Variational operator in the least action principle

[Haorong Wu]

TL;DR Summary

the difference between the variational operator $\delta$ in $\delta S=0$ and the differential operator $d$

Hello. Since I learned the least action principle several years ago, I cannot figure out the difference between the variational operator $\delta$ in $\delta S=0$ and the differential operator $d$ in, say $dS$.

Everytime I encountered the variational operator, I just treated it as a differential operator and everything worked out.

I understand that $\delta S=\tilde S - S$ means the change due to a small variation of the action $S$, but isn't that just the meaning of differentiation?

 

[PeroK]

Summary:: the difference between the variational operator $\delta$ in $\delta S=0$ and the differential operator $d$

Hello. Since I learned the least action principle several years ago, I cannot figure out the difference between the variational operator $\delta$ in $\delta S=0$ and the differential operator $d$ in, say $dS$.

Everytime I encountered the variational operator, I just treated it as a differential operator and everything worked out.

I understand that $\delta S=\tilde S - S$ means the change due to a small variation of the action $S$, but isn't that just the meaning of differentiation?

$\delta S$ is a functional differential. See here

https://mathworld.wolfram.com/FunctionalDerivative.html

http://www.physics.usu.edu/Wheeler/QFT2016/Notes/QFT09FunctionalDerivatives.pdf

 

[Haorong Wu]

$\delta S$ is a functional differential. See here

https://mathworld.wolfram.com/FunctionalDerivative.html

http://www.physics.usu.edu/Wheeler/QFT2016/Notes/QFT09FunctionalDerivatives.pdf

Hi, @PeroK. From the definition of functional derivative, could I treat a functional as a function, and a function in the functional as a variable so that I could apply every rules for function derivatives to functional derivatives? It seems to work quite well to me.

 

[PeroK]

Hi, @PeroK. Could I treat a functional as a function, and a function in the functional as a variable so that I could apply every rules for function derivatives to functional derivatives? It seems to work quite well to me.

Yes, the idea is that all the familiar rules of differentiation apply to the functional derivative. One point to note is that for discrete variables we have:
$$\frac{\partial x_i}{\partial x_j} = \delta_{ij}$$
And for a continuous functional derivative we have:
$$\frac{\delta f(x)}{\delta f(y)} = \delta(x-y)$$

 

[Haorong Wu]

Yes, the idea is that all the familiar rules of differentiation apply to the functional derivative. One point to note is that for discrete variables we have:
$$\frac{\partial x_i}{\partial x_j} = \delta_{ij}$$
And for a continuous functional derivative we have:
$$\frac{\delta f(x)}{\delta f(y)} = \delta(x-y)$$

I get it now. Thanks, @PeroK.