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Einstein Velocity Addition for a Moving Charge in a Wire

  • #1
CDL
20
1

Homework Statement



I am reading through Griffiths' Electrodynamics, and I have come to the scenario in the Relativity chapter where in an inertial reference frame ##S##, we have a wire, with positive charges (linear density ##\lambda##) moving to the right at speed ##v##, and negative charges (##\rho = - \lambda##) moving to the left also at speed ##v##.

There is also charged particle moving to the right at speed ##u < v##, a perpendicular distance ##s## from the wire.

Griffiths writes that the velocities of the right and left moving wires, ##v_+## and ##v_-## respectively, according to an observer moving along with the charge moving with speed ##u## (let's say in frame ##\bar{S}##) are ##v_{\pm} = \frac{v \mp u}{1 \mp \frac{vu}{c^2}}.## This doesn't agree with my attempt.

Homework Equations



Suppose we have two inertial frames of reference ##S## and ##\bar{S}## whose origins in spacetime coincide. ##\bar{S}## is moving at a constant velocity ##u## along the ##x##-axis according to an observer in ##S##. For a 4-vector ##X^\mu## in ##S##, the corresponding 4-vector ##\bar{X}^{\mu}## in ##\bar{S}## is given by ##\bar{X}^{\mu} = \begin{pmatrix}\gamma & -\beta\gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} X##, where ##\beta = \frac{u}{c}## and ##\gamma = \frac{1}{\sqrt{1-\beta^2}}.##

3. The Attempt at a Solution


Without a loss of generality, we can make ##t = 0, x = 0## in ##S## correspond to ##\bar{t} = 0## and ##\bar{x}=0## in ##\bar{S}##. We can also make all motion happen along a common ##x##/##\bar{x}##-axis. Suppose one of the positive charges in the positive line density traverses a displacement ##\Delta x## along the ##x##-axis in a time interval ##\Delta t## in ##S##, then ##\pm v = \frac{\Delta x}{\Delta t}##, plus corresponding to right-moving positive charge in ##S##, minus corresponding to left-moving negative charge in ##S##. By considering a Lorentz transformation from ##S## to ##\bar{S}##, we have ##\begin{pmatrix}c \Delta \bar{t} \\ \Delta \bar{x}\end{pmatrix} = \begin{pmatrix} \gamma & - \beta \gamma \\ -\beta \gamma & \gamma\end{pmatrix} \begin{pmatrix} c\Delta t \\ \Delta x \end{pmatrix},## where ##\beta = \frac{u}{c}## and ##\gamma = \frac{1}{\sqrt{1 - {u^2 \over c^2}}}.## This implies that ##\frac{\Delta{\bar{x}}}{\Delta \bar{t}} = c\frac{-\beta \gamma c \Delta t + \gamma \Delta x}{\gamma c \Delta t - \beta \gamma \Delta x}.## Dividing top and bottom by ##\Delta t## and cancelling the ##\gamma##'s yields ## v_{\pm} = c\frac{-\beta c \pm v}{c\mp \beta v}## and so ##v_{\pm} = \frac{\pm v - u}{1 \mp \frac{uv}{c^2}}.##

Edit: I think I see the issue here, the magnitudes of my results match that of Griffiths', he may have been referring to speed. Still, it would be great for someone to check whether or not my analysis is correct :)
 
Last edited:

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,530
2,951
Your work looks correct to me.
 

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