= Solution
The product of the magnetic field with the velocity in this question is the three-dimensional <cross product>. Fix $\epsilon_{123}=1$ and write $K^i{}_j=\epsilon^i{}_{kj}B^k$, 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>:
$$
\boxed{\Gamma^i{}_{00}=-E^i,\qquad\Gamma^i{}_{0j}=\Gamma^i{}_{j0}=-K^i{}_j=\epsilon^i{}_{jk}B^k,}
$$
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 $\lambda$, satisfies
$$
t''=0,\qquad \mathbf x''-\mathbf E(t')^2-2t'\mathbf B\times\mathbf x'=0.
$$
On the branch $t'=c\ne0$, use $t$ itself as an <affine parameter>. Dividing the spatial equation by $c^2$ gives
$$
\boxed{\frac{d^2\mathbf x}{dt^2}=\mathbf E+2\mathbf B\times\frac{d\mathbf x}{dt}.}
$$
Conversely, 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 $t'=0$ 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 form>
$$
F=\iota_{\mathbf B}(dx^1\wedge dx^2\wedge dx^3)=\frac12F_{ij}\,dx^i\wedge dx^j,\qquad F_{ij}=\epsilon_{ijk}B^k.
$$
Then $dF=(\nabla\cdot\mathbf B)\,dx^1\wedge dx^2\wedge dx^3=0$. The global <Poincare lemma> on the contractible space $\mathbb R^3$ supplies a one-form $A=A_i dx^i$ 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> is
$$
\boxed{A_i(\mathbf x)=\int_0^1 s x^jF_{ji}(s\mathbf x)\,ds,\qquad\mathbf A(\mathbf x)=\int_0^1s\,\mathbf B(s\mathbf x)\times\mathbf x\,ds.}
$$
This is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf A=\tfrac12\mathbf B\times\mathbf x$.
Consider the <Eisenhart-Duval lift> with the <Lorentzian metric>
$$
g=d\mathbf x^2+2dt\bigl(du-2A_i dx^i-Udt\bigr).
$$
The independent one-forms $dx^1,dx^2,dx^3,dt,du-2A_i dx^i-Udt$ 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> $\lambda$ is
$$
L=\frac12|\mathbf x'|^2-U(t')^2-2A_i x'^i t'+t'u'.
$$
The cyclic coordinate $u$ gives the <conserved quantity> $p_u=\partial L/\partial u'=t'$. Work at a nonzero value of $p_u$ and rescale the <affine parameter> to set $t'=1$. 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 $t'=1$, are
$$
x''_i+\partial_iU(t')^2+2(\partial_iA_j-\partial_jA_i)x'^j t'-2A_i t''=0.
$$
Since $t''=0$ and $F_{ij}v^j=-(\mathbf B\times\mathbf v)_i$, their reduction is
$$
\boxed{\ddot{\mathbf x}=-\nabla U+2\mathbf B\times\dot{\mathbf x}.}
$$
Equivalently, after taking $t$ as the <affine parameter>, the term $\dot u$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\tfrac12|\dot{\mathbf x}|^2-2\mathbf A\cdot\dot{\mathbf x}-U$. Its <Euler-Lagrange equations> give the same sign and factor two.
There is also an explicit converse using null <geodesics>. For any physical trajectory, set
$$
\boxed{\dot u=U+2\mathbf A\cdot\dot{\mathbf x}-\frac12|\dot{\mathbf x}|^2.}
$$
This makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomes
$$
p_t=-2U-2\mathbf A\cdot\dot{\mathbf x}+\dot u=-\left(U+\frac12|\dot{\mathbf x}|^2\right).
$$
The 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. \b[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 $du-2A_i dx^i-Udt$ and the five-dimensional metric are unchanged. These <gauge transformations of an Eisenhart-Duval lift> alter the reduced Lagrangian only by the total derivative $-2d\chi/dt$.
Back to article page
