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An electric field is defined through the electric force on a stationary electric charge: $\mathbf{F}=q\mathbf{E}$. Therefore the electric-field dimension is

By cylindrical symmetry the electric field of an infinite line charge is radial, independent of $z$ and angle. Use a coaxial cylindrical Gaussian surface of radius $r$ and length $L$. The end caps have zero flux, so Gauss's law gives $2\pi rL{E}_r=\mu L/{\epsilon }_0$. Hence the field and electrostatic potential areHere $r_0>0$ is an arbitrary reference length, and $\mathbf{E}=-\nabla \varphi$. One cannot choose zero electrostatic potential at infinity for a single infinite line charge, because its logarithmic potential diverges there; the arbitrary reference handles the additive constant.

For the conductor in electrostatic equilibrium, the interior electric field is zero and the electrostatic potential is constant. Earthing fixes that constant to zero. Thus the conductor boundary conditions at $x=d$ areThe normal electric field need not vanish; it is supported by a surface charge density. The outward normal from this conductor into the exterior is $\mathbf{n}=-{\mathbf{e}}_x$, so the normal jump condition isThis sign convention will determine the sign of the induced surface charge density below.

For a moving closed circuit $C(t)$ bounding an oriented surface $S(t)$, Faraday's law isHere $\mathbf{E}$ is the electric field, $\mathbf{B}$ the magnetic field, $\mathbf{v}$ the local velocity of a circuit element, and ${\mathcal{E}}_{\mathrm{em}}$ the electromagnetic electromotive force per unit charge around the circuit. The line orientation and normal $\mathbf{n}$ obey the right-hand rule. The flux derivative is a total derivative and includes both changing field and changing surface. The term $\mathbf{v}\times \mathbf{B}$ is the motional electromotive force. A battery contributes an additional non-electromagnetic source electromotive force.

For the oscillatory magnetic field, Faraday's law gives $\mathbf{B}=(1/\omega )\mathbf{k}\times \mathbf{E}$. Indeed its time derivative equals $-\nabla \times \mathbf{E}$, and the dispersion relation makes it satisfy the remaining Maxwell equations. ThusFor $\omega =ck$, its amplitude is $E_0/c$. The magnetic field, electric field and direction of propagation are mutually perpendicular. A constant background magnetic field could be added; the expression above is the wave-associated field.

In electrostatics, Gauss's law is ${\oint }_{\partial \Omega }\mathbf{E}\cdot \mathbf{n}dS=Q_{\mathrm{enclosed}}/{ϵ}_0$, or locally $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, where ${\rho }_e$ is charge density and ${ϵ}_0$ is vacuum permittivity. Spherical symmetry makes the electric field radial, and Gauss's law givesChoose the additive constant of the electrostatic potential by $\varphi (\infty )=0$. Using $\mathbf{E}=-\nabla \varphi$ and continuity of potential across each charged shell givesThe field jumps at the shell surfaces, so the displayed formulas are for the requested open regions. The potential difference of the spherical capacitor is $\Delta \varphi =Q(1/a-1/b)/(4\pi {ϵ}_0)$. Thus its capacitance and electrostatic energy areThe energy follows either by charging work ${\int }_0^{Q}q/Cdq$, or directly from $W=({ϵ}_0/2)\int |\mathbf{E}{|}^2{d}^{3}x=Q^2(1/a-1/b)/(8\pi {ϵ}_0)$.

Write $R=|\mathbf{p}|>a$. In SI units, the exterior electrostatic potential solveswith the point-charge singularity $q/(4\pi {\epsilon }_0|\mathbf{x}-\mathbf{p}|)$. The image charge for a grounded conducting sphere choiceslie inside the sphere. On $|\mathbf{x}|=a$, direct squaring gives $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$. Thereforesatisfies the source equation, grounded boundary value and decay at infinity. The Uniqueness of the Dirichlet problem, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior electric field.

The charge feels the induced field, equivalently the field of the image charge, excluding its own singular field. Their separation is $(R^2-a^2)/R$, so Coulomb's law givesIts magnitude is $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$ and its direction is toward the sphere's center, independent of the sign of $q$. No factor one-half belongs in this force; that factor occurs in induced electrostatic energy instead.

Use $t_{\mu }=(1,-1,-1,-1{)}_{\mu }$ without summing in individual component transformations. The Dirac current transforms asSince the coupling is real, invariance of its contraction fixes the time reversal of an electromagnetic gauge field:These equations use a compatible gauge; an additional pure gauge transformation would not change the gauge field strength.

Coordinate differentiation at $x_T$ introduces $r_{\mu }=-t_{\mu }$, because the time coordinate is reversed and the spatial coordinates are unchanged. HenceThus the electric field is even and the magnetic field is odd. The chirality matrix is even by part (a). The two tensor signs cancel in the dipole contraction, but antiunitarity conjugates its explicit $i$. Therefore the time-reversal parity of a fermion electric dipole operator isIt is a time-reversal symmetry violating interaction, and, under the usual local relativistic quantum field theory hypotheses of the CPT theorem, it violates CP symmetry. A nonzero coefficient is absent from the renormalizable tree-level Standard Model Lagrangian: the broken-phase fermion electric dipole moment operator has mass dimension five, and its electroweak-invariant completion requires a Higgs field and has dimension six. However, it can arise as a radiatively induced effective interaction in the Standard Model, whose CKM matrix contains a CP-violating phase. It is not forbidden to all orders. An explicit primary calculation of quark dipoles is Czarnecki and Krause's Standard Model calculation.

Use metric $(+,-,-)$ and the paper's normalization. A static multi-vortex solution is a smooth finite-energy field configuration of Abelian Higgs vortices whose complex Higgs field has winding at spatial infinity and zeros in the plane. In a positive $N$-vortex sector, the zeros have total multiplicity $N$ and the magnetic flux is $2\pi N$; they may be separated or coincident. Finite energy requires $|\varphi |\to 1$, $D_i\varphi \to 0$ and $B=f_{12}\to 0$ at infinity. Then $a_i$ asymptotically compensates the phase gradient, so Stokes' theorem relates flux to the phase winding. The positive Bogomolny vortex equations describe this same-sign sector; negative flux uses the opposite signs and is not covered by simply mixing positive and negative zero multiplicities.

For the critically coupled Abelian Higgs field equations, vary $\varphi$ and $\bar{\varphi }$ independently. After covariant integration by parts,Thus the scalar equation and its complex conjugate areVarying the gauge field gives ${\partial }_{\mu }{f}^{\mu \nu }\delta a_{\nu }$ from the Maxwell term. The matter variation isConsequentlyThese factors use the scalar kinetic coefficient $1/2$; importing the current from a convention with coefficient one would give an incorrect factor two. The current is invariant under gauge transformations, and its conservation follows from the field equations and antisymmetry of $f^{\mu \nu }$.

For the static purely magnetic solutions take temporal gauge $a_0=0$. The scalar equation reduces toSince $[D_1,D_2]=-iB$, applying $D_1-i{D}_2$ to the first Bogomolny vortex equation givesThe second equation sets $B=(1-|\varphi {|}^2)/2$, proving the scalar equation, including at smooth vortex cores without dividing by $\varphi$.

For the spatial gauge equations, put $k_i=\mathrm{Im}(\bar{\varphi }{D}_i\varphi )$. The first equation implies $D_1\varphi =-i{D}_2\varphi$, and ${\partial }_i|\varphi {|}^2=2\mathrm{Re}(\bar{\varphi }{D}_i\varphi )$. ThereforeWith the stated metric, $j^i=-k_i$ and the two static Maxwell equations are ${\partial }_{2}B=k_1$ and ${\partial }_{1}B=-k_2$. Both follow by differentiating $B=(1-|\varphi {|}^2)/2$. The time component is the Gauss law constraint in gauge theory; it is identically satisfied because both the electric field and $j^0$ vanish in this static temporal gauge. HenceAs an energy check, the Bogomolny square completion for an Abelian Higgs vortex gives $E\ge \pi N$ in this normalization, saturated by these equations. On the plane, the Abelian Higgs vortex moduli space admits arbitrary vortex positions, including coincidence; its static energy is $\pi N$, so there is no position-dependent static potential. Their slow dynamical interactions are a different question.

The logarithmic form of a Bogomolny vortex is only local away from zeros. Write $\varphi =e^{u+i\chi }$ there, with $u=\log |\varphi |$. ThenSeparating the real and imaginary parts of $(D_1+i{D}_2)\varphi =0$ yields $u_1-{\chi }_2+a_2=0$ and ${\chi }_1-a_1+u_2=0$. Hencewhere $J(v_1,v_2)=(-v_2,v_1)$ is the positive quarter-turn. Under a gauge transformation, $\chi \mapsto \chi +\omega$ and $\mathbf{a}\mapsto \mathbf{a}+\nabla \omega$, as this relation requires.

At regular contour points, the stipulated orthogonality means $\nabla \chi \cdot \nabla u=0$. Since $J\nabla u$ is also perpendicular to $\nabla u$,Thus, in that gauge, the gauge potential is tangent to contours of $u$, equivalently contours of $|\varphi |$, and is parallel to the phase gradient wherever that gradient is nonzero. This is a gauge-dependent statement about the vector potential, not an assertion that the magnetic field is tangent to these curves, nor that the potential is a pure gradient or divergence-free. The two gradient magnitudes are not fixed by contour orthogonality.

On a phase patch $B={\partial }_1{a}_2-{\partial }_2{a}_1=-\Delta u$. Combining with the second equation gives $\Delta u+(1-e^{2u})/2=0$ away from the cores. At a zero of multiplicity $n_j$, $u\sim n_j\log |x-X_j|$, and the phase has winding $2\pi n_j$; $u$ is not a finite real smooth field there. Distributionally, the full equality isThis is the Taubes equation in the paper's amplitude logarithm, rather than the often-used $\log |\varphi {|}^2=2u$ convention. No single smooth globally defined phase or regular contour description at every zero has been assumed.

The normalized Bloch states $|k⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{ikna}|n⟩$ satisfy periodicity when $k=2\pi j/(Na)$. Acting with the hopping terms shifts the coefficients by one site, givingThe first Brillouin zone is a fundamental cell of reciprocal space, for example $[-\pi /a,\pi /a)$; wavevectors differing by $2\pi /a$ label the same state. It contains exactly $N$ distinct allowed wavevectors and hence $N$ orbital states. The written Hamiltonian has no spin index; if electron spin degeneracy is included there are $2N$ states.

For a narrow wave packet, group velocity and band-curvature effective mass areFor the usual $J>0$, the mass is negative in the outer half of the zone, $\pi /(2a)<|k|\le \pi /a$, taking one endpoint representative. At $|k|=\pi /(2a)$ the inverse mass vanishes and the mass diverges. For arbitrary nonzero $J$, the sign criterion is $J\cos ka<0$; $J=0$ is a flat band.

For electron charge $-e$ with $e>0$ in a constant electric field $\mathcal{E}$, the single-band semiclassical equations are $\hslash \dot{k}=-e\mathcal{E}$, $\dot{x}=v(k)$. Thus $k(t)=k_0-e\mathcal{E}t/\hslash$, and for $\mathcal{E}\ne 0$,These are Bloch oscillations of period $2\pi \hslash /(|e\mathcal{E}|a)$. The zero-field limit is $x-x_0=v(k_0)t$. This assumes a narrow packet and neglects scattering and interband transitions.

For many electrons obeying the Pauli exclusion principle, a partially filled band has available nearby states and can carry current under a displaced occupation distribution: it describes a metal within band theory. A completely filled band has cancelling velocities over the zone, so its current is zero; if separated from empty higher bands by a nonzero gap, it describes an insulator at sufficiently low temperature and weak electric field. With spin degeneracy this isolated band is filled at two electrons per site. The single-electron Hamiltonian alone does not specify the filling or a gap to a next band, so those extra physical assumptions are required for classifying a material.

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.

For variations vanishing at both endpoints, differentiating the action and using integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationsFor the specified Lagrangian, the gauge transformation changes it byFor arbitrary $F$, it is a total derivative precisely when $\boxed{\alpha =-1}$. The corresponding change in action is then an endpoint term whose variation is zero. In general time-dependent notation, the equations areThe electric field $-\nabla \varphi -{\partial }_t\mathbf{A}$ and magnetic field $\nabla \times \mathbf{A}$ are unchanged by the gauge transformation with $\alpha =-1$. Although the original potentials have no time dependence, their transformed versions generally do; allowing this is essential to the stated invariance.

A scalar coupling linear in the nematic order parameter requires a second-rank anisotropy. A single electric field supplies no such tensor, while $E_i{E}_j$ does. Its isotropic part gives $E^2\mathrm{Tr}Q=0$. Hence the lowest joint order in $E$ and $Q$ of the electric-field coupling to nematic order isTerms depending only on $E$ do not affect minimization over $Q$. With $\vartheta$ the angle between the nematic director and the electric field,For $\chi \lambda >0$, the lowest free energy occurs at ${\cos }^2\vartheta =1$, so the nematic director aligns along either $+\mathbf{E}$ or $-\mathbf{E}$. Then $E_i{Q}_{ij}{E}_j=\lambda E^2$, givingUnlike the linear composition term in a conserved binary fluid mixture, this term changes the equilibrium nematic order parameter: its spatial integral is not fixed.

Within this analytic quartic Landau theory, equilibrium of a nonconserved scalar order parameter requires $f^{\prime }=0$. At a continuous phase transition, its restoring curvature vanishes, so $f^{\prime \prime }=0$. A nonzero $f^{\prime \prime \prime }$ would leave a cubic leading term in the Taylor expansion about that state and prevent it from being a local minimum. Thus a stable critical endpoint of a quartic Landau free energy requires all three derivatives to vanish, with $f^{\prime \prime \prime \prime }>0$. The condition $f^{\prime \prime }=0$ alone can instead identify a spinodal point.

For $f=\bar{b}{\lambda }^4+\bar{c}{\lambda }^3+\bar{a}{\lambda }^2+\bar{d}\lambda$, $\bar{b}>0$, the successive equations areThey giveAt these values, the free energy is exactly $f({\lambda }_c)+\bar{b}(\lambda -{\lambda }_c{)}^4$, proving stability. With no electric field, $\bar{d}=0$, so necessarily $\bar{c}=0$, followed by ${\lambda }_c={\bar{a}}_c=0$.

For $\bar{c}<0$ and $\chi >0$, the electric-field coupling to nematic order supplies the required negative ${\bar{d}}_c$ atTuning the quadratic coefficient to ${\bar{a}}_c$ simultaneously reaches the field-induced critical endpoint of the isotropic-nematic transition. The PDF's $\tilde{c}$ in the last clause is undefined; it is interpreted here as the previously defined $\bar{c}$. The endpoint is a prediction of the stable quartic, small-field mean-field approximation; higher-order terms or a field beyond the expansion's validity can shift it.

Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrodinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

The electric field is force per unit electric charge, so

The operator $ia\bar{\psi }{\sigma }^{\mu \nu }{\gamma }^5\psi F_{\mu \nu }$ is the relativistic fermion electric dipole moment operator. Its nonrelativistic limit contains $\mathbf{S}\cdot \mathbf{E}$: spin angular momentum is an axial vector, whereas the electric field is a polar vector. It is consequently odd under parity symmetry in quantum field theory. Both the pseudotensor fermion bilinear and the electromagnetic field tensor are odd under charge conjugation, so their product is even. ThereforeThe CPT theorem then makes it odd under time-reversal symmetry. Such an interaction can arise only from CP violation. The Cabibbo-Kobayashi-Maskawa matrix therefore induces a nonzero Standard Model electric dipole moment at sufficiently high loop order, but its flavor structure and loop suppressions make the result extraordinarily small.