Relativistic interpretation of magnetic effect on a charge due to a perpendicular current carrying wire

Question

Assume that we have a current-carrying conductor and a $+v_e$ test charge moving along the current. In the test charge's rest-frame, the electrons in the wire are length-expanded, and the positive ions of the metal are length-contracted. This makes the length charge density net positive, resulting in a repulsive force on the test charge, perpendicular to the wire. This explanation also concurs with the magnetic force $\mathbf{F}_L=q({\bf v} \times {\bf B})$ on the charge in lab-frame.

Now, let's assume that the test charge is moving perpendicular to the wire. We know from elementary magnetism that the magnetic force on test charge is now along the wire.

But how do I get the same result via relativistic length-contraction/expansion in the test charge's rest frame?

From what I understand, the moving test charge should observe no length contraction of any elements of wire, since it is moving perpendicular to the direction of relative motion.

Answer 1

Let's say that at first the current is on, but the wire is not moving relative to us. The electric field of each electron is length contracted. For some reason this is easier if the electrons move quite fast, so let's say they do move quite fast.

Then the wire starts moving towards us, quite slowly. So basically what happens is that the velocity vector of each electron turns a few degrees. So also the electric field around each electron turns a few degrees, the fields are not perfectly round so this turning actually means something.

It means that our test charge experiences a different force when the fields are turned.

(I mean the shape of the field turns, not necessarily the field itself)

See chapter "Charge moving Perpendicular to the Wire" here: https://physics.weber.edu/schroeder/mrr/MRRtalk.html

Answer 2

From The Mathematics of Gravity and Quanta

The 4-vector describing a charged particle is current, $J$. $J$ contains information about both the charge and the motion of the particle. For a stationary charge the 3-current is zero, so the 4-current is $(q,\mathbf 0)$. The general form of $J$ is found by Lorentz transformation and has the form $J=(\gamma q, \gamma q\mathbf v)$. Since covariance requires that physical laws are represented by tensors, to describe the electromagnetic force acting on a charged particle we need to contract $J$ with a tensor, $F$, representing the electromagnetic field. $F$ is the Faraday tensor. The result of contracting $J$ with $F$ is a vector, force. So, Faraday is a rank-2 tensor. Faraday should express both the force acting on a charged particle, and the equal and opposite reactive force exerted by the particle on its environment. If Faraday is an antisymmetric tensor, contracting with one index will give the force on the particle and contracting with the other will give the reactive force. We write down the 4-vector law of force,

$$(\mathrm {Force})^i = F^{ij}J_j.$$ Although it does not at first look like it, this is precisely the Lorentz Force law, which is usually expressed in terms of two 3-vector fields, the electric field $\mathbf E$ and the magnetic field, $\mathbf B$, $$\mathbf {Force} = e(\mathbf E + \mathbf v \times \mathbf B),$$ where we have $$ F^{ij} = \begin{bmatrix} 0 & \mathbf E_x & \mathbf E_y & \mathbf E_z \\ -\mathbf E_x & 0 & \mathbf B_z & -\mathbf B_y \\ -\mathbf E_y & -\mathbf B_z & 0 & \mathbf B_x\\ -\mathbf E_z & \mathbf B_y & -\mathbf B_x &0 \end{bmatrix}. $$ To see this, we need simply consider a static electric field. In this case the 3-vector force is $$ q\mathbf E = (q\mathbf E_x, q\mathbf E_y, q\mathbf E_z) $$ We can then write the Faraday tensor, using antisymmetry to determine the other components, $$ F^{ij} = \begin{bmatrix} 0 & \mathbf E_x & \mathbf E_y & \mathbf E_z \\ -\mathbf E_x & 0 & 0 & 0 \\ -\mathbf E_y & 0 & 0 & 0\\ -\mathbf E_z & 0 & 0 &0 \end{bmatrix}. $$ For a field due to a moving charge, Faraday is found from Lorentz transformation, acting on both indices, $$ F^{m'n'} = k^{m'}_i k^{n'}_j F^{ij}.$$ This introduces the magnetic fields for a moving charge in terms of the electric field of a static charge. Anti-symmetry of $F^{m'n'}$ follows directly from antisymmetry of $F^{ij}$. Generally, Faraday is not represented simply through boosting a static field, but it is the result of summing the fields generated by many particles, each one of which could be regarded as static in the rest frame of that particle. This would lead to a complicated expression, retaining antisymmetry, and summarised by the resultant electric and magnetic fields, $\mathbf E$ and $\mathbf B$.

Answer 3

But how do I get the same result via relativistic length-contraction/expansion in test charge's rest frame?

From what I understand, the moving test charge should observe no length contraction of any elements of wire, since it is moving perpendicular to them.

You have answered your own question. You don’t get length contraction with this setup. Length contraction is directionally dependent and only is important for the parallel configuration.

The perpendicular configuration has no length contraction. The current density and the charge density of the wire are unchanged in the test charge’s frame. Per Faraday’s law, the electric field that pushes the test charge is induced from the time varying B field as the wire moves.

The parallel configuration is of interest because for that case a relativistic explanation is simple and interesting. It should not be taken as a general approach.