How to compute the variation of functionals

Question

During my physics classes I encountered various definitions on how to calculate the variation of a functional/function for certain "boundary conditions" and the thing is, I don't really understand if they are all equal or if I have to use some in certain situations...

Lets assume I have a functional of the form $$S[f] = \int_{x_i}^{x_f} d^4x\, \mathcal{L}(f(x), \partial_\mu f(x),x).$$ The first definition I ever saw was $$\delta S := \frac{d}{d \varepsilon} S[f+ \varepsilon \delta f]\Big|_{\varepsilon =0}\quad \text{where}\quad \delta f(x_i)=\delta f(x_f)=0.\label{1} \tag{1}$$ Sometime later I found $$\delta \mathcal{L}= \delta_o\mathcal{L} +\delta x^\mu \partial_\mu \mathcal{L}\label{2}\tag{2},$$ where $$\delta x^\mu = (x^\mu)'-x^\mu,\quad \delta_o f= f'(x)-f(x),\quad \delta f = f'(x')-f(x),$$ for $x\mapsto x'$ some kind of transformation. I also saw $$\delta\mathcal{L} = \frac{\delta \mathcal{L}}{\delta f}\delta f + \frac{\delta \mathcal{L}}{\delta(\partial_\mu f)}\partial_\mu \delta f \quad\text{where} \quad [\delta, \partial_\mu]=0\label{3}\tag{3}$$ was stated without any explanation.

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The only thing that the above "$\delta$" all had in common was, that no at single of the professors really explaind what I was supposed to do with them.

TL;DR: How is this $\delta$ defined, is $\delta \mathcal{L}$ different from $\delta S$ and $\frac{\delta \mathcal{L}}{\delta f}$? And if someone tells me to calculate $\delta \mathcal{L}$, how am I supposed to do this?

Preferably, I'd like to see an example for a global symmetry and a local symmetry transformation and how to calculate $\delta\mathcal{L}$ (it seems that different rules apply here).

Answer 1

Well I can see that your first and third definitions are compatible.But the second one I don't understand. Start from your first definition: $$ \delta S = \frac{d}{d\varepsilon}(f+\varepsilon \delta f) = \lim_{\varepsilon\rightarrow0}\frac{1}{\varepsilon}\int_{x_i}^{x_f}\left[\mathcal{L}((f+\varepsilon \delta f),\partial_{\mu}(f+\varepsilon \delta f),x)-\mathcal{L}(f,\partial_{\mu}f,x)\right]d^4x = \int_{x_i}^{x_f}(\frac{\partial \mathcal{L}}{\partial f}\delta f+\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}f)})\delta (\partial_{\mu}f))d^4x $$ Because the operator $\delta$ performs on $f$ but the operator $\partial_{\mu}$ performs on $x$, $[\delta,\partial_{\mu}]=0$.

Then from your third definition: $\delta S= \int_{x_i}^{x_f}\delta \mathcal{L}d^4x$

Answer 2

Premises

For the reasons briefly exposed in this Q&A, you should not rely too much on the Wikipedia entry on functional derivatives: in the sequel, I also assume (just for reference, since this choice does not significantly impact on the mathematical developments involved) that the integral functional $S(f)$ is defined as follows $$ S[f] = \int\limits_{x_i}^{x_f}\!\! \mathcal{L}\big(x,f(x), \partial_\mu f(x)\big) \mathrm{d}^4x\,\label{A}\tag{A} $$ Said that, I'll try to give ananswer to your questions, following mainly the two original source [1].

Questions and answers

TL;DR: How is this $\delta$ defined, is $\delta \mathcal{L}$ different from $\delta S$ and $\frac{\delta \mathcal{L}}{\delta f}$? And if someone tells me to calculate $\delta \mathcal{L}$, how am I supposed to do this?

Strictly speaking, $\delta$ represents the increment of a given quantity defined on a function space, say $X$ (it could be $X=C^k([a,b])$, for example).

• If $f\in X$ is intended as an independent variable, then the variation of $f$ is defined as ([1] §I.II.1.26 p. 22) $$ \delta f(x)=\varepsilon \varphi(x) \quad \varepsilon>0\label{B}\tag{B} $$ where $\varphi\in X$ as well.
• In a similar, but not identical, way if $S: X\to \Bbb R$ is a functional (or an operator between function spaces $X\to Y$ like $\mathcal{L}$), its variation is defined as ([1] §I.II.1.28 p. 24) $$ \begin{split} \delta S(\delta f)=\delta S(\varepsilon\varphi) & = S(f+\delta f) - S(f)\\ & = S(f+\varepsilon\varphi) - S(f) \end{split}\quad\left( \begin{split} \delta \mathcal{L}(\delta f)=\delta \mathcal{L}(\varepsilon\varphi) & = \mathcal{L}(f+\delta f) - \mathcal{L}(f)\\ & = \mathcal{L}(f+\varepsilon\varphi) - \mathcal{L}(f) \end{split}\right) $$ Clearly, if we assume \eqref{A}, we have $\delta S(\varphi)\neq\delta \mathcal{L}(\varphi)$
• The functional derivative of a functional (or an operator) is defined as $$ \begin{split} \frac{\delta S}{\delta f}= \frac{\delta S}{\delta f}(\varphi) &:= \frac{\mathrm d}{\mathrm d \varepsilon} S[f+ \varepsilon \varphi]\Big|_{\varepsilon =0}\\ &=\lim_{x\to 0}\frac{S(f+\delta f) - S(f)}{\varepsilon}\\ &=\lim_{x\to 0}\frac{S(f+\varepsilon\varphi ) - S(f)}{\varepsilon} \end{split}\label{1'}\tag{1'} $$ In your equation \eqref{1}, $\varphi(x)\equiv\delta f(x)$ by (though customary) abuse of notation: also, the condition $\delta f(x_i)=\delta f(x_f)$ is only required when your aim is to deduce the Euler-Lagrange equations for a functional of integral type like $S(f)$ \eqref{A} is.

From the simple examples given, but more properly from the definition \eqref{1'}, it is clear that the functional derivative is the extension to function spaces of the directional derivative (and thus also of the partial derivative) concept: as we can see, if we assume $X=\Bbb R^n$, then we have that $\varphi(x)=(0,\ldots,0,x^\mu,0\ldots,0)$ we have $$ \delta f(x)=\varepsilon(0,\ldots,0,x^\mu,0\ldots,0):=\delta x^\mu=\partial x^\mu\label{C}\tag{C} $$

I think this is the problem behind the confusion in the example shown: by abuse of notation, the same $\delta$ symbol is adopted for both functionals and variations defined on finite and infinite dimensional spaces, mixing up the two (however closely related) concepts. Thus

• $\delta x^\mu= (x^\mu)'-x^\mu$ is the variation of an independent variable in a finite dimensional space, in the spirit of \eqref{C}: the $\varepsilon$ parameter is implicitly considered in the transformation $(\:)^\prime:\Bbb R^n \to \Bbb R^n$ (if we are working on Euclidean spaces).
• $\delta_o f=f^\prime(x) -f(x)$ is the variation of an independent variable in a function space, in the spirit of \eqref{B}, and again the $\varepsilon$ parameter is implicitly considered in the transformation $(\:)^\prime:\Bbb R^n \to \Bbb R^n$.
• Finally, $\delta f=f^\prime(x^\prime)-f(x)$ is a mix of the two variation concepts \eqref{B} and \eqref{C} in the sense that in this case the function varies both as an element of a function space and bot as a function of its independent variable $x$.

These observation give meaning to the commutator relation $[\delta,\partial_\mu]=0$, and perhaps, by using the formulas above it seems possible to deduce \eqref{3} from \eqref{2}.

Unfortunately I cannot offer you an example calculation involving the formalism of symmetries as used by physicists, since I am not accustomed to it.

References

[1] Volterra, Vito, Theory of functionals and of integral and integro-differential equations. Dover edition with a preface by Griffith C. Evans, a biography of Vito Volterra and a bibliography of his published works by Sir Edmund Whittaker. Unabridged republ. of the first English transl, New York: Dover Publications, Inc. pp. 39+XVI+226 (1959), MR0100765, ZBL0086.10402.