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Consider a system of two relativistic charged point particles 1 and 2 which interact through their electric and magnetic fields. The equation of motion for the first particle is then given by the Lorentz force.

$$\vec{F_1} = \dot{\vec{p_1}} = m_1 \dfrac{\mathrm{d}}{\mathrm{d} t} \left(\gamma\left(\dot{r_1}\right) \dot{\vec{r_1}}\right) = q_1 \left(\vec{E_2} \left(\vec{r_1}\right) + \dot{\vec{r_1}} \times \vec{B_2} \left(\vec{r_1}\right) \right) = q_1 \left(- \left(\vec{\nabla} \phi_2 \right) \left(\vec{r_1}\right) -\dot{\vec{A_2}} \left(\vec{r_1}\right) + \dot{\vec{r_1}} \times \left(\vec{\nabla} \times \vec{A_2}\right) \left(\vec{r_1}\right) \right)$$

$\phi_2$ and $\vec{A_2}$ are the retarded potentials of particle 2. This equation can also be derived by using the hamiltonian of a relativistic charged particle in an electromagnetic field as shown here https://en.wikipedia.org/wiki/Hamiltonian_mechanics#Relativistic_charged_particle_in_an_electromagnetic_field.

But actually this equation is not correct, because it does not take the radiation of the accelerated particle into account. Or more specific

$$\vec{F_1} \neq \dot{\vec{p_1}} = \vec{F_1} + \vec{F_{1,rad}}$$

as described here https://en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force.

Does that mean, that the hamiltonian of a relativistic charged particle in an electromagnetic field is actually not correct but only an approximation?

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    $\begingroup$ That's right - it is an approximation. Read Feynman lectures about electromagnetic mass. $\endgroup$ – Vladimir Kalitvianski Jun 28 '18 at 14:05
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Does that mean, that the hamiltonian of a relativistic charged particle in an electromagnetic field is actually not correct but only an approximation?

If the charged particle's charge is distributed in a space of non-zero dimensions, then yes, the Hamiltonian is only approximate because it does not take into account the mutual self-forces inside the particle (EM forces between different parts of the particle).

If the charged particle's charge is concentrated at a point, there is no reason to introduce Lorentz-Abraham self-forces, the model is naturally formulated only with interaction forces between different particles. Then, the equation of motion

$$ m_1\frac{d (\gamma_1\mathbf v_1)}{dt} = q\mathbf E(\mathbf r_1,t) + q\mathbf v_1\times\mathbf B(\mathbf r_1,t) $$

where $\mathbf E(\mathbf x,t),\mathbf B(\mathbf x,t)$ are total external fields to the particle 1, is exact. This model is not a limit of the above model where particle size goes to zero. It is a different model, free of self-interaction and infinities. Variants of this model were investigated in the past by Tetrode, Fokker, Frenkel, Wheeler and Feynman and others; see the paper

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

In English, this article also explains it in short:

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

If the external field is only due to one other particle 2, we can write this as

$$ m_1\frac{d (\gamma_1\mathbf v_1)}{dt} = q\mathbf E_2(\mathbf r_1,t) + q\mathbf v_1\times\mathbf B_2(\mathbf r_1,t) $$ and similar equation for the particle 2: $$ m_2\frac{d (\gamma_2\mathbf v_2)}{dt} = q\mathbf E_1(\mathbf r_2,t) + q\mathbf v_2\times\mathbf B_1(\mathbf r_2,t). $$

This system of equations (with retarded fields) was studied by Synge; he calculated numerically motion of point proton and electron, influenced by mutual EM interaction. Since there is no Lorentz-Abraham term, the system loses energy more slowly than the Larmor formula would suggest. (The Larmor formula is based on Poynting's formulae; neither are valid for point particles in this theory).

With protons and atoms it is known they are not point objects, so this equation of motion cannot be exact, only approximate. However, in macroscopic settings (accelerators, charged particles moving in Earth's magnetosphere) where this equation is usually used, the expected error due to neglecting the self-interaction is usually so tiny that it is safely ignored together with other small details, such as magnetic moment interacting with magnetic field gradient.

With electrons, there is a possibility these are in fact point particles (based on low-energy experiments, the size is thought to be less than 1e-18 m) so the above equation could be exact. In any case, there is no evidence that the equation is wrong in macroscopic settings; the radiation damping observed in accelerators can be explained as a result of mutual interaction of electrons in the bunch. (The Lorentz-Abraham force being observed for the whole bunch - extended particle with charged parts).

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  • $\begingroup$ If we consider a motion of a point-like charge in a unifrom magnetic field, then your model equation will provide a perpetulum mobile thing. $\endgroup$ – Vladimir Kalitvianski Jun 28 '18 at 14:01
  • $\begingroup$ @VladimirKalitvianski, Frenkelian theories (point particle theories with no self-interaction of single point particle) obey local conservation of energy. The energy is just defined differently - single particle in uniform magnetic field will circle indefinitely, it will have radiation field, but if the source of the magnetic field is fixed and no other particles get into picture, net energy radiated will be 0. $\endgroup$ – Ján Lalinský Jun 28 '18 at 14:42
  • $\begingroup$ ... if the source of the magnetic field is fixed and no other particles get into picture, net energy radiated will be 0. That's the problem! In order to not have radiation, one has to modify the Maxwell equations too, and certainly into a wrong direction. $\endgroup$ – Vladimir Kalitvianski Jun 28 '18 at 15:23
  • $\begingroup$ I don't follow what you mean. Charge orbiting in magnetic field will produce radiation according to Maxwell's equations. The good property of this theory is that the particle circles indefinitely and the radiation it produces is carrying 0 energy to infinity, in the special case of fixed uniform magnetic field. If the source of the magnetic field moves, well then some energy may escape to infinity and the particle may slow down, but that is OK too. There is no modification of Maxwell's equations and no violation of energy conservation, if that is your concern. So where is the problem? $\endgroup$ – Ján Lalinský Jun 28 '18 at 15:40
  • $\begingroup$ You state that the net radiated energy is zero. In other words, no radiation at all. Thus the variable EMF is purely reactive in this case. For that one has to modify the Maxwell equations since here we have a time-dependent dipole radiation. The best classical fit to this experimental situation ares equations or radiation friction force by Landau-Lifshitz, which are reduced to $\dot{\vec{F}}_{\rm{ext}}$, à la Lorentz radiative friction that gives the line width. $\endgroup$ – Vladimir Kalitvianski Jun 28 '18 at 15:55

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