Help with a formula in Weinberg

  • Context: Graduate 
  • Thread starter Thread starter Jim Kata
  • Start date Start date
  • Tags Tags
    Formula Weinberg
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
8 replies · 4K views
Jim Kata
Messages
198
Reaction score
10
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}<br /> {\bar \upsilon _e } \\<br /> {\bar e} \\<br /> <br /> \end{array} } \right)\sum\limits_\alpha {\gamma _\mu A^\mu _\alpha } t_\alpha \left( {\begin{array}{*{20}c}<br /> {\upsilon _e } \\<br /> e \\<br /> <br /> \end{array} } \right)[/tex]

but I think he meant to write

[tex] iL_e = - \left( {\begin{array}{*{20}c}<br /> {\bar \upsilon _e } \\<br /> {\bar e} \\<br /> <br /> \end{array} } \right)\sum\limits_\alpha {\gamma _\mu \left( {A^\mu _\alpha t_{\alpha L} + B^\mu y} \right)} \left( {\begin{array}{*{20}c}<br /> {\upsilon _e } \\<br /> e \\<br /> <br /> \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}<br /> {2}\left( {1 - \gamma _5 } \right)[/tex]

but whatever

From this I was able to derive next line

Namely

[tex] \begin{gathered}<br /> - \left( {\begin{array}{*{20}c}<br /> {\bar \upsilon _e } \\<br /> {\bar e} \\<br /> <br /> \end{array} } \right)[\frac{1}<br /> {{\sqrt 2 }}\gamma _\mu W^\mu \left( {t_{1L} - it_{2L} } \right) + \frac{1}<br /> {{\sqrt 2 }}\gamma _\mu W^{*\mu } \left( {t_{1L} + it_{2L} } \right) \hfill \\<br /> + \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}<br /> {\upsilon _e } \\<br /> e \\<br /> <br /> \end{array} } \right) \hfill \\ <br /> \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}<br /> - \frac{g}<br /> {{\sqrt 2 }}\left( {\bar e\gamma _\mu W^\mu \left( {\frac{{1 + \gamma _5 }}<br /> {2}} \right)\upsilon _e } \right) - \frac{g}<br /> {{\sqrt 2 }}\left( {\bar \upsilon _e \gamma _\mu W^{*\mu } \left( {\frac{{1 + \gamma _5 }}<br /> {2}} \right)e} \right) \hfill \\<br /> + \frac{1}<br /> {2}\sqrt {g^2 + g'^2 } \bar \upsilon _e \gamma _\mu Z^\mu \left( {\frac{{1 + \gamma _5 }}<br /> {2}} \right)\upsilon _e - \frac{1}<br /> {2}\frac{{\left( {g^2 - g'^2 } \right)}}<br /> {{\sqrt {g^2 + g'^2 } }}\bar e\gamma _\mu Z^\mu \left( {\frac{{1 + \gamma _5 }}<br /> {2}} \right)e \hfill \\<br /> - g'\sin \theta _W \bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}<br /> {2}} \right)e - \left( {\begin{array}{*{20}c}<br /> {\bar \upsilon _e } \\<br /> {\bar e} \\<br /> <br /> \end{array} } \right)\gamma _\mu A^\mu \left( { - t_{3L} \sin \theta _W + y\cos \theta _W } \right)\left( {\begin{array}{*{20}c}<br /> {\upsilon _e } \\<br /> e \\<br /> <br /> \end{array} } \right) \hfill \\ <br /> \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 }}<br /> {2}} \right)e + e\left( {\bar e\gamma _\mu A^\mu e} \right)[/tex]

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

even with

[tex] q = - \sin \theta _W t_3 + \cos \theta _W y[/tex]
 
Physics news on Phys.org
trying to learn EW theory from Weinberg?... not a good choice! :smile:
sorry don't have a copy of Weinberg at hand
 
Last edited:
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 }}<br /> {2}} \right)e[/tex]

It should read:

[tex]- g'\sin \theta_W \bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}<br /> {2}} \right)e[/tex]
 
mjsd said:
trying to learn EW theory from Weinberg?... not a good choice! :smile:
The funny thing is that Weinberg is the guy who discovered EW theory. Moreover, he received the Nobel prize for it. :smile:
 
Demystifier said:
The funny thing is that Weinberg is the guy who discovered EW theory. Moreover, he received the Nobel prize for it. :smile:

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:
 
mjsd said:
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:
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!
 
Jim Kata said:
Anybody? Help!

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.
 
kdv said:
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.

I think there is a mistake, take a look