Help with a formula in Weinberg

In summary: It should be:- g'\sin \theta_W \bar e\gamma _\mu Z^\mu \left( {\frac{{1 - \gamma _5 }}{2}} \right)e
  • #1
Jim Kata
197
6
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]
 
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  • #2
Anybody? Help!
 
  • #3
trying to learn EW theory from Weinberg?... not a good choice! :smile:
sorry don't have a copy of Weinberg at hand
 
Last edited:
  • #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]
 
  • #5
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:
 
  • #6
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:
 
  • #7
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!
 
  • #8
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.
 
  • #9
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
 

1. How do I calculate the Weinberg angle?

The Weinberg angle is calculated using the equation sin^2(θw) = 1 - [mW/mZ]^2, where mW is the mass of the W boson and mZ is the mass of the Z boson. This can be obtained from experimental data or theoretical calculations.

2. What is the significance of the Weinberg angle in particle physics?

The Weinberg angle, also known as the weak mixing angle, is a fundamental parameter in the Standard Model of particle physics. It governs the strength of the electroweak interaction, which is responsible for the radioactive decay of particles and plays a crucial role in the behavior of subatomic particles.

3. How is the Weinberg angle related to the Higgs mechanism?

The Higgs mechanism is a process proposed by Peter Higgs and others to explain how particles acquire mass. The Weinberg angle is a key component of this mechanism, as it is used to determine the masses of the W and Z bosons, which are crucial for the Higgs mechanism to work.

4. Can the Weinberg angle be measured experimentally?

Yes, the Weinberg angle can be measured experimentally through high-energy particle collisions. By analyzing the behavior of particles produced in these collisions, scientists can determine the value of the Weinberg angle and test its predictions in the Standard Model.

5. Are there any alternative equations or theories that can explain the Weinberg angle?

There are various extensions to the Standard Model that propose alternative equations or theories to explain the Weinberg angle. Some of these include Grand Unified Theories (GUTs) and Supersymmetry. However, these theories have not yet been experimentally proven and the Standard Model remains the most widely accepted explanation for the Weinberg angle.

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