Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section B / 6

Fix coordinates $x^{\mu }=(ct,x,y,z)$ and use the following convention for the contravariant electromagnetic field tensor:For $O^{\prime }$ moving with velocity $+v\hat{\mathbf{x}}$ relative to $O$, the Lorentz transformation hasThe rank-two tensor transformation rule isFor example,Similarly, $F^{\prime 03}=-\gamma (E_z+v{B}_y)/c$ and $F^{\prime 13}=\gamma (B_y+v{E}_z/c^2)$. The $01$ component is unchanged because ${\gamma }^2(1-{\beta }^2)=1$, and the $23$ component is unchanged since neither index lies in the boost plane. Reading off the six field components givesAn alternative overall sign convention for $F^{\mu \nu }$ gives the same physical Lorentz transformation of electromagnetic fields when its identifications with $\mathbf{E}$ and $\mathbf{B}$ are used consistently.

Write $a_y=E_{A,y}^0$, $a_z=E_{A,z}^0$, $b_y=E_{B,y}^0$ and $b_z=E_{B,z}^0$, evaluated at the same transverse coordinates $(y,z)$. A superscript $0$ labels each filament's own rest frame, rather than a single common frame in all cases. The filaments are infinite and uniform along $x$, so the change of longitudinal coordinate under a Lorentz boost creates no extra dependence in these functions. After the acceleration transients have passed, the magnetic field of each filament vanishes in its own rest frame.

For a filament moving at $+v\hat{\mathbf{x}}$, use the inverse of part (a): its transverse electric field in the laboratory is $\gamma {\mathbf{E}}^0$ and its magnetic field is $\mathbf{v}\times \mathbf{E}/c^2$. Superposition for the two co-moving filaments therefore gives, in $O$,In the common co-moving frame $O^{\prime }$ both filaments are at rest, soFor example, $B_y^{\prime }=\gamma [-\gamma v(a_z+b_z)/c^2+v\gamma (a_z+b_z)/c^2]=0$, a useful sign check. The supplied $\eta$ is the proper linear charge density. Thus each laboratory density is $\gamma \eta$ and each laboratory electric current is $\gamma \eta v$, consistently with length contraction and the four-current transformation. In $O^{\prime }$ each density is $\eta$ and both currents vanish.

Retain $a_y,a_z,b_y,b_z$ as defined in Case 1. In the laboratory, A moves and B is stationary. Transform A's fields and add B's unchanged electrostatic field:The laboratory linear charge densities are ${\lambda }_A=\gamma \eta$ and ${\lambda }_B=\eta$, with electric currents $I_A=\gamma \eta v$ and $I_B=0$.

In $O^{\prime }$, A is stationary while B moves with velocity $-v\hat{\mathbf{x}}$. Thus A contributes its rest electrostatic field, and B contributes the boosted electric field and the opposite-sign convective magnetic field:The densities in this frame are ${\lambda }_A^{\prime }=\eta$, ${\lambda }_B^{\prime }=\gamma \eta$, with $I_A^{\prime }=0$ and $I_B^{\prime }=-\gamma \eta v$. Substituting the laboratory result into part (a) gives the same fields; for instance $E_y^{\prime }=\gamma [(\gamma a_y+b_y)-v\gamma v{a}_y/c^2]=a_y+\gamma b_y$. The magnetic contribution attributed to A disappears in its own frame, but B's contribution does not.

The infinitely long filaments have no finite total force; the meaningful result is force per unit length in the laboratory. Evaluate only A's field at B, excluding B's singular self-field. B is at $y=0,z=-d/2$, so defineIn both cases A moves at $+v\hat{\mathbf{x}}$, hence at BFor a filament with linear charge density ${\lambda }_B$ and convective electric current $I_B$, the Lorentz force per laboratory length isIn Case 1, ${\lambda }_B=\gamma \eta$ and $I_B=\gamma \eta v$. Electric repulsion and magnetic attraction givesoThe cancellation uses ${\gamma }^2(1-{\beta }^2)=1$. In Case 2, B is stationary, ${\lambda }_B=\eta$ and $I_B=0$, givingBoth results have units $\mathrm{N}{\mathrm{m}}^{-1}$ and are repulsive, away from A, irrespective of the common sign of $\eta$. In Case 1 the velocity independence of force per laboratory length does not violate the transverse force transformation: a material segment of proper length ${\ell }_0$ occupies laboratory length ${\ell }_0/\gamma$, so its force is the rest-frame force divided by $\gamma$. The calculation interprets the stated $\eta$ as each filament's final proper density; holding a laboratory density fixed during acceleration would be a different preparation.

Spin-orbit coupling is a relativistic interaction between a particle's spin and its orbital angular momentum or, more generally, its motion through an electric field. A moving Electron experiences a transformed magnetic field, which couples to its magnetic dipole moment. For an electron bound in a central electrostatic potential energy $U(r)$, the leading interaction, including the Thomas precession correction, isHere $\mathbf{L}$ is orbital angular momentum and $\mathbf{S}$ is spin. The factor of one-half relative to the simplest rest-field argument comes from Thomas precession of the accelerated Electron frame.^[ref]

For example, in the hydrogen atom, the $2p$ states with total angular momentum $j=1/2$ and $j=3/2$ have different $\mathbf{L}\cdot \mathbf{S}$ eigenvalues. Their splitting is part of the atomic fine structure. The interaction redistributes and splits spin-dependent energies; it is not the same physical effect as the spin-independent convective magnetic force between charged filaments.

The information in the original question is insufficient to determine a unique spin-induced change of charge density. It specifies ideal dielectric filaments with a prescribed excess linear charge density, but not the charge carrier species, their spin state, charge carrier mobility, binding potential or relaxation mechanism. A classical line charge has no spin variable at all. The following conditional conclusions are supported by the fields already calculated.

For a carrier moving with its filament, the local magnetic field in its instantaneous rest frame is, for the transverse fields here,where $\mathbf{u}$ is that carrier's laboratory velocity. Exclude its filament's singular self-field and first retain only the prescribed straight convective motion.

In Case 1 there is a common rest frame in which both filaments and the excess charges are stationary and the interfilament magnetic field vanishes. Equivalently, in the laboratory ${\mathbf{B}}_A=\mathbf{v}\times {\mathbf{E}}_A/c^2$, so B's co-moving charges have ${\mathbf{B}}_{A,\mathrm{rest}}=0$. Thus uniform common translation alone creates no extra mutual rest-field spin splitting. This does not exclude intrinsic spin-orbit coupling of electrons moving within their atomic binding potentials; that material physics was never specified.

In Case 2 there is relative motion and no frame in which both source distributions are stationary. Put $E_{\ast }=\eta /(2\pi {ϵ}_{0}d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame areFor the first expression B is at rest in $O$; for the second A is at rest in $O^{\prime }$ and the field comes from B moving at $-v\hat{\mathbf{x}}$. For a carrier with magnetic dipole moment ${\mathbf{\mu }}_s$, the interaction $H_s=-{\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}}$ distinguishes opposite transverse spin orientations. A field gradient could exert a spin-dependent force $\nabla ({\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}})$ on mobile carriers; spin relaxation could instead produce a spin polarization. These are possible responses, not consequences of charge density alone. In B's frame the displayed interaction is directly a magnetic-moment coupling to the moving source's field; treating the moving carrier's electrostatic orbital term in another frame also requires consistent binding and acceleration dynamics, including any applicable Thomas precession.

For example, if $\eta >0$ and mobile electron spins can relax, the lower magnetic-energy orientation has the electron's magnetic moment toward $+\hat{\mathbf{y}}$ and its spin toward $-\hat{\mathbf{y}}$. Oppositely oriented moments would experience opposite gradient forces. The sign changes if the source charge changes sign. None of these statements fixes a net charge redistribution without a confinement and response model, and the stipulated zero transverse size supplies no resolved “sides” on which charge could accumulate.

Therefore Case 1 has no new mutual spin effect from common bulk motion alone; Case 2 permits motion-induced spin-dependent interactions, but neither case determines an additional charge-density profile from the supplied data. In particular, the factor $\gamma$ relating proper and laboratory linear charge densities in part (b) follows from length contraction and is not an effect of spin-orbit coupling. If the charges remain pinned in an ideal dielectric, their imposed density remains the prescribed density in either case.