# Help with a formula in Weinberg

1. Jan 25, 2008

### Jim Kata

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:

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

but I think he meant to write

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

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

Namely $$e_L = \frac{1}{2}\left( {1 + \gamma _5 } \right)$$ and $$e_R = \frac{1} {2}\left( {1 - \gamma _5 } \right)$$

but whatever

From this I was able to derive next line

Namely

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

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

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

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

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

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

even with

$$q = - \sin \theta _W t_3 + \cos \theta _W y$$

2. Jan 30, 2008

### Jim Kata

Anybody? Help!!!!

3. Jan 30, 2008

### mjsd

trying to learn EW theory from Weinberg?... not a good choice!
sorry don't have a copy of Weinberg at hand

Last edited: Jan 30, 2008
4. Jan 30, 2008

### Jim Kata

Plain and simple, I think his formula is wrong. He got his signs backwards and he is missing a
$$\sin \theta _W$$ term in

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

$$- g'\sin \theta_W \bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }} {2}} \right)e$$

5. Feb 7, 2008

### Demystifier

The funny thing is that Weinberg is the guy who discovered EW theory. Moreover, he received the Nobel prize for it.

6. Feb 7, 2008

### mjsd

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

7. Feb 8, 2008

### Demystifier

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!

8. Feb 8, 2008

### kdv

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.

9. Feb 9, 2008

### Jim Kata

I think there is a mistake, take a look