Is the magnetic field of a moving electron caused by length contraction in the direction of motion?

Question

Consider an electron moving relative to us. Because the space in the electron's rest frame is contracted relative to us in the direction of the electron's velocity, the electric field lines are squeezed in the same direction, so the electric field "density" is bigger perpendicular to the electron's motion (but smaller (zero?) in the direction parallel to its motion). Is this the qualitative source of the magnetic field?

Answer 1

So consider your electron moving parallel to another electron moving at the same velocity. In the stationary frame, the electric fields of both electrons would be affected in the way you suggest. Therefore, when considered in the stationary frame, the repulsion between the electrons would be much greater than given by Coulomb's law. The effect would increase the greater the velocity of the electrons.

We can't use Coulomb's law when the charges are moving but we can introduce the covariant Lorentz force, but only by introducing a new force , besides that due to the electric field, which acts in the stationary frame (or any other frame but the rest frame of the electrons), and which depends on the velocity of the electron, the size of its electric charge and which acts in a direction between the two electrons and perpendicular to their velocity.

This is of course the direction of the magnetic part of the Lorentz force, which forms the basis for a definition of what we mean by the magnetic field (or magnetic flux density $\vec{B}$ to be precise). It is the field that produces a force of $q\vec{v}\times\vec{B}$.

EDIT: The maths goes something like this.

In the frame of the electron(s) (we will label the laboratory frame as the primed frame), the electric field is given by Coulomb's law and so the force on another electron at a separation $R$ would be $$\vec{F} = \frac{e^2}{4\pi \epsilon_0 R^2} \hat{R},$$ where $\hat{R}$ is a unit vector in the direction between the two electrons.

Now observed from the laboratory frame there will be a transformed electric field and a magnetic field. According to the relativistic transforms of the E- and B-fields, the electric field along a line between the two electrons is $$ \vec{E}' = \gamma \vec{E} = -\gamma \frac{e}{4\pi \epsilon_0 R^2} \hat{R},$$ and so the observed Coulomb force between the electrons in the laboratory frame would be increased by a factor of $\gamma$.

However, reversing the direction of the question - if we now suppose the existence of a B-field (and the correctness of electromagnetism as a covariant theory) then the B-field in the laboratory frame is given by $$\vec{B}' = -\frac{\gamma}{c^2} \vec{v} \times \vec{E} = -\frac{\gamma}{c^2}\frac{ve}{4\pi \epsilon_0 R^2} \ \hat{z} \times \hat{R} = -\frac{\gamma v e}{4\pi \epsilon_0 c^2 R^2}\ \hat{\phi}$$ where the lab frame moves at $\vec{v} = -v \hat{z}$ with respect to the electrons and $\hat{\phi}$ is the usual cylindrical coordinate unit vector.

We can now combine these fields in the moving frame to find the net force on the other electron. $$\vec{F}' = -e\vec{E}' -e\vec{v}\times \vec{B}' = \gamma \frac{e^2}{4\pi \epsilon_0 R^2} \hat{R} +ev\frac{\gamma v e}{4\pi \epsilon_0 c^2 R^2}\ \hat{z} \times \hat{\phi}$$ and where this time $\vec{v} = +v\hat{z}$ according to an observer in the lab frame, so $$\vec{F}' = \frac{e^2}{4\pi \epsilon_0 R^2} \left( \gamma -\gamma\frac{v^2}{c^2} \right) \ \hat{R} = \frac{\vec{F}}{\gamma}.$$

This is exactly how a force should transform according to special relativity.

Answer 2

The electric field for a charged particle moving in the $x$ direction of motion is Lorentz transformed as $$ E'_x~=~\gamma E_x,~E'_y~=~\gamma(E_y~-~vB_z/c^2),~E'_z~=~\gamma(E_Z~+~vB_y/c^2). $$ The motion of charge transforms the electric field components into magnetic field components.

Answer 3

The idea is good but it is a little more complicated.
You have to work with the tensor form of the electromagnetic field $ F^{\mu \nu}$and the 4 dimensions of space-time.
The Lorentz transformation that takes an electron from rest to an electron with a constant velocity can be seen as a rotation in the 4 dimensions of space time (${\Lambda^{\nu'}}_\nu $).
All tensors will change, "rotate", according to this rotation. Like 3 dimensional vectors rotate when you rotate a frame. Because the electromagnetic tensor has two indices you have to apply the rotation on each index as described here .
Basically the tensor transforms like this:
$F^{\mu'\nu'} = {\Lambda^{\mu'}}_\mu F^{\mu\nu} {\Lambda^{\nu'}}_\nu $
And you get the new electric and magnetic field inside the new tensor $F^{\mu'\nu'}$.

You could also use the electromagnetic potential vector $A^{\mu}$. Its transformation is simpler because it has only one index (it's a vector).

So basically your idea is good. Contraction (which is a rotation in fact) of space and time will "rotate" the electric and magnetic fields. The math involves tensor transformation.

Answer 4

There is the effect called magnetism, and then there is the effect that you described.

Some basic things about magnetism: There's no magnetism between a moving charge and a still standing charge. There's no magnetism between charges that move along a line.

And now some basic things about the effect you described: There is an effect between a moving charge and a still standing charge. There is an effect between charges that move along a line.

It's reasonable to say that those are two different effects.

Let's consider a current carrying wire that contains an equal amount of plus and minus charges. There is no electric field next to that wire, although every electron's field lines are squeezed in the direction perpendicular to the motion. All those electrons' field lines added together results in identical field lines as in the case when there is no current.