In Landau's Classical Field Theory, the action function of the electromagnetic field is constructed based on the following facts (Chapter 4,$\S 27$):
- $S_f$ is the integration of sum scalar function over the whole space time $\implies S_f = \iint (\text{Lorentz scalar}) \mathrm{d}V \mathrm{d}t$
- EM field satisfies superposition principle $\implies$ the integrand must be quadratic in fields.
- potentials cannot enter into the expression of $S_f$ since they are not uniquely determined.
- derivative of $F_{\mu\nu}$ cannot enter into the expression of $S_f$ since it can only contains coordinates $A_\mu$ and its first time derivative.
based on (3) and (4), we have $$S_f = a \iint F_{\mu\nu} F^{\mu\nu} \mathrm{d}V\mathrm{d}t = 2a \iint (\mathbf{B}^2 - \mathbf{E}^2) \mathrm{d}V\mathrm{d}t.$$
- since $\mathbf{E}^2$ contains the term $(\partial \mathbf{A} / \partial t)^2$, it must appear in the action with a plus sign since if it follows a minus sign, we can let it change rapidly to make $S_f$ a negative quantity with arbitrary large absolute value. $\implies$ $a$ is negative, and in Gaussian system, $a = -\frac{1}{16\pi}$
Here are my questions:
- Question about (3): Even though $A_\mu$ is not uniquely determined, it enters into the expression of $S_{mf} = -\frac{q}{c} \int A_\mu \mathrm{d}x^\mu$. Under a gauge transformation, $S_{mf}$ will pick up a term which depends only on the boundary condition. So why can't $S_f$ behave like this?(i.e. the Lagrangian of EM field is not gauge invariant itself but when integrated, the action $S_f$ is gauge invariant)
- Question about (4): Is this just a empirical fact or there are some deeper reasons?
- Question about (5): I think this is most confusing. I can apply Landau's argument in another way: Since $\mathbf{B}^2$ contains the term $(\nabla \times \mathbf{A})^2$, if it follows a minus sign, we can let $\mathbf{A}$ be "arbitrarily curly" to make $S_f$ a negative quantity with arbitrary large absolute value. So it seems that $a$ must be positive?
