Derivative of a Lorentz-Transformed Field

[looseleaf]

Please help me understand this line from P&S, or point me towards some resources:

Why is there another Lorentz transformation acting on the derivative on the RHS?

Thanks

 

[looseleaf]

Oh wait, I realize now it comes from the covariant transformation rules. Is it because we have to boost both the coordinate and the derivative operator?

 

[vanhees71]

It's just using the chain rule. A more careful notation is
$$\phi'(x')=\phi(x)=\phi(\hat{\Lambda}^-1 x').$$
Then it's clear that
$$\partial_{\mu}' \phi'(x')=\frac{\partial x^{\nu}}{\partial x^{\prime \mu}} \partial_{\nu} \phi(x)=\left [{(\Lambda^{-1})^{\nu}}_{\mu} \partial_{\nu} \phi(x) \right]_{x=\hat{\Lambda}^{-1} x'},$$
which shows that $\partial_{\mu}$ transforms as covariant vector components, i.e., $\partial_{\mu} \phi$ transforms as the covariant components of a vector field.