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Anyone can help me how to argue that interaction lagrangian is invariant under gauge transformation?
[tex]\delta \int_{ D } d^{ 4 } x \ A_{ \mu } J^{ \mu } = \int_{ D } d^{ 4 } x \ \partial_{ \mu } ( \Lambda J^{ \mu } ) = \int_{ \partial D } d S_{ \mu } \ \Lambda J^{ \mu } = 0[/tex]Anyone can help me how to argue that interaction lagrangian is invariant under gauge transformation?
Vanishing of ##J_\mu## at spatial infinity is uncontroversial. But... why should it vanish as ##|t|\to\infty## ?[tex]\delta \int_{ D } d^{ 4 } x \ A_{ \mu } J^{ \mu } = \int_{ D } d^{ 4 } x \ \partial_{ \mu } ( \Lambda J^{ \mu } ) = \int_{ \partial D } d S_{ \mu } \ \Lambda J^{ \mu } = 0[/tex]
The first equality follows from the fact that gauge fields couple to conserved (matter) current, the second equality is just the divergence theorem, and the last one follow because [itex]J_{ \mu }[/itex] vanishes at the boundary [itex]\partial D[/itex] at infinity.
Hi,Hi Sam,
(Since the OP seems happy with the answers, I'll venture a clarification question...)
Vanishing of ##J_\mu## at spatial infinity is uncontroversial. But... why should it vanish as ##|t|\to\infty## ?
I had presumed that gauge transformations are required to approach the identity in that limit, but is there perhaps another reason?
OK, thanks.Hi,
[...] If [itex]\Lambda = \lambda[/itex] is constant at [itex]\partial D[/itex], then the statement reduces to that of charge conservation.
[...]
Thus
[tex]\delta \int_{ D } d^{ 4 } x \ A^{ \mu } J_{ \mu } = \lambda \left( \int_{ \partial D_{ 1 } } d^{ 3 } x \ J_{ 0 } ( t_{ 1 } , x ) - \int_{ \partial D_{ 2 } } d^{ 3 } x \ J_{ 0 } ( t_{ 2 } , x ) \right) = 0 .[/tex]