Least-action classical electrodynamics without potentials

Question

Is it possible to formulate classical electrodynamics (in the sense of deriving Maxwell's equations) from a least-action principle, without the use of potentials? That is, is there a lagrangian which depends only on the electric and magnetic fields and which will have Maxwell's equations as its Euler-Lagrange equations?

Answer 1

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 $$\tag{1} {\rm eom}_i = 0, \qquad i\in\{1, \ldots, n\},$$ in a brute force manner with the help of Lagrange multipliers $$\tag{2}\lambda^i, \qquad i\in\{1, \ldots, n\},$$ so that the Lagrangian density simply reads $$\tag{3}{\cal L}~=~\sum_{i=1}^n\lambda^i ~{\rm eom}_i.$$

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

$$\tag{5} m\dot{\bf v}~=~ q{\bf E}? $$

The answer is well-known

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

with potential energy

$$\tag{7}V~=~ q\phi, $$

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

$$\tag{9}{\cal V}~=~ \rho\phi, $$

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

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

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

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

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 ${\bf r}_i)$.

Answer 2

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

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

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

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

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

$$\mathcal{L}=-\frac{1}{\mu_0}F^{\mu\nu}F_{\mu\nu} $$

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 $F_{\mu\nu}$ tensor, choosing more carefully the independent fields?

Answer 3

Why not try? for example with a Lagrangian of the specific form "polynomial of second order" $$\mathcal{L}(x^{\mu},X,\partial_{\mu}X):= \sum_{i,j=1}^6\left( A_{ij} X^i X ^j + \sum_{\mu=0}^3 B^{\mu}_{ij} \partial_{\mu}X^i X ^j +\sum_{\mu,\nu=0}^3 C^{\mu\nu}_{ij} \partial_{\mu} X^i \partial_{\nu} X^j \right) \tag{1}\label{1}$$ 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": $$ \left\lbrace \begin{aligned} \vec{\nabla}\cdot \vec{\mathbf{E}} &= 0 \\ \vec{\nabla}\wedge \frac{\vec{\mathbf{B}}}{\mu} &= \epsilon_0 \frac{\partial \vec{\mathbf{E}}}{\partial t} \end{aligned} \right. \tag{2}\label{2}$$ 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 ) $$ \left\lbrace \begin{aligned} \alpha(\partial_{\mu}X)&:= \sum_{i=1}^3\sum_{\mu=0}^3 \alpha^{\mu}_{i} \partial_{\mu}X^i = \sum_{i=1}^3\sum_{\mu=0}^3 \delta^{\mu}_{i} \partial_{\mu}X^i = \sum_{i=1}^3 \partial_i X^i \\ \beta(\partial_{\mu}X)&:= \sum_{i=1}^6\sum_{\mu=0}^3 \beta^{\mu}_{i} \partial_{\mu}X^i = \frac{1}{\mu} \left(\partial_y B_z - \partial_z B_y \right) - \epsilon_0 \frac{\partial E_x}{\partial t} = \frac{1}{\mu} \left(\partial_y X^6 - \partial_z X^5 \right) - \epsilon_0 \frac{\partial X^1}{\partial t}\\ \vdots\quad &= \quad \vdots \end{aligned} \right. \tag{3}\label{3} $$ 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 $$ \frac{\partial \mathcal{L}}{\partial X^i} = \sum_{j=1}^6\left(2 A_{ij} X ^j + \sum_{\mu=0}^3 B^{\mu}_{ji} \partial_{\mu}X^j \right) $$ and r.h.s. (with Einstein's summation convention) $$ \partial_{\mu}\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}X^i)} = \partial_{\mu} \left[\sum_{j=1}^6\left( \sum_{\mu=0}^3 B^{\mu}_{ij} X ^j + 2 \sum_{\mu,\nu=0}^3 C^{\mu\nu}_{ij} \partial_{\nu} X^j \right) \right] = \sum_{j=1}^6\left( \sum_{\mu=0}^3 B^{\mu}_{ij} \partial_{\mu}X^j + 2 \sum_{\mu,\nu=0}^3 C^{\mu\nu}_{ij} \partial_{\mu\nu} X^j \right) $$ One sees that in order to get (\ref{3}), the Lagrangian must not have terms in $A_{ij}$ and $C^{\mu\nu}_{ij}$. Then $$ \frac{\partial \mathcal{L}}{\partial X^i} - \partial_{\mu}\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}X^i)} = \sum_{j=1}^6 \sum_{\mu=0}^3\left( B^{\mu}_{ji} - B^{\mu}_{ij} \right)\partial_{\mu}X^j $$ (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 $i=4,\ 0 \leq \mu \leq 3$ $$ B^{\mu}_{4j}= - B^{\mu}_{j4}= \delta_j^{\mu}\enspace \text{if}\ 1\leq j \leq 3\quad \text{and}\quad B^{\mu}_{4j}= - B^{\mu}_{j4} = 0\enspace \text{if}\ 4\leq j \leq 6 \tag{4}\label{4}$$ (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 $$ B_{26}^2 = - B_{62}^2= \frac{-1}{2\mu}\ ,\quad B_{25}^3 = - B_{52}^3= \frac{1}{2\mu}\ ,\quad B_{21}^0 = - B_{12}^0= \frac{\epsilon_0}{2} \tag{5}\label{5}$$ 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: $$ -\frac{1}{2\mu}\left( \partial_y E_y\, B_z - \partial_y B_z\, E_y\right) +\frac{1}{2\mu}\left( \partial_z E_y\, B_y - \partial_z B_y\, E_y\right) +\frac{\epsilon_0}{2}\left( \partial_t E_y\, E_x - \partial_t E_x\, E_y\right) $$ )

Partial conclusion: at this point there seems to be no contradiction. For each fixed $\mu,\ (B^{\mu}_{ij})$ is a $6\times 6$ 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.

Answer 4

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.

Answer 5

We'll take, as the vacuum equations $$∇×𝐇 - ε_0\frac{∂𝐄}{∂t} = +𝐉, \hspace 1em ∇×𝐄 + μ_0\frac{∂𝐇}{∂t} = -𝐊,$$ coupled to external sources $𝐉$ and $𝐊$, respectively for electric and magnetic current densities. Adopting an approach somewhat analogous to Hertz, we'll leave the divergence laws aside; instead, just treating $ε_0∇·𝐄$ and $μ_0∇·𝐇$, respectively, as the definitions of the densities for electric charge $ρ$ and magnetic charge $σ$; so that the equations $$ρ ≡ ε_0∇·𝐄, \hspace 1em σ ≡ μ_0∇·𝐇,$$ become tautological. This provides a way to directly factor in the pre-conditions that the charges and currents satisfy their respective continuity equations: $$\frac{∂ρ}{∂t} + ∇·𝐉 = 0, \hspace 1em \frac{∂σ}{∂t} + ∇·𝐊 = 0.$$

The action $S = \int 𝔏 dt dx dy dz$ is formulated with a Lagrangian density $𝔏$ whose variational with respect to the fields - after integration by parts - shall be written as: $$\begin{align} Δ𝔏 &= λ ε_0Δ𝐄·\left(∇×𝐄 + μ_0\frac{∂𝐇}{∂t} + 𝐊\right) \\ &+ λ μ_0 Δ𝐇·\left(∇×𝐇 - ε_0\frac{∂𝐄}{∂t} - 𝐉\right) \\ &+ \frac{∂}{∂t}(⋯) + ∇·(⋯). \end{align}$$ The coefficient $λ$ must, then, have the dimension of length. This may be satisfied with the following Lagrangian density $$𝔏 = λ \left(\frac{ε_0𝐄·∇×𝐄 + μ_0𝐇·∇×𝐇}{2} + \frac{ε_0μ_0}{2}\left(𝐄·\frac{∂𝐇}{∂t} - 𝐇·\frac{∂𝐄}{∂t}\right) + ε_0𝐄·𝐊 - μ_0𝐇·𝐉\right).$$

Ignoring the contribution from the external sources ...

The Lagrangian density for the purely field part of the Lagrangian density is "on-shell" equal to 0. The situation is entirely analogous to that for the Dirac Lagrangian. By "entirely" I mean: that the conversion can be applied in the opposite direction to the Dirac field, pulling out the "fermionic potentials" analogous to the $𝐀$ and $φ$ potentials of electromagnetism, a "fermionic gauge transform" analogous to that for $(φ,𝐀)$, and a Lagrangian density that is quadratic in the gradients of the fields, just like the Maxwell-Lorentz or Yang-Mills Lagrangian densities are, and which is not on-shell equal to 0.

Edit:
If the charge densities are also inputs to the field law, then we can't treat them tautologically. Then, you'll just have to add them with Lagrange multipliers as constraints. Effectively, you're writing in Gaussian constraints by hand, and the respective continuity equations become pre-conditions for the sources. Denoting the previous Lagrangian density $𝔏$ by $𝔏_λ$, and introducing Lagrange multipliers $Λ_E$ and $Λ_H$, the new Lagrangian density, now denoted $𝔏$, would be modified to: $$𝔏 = 𝔏_λ + \frac{Λ_E (ρ - ε_0 ∇·𝐄)^2 + Λ_H (σ - μ_0 ∇·𝐇)^2}{2}.$$