Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with a formula in Weinberg

  1. Jan 25, 2008 #1
    Alright, I'm trying to follow Weinberg's derivation of the matter part of the Lagrangian for electroweak theory, and I am all confused. This is equation (21.3.20) in volume II.

    He writes:

    [tex]
    iL_e = - \left( {\begin{array}{*{20}c}
    {\bar \upsilon _e } \\
    {\bar e} \\

    \end{array} } \right)\sum\limits_\alpha {\gamma _\mu A^\mu _\alpha } t_\alpha \left( {\begin{array}{*{20}c}
    {\upsilon _e } \\
    e \\

    \end{array} } \right)
    [/tex]

    but I think he meant to write

    [tex]
    iL_e = - \left( {\begin{array}{*{20}c}
    {\bar \upsilon _e } \\
    {\bar e} \\

    \end{array} } \right)\sum\limits_\alpha {\gamma _\mu \left( {A^\mu _\alpha t_{\alpha L} + B^\mu y} \right)} \left( {\begin{array}{*{20}c}
    {\upsilon _e } \\
    e \\

    \end{array} } \right)
    [/tex]

    He defines his left handed and right handed parts different than Itzykson and Zuber

    Namely [tex]e_L = \frac{1}{2}\left( {1 + \gamma _5 } \right)[/tex] and [tex]
    e_R = \frac{1}
    {2}\left( {1 - \gamma _5 } \right)
    [/tex]

    but whatever

    From this I was able to derive next line

    Namely

    [tex]
    \begin{gathered}
    - \left( {\begin{array}{*{20}c}
    {\bar \upsilon _e } \\
    {\bar e} \\

    \end{array} } \right)[\frac{1}
    {{\sqrt 2 }}\gamma _\mu W^\mu \left( {t_{1L} - it_{2L} } \right) + \frac{1}
    {{\sqrt 2 }}\gamma _\mu W^{*\mu } \left( {t_{1L} + it_{2L} } \right) \hfill \\
    + \gamma _\mu Z^\mu \left( {t_{3L} \cos \theta _W + y\sin \theta _W } \right) + \gamma _\mu A^\mu \left( { - t_{3L} \sin \theta _W + y\cos \theta _W } \right)]\left( {\begin{array}{*{20}c}
    {\upsilon _e } \\
    e \\

    \end{array} } \right) \hfill \\
    \end{gathered}
    [/tex]

    but when I tried to go to the final line of his derivation I got an opposite sign and some different stuff, namely I got

    [tex]
    \begin{gathered}
    - \frac{g}
    {{\sqrt 2 }}\left( {\bar e\gamma _\mu W^\mu \left( {\frac{{1 + \gamma _5 }}
    {2}} \right)\upsilon _e } \right) - \frac{g}
    {{\sqrt 2 }}\left( {\bar \upsilon _e \gamma _\mu W^{*\mu } \left( {\frac{{1 + \gamma _5 }}
    {2}} \right)e} \right) \hfill \\
    + \frac{1}
    {2}\sqrt {g^2 + g'^2 } \bar \upsilon _e \gamma _\mu Z^\mu \left( {\frac{{1 + \gamma _5 }}
    {2}} \right)\upsilon _e - \frac{1}
    {2}\frac{{\left( {g^2 - g'^2 } \right)}}
    {{\sqrt {g^2 + g'^2 } }}\bar e\gamma _\mu Z^\mu \left( {\frac{{1 + \gamma _5 }}
    {2}} \right)e \hfill \\
    - g'\sin \theta _W \bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}
    {2}} \right)e - \left( {\begin{array}{*{20}c}
    {\bar \upsilon _e } \\
    {\bar e} \\

    \end{array} } \right)\gamma _\mu A^\mu \left( { - t_{3L} \sin \theta _W + y\cos \theta _W } \right)\left( {\begin{array}{*{20}c}
    {\upsilon _e } \\
    e \\

    \end{array} } \right) \hfill \\
    \end{gathered}
    [/tex]

    The first four terms I got are similar to his, but with a different sign, and I understand he uses a Gell Mann Nishijima equation to get last part, but how do you get

    [tex]
    - g'\bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}
    {2}} \right)e + e\left( {\bar e\gamma _\mu A^\mu e} \right)
    [/tex]

    from
    [tex]
    - \left( {\begin{array}{*{20}c}
    {\bar \upsilon _e } \\
    {\bar e} \\

    \end{array} } \right)[\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}
    {2}} \right)\sin \theta _W + \gamma _\mu \left( { - t_{3L} \sin \theta _W + y\cos \theta _W } \right)\left( {\begin{array}{*{20}c}
    {\upsilon _e } \\
    e \\

    \end{array} } \right)
    [/tex]

    even with

    [tex]
    q = - \sin \theta _W t_3 + \cos \theta _W y
    [/tex]
     
  2. jcsd
  3. Jan 30, 2008 #2
    Anybody? Help!!!!
     
  4. Jan 30, 2008 #3

    mjsd

    User Avatar
    Homework Helper

    trying to learn EW theory from Weinberg?... not a good choice! :smile:
    sorry don't have a copy of Weinberg at hand
     
    Last edited: Jan 30, 2008
  5. Jan 30, 2008 #4
    Plain and simple, I think his formula is wrong. He got his signs backwards and he is missing a
    [tex]\sin \theta _W[/tex] term in

    [tex]- g'\bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}
    {2}} \right)e[/tex]

    It should read:

    [tex]- g'\sin \theta_W \bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}
    {2}} \right)e[/tex]
     
  6. Feb 7, 2008 #5

    Demystifier

    User Avatar
    Science Advisor

    The funny thing is that Weinberg is the guy who discovered EW theory. Moreover, he received the Nobel prize for it. :smile:
     
  7. Feb 7, 2008 #6

    mjsd

    User Avatar
    Homework Helper

    And the most amusing part is that that book assumes that you know "the entire thing" before you start reading... I guess that's what make him the Nobel laureate. you need to be ahead of your time :smile:
     
  8. Feb 8, 2008 #7

    Demystifier

    User Avatar
    Science Advisor

    I have a general rule: If you want to learn the basics of something, never ask the best experts for that to explain it to you!
     
  9. Feb 8, 2008 #8

    kdv

    User Avatar

    Jim, have you resolved the issue to your satisfaction? Do you agree with him now or are you now convinced there is a mistake? I am asking because I was considering double checking this.
     
  10. Feb 9, 2008 #9
    I think there is a mistake, take a look
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with a formula in Weinberg
Loading...