Gauge invariance of interaction lagrangian

[elsafo]

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

 

[ShayanJ]

By construction!

 

[samalkhaiat]

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

\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
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 J_{ \mu } vanishes at the boundary \partial D at infinity.

 

[dextercioby]

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

 

[elsafo]

Thanks all

 

[strangerep]

Hi Sam,

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

\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
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 J_{ \mu } vanishes at the boundary \partial D 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]

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 \Lambda is bounded at \partial D. The case will be trivial if \Lambda |_{ \partial D } = 0. If \Lambda = \lambda is constant at \partial D, then the statement reduces to that of charge conservation. Let me explain this: we consider the space-time region D to be a (world) tube containing the field, i.e., J_{ \mu } = 0 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 (\partial D_{ 1 } at t_{ 1 } and \partial D_{ 2 } at t_{ 2 }) plus time-like surfaces at infinity (\Sigma^{ 3 }) that join the two surfaces together. If J_{ \mu } dies out rapidly enough at spatial infinity, then the surfaces at infinity (\Sigma^{ 3 }) do not contribute to the integral \int_{ \partial D } d^{ 3 } x \ n^{ \mu } J_{ \mu }, where n^{ \mu } is the unit outer normal to \partial D:
\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 } .
Thus
\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 .

 

[julian]

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

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

is invariant, i.e.,

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

This demands that

{\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}).

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

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

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

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

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

 

[strangerep]

Hi,
[...] If \Lambda = \lambda is constant at \partial D, then the statement reduces to that of charge conservation.
[...]
Thus
\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 .

OK, thanks.