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There is a Lagrangian for a particle of mass $m$ and charge $q$ $$\mathcal{L}_1 = \mathcal{L}_k(m, \vec{v}) - q\phi + q\vec{v}\cdot\vec{A}$$ where $\mathcal{L}_k(m, \vec{v})$ is either $\frac{1}{2}m\vec{v}\cdot\vec{v}$ for classical mechanics or $-mc^2\sqrt{1-\vec{v}\cdot\vec{v}/c^2}$ for special relativity. The Euler-Lagrange equations for $\mathcal{L}_1$ give the particle's equations of motion under the Lorentz force $$\frac{d\vec{p}_k}{dt}=q\vec{E}+q(\vec{v}\times\vec{B})$$ where $\vec{p}_k$ is either $m\vec{v}$ or $(1-\vec{v}\cdot\vec{v}/c^2)^{-1/2}m\vec{v}$ based on which $\mathcal{L}_k$ is used. Here $\vec{E}=-\nabla\phi-d\vec{A}/dt$ and $\vec{B}=\nabla\times\vec{A}$.

On the other hand, there is a field Lagrangian density $$\mathcal{L}_2 = \frac{\epsilon_0}{2}\vec{E}\cdot\vec{E} - \frac{1}{2\mu_0}\vec{B}\cdot\vec{B} - \rho\phi + \vec{j}\cdot\vec{A}$$ where this time $\rho$ is the charge density and $\vec{j}$ current density. The Euler-Lagrange equations for $\mathcal{L}_2$ give the field equations $$\nabla\times\vec{B}-\frac{1}{c^2}\frac{\partial\vec{E}}{\partial t} = \mu_0 \vec{j}\quad\quad\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$ with the other field equations $\nabla\times\vec{E}=-\partial\vec{B}/\partial t, \nabla\cdot\vec{B}=0$ following from $\vec{E},\vec{B}$ being derived from the potentials $\phi,\vec{A}$.

Is there a combined Lagrangian that gives both the Lorentz force and Maxwell equations via the Euler-Lagrange equations?

I think that for a single particle (or finitely many), it's possible to combine them by taking $\rho=q\delta^3(\vec{r}-\vec{\gamma}(t)), \vec{j}=\rho\vec{v}(t)$ where $\vec{\gamma}(t)$ describes the path of the particle and $\vec{v}=d\vec{\gamma}/dt$. Is this correct?

Can the Lagrangians be combined for a continuous density $\rho,\vec{j}$? The Lorentz force density is $\rho\vec{E}+\vec{j}\times\vec{B}$. There should be some mass density and velocity field for the kinetic term $\mathcal{L}_k$ and they have to be related to $\rho,\vec{j}$ somehow. Can this be done?

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    $\begingroup$ If you take the QED Lagrangian and ignore their quantum nature, you would get a set of classical fields with what you want. $\endgroup$ Dec 31, 2023 at 10:08
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    $\begingroup$ @naturallyInconsistent: I have no idea how to go from bispinor fermion field to anything classical, or why Dirac gamma matrices should play a role in the classical theory. Or where the particlar mass $m$ and charge $q$ of a single type of fermion fit in the classical theory unless they somehow translate to $\epsilon_0,\mu_0$. $\endgroup$
    – Chad K
    Dec 31, 2023 at 10:30

6 Answers 6

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Is there a combined Lagrangian that gives both the Lorentz force and Maxwell equations via the Euler-Lagrange equations?

There is the standard textbook Lagrangian for $N$ charged particles:

$$ L(A^\nu,\partial_\mu A^\nu,\{\mathbf r_a,\mathbf v_a\}_{a=1}^{N}) = \int_V -\frac{1}{4\mu_0}F^{\mu\nu}F_{\mu\nu} d^3 \mathbf x~~~+$$ $$- ~~\sum_a q_a \varphi(\mathbf r_a,t) + \sum_a q_a\mathbf v_a \cdot \mathbf A(\mathbf r_a,t) ~~~+ $$ $$ -~~~\sum_a \sqrt{1-v_a^2/c^2}m_a c^2, $$ where $\varphi,\mathbf A, F$ refer to total EM field. Thus this is a function of $N$ positions, $N$ velocities, and a functional of the fields $\varphi,\mathbf A$.

This Lagrangian is in textbooks used to "derive" (e.g. in Landau&Lifshitz) both the Maxwell equations for total fields in presence of the current density $\sum_a q_a\mathbf v_a \delta(\mathbf x - \mathbf r_a)$ and charge density $\sum_a q_a\delta(\mathbf x - \mathbf r_a)$, and also to "derive" the equations of motion for all the particles, with each particle experiencing the Lorentz force $q_a\mathbf E(\mathbf r_a,t) + q_a\mathbf v_a\times \mathbf B(\mathbf r_a,t)$.

The problem with this is that when $\varphi,\mathbf A$ refer to total fields (due to all particles), expressions such as $\varphi(\mathbf r_a,t)$ and $\mathbf v_a \cdot \mathbf A(\mathbf r_a,t)$ are undefined, because the fields $\varphi,\mathbf A$ are singular at positions of the particles. The equations of motion appear to be correct, but are not, because they refer to undefined forces $q_a\mathbf E(\mathbf r_a,t) + q_a\mathbf v_a\times \mathbf B(\mathbf r_a,t)$, because fields $\mathbf E,\mathbf B$ too are undefined at the positions of the particles.

One could try to regard the field functions $\varphi,\mathbf A,\mathbf E,\mathbf B$ as "doctored", to have just the right values at positions of the particles to get the actual forces acting on the particles. But even if this was mathematically possible (the functions are necessarily discontinuous), the "doctored" values cannot be found from the Lagrangian function, the resulting equations of motion and initial conditions; thus the Lagrangian is not predictive of the resulting motion. Without knowing the actual motion of the particles, the completion values can be anything and produce arbitrary forces.

The Lagrangian function above is mathematically valid and predictive when $\varphi, \mathbf A, F$ refer to external fields not due to particles themselves, e.g. a free EM field, or a field of distant sources not considered part of the system. But then such Lagrangian can be physically valid only approximately, because it ignores interaction between the particles in the system. This is fine if the particles are far from each other and interact weakly, so the dominant force is the one due to the external field, but is not if there are strongly interacting particles.

If we want to have a valid Lagrangian (for a system of point particles) which takes into account mutual interactions of the particles, we have to formulate it in a way which makes all the quantities in it defined and which produces equations of motions with all values of forces there defined.

This is the case for the Frenkel theory, where 1) all particles produce their own EM field obeying the Maxwell equations with point sources; this field is singular at the particle producing it, and thus different from any other EM field produced by the other particles; 2) every particle experiences Lorentz force due to all EM fields except the one it itself produces; no self-interaction of particles is allowed.

Thus we have $N$ potentials $\varphi_b$, and $N$ vector potentials $\mathbf A_b$. The Lagrangian respecting 1), 2) is

$$ L(\{A_b^\nu,\partial_\mu A_b^\nu\}_{b=1}^{N},\{\mathbf r_a,\mathbf v_a\}_{a=1}^{N}) ~~~= $$ $$ =~~~ \int_V -\sum_{a,b}'\frac{1}{2\mu_0}F_a^{\mu\nu}F_{b,\mu\nu} ~d^3 \mathbf x~~~+$$ $$- ~~\sum_{a,b}' q_a \varphi_b(\mathbf r_a,t) + \sum_{a,b}' q_a\mathbf v_a \cdot \mathbf A_b(\mathbf r_a,t) ~~~+ $$ $$ -~~~\sum_a \sqrt{1-v_a^2/c^2}~m_a c^2. $$ The summation is over all pairs of indices $a,b$, where each pair is present only once; the prime symbol next to the summation symbol means the case $a=b$ is to be omitted.

The Euler-Lagrange equations produce the expected equations of motion, free of self-interaction:

$$ \frac{d}{dt}\bigg(\gamma_a m_a \mathbf v_a\bigg) = q_a \sum_{b}' \mathbf E_b(\mathbf r_a,t) + q_a\mathbf v_a \times \sum_{b}' \mathbf B_b(\mathbf r_a,t). $$ where $\mathbf E_b:=-\nabla\varphi_b -\partial_t \mathbf A_b,\mathbf B_b:=\nabla\times\mathbf A_b$, and also produces the time-dependent Maxwell equations for every pair $\mathbf E_b,\mathbf B_b$. The other Maxwell equations without time (the Gauss law for electric field, and the Gauss law of magnetism) are satisfied too, but they don't come out as consequence of the Euler-Lagrange equations, but because all $q_a$ are assumed to be constant in time, so four-current is locally conserved, and because of the definition $\mathbf B_b :=\nabla \times \mathbf A_b$.

See also the short paper

R. C. Stabler, A Possible Modification of Classical Electrodynamics, Physics Letters, 8, 3, (1964), p. 185-187. http://dx.doi.org/10.1016/S0031-9163(64)91989-4

and the original paper by Frenkel

J. Frenkel, Zur Elektrodynamik punktförmiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534. http://dx.doi.org/10.1007/BF01331692

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  • $\begingroup$ Is the summation being taken twice over each pair the reason the $\frac{1}{4\mu_0}$ changed to $\frac{1}{2\mu_0}$? $\endgroup$
    – Chad K
    Dec 31, 2023 at 16:49
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    $\begingroup$ Excellent post! The only time one would ever need such a Lagrangian as the OP desires is in this very case where there are N interacting particles. Lagrangians must correspond to actual physical problems; they are not magic tricks that produce a list of useful formulae. $\endgroup$ Dec 31, 2023 at 16:49
  • $\begingroup$ Is your lagrangian for N interacting particles, or is it one for N-1 interacting particles in an external field? $\endgroup$ Dec 31, 2023 at 16:50
  • $\begingroup$ @AlbertusMagnus: I think $\phi_b, \vec{A}_b$ are the fields generated by each of the $N$ particles, so this doesn't include an external field. $\endgroup$
    – Chad K
    Dec 31, 2023 at 16:55
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    $\begingroup$ @ChadK The actual reason for the divisor in front of the field Lagrangian term, in general, is to get the correct Maxwell equations from the Euler-Lagrange rules. In the usual field Lagrangian term, the divisor has to be 4 to accomplish that. Here, because the sum over the field pairs has each pair only once, the divisor needed is half of that, that is, 2. $\endgroup$ Dec 31, 2023 at 18:17
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In Proc. Roy. Soc. London A 209, 291 (1951), Dirac considers the following conditions of stationary action for the free electromagnetic field Lagrangian subject to the constraint $A_\mu A^\mu=k^2$: $$\Box A_\mu-A^\nu_{,\nu\mu}=\lambda A_\mu, (1)$$ where $A^\mu$ is the potential of the electromagnetic field, and $\lambda$ is a Lagrange multiplier. The constraint represents a nonlinear gauge condition. One can assume that the conserved current in the right-hand side of Eq. (1) is created by particles of mass $m$, charge $e$, and momentum (not generalized momentum!) $p^\mu=\zeta A^\mu$, where $\zeta$ is a constant. If these particles move in accordance with the Lorentz equations $$\frac{dp^\mu}{d\tau}=\frac{e}{m}F^{\mu\nu}p_\nu, (2)$$ where $F^{\mu\nu}=A^{\nu,\mu}-A^{\mu,\nu}$ is the electromagnetic field, and $\tau$ is the proper time of the particle ($(d\tau)^2=dx^\mu dx_\mu$), then $$\frac{dp^\mu}{d\tau}=p^{\mu,\nu}\frac{dx_\nu}{d\tau}=\frac{1}{m}p_\nu p^{\mu,\nu}=\frac{\zeta^2}{m}A_\nu A^{\mu,\nu}. (3)$$ Due to the constraint, $A_\nu A^{\nu,\mu}=0$, so $$A_\nu A^{\mu,\nu}=-A_\nu F^{\mu\nu}=-\frac{1}{\zeta}F^{\mu\nu}p_\nu. (4)$$ Therefore, Eqs.(2,3,4) are consistent if $\zeta=-e$, and then $p_\mu p^\mu=m^2$ implies $k^2=\frac{m^2}{e^2}$ (so far the discussion is limited to the case $-e A^0=p^0>0$).

Thus, Eq. (1) with the gauge condition $$A_\mu A^\mu=\frac{m^2}{e^2} (5)$$ describes both independent dynamics of electromagnetic field and consistent motion of charged particles in accordance with the Lorentz equations. The words "independent dynamics" mean the following: if values of the spatial components $A^i$ of the potential ($i=1,2,3$) and their first derivatives with respect to $x^0$, $\dot{A}^i$, are known in the entire space at some moment in time ($x^0=const$), then $A^0$, $\dot{A}^0$ may be eliminated using Eq. (5), $\lambda$ may be eliminated using Eq. (1) for $\mu=0$ (the equation does not contain second derivatives with respect to $x^0$ for $\mu=0$), and the second derivatives with respect to $x^0$, $\ddot{A}^i$, may be determined from Eq. (1) for $\mu=1,2,3$.

The above is just my paraphrase of the Dirac's results. Please look at the freely accessible original article. In Dirac's words: "...the new theory requires that infinitesimal charges in a given field shall move in a way which agrees with Lorentz’s equations for electrons moving in this field, with neglect of the influence of the field produced by the electrons."

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Suppose instead of just one particle we have several particles (with mass $m_i$, charge $q_i$, at position $\mathbf{r}_i(t)$) .

Then the complete Lagrangian looks like this: $$\begin{align}\mathcal{L}&= \sum_i \left( \frac{1}{2}m_i\dot{\mathbf{r}}_i(t)^2 -q_i \Phi(\mathbf{r}_i(t),t) +q_i\dot{\mathbf{r}}_i(t)\cdot\mathbf{A}(\mathbf{r}_i(t),t) \right) \\ &+\int d^3r \left(\frac{\epsilon_0}{2}\mathbf{E}(\mathbf{r},t)^2 -\frac{1}{2\mu_0}\mathbf{B}(\mathbf{r},t)^2 \right) \end{align}\tag{1}$$

We can write the charge density and current density field made up by the particles as $$\begin{align} \rho(\mathbf{r},t)&=\sum_i q_i\ \delta^3(\mathbf{r}-\mathbf{r}_i(t)) \\ \mathbf{j}(\mathbf{r},t)&=\sum_i q_i \dot{\mathbf{r}}_i(t)\ \delta^3(\mathbf{r}-\mathbf{r}_i(t)) \end{align} \tag{2}$$

Using these densities we can rewrite the Lagrangian from (1) as $$\begin{align}\mathcal{L}&= \sum_i \left( \frac{1}{2}m_i\dot{\mathbf{r}}_i(t)^2\right) \\ &+\int d^3r \underbrace{\left( -\rho(\mathbf{r},t)\Phi(\mathbf{r},t) +\mathbf{j}(\mathbf{r},t)\cdot\mathbf{A}(\mathbf{r},t) +\frac{\epsilon_0}{2}\mathbf{E}(\mathbf{r},t)^2 -\frac{1}{2\mu_0}\mathbf{B}(\mathbf{r},t)^2 \right)}_{\mathcal{L}_2} \end{align}\tag{3}$$

From Lagrangian (1) or (3) we get the equations of motion.

From Lagrangian (1) the Euler-Lagrange equations with respect to the particles' motion $\mathbf{r}_i(t)$ give the Lorentz force: $$m_i\ddot{\mathbf{r}}_i=q_i(\mathbf{E}+\dot{\mathbf{r}}_i\times\mathbf{B}) \tag{4}$$

From Lagrangian (3) the Euler-Lagrange equations with respect to the field $\Phi(\mathbf{r},t)$ give Gauss' law $$\epsilon_0\nabla\cdot\mathbf{E}=\rho \tag{5}$$ and with respect to the field $\mathbf{A}(\mathbf{r},t)$ give the Ampere-Maxwell law $$\frac{1}{\mu_0}\nabla\times\mathbf{B}+\epsilon_0\frac{\partial\mathbf{E}}{\partial t}=\mathbf{j} \tag{6}$$

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  • $\begingroup$ This is the usual textbook exposition using only single total field $\phi,\mathbf A$, but it is mathematically invalid, because $\mathbf j \cdot \mathbf A$ and its integral are not defined at points where the point particles are, and $\mathbf E^2 - \mathbf B^2$ is too singular at the particles and can't be integrated at those points as well. Blind application of the usual manipulations to this invalid Lagrangian produces equations of motion with undefined forces, and leads to infinite Poynting energy. $\endgroup$ Dec 31, 2023 at 13:46
  • $\begingroup$ See my answer for a possible mathematically valid Lagrangian. $\endgroup$ Dec 31, 2023 at 15:41
  • $\begingroup$ @JánLalinský Well, you are right. The Lagrangian given here would be fine, if $\rho$ and $\mathbf{J}$ were smooth functions of $\mathbf{r}$, but unfortunately they are not. So the fact that the Lagrangian nevertheless yields the "correct" equations of motions seems like a lucky accident (which also can be found in many EM text-books). $\endgroup$ Dec 31, 2023 at 16:48
  • $\begingroup$ @JánLalinský: An integral may be perfectly valid even if the integrand is undefined at a countable number of points, cf. the classical field theory, the gravitational potential of a finite body etc. $\endgroup$
    – auxsvr
    Jan 1 at 17:15
  • $\begingroup$ @auxsvr that is not the case here. Integral of $\rho \varphi$ or $\mathbf j \cdot \mathbf A$ is not defined when the point particle is in the integration domain. $\endgroup$ Jan 1 at 17:37
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Yes, a (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

$$ S[A^{\mu}; {\bf r}_i] ~=~\int \! dt ~L,\tag{1} $$

where the Lagrangian is

$$ L ~=~ -\frac{1}{4\mu_0}\int \! d^3r ~F_{\mu\nu}F^{\mu\nu} -\sum_{i=1}^n \left(\frac{m_{0i}c^2}{\gamma({\bf v}_i)} +q_i\{\phi({\bf r}_i) - {\bf v}_i\cdot {\bf A}({\bf r}_i)\} \right). \tag{2}$$

The corresponding Euler-Lagrange (EL) eqs. yields the correct EOMs:

  • $4$ Maxwell eqs. with sources (when varying $A_{\mu})$, and
  • $n$ (special relativistic) Newton's 2nd laws with Lorentz forces (when varying ${\bf r}_i)$.

NB: In the derivation of EL equations, it is implicitly assumed that the field configurations are sufficiently many times differentiable. However, if we try to put the EM field $A_{\mu}$ on-shell (wrt. EOMs), they contain singularities at the positions of the point charges. In particular, the action becomes mathematically ill-defined on-shell, and hence some regularization of self-energies is strictly speaking needed, as pointed out in Jan Lalinsky's answer.

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  • $\begingroup$ This does not work for point particles, see my comment to Thomas Fritsch's answer and also my answer. $\endgroup$ Dec 31, 2023 at 15:31
  • $\begingroup$ I updated the answer accordingly. $\endgroup$
    – Qmechanic
    Dec 31, 2023 at 16:29
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An easy way, classically relativistic: Gauge theories. The local 4d matter Lagrangian density

$$L = \sum_{ik} g(x)_{ik} u^i u^k d^4x $$

Remove trivial rigid coupling of 4-velocities and 4-momentum by a locally varying translation in momentum space

$$L = \sum_{ik} g(x)_{ik} u^i (u + e A) ^k d^4x $$ yielding the canonical 4-momentum

$$p_k = \frac{\delta L}{\delta u^m} = g_{m k}(u^k + e A^k)$$

yielding the invariant 4d mass-Hamiltonian

$$H = g_{ik}(p-e A)^i (p -e A)^k$$

In modern differential geometry, the choice of the local and cotangent bases is free. The fundamental equations have to be formulated basis independent. Gauge theories decouple tangent vector and momentum duality $m u = p =dL/du$.

If $\nabla \times A=0$ the theory is a pure gauge, if not, then $ A$ is a dynamic field, carrying and transferring momentum itself. In order to maintain gauge invariance for such a case, the Lagrangian has to be supplemented by the free Lagrangian in the derivatives of $A$, that is

$$ L = \sum_{ik} g(x)_{ik} u^i (u + e A) ^k d^4x + c * (\partial_m A^k)( \partial_k A^m) $$

This ansatz, thought to describe movements of massive charges in quasistatic em- fields, immediately demands matter-charge- and field quantization, because the classical Maxwell-Lorentz-Einstein approach yields runaway solutions. The gravitational part in the metric tensor - again is trivial, only if the Riemann curvature of g is zero, that is, g is the metric tensor of any flat coordinate system.

Up to today QFT has not overcome the difficulties of quantizing g, too. Mainly, because an arbitrarily gauged metric $g$ kills translation and rotation invariance and the standard technique to construct Hilbert spaces as representation spaces of the generators of some simple groups.

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The standard lagrangian leads via the Noether theorem to the canonical energy-momentum tensor distribution. Its divergence is the Lorentz force. You can also use the symmetrised Belinfante-Rosen tensor with same result. See https://en.m.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

I prefer the Lagangian $$\frac{\epsilon_0}{2} \partial_\mu A_\nu \partial^\mu A^\nu - \sum_i \left( m_i\sqrt{1-v_i^2} - q_i u_{i\mu} u_i^\mu \right)\delta(\vec{x}-\vec{x}i) \,,$$ but convention prefers $$\frac{\epsilon_0}{4} F_{\mu\nu}F^{\mu\nu} etc. \,.$$

See my paper at https://arxiv.org/abs/physics/0106078.

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