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?

  • $\begingroup$ 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 squar $\endgroup$
    – TMS
    Feb 4, 2013 at 17:33
  • $\begingroup$ Going through the answers, Occam's razor tells me this is not the way to go. $\endgroup$
    – my2cts
    Feb 28, 2021 at 10:47
  • $\begingroup$ Related: physics.stackexchange.com/a/469672/226902 $\endgroup$
    – Quillo
    Apr 2, 2021 at 13:49

4 Answers 4


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


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 ${\bf r}_i)$.

  • 1
    $\begingroup$ Thank you @Qmechanic! Does it mean that even given the equations of motion (such as Maxwell equations for EM field dynamics and Lorentz force equations for charge dynamics) we cannot reproduce the Lagrangian from equations of motion? For instance, if we allow for non-zero magnetic charges, then we cannot use potentials, but we still have both Maxwell and Lorentz force equations valid (of course, adjusted to existence of magnetic charges). Can we formulate least action principle in this case? $\endgroup$ Feb 5, 2013 at 6:49
  • 1
    $\begingroup$ I updated the answer. The question concerning action principles for magnetic monopoles is a huge separate topic, and should be posted as a separate question. Besides OP's other Phys.SE question, a related topic of potentials in presence of magnetic monopoles has also been asked in this Phys.SE post and their links. $\endgroup$
    – Qmechanic
    Feb 5, 2013 at 14:48

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:

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


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

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?

  • 1
    $\begingroup$ The EM lagrangian is known to contain additional term $A^\mu J_\nu$. $\endgroup$
    – Muphrid
    Feb 4, 2013 at 18:12

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.

  • $\begingroup$ I'm pretty sure I read in Arxiv a paper where something like this was already done. Unfortunately, I forgot the name and I'm been unable to find it with a quick search. $\endgroup$
    – AndresB
    Sep 30, 2020 at 20:46
  • $\begingroup$ Ok, i'll look for it if I have time and replace this comment by a link to that article $\endgroup$
    – Noix07
    Oct 1, 2020 at 11:59

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.

  • 3
    $\begingroup$ How does this address "...but using [a] least action principle?" $\endgroup$
    – Muphrid
    Feb 4, 2013 at 15:24
  • $\begingroup$ Thanks, @RobertFilter. However I think that sigificance of Aharonov-Bohm effect is overstated. All speculations about the physical role of potentials are only valid if particles considered point-like. $\endgroup$ Feb 4, 2013 at 15:26
  • 1
    $\begingroup$ The coupling to matter explicitly involves the potentials. $\endgroup$ Feb 4, 2013 at 18:17
  • $\begingroup$ @JerrySchirmer This is true, if we write the interaction term in the Lagrangian as the usual $\partial_\mu A^\mu$. However, if you come back to Maxwells equations and Lorentz force, $\propto j_\nu F^{\mu \nu}$, the potentials are not included in any equations of motion in CED. Btw: Who made my comment an answer?! $\endgroup$ Feb 5, 2013 at 8:57
  • $\begingroup$ @RobertFilter: Maxwell's equations and the Lorentz force aren't part of the action principle. The lagrangian is $\frac{1}{4}F_{ab}F^{ab} + A_{\mu}j^{\mu}$. Most matter lagrangians will break the gauge invariance, so this is not trivial. $\endgroup$ Feb 10, 2013 at 2:36

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