Electric field of a current in a wire

Question

The electric field of an electron moving in $x$ direction has the field $E_y'=\gamma E_y$ in the $y$ direction (where $E_y$ is the field at rest and $\gamma$ the Lorentz factor).

Why is this stronger field $E_y'$ not observed for a current in a wire?

Problem 5.5 of Purcell: "Electricity and Magnetism" (3rd edition) calculates the $E$ field of an infinitely long current as seen from a point. He states:

However, it is by no means obvious that the sum of the nonspherically symmetric fields in Eq. (5.15), from all the individual charges, equals $λ/2π\epsilon_0 r$ for any value of $β$. The task of Problem 5.5 is to demonstrate this explicitly.

By integrating the $E'$ field of a moving charge

$E' = \frac{q}{4π\epsilon_0l^2} \frac{1 − β^2}{(1 − β^2 \cos^2 θ)^\frac32}$

for all charge segments from $-$ to $+$ infinity ($-90°$ to $+90°$) he shows, that the amplification of the $E'$ field in the $y$ direction is compensated by its attenuation in the $x$ direction. So the net $E'$ field from moving and stationary charges are the same. However, for a finite wire of a few cm seen from a distance of a few cm under a reduced angle from $-45°$ to $+45°$ one would expect a different result, since the stronger $E_y$ field will not completely canceled out.

Answer 1

This is one of those things that you should try to not be analysing from Coulomb's law, precisely because it would be confusing like this.

The transformation of the fields is correct as you saw. I also looked at what happens to the charge-current density 4-vector, and it also transforms correctly, as you expect, but it does so because the volume element is transformed, not the charge part.

Instead, you should be using Maxwell's equations directly. Here, it is Gauß's law, which is not just easier, but that it covers all cases, being that we derive Lorentz transformations from Maxwell's equations. After all, the system of charges and fields will do its internal balancing act, and we get to see the steady state behaviour and deduce backwards.

Gauß's law will tell you that the wire with constant electron flow, will have the same number of charges, in order to have cancellation of the electric field outside of the wire. To have electrical neutrality. The complicated effect of the velocity, is to generate a magnetic field. The behaviour is frame-dependent: If you do a Lorentz boost, you will instead see that the charge is not neutral in the wire any more.

Answer 2

Suppose that the wire as a whole is electrically neutral and parallel to the x-axis, and for every segment of the wire that contains 100 travelling electrons, there are also 100 stationary protons. We know that the electric field around the wire is supposed to be zero. How can the protons' electric fields cancel out those of the electrons, if the fields produced by the electrons have their y-components stretched out (magnified) by a factor of $\gamma$, while the fields produced by the protons do not, and there are the same number of electrons and protons?

If $(x, y)$ are the displacements of point P from a point of interest in the wire, the contribution of a proton in the wire to $E_y$ is

$$ E_y^+ = \frac{e}{4\pi \epsilon_0} \frac{y}{(x^2 + y^2)^{3/2}} $$

and the contribution of an electron is

$$ E_y^- = -\frac{e}{4\pi \epsilon_0} \frac{\gamma y}{(\gamma^2 x^2 + y^2)^{3/2}} $$

The formula for $E^+_y$ was described in words in the linked answer: the fields of the electron are "compressed" by a factor of $\gamma$ in the x-direction, i.e., the effective x-displacement is $\gamma$ times greater than the actual x-displacement. But the y-component of the resulting field is also "stretched" by a factor of $\gamma$.

                    P
                    .

... + + + + + + + + + + + + + + + + + ...
... - - - - - - - - - - - - - - - - - ...
    =================================>

In the above diagram, the point labelled P is directly above a proton (represented by +) and an electron (represented by -) that are at the same point (unfortunately, it's hard to show this in the diagram). So here, $x = 0$. In that case, $E^-_y$ is greater in magnitude than $E^+_y$ by a factor of $\gamma$. However, when $x$ is much greater than $y$ (as for an electron or proton near one end of the diagram), notice that $E^-_y$ is reduced in magnitude by an overall factor of $\gamma^2$ compared with $E^+_y$.

So, close to P, the moving charges contribute more than the stationary charges. Far away from P, it's the reverse. When we add up all the contributions from the electrons and protons both close to P and far away, along an infinitely long wire, the result is zero.