Relativistic E and B fields of an infinitely long wire

[JulienB]

Homework Statement

Hi everybody! I have the following problem to solve:

An infinitely long and thin straight wire carries a constant charge density $\lambda$ and moves at a constant relativistic speed $\vec{v}$ perpendicularly ($\beta$) and parallel ($\alpha$) to its axis.
a) Determine the $\vec{E}$ and $\vec{B}$ fields in the whole space.
b) Through which charge density $\rho$ and current density $\vec{j}$ are those fields generated?
Tip: previously we had the electrostatic potential $\phi = - \lambda \ln(x^2+y^2)$ for an infinitely long straight wire aligned with the $z$-axis and with charge density $\lambda$.

Homework Equations

$\vec{E} = -\nabla \phi - \partial_t \vec{A}$
$\vec{A} = \frac{\mu_0}{4 \pi} \int d^3x'\ \frac{\vec{j} (\vec{x}', t_r)}{|\vec{x} - \vec{x}'|}$
where $t_r = t - \frac{|\vec{x} - \vec{x}'|}{c}$
$j_{\mu} = (j^0, j^i) = (c \rho, \rho \dot{\vec{x}})$

The Attempt at a Solution

Not sure what "parallel" and "perpendicular" here mean. Because of the given potential, I assumed that the wire is moving in the $x$-direction at speed $\alpha$ and in the $y$-direction at speed $\beta$. Would you agree to this interpretation? It could also be that the wire moves in the $z$-direction though ("parallel" to its axis), but then I imagine that the scalar potential should not be independent of $z$. Is that right?

If my interpretation is correct I can calculate $-\nabla \phi$ which yields

$-\nabla \phi = \frac{2 \lambda}{ x^2+y^2} (x,y,0) = \frac{2 \lambda t}{t^2 (\alpha^2 + \beta^2)} (\alpha, \beta, 0)$

where I set $x=\alpha t$ and $y=\beta t$. I am now not sure how to proceed regarding $\vec{A}$ since we didn't study the retarded time/potential yet (I could give it a try though). Should I try to calculate $t_r$ or is there another method I can use?Thanks a lot in advance for your suggestions.Julien.

 

[JulienB]

Hi again! Actually I got another idea which sounds better to me even though I still don't manage to make it work.

I first calculate $\vec{E}'(\vec{x}')$ in the frame $\Sigma'$ of the moving wire so that I find myself in an electrostatics situation, therefore allowing me to calculate $\vec{E}'(\vec{x}') = - \nabla \phi (\vec{x})$ and having $\vec{B}=0$ since there is no current in that inertial frame. Then I'd perform a Lorentz transformation on both fields so that I retrieve the electric and magnetic fields in the not moving frame $\Sigma$. Would that make sense?

Thanks a lot in advance for your answers.Julien.

 

[TSny]

I first calculate $\vec{E}'(\vec{x}')$ in the frame $\Sigma'$ of the moving wire so that I find myself in an electrostatics situation, therefore allowing me to calculate $\vec{E}'(\vec{x}') = - \nabla \phi (\vec{x})$ and having $\vec{B}=0$ since there is no current in that inertial frame. Then I'd perform a Lorentz transformation on both fields so that I retrieve the electric and magnetic fields in the not moving frame $\Sigma$. Would that make sense?

Hi, Julien. That sounds like a good way to go.

 

[JulienB]

Hi @TSny and thanks for your answer. I give it a try:

I will redefine the velocity components because I want to keep $\beta$ for the Lorentz transformation. Therefore, the velocity is simply $\vec{v}=(0, v_y, v_z)$. The frame of the moving wire is $\Sigma'$ while the non moving frame is $\Sigma$. Another way to describe the latter is to say that it moves with velocity $-\vec{v}$ relative to the wire.

• In $\Sigma'$:

$\vec{v'}=0 \implies \begin{cases} \vec{E}' (\vec{x}') = - \nabla \phi' (\vec{x}') = \frac{2 \lambda}{x'^2+y'^2} (x',y',0) \\ \vec{B}'(\vec{x}')= 0 \end{cases}$

• In $\Sigma$:

$\vec{v} = (0,-v_y,-v_z) \implies x'=x, y'=\gamma(y+v_y t), z' = \gamma v_z t$
$\implies \vec{E}(\vec{x}) = \gamma (\vec{E}'(\vec{x}') + \frac{1}{c} (\vec{v} \times \vec{B}'(\vec{x}')) + \frac{\gamma^2}{c^2(1+\gamma)} (\vec{v} \cdot \vec{E}'(\vec{x}')) \vec{E}'(\vec{x}')$
$= \frac{2 \lambda \gamma}{x^2+\gamma^2(y+v_y t)^2} \bigg(1 + \frac{\gamma^2}{c^2(1+\gamma)} \frac{-v_y (y+ v_y t) - v_z^2 t}{x^2+\gamma^2(y+v_y t)^2} \bigg) (x, \gamma(y+v_y t), \gamma v_z t)$.

Well well that looks strange, especially since I was expecting that $E_x'(\vec{x}') = E_x (\vec{x})$.. Am I doing something wrong with the Lorentz transformation? Did I transform for example the $z$-component in the right way?Thank you very much in advance.Julien.

 

[TSny]

Your Lorentz transformation equations for $y'$ and $z'$ don't appear to be correct.
https://en.wikipedia.org/wiki/Lorentz_transformation#Vector_transformations

Also, unless I'm mistaken, there is a complication due to the fact that the orientation of the rod changes as you change frames. You have the rod oriented along the z'-axis in the primed frame. However, in that case, the rod will generally not be oriented along the z-axis in the unprimed frame. Thus, your boost velocity component $v_z$ will not be parallel to the rod in the unprimed frame. Likewise, $v_y$ will not be perpendicular to the rod in the unprimed frame. So, $v_z$ and $v_y$ would not represent the quantities $\alpha$ and $\beta$ given in the problem statement.

At first, I thought the problem might be asking you to treat two cases separately:
(1) boost velocity perpendicular to rod
(2) boost velocity parallel to rod

That would be much easier than the case where the boost velocity has a nonzero component perpendicular to the rod and a nonzero component parallel to the rod. But it looks like they want you to do the harder problem.

 

[JulienB]

@TSny Thank you very much for your answer. I've given my homework back and am waiting for a feedback now. There was also an error in the script of my teacher for the Lorentz transformation of the electric field, that's why I was getting such a complicated equation. I'll let you know what the solution was once I get my homework back.

 

[TSny]

@TSny Thank you very much for your answer. I've given my homework back and am waiting for a feedback now. There was also an error in the script of my teacher for the Lorentz transformation of the electric field, that's why I was getting such a complicated equation. I'll let you know what the solution was once I get my homework back.

OK. Thanks.