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I read Quantum Field Theory, Ryder, second edition. Relation (8.86) brings us the famous result:

$e = g \sin \theta_W$

Here Ryder says tht $e$ is the proton charge. However, according to what I understand from the book, that should be the electron charge (which is negative). This is because in relation (8.85) I read that the lagrangian becomes

$L_i= - g \sin \theta_W (\bar{e}\gamma^\mu e) A_\mu +....$

This QED lagrangian can be found on (7.100) as:

$L_i= - e \bar{\psi}\gamma^\mu \psi A_\mu +....$

Just below on the page Ryder states that here e<0 is the electron charge, which would also agree with the QED form of the lagrangian in the (+,-,-,-) convention, as presented in Gauge covariant derivative in different books.

From these I conclude that in the relation $e = g \sin \theta_W$ above $e$ is the electron charge and not the proton charge as stated by Ryder. Where do I make a mistake?

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cristian
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  • You are right. I think it is a typo. – Trimok Sep 01 '13 at 13:59
  • I doubt that any of the standardized constants $e,g$ etc. is ever meant to be negative. When they talk "charge of the electron", they mean the absolute value - or the positron. – Luboš Motl Sep 05 '13 at 16:37

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