In Fig. , we schematically represent the vector potential in Cartesian coordinates as a vector field that somehow "rotates" around the z-axis. The magnitude of the vector potential appears to increase as the distance from the nucleus (a proton, in Fig. , since Z = 1) which is in agreement with the interpretation of A as the current density divided by the density in our gauge. Moreover, since A ϕ only depends on the radial distance r and the polar angle ϑ, the left hand side of Eq. ( ) is itself zero, implying that, in this case, not only the divergence of the current density is zero, but the electronic vector potential (or the related current density) is orthogonal to the gradient of the density