Wanting More, All Over And Twisted

In the last post the expression,

$\boxed{-i\cfrac{\partial E_i}{\partial x^{'}} =} \cfrac{\partial( i.-E_i )}{\partial x^{'}}=-i\cfrac{\partial E_i}{\partial x}=-i\cfrac{1}{2\pi\varepsilon_o r}.\cfrac{-\partial\lambda}{\partial x}=-i\cfrac{J}{2\pi\varepsilon_o r}=B$

$-i\cfrac{\partial E_i}{\partial x^{'}}$ 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 90^o 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 90^o 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.