Loss, So What?

Consider an EMW propagating down a sharp point of half angle $\theta$,

Electric displacement,  $\varepsilon E$  perpendicular to the slop boundary is the same.  The tangential component of  $E$  is also the same.  So, the  $E$  field after the boundary is,

$E^2_f=\left\{ Esin(\theta ) \right\} ^{ 2 }+\left\{ \cfrac { \varepsilon_a  }{ \varepsilon _{ o } } Ecos(\theta ) \right\} ^{ 2 }$

Loss in  $E$ is,

$E^{ 2 }_{ loss }=E^{ 2 }-E^2_f$

$E^{ 2 }_{ loss }=E^{ 2 }-(\left\{ Esin(\theta ) \right\} ^{ 2 }+\left\{ \cfrac { \varepsilon_a  }{ \varepsilon _{ o } } Ecos(\theta ) \right\} ^{ 2 })$

$E^{ 2 }_{ loss }=E^{ 2 }\left\{ 1-sin^{ 2 }(\theta )-(\cfrac { \varepsilon_a  }{ \varepsilon _{o } } )^{ 2 }cos^{ 2 }(\theta ) \right\}$

$E^{ 2 }_{ loss }=E^{ 2 }cos^{ 2 }(\theta )\left\{ 1-(\cfrac { \varepsilon _{ a } }{ \varepsilon _{ o } } )^{ 2 } \right\}$

The perpendicular component of the  $B$  field is the same, and the tangential  $H$ field is the same

$\cfrac{B_c}{\mu_o}=\cfrac{Bsin(\theta)}{\mu_a}$

The  $B$  field after the boundary is,

$B^2_f=\left\{Bcos(\theta)\right\}^2+\left\{\cfrac{\mu_o}{\mu_a}Bsin(\theta)\right\}^2$

Loss in  $B$ is,

$B^{ 2 }_{ loss }=B^{ 2 }-B^2_f$

$B^{ 2 }_{ loss }=B^{ 2 }-(\left\{Bcos(\theta)\right\}^2+\left\{\cfrac{\mu_o}{\mu_a}Bsin(\theta)\right\}^2)$

$B^{ 2 }_{ loss }=B^{ 2 }\left\{ 1-cos^{ 2 }(\theta )-(\cfrac { \mu _{ o } }{ \mu _{ a } } )^{ 2 }sin^{ 2 }(\theta ) \right\}$

$B^{ 2 }_{ loss }=B^{ 2 }sin^{ 2 }(\theta )\left\{ 1-(\cfrac { \mu _{ o } }{ \mu _{ a } } )^{ 2 } \right\}$

Total loss in energy crossing the boundary,

$U_{loss}=\cfrac{1}{2}\varepsilon_oE^2_{loss}+\cfrac{1}{2}\mu_oB^2_{loss}$

$U_{loss}=\cfrac{1}{2}\left[\varepsilon_oE^{ 2 }cos^{ 2 }(\theta )\left\{ 1-(\cfrac { \varepsilon _{ a } }{ \varepsilon _{ o } } )^{ 2 } \right\}+\mu_oB^{ 2 }sin^{ 2 }(\theta )\left\{ 1-(\cfrac { \mu _{ o } }{ \mu _{ a } } )^{ 2 } \right\}  \right]$

So what?