In Fermi V-A theory, where does the Weak Interaction occur

  • #1
charlesmartin14
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Where is the electron-nucleon interaction? On the surface of the nucleus, or at the origin (R-0) ?
 

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  • #2
dextercioby
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Can you write down the interaction term or, even better the whole, Lagrangian density?
 
  • #3
charlesmartin14
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The Lagrangian contains a 'contact potential' , which Fermi assumed is a contact coupling of two vector currents. (prior to when parity breaking was discovered. This is now part of the Standard Model, in some cryptic form.

IMHO, you can not see it obviously in the Lagrangian until you specify the Dirac Spinors explicitly and evaluate the invariant matrix elements The Weak Interaction Hamiltonian element is

[tex]\mathcal{H}(x)=-\dfrac{G_{F}}{\sqrt{2}}\left[J^{\mu}(x)L^{+}_{\mu}(x)+h.c.\right][/tex]


where

[tex]J^{\mu}[/tex] and [tex]L_{\mu}[/tex]

are (in modern parlance) the Hadron and Lepton currents, resp.,

The question is really how to evaluate the invariant matrix element -- at the surface of the nucleus, or at the origin ? That is, say, given some expression that appears in a cross section or a rate

[tex] \sum_{fi}\big\vert\mathcal{M}_{fi}\left[p_{ep}\rightarrow i\nabla)\right]\psi_{ep}(\mathbf{x})\big\vert_{\mathbf{x}=0}\big\vert^{2}[/tex]

or would we compute the electron density at the surface ?

for example, we do compute the rate of Electron Capture , a typical weak process, we can write the capture rate as

[tex]\Gamma_{EC}=\big\vert\Psi_{ep}(0)\big\vert^{2}\mathbf{v}^{in}_{ep}\sigma_{EC}[/tex]

where [tex]\big\vert\Psi_{ep}(0)\big\vert^{2}[/tex] is the electron density at the origin, representing the nuclear charge.

Or , again, should this be the charge on the surface of the nucleus ?
 
  • #5
charlesmartin14
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Whatever the experimentalists tell me their best estimate it.
The nuclear radius can be estimated using a simple formula

[tex]R\sim A^{\frac{1}{3}}[/tex]

where A is the Atomic mass number.

I am asking how the Fermi VA theory is typically used in very detailed calculations.
 
  • #6
dextercioby
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Me neither, I have only seen a „finite nuclear radius” hypothesis only in a Dirac equation context (as a perturbative effect). AFAIK, it was described only in section 13.4 of Akhiezer and Berestetskiis book on QED. A finite nuclear hypothesis in case of a weak coupling is not known to me.
 
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  • #7
vanhees71
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Fermi theory is an effective low-energy theory which treats the nucleons as elementary fields.

To define a "proton radius" you need a context, how you measure it. One way is to use electron scattering, defining a charge radius. Another way is to consider the fine structure of the atomic hydrogen transition lines, also defining a charge radius. There's some "crisis" there, because measuring the charge radius with the transitions in muonic atoms, you get a different (smaller result). For a review, see this recent talk from our Nuclear Physics Colloquium:

http://th.physik.uni-frankfurt.de/~hees/np-colloquium/ss17/talk-gao.pdf
 

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