What is the classical description of a system consisting of two point charges moving under the influence of the fields generated by their presence (no additional external fields are assumed)? In the lab frame, these point-charge particles will in general be moving, so both (relativistically transformed) electric and magnetic fields will be present. In addition, they will be accelerated by the forces caused by these fields, so will radiate, I guess.

Is there a complete description of the equations of motion of such a system? In particular, what would the Lagrangian and Hamiltonian for such a system be?

EDIT: The reason for asking the question is that I wondered why the Hamiltonian in the Schrödinger equation for the hydrogen atom is given by $H = \frac{\mathbf{p}_p^2}{2m_p} + \frac{\mathbf{p}_e^2}{2m_e} - \frac{e^2}{4 \pi \epsilon_0 \left|\mathbf{r}_e - \mathbf{r}_p\right|}$ where the subscripts denote the proton and electron respectively. To me, this seems to be the Hamiltonian for two particles moving only under the influence of their electrostatic field, where this electrostatic field takes the form that would be seen in the frame where one charge is stationary. I am aware that the Schrödinger equation is non-relativistic. However, I would have assumed that this Hamiltonian can be rigorously derived as the low-velocity limit of the fully-relativistic classical Hamiltonian describing two interacting charges. Also, it is not immediately obvious to me that the velocities of the two particles should be small in general. Is there any justification based on classical mechanics in assuming from the start that a non-relativistic Hamiltonian should be sufficient? Or is this just done because it turns out to be close enough to the true (Dirac equation) result?

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    $\begingroup$ Even a full classical description of the fields and dynamics of a single point charged object is actually tricky. Sooner or later you'll find yourself facing the issues of self-energy and requiring the need to renormalize. Some may claim that only QED can give a consistent description. Nevertheless, from the classical point of view there have been efforts to treat this problem. Take a look of this for instance, and references therein. The introduction does a nice job explaining some of the issues. $\endgroup$
    – secavara
    Jan 18, 2018 at 16:15
  • $\begingroup$ Since there has been an edit of the question I am commenting further. You can start from the Dirac equation and consider an expansion of the effective Hamiltonian in powers of $v/c$ ,with $v$ the velocity of the electron, and you can see explicitly how the dominant effect is the electrostatic potential, followed by other sub-leading contributions that include the effect of the magnetic field of the nucleus felt by the electron. Take a look of chapter 12 in Cohen-Tannoudji, Diu and Laloe's Quantum Mechanics book, volume 2. $\endgroup$
    – secavara
    Jan 20, 2018 at 20:40
  • $\begingroup$ I am aware of this expansion. However, most Hamiltonians in quantum mechanics are derived from the corresponding classical Hamiltonians, with suitable commutation relations then imposed on the canonical variables. I was therefore interested what the classical Hamiltonian for this system is, from which the quantum Hamiltonian can then be derived. I had assumed that even classically a low-velocity Hamiltonian can be derived from the general fully relativistic Hamiltonian. $\endgroup$
    – Quantum
    Jan 21, 2018 at 15:21
  • $\begingroup$ You could do that partly here. In a way this is the reverse of the process done there, in which they perform a "classical identification" of the term by term contribution. But notice that, following your path, you would immediately be missing fundamental quantum contributions coming, for instance, from the spin(s), which are even dominant in comparison. In fact, this is a great example of the limitations of the correspondence principle for operators that you mention. Beyond this, I don't really see what would satisfy you. $\endgroup$
    – secavara
    Jan 21, 2018 at 15:45
  • $\begingroup$ I guess there are quite a few other questions that are the reason why I posted this question: What is the justification for using the standard Schrödinger hydrogen Hamiltonian? Starting from the most general situation in classical physics (hence this question), can one rigorously show that the static-potential Hamiltonian is sufficient at low velocities? Is there a priori any reason why the two charges should have low velocities? Is the most general quantum Hamiltonian (QED) based on the most general classical Hamiltonian or is it derived in another way? $\endgroup$
    – Quantum
    Jan 22, 2018 at 22:51

3 Answers 3


Classical electromagnetism is not consistent with the existence of pointlike charges. The theory is inherently relativistic, and a classical pointlike particle has infinite energy in its static field, so by $E=mc^2$ it should have infinite inertia. An electron can't even have a size less than the "classical electron radius," since then its field would have an energy greater than the electron's mass.

If you insist on having pointlike charges, then you run into all kinds of problems. For example, you get preacceleration (charges accelerating before a force is applied) and pathological solutions involving exponential motion.

You can certainly model charges as finite sphere, but then you're not talking about pure classical electrodynamics. There is now some other force that is holding your spheres together against their own electrostatic repulsion.

Further reading

Brown, http://www.mathpages.com/home/kmath528/kmath528.htm

Poisson, http://arxiv.org/abs/gr-qc/9912045

Rohrlich: The dynamics of a charged sphere and the electron Am J Phys 65 (11) p. 1051 (1997), http://www.lepp.cornell.edu/~pt267/files/teaching/P121W2006/ChargedSphereElectron.pdf

Medina, Radiation reaction of a classical quasi-rigid extended particle, J.Phys. A39 (2006) 3801-3816, http://arxiv.org/abs/physics/0508031

Morette-DeWitt, "Falling Charges," Physics, 1,3-20 (1964), http://www.scribd.com/doc/100745033/Dewitt-1964

  • $\begingroup$ I have added an edit to the original question. If you could add an explanation addressing these points, I will accept this as the answer. $\endgroup$
    – Quantum
    Jan 20, 2018 at 20:14
  • $\begingroup$ Also, could you maybe comment on Jan Lalinsky's answer that point charges are compatible with classical electrodynamics? $\endgroup$
    – Quantum
    Jan 22, 2018 at 22:35

Is there a complete description of the equations of motion of such a system? In particular, what would the Lagrangian and Hamiltonian for such a system be?

It depends on whether the particles are assumed to be points, or extended charged bodies.

If they are extended charged bodies, there is, as far as I know, no knowledge of a unique model of the system. The problem is that due to relativity the charged body cannot be idealized as a rigid body, but it is a system with infinity of degrees of freedom, like ball made of jelly. Mathematically complete description would require a model of motion and mutual internal forces between the charged parts of the particle. We do not have any convincing model of this. There are some published works that work with a more simplistic model, where the charged body is very regular ellipsoid that undergoes little or no deformation (Lorentz, Abraham, more recently Yaghijan and Medina are often cited) and are able to derive some conclusions about it, but all these calculations are of approximate character.

If the charged particles are points, point particles have only a handful of degrees of freedom and can be described by single position and velocity vectors. The situation is much simpler and this makes this kind of model much more attractive. There were papers by Fokker and Tetrode at the beginning of 20th century that show how a particular model of interacting particles, fully relativistic and in agreement with Maxwell's equations, may be formulated. Their formulation was focused on eliminating the EM fields from the description and used a variational principle to obtain the equations of motion of the particles directly from the action, without any middleman in form of EM field. However, this approach restricts the solutions to a highly special solutions of the Maxwell equations - so called half-retarded half-advanced fields.

More general formulation that does not require such restriction on the fields was first published, I think, by J. Frenkel in his paper

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

For a shorter, more easy-to-read account, see also

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

It is true Frenkel too proposes half-retarded half-advanced solutions as particularly interesting since they allow for stable motion of hydrogen atom particles, but his formalism actually does not require them, it allows for any EM field that obeys Maxwell's equations.

The core idea is that particles act on other particles but never on themselves. The reason for this assumption for Frenkel was that self-action of a point on itself is contradictory and leads nowhere.

A particle acts on other particles via electromagnetic field of its own,so each field acquires an index that indicates which particle the field 'belongs to'. For example, particle $a$ generates electric field and its value at point $\mathbf r_b$ is $\mathbf E_a(\mathbf r_b)$. This is introduced so we can keep track of which field acts on which particle.

The fields obey the Maxwell equations with the owning particle as source:

$$ \nabla \cdot \mathbf E_a = \rho_a/\epsilon_0 $$

$$ \nabla \cdot \mathbf B_a = 0 $$

$$ \nabla \times \mathbf E_a = - \frac{\partial \mathbf B_a}{\partial t} $$

$$ \nabla \times \mathbf B_a = \mu_0 \mathbf j_a + \mu_0\epsilon_0 \frac{\partial \mathbf E_a}{\partial t} $$

Superposition of the elementary fields of all particles still obeys the Maxwell equations (thanks to their linearity), so this superposition is a good candidate for macroscopic total EM field.

The equation of motion of a charged particle $b$ is

$$ m_b \frac{d(\gamma_b \mathbf v_b)}{dt} = \sum_a' q_b \mathbf E_a(\mathbf r_b) + q_b\mathbf v_b \times \mathbf B_a(\mathbf r_b) $$

(the prime near the sum sign means in case $a$ = $b$ the term is to be omitted) This is a general formulation, fully relativistic and obeying both the Maxwell equations and the Lorentz force formula.

This direct formulation of equations of motion can be used to infer and check a variational Lagrangian formulation, where both field and particle variables are Lagrangian variables. The Lagrangian is

$$ L = \int d^3\mathbf x \mathcal{L} $$


$$ \mathcal{L} = \sum_a\sum_b' -\frac{1}{4} F_a^{\mu\nu}F_{b,\mu\nu} + \sum_a\sum_b' j_a^\mu A_{b,\mu} - \sum_a m_a c^2\sqrt{1-v_a^2/c^2} \delta(\mathbf x - \mathbf r_a) $$

Back to the case with two particles. One can make an approximation: the potentials are as if the particles were static or moving with speeds much lower than speed of light. These potentials can be inserted into the Lagrangian and then another Lagrangian, a function of particle positions and their derivatives, with no field variables, can be obtained.

In this way, the effect of fields is approximately expressed as a function of the particle variables. In the simplest case, this is the Coulombic term $\frac{q_aq_b}{4\pi\epsilon_0|\mathbf r_a - \mathbf r_b|}$. If the particle kinetic term of $L$ is linearized, one obtains non-relativistic Lagrangian of two particles interacting via static electric forces. For this approximate Lagrangian, one one can do the Legendre transformation and derive the common Hamiltonian function for the hydrogen atom. From the derivation it is clear that all this ignores magnetic interactions, retardation of interaction and is valid only for speeds much lower than the speed of light.

If better approximation is desired, one may insert potentials of particles moving rectilinearly with low speed and then the resulting Lagrangian function contains, in addition to the Coulombic term, a term that describes magnetic interaction. The term is called the Darwin interaction term and the whole Lagrangian the Darwin Lagrangian.

  • $\begingroup$ Could you maybe comment on how this description is compatible with the statement in Ben Crowell's answer above that classical electromagnetism is not compatible with the existence of point-like charges? $\endgroup$
    – Quantum
    Jan 21, 2018 at 15:18
  • $\begingroup$ What Ben Crowell's answer states is a common sentiment based on century of failure of all attempts to incorporate the effect of radiated Poynting energy and momentum on the motion of point particle (the infamous Lorentz - Abraham - Dirac equation and its derivatives). The usual point of view is that this means point particles are not consistent with the EM theory. But actually there is a consistent theory of charged point particles, obeying relativity and Maxwell's equations. In this theory, the Poynting expressions do not give energy so there is no inconsistency. $\endgroup$ Jan 21, 2018 at 20:48
  • $\begingroup$ In other words, classical equations of motion and Maxwell equations are consistent with point charges (one can find solutions to the equations). The reason why people think this is not possible is that they implicitly assume that the Poynting expressions for energy should be valid even for point particles, which then leads to contradiction (infinite energy of point charge, undefined transfer of energy from field to particle etc.). $\endgroup$ Jan 21, 2018 at 20:53
  • $\begingroup$ It will take me some time to read through the papers you linked to. Does the approximation that the potentials are treated as if the particles were static come out explicitly as the low-velocity limit? Or would the low-velocity limit still contain terms in the potentials that are non-static? In other words, is the assumption that the potentials are those of static particles a stronger approximation than the assumption of low velocities? $\endgroup$
    – Quantum
    Jan 22, 2018 at 22:33
  • $\begingroup$ The static approximation means that the potentials are functions of source particles positions only, they are not functions of particle velocities. However, since the particle positions change in time, these approximative potentials at any point of space also change in time. In this sense the potentials are non-static even if they are given by static formulae. The static approximation can be obtained mathematically as a limit of the Liénard-Wiechert solution when $v$ of the particle goes to zero. If we want to get more accurate expression for potentials,we must not perform the mentioned limit. $\endgroup$ Jan 23, 2018 at 22:25

In the non-relativistic limit, for the Lagrangian, $L=T-V$. Let $\vec{x_1}$ be the position of charge $q_1$ and $\vec{x_2}$ be the position of charge $q_2$. Let $\vec{A_1}(\vec{x},t)$ be the vector potential created by the motion of charge $q_1$ (which in general can be described by the Lienard-Wiechert potential). We know that





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    $\begingroup$ I should probably note that much of the difficulty of this problem is hidden in the Lienard-Wiechert potentials above. $\endgroup$ Jan 18, 2018 at 16:32
  • $\begingroup$ Shouldn't the coulomb potential have a $1/r$ rather than a $1/r^2$? $\endgroup$ Jan 18, 2018 at 17:34
  • $\begingroup$ Oops, yeah, should be fixed now. $\endgroup$ Jan 18, 2018 at 17:36

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