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I have a question on the variational formulation of electromagnetism.

I read that the action is given by

$\mathcal{S}_{EM} = \int d^4x\ F_{\alpha \beta} F^{\alpha \beta}$ (1)

and the source term is given by

$\mathcal{S}_{source} = \int d^4x\ A_{\alpha} J^{\mu}$ (2)

My question is on how one obtains the above two actions. Here is the explanation that I read.

A.) For expression (1), by the Lorentz invariant of electromagnetism, the Lagrangian must be of the form

$c_1 F_{\alpha \beta} F^{\alpha \beta} + c_2(F_{\alpha \beta} F^{\alpha \beta})^2 + c_3(F_{\alpha \beta} F^{\alpha \beta})(F_{\sigma \tau} F^{\sigma \tau})+...$ (3)

Furthermore, we assert that the higher order terms (i.e. the terms including and after the second term) are all small, experimentally. (Alternatively, we can assert that they are small when we work in Plank units.) Therefore, we are left with the form of the action as in equation (1).

B.) We get equation (2) as follows. We want $\mathcal{S}_{EM} + \mathcal{S}_{source}$ to be invariant under the gauge transformation $A_{\alpha} \to A_{\alpha} + \partial \Lambda$ for any smooth scalar function $\Lambda$. Furthermore, under this gauge transformation, we get $\delta A_{\mu} = \partial_{\mu} \Lambda$, so

$\delta(\mathcal{S}_{source}) = -\int d^4x\ \Lambda \partial_{\mu} J^{\mu} + \text{boundary terms}$

We get gauge invariance if $\partial_{\mu} J^{\mu} = 0$ holds, i.e. if the charge conservation holds. Thus, we get equation (2).

In summary, equation (1) follows from Lorentz invariance and experimental evidence. Equation (2) follows from gauge invariance and conservation of charge.

The above formulation seems unsatisfying to me for the following reasons.

  • Argument A.) relies on experimental evidence which seems to not follow from general principles. Is there a way of coming up with this using only knowledge in, say Purcell?

  • All argument B.) seems to do is check whether equation (2) satisfies gauge invariance and charge conservation, and in particular, it does not help us come up with equation (2).

My question: is there a more intuitive way of "seeing" equation (1) and (2)? (For instance, in classical mechanics, the Lagrangian of a system is fairly easy to see since we can just take the difference of the kinetic and potential energies which in turn follows from undergraduate Newtonian mechanics.)

What I have tried: I noticed that $F_{\alpha\beta}F^{\alpha\beta} = \frac{1}{2}(\mathbf{E}^2 - \mathbf{B}^2)$, and I also recalled that the energy density of an electromagnetic field is $\frac{1}{2}(\mathbf{E}^2 + \mathbf{B}^2)$. This seems to be analogous to $L=T-V$ and $E = T+V$ from classical mechanics. I am not sure if this analogy is just a coincidence.

I tried Googling all of the above of course, but I could not find anything that satisfied me.

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    $\begingroup$ Your $c_3$ terms is the same as the $c_2$ one, You presumably meant it to be $(F^2)^3$. $\endgroup$
    – G. Smith
    Commented Feb 29, 2020 at 0:18
  • $\begingroup$ Chapters 1 to 4 of this book explain how to get these things from first principles. $\endgroup$
    – bolbteppa
    Commented Feb 29, 2020 at 0:19
  • $\begingroup$ There is another invariant, $(\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta})^2$ proportional to $(\mathbf E\cdot\mathbf B)^2$, and you can include powers of it as well. $\endgroup$
    – G. Smith
    Commented Feb 29, 2020 at 0:29
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/55291/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Feb 29, 2020 at 4:08

1 Answer 1

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Arguments like A) or B) are formalistic exercises that start with "simple assumptions" about functional and transformation properties of field action and pretend that those inevitably lead to the correct equations of motion for the field. In reality, those arguments assume additional things that are rarely mentioned, for example that there is only one universal field $A^\mu$, or that the field is not too singular function of position (no point particles), or that the Lagrangian density is Lorentz invariant (not necessary). These arguments have no predictive value in case of EM theory, they are created in reverse, with knowledge of what the correct result is.

The actual first principles based on generalization of experience are that 1) EM field obeys Maxwell's equations and 2) particles experience the field according to the equations of motion including the Lorentz force formula. Then the simplest form of action that results in Maxwell's equations and particle equations of motion is sought.

The action terms look like what you wrote above because that is one of the forms that results in Maxwell's equations for fields for prescribed behaviour of charge and current distribution. It does not give equations for the charge and current distributions, those are assumed to be known.

I wrote one of the forms, because there are different ways to formulate the Hamilton principle for the EM field obeying Maxwell's equations. Just as in classical mechanics, many different Lagrangians can give the same result.

The formulations can also differ in what the independent variables are, and what the action function is. Some are gauge-invariant like $F^{\mu\nu}F_{\mu\nu}$, some are gauge-variant such as $\partial_\mu \! A_\nu~~\partial^\mu\! A^\nu$.

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  • $\begingroup$ the Lagrangian density is Lorentz invariant (not necessary) Can you really get the Lorentz-covariant Maxwell equations from a non-invariant Lagrangian density? If so, what is one such? $\endgroup$
    – G. Smith
    Commented Feb 29, 2020 at 4:42
  • $\begingroup$ Indeed we can. Let $A^\mu$ be a 4-vector field, then we can define field $P^\mu = A^\mu + \partial^\mu \chi$ where $\chi$ is not a Lorentz invariant field, for example, $\chi = v^2z$ where $v$ is velocity of the inertial frame with respect to some preferred inertial frame. Then $P^\mu$ is 4 fields, but not a 4-vector field. $\endgroup$ Commented Feb 29, 2020 at 17:42
  • $\begingroup$ I see. Thanks for the example! $\endgroup$
    – G. Smith
    Commented Feb 29, 2020 at 17:47
  • $\begingroup$ This does not make the standard field Lagrangian (FF) Lorentz variant, but it makes the interaction term $P_\mu j^\mu$ Lorentz variant. $\endgroup$ Commented Feb 29, 2020 at 17:49