# What is the Lagrangian for a relativistic charge that includes the self-force?

The usual Lagrangian for a relativistically moving charge, as found in most text books, doesn't take into account the self force from it radiating EM energy. So what is the Lagrangian for a relativistic charge that includes the self-force?

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– Arun Nanduri Jul 6 '11 at 15:34
To properly address self-energy issues, one has to abandon a classical description and consider full-fledge quantum-electro-dynamics (QED) and renormalization. – Qmechanic♦ Jul 6 '11 at 15:50

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I wouldn't go as far as what Qmechanic says, but it is clear that one must abandon classical Maxwellian description of electromagnetism. It is actually quite an active research topic whether there can be non-linear theories of classical electrodynamics that can treat self-force. For an example, see M. Kiessling, J. Stat. Phys. (2004) vol 116 p1057. Kiessling has a second paper in the same issue of the journal starting on p1123 dealing with the quantum version of the procedure. – Willie Wong Jul 6 '11 at 16:38
@Willie is the problem that an acceleration is modelled as the cause of the proper self force? Usually, proper force is the cause of acceleration, and independent of the acceleration it causes. – Larry Harson Jul 7 '11 at 10:40

## 5 Answers

classical electrodynamics mainly deals with two kinds of proplems: a) The action of a field on a charged particle and b) the fields arising from the motion of such a field. Of course, this can only be approximative but it turns out that a lot of phenomena can be described in this way.

However, you are right, an entire treatment would include a) and b) simultaniously - including the whole dynamics of such a system with radiative reaction (or, the Abraham-Lorentz-Force and its relativistic counterpart).

But as Qmechanic pointed out in a comment, there may be no fully consistent way to do so within the framework of classical electrodynamics. Jackson (Chapter 16) states:

The difficulties presented by this problem touch one of the most fundamental aspects of physics, the nature of an elementary particle. Although partial solutions, workable within limiting areas, can be given, the basic problem remains unsolved.

Thus, you will have to search for a really satisfactory answer within the description of Quantum Electrodynamics. Otherwise, you may include the effect phenomenologically. This was done e.g. by Barone and Mendes in Lagrangian description of the radiation damping (PRA, 2007) and they give a Lagrangian of the form

$$\mathcal{L}=\frac{m}{2}\dot{\mathbf{r}}_1\cdot\dot{\mathbf{r}}_2 - \frac{\gamma}{2}\epsilon\left(\dot{\mathbf{r}}_1,\ddot{\mathbf{r}}_2\right) -V\left(\mathbf{r}_1,\mathbf{r}_2\right),$$

with $\gamma := 2e^2/3c^3$, with $\epsilon = \epsilon_{ij}dx^idx^j$ beeing the Levi-Cevita tensor, and the $\mathbf{r}_i$ arise from the special treatment of the problem used in the paper employing some kind of image phase-space representation of the system. Furthermore, $V\,$ is the potential related to the Abraham–Lorentz–Dirac-Force which is given explicitely in the paper.

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Solving both, Lorentz force and Maxwell's equation, does include radiation reaction. Radiation reaction emerges from the interaction of the charge with the field it has emitted itself, the self-field. If you do not want to solve Maxwell's equations and treat the fields as external, you can use Lorentz-Abraham-Dirac equation with some known inconsistencies for unphysical initial conditions, or the Landau-Lifshitz equation as approximation. Some authors claim that Landau-Lifshitz equation is not an approximation but agrees with solutions Lorentz-Abraham-Dirac equation for physical initial conditions. Lorentz-Abraham-Dirac and Landau-Lifshitz equations cannot be derived from Hamilton's principle because radiation emission is a dissipative process. For a pedagogical review see http://arxiv.org/abs/gr-qc/9912045.

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It is the usual CED Lagrangian for a charge and electromagnetic field; it includes the free field Lagrangian, a free particle Lagrangian and the interaction Lagrangian $jA$. Self-action is encoded in it when you consider the system as a whole, as coupled unknown variables. Of course, it yields rubbish: an infinite "electromagnetic mass" that you need to discard.

I say "rubbish" because, on one hand, we define the charge as a stable, fixed entity; on the other hand we introduce a self-force to get some dynamics due to it. Our definition contradicts our intentions.

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I would say that it is already in the Lagrangian, problem being: in the classical ED lagrangian you describe point-like particles, and the self interaction is classically described as the classical lorentz force a charge creates on itself if it is a sphere-like particle, considering retarded potentials. A classical solution would be to include the shape in the lagrangian, but the lagrangian is local, so it deals with points. If someone talks to you about QED, it is because classically you reach the conclusion that the reaction creates a mass like term opposing acceleration, and this mass can become infinite, in QED you also have some infinites, but you can remove them. Nonetheless I would say you remove it because what actually interests QED is the measure you make, and you never really reach the size of a point while measuring, but it might be an over interpretation.

BUT I have a good news for you there exists a classical formulation that includes the retroaction, to the expense of retrocausality, or waves flowing backward in time, or a kind of action at a distance: it is the lagrangian you can find in the Feynman-Wheeler absorber theory. very interesting but Feynman abandoned it.

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If the charge is point-like then its classical self-energy is infinite. As the self-force comes from the gradient of potentials, the result is undetermined.

Consider a theory with a local interaction. Then we must take into account some field, e.g. the electromagnetic field. Now for a massive relativistic particle moving at constant velocity there is always an inertial frame of reference where the particle is not moving. Therefore the magnetic field vanishes and the electric field is that given by Coulomb's law, i.e. $$\mathbf E(\mathbf r)=\frac q{4\pi\epsilon}\frac{\mathbf r}{\Vert \mathbf r\Vert^3}.$$ Since the particle stays at $\mathbf r=0$, the force is undefined.

So far the only massless particle known is the photon, which is neutral.

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-1 The OP question is about Lagrangian. – Vladimir Kalitvianski Jul 6 '11 at 22:55

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Potential energy enters into the Lagrangian... – Phoenix87 Jul 7 '11 at 8:25

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Yes, the potential energy is a part of the Lagrangian but it depends on not-yet-determined variables there. One does not use it for calculations because the Lagrangian serves first of all to get the equations. In particular, from CED Lagrangian and the Noether theorem it follows that the total energy is conserved, and this energy includes a self-action contribution. The force is a term from equations of motion, not from the Lagrangian. – Vladimir Kalitvianski Jul 7 '11 at 14:04
– Ben Crowell Aug 6 '11 at 17:46
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