Gauge invariance of interaction lagrangian

elsafo

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

Related Beyond the Standard Model News on Phys.org

ShayanJ

Gold Member
By construction!!!

• dextercioby

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
$$\delta \int_{ D } d^{ 4 } x \ A^{ \mu } J_{ \mu } = \lambda \left( \int_{ \partial D_{ 1 } } d^{ 3 } x \ J_{ 0 } ( t_{ 1 } , x ) - \int_{ \partial D_{ 2 } } d^{ 3 } x \ J_{ 0 } ( t_{ 2 } , x ) \right) = 0 .$$

• strangerep

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.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving