# Electric field energy of two parallel moving charges at relativity speeds

Assume two particles of the same charge $q$ moving in parallel trajectories on the $x,y$ plane. The distance between them is $d$ and it's defined in the $y$ axis. The problem is to calculate the energy of the electric field when the speed of the charges is relative to the speed of light.

I would say that the energy of the electric field should be given by $$U_E = {\epsilon_0 \over 2} \int |E|^2 dV ~,$$ where $dV$ is for integration in volume $V$.

Also I would say that the field of the particles at time $t$ is defined by the position of the particles at a time $t'$ before $t$. The field is a long equation one can find if he uses the Liénard-Wiechert potentials.

And here I arrive to my problem: The expression of the electric field is too complicated and I have $E=E_1 +E_2$ by the two charges. So it seems that a direct calculation of the integral would be quite the situation.

So, the question is: Is there something I am missing that could simplify the calculations or is there a simple answer that I don't see? If no, and the answer must come from the above integral, is there some easy way to calculate (don't solve the integral, please. Just give a direction- mostly I ask if there is a physics admission that would simplify the problem).

Thank you.

Note: I will give here the expression for the fields:

$$\bar E(\bar r ,t ) = {q \over 4 \pi \epsilon_0} {i \over (\bar i \cdot \bar u)^3 } [(c^2 - v^2) \bar u + i \times (u \times a)]$$ and

$$\bar B( \bar r ,t )={1 \over c} i \times \bar E( \bar r, t)$$ where, $i$: the distance between the charge at $t'$ and a point calculating the field at $t$.

$v$: the speed of the particle

$\bar u =c \hat i - \bar v$

$a$: the acceleration, here zero (0)

EDIT

Note2 (6/25/2015): The solution given to this particular problem by my professor at the university goes as follows:

The electric energy is given by $U_E =\frac{1}{2} qV_1 + \frac{1}{2} q V_2$. Because the trajectories are parallel the potentials are equal to a specific point in space so: $U_E =qV$. Also, because the speeds are very high (relativistic) we take the Liénard-Wiechert electric potential. Thus the electric energy is: $U_E ={ q^2 \over e \pi \epsilon_0 d } {1 \over \sqrt{1-{u^2 \over c^2}} }$

I disagree with this approach. Because we have two charges moving, they emit both electric and magnetic fields. So the electric field as a function of the above-mentioned potential is given : $E= - \nabla V - {\partial \bar A \over \partial t}$, and the vector potential or its time derivative isn't zero. So we cannot just write that the energy of the electric field is equal to charge times the electric potential.

Is there something I haven't understood or is this approach truly mistaken?

• Constantine, I have to go, but for now: the electron doesn't have an electric field, it has an electromagnetic field. The E denotes the linear force between two electrons with no initial relative motion. Also note that motion is relative - think about what you'd see if you flashed past those two electrons at some relativistic speed. – John Duffield Jun 24 '15 at 13:14
• @JohnDuffield Thanks for the reply. I understand and note that part of my thinking at the beginning was like yours. But the problem here is that we wish to find only the energy of the electric field. You are saying that this task is impossible, or wrong? – Constantine Black Jun 24 '15 at 16:14
• I'd say it's confusing, because the electron's electromagnetic field is what it is. So the electron energy is 511keV. When the electron is attracted to the proton, there's a mass deficit of 13.6eV unequally shared between the electron and proton. When you push two electrons together, there's a mass "surplus" as it were. That relates to the energy you've added by exerting an increasing force for a given distance. – John Duffield Jun 24 '15 at 17:21
• See Jackson: "one should properly speak of the electromagnetic field F$_{\mu\nu}$ rather than E or B separately". Then see Wikipedia: "An electric field is a vector field that associates to each point in space the Coulomb force experienced by a test charge". The E isn't a field that stores energy. It denotes the linear force that results from the interaction of two F$_{\mu\nu}$ fields. – John Duffield Jun 25 '15 at 7:12
• @JohnDuffield I, again, agree with you. Let me make a suggestion: Could we say that the electic field energy is $U=qV$ ? I'd say no because here we have moving charges with relativistic speeds, so we have the Lenard-Wiearchat potentials, V and A. So we can't say that the electric energy is given by the above equation. May I ask your opinion on this. As for the references you are making in your comment, I would say this is my though also. But the problem is to find the electric energy. Thanks a lot. – Constantine Black Jun 25 '15 at 11:05

Constantine, take a look at what Minkowski said in Space and Time:

"In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy to the force-screw in mechanics; the analogy is, however, imperfect."

Note how he talked about THE field, singular. There's only one field, and it's the electromagnetic field $F_{\mu\nu}$. This field is associated with two forces, and to understand why, you should combine typical depictions of the electric field and the magnetic field to depict the electromagnetic field like this: Note how the resulting "spinor" is reminiscent of vector fields? Now imagine you have two of them, with no initial relative motion. How are they going to move? Remember this: co-rotating vortices repel, counter-rotating vortices attract. They're going to move like this, linearly apart: The linear "electric" force E between them is given by Coulomb's law, see Wikipedia where you can read this: "an electric field is a vector field that associates to each point in space the Coulomb force experienced by a test charge". But note that there is no actual electric field present, merely a linear force resulting from the interaction of two electromagnetic fields. These fields can also cause rotational force, because if you sling one spinor past the other, they don't just repel linearly, they go round one another too, something like this: Now you can also see rotational "magnetic" force B, and you can perhaps appreciate why we talk of rot which is short for rotor. But note that the spinors have to have some initial motion relative to each other for this, and not just linearly towards each other or away from each other. Hence your two electrons only move linearly apart. Their motion relative to you doesn't change their motion relative to each other. They still move linearly apart even though you're passing them at some relativistic speed. Or they're passing you at some relativistic speed. And there is no electric field, just electric force resulting from $F_{\mu\nu}$ field interactions, so the energy associated with this is the integral of the force x distance. Then you chuck in the Lorentz factor for the relativistic motion with respect to you. What your professor said sounds right to me.

• Thanks for the answer. I will only ask this- maybe I didn't understood where you mention it in your answer. How are the trajectories of the two charges parallel with constant velocity on the x axis? Again, thank you for all the effort to clear things up for me. – Constantine Black Jun 27 '15 at 13:04
• Because they're moving like this Constantine. – John Duffield Jun 27 '15 at 13:39

The result given to you by your professor is OK. Special relativity allows you to solve the problem in any reference system, and then going back to the original reference system.

So the easiest way to solve your problem is going to the reference frame where the two electrons are at rest. There you have only a electric field $$\vec{E}'(\vec{r}) = \frac{q}{4\pi\varepsilon_0 d_1^2}\hat{d}_1 + \frac{q}{4\pi\varepsilon_0 d_2^2}\hat{d}_2$$ where $\vec{d}_1$ and $\vec{d}_2$ are the vectors between the point $\vec{r}$ and the location of the charges.

And the electric energy is $$U'_E = \frac{\varepsilon_0}{2}\int |\vec{E}'(\vec{r})|^2 d^3V = \frac{1}{2}qV_1 + \frac{1}{2}qV_2$$ Now back to your original reference frame. Energy is the 0th component of a four-vector, and thus it transforms according to $$U_E = \gamma U'_E - \gamma \beta p'$$ where $\gamma$ is the Lorentz factor and $p'$ the momentum in the second reference frame (wich is, of course, zero).

As I said, this is the easiest way to prove it, but not the only one. You can also calculate the electromagnetic tensor in the second reference frame $$F'^{\mu\nu} = \begin{pmatrix}0 & -E'_x/c & -E'_y/c & -E'_z/c\\ E'_x/c & 0 & 0 &0 \\ E'_y/c & 0 & 0 & 0 \\ E'_z/c & 0 & 0 & 0 \end{pmatrix}$$and then perform a Lorentz transform to bring it back to the original reference frame $$F^{\rho\sigma} = \Lambda_\mu^\rho \Lambda_\nu^\sigma F'^{\mu\nu} = \begin{pmatrix}0 & -E_x/c & -E_y/c & -E_z/c\\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_z \\ E_z/c & -B_y & B_x & 0 \end{pmatrix}$$ If you calculate explicitly the Lorenz transform, you'll find the equations for the electric and magnetic fields that you posted in your question.

Now you can calculate the energy as $$U = \frac{\varepsilon_0}{2}\int |\vec{E}|^2 dV + \frac{1}{2\mu_0}\int |\vec{B}|^2 dV$$ as in the original reference frame you have also a magnetic field, you have to account its energy too. After some tedious calculation, with the second method you have to get the same results as with the first one.

• Thanks for the answer. I also thought of changing reference frame but, as the problem was stated, I wasn't sure if I should. Note that my professor, at his solution, doesn't specify in which reference frame he solved the problem. If he solved it in the reference frame of one charge then all is OK, I also agree with him. But in the RF as in the end of your answer, you must also consider the magnetic field. That means that we don't calculate the "electric energy field" but the energy- in this RF is electromagnetic. Thank you. – Constantine Black Jun 28 '15 at 8:37