Is there an answer to why a charge placed next to a current carrying conductor only experiences a force if it moves relative to the wire? [duplicate]

Question

The idea that special relativity provides the link between electrical force and magnetic force is very satisfactory. Unfortunately with only one exception have I found anything that explains why a test charge stationary in the laboratory frame next to a current carrying conductor experiences no force. This includes the excellent Purcell 3rd Ed.

It is not acceptable to simply quote the Lorenz equation without any explanation.

Because we are talking about motion we can use relativity. Let's have an observer move at the same velocity as the electrons in the conductor.

In the observers frame the positively charged nucleons will be moving in the opposite direction. Hence we have a current hence we have magnetic fields.

But the test charge is now moving through this magnetic field so must experience a force as a result of the Lorenz equation.

Bit of a contradiction?

What is wrong with this crazy view?

Back in the lab frame the electrons in the conductor are moving as a result of some applied voltage. So there is a moving negative charge from, say, left to right.

But as each electron moves it leaves behind a "hole" that will be filled when the next in line electron moves.

As we have negative charges moving from left to right we have the positive holes moving from right to left.

In the lab frame this means that everything cancels out provided the test charge isn't moving.

Once the test charge starts to move there will be differential velocities between the positive hole stream and the electron stream. Due to SR a net charge will result that will either attract or repel the moving test particle.

This is the only explanation that appears to tick all the boxes.

What's wrong with it please?

Answer 1

It isn't true that charged particle next to current-carrying conductor experiences no force. It experiences electric force due to electric field of the conductor. This field is present because 1) the conductor has surface charges that maintain electric field inside and outside the conductor 2) even if 1) is negligible, the particle next to the conductor induces modified charge distribution on the conductor's surface that creates attractive electric force on the particle (electrostatic induction).

When frame changes, total force may change as well. Total force is no longer purely electric, but is partly electric and partly magnetic. But it can't be zero in one frame and non-zero in another.

Answer 2

This is an observational fact. The theory was crafted to match this and other observations. The math doesn't really explain anything.

But I think the misconception in your model is "As we have negative charges moving from left to right we have the positive holes moving from right to left. In the lab frame this means that everything cancels out provided the test charge isn't moving."

Electrons (of negative charge) flowing from left to right make a current from right to left. A current of holes (of positive charge) flowing from right to left also make a current from right to left. There is no cancellation. But you shouldn't add them anyway: that is "double counting".

Answer 3

Unfortunately with only one exception have I found anything that explains why a test charge stationary in the laboratory frame next to a current carrying conductor experiences no force.

This fact cannot be explained by relativity. It is entirely due to your problem setup. There is no force simply because the wire is uncharged in the lab frame. That is it. This fact cannot be derived from relativity or anything else. It is and must be supplied by the problem setup.

It is certainly possible, instead, to specify that the wire is charged in the lab frame. In that case a stationary charge in the lab frame will experience a non-zero force. That would also be perfectly acceptable and completely compatible with relativity.

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 there is no force on the red charge you are simply stating that $V=0$.

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.

Let's have an observer move at the same velocity as the electrons in the conductor.

In this frame we can obtain the charge and current density as follows:

Let $v=J_0/\rho_0$ be the velocity of the observer in 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 charge density of the protons in the lab frame (the charge density of the electrons in the lab frame is $-\rho_0$).

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 observer's 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 observer's frame there is both a charge density and a current density. The charge density produces an E field and the current density produces a B field. Since the test particle is both charged and moving in that frame it experiences both an electrical force and a magnetic force. These forces, in turn, are equal and opposite and cancel out. Therefore there is no net force in any frame.

As we have negative charges moving from left to right we have the positive holes moving from right to left.

This would give you twice the current density as actually observed. This is not a good way to think of a wire.