Magnetic Forces from Special Relativity

Introduction

The idea here is to show that the forces on a charged test particle in the vicinity of a wire carrying a current can be computed without reference to a magnetic field and to show that these have the same value as those computed from the magnetic field. That is done by choosing a frame of reference where the particle is at rest while positive and negative the charge speeds in the wire are modified to accommodate this particle frame of reference.

In the laboratory frame let the particle speed be v_o. And let the negative charge speed in the wire be v_c while the positive charge speed is 0

Then in the particle at rest frame the negative charge speed is -v_o+v_c while the positive charge speed is -v_o-v_cwhich results in the beta values:

$\begin{array}{l} β_{-} = - \frac{- v_{o} + v_{c}}{c} \\ β_{+} = - \frac{- v_{o} - v_{c}}{c} \end{array}$

(1.1)

This expression for the betas gives the expected result, that the total charge density is zero when either vo=0 or vc=0.

For the total charge density Feynman would get a result like the following:

$\begin{array}{l} ρ = ρ_{0} (\frac{1}{\sqrt{1 - β_{+}^{2}}} - \frac{1}{\sqrt{1 - β_{-}^{2}}}) \\ = ρ_{0} (\frac{\sqrt{1 - β_{-}^{2}} - \sqrt{1 - β_{+}^{2}}}{\sqrt{1 - β_{+}^{2}} \sqrt{1 - β_{-}^{2}}}) \\ \approx \frac{ρ_{0}}{2} (\frac{β_{+}^{2} - β_{-}^{2}}{\sqrt{1 - β_{+}^{2}} \sqrt{1 - β_{-}^{2}}}) \end{array}$

(1.2)

Expanding the result in equation (1.2) we get:

$ρ \approx \frac{ρ_{0}}{2 c^{2}} (\frac{{(v_{c} + v_{o})}^{2} - {(v_{c} - v_{o})}^{2}}{\sqrt{1 - β_{+}^{2}} \sqrt{1 - β_{-}^{2}}})$

(1.3)

$ρ \approx \frac{ρ_{0}}{c^{2}} (\frac{2 v_{c} v_{o}}{\sqrt{1 - β_{+}^{2}} \sqrt{1 - β_{-}^{2}}})$

(1.4)

The current density, since positive charge moves in +x direction and negative charge moves in the -x direction, is just

$J = 2 ρ_{0} v_{c}$

(1.5)

so we can express equation (1.4) in terms of the current, I, through the wire, I=JA:

$I = J A$

(1.6)

We can also express equation (1.4) as charge per length

$λ = ρ A \approx \frac{ρ_{0} A}{c^{2}} (\frac{2 v_{c} v_{o}}{\sqrt{1 - β_{+}^{2}} \sqrt{1 - β_{-}^{2}}})$ (1.7)

where λ is the charge per unit length and A is the cross sectional area of the wire.

Now the task is to show that the force on our test particle is the same whether we use I to compute the magnetic force or we use λ to compute the electrostatic force.

The electrostatic force is:

$F_{E} = \frac{q λ}{2 π ε_{0} r} \approx \frac{2 q ρ_{0} A v_{c} v_{o}}{2 π ε_{0} r c^{2}}$

(1.8)

where q is the charge of the test particle and r is its distance from the wire.

The magnetic force is:

$F_{B} = \frac{q μ_{0} J A}{2 π r} = \frac{q J A v_{o}}{2 π ε_{0} r c^{2}} = \frac{2 q ρ_{0} A v_{o} v_{c}}{2 π ε_{0} r c^{2}}$

(1.9)

Where I have set the gamma factors in (1.7) to unity.

Note that the forces are the same whether computed from magnetic fields or from electric fields.

http://www.feynmanlectures.caltech.edu/II_13.html#Ch13-S6