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Why the topological term F\til{F} is scale independent? 
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#1
Jun2014, 07:07 AM

P: 4

why the topological term in gauge theory, [itex]ε_{\mu\nu ρσ}F^{\mu \nu} F^{ρσ} [/itex] ,is scaleindependent?



#2
Jun2014, 09:55 AM

Sci Advisor
P: 908




#3
Jun2214, 03:29 AM

P: 4




#4
Jun2214, 01:29 PM

Sci Advisor
P: 908

Why the topological term F\til{F} is scale independent?



#5
Jun2214, 09:29 PM

P: 4

We are considering the most general rescaling, so space and time scale differently. This is especially true for nonrelativistic case.



#6
Jun2314, 06:34 AM

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P: 908

Any way, in relativistic field theories, the coordinates scale according to [tex]x^{ \mu } \rightarrow \bar{ x }^{ \mu } = e^{  \lambda } x^{ \mu } ,[/tex] and the field transforms as [tex]F ( x ) \rightarrow \bar{ F } ( \bar{ x } ) = e^{ \lambda \Delta } F ( x )[/tex] where [itex]\Delta[/itex] is the scaling dimension of the field. In Ddimensional spacetime: [tex]\Delta = \frac{ D  2 }{ 2 }, \ \ \mbox{ for } \ \ A_{ \mu } (x) ,[/tex] and [tex]\Delta = \frac{ D }{ 2 } , \ \ \mbox{ for } \ \ F_{ \mu \nu } ( x ) .[/tex] See (for more detailed description) the link below www.physicsforums.com/showthread.php?t=172461 


#7
Jun2314, 09:51 PM

P: 4

Thank you for your reply. The link given by you is so long that I need some time to follow. However, I donn't agree with your point that
"Time and space scale differently ONLY in nonrelativistic theory". It is the textbook's convention that space and time scale the same in relativistic region, but it is not the principle law. Anyway, I don't want to get invoked into this aspect. There is quick question I want to ask you: How do you determine the scaling dimension of a field? By dimension analysis and keep kinetic term dimensionD? This textbook's answer doesn't make sense... 


#8
Jun2414, 06:20 AM

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P: 908




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