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What fields am I allowed to use as variables of variation for extremizing the action?

For example, using the free EM action: \begin{equation} S_{EM}=-\frac{1}{4\mu_{o}}\int d^4x F_{\mu\nu}F^{\mu\nu} \end{equation} where $F_{\mu\nu}[(A_{\mu}(x,t)]$. Is it a legal move to vary the action w.r.t. $F_{\mu\nu}$ and $\partial_{\lambda}F_{\mu\nu}$ instead of $A_{\mu}$ and $\partial_{\lambda} A_{\mu}$?

ie: $\delta S = \int d^4 x \left( \frac{ \partial \mathcal{L} }{ \partial F_{\mu\nu}} \delta F_{\mu\nu} + \frac{ \partial \mathcal{L} }{ \partial ( \partial_\lambda F_{\mu\nu} ) } \partial_\lambda \delta F_{\mu\nu} \right)$

So long as the fields or field derivatives are present in the action, what dictates which variables are allowed to be used for variation?

Edit: The above example is used to illustrate my question. I understand that it gives a trivial solution.

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    $\begingroup$ Don't you just get $F_{\mu \nu} = 0$ if you do this? That would seem to be an argument against using $F_{\mu \nu}$ as your configuration variable. $\endgroup$ Jul 4, 2023 at 2:27
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    $\begingroup$ This was just to show what I meant by changing the variation variable. Disregarding getting a trivial solution, is it still legal to choose $F_{\mu\nu}$ as the configuration variable? Besides getting something sensible, what else defines our choice of variation variable? $\endgroup$ Jul 4, 2023 at 3:33
  • $\begingroup$ You’ve got it backwards. It’s not that we have arbitrary actions, and we come up with some mathematical rule to decide which fields can be varied. Instead, we start by thinking about which fields are the appropriate dynamical variables, and then construct actions which are functions of those fields. The “right” thing to vary with respect to is actually built in from the start. $\endgroup$
    – knzhou
    Jul 4, 2023 at 10:41
  • $\begingroup$ Hmm, but if we derive a new equation of motion from combining/modification of others, we would have to guess the form of the action for the newly derived EOM. More than one action may result in the same EOM, especially if they are varied with different dynamical variables. How can we decide which dynamical variables are correct if they lead to the same equations of motion? $\endgroup$ Jul 4, 2023 at 11:21

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That's quite an open question. Mathemtically, I think, you can define any old functional and calculate a variation of it.

In physics it depends on the context. You want the results to be in accordance with physical principles and observations. And in some way one might say you are putting the cart before the horse with your question because often one would start with a field and ask how the Lagrangian for it would look like.

But ok, I think what you might be after is this: Given an action functional and a Lagrangian like $$ S = \int \mathcal{L}\, \text{d}x \,, \\ \mathcal{L}=\frac{1}{2}\varepsilon_0 E^2 - \rho\varphi\,, $$ do you take the variation of $S$ with respect to the electric field $E$ or the potential $\varphi$? (I'm using electrostatics in one dimension for simplicity, the principles should generalize, of course.) Add to that the fact that $E = -\partial_x \varphi$. This is not contained in the Lagrangian, but it is quite important because in consequence you cannot calculate $\frac{\delta S}{\delta \varphi} = -\rho$ (incorrect! for demonstration) while ignoring the fact that $E$ depends on $\varphi$.

What you usually do is to calculate $$ \frac{d}{d\varepsilon}S\left(\varphi + \varepsilon \delta\varphi\right) = \frac{d}{d\varepsilon} \int \frac{1}{2}\varepsilon_0 \left(\partial_x \varphi(x) + \varepsilon \partial_x \delta\varphi(x)\right)^2 - \rho \left(\varphi(x) + \varepsilon \delta\varphi(x)\right)\text{d}x \\ = \int \left(- \varepsilon_0 \partial_x^2 \varphi(x) - \rho \right) \delta\varphi(x)\text{d}x + \mathcal{O}(\varepsilon) $$ and you get your Euler-Lagrange equation, the (here one-dimensional) Poisson equation $$ \partial_x^2 \varphi = - \frac{\rho}{\varepsilon_0}\,. $$

Now, let's have a look at what the variation with respect to $E$ yields: $$ \frac{d}{d\varepsilon}S\left(E + \varepsilon \delta E\right) = \frac{d}{d\varepsilon} \int \frac{1}{2}\varepsilon_0 \left(E(x) + \varepsilon \delta E(x)\right)^2 + \left(\int_{-\infty}^x E(x') + \varepsilon \delta E(x') \text{d}x'\right) \rho(x) \text{d}x\\ = \int \varepsilon_0 E(x) \delta E(x) + \rho(x) \left(\int_{-\infty}^x \delta E(x') \text{d}x'\right) \text{d}x + \mathcal{O}(\varepsilon)\\ = \int \left( \varepsilon_0 E(x) - \int_{-\infty}^x \rho(x')\text{d}x'\right) \delta E(x)\text{d}x + \mathcal{O}(\varepsilon) $$ and you get your Euler-Lagrange equation $$ \varepsilon_0 E(x) = \int_{-\infty}^x \rho(x')\text{d}x'\,. $$ Differentiate once to find that it is Gauß' law in one dimension and insert $E=-\varphi$ to find that it is equivalent to the Euler-Lagrange equation we found above from $\frac{\delta S}{\delta \varphi}$.

So as long as you do it right, it doesn't matter whether you take the variation with respect to $\varphi$ or $E$. That being said, usually you would use $\varphi$, i.e. such that the action functional depends on the function and its derivatives, not on its antiderivatives. (In fact, I've never seen the second calculation before and had to do it myself, which means I learnt something interesting from your question. Thanks.)

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  • $\begingroup$ Thank you for your amazing answer! It helped clear up a lot of things. I'm glad that it could also give you new insight! $\endgroup$ Jul 4, 2023 at 11:13
  • $\begingroup$ As for your last remark about how the action functionals depending on the function and its derivatives, is it allowed for the action functional to only depend one or the other? ie: only depending on a derivative vs only depending on the function? $\endgroup$ Jul 4, 2023 at 11:17
  • $\begingroup$ Oh never mind, I have thought of several examples. It is early in the morning and the obvious solutions slipped my mind. Please disregard the last comment. $\endgroup$ Jul 4, 2023 at 11:45
  • $\begingroup$ Just to muddy the waters: while you can't vary $\phi$ without varying $\vec{E}$ here, in general relativity it is permissible to vary the metric and the derivative of the metric separately (the so-called "Palatini formulation") — you get the same equations of motion overall. $\endgroup$ Jul 4, 2023 at 12:22

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