# Gauge invariance of interaction lagrangian

#### elsafo

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

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#### ShayanJ

Gold Member
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

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

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

Gold Member
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 $A^I_\mu$.

Last edited:

#### 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.

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