# A Field transformation under an active transformation

1. Apr 22, 2016

### spaghetti3451

Under the infinitesimal translation $x^{\nu} \rightarrow x^{\nu}-\epsilon^{\nu}$,

the field $\phi(x)$ transforms as $\phi_{a}(x) \rightarrow \phi_{a}(x) + \epsilon^{\nu}\partial_{\nu}\phi_{a}(x)$.

I don't understand why the field transforms as above. Let me try to do the math.

The Taylor expansion of $f(x+\delta x)$, where the argument $x+\delta x$ is a $4$-vector and $f$ is a scalar, is $f(x+\delta x)=f(x)+\frac{\partial f}{\partial x^{\nu}}(\delta x)^{\nu} + \dots$

So, $\phi_{a}(x) \rightarrow \phi_{a}(x-\epsilon) = \phi_{a}(x) + (-\epsilon^{\nu})\partial_{\nu}\phi_{a}(x)$.

Now, where did I go wrong?

2. Apr 22, 2016

### JorisL

The translation you applied is $x^\nu\to x^\nu - \epsilon^\nu$.
That's the only difference.

The field transformation you are trying to prove is $\phi_a(x) \to \phi^{\prime}_a(x) \equiv \phi_a(x^\prime)=\phi_a(x+\epsilon)=\phi_a(x) +\epsilon^\nu\partial_\nu\phi_a(x)$.

I find the treatment in Chapter 1 of Freedman and Van Proeyen's book very clear.