# I Derivative of a Variation vs Variation of a Derivative

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1. Jul 13, 2018 at 4:06 PM

### 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 $\delta (\partial _{\mu }\phi )$?

2. Jul 13, 2018 at 4:49 PM

### 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.

3. Jul 13, 2018 at 5:07 PM

### quickAndLucky

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

4. Jul 13, 2018 at 6:09 PM

### samalkhaiat

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
$$\delta (\partial_{\mu}\phi )(x) = \partial_{\mu}\phi^{'}(x) - \partial_{\mu}\phi(x) = \partial_{\mu}\left(\phi^{'} - \phi \right) (x) = \partial_{\mu}( \delta \phi )(x) .$$

5. Jul 13, 2018 at 6:58 PM

### quickAndLucky

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