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The electromagnetic Lagrangian (density) is given by

$$\mathcal{L}=-\frac{1}{4}F_{\mu \nu}F^{\mu\nu}$$

It is said that the Maxwell's equations (and hence this Lagrangian) are $\text{SO}(1,3)$ invariant. This can be checked by explicitly finding the transformation laws of $F_{\mu\nu}$ and plugging them into Maxwell's equations.

How does one formally express the invariance in terms of principal and associated bundles? Would it be wrong to assume that one can express this as invariance of the Lagrangian with respect to the left action on the associated bundle of (covector) densities? How would one write this down?

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  • $\begingroup$ Related : Deriving Lagrangian density for electromagnetic field $\endgroup$ – Frobenius Jul 18 '18 at 12:22
  • $\begingroup$ You don't need to consider the answer in terms of principal bundles, etc, you can simply do the algebra and see that $\mathcal{L}$ remains invariant under the transformation. Is your question actually on seeking the rigorous mathematical side of this invariance? $\endgroup$ – Oktay Doğangün Jul 18 '18 at 19:42
  • $\begingroup$ @OktayDoğangün Yes. I am sufficiently familiar with the algebraic proof but I would like to see how this is implemented from a mathematically rigorous point of view $\endgroup$ – TheBro21 Jul 18 '18 at 19:47
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    $\begingroup$ OK, I will write a suitable answer (not the full thing) in a few minutes. $\endgroup$ – Oktay Doğangün Jul 18 '18 at 19:50
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I am going to give you some basic things you should consider as a couple of hints. I think proving the invariance in a rigorous way would be clearer and probably trivial after making the following correspondences.

First of all in mathematics, a principal $G$-bundle on a manifold, $\mathcal{M}$, is a fiber bundle, $\pi : P\rightarrow \mathcal{M}$, associated with a group action, $G \times P \rightarrow P$, that preserves the fibers. So, a point $p \in P_x$ should map to a point $gp \in P_x$ in the same fiber.

In terms of your question, the group is $SO(1,3)$, the manifold is the flat Minkowski spacetime, $\mathbb{R}^{1+3}$, and the group transformation is the Lorentz transformations. In your case, the fiber over a point $x \in \mathcal{M}$ could be regarded as the set of all frames (i.e. ordered bases) for the tangent space, $T_x \mathcal{M}$.

A vector, $A_\mu (x)$, on the manifold should transform by the transformation, $$ A_\mu (x) \mapsto \Lambda_\mu^\nu (x) \, A_\nu (x) $$ where $\Lambda (x)$ is the Lorentz transformation, but the frame fibers, i.e. the frames, should be preserved.


If the question were about the gauge transformations, instead of Lorentz transformations, then the group associated with the principal bundle would be $U(1)$ symmetry group and each fiber at a point would be the set of all phase configurations (i.e., a smooth real function), and the transformation would be as follows: $$ A_\mu (x) \mapsto A_\mu (x) - \partial_\mu \theta (x) $$ where $\theta (x)$ is the phase parameter. Also here $A_\mu (x)$ would be the connection, and $F_{\mu\nu} (x)$ would be the curvature (since $F=dA$).

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Hint :

It's not good practice to express the scalar $\:F_{\mu\nu}F^{{\mu\nu}}$ as $\:\left(F_{\mu\nu}\right)^{2}$. I think you know that in the former expression we use the summation Einstein's convention, that is : \begin{equation} F_{\mu\nu}F^{{\mu\nu}}\equiv \sum\limits_{\rho,\sigma=1}^{\rho,\sigma=4}F_{\rho\sigma}F^{{\rho\sigma}} \tag{01} \end{equation} Now the electromagnetic field tensor $\:F_{\mu\nu}\:$ is transformed to another reference frame by \begin{equation} F'^{\alpha\beta} = \Lambda^{\alpha}_{\mu}F^{\mu\nu}\Lambda^{\beta}_{\nu} \tag{02} \end{equation} where $\:\Lambda\:$ the Lorentz transformation. So, what you must prove is that \begin{equation} F'_{\alpha\beta}F'^{\alpha\beta} = F_{\mu\nu}F^{{\mu\nu}} \tag{03} \end{equation}


For your information : a representation of $\:F^{\mu\nu}\:$ in terms of the vectors $\:\mathbf{E},\mathbf{B}\:$ is \begin{equation} F^{\mu\nu}= \begin{bmatrix} 0 & \boldsymbol{-}E_{1} & \boldsymbol{-} E_{2} & \boldsymbol{-}E_{3} \\ E_{1} & \hphantom{\boldsymbol{-}} 0& \boldsymbol{-}cB_{3} & \hphantom{\boldsymbol{-}}cB_{2} \\ E_{2} & \hphantom{\boldsymbol{-}} cB_{3} & \hphantom{\boldsymbol{-}} 0 & \boldsymbol{-}cB_{1} \\ E_{3} & \boldsymbol{-}cB_{2} & \hphantom{\boldsymbol{-}} cB_{1} & \hphantom{\boldsymbol{-}} 0 \end{bmatrix} \tag{04} \end{equation} so that \begin{equation} \boldsymbol{-}\frac{1}{2}F_{\mu\nu}F^{{\mu\nu}}=\left|\!\left|\mathbf{E}\right|\!\right|^{2} \boldsymbol{-}c^{2}\left|\!\left|\mathbf{B}\right|\!\right|^{2} \tag{05} \end{equation} So what you are called to prove is that $\:\left(\left|\!\left|\mathbf{E}\right|\!\right|^{2} \boldsymbol{-}c^{2}\left|\!\left|\mathbf{B}\right|\!\right|^{2}\right)\:$ is a Lorentz invariant scalar of the electromagnetic field \begin{equation} \left|\!\left|\mathbf{E'}\right|\!\right|^{2} \boldsymbol{-}c^{2}\left|\!\left|\mathbf{B'}\right|\!\right|^{2}=\left|\!\left|\mathbf{E}\right|\!\right|^{2} \boldsymbol{-}c^{2}\left|\!\left|\mathbf{B}\right|\!\right|^{2} \tag{06} \end{equation} There exists a second Lorentz invariant scalar of the electromagnetic field and this is \begin{equation} \mathbf{E'}\boldsymbol{\cdot}\mathbf{B'}=\mathbf{E}\boldsymbol{\cdot}\mathbf{B} \tag{07} \end{equation} I suggest : if you could prove (03) then try to prove (07).

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    $\begingroup$ Oh, my apologies. I copied the LaTeX code from elsewhere. Apart from that, this does not answer my question as to how to formalise the invariance on bundles $\endgroup$ – TheBro21 Jul 18 '18 at 19:01
  • $\begingroup$ Your question is homework-like. By site policy we must not give full answers to questions of this kind. On the other hand and for your information probably your question will be voted to close since you don't show any effort to solve it. $\endgroup$ – Frobenius Jul 18 '18 at 19:12
  • $\begingroup$ @TheBro21 I think that is important to add any part of my question: physics.stackexchange.com/questions/324041/… $\endgroup$ – Sebastiano Jul 18 '18 at 19:31
  • $\begingroup$ This lagrangian is manifestly invariant. $\endgroup$ – my2cts Jul 18 '18 at 19:47

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