1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic E and B fields of an infinitely long wire

  1. Jan 23, 2017 #1
    1. The problem statement, all variables and given/known data

    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##.

    2. Relevant 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}})##

    3. 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.
     
  2. jcsd
  3. Jan 23, 2017 #2
    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.
     
  4. Jan 23, 2017 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hi, Julien. That sounds like a good way to go.
     
  5. Jan 24, 2017 #4
    Hi @TSny and thanks for your answer. I give it a try:

    I will redefine the velocity components because I wanna 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.
     
  6. Jan 24, 2017 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  7. Jan 28, 2017 #6
    @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.
     
  8. Jan 28, 2017 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Relativistic E and B fields of an infinitely long wire
  1. B-field from E-field (Replies: 1)

Loading...