\(B=-i\cfrac{\partial E}{\partial x^{'}}\)
the negative sign is the result of \(E\) pointing towards the negative charge in circular motion. And the complex imaginary \(i\) is the result of rotation \(\cfrac{\pi}{2}\) anti-clockwise. And \(x^{'}\) is in the plane perpendicular to direction \(x\)
In case of photons in motion,\(B= i\cfrac{\partial E}{\partial x^{'}}\)
These equations prove one thing, I am a complete WHACKO. Because,
\(B=i\cfrac { \partial E }{ \partial x^{'} } \)
\( \cfrac { \partial B }{ \partial t } =i\cfrac { \partial ^{ 2 }E }{ \partial x^{'}\partial t } \)
\( \cfrac { \partial B }{ \partial t } =i\cfrac { \partial ^{ 2 }E }{ \partial x^{'} \partial x^{'}}\cfrac{\partial x^{'}}{\partial t } \)
Since substituting for \(B\) again,
\(i\cfrac { \partial }{ \partial t } \cfrac { \partial E }{ \partial x^{'} } =i\cfrac { \partial ^{ 2 }E }{ {\partial x^{'}}^{ 2 } } \cfrac{\partial x^{'}}{\partial t }\)
Multiplying both sides by \(i\) rotates both \(B\) and \(E\) again by \(\cfrac{\pi}{2}\) anti-clockwise, into the direction of \(x\), ie \(x^{'} \rightarrow x\),
\(-\cfrac { \partial }{ \partial t } \cfrac { \partial E }{ \partial x } =-\cfrac { \partial ^{ 2 }E }{ {\partial x}^{ 2 } } \cfrac{\partial x}{\partial t }\)
since \(\cfrac{\partial x}{\partial t } = c\)
\(\cfrac { \partial ^{ 2 }E }{ \partial t \partial t } \cfrac { \partial t }{ \partial x } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } c\)
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } \cfrac { 1 }{ c } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } c\)
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } c^{ 2 }\)
Which is the wave equation for photons at speed \(c\). This wave equation was derived from one relationship between \(B\) and \(E\), this suggest that a dipole with one end spinning is all that is required for wave propagation. In this case of a photon, the positive end is in circular motion behind the negative charge, both moving against a E-field. Furthermore, for the case of charge/dipole
\(B=-i\cfrac { \partial E }{ \partial x^{'} } \)
\( -\cfrac { \partial B }{ \partial t } =i\cfrac { \partial ^{ 2 }E }{ \partial x^{'}\partial t } \)
\( -\cfrac { \partial B }{ \partial t } =i\cfrac { \partial ^{ 2 }E }{ \partial x^{'} \partial x^{'}}\cfrac{\partial x^{'}}{\partial t } \)
Substituting for \(B\) again,
\(-i\cfrac { \partial }{ \partial t }(- \cfrac { \partial E }{ \partial x^{'} }) =i\cfrac { \partial ^{ 2 }E }{ {\partial x^{'}}^{ 2 } } \cfrac{\partial x^{'}}{\partial t }\)
\(i\cfrac { \partial }{ \partial t } \cfrac { \partial E }{ \partial x^{'} } =i\cfrac { \partial ^{ 2 }E }{ {\partial x^{'}}^{ 2 } } \cfrac{\partial x^{'}}{\partial t }\)
Multiplying both sides by \(i\) rotates both \(B\) and \(E\) again by \(\cfrac{\pi}{2}\) anti-clockwise, into the direction of \(x\), ie \(x^{'} \rightarrow x\) and since \(\cfrac{\partial x}{\partial t }=v\)
\(\cfrac { \partial ^{ 2 }E }{ \partial t \partial t } \cfrac { \partial t }{ \partial x } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v\)
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } \cfrac { 1 }{ v } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v\)
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v^{ 2 }\)
This is the wave equation for electromagnetic wave at speed \(v\). In this case, the negative end of the dipole is in circular motion behind the positive charge, both moving in the direction of the E field. The initial negative sign cancels when \(B\) is substituted again into the derivation. The negative sign, however indicates that
\(B=-i\cfrac{\partial E}{\partial x}\)
obeys Lenz's Law, that work is done against the change taking place; that electromagnetic wave established between two point requires continued energy input. The wave disappears when power is lost. In the case of photons
\(B= i\cfrac{\partial E}{\partial x}\)
runs away. Photons continues to propagate after source power has been turned off. Photons are free from the source once it leaves the source.
