Least-action classical electrodynamics without potentials

In classical electrodynamics, the physical quantities of interest are the fields. The theory is already formulated "without" potentials if you think of Maxwell's equations.

The potentials come into play later if you want to simplify the equations and find solutions using i.e. Green's functions etc. However, in quantum electrodynamics, the potentials acquire a real physical role, see e.g. the Aharanov-Bohm-effect.

I really can't see the connection between the derivation of Maxwell Eq. from the least action principle and to not use potentials, we can include or remove potentials at any point, because in classical physics (not quantum) this is just a matter of definitions.

Regarding the least action derivation, to use it we need to have some Lagrangian to start with, if you suppose it given then off course you can derive all Maxwell Equations.

Anyway an interesting thing I know, that if you will just suppose that we have some force that behaves like gravity (central and proportional to the distances squared) but admits different signs of mass (which is actually charges), in other words if you suppose that Coulombэы law is given, then by applying least action principle on the Lagrangian of special relativity, you can show that there should be definitely another coupling "force", which is magnetism.

I don't know if other approach is possible but this one does not workd, We start with tensor $F_{munu}$:

$$F_{munu}=partial_{mu}A_{nu}-partial_{nu}A_{mu} $$

but forget about the 4-potential and define it to be:

$$F^{munu} = begin{bmatrix} 0 &-E_x/c & -E_y/c & -E_z/c E_x/c & 0 & -B_z & B_y E_y/c & B_z & 0 & -B_x E_z/c & -B_y & B_x & 0 end{bmatrix} $$

and write the Lagrangian density as a function of the Cartesian components of the fields, say:

$$mathcal{L}=mathcal{L}(E_{x},..,B_{z}) $$

and

$$mathcal{L}=-frac{1}{mu_0}F^{munu}F_{munu} $$

Then Euler Lagrange equations gives you (for example applied to $E_{x}$) $displaystyle frac{partial mathcal{L}}{partial{E_{x}}}=0$ so this is not consistent.

How we can solve the problem? Thinking in another Lagrangian density, defining a new $F_{munu}$ tensor, choosing more carefully the independent fields?

1) Well, at the classical level, if we are allowed to introduce auxiliary variables, we can always trivially encode a set of equations of motion $$tag{1} {rm eom}_i = 0, qquad iin{1, ldots, n},$$ in a brute force manner with the help of Lagrange multipliers $$tag{2}lambda^i, qquad iin{1, ldots, n},$$ so that the Lagrangian density simply reads $$tag{3}{cal L}~=~sum_{i=1}^nlambda^i ~{rm eom}_i.$$

This is for many reasons not a very satisfactory solution. (Especially if we start to think about quantum mechanical aspects. However, OP only asks about classical physics.) Nevertheless, the above trivial rewritings (3) illustrates how it is hard to formulate and prove no-go theorems with air-tight arguments.

2) To proceed, we must impose additional conditions on the form of the action principle. Firstly, since we are forbidden to introduce gauge potentials $A_{mu}$ as fundamental variables (that we can vary in the action principle), we will assume that the fundamental EM variables in vacuum should be given by the ${bf E}$ and ${bf B}$ field. Already in pure EM, it is impossible to get the $1+1+3+3=8$ Maxwell eqs. (in differential form) as Euler-Lagrange eqs. by varying only the $3+3=6$ field variables ${bf E}$ and ${bf B}$. So we would eventually have to introduce additional field variables, one way or another.

3a) It doesn't get any better if we try to couple EM to matter. In decoupling corners of the theory, we should be able to recover well-known special cases. E.g. in the case of EM coupled to charged point particles, say in a non-relativistic limit where there is no EM field, the Lagrangian of a single point charge should reduce to the well-known form

$$tag{4}L~=~frac{1}{2}mv^2$$

of a free particle. A discussion of eq. (4) can be found e.g. in this Phys.SE post. Here we will assume that eq. (4) is valid in what follows.

3b) Next question is what happens in electrostatics

$$tag{5} mdot{bf v}~=~ q{bf E}? $$

The answer is well-known

$$tag{6} L~=~frac{1}{2}mv^2 - V $$

with potential energy

$$tag{7}V~=~ qphi, $$

where $phi$ is the scalar electric potential. However, since we are forbidden to introduce the potential $phi$ as a fundamental variable, we must interpret it

$$tag{8}phi({bf r})~:=~-int^{bf r} !d{bf r}^{prime}cdot{bf E}({bf r}^{prime}) $$

as a functional of the electric field ${bf E}$, which in turn is taken as the fundamental field. Note that eqs. (6)-(8) correspond to a non-local action.

3c) The straightforward generalization (from point mechanics to field theory) of eq. (7) is a potential density

$$tag{9}{cal V}~=~ rhophi, $$

where $rho$ is an electric charge density. Readers familiar with the usual action principle for Maxwell's eqs. will recognize that we are very close to argue that the interaction term between pure EM and matter must be of the form

$$tag{10} {cal L}_{rm int}~=~J^{mu}A_{mu},$$

even if we haven't yet discussed what should replace the standard Lagrangian

$$tag{11} {cal L}_{rm EM} ~=~-frac{1}{4}F_{munu}F^{munu}$$

for pure EM.

3d) Staying in electrostatics, let us ponder our prospects of deriving Gauss' law in differential form

$$tag{12} nabla cdot {bf E} ~=~ rho. $$

Obviously, the rhs. of the single eq. (12) should appear by varying the potential density (9) wrt. one of the three $E$ fields, but which one? The counting is not right. And because eq. (9) is non-local, we will in any case get an integrated version of $rho$ rather than $rho$ itself, which appears on the rhs. of eq. (12), and which we set out to reproduce.

3e) In conclusion, it seems hopeless to couple a EM theory (with ${bf E}$ and ${bf B}$ as fundamental variables) to matter, and reproduce standard classical eqs. of motion.

4) The standard remedy is to introduce $4$ (globally defined) gauge potentials $A_{mu}$ as fundamental variables. This makes $1+3=4$ source-less Maxwell eqs. trivial, and the remaining $1+3=4$ Maxwell eqs. with sources may be derived by varying wrt. the $4$ fundamental variables $A_{mu}$.

For instance, the standard (special relativistic) action for EM coupled to $n$ massive point charges $q_1, ldots, q_n$, at positions ${bf r}_1, ldots, {bf r}_n$, is given as

$$tag{13} S[A^{mu}; {bf r}_i] ~=~int ! dt ~L, $$

where the Lagrangian is

$$ tag{14} L ~=~ -frac{1}{4}int ! d^3r ~F_{munu}F^{munu} -sum_{i=1}^n left(frac{m_{0i}c^2}{gamma({bf v}_i)} +q_i{phi({bf r}_i) - {bf v}_icdot {bf A}({bf r}_i)} right). $$

The corresponding Euler-Lagrange eqs. are $4$ Maxwell eqs. with sources (when varying $A_{mu})$, and $n$ (special relativistic) Newton's 2nd laws with Lorentz forces (when varying ${bf r}_i)$.

Why not try? for example with a Lagrangian of the specific form "polynomial of second order" $$mathcal{L}(x^{mu},X,partial_{mu}X):= sum_{i,j=1}^6left( A_{ij} X^i X ^j + sum_{mu=0}^3 B^{mu}_{ij} partial_{mu}X^i X ^j +sum_{mu,nu=0}^3 C^{munu}_{ij} partial_{mu} X^i partial_{nu} X^j right) tag{1}label{1}$$ in $X=big(E_x, E_y,E_z, B_x, B_y, B_z big)$. (Since the classical fields $X^i X^j= X^j X^i$ commute, only the symmetric part of $A$ and $C$ play a role, so we can assume they are symmetric.) Taking into account the answer of Qmechanics, we'll not try to recover all of the Maxwell equations, but let us for example try the ones with possible "source terms": $$ leftlbrace begin{aligned} vec{nabla}cdot vec{mathbf{E}} &= 0 vec{nabla}wedge frac{vec{mathbf{B}}}{mu} &= epsilon_0 frac{partial vec{mathbf{E}}}{partial t} end{aligned} right. tag{2}label{2}$$ With the notations of (ref{1}), these corresponds to the vanishing of 4 linear forms (1 for the first "scalar" equation, 3 for the second vector equation; $X^i$ i-th component, not something to the power i ) $$ leftlbrace begin{aligned} alpha(partial_{mu}X)&:= sum_{i=1}^3sum_{mu=0}^3 alpha^{mu}_{i} partial_{mu}X^i = sum_{i=1}^3sum_{mu=0}^3 delta^{mu}_{i} partial_{mu}X^i = sum_{i=1}^3 partial_i X^i beta(partial_{mu}X)&:= sum_{i=1}^6sum_{mu=0}^3 beta^{mu}_{i} partial_{mu}X^i = frac{1}{mu} left(partial_y B_z - partial_z B_y right) - epsilon_0 frac{partial E_x}{partial t} = frac{1}{mu} left(partial_y X^6 - partial_z X^5 right) - epsilon_0 frac{partial X^1}{partial t} vdotsquad &= quad vdots end{aligned} right. tag{3}label{3} $$ Left-hand side of the Euler-Lagrange equation in the form $frac{partial mathcal{L}}{partial X^i} = partial_{mu}frac{partial mathcal{L}}{partial (partial_{mu}X^i)}$ reads $$ frac{partial mathcal{L}}{partial X^i} = sum_{j=1}^6left(2 A_{ij} X ^j + sum_{mu=0}^3 B^{mu}_{ji} partial_{mu}X^j right) $$ and r.h.s. (with Einstein's summation convention) $$ partial_{mu}frac{partial mathcal{L}}{partial (partial_{mu}X^i)} = partial_{mu} left[sum_{j=1}^6left( sum_{mu=0}^3 B^{mu}_{ij} X ^j + 2 sum_{mu,nu=0}^3 C^{munu}_{ij} partial_{nu} X^j right) right] = sum_{j=1}^6left( sum_{mu=0}^3 B^{mu}_{ij} partial_{mu}X^j + 2 sum_{mu,nu=0}^3 C^{munu}_{ij} partial_{munu} X^j right) $$ One sees that in order to get (ref{3}), the Lagrangian must not have terms in $A_{ij}$ and $C^{munu}_{ij}$. Then $$ frac{partial mathcal{L}}{partial X^i} - partial_{mu}frac{partial mathcal{L}}{partial (partial_{mu}X^i)} = sum_{j=1}^6 sum_{mu=0}^3left( B^{mu}_{ji} - B^{mu}_{ij} right)partial_{mu}X^j $$ (Caution: $B^{mu}_{ij}$ notation from (ref{1}), not the magnetic field...). Only its antisymmetric part plays a role so let us choose it antisymmetric to have unicity. By hand, let us set for $i=4, 0 leq mu leq 3$ $$ B^{mu}_{4j}= - B^{mu}_{j4}= delta_j^{mu}enspace text{if} 1leq j leq 3quad text{and}quad B^{mu}_{4j}= - B^{mu}_{j4} = 0enspace text{if} 4leq j leq 6 tag{4}label{4}$$ (i.e. very explicitly the following contribution $ vec{mathbf{E}}cdot vec{nabla} B_x - B_x big(vec{nabla} cdot vec{mathbf{E}} big)$ to the Lagrangian (ref{1}))

This does yield the first equation of (ref{3}) ($-1$). For $i=2, 0 leq mu leq 3$, let us try $$ B_{26}^2 = - B_{62}^2= frac{-1}{2mu} ,quad B_{25}^3 = - B_{52}^3= frac{1}{2mu} ,quad B_{21}^0 = - B_{12}^0= frac{epsilon_0}{2} tag{5}label{5}$$ and $B_{2j}^{mu}= B_{j2}^{mu}=0$ for all other $(i=2,mu,j)$ different of the ones above.

(Explicit contribution to the Lagrangian: $$ -frac{1}{2mu}left( partial_y E_y, B_z - partial_y B_z, E_yright) +frac{1}{2mu}left( partial_z E_y, B_y - partial_z B_y, E_yright) +frac{epsilon_0}{2}left( partial_t E_y, E_x - partial_t E_x, E_yright) $$ )

Partial conclusion: at this point there seems to be no contradiction. For each fixed $mu, (B^{mu}_{ij})$ is a $6times 6$ antisymmetric matrix, so 5 independent coefficients and $20$ all in all. Imposing to recover (ref{2}-ref{3}) should naively reduce the possibilities to a vector space of dimension $20-4=16$ but the examples (ref{4}-ref{5}) show that one should probably think twice.