Derivative of a Variation vs Variation of a Derivative

In summary, the commutation of variations and derivatives is justified by keeping the coordinates fixed during the variation, where the variation measures the change in the functional form of a field at a fixed coordinate value. This is evident in the derivation of the Euler-Lagrange equations, where the field variations involve functional variations. This concept can also be understood by thinking of the partial derivative of a field as an independent vector field that itself varies.
  • #1
quickAndLucky
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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 )##?
 
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  • #2
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.
 
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  • #3
haushofer said:
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?"
 
  • #4
quickAndLucky said:
I guess my question is "why do variations and derivatives commute?"
haushofer answered your question correctly. The variation [itex]\delta[/itex] measures the change in the functional form of a field at a fixed coordinate value. So, if you define the field [itex]\psi_{\mu}(x) = \partial_{\mu}\phi (x)[/itex], then it follows from the definition of [itex]\delta[/itex] that [tex]\delta \psi_{\mu}(x) = \psi_{\mu}^{'}(x) - \psi_{\mu}(x),[/tex] or
[tex]\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) .[/tex]
 
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  • #5
Thinking of ##\partial _{\mu}\phi ## as an independent vector field that itself varies seemed to help! Thanks haushofer and samalkhaiat!
 

1. What is the difference between the derivative of a variation and the variation of a derivative?

The derivative of a variation refers to the rate of change of a variable with respect to another variable, while the variation of a derivative refers to the change in the derivative itself.

2. How are the derivative of a variation and the variation of a derivative related?

The derivative of a variation and the variation of a derivative are inversely related. This means that as one increases, the other decreases.

3. Can you provide an example of the derivative of a variation and the variation of a derivative?

For example, if y = 3x^2, the derivative of the variation would be dy/dx = 6x, while the variation of the derivative would be d(dy/dx)/dx = d(6x)/dx = 6.

4. What is the importance of understanding the derivative of a variation and the variation of a derivative?

Understanding these concepts is important in calculus and physics, as they help us understand the relationship between variables and their rates of change.

5. Are there any real-life applications of the derivative of a variation and the variation of a derivative?

Yes, these concepts are used in fields such as economics, engineering, and statistics to analyze data and make predictions based on the rate of change of variables.

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