look simple enough that I'm kicking myself for not having thought of that. –  Simon Jan 4 '11 at 2:32
@Harvey: While the ratio of the weak to the strong force between two nucleons is $e^{-640}$, there's still the weak force acting upon a single nucleon as in beta decay. If the resulting nucleus is unstable due to beta decay, this would show up as a broadening of its mass spectrum with a Breit-Wigner distribution. The contribution is still small, but is no longer that exponentially suppressed. –  QGR Jan 13 '11 at 23:07

The answer is, there is a very small affect. If you take a look at the Wikipedia article, you will note that the weak force is about 10^-13 weaker then the strong force. Therefore it's just not going to have much of an affect.

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I think that 10^-13 is compared to the strong force that binds quarks, not to the residual strong force that binds nuclei... –  Simon Jan 3 '11 at 21:02

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