\(\boxed{-i\cfrac{\partial E_i}{\partial x^{'}} =} \cfrac{\partial( i.-E_i )}{\partial x^{'}}=-i\cfrac{\partial E_i}{\partial x}=-i\cfrac{1}{2\pi\varepsilon_o r}.\cfrac{-\partial\lambda}{\partial x}=-i\cfrac{J}{2\pi\varepsilon_o r}=B\)
\(-i\cfrac{\partial E_i}{\partial x^{'}}\) is wrong.
is really clumsy because we know \(E\) along \(r\) to be orthogonal to \(x\) and so orthogonal to \(J\), at the same time \(B\) is orthogonal to both \(E\) and \(J\). All conforming to the right hand rule. But these relationships are expressed in part as the partial derivatives and the term \(-i\) which rotates everything clockwise 90o on the side of the equation it is applied to. The negative sign in \(\cfrac{-\partial\lambda}{\partial x}\) is because of \(E\) being negative in the original formulation. And the whole term is rotated by \(-i\) to be along \(B\) (ie. \(\partial x\rightarrow\partial x^{'}\)), eventually. So it might be clearer, to express
\(-i\cfrac{\partial E_i}{\partial x^{'}}\rightarrow \cfrac{\partial ( i. -E_i )}{\partial x^{'}}\) This is why.
instead where \(i\) is applied to \(-E_i\) to be in the direction of \(B\) along \(x^{'}\). And so
\( \cfrac{\partial ( i. -E_i )}{\partial x^{'}} = -i.\left\{\cfrac{\partial E_i}{\partial x}\right\}\)
where \(x^{'}\) has been rotated by \(i\) to \(\partial x\) in reverse as \(-i\) is brought out to apply to both denominator and numerator. (\(i .-i = 1\)). If, however \(i\) or \(-i\) were to apply to the numerator only by convention, then an expression like \(i\cfrac { \partial ^{ 2 }E }{ \partial x^{'} \partial x^{'}}\cfrac{\partial x^{'}}{\partial t }\) will be confusing because \(x^{'}\) is in both the numerator and denominator. All such \(i\) rotations, jump from axis to axis on the right hand frame.
So we have, hopefully in its final form,
\(B=\cfrac{\partial ( i.E )}{\partial x^{'}}=i\cfrac{\partial E}{\partial (-x)}\neq -i\cfrac{\partial E}{\partial x } \)
but
\(B= -i\cfrac{\partial (-E)}{\partial x } \)
Both \(E\) and \(-x\) are rotated by \(i\) to the \(x^{'}\) direction along \(B\). In the second instance, \(-E\) and \(x\) are rotated by \(-i\) to the \(x^{'}\) direction. Still \(i*i=-1\), which should be anticlockwise 90o rotation twice and not reverse, in the negative direction. This ambiguity means \(i\) should not be used as a notation for rotation. So,
\(B=-\cfrac{\partial E}{\partial x'}\)
and all directions are discussed separately. \(B\) is perpendicular to \(E\), \(x'\) is along \(B\) and \(B \times E\) gives the \(x\) direction that is perpendicular to \(x'\)
The cross product, does not lend itself to manipulation. The two terms involved are stuck on each other. The \(i\) term at least allows a partial derivative in between. But the cross product allows for non-orthogonal vectors with a \(cos(\theta)\) term, where \(\theta\) is the angle between the two vectors.
There has to be a better way that all this is represented, specially when they are written out as a computer program.
