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The Coulomb gauge imposeson the magnetic vector potential. A gauge transformation $A\mapsto A+\nabla \chi$ reaches this gauge when $\chi$ solves

The four-potential combines the electric potential and magnetic vector potential as $A^{\mu }=(\varphi /c,\mathbf{A})$. The electromagnetic four-potential $A_{\mu }$ transforms underwhile the antisymmetric field tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is unchanged.

For a thin wire carrying current $I$, the given magnetic vector potential reduces to $\mathbf{A}=({\mu }_{0}I/4\pi ){\oint }_{C}d{\mathbf{r}}^{\prime }/|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Taking its curl and using ${\nabla }_{\mathbf{r}}(1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart lawThis is exactly the sign convention in the printed form using ${\mathbf{r}}^{\prime }-\mathbf{r}$.

Let $\mathbf{v}=(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$. Away from the wire, $\nabla \cdot \mathbf{v}=0$. The supplied divergence and curl of a cross product therefore gives $\nabla \times (d{\mathbf{r}}^{\prime }\times \mathbf{v})=-(d{\mathbf{r}}^{\prime }\cdot {\nabla }_{\mathbf{r}})\mathbf{v}$. But ${\nabla }_{\mathbf{r}}\mathbf{v}=-{\nabla }_{{\mathbf{r}}^{\prime }}\mathbf{v}$, so the integrand is the total differential of $\mathbf{v}$ along the source loop. Its closed-loop integral is zero. Thus $\nabla \times \mathbf{B}=0$ at every point outside $C$, as demanded by the current-free magnetostatic Ampère's law.

For $r^{\prime }=f(\theta )>0$, $z^{\prime }=0$, and counterclockwise current, $d{\mathbf{r}}^{\prime }=(f^{\prime }{\mathbf{e}}_r+f{\mathbf{e}}_{\theta })d\theta$. At the origin, $\mathbf{v}=-{\mathbf{e}}_r/f^2$, soFor the ellipse expressed relative to a focus, $1/f=(1-e\cos \theta )/\ell$. The cosine integrates to zero, givingHere $\ell$ is the semilatus rectum. Reversing the current reverses the field direction.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincare lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.

The MHD induction equation can be written in material form asIts $y$ component immediately givesFor the Cartesian magnetic flux function, the $x,z$ components of the MHD induction equation, or the $y$ component of the magnetic vector potential equation, giveThe additive function of time in $\psi$ has no effect on $\mathbf{B}$. Choose this gauge so that $C=0$. Then the flux label is materially conserved:This is magnetic flux freezing in the two-dimensional geometry.

To prove the absence of axial motion, construct an invariant solution with $u_y=0$. Along each Lagrangian trajectory, $\psi$ is fixed, and the assumed divergence givesThus $B_y$ remains a function of $\psi$ and time alone. Its gradient stays parallel to $\nabla \psi$, so the $y$ component of the Lorentz force remains zero. The axial momentum equation isbecause $(B_x{\partial }_x+B_z{\partial }_z)\psi =0$. With $u_y=0$ initially, it remains zero. Hence neither axial force nor axial motion is generated. This flux-surface preservation during magnetic-tube expansion argument assumes the smooth ideal evolution for which the initial-value problem is unique; the particular in-plane rising motion need not be calculated.

Choose an origin near the bounded loop and write ${\mathbf{r}}^{\prime }$ for the source position. The magnetic vector potential and its far-field multipole expansion arewith the loop held fixed as $r=|\mathbf{r}|\to \infty$. The first term integrates to zero since $C$ is closed. Define its oriented vector area byThe second equality is Stokes theorem, applied componentwise to any oriented spanning surface. In particular the vector area is independent of the choice of spanning surface. From ${\oint }_{C}d(r_i^{\prime }{r}_j^{\prime })=0$, the matrix ${\oint }_C{r}_i^{\prime }d{r}_j^{\prime }$ is antisymmetric, which gives ${\oint }_C(\mathbf{r}\cdot {\mathbf{r}}^{\prime })d{\mathbf{r}}^{\prime }=\mathbf{S}\times \mathbf{r}$. HenceThe orientation follows the current by the right-hand rule. For $r\ne 0$, taking the curl of this leading magnetic vector potential gives the magnetic dipole fieldFor a planar loop, $\mathbf{S}$ is the signed area times the unit normal vector. For a nonplanar loop it is the oriented vector area, rather than the scalar area of an arbitrarily chosen surface.

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.