Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 56 / 3

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 $-2d\chi /dt$.