\(\nabla .E=4\pi \rho _{ e }\)
\( \nabla .B=4\pi \rho _{ m }\)
\( -\nabla \times E=\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +\cfrac { 4\pi }{ c } j_{ m }\)
\( \nabla \times B=\cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t } +\cfrac { 4\pi }{ c } j_{ e }\)
Let's see, maybe we should start with,
\(\nabla .E=4\pi \rho _{ e }\)
\( \nabla .B=X\)
\( -\nabla \times E=-i\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +Y\)
\( -i\nabla \times B=\cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t } +\cfrac { 4\pi }{ c } j_{ e }\)
and solve for \(X\) and \(Y\). \(-i\) has been added to explicitly denote the directional relationship between \(E\) and \(B\).
First of all,
\( \nabla .\nabla \times E=0=\nabla.(i\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +Y)\)
\(i\cfrac { 1 }{ c } \cfrac { \partial \nabla.B }{ \partial t } +i\nabla.\cfrac{ 1 }{ c } \cfrac { \partial B }{ \partial t } +\nabla.Y=0\)
since, \(\nabla.\cfrac{ 1 }{ c } =0\)
\(i\cfrac { 1 }{ c } \cfrac { \partial X }{ \partial t } = \nabla.Y\) --- (1)
and
\( \nabla\times(-i\nabla \times B)=\nabla\times(\cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t } +\cfrac { 4\pi }{ c } j_{ e })\)
since,
\(\nabla\times\cfrac { 1 }{ c }=0\) and \(\nabla\times\cfrac { 4\pi }{ c }=0\)
\(-i\nabla(\nabla.B)+i\nabla^2B=\cfrac { 1 }{ c } \cfrac { \partial(\nabla\times E) }{ \partial t }+\cfrac { 4\pi }{ c }\nabla\times j_{ e }\)
since
\( \nabla .B=X\)
\(-i\nabla X+i\nabla^2B=\cfrac { 1 }{ c } \cfrac { \partial}{ \partial t }\left\{i\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +Y\right\}+\cfrac { 4\pi }{ c }\nabla\times j_{ e }\)
\(-i\nabla X+i\nabla^2B=i\cfrac { 1 }{ c^2 } \cfrac { \partial^2 B }{ \partial t^2 } +\cfrac { 1 }{ c }\cfrac { \partial Y}{ \partial t }+\cfrac { 4\pi }{ c }\nabla\times j_{ e }\)
We recognize that,
\(\nabla^2B=\cfrac { 1 }{ c^2 } \cfrac { \partial^2 B }{ \partial t^2 }\)
is the wave equation that was derived in the post "Whacko and the Free Photons".
\(-i\nabla X=\cfrac { 1 }{ c }\cfrac { \partial Y}{ \partial t }+\cfrac { 4\pi }{ c }\nabla\times j_{ e }\)
\(\nabla\times j_{ e }\) suggests that a looped charge density creates a \(\nabla.X\) component, which is consistent with a \(B\) field being created by a circular current. We can set \(j_e\) to zero and use superposition later.
\(-i\nabla X=\cfrac { 1 }{ c }\cfrac { \partial Y}{ \partial t }\) --- (2)
From (1),
\(i\cfrac { 1 }{ c } \cfrac { \partial \nabla.B }{ \partial t } = \nabla.Y\)
\(-i\cfrac { 1 }{ c }(- \cfrac { \partial B }{ \partial t }) = Y\) --- (*)
From (2),
\(-i\nabla^2.B=\cfrac { 1 }{ c }\cfrac { \partial }{ \partial t }\left\{i\cfrac { 1 }{ c }\cfrac { \partial B }{ \partial t } \right\}\)
\(\nabla^2.B=-\cfrac { 1 }{ c^2 }\cfrac { \partial^2B }{ \partial t^2 }\)
\(\nabla^2.B=\cfrac { 1 }{ (ic)^2 }\cfrac { \partial^2B }{ \partial t^2 }\)
A wave in the \(ic\) direction, ie. a wave perpendicular to \(c\), the wave in the previous wave equation. If \(X\) exist, then there is a second wave perpendicular to the normal electromagnetic wave.
Furthermore,
\(\nabla^2.(\nabla.B)=\cfrac { 1 }{ (ic)^2 }\cfrac { \partial^2\nabla.B }{ \partial t^2 }\)
\(\nabla^2 X=\cfrac { 1 }{ (ic)^2 }\cfrac { \partial^2X }{ \partial t^2 }\)
There is also a \(X\) wave in the perpendicular direction. A third wave!
Differentiating (2) wrt \(t\),
\(-i\nabla .\cfrac { \partial X }{ \partial t } =\cfrac { 1 }{ c } \cfrac { \partial ^{ 2 }Y }{ \partial t^{ 2 } } \)
Substitute (1) in,
\(-\nabla .(c\nabla .Y)=\cfrac { 1 }{ c } \cfrac { \partial ^{ 2 }Y }{ \partial t^{ 2 } } \)
\(\nabla ^{ 2 }Y=\cfrac { 1 }{ (ic)^{ 2 } } \cfrac { \partial ^{ 2 }Y }{ \partial t^{ 2 } } \)
We only have a third wave because \(Y\) and \(X\) are related.
Consider,
\( \nabla\times(-\nabla \times E)=\nabla\times(-i\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +Y)\)
\(-\nabla (\nabla .E)+\nabla ^{ 2 }E=-i\cfrac { 1 }{ c } \cfrac { \partial \nabla \times B }{ \partial t } +\nabla \times Y\)
since,
\(\nabla .E=4\pi \rho _{ e }\)
\( -4\pi \nabla \rho _{ e }+\nabla ^{ 2 }E=-i\cfrac { 1 }{ c } \cfrac { \partial }{ \partial t } \left\{ i\cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t }+i\cfrac { 4\pi }{ c } j_{ e } \right\} +\nabla \times Y\)
\(-4\pi \nabla \rho _{ e }+\nabla ^{ 2 }E=\cfrac { 1 }{ c^{ 2 } } \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } }+\cfrac { 4\pi }{ c^2 }\cfrac{\partial \, j_{ e }}{\partial t} +\nabla \times Y\)
Since,
\(\nabla ^{ 2 }E=\cfrac { 1 }{ c^{ 2 } } \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } \)
\(\nabla^2.(\nabla.B)=\cfrac { 1 }{ (ic)^2 }\cfrac { \partial^2\nabla.B }{ \partial t^2 }\)
\(\nabla^2 X=\cfrac { 1 }{ (ic)^2 }\cfrac { \partial^2X }{ \partial t^2 }\)
There is also a \(X\) wave in the perpendicular direction. A third wave!
Differentiating (2) wrt \(t\),
\(-i\nabla .\cfrac { \partial X }{ \partial t } =\cfrac { 1 }{ c } \cfrac { \partial ^{ 2 }Y }{ \partial t^{ 2 } } \)
Substitute (1) in,
\(-\nabla .(c\nabla .Y)=\cfrac { 1 }{ c } \cfrac { \partial ^{ 2 }Y }{ \partial t^{ 2 } } \)
\(\nabla ^{ 2 }Y=\cfrac { 1 }{ (ic)^{ 2 } } \cfrac { \partial ^{ 2 }Y }{ \partial t^{ 2 } } \)
We only have a third wave because \(Y\) and \(X\) are related.
Consider,
\( \nabla\times(-\nabla \times E)=\nabla\times(-i\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +Y)\)
\(-\nabla (\nabla .E)+\nabla ^{ 2 }E=-i\cfrac { 1 }{ c } \cfrac { \partial \nabla \times B }{ \partial t } +\nabla \times Y\)
since,
\(\nabla .E=4\pi \rho _{ e }\)
\( -4\pi \nabla \rho _{ e }+\nabla ^{ 2 }E=-i\cfrac { 1 }{ c } \cfrac { \partial }{ \partial t } \left\{ i\cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t }+i\cfrac { 4\pi }{ c } j_{ e } \right\} +\nabla \times Y\)
\(-4\pi \nabla \rho _{ e }+\nabla ^{ 2 }E=\cfrac { 1 }{ c^{ 2 } } \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } }+\cfrac { 4\pi }{ c^2 }\cfrac{\partial \, j_{ e }}{\partial t} +\nabla \times Y\)
Since,
\(\nabla ^{ 2 }E=\cfrac { 1 }{ c^{ 2 } } \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } \)
\(\nabla \times Y=-4\pi \nabla \rho _{ e }-\cfrac { 4\pi }{ c^2 }\cfrac{\partial \, j_{ e }}{\partial t}\)
The first factor suggests \(Y\) is because the distribution of \(\rho_e\) form a loop or vortex. The second factor suggests the current density flow forms a loop or vortex. Both of which establishes a \(B\) field.
From (2),
\(\nabla X=i\cfrac { 1 }{ c }\cfrac { \partial Y}{ \partial t }\)
And the spread of \(X\) is the result of changing \(Y\) over time as the wave passes.
No magnetic monopole here just loops and vortexes,
The first factor suggests \(Y\) is because the distribution of \(\rho_e\) form a loop or vortex. The second factor suggests the current density flow forms a loop or vortex. Both of which establishes a \(B\) field.
From (2),
\(\nabla X=i\cfrac { 1 }{ c }\cfrac { \partial Y}{ \partial t }\)
And the spread of \(X\) is the result of changing \(Y\) over time as the wave passes.
No magnetic monopole here just loops and vortexes,
