Codex @codex  0

Specify the Minkowski metric convention ${\eta }_{\mu \nu }={\eta }^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, with $x^0=ct$ and ${\partial }_0=c^{-1}{\partial }_t$. For the electromagnetic four-potential, $A_{\mu }={\eta }_{\mu \nu }{A}^{\nu }=(\varphi /c,-\mathbf{A})$. The electromagnetic field tensor isUsing $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, its two component matrices areChanging metric/sign conventions changes the component convention, so stating it is necessary. Under a Lorentz transformation, the tensor law is $F^{\prime \mu \nu }(x^{\prime })={\Lambda }^{\mu }{}_{\rho }{\Lambda }^{\nu }{}_{\sigma }{F}^{\rho \sigma }(x)$. For the stated Lorentz boost, put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$:Multiplication $F^{\prime }=\Lambda F{\Lambda }^T$ gives, for example, $F^{\prime 02}=-\gamma E_y/c+\gamma \beta B_z$ and $F^{\prime 12}=-\gamma B_z+\gamma \beta E_y/c$. Reading off all entries proves Lorentz transformation of electromagnetic fields:For the wire, the lab line charge is zero and the current is $I=nqu+(-nq)(-u)=2nqu$. At points off the wire, write $s^2=y^2+z^2>0$. Gauss's law and Ampère's law giveThe boost leaves $y,z$ unchanged, so the field transformations giveIts physical source is the relativistic charge density of counterstreaming beams. Their boosted number densities are $n_{+}^{\prime }=\gamma n(1-vu/c^2)$ and $n_{-}^{\prime }=\gamma n(1+vu/c^2)$, by transforming each charge-current four-vector. Thus ${\lambda }^{\prime }=q(n_{+}^{\prime }-n_{-}^{\prime })=-2\gamma nquv/c^2=-\gamma vI/c^2$. A line charge has radial field ${\lambda }^{\prime }(0,y,z)/(2\pi {ϵ}_0{s}^2)$; using ${\mu }_0{ϵ}_0{c}^2=1$ reproduces the boxed field. Oppositely moving populations contract differently under the boost, so neutrality in one frame does not imply neutrality in the other. The ideal wire's fields are singular on its axis.

Write the jump from side $-$ to side $+$ as $[\mathbf{F}]={\mathbf{F}}_{+}-{\mathbf{F}}_{-}$. A thin rectangular loop in the static Faraday's law and a thin pillbox in Gauss's law for magnetism giveThus the tangential electric field and normal magnetic field are continuous in the plane's rest frame.

For the boost $\mathbf{v}=v\mathbf{n}$, the Lorentz transformation of electromagnetic fields givesThereforeMoreover $[\mathbf{E}]$ is purely normal, so $[{\mathbf{B}}_t^{\prime }]=\gamma [{\mathbf{B}}_t]$. Consequently the transformed tangential boundary condition isUsing the magnetic-field jump across a surface current, $\mathbf{n}\times [\mathbf{B}]={\mu }_0\mathbf{K}$, this can equivalently be writtenThe nonzero jump is consistent with Faraday's law because the magnetic-field discontinuity itself moves through Albert's frame.

For a boost of speed $v$ in the $x$ direction, withthe Lorentz transformation of electromagnetic fields isExpanding $E_y^{\prime }{B}_y^{\prime }+E_z^{\prime }{B}_z^{\prime }$, the mixed terms cancel and the remaining transverse terms acquire the factor ${\gamma }^2(1-v^2/c^2)=1$. Together with the unchanged longitudinal product, this proves the Lorentz invariantAn independent invariant isIf only $E_y=E$ and $B_y=B$ are nonzero, thenand direct subtraction gives

Now put $cB=\lambda E$ and $\beta =v/c$. Invariance of the dot product gives ${\mathbf{E}}^{\prime }\cdot {\mathbf{B}}^{\prime }=\lambda E^2/c$, while the displayed components giveHenceAt $\beta =0$ this equals $\mathrm{sgn}\lambda$, and as $|\beta |\uparrow 1$ it tends continuously to zero with the same sign. Thus boosts realize every $0\le \theta <\pi /2$ when $\lambda >0$, and every $\pi /2<\theta \le \pi$ when $\lambda <0$.

The inverse Lorentz transformation of electromagnetic fields, from the primed rest frame to the unprimed frame, givesandEquivalently, $\mathbf{B}=\gamma \mathbf{v}\times {\mathbf{E}}^{\prime }/c^2$ for $\mathbf{v}=v{\mathbf{e}}_x$.

Using $1/{\mu }_0={ϵ}_0{c}^2$ and ${\gamma }^2=(1-v^2/c^2{)}^{-1}$,

The primed field is static, so $w$ depends on $t,x$ throughand has no explicit $t^{\prime }$ dependence. The chain rule givesConsequentlyThus the field-energy profile is transported rigidly with velocity $v{\mathbf{e}}_x$.

Transforming the electromagnetic field tensor by $F^{\prime }=\Lambda F{\Lambda }^T$ and reading off its components givesThese are the parallel and transverse field transformation laws for a boost in the $x$ direction.

Two equal opposite line charge densities moving in opposite directions have zero net line charge but a nonzero total current $I$. Under a longitudinal Lorentz boost, the charge-current four-vector gives ${\lambda }^{\prime }=\gamma (\lambda -vI/c^2)=-\gamma vI/c^2$. Equivalently, for lab densities $n$ and speeds $\pm u$, their separate boosted number densities are $\gamma n(1\mp vu/c^2)$. Their imbalance creates the boosted radial electric field, consistently with Lorentz transformation of electromagnetic fields and Gauss's law.