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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.

Use a Lorentz boost along positive $x$ with $v=E/B$. Since $E<cB$, its dimensionless speed $\beta =E/(cB)$ is less than one. The given transformation of the electric field and magnetic field yields $E_y^{\prime }=\gamma (\beta )(E-vB)=0$ andAll other electric components vanish. In this frame a magnetic field does no work, so the particle's speed $u<c$ and Lorentz factor ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$ are constant. In proper time, the Lorentz force equations are $d({\gamma }_u{u}_x^{\prime })/d\tau =(q{B}^{\prime }/m){\gamma }_u{u}_y^{\prime }$ and $d({\gamma }_u{u}_y^{\prime })/d\tau =-(q{B}^{\prime }/m){\gamma }_u{u}_x^{\prime }$. Thus, choosing the phase and origin,This is the signed proper-time cyclotron frequency; the frequency with respect to $t^{\prime }$ is $\omega /{\gamma }_u$. The inverse Lorentz boost givesIn particular the average drift speed is $\beta c=E/B$ along $\mathbf{E}\times \mathbf{B}$.

For $u>0$, $q\ne 0$, set $s=\omega \tau$ and $d=\beta c/u$. The dimensionless spatial curve isThe drift between successive cycles is $2\pi \gamma (\beta )d$. When $2\pi \beta \ll u/c$, this is small compared with the oscillation size: the curve consists of nearly closed loops with slow rightward drift. When $2\pi \beta \gg u/c$, the longitudinal drift is large, producing a progressing oscillatory curve. More precisely $d\tilde{x}/ds=\gamma (\beta )(d+\cos s)$: loops occur for $d<1$, cusps at $d=1$, and monotone progression for $d>1$. The stated large-drift inequality compares the drift over a full cycle, so an example of that regime with $d>1$ is shown; that comparison alone does not replace the exact loop criterion.

Reversing $q$ reverses the cyclotron orientation; the drift remains along positive $x$. If $u=0$ the trajectory is pure straight drift and the dimensionless coordinates are undefined. If $q=0$ the particle is inertial and the displayed $1/\omega$ parametrization must be replaced by its inertial limit.

A Bogomolny bound expresses a static field energy as nonnegative squares plus a term fixed by the topological charge or boundary data. Setting the squares to zero gives the first-order Bogomolny equations. Their solutions minimize the energy in that sector and satisfy the second-order Euler-Lagrange field equations, although a general stationary solution need not attain the bound.

For one real scalar in one spatial dimension, take the Lagrangian densityA finite-energy field configuration in this static sector must approach scalar-field vacua ${\varphi }_{\pm }$ with $W^{\prime }({\varphi }_{\pm })=0$ at the two ends. Completing the square gives the square completion for a one-dimensional kink:Choose the sign for which the boundary term is nonnegative. ThusTaking the derivative of the equality equation gives ${\varphi }_{xx}=W^{\prime }{W}^{\prime \prime }=d([W^{\prime }{]}^2/2)/d\varphi$, the static Euler-Lagrange field equation. The boundary term is invariant under deformations keeping the asymptotic scalar-field vacua fixed; it is not a contribution from the local shape of the kink.

For a phi-four kink, let $c>0$, $W(\varphi )=c^2\varphi -{\varphi }^3/3$, and select the sector ${\varphi }_{-}=-c$, ${\varphi }_{+}=c$. Then $W^{\prime }=c^2-{\varphi }^2$, and the increasing Bogomolny equation is ${\varphi }_x=c^2-{\varphi }^2$. Separating variables yieldsThe integration constant $X$ is the translational collective coordinate. Directly, ${\varphi }_K^{\prime }=c^2{\mathrm{sech}}^2(c(x-X))$ and its energy density is $c^4{\mathrm{sech}}^4(c(x-X))$, whose integral is $4c^3/3$. The decreasing antikink has the reversed boundary values, profile $-{\varphi }_K$, and the same energy. A Lorentz boost produces the exact uniformly moving kink $c\tanh [c\gamma (x-X_0-vt)]$, with Lorentz factor $\gamma =(1-v^2{)}^{-1/2}$ and energy $\gamma M$.

In two spatial dimensions, choose the Abelian Higgs model at critical coupling, with $D_i={\partial }_i-i{A}_i$, $B={\partial }_1{A}_2-{\partial }_2{A}_1$, and energyThe gauge covariant derivative is used throughout this normalization. Integration by parts, using $[D_1,D_2]=-iB$, gives $\int |D_i\varphi {|}^2=\int |(D_1+i{D}_2)\varphi {|}^2+\int B|\varphi {|}^2$, with a vanishing boundary divergence for the decaying vortex fields. Combining this identity with the magnetic and potential terms gives the Bogomolny square completion for an Abelian Higgs vortex:Finite-energy field configurations have $|\varphi |\to 1$ at infinity, and $D_i\varphi \to 0$ makes the magnetic flux equal to the phase winding $2\pi N$. For $N>0$,These are the Bogomolny vortex equations for an Abelian Higgs vortex. Opposite signs give antivortices and the bound $E\ge \pi |N|$. The coefficient $\pi$ follows from the explicit energy normalization above; other conventions can give $2\pi$. Static solutions of fixed positive vortex number have equal energy independent of their positions, giving the Abelian Higgs vortex moduli space used for slow dynamics.

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.

For ${\varphi }_K(x-X)=c\tanh (c(x-X))$ in the unit kinetic normalization, $\int ({\varphi }_K^{\prime }{)}^{2}dx=M=4c^3/3$. Substitution of $X(t)$ therefore gives the free-particle collective-coordinate effective Lagrangian $-M+M{\dot{X}}^2/2$. Its quantization has momentum $p$ and energy $M+p^2/(2M)$ to this order. The exact uniformly moving classical kink is a Lorentz boost of the static one, with energy $M/\sqrt{1-v^2}$.