Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B4

Use a boost along the wire axis when comparing frames. The current variable in a four-current must be current density, in $\mathrm{A}{\mathrm{m}}^{-2}$, rather than a scalar total current in amperes. For the wire-force calculation it is convenient to integrate charge and current over each cross-section and work with line quantities.

A Lorentz four-vector is a set of four components transforming as $A^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }$ under a Lorentz transformation. Examples include the four-position and four-velocity. A Lorentz invariant is a quantity unchanged by that transformation. With metric signature $(+---)$, the scalar product $A_{\mu }{B}^{\mu }=A^0{B}^0-\mathbf{A}\cdot \mathbf{B}$ is invariant. The components of a four-vector are generally not individually invariant.

The proton's total electric charge is taken to be a scalar: it does not acquire a factor $\gamma$ when the proton moves. Experimental evidence comes from sensitive charge neutrality tests of matter. Hydrogen molecules and helium atoms remain neutral despite different binding and internal Electron and nuclear motions. If constituent charge scaled with kinetic energy as relativistic mass does, the same cancellation of fixed numbers of proton and Electron charges would generally fail in different bound systems.

In a gas-efflux test of charge neutrality, gas escapes from an electrically isolated vessel connected to an electrometer. Any net charge per escaping molecule would leave a measurable opposite vessel charge, after residual ions and charged contaminants are removed. The hydrogen/helium efflux experiment supports velocity-independent additive charge. ^[ref]

This is evidence for charge invariance, not merely conservation of electric charge: the latter concerns constancy during a process, while invariance compares descriptions or internal motions without rescaling charge by a boost factor. Net-matter neutrality constrains combinations of constituent charges; it should not be described as an exact proof covering every possible velocity-dependent microscopic model.

Let an object be at rest in $S_0$, with proper charge density ${\rho }_0$ and no rest-frame current. In a frame moving at $\mathbf{v}$ relative to it, the object moves with velocity $-\mathbf{v}$. A simultaneous volume element is Lorentz-contracted along the motion, $dV=d{V}_0/\gamma$, while its charge is invariant. ThereforeThe object's four-velocity in this frame is $U^{\mu }=\gamma (c,-\mathbf{v})$, soA scalar proper density times a Lorentz four-vector transforms as a Lorentz four-vector, demonstrating the claim. With the alternative convention that $\mathbf{v}$ denotes the object's velocity in the observer frame, the spatial component is $+\gamma {\rho }_0\mathbf{v}$ instead. Summing the four-currents of different carrier species preserves this transformation law even when the combined current has no rest frame, as for a lab-neutral current-carrying wire.

A long straight wire carrying current $I_1$ produces field magnitude $B_1={\mu }_0|I_1|/(2\pi d)$ at a parallel wire a distance $d$ away. The magnetic force on a current-carrying wire is $d\mathbf{F}=I_{2}d\mathbf{\ell }\times {\mathbf{B}}_1$. Hence for equal currents,The two forces have equal magnitude and opposite directions. Net electrical neutrality removes a lab-frame external electrostatic force between ideal uniform wires, but does not remove their magnetic interaction.

Put wire 1 on the $z$ axis and wire 2 at $x=d$, with positive $x$ pointing away from wire 1. Let each wire have stationary positive linear charge density $+\lambda$ and moving Electron line density $-\lambda$ in the lab. For equal currents, both Electron populations have drift velocity $+v\hat{\mathbf{z}}$, so the signed conventional current is $I=-\lambda v$.

First consider wire 2's positive charges in their own rest frame, which is the lab. Wire 1 has zero total line charge there, so its external electrostatic field is zero. The positive-charge force per length is therefore $\boxed{f_{+}=0}$.

Next use the Electron rest frame $S^{\prime }$ moving at $+v\hat{\mathbf{z}}$. Wire 1's ion line density becomes ${\lambda }_{+}^{\prime }=\gamma \lambda$. Its Electron density becomes ${\lambda }_{-}^{\prime }=-\lambda /\gamma$: the Electron rest density is lower than its lab density. The transformed total isEquivalently this follows from the four-current transformation ${\lambda }_1^{\prime }=\gamma ({\lambda }_{1,\mathrm{net}}-v{I}_1/c^2)$, with ${\lambda }_{1,\mathrm{net}}=0$ and $I_1=-\lambda v$. By the electrostatic field of a line charge,A wire-2 Electron is at rest in $S^{\prime }$ and feels only the electric force, $F_{e,x}^{\prime }=-e{\lambda }_1^{\prime }/(2\pi {ϵ}_{0}d)$, toward wire 1.

Apply the supplied transverse relativistic force transformation with the Electron's lab velocity $\mathbf{u}=\mathbf{v}$. Its denominator is $\gamma (1-v^2/c^2)=1/\gamma$, so $F_{e,x}^{\prime }=\gamma F_{e,x}$, or $F_{e,x}=F_{e,x}^{\prime }/\gamma$. There are $\lambda /e$ Electrons per unit lab length. ThusAdding $f_{+}=0$ gives exactly the attractive magnetic result. This electrostatic-frame derivation of parallel-wire forces uses electrostatics in each test carrier's own rest frame; no magnetic force acts on that stationary test carrier there. The other wire can still carry current in that frame. Transform the force per Electron first and then count Electrons per lab length: transforming a force per unit length as if it were a particle force would introduce an incorrect extra contraction factor.

Keep wire 1's Electrons moving at $+v$, but let wire 2's Electrons move at $-v$, so $I_2=-I_1$. The positive charges remain lab-stationary and still feel zero external electrostatic force in the lab. The appropriate rest frame for wire 2's Electrons now moves at $-v$.

In this frame, wire 1's total line charge isIt produces an electric field pointing toward wire 1. The stationary test Electron is negatively charged, so its electric force points away from wire 1. Transforming back again divides the individual transverse force by $\gamma$, and multiplying by $\lambda /e$ givesThe changed sign arises from the changed transformed charge imbalance, without needing a magnetic-force calculation. More generally the signed result is $F_x/\ell =-{\mu }_0{I}_1{I}_2/(2\pi d)$ for the chosen outward radial direction.