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I am currently studying the behavior of a scalar field $\phi$ under a Lorentz transformation $\Lambda$. However I am having trouble understanding why the following holds true:

$$\partial_{\mu}\left(\phi(\Lambda^{-1}x)\right) = (\Lambda^{-1})^{\nu}_{~~\mu}~(\partial_{\nu}\phi)\left(\Lambda^{-1}x\right).$$

This should should probably follow from the chain rule with $\Lambda$ being a linear transformation but I can't quite figure out why this is.

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  • $\begingroup$ Your using two different notations, do you mean $\left(\partial_{\mu}\phi\right)(\Lambda^{-1}x) = (\Lambda^{-1})^{\nu}_{~~\mu}~(\partial_{\nu}\phi)\left(\Lambda^{-1}x\right)$? $\endgroup$ May 9, 2016 at 20:04
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    $\begingroup$ Possible duplicate of A few questions on passive vs active Lorentz transformations $\endgroup$ May 9, 2016 at 20:08
  • $\begingroup$ No I am meaning it as noted in the question which is also consistent to the notation in "An Introduction to Quantum Field Theory" by Peskin, Schroeder. Note that $\vec{\nabla}(f\circ g)(x)\ne (\vec{\nabla} f)(g(x))$. $\endgroup$ May 9, 2016 at 20:11
  • $\begingroup$ Use index notation, remembering that $\partial^{\mu}x_{\nu}=\delta^{\mu}_{\nu}.$ $\endgroup$
    – user82635
    May 9, 2016 at 20:17

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This is a simple application of the chain rule $$ \frac{\partial}{\partial x^\mu} \phi \big( \Lambda_\nu{}^\mu x^\nu \big) = \frac{\partial}{\partial x^\mu} \big( \Lambda_\rho{}^\nu x^\rho \big) \left[ \frac{ \partial }{ \partial y^\nu } \phi (y) \right]_{y^\mu = \Lambda_\nu{}^\mu x^\nu} = \Lambda_\mu{}^\nu \left[ \frac{ \partial }{ \partial y^\nu } \phi (y) \right]_{y^\mu = \Lambda_\nu{}^\mu x^\nu} ~. $$

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