Gauge invariance of interaction lagrangian

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Anyone can help me how to argue that interaction lagrangian is invariant under gauge transformation?
 

samalkhaiat

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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]
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.
 

dextercioby

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Either through Deser's Noether procedure, or directly in the BRST formulations, interactions (couplings) among classical fields are automatically gauge invariant.
 
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Thanks all
 

strangerep

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Hi Sam,

(Since the OP seems happy with the answers, I'll venture a clarification question...)

[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.
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?
 

samalkhaiat

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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?
Hi,
I think you do have the correct picture. We always assume that [itex]\Lambda[/itex] is bounded at [itex]\partial D[/itex]. The case will be trivial if [itex]\Lambda |_{ \partial D } = 0[/itex]. If [itex]\Lambda = \lambda[/itex] is constant at [itex]\partial D[/itex], then the statement reduces to that of charge conservation. Let me explain this: we consider the space-time region [itex]D[/itex] to be a (world) tube containing the field, i.e., [itex]J_{ \mu } = 0[/itex] outside the tube. We also assume that the tube is very “fat”. That is, its boundary consists of two (space-like) hypersurfaces taken at constant times ([itex]\partial D_{ 1 }[/itex] at [itex]t_{ 1 }[/itex] and [itex]\partial D_{ 2 }[/itex] at [itex]t_{ 2 }[/itex]) plus time-like surfaces at infinity ([itex]\Sigma^{ 3 }[/itex]) that join the two surfaces together. If [itex]J_{ \mu }[/itex] dies out rapidly enough at spatial infinity, then the surfaces at infinity ([itex]\Sigma^{ 3 }[/itex]) do not contribute to the integral [itex]\int_{ \partial D } d^{ 3 } x \ n^{ \mu } J_{ \mu }[/itex], where [itex]n^{ \mu }[/itex] is the unit outer normal to [itex]\partial D[/itex]:
[tex]\int_{ D } d^{ 4 } x \ \partial^{ \mu } ( \Lambda J_{ \mu } ) = \lambda \int_{ \partial D } d^{ 3 } x \ n^{ \mu } J_{ \mu } = \lambda \int_{ \partial D_{ 1 } } d^{ 3 } x \ n^{ \mu } J_{ \mu } + \lambda \int_{ \partial D_{ 2 } } d^{ 3 } x \ n^{ \mu } J_{ \mu } .[/tex]
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]
 

julian

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There is a matter field which is invariant under the kind of familiar idea [itex]\Psi \rightarrow \Psi' = \hat{U} \Psi(x) = exp (i {\bf a}(x) \cdot \hat{T}) \Psi(x)[/itex]. How the Yang-Mills field [itex]A_\mu^I[/itex] transforms can be derived by demanding the minimal coupling interaction term

[itex]L_{int} = i \overline{\Psi} \gamma^\mu \partial_\mu \Psi + g \overline{\Psi} \gamma^\mu {\bf A}_\mu \cdot \hat{T} \Psi[/itex]

is invariant, i.e.,

[itex]L_{int} = L_{int}' = i \overline{\Psi}' \gamma^\mu \partial_\mu \Psi' + g \overline{\Psi}' \gamma^\mu {\bf A}_\mu' \cdot \hat{T} \Psi'[/itex]

This demands that

[itex]{\bf A}_\mu' \cdot \hat{T} = \hat{U} {\bf A}_\mu \cdot \hat{T} \hat{U}^{-1} + {i \over g} \hat{U} (\partial_\mu \hat{U}^{-1})[/itex].

Note for electrodynamics [itex] \hat{U} = exp (i a(x))[/itex] the gauge transformation is then

[itex]A_\mu' (x) = A_\mu (x) + {1 \over g} \partial_\mu a(x)[/itex].

It can be shown that the kinematic energy term for the field [itex]{\bf A}[/itex] of the Lagrangian (analogous to Maxwell's theory)

[itex]L_A = - {1 \over 2} Tr \{ ({\bf F}_{\mu \nu} \cdot \hat{T}) ({\bf F}^{\mu \nu} \cdot \hat{T}) \} [/itex]

is invariant under such gauge transformations where we have defined the field strength operator [itex]{\bf F}_{\mu \nu} = {i \over g} [D_\mu , D_\nu][/itex] where [itex]D_\mu[/itex] is the covariant derivative defined by the connection [itex]A^I_\mu[/itex].
 
Last edited:

strangerep

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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]
OK, thanks.
 

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