Past exam of the symmetry and relativity course of the University of Oxford / 2017 / 4 / d

Use a neutral, infinitely long wire on the $x$ axis. Denote the total current by $I=\int j_{x}d{A}_{\perp }$, and write $r^2=y^2+z^2$. Outside the wire the magnetic field and electric field areThese follow from cylindrical symmetry and Ampère's law. The zero temporal component of the given total four-current specifies lab neutrality; the $\rho$ in $j_x=\rho v_0$ must refer to a carrier density, not that zero net charge density.

TODO: normalization from a literal volume current density requires the unspecified wire cross-section. The source does not supply its area or profile. The exterior field family below is fully determined in terms of $I$; it cannot be reduced to $j_x$ alone without that missing datum. If $j_x$ was intended as a cross-section-integrated line current instead, set $I=j_x$ and interpret its accompanying charge density as charge per length. No inside-wire profile is invented.

For a boost $u=v_0$ along positive $y$, let $\Gamma =(1-u^2/c^2{)}^{-1/2}$. Apply the Lorentz transformation of electromagnetic fields at corresponding events:Here the subscript labels the perpendicular-to-wire boost, not the direction of the transformed vector. In these formulas evaluate the original field at $y=\Gamma (y^{\prime }+u{t}^{\prime })$, $z=z^{\prime }$ and $x=x^{\prime }$. The wire moves in this frame, so the fields are time-dependent. The local current transforms to $J^{\prime 0}=0$, $J^{\prime x}=j_x$, $J^{\prime y}=J^{\prime z}=0$. A nonzero transformed electric field is possible even with zero net charge density: it is not a purely electrostatic field.

For a boost $u=v_0$ along positive $x$, the transverse coordinates are unchanged. The transformed fields areThe transformed line charge and current are ${\lambda }^{\prime }=-\Gamma uI/c^2$ and $I^{\prime }=\Gamma I$. These reproduce the radial electric field through Gauss's law. A boost comoving with the electrons removes their individual current contribution, but the positive background then moves and the total current is not zero.

For two identical wires, their lab interaction is attractive, with magnitude ${\mu }_0{I}^2/(2\pi d)$ per length. The total four-current is spacelike, $J^2=-j_x^2<0$, so no inertial frame makes it a pure charge density. Also $B^2-E^2/c^2=B^2>0$ forbids a frame in which the exterior field is purely electric. For an arbitrary boost along the wires, the electrical part is repulsive and the magnetic part attractive:The latter always dominates, and their sum remains $-{\mu }_0{I}^2/(2\pi d)$.

Even a more general boost cannot make the entire-wire force electrical alone. At the second wire choose $y=d,z=0$, so $\mathbf{B}=B{\mathbf{e}}_z$ and $\mathbf{j}=j{\mathbf{e}}_x$ with $jB>0$. The physical Lorentz force density is $f=(0,0,-jB,0)$. If the boosted magnetic force vanished, ${\mathbf{j}}^{\prime }\times {\mathbf{B}}^{\prime }=0$, then ${\mathbf{E}}^{\prime }\cdot {\mathbf{B}}^{\prime }=0$ would also give ${\mathbf{j}}^{\prime }\cdot {\mathbf{E}}^{\prime }=0$. Thus $f^{\prime 0}=0$, which from the boost of $f$ requires $V_y=0$. But for such a boost $f^{\prime y}=-jB<0$, whereas ${\rho }^{\prime }{E}_y^{\prime }={\Gamma }_V^2{V}_x^{2}jB/c^2\ge 0$. The electrical term alone cannot equal the actual force. Therefore there is no such frame for both complete wires, although the force on electrons considered separately can be electrical in their rest frame.