# Gauge transformation of Yang-Mills field strength

1. May 8, 2013

### center o bass

Hi. I'm reading about non-abelian theories and have thus far an understanding that a gauge invariant Lagrangian is something to strive for. I previously thought that the Yang-Mills gauge boson free field term $-1/4 F^2$ was gauge invariant, but now after realizing that the field strength transform homogeneously under gauge transformation U(x) s.t

$$F \to UFU^\dagger$$

it seems like this term also transform like

$$F^2 \to UF^2U^\dagger.$$

Am I right there? If so isn't this a problem for the physics? Why, why not?

2. May 8, 2013

### The_Duck

The actual lagrangian term is the trace of $F^2$:

${\rm tr} (F^{\mu \nu} F_{\mu \nu})$

This is gauge invariant. Sometimes this gets written as

$F^{a \mu \nu} F^a_{\mu \nu}$

where $F^a_{\mu \nu}$ is defined by $F_{\mu \nu} \equiv F^a_{\mu \nu} T^a$. Here $a$ is a gauge group adjoint representation index, the $T^a$ are the generator matrices of the adjoint representation and $F^a_{\mu \nu}$ is a set of numbers (not matrices). This is just decomposing the field strength matrix as a linear combination of the generators.

The second form of the Lagrangian is equivalent because ${\rm tr} (T^a T^b) = \delta^{a b}$. So if you plug the decomposition of $F^{\mu \nu}$ into the first form of the Lagrangian, you get the second form.

Last edited: May 8, 2013
3. May 8, 2013

### center o bass

Ah, that's true. So the trace is actually essential. I've heard that it was nothing but compact notation :)

4. May 9, 2013

### andrien

It should most probably be seen like this after gauge transformation
$UFU^\dagger UFU^\dagger= UF^2U^\dagger$