Derivative of a Variation vs Variation of a Derivative

[quickAndLucky]

When a classical field is varied so that $\phi ^{'}=\phi +\delta \phi$ the spatial partial derivatives of the field is often written $$\partial _{\mu }\phi ^{'}=\partial _{\mu }(\phi +\delta \phi )=\partial _{\mu }\phi +\partial _{\mu }\delta \phi $$. Often times the next step is to switch the order of the variation and the partial derivative to get $\partial _{\mu }\phi ^{'}=\partial _{\mu }\phi +\delta (\partial _{\mu }\phi )$. What justifies the replacement of $\partial_{\mu }(\delta\phi )$ by $\delta (\partial _{\mu }\phi )$?

 

[haushofer]

Variations and derivatives commute if you keep your coordinates fixed during the variation. In deriving the Euler Lagrange eqns e.g. this is the case: the field variations involve functional variations.

 

[quickAndLucky]

Variations and derivatives commute if you keep your coordinates fixed during the variation. In deriving the Euler Lagrange eqns e.g. this is the case: the field variations involve functional variations.

I guess my question is "why do variations and derivatives commute?"

 

[samalkhaiat]

I guess my question is "why do variations and derivatives commute?"

haushofer answered your question correctly. The variation \delta measures the change in the functional form of a field at a fixed coordinate value. So, if you define the field \psi_{\mu}(x) = \partial_{\mu}\phi (x), then it follows from the definition of \delta that \delta \psi_{\mu}(x) = \psi_{\mu}^{'}(x) - \psi_{\mu}(x), or
\delta (\partial_{\mu}\phi )(x) = \partial_{\mu}\phi^{'}(x) - \partial_{\mu}\phi(x) = \partial_{\mu}\left(\phi^{'} - \phi \right) (x) = \partial_{\mu}( \delta \phi )(x) .

 

[quickAndLucky]

Thinking of $\partial _{\mu}\phi$ as an independent vector field that itself varies seemed to help! Thanks haushofer and samalkhaiat!