Either \(i\) or \(-i\) can be used to denote the change in direction of \(B\) relative to \(E\) as \(B\) is circular around \(E\).
What if \(\nabla.\cfrac{1}{c}\ne0\)? Then we have changing space density and thing gets very interesting.
From the post "Pointing Where? No Side Issue, Please", a wire was shown to radiate from one of its end in semi-spherical wave fronts. The maths strictly speaking, models the wire without any other dimension but along \(x\). It is an infinitely thin piece of wire. The wave should be an equally thin beam shooting off the end of the wire. A second wave as suggested by the presence of \(Y\) spreads this thin line of wave into a inverted cone with a planar wave front.
Which might just be the actual case in reality. The wave equation constrained to \(x\)
\(\nabla ^{ 2 }B=\cfrac { 1 }{ c^{ 2 } } \cfrac { \partial ^{ 2 }B }{ \partial t^{ 2 } } \)
applies right up the end of the wire, thereafter continuity conditions apply at the air-wire boundary. The tangential component of \(E\) field must be equal before and after the boundary and \(B\) is continuous. So, both \(E_1=E_2\) and \(\cfrac{B_1}{\mu_1}=\cfrac{B_2}{\mu_2}\) is continuous across a smooth boundary perpendicular to the direction of the wave, since both components are parallel to the boundary surface. There is really no reason for the wave to spread into a semi-spherical front. If this second wave is the reason for EMW to spread, then \(X\) and \(Y\) must exist. They are charge loops and vortexes.
Note: The boundary conditions at the radiating end of the wire determines the wave front so formed.
