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Physics
High Energy, Nuclear, Particle Physics
##A_\mu^a=0## in global gauge symmetries ?
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[QUOTE="JuanC97, post: 5999551, member: 532191"] Hi, this question is related to global and local SU(n) gauge theories. First of all, some notation: ##A## will be the gauge field of the theory (i.e: the 'vector potential' in the case of electromagnetic interactions) also known as 'connection form'. In components: ##A_\mu## can be expanded in terms of the basis given by the generators $F_a$ of a SU(n) group using some coefficients (or components) ##A_\mu^a## so one can write ##A_\mu=A_\mu^aF_a##. As you can see in the image I uploaded (from the book "An elementary primer for Gauge theory") one can identify ##A_\mu^a## wih the partial derivatives of the parameters ## \theta^a ## (in the last line you have ##A_\mu=\partial_\mu\theta^a\,F_a##) which parametrize the infinitesimal amount of transformation that is made by the operator ## U(dx) ## on an arbitrary vector [B]u[/B]. The question is: If the transformation is global, i.e: ## \theta^a\neq f(x) ##, then... ## A_\mu^a=0 ## implies that is impossible to deffine a connection?. [/QUOTE]
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High Energy, Nuclear, Particle Physics
##A_\mu^a=0## in global gauge symmetries ?
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