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The answer to this question might seem obvious, but I'm struggling to find good, accessible sources on the calculus of variations generally, and this pesky detail in particular, so I thought I might as well try my luck and possibly help others struggling to get a firm hold of some of the basics.

My question is largely one about "proper" notation: Reading physics literature (I'm a physicist before a mathematician) and taking various classes, I've seen conflicting use of what the "$\delta$" in e.g. "$\delta F$" means.

Specifically, some sources (including https://cds.cern.ch/record/1383342/files/978-3-642-14090-7_BookBackMatter.pdf and Anthony Duncan's book "The Conceptual Framework of Quantum Field Theory") seems to label this "the variation of $F$", defined by \begin{equation} \tag{1}\label{1}\delta F[\rho] \equiv F[\rho + \delta\rho] - F[\rho], \end{equation} where $\delta\rho \mathop{\widehat{=}} \varepsilon\eta$, and, as far as I understand, $\eta$ belongs to the same space of functions as $\rho$, and $\varepsilon$ is a small number.

Other sources, however (for example Wikipedia: https://en.wikipedia.org/wiki/First_variation) label this "the first variation of $F$", defined by \begin{equation} \tag{2}\label{2} \delta F[\rho]_{\eta} \equiv \lim_{\varepsilon \to 0} \frac{F[\rho + \varepsilon\eta] - F[\rho]}{\varepsilon} = \int \frac{\delta F[\rho]}{\delta\rho(x)}\, \eta(x)\, dx, \end{equation} where the last equation holds only when the functional derivative $\delta F/\delta\rho$ is defined.

Now, I do understand that these conflicting definitions often will be asymptotically equal, in the sense that \begin{equation} \tag{3}\label{3} F[\rho + \varepsilon\eta] \simeq F[\rho] + \varepsilon\int \frac{\delta F[\rho]}{\delta\rho(x)}\, \eta(x)\, dx + \cdots, \end{equation} but I nonetheless wonder which might be seen as most pedantically "correct" ...

If there's anything I might have misunderstood, feel free to point it out.

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These definitions are not really conflicting, the first one is technically imprecise however.

The proper definition would be the "differentiation" definition.

In the first case, $\delta$ isn't what maps $F[\rho]$ to $F[\rho+\delta\rho]-F[\rho]$, but what maps $F[\rho]$ to $F[\rho+\delta\rho]-F[\rho]$ with second-order and higher terms neglected in the Taylor-expansion of $F[\rho+\delta\rho]$. A differentiation is essentially taking the first order coefficient of a Taylor-expansion, so technically these two are the same.

Of course in old-school physics literature, they will make it sound the Taylor-expansion definition is just some sort of "approximation", while the "differentiation" definition seems exact, but both essentially give the linear approximation to the change in the functional in the direction of the function $\delta\rho$, which is what differentiation does.

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  • $\begingroup$ Thanks! That was my inkling. However, isn't there missing a $\varepsilon$-factor in definition (1) for (1) and (2) to be asymptotically equal? (And I suppose I should have added it in into (3) as well, somehow ...). $\endgroup$
    – B. Bergtun
    Aug 3, 2016 at 14:07
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    $\begingroup$ That depends on whether you take the variation of the field itself to be $\delta\psi$, where $\delta\psi$ itself is "infinitesimal", or you take it to be $\epsilon\delta\psi$, where you only adjust the value of $\epsilon$. The latter is better imo. Epsilon makes bookkeeping on the orders easier. After all if $F[\psi]=\int L(\psi,\nabla\psi)\ d^nx$, then $F[\psi+\epsilon\delta\psi]=\int L(\psi,\nabla\psi)\ d^nx+\int\delta F/\delta\psi\cdot\epsilon\delta\psi\ d^nx+O(\epsilon^2)$, then if you substract the zeroth order term and divide by epsilon then take the limit, then it becomes clear that... $\endgroup$ Aug 3, 2016 at 14:13
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    $\begingroup$ @B.Å.Bergtun ...$\int\delta F/\delta\psi\cdot\delta\psi\ d^nx$ survives as the first order term, while the higher order terms get annihilated because assimptotically $O(\epsilon^2)/\epsilon$ is vanishing (as $\epsilon\rightarrow 0$). This is harder to see if you "bake" the $\epsilon$ into $\delta\psi$. $\endgroup$ Aug 3, 2016 at 14:15
  • $\begingroup$ That makes sense. I've edited the notation in my original post accordingly, to avoid confusing others seeking answers. $\endgroup$
    – B. Bergtun
    Aug 3, 2016 at 23:47

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