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This question already has an answer here:

We all know that the lagrangian for the free electromagnetic field is given by $$ \mathscr{L} = -\frac{1}{4}F^{\mu \nu}F_{\mu \nu} $$ where $ F^{\mu \nu} = \partial^\mu A^\nu -\partial^\nu A^\mu $ is the electromagnetic field tensor. But also we know that

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Lets consider $c=1$ for simplicity. Then, doing the math, the lagrangian can be written as $$ \mathscr{L} = \frac{1}{2} (|\vec{E}|^2 - |\vec{B}|^2) $$

By applying Euler-Lagrange, i.e. $$ \partial_\mu \left( \frac{\partial \mathscr{L}}{\partial (\partial_\mu \phi_i)} \right) - \frac{\partial \mathscr{L}}{\partial \phi_i} = 0 $$ where $\phi_i$ is each of the components of each field, I find $$ \vec{E} = 0 $$ and $$ \vec{B} = 0 $$ but not the Maxwell equations... What is going on?

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marked as duplicate by Qmechanic lagrangian-formalism Aug 31 '18 at 3:49

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    $\begingroup$ You could do the same thing for, say, the Klein-Gordon Lagrangian: define $F_\mu=\partial_\mu\phi$, so that $L=\frac12F^2$. The Euler-Lagrange equations with respect to $F$ yield $F=0$, which is not equivalent to $\partial^2\phi=0$. In conclusion, you cannot freely redefine your configuration-space variables when using the Euler-Lagrange equations. For the EM field, with Lagrangian $L=\frac14F^2$, the variables are $A^\mu$, not $E,B$. This is particularly clear if you write the action explicitly: it is a functional of $A$, not of $E,B$. Minimising with respect to the latter is not meaningful. $\endgroup$ – AccidentalFourierTransform Aug 31 '18 at 0:21
  • $\begingroup$ Does this mean that $A$ is more fundamental than $E$ and $B$? $\endgroup$ – user171780 Aug 31 '18 at 0:38
  • $\begingroup$ It should also be noted that this form of the field stregth tensor relies on using Maxwell's equations, meaning that the Lagrangian written only applies to stationary field configurations, and can't be used for variational calculations. $\endgroup$ – Bob Knighton Aug 31 '18 at 1:08
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    $\begingroup$ That's now how Euler-Lagrange works. For example, consider the good old $L = m \dot{x}^2 / 2 - V(x)$. If you defined $y = \dot{x}$ and considered $x$ and $y$ as independent variables, you would get trivial equations. It's wrong, because they're not independent. $\endgroup$ – knzhou Aug 31 '18 at 1:44
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    $\begingroup$ @BobakHashemi No, without charges you should get Maxwell’s equations. The field is free, but that doesn’t mean the field is zero. And not every choice of variable is valid, see my example. $\endgroup$ – knzhou Aug 31 '18 at 2:41
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$\vec{E}(x,t)$ and $\vec{B}(x,t)$ are not totally independent variables.


I'm not familar with field theory, but have simpler example with LC circuit, a 0-dimensional field theory:

$$ H=T+V=\frac{L}{2}I^2+\frac{1}{2C}Q^2 \quad \sim \quad \frac{1}{2}(|\vec{B}|^2+|\vec{E}|^2) $$

$$ L=T-V=\frac{L}{2}\dot{Q}^2-\frac{1}{2C}Q^2 \quad \sim \quad \frac{1}{2}(|\vec{B}|^2-|\vec{E}|^2) $$

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Charge $Q$ is the dynamic variable. $|\vec{E}|$ is propotional to $Q$, $|\vec{B}|$ is propotional to $\dot{Q}=I$,

You can't take current $I$ as independent from $Q$. This is the constraint of the system.



If you insist to take $I$ and $Q$ as independent variable, then you get a different system: a capacitor and a inductor seperately

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