Proof that the E.M Field is invariant under guage transformation.

  • Thread starter hob
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  • #1
hob
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To prove:

[tex]F[/tex] [tex]\overline{} \mu\nu[/tex] = [tex]\nabla [/tex][tex]\overline{} \mu[/tex][tex]A[/tex] [tex]\overline{} \nu[/tex] - [tex]\nabla [/tex][tex]\overline{} \nu[/tex][tex]A[/tex] [tex]\overline{} \mu[/tex]

is invariant under the gauge transformation:

[tex]A[/tex] [tex]\overline{} \mu[/tex] [tex]\rightarrow[/tex] [tex]A[/tex] [tex]\overline{} \mu[/tex] + [tex]\nabla [/tex][tex]\overline{} \mu[/tex][tex]\Lambda[/tex]


I end up with:

[tex]F[/tex] [tex]\overline{} \mu\nu[/tex] = [tex]F[/tex] [tex]\overline{} \mu\nu[/tex] + [[tex]\nabla [/tex][tex]\overline{} \mu[/tex],[tex]\nabla [/tex][tex]\overline{} \nu[/tex]][tex]\Lambda[/tex]

Which I guess is invariant provided [tex]\nabla [/tex][tex]\overline{} \mu[/tex] & [tex]\nabla [/tex][tex]\overline{} \nu[/tex] commute?

Do they commute? and if so why?

Many thanks.
 

Answers and Replies

  • #2
525
7
Yes, they commute. In the case of normal minkowski space time and the Abelian gauge group U(1) the differential operators reduce to ordinary derivatives.
 

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