Lorentz transformation of a scalar field

1. Nov 18, 2015

spaghetti3451

Hi, the following is taken from Peskin and Schroeder page 36:

$\partial_{\mu}\phi(x) \rightarrow \partial_{\mu}(\phi(\Lambda^{-1}x)) = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)$

It describes the transformation law for a scalar field $\phi(x)$ for an active transformation.

I would like to work out the intermediate steps by myself as they are missing from the textbook. Can you please correct any mistakes I make?

$\partial_{\mu}\phi(x) \rightarrow \partial_{\mu}(\phi(\Lambda^{-1}x)) = \frac{\partial(\phi(\Lambda^{-1}x))}{\partial x^{\mu}} = \frac{\partial (\Lambda^{-1}x)^{\nu}}{\partial x^{\mu}} \frac{\partial \phi((\Lambda^{-1}x))}{\partial (\Lambda^{-1}x)^{\nu}} = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)$.

Am I correct?

2. Nov 18, 2015

haushofer

Yes. Personally I find this convention very confusing; a scalar transforms under x --> x' as

$\phi(x) \rightarrow \phi'(x')$
So
$\partial_{\mu} \phi(x) \rightarrow \partial'_{\mu} \phi'(x') = \frac{\partial x^{\rho}}{\partial x^{'\mu}} \partial_{\rho} \phi'(x')$

But whatever suits you, of course; I guess it comes down to the 'passive v.s. active'-discussion.

3. Nov 18, 2015

spaghetti3451

All right, so $\partial_{\mu}(\phi(\Lambda^{-1}x)) = \frac{\partial(\phi(\Lambda^{-1}x))}{\partial x^{\mu}}$ because only $\phi$ is a function of $\Lambda^{-1}x$.

But then, $(\partial_{\nu}\phi)(\Lambda^{-1}x) = \frac{\partial \phi(\Lambda^{-1}x)}{\partial (\Lambda^{-1}x)^{\nu}}$, all of $\partial_{\nu} \phi$ is a function of $(\Lambda^{-1}x)$.

Am I correct?

4. Nov 18, 2015

spaghetti3451

Was your computation from the passive point of view?

5. Nov 18, 2015

haushofer

I use x- primes for your Lamda-x. And yes, my approach is usually called the passive one.