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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.


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).

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    $\begingroup$ $\delta J$ just represents the first variation of $J$, an integral of some functional $L$. I think this will answer your questions. Just note that$$\delta\mathcal{L} = \frac{\delta \mathcal{L}}{\delta f}\delta f + \frac{\delta \mathcal{L}}{\delta(\partial_\mu f)}\partial_\mu \delta f \quad\text{where} \quad [\delta, \partial_\mu]=0$$ is the Euler-Lagrange equation which is discussed in both linked wiki articles. $\endgroup$
    – mattos
    Oct 11, 2019 at 13:34
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    $\begingroup$ What physicists do with integrals without a shred of proof that it actually works is something that makes mathematicians either laugh or cringe. Also how vague their terminology is (they will say "$dx$ is a miniscule change in the value of $x$", and mean it quite seriously and not just as an intuitive visualisation of something more abstract and rigorous). What your describe here reflects my own impressions from taking courses in general relativity and quantum mechanics. $\endgroup$
    – Arthur
    Oct 11, 2019 at 13:34
  • $\begingroup$ @mattos Thank you for the links, I will look into them. I noticed that the equation kind of has the form of the E-L-equation, but there are still these weird "$\delta$'s"... $\endgroup$
    – Sito
    Oct 11, 2019 at 13:52
  • $\begingroup$ If $J$ is the variation you want to minimise/maximise, then $\delta J$ is the Gateaux derivative of $J$, so the notation $\delta$ just says 'compute the Gateaux derivative of'. To actually compute the variation, you literally just compute the derivative of the perturbed functional $L$ with respect to $\epsilon$ and evaluate the result at $\epsilon = 0$. You can follow the procedures here. $\endgroup$
    – mattos
    Oct 11, 2019 at 14:04
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    $\begingroup$ @Sito I deleted my answer because I think it was not useful, but I'm thinking on this problem. $\endgroup$
    – Botond
    Oct 11, 2019 at 18:11

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much for the answer. I will go through it in the next couple of days and see if I can calculate a couple of examples. If questions arise, I hope you don't mind me addressing them directly here. $\endgroup$
    – Sito
    Apr 22, 2020 at 18:17
  • $\begingroup$ @Sito, this is absolutely no problem. Comments exists for this purpose. $\endgroup$ Apr 22, 2020 at 18:25
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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: $\delta S= \int_{x_i}^{x_f}\delta \mathcal{L}d^4x$

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