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  Lagrangian for Maxwell equation in terms of $F_{\mu\nu}$

+ 3 like - 0 dislike

I was thinking many times about the action for Maxwell  vacuum equations written in terms of $F_{\mu\nu}$. I.e., the action $S$ whose variation with respect to $F_{\mu\nu}$ gives Maxwell equations $\partial_{\mu}F^{\mu\nu} = 0$ and $\epsilon^{\mu\nu\alpha\beta}\partial_{\nu}F_{\alpha\beta} = 0$.

I can't construct it, and as for me there is a reason for this: the Maxwell equations are scale invariant, and since $F_{\mu\nu}$ is dimension-2 in natural units, the only scale invariant quantity which can be written in the action is $F_{\mu\nu}F^{\mu\nu}$ (aside from irrelevant for the EOMs term $\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}$). But this only gives a static term in EOMs, $F_{\mu\nu} = 0$.

This reason, however, isn't convincing for me. So, do you know what is the underlying reason for impossibility to write down Maxwell equations as the variation of some action with respect to $F_{\mu\nu}$?

asked Aug 22, 2017 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]

2 Answers

+ 2 like - 0 dislike

Well, there is of course the cheating way to use constraint fields

$$\mathcal{L} = \alpha_\nu \partial_\mu F^{\mu \nu} + \beta_\nu \partial_\mu *\!F^{\mu\nu}$$

The variation with respect to $\alpha$ and $\beta$ then give you Maxwell's equations, the variation with respect to F will give you equations for the constraint fields which are solved after solving the Maxwell equations (and we do not really care about them).

But the reason why this cannot be done without using auxiliary fields seems to me to be the more general fact that one cannot take a single bosonic field (that is, not a field and its antiparticle comrade) and give a Lagrangian generating a first-order differential equation. Technically, this has to do with how the gamma matrices are able to "double the index" on the gradient for fermions, $\partial_\mu \to \partial_\mu \gamma^\mu_{ab}$.

Actually, $F^{\mu\nu}, R^\mu_{\;\nu \kappa \lambda}$ are "the" proper covariant representations of fields of helicity $h=\pm 1, \pm 2$ and the fact that they do not have standalone Lagrangian principles can be seen as the reason why we need Gupta-Bleuler and similar business.

answered Aug 22, 2017 by Void (1,645 points) [ revision history ]
edited Sep 13, 2017 by Void

Outside a still point-like charge the field $F_{\mu\nu}$ is pure Coulomb, isn't it? Thus your words about helicity only describe a part of the total field $F_{\mu\nu}$, namely, the propagating to infinity part.

I do not know why one wants to have a Lagrangian "in terms of $F_{\mu\nu}$". Lagrangian or vacuum equations, say, $\square F= 0$, are not everything. One needs the boundary conditions too, and badly. Otherwise the solutions are unknown. They (the boundary conditions) do not follow from the least action principle.

@VladimirKalitvianski The non-virtual excitation will always be of helicity 1, though. Anyways, the word "helicity" is used here instead of "spin of massless particles" because one finds out from a Weinbergian group-theoretic analysis that massive and massless particles are nothing alike and helicity is a different quantity defined more carefully than spin. When you then go forward and construct causal field operators from these, you find out that helicity $\pm 1$ causal fields do not "fit" into the vector representation of the Lorentz group $D^{(\frac{1}{2}|\frac{1}{2})} \sim A_\mu$, the simplest possibility is $D^{(1|0)} \bigoplus D^{(0|1)} \sim F_{\mu\nu}$. I.e., we cannot expect $A_\mu$ to have causal commutation relations, propagation and so on and so on.

I do not get your point. Does it mean that $D^{(\frac{1}{2}|\frac{1}{2})} \sim A_\mu$ is not full solution, i.e., it is incomplete without the near (Coulomb-like) field? Does $D^{(1|0)} \bigoplus D^{(0|1)} \sim F_{\mu\nu}$ guarantee the presence of the near field in it?

@VladimirKalitvianski The point is not about the classical field theory at all, or even about a theory which is coupled to other stuff such as charges. The point is about completely forgetting about classical fields and building QFT from a Fock space for quantum particles with sharp momenta. During the construction you are free to choose an integer parameter of the representation of the Lorentz group  which is called "helicity". From this point of view, the quantum "helicity" is assigned simply to a quantum field/particle without any reference to the classical solutions and the historical meaning of "helicity", which would be, of course, defined only for classical plane-wave solutions.

I have nothing against helicity and Weinbergian classifications of propagating excitations. Note, although you want to decouple the quanta from charges, this decoupling looks artificial and is dangerous. Artificial, because a "certain" EMF only has the meaning when it is the external field in the equations of motion of some changes in order to be observed and to be spoken of with certainty. And the Lorentz group is also about equations of motion of charges and fields, i.e., it has a practical meaning rather than being an abstract mathematical requirement. Anyway, the real solutions involve the total field sourced with charges and felt with other charges. The Lorentz group representations must deal with the total field to be meaningful. The propagating quanta do not form the total solution, so concentrating on their properties and forgetting the rest is dangerous, in my humble opinion.

"can be seen as the reason why we need Gupta-Bleuler and similar business."

I don't understand how the Gupta-Bleuler and similar are related to $F_{\mu\nu}$ itself. The quantization in terms of $F_{\mu\nu}$ is perfectly defined at least if we're talking about spinorial degrees of freedom. The propagators are also well-defined.

@NAME_XXX Well, this is only a point addressing the discussion "why these laws" or "why these mathematical structures" show up and as such it will necessarily be hand waivy and appealing to some sense of metaphor. The point is simply that $F_{\mu\nu}$ carries the "correct" degrees of freedom in the "correct" representation of the Lorentz group and the fact that we need the "incorrect" $A_\mu$ to formulate dynamics means that we have to take care of the "incorrectness" with a special quantization procedure. That is all.

+ 0 like - 0 dislike

Wat you want: Obtaining all four Maxwell equations from a Lagrangian in $F$, works perfectly well if you use the bi-spinor generators to define $F$ and the other electromagnetic fields instead. Using the chiral gamma matrices and the bi-spinor boost and rotation generators we get:

\[\begin{equation} \begin{array}{lrcl} \mbox{mass dimension 1:}~~~~ & \mathbf{A} &=& \gamma^\mu A^\mu \\ \mbox{mass dimension 2:}~~~~ & \mathbf{F} &=& \vec{K}\cdot\vec{E}-\vec{J}\cdot\vec{B} \\ \mbox{mass dimension 3:}~~~~ & \mathbf{J}\, &=& \gamma^\mu\,j^\mu \\ \end{array} \end{equation}\]

We can now write the laws of the Electromagnetic field, going from one mass dimension to another, by applying the differential operator matrix $/\!\!\!{\partial} =\gamma^\mu\partial_\mu = \sqrt{\Box}$ on the operator field matrices defined above.

\[\begin{equation} /\!\!\!{\partial}\mathbf{A}^{\!\dagger} = \mathbf{F}~~~~~~ ~~~/\!\!\!{\partial}\mathbf{F} = \mathbf{J}^\dagger \end{equation}\]

The complex conjugate transpose $\mathbf{A}^{\!\dagger}=\gamma^\mu A_\mu $ is the covariant form of $\mathbf{A}$.

In the first step we have applied the conservation law $\partial_\mu A^\mu\!=\!0$ on the diagonal and in the second step we find all four of Maxwell's laws, the inhomogeneous $\partial_\mu F^{\mu\nu}\!=\!j^\nu$ as well as the homogeneous $~\partial_\mu\! *\!\!F^{\mu\nu}\!=j^\nu_{^A}=\!0$.

We can write for the Lagrangian of the Electromagnetic field in vacuum:

\[\begin{equation} \mathcal{L} ~=~ \tfrac12\mathbf{F}\mathbf{F} \end{equation}\]

Where $\mathcal{L}$ is a matrix operator field invariant under Lorentz transform. To find the equations of motions: The four Maxwell equations. We write:

\[\begin{equation} \mathcal{L} ~=~ \tfrac12/\!\!\!{\partial}\mathbf{A}^{\!\dagger}/\!\!\!{\partial}\mathbf{A}^{\!\dagger}~~~~ \underrightarrow{\mbox{ Euler Lagrange }}~~~~ /\!\!\!{\partial}/\!\!\!{\partial}\mathbf{A}^{\!\dagger} ~=~ /\!\!\!{\partial}\mathbf{F} ~=~0 \end{equation}\]

Thus the equations of motion, the Maxwell equations, are given by $/\!\!\!{\partial}\mathbf{F}=0$.  If we work out the Lorentz invariant Lagrangian operator field we get.

\[\begin{equation} \mathcal{L} ~~=~~ \tfrac12\mathbf{F}\mathbf{F} ~~=~~ \tfrac12\left(E^2-B^2\right)\!I ~+~ \left(\vec{E}\cdot\vec{B}\right)i\gamma^5\end{equation}\]

The Lorentz scalar $\tfrac12(E^2-B^2)$ of the electromagnetic field is associated with the diagonal matrix $I$ and gives rise to the inhomogenious Maxwell equations. The pseudo scalar $\vec{E}\cdot\vec{B}$ of the electromagnetic field is associated with the pseudo scalar generator $i\gamma^5$ and gives rise to the homogenious Maxwell equations.

There is more information in the PDF here and there is a Mathematica file here.

This is all part of a much larger project: https://thephysicsquest.blogspot.com/ 

answered Nov 23, 2018 by Hans de Vries (90 points) [ revision history ]
edited Nov 23, 2018 by Hans de Vries

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