3
$\begingroup$

I have a lot of trouble in attempting to understand several aspects of the relativistic explanation of the magnetic field produced by electric currents. To be more specific, my problem is not in the conceptual understanding of the relativistic origin of magnetism, nor in following the derivation of the strength of magnetic force acting on a moving charge next to an infinite, neutral and current carrying wire. My misunderstanding is more basic — suppose we have a stationary charge next to such a wire. The stationary charge views the distances between the moving electrons in the wire contracted by $1/\gamma = \sqrt{1-\beta^2}$, while the stationary positive ions remain in their natural density. Therefore, from the charge's point of view, the wire should be negatively charged.

Apparently, this alone implies that the charge should experience electric force (and this was my first indication that I have a confusion here, since the electric field outside the wire must be zero), but one should not forget that the electric field produced by a moving charge (here the moving particles are the electrons) doesn't have radial symmetry; it is rather given by the formula (which is a generalization of Coulomb's law to a moving charge): $$\vec{E}(r,\theta) = \frac{kq}{r^2}\frac{1-\beta^2}{\left(1-\beta^2 \sin^2\theta\right)^{3/2}}\hat{r},$$ so one should integrate over the entire infinite wire to get the radial field of the infinite wire (taking into account both the radial and angular dependences of the electric fields of the moving electrons). At first I thought that after making the integration, this paradox will be resolved, but the integral gave exactly the same paradoxical result — that the electric field experienced by the external charge is $E = \frac{\lambda}{2\pi\epsilon_0 r}(\gamma - 1)$, where $\lambda$ is the charge density of the ions (and electron) when there is no current in the wire (the electrons are put into rest).

Where is the mistake in my calculations? or had I calculated correctly but misinterpreted the results?

$\endgroup$
7
  • $\begingroup$ Thats wrong.The electric field outside the wire depends only on the amount of charge inside the wire. $\endgroup$
    – Miss Mulan
    Apr 15 at 20:34
  • $\begingroup$ @MissMulan - I know this must be wrong, but I just want to settle the "common sense" with the principles of special relativity. $\endgroup$
    – user2554
    Apr 15 at 20:37
  • $\begingroup$ Special Relativity doesnt explain electromagnetism because a magnetic field is as real as a electric field $\endgroup$
    – Miss Mulan
    Apr 15 at 20:40
  • $\begingroup$ How are you finding a force on a stationary charge outside a neutral wire?? In the frame where the charge is stationary, there is no net charge density on the wire, so no E-field, and the B-field is irrelevant because the charge has no velocity; thus there is no force. (A stationary charge "sees" the same thing we do; its rest frame is the same as our lab frame!) (If the outside charge were moving, moving to its rest frame would show a charge density on the wire... @MissMulan a Lorentz boost changes the charge/current density inside the wire.) $\endgroup$
    – HTNW
    Apr 15 at 20:52
  • $\begingroup$ The wire was neutral before the electrons began to move (it was neutral when there was no current). Now the electrons begin to move, so the stationary charge sees the distances between them contracted by $1/\gamma$. Do you agree about that? $\endgroup$
    – user2554
    Apr 15 at 21:04

1 Answer 1

2
$\begingroup$

an infinite, neutral and current carrying wire.

Note that you began the problem statement with the fact that the wire is uncharged. This fact cannot be explained by relativity. It is entirely due to your problem setup. Nor can this fact be contradicted by relativity. It is a given, an assumption of the problem.

It is certainly possible, instead, to specify that the wire is charged in the lab frame. That would also be perfectly acceptable and completely compatible with relativity, but it would be a physically different scenario.

For example, consider the following circuit where the red dot indicates the location of the test charge, and $C$ represents the self-capacitance of the wire. With this setup the wire is uncharged if $V=0$, and therefore there is no force on the test charge. But if $V$ is some positive value then the wire is positively charged and there is an upwards force on the test charge. By stating in the problem that the wire is neutral you are simply stating that $V=0$.

Circuit for question

Thus, once you have specified the scenario in one frame, then you can use relativity to determine how the situation works in another frame. But relativity alone cannot determine the situation in the initial frame. The situation in the initial frame is determined by the circuit. Relativity cannot explain why you chose $V=0$, but given that you chose $V=0$ relativity can explain the fields and forces in other frames.

The stationary charge views the distances between the moving electrons in the wire contracted by, while the stationary positive ions remain in their natural density.

Here is the heart of your mistake. The positive lattice charges have a “natural density”. You implicitly assume that the electrons also have a natural density and that their proper density is equal to the natural density of the positive lattice. This assumption is incorrect.

The free electrons do not have anything that gives them a natural density. Instead, they have whatever density is imposed upon them by the circuit. Since your circuit has $V=0$ the proper density is whatever density is required to make the wire uncharged in the lab frame, as given in your problem description.

Once you have determined the charge density in one frame, we can use relativity to determine the charge density in any other frame. In this problem we can obtain the charge and current density as follows:

Let $v=J_0/\rho_0$ be the velocity of the electron’s proper frame relative to the lab frame, where $J_0$ is the current density of the electrons in the wire in the lab frame (the current density of the protons is 0 in the lab frame), and $\rho_0$ is the natural charge density of the protons in the lab frame (the charge density of the electrons in the lab frame is $-\rho_0$, as given).

The proton four-current in the lab is $\mathbf{J_+}=(c \rho, \vec J) = (c \rho_0,0,0,0)$. The electron four-current in the lab is $\mathbf{J_-}=(-c \rho_0,J_0,0,0)$. The total four-current in the lab is $\mathbf{J}=\mathbf{J_+}+\mathbf{J_-}=(0,J_0,0,0)$

Now, in the proper frame we have $\mathbf{J'_+}=\Lambda \mathbf{J_+} = (\gamma c \rho_0, \gamma J_0,0,0) $ and $\mathbf{J'_-}=\Lambda \mathbf{J_-} = (\gamma v J_0/c-\gamma c \rho_0, 0,0,0) $ and $\mathbf{J'}=\Lambda \mathbf{J} = \mathbf{J'_+} + \mathbf{J'_-} = (\gamma v J_0/c, \gamma J_0,0,0) $

So, in the proper frame there is both a charge density and a current density. More importantly for your question is the fact that the electrons have a smaller charge density in the proper frame than in the lab frame. This means that the proper spacing between electrons is larger than the natural proper spacing between protons. It is this larger proper spacing that is length contracted in order to produce the given neutral wire in the lab frame.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.