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I have a question about a rule from the calculus of variations.

Assume we consider the space of differentiable functions on $C^1(\mathbb{R})$ (or for the sake of simplicity the smooth functions $C^{\infty}(\mathbb{R})$.

My question is why the "variational derivative" commutes with a total derivative, namely why for $q \in C^1(\mathbb{R})$ holds

$$\frac{\delta}{\delta q(\tilde{t})} \frac{d}{dt} = \frac{d}{dt} \frac{\delta}{\delta q(\tilde{t})} ~?$$

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OP is essentially asking the following.

Why the total time derivative $$ \frac{d}{dt}~=~\frac{\partial}{\partial t} + \sum_{m=0}^{\infty}\sum_{i=1}^n q^{i(m+1)}(t)\frac{\partial}{\partial q^{i(m)}(t)} \tag{1} $$ and the functional/variational derivative $$ \frac{\delta}{\delta q^j(t^{\prime})} \tag{2}$$ commute?

That's a good question. The intuitive reason is that the differentiations refer to different variables. But actual calculations make it less obvious (cf. e.g. eq. (3) below). Assume that the derivatives (1) & (2) act on some space ${\cal F}$.

  1. For instance, say that ${\cal F}$ is the space of functions of the form $f(q^{i(m)}(t),t)$. Then we may write the functional derivative (2) as $$ \frac{\delta}{\delta q^j(t^{\prime})} ~=~ \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)},\tag{2'}$$ because an infinitesimal variation is of the form $$\begin{align}\int\! dt^{\prime} \sum_{j=1}^n \frac{\delta f(q^{i(m)}(t),t)}{\delta q^j(t^{\prime})}\delta q^j(t^{\prime}) &~=~\delta f(q^{i(m)}(t),t) ~=~\sum_{j=1}^n \sum_{\ell=0}^{\infty}\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)}\delta q^{j(\ell)}(t)\cr &~=~\int\! dt^{\prime} \sum_{j=1}^n \sum_{\ell=0}^{\infty}\delta^{(\ell)}(t\!-\!t^{\prime})\frac{\partial f(q^{i(m)}(t),t)}{\partial q^{j(\ell)}(t)} \delta q^j(t^{\prime}). \end{align} \tag{2"} $$ When we calculate the commutator $$\begin{align}\left[\frac{\delta}{\delta q^j(t^{\prime})}, \frac{d}{dt}\right] &\stackrel{(1)+(2')}{=}~ \sum_{m=0}^{\infty}\sum_{i=1}^n \left[\frac{\delta}{\delta q^j(t^{\prime})},q^{i(m+1)}(t)\right]\frac{\partial}{\partial q^{i(m)}(t)} -\sum_{\ell=0}^{\infty}\left[\frac{d}{dt}, \delta^{(\ell)}(t\!-\!t^{\prime})\right]\frac{\partial}{\partial q^{j(\ell)}(t)} \cr &~~~=~~~ \sum_{m=0}^{\infty} \delta^{(m+1)}(t\!-\!t^{\prime}) \frac{\partial}{\partial q^{i(m)}(t)} - \sum_{\ell=0}^{\infty} \delta^{(\ell+1)}(t\!-\!t^{\prime})\frac{\partial}{\partial q^{j(\ell)}(t)}~=~0,\end{align} \tag{3} $$ we get zero!

  2. If we extend ${\cal F}$ with functions of finite many other times, or with functionals (say, with internal time-integrations), or both, similar calculations show that the derivatives (1) & (2) commute.

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  • $\begingroup$ Hi. Thank you for your enlightening explanations. One point is unclear: how do you obtain the expression $(2')$ for $\frac{\delta}{\delta q^j(t^{\prime})} $? $\endgroup$ – KarlPeter Apr 6 at 20:40
  • $\begingroup$ the only applyable tool to calculate $\frac{\delta}{\delta q^j(t^{\prime})}$ I found at your linked wiki page en.wikipedia.org/wiki/Functional_derivative. Here it is stated that for every functional $F\colon M \rightarrow \mathbb{R} $ by definition holds (*)$\begin{align} \int \frac{\delta F}{\delta\rho}(x) \phi(x) \; dx &= \lim_{\varepsilon\to 0}\frac{F[\rho+\varepsilon \phi]-F[\rho]}{\varepsilon} \\&= \left [ \frac{d}{d\epsilon}F[\rho+\epsilon \phi]\right ]_{\epsilon=0},\end{align}$ $\endgroup$ – KarlPeter Apr 7 at 2:26
  • $\begingroup$ I'm not sure which "simple cases" you have in mind but I think that we can take for example the "evaluation functional" $ F_{t_0}: M \to R, f \mapsto f(x_0)$ at $t_0$. Then the right hand side of (*) is $\phi(x_0)$ and on the left side we are looking for a $\frac{\delta F}{\delta\rho}(t) $ such that the integral equals $\phi(t_0)$. We see that $\frac{\delta F}{\delta\rho}(t):= \delta(t -t_0) $ solves the problem. But $\frac{\delta F}{\delta\rho(t')}(t) \neq \delta(t -t_0) \frac{\partial}{\partial \rho(t)} F$. $\endgroup$ – KarlPeter Apr 7 at 2:26
  • $\begingroup$ So this contradicts (2') or do I have overseen something? Futhermore - assume that my calculations are just wrong - how to verify that (2') holds in general? $\endgroup$ – KarlPeter Apr 7 at 2:26
  • $\begingroup$ For starters, eq. (2') applies to functions, not functionals. $\endgroup$ – Qmechanic Apr 7 at 10:57

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