# How is energy conserved in this electromagnetic scattering?

I have been pondering over an energy conservation issue for a few weeks and would like to share it with the hope that its interesting and someone will be able to answer.

Suppose I have two positive charges at $(-a,0,0)$ and $(a,0,0)$ connected by a spring in such a way the the above positions are equilibrium positions. Now suppose I shoot a massless positively charged particle moving along the y axis in such a way that it passes the x-axis at time t=0.

According to Jackson problem 11.18 the electric and magnetic fields of the massless particle exist only in the x-z plane passing through the instantaneous location of the particle and are given by

$$\vec E = 2q \frac{ \vec r_\perp }{|r_\perp|^2} \delta(ct-z)$$

and

$$\vec B = 2q \frac{ \vec v \times \vec r_\perp }{|r_\perp|^2} \delta(ct-z)$$ where v is the 3-velocity of the particle so in our case $(0,0,1)$

Thus, while there is a force on the massless particle when approaching the x-z plane, there is no force by the massless charge on the massive charges connected by springs except for an impulse at $t=0$. This impulse gives an equal and opposite momentum to both the charged particle.

$$|p| = \frac{2q}{a}$$

When the massless charge moves away from the x-z plane the electric and magnetic fields it "feels" is equal and opposite to what it felt when approaching because any change produced by the massive charges moving will not reach the massless particle. Thus the energy in the massless particle is the same at very early and very late times.

We are left with the two massive charges oscillating. Where does the energy come from? It is tempting to say the energy comes from the field but how? At the end of the day, due to radiation losses the charged particles will go back to being at rest at $(-a,0,0)$ and $(a,0,0)$ and the only difference would be the field configuration produced by them. The massless particle seems to have just come along for a ride.

Is there a true degeneracy in that there are two field configurations with charges at $(-a,0,0)$ and $(a,0,0)$ with the exact same energy?

• It appears that you want the massless particle moving at the speed of light, which you should make more explicit. Then you should question why you are applying classical physics to a relativistic problem. – Jon Custer Feb 16 '16 at 13:51
• One wonders why nature hasn't made any charged massless particles, does one not? Well, probably not so much. – CuriousOne Feb 16 '16 at 13:56
• @Jon Custer. Well a massless particle by definition moves at speed of light so I didn't say it explicitly. I am not sure what you mean by the latter? I am using classical relativity. The electro-magnetic fields I wrote satisfy the covariant Maxwell's equations. Check Jackson problem 11.18. Infant chapter is about classical relativity. – Borun Chowdhury Feb 16 '16 at 13:56
• @CuriousOne I am not sure the masslessness is the key point here. What masslessness does is make the problem clean. – Borun Chowdhury Feb 16 '16 at 14:03
• I would also note that Jackson has different editions - 11.18 in my edition (second) is about the decay of a particle of mass M. The 'Transformation of Electromagnetic Fields' is covered in section 11.10, where a massless particle at c is not covered. – Jon Custer Feb 16 '16 at 14:59

Jon has a point when he says that you shouldn't be treating the motion of the massless particle as a classical (i.e. non-relativistic, non-quantistic) one, except that he left out the reason. I'll try to explain why it is so in what follows.

Of course, massless particles do not obey Newton's classical second law,

$$\vec{F}=m\vec{a}$$ not even in the relativistic approximation, as $m=0$ in this case and there is really no non-relativistic approximation for massless particles. In passing to the relativistic (exact) formulation of the motion of a massless particle, one should derive the law from a variational principle. I hope you have at least heard of such a principle. It says that a point-like particle moves in orbits that minimize the value of some integral taken along the orbit. For example, in the classical case, the integral $\mathcal{S}$, called the Action Integral, is $$\mathcal{S}=\int_{t_{1}}^{t_{2}} dt\ \ \mathscr{L}(\vec{r}(t),\dot{\vec{r}}(t))=\int_{t_{1}}^{t_{2}} dt\ \ \left[\frac{m\,\dot{r}^{\,2}(t)}{2}-V(\vec{r}(t))\right]$$ where $V(\vec{r})$ is the potential energy of the particle. Applying to the "Lagrangian" $\mathscr{L}$ what are called the "Euler-Lagrange equation", $$\frac{d}{dt}\frac{\partial\mathscr{L}}{\partial\dot{\vec{r}}}=\frac{\partial\mathscr{L}}{\partial\vec{r}}$$ you recover the Newton equation $$m\ddot{\vec{r}}=-\vec{\nabla} V$$ As for massive relativistic particles in an electromagnetic field, the action is (it may be missing some factors of $c$)

$$\mathcal{S}=\int_{a}^{b}\ ds\ \left\{-m\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}+eA_{\mu}\frac{dx^{\mu}}{ds}\right\}$$ where $s$ is an arbitrary parameter (you can't use time as a parameter in relativity. This action is valid even in curved spacetimes, by using a metric $g_{\mu\nu}$ different from the Minkowski metric). Applying the Euler-Lagrange equations, you get the relativistic equation for a massive particle, namely $$\frac{d}{dt}\ \frac{m\vec{v}}{\sqrt{1-v^{2}/c^{2}}}=e(\vec{E}+\vec{v}\times\vec{B})$$ (the derivation is somewhat complicated by the fact that you're not using the time $t=x^{0}/c$ as an independent variable, I'd rather not write it all here, so just trust me). This action doesn't work for massless particles as much as the classical one doesn't, too. In fact, should you set $m=0$, the kinetic term of the Lagrangian would cancel out, and this is not allowed. How should one then describe the motion of a massless particle in special relativity? One simply doesn't. Special relativity is written precisely in order to prevent particles from reaching the speed of light. Notice that photons are not massless particles, at least not in the classical sense. That of photon is a quantum concept. Photons are quantum massless particles, and no quantum particle has a definite position (at least in the sense in which a single photon is intended to be a particle), hence no definite trajectory in physical space. There is one way out to derive massless trajectories in classical (relativistic) physics, which reproduces what you tried to do. Write the action as follows: $$\mathcal{S}=\int_{a}^{b}\ ds\ \left\{-\mu\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}+eA_{\mu}\frac{dx^{\mu}}{ds}\right\}$$ and use the Euler-Lagrange equation $$\frac{d}{ds}\frac{\partial\mathscr{L}}{\partial\dot{\mu}}=\frac{\partial\mathscr{L}}{\partial\mu}$$ You simply get $$\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}=0$$ i.e. $$\left(\frac{dx^{0}}{ds}\right)^{2}-\sum_{i=1}^{3}\left(\frac{dx^{i}}{ds}\right)^{2}=0$$ or, dividing by $\left(dx^{0}/ds\right)^{2}$ $$|\vec{v}|=c$$ Notice that the electromagnetic field does not appear here. This is because the electromagnetic field cannot modify the motion of the massless particle. In terms of the particle's dynamics, the former action is completely equivalent to $$\mathcal{S}=\int_{a}^{b}\ ds\ \left\{-\mu\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}\right\}$$ which is the action of an uncharged massless particle, i.e. the particle is not charged. On the other hand, the second term couples to the electromagnetic field, i.e. the particle produces an electromagnetic field. So you get a particle which is both interacting and not interacting with the EM field: the interaction is not two-way as it should be - this violates Newton's third law, too; moreover, the massless particle (in this formulation) does not carry a kinetic energy, nor it has a well-defined momentum. It really is some strange kind of ghost particle, that violates nearly every physical law that we know. This is not something remarkable, it is just the sign that something is terribly wrong. On the other hand, quantum massless particles can be charged (even there, they come with their own problems, but these can be dealt with). How it is so would require a pretty long discussion. Let's just say that the mass value $m$ in the quantum (field) Lagrangian of a particle doesn't multiply the kinetic term; then it can be freely set to zero. This is why massless particles should not be dealt with in classical physics.

P.S.: As for Jackson's formula for the EM field produces by a massless particle, it should be right; however, you should interpret it as the unphysical limit of a perfectly reasonable (for $m>0$) formula.

• Thanks Giorgio. I am a professional theoretical physicist trained in GR and string theory so rest assured all the details were not important. Now why did you take a derivative wrt $\dot \mu$ when talking about massless particles? $\mu$ is not a dynamical variable. There should be a relativistic generalisation of Lorentz force that talks about change in momentum directly instead of acceleration. Also I don't agree you cannot talk about massless particles in relativity. You cannot accelerate a massive one to speed of light but of course you can talk about massless particles. – Borun Chowdhury Feb 17 '16 at 10:09
• Intuitively the energy (and hence momentum) of massless particles should be changeable. If it moves a distance d in a constant electric field its energy should change by qEd. Finally, Jackson's answer may have been derived as a limit (don't know why you call it unphysical) but you can verify it satisfies Maxwell's equations for four-current $J^\mu = q c v^\alpha \delta^{(2)} (r_\perp) \delta(ct-z)$. This is part (b) of the problem. So it doesn't matter how you get there, it is a solution to the equations of motion of the EM fields. – Borun Chowdhury Feb 17 '16 at 10:09
• In fact if you look at Kiritsis's string theory book chapter 2 equation 2.1.8 onwards (this just concerns particles so don't let it bother if you don't know string theory) you will find an action that makes sense in the massless limit. You just add a piece $\int dx^\mu A_\mu$ to it for minimal coupling and you can then derive the equivalent of Lorentz force for massless particles $\partial_\tau ( \frac{\dot x^\mu}{e(\tau)} ) = F^{\mu \nu} \dot x_\nu$ where $\tau$ is now an affine parameter and $e(\tau)^{-1}$ has the meaning of inverse energy. This is exactly as one would have expected. – Borun Chowdhury Feb 17 '16 at 10:33
• In the highly symmetric case I am considering, if I write the four momentum as $p^\mu=p(1,0,0,1)$ this simply gives $\frac{dp}{dt} = q E$ as one would expect. – Borun Chowdhury Feb 17 '16 at 10:35
• Great! It wasn't clear from the question, now I guess we'll be able to talk freely. I'm just a student, but I'll see if I can help all the same. Let me first of all have a look at Kiritsis book and I'll get back to you – Giorgio Comitini Feb 17 '16 at 11:37