\(B=-i\cfrac { \partial \, E }{ \partial \, x^{ ' } } \)
\( B=-i\cfrac { i }{ i } \cfrac { \partial \, E }{ \partial \, x^{ ' } } =\cfrac { \partial \, E }{ \partial \, x } \)
\( \cfrac { \partial \, B }{ \partial \, t } =\cfrac { \partial ^{ 2 }E }{ \partial \, x\partial \, t } =\cfrac { \partial ^{ 2 }E }{ \partial \, x\partial \, x } \cfrac { \partial \, x }{ \partial \, t } \)
Since, \(\cfrac{\partial\,x}{\partial\,t}=c\) and substitute for \(B\) again,
\( \cfrac { \partial \, }{ \partial \, t } \left\{ \cfrac { \partial \, E }{ \partial \, x } \right\} =\cfrac { \partial ^{ 2 }E }{ \partial \, x^{ 2 } } c\)
\( \cfrac { \partial \, }{ \partial \, x } \left\{ \cfrac { \partial \, E }{ \partial \, t } \right\} =\cfrac { \partial \, }{ \partial \, t } \left\{ \cfrac { \partial \, E }{ \partial \, t } \right\} \cfrac { \partial \, t }{ \partial \, x } =\cfrac { \partial ^{ 2 }E }{ \partial \, x^{ 2 } } c\)
\( \cfrac { \partial ^{ 2 }E }{ \partial \, t^{ 2 } } \cfrac { 1 }{ c } =\cfrac { \partial ^{ 2 }E }{ \partial \, x^{ 2 } } c\)
\( \cfrac { \partial ^{ 2 }E }{ \partial \, t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial \, x^{ 2 } } c^{ 2 }\)
we have a wave from the relationship between \(B\) and \(E\) alone. This was previously derived in the post "Whacko and the Free Photons".
Similarly for a dipole with the positive end spinning,
\(B=i\cfrac { \partial \, E }{ \partial \, x^{ ' } } \)
\( B=i\cfrac { i }{ i } \cfrac { \partial \, E }{ \partial \, x^{ ' } } =-\cfrac { \partial \, E }{ \partial \, x } \)
\( \cfrac { \partial \, B }{ \partial \, t } =-\cfrac { \partial ^{ 2 }E }{ \partial \, x\partial \, t } =-\cfrac { \partial ^{ 2 }E }{ \partial \, x\partial \, x } \cfrac { \partial \, x }{ \partial \, t } \)
Since, \(\cfrac{\partial\,x}{\partial\,t}=c\) and substitute for \(B\) again,
\( \cfrac { \partial \, }{ \partial \, x } \left\{ \cfrac { \partial \, E }{ \partial \, t } \right\} =\cfrac { \partial \, }{ \partial \, t } \left\{ \cfrac { \partial \, E }{ \partial \, t } \right\} \cfrac { \partial \, t }{ \partial \, x } =\cfrac { \partial ^{ 2 }E }{ \partial \, x^{ 2 } } c\)
\( \cfrac { \partial ^{ 2 }E }{ \partial \, t^{ 2 } } \cfrac { 1 }{ c } =\cfrac { \partial ^{ 2 }E }{ \partial \, x^{ 2 } } c\)
\( \cfrac { \partial ^{ 2 }E }{ \partial \, t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial \, x^{ 2 } } c^{ 2 }\)
we also have a wave. And
\(\cfrac { \partial \, B }{ \partial \, t } =i\cfrac { \partial \, E }{ \partial x^{ ' } } \)
\( B=i\int {\cfrac { \partial \, t }{ \partial x^{ ' } } \cfrac { \partial \, E }{ \partial x^{ ' } } } \cfrac { \partial \, x^{ ' } }{ \partial \,t }d\, t\)
\( B=i\int { \cfrac { \partial \, t }{ \partial x^{ ' } }\partial \, E } \)
\( B=\cfrac { i\,E }{ c } \) assuming \(B=0\) when \(E=0\)
where \(i\) rotates \(E\) into the \(x^{'}\) direction which is also the \(B\) and \(c\) direction. \(E\) is due to a positive charge. This expression is the same as that for a spinning negative dipole when we also consider the direction of \(E\) that has been reversed because the spinning charge is now positive. If we denote \(B_{ph}\) as the B field due to a spinning positive charge dipole and \(B_{em}\) as the B field due to a spinning negative charge dipole,
\(B_{ph}=-B_{em}\) --- (*)
Given,
\(\cfrac{1}{c}=\sqrt{\mu_o\varepsilon_o}\), we have
\( B^2=\cfrac { (i)^2}{ c^2 }E^2 \)
\( B^2=-\mu_o\varepsilon_oE^2 \)
\( B^2=\mu_o\varepsilon_oE^2e^{i\pi} \)
or,
\( B^2=\mu_o\varepsilon_o(Ee^{i\pi/2})^2 \)
that \(B\) and \(E\) are \(\pi/2\) out of phase. And so,
\(\cfrac{1}{2\mu_o}\left|B\right|^2=\cfrac{1}{2}\varepsilon_o\left|E\right|^2 \)
which is the same as the expression for a spinning negative charge dipole. The expression for energy density, \(\psi\) in the case of a spinning positive charge dipole is also the same as that for a spinning negative charge dipole. Which means both the \(\psi\) waves discovered previously are also found around a spinning positive charge dipole.
How does then, a spinning negative charge dipole interact with a spinning positive charge dipole when they both have the same expression for \(\psi\) but a reversed \(E\) ?
We consider their interaction through their \(B\) fields because the charges are all in motion. From (*), when the dipoles are in parallel with velocity in the same direction they repel each other and when they are in parallel but travelling in opposite directions they attract each other.
This is the reverse of parallel moving charges of the same polarity. This might open up the possibility of manipulating photons with EMW.
