Update textbook.fr.md

12.temporary_ins/96.electromagnetism-in-media/20.reflexion-refraction-at-interfaces/10.boundary-conditions/10.main/textbook.fr.md

@@ -39,7 +39,7 @@ while the third and fourth
 @@@@@@@@@@@@
 
 will give information on the tangential components of $`\overrightarrow{E}`$ and $`\overrightarrow{H}`$.
->
+
 To obtain the boundary conditions we consider a surface separating two
 Linear Homogeneous and Isotropic (LHI) media whose properties such as that
 the permittivity and permeability are different. We denote as
@@ -47,159 +47,152 @@ $`\overrightarrow{E}_1\,,\overrightarrow{B}_1\,,\overrightarrow{D}_1`$ and $`\ov
 material 1 close to its surface. Likewise an index 2 will be used for
 the fields in the second material.
 
-Figure 3.1: []{#_bookmark59 .anchor}Scheme for deriving boundary
-conditions for perpendicular field components. S~1~, S~2~ and S^t^
-represent respectively the surface at the top, bot- tom and interface.
+![](electromag-in-media-reflexion-transmission-fig-31.jpg)   
+_Figure 3.1 : Scheme for deriving boundary conditions for perpendicular field components._
+_$`S_1\,, S_2`$ and $`S'`$ represent respectively the surface at the top, bot- tom and interface._
 >
 
-__**Normal components**__
+##### Normal components
 
 __D vector__
 
 Let's apply Maxwell equation (i) to the small cylinder showed in
-figure [3.1](#_bookmark59) which extends from one side to the other on
+figure [3.1] which extends from one side to the other on
 the separation surface. This box has a base surface A and an
-infinitesimally small thickness *δ*. We get:
->
-tS $`\overrightarrow{D}`$ · d**a** = {S
->
-$`\overrightarrow{D}`$~1~ · d**a**~1~ + {S
->
-$`\overrightarrow{D}`$~2~ · d**a**~2~ + {S
+infinitesimally small thickness $`\delta`$. We get:
 
-side
+@@@@@@@@@@@@@@@@@@@@@@@@@
 
-$`\overrightarrow{D}`$ · d**a** = {*V*
->
-*ρ~c~* d*V.*
->
-If we now let *δ* 0 symmetrically with respect to the separation
+If we now let $`\delta\longrightarrow 0`$ symmetrically with respect to the separation
 surface such that the cylinder gets "pressed" onto the surface:
 
--   The third term on the left hand side of the previous equation will
+* The third term on the left hand side of the previous equation will
   be negligible as the flux of the vector $`\overrightarrow{D}`$, which is a finite
     quantity, through an infinitesimally small surface will approach
     0.
 
--   The volume charge will be reduced to a surface charge at the
+* The volume charge will be reduced to a surface charge at the
     interface only as the volume will approach 0. The volume charge
-    density is re- placed by a surface charge density *σ~s~* and the
+    density is replaced by a surface charge density $`\sigma`$ and the
     volume integral is replaced by surface integral over the middle
-    surface S^t^.
+    surface $`S'`$.
 
 We obtain:
 
-{S1 {S {Sl
+@@@@@@@@
 
-$`\overrightarrow{D}`$~1~ · d**a**~1~ + $`\overrightarrow{D}`$~2~ · d**a**~2~ = *σ~s~* d*a.* (3.7)
->
-Now, considering that d**a**~2~ = −d**a**~1~ we can write:
+Now, considering that $`d\overrightarrow{a_2}=-d\overrightarrow{a_1}`$ we can write:
 
-{S ($`\overrightarrow{D}`$~1~ − $`\overrightarrow{D}`$~2~) · d**a**~1~ = {Sl *σ~s~* d*a.* (3.8)
+@@@@@@@@@@@@@
 
-Finally, as S~1~ = S^t^ and d**a**~1~ = **nˆ**~2→1~d*a*~1~ we can
+Finally, as $`S_1=S'`$ and $`d\overrightarrow{a_2}=\overrightarrow{n}_{2\rightarrow 1}\,da_1`$ we can
 write:
->
-**nˆ**~2→1~ · ($`\overrightarrow{D}`$~1~ − $`\overrightarrow{D}`$~2~) = *σ~s~* (3.9)
->
+
+@@@@@@@@@
+
 or
->
-*D*~1*n*~ − *D*~2*n*~ = *σ~s~* (3.10)
->
+
+@@@@@@@@@@@
+
 The normal component of the vector $`\overrightarrow{D}`$ is in general discontinuous.
 It is continuos only if there are no conduction charges at the
 separation surface.
 
-__chap5 B vector__
+__B vector__
 
 The situation is identical for the vector $`\overrightarrow{B}`$, the only difference
 being that the right hand side of the equation is always 0. We
 conclude that:
->
-**nˆ**~2→1~ · ($`\overrightarrow{B}`$~1~ − $`\overrightarrow{B}`$~2~) = 0 (3.11)
->
+
+@@@@@@@@@
+
 or
->
-*B*~1*n*~ − *B*~2*n*~ = 0 (3.12)
->
+
+@@@@@@@@@
+
 The normal component of $`\overrightarrow{B}`$ is always conserved.
 
-__**Tangential components**__
 
-![](media/image171.jpeg)
+##### Tangential components
 
-Figure 3.2: []{#_bookmark60 .anchor}Contour for deriving boundary
-conditions for parallel field com- ponents.
+![](electromag-in-media-reflexion-transmission-fig-32.jpg)   
+_Figure 3.2 : Contour for deriving boundary conditions for parallel field components._
 
 __chap5 E vector__
 
 We integrate the third Maxwell equation around the rectangular contour
-C that straddles the boundary of width W and thickness *δ* as shown in
-figure [3.2.](#_bookmark60) We chose to integrate the line integral
+C that straddles the boundary of width W and thickness $`\delta`$ as shown in
+figure [3.2.]. We chose to integrate the line integral
 following the right-hand sense relative to the surface normal
-**nˆ***~a~*. By letting *δ* → 0, we get
->
-t $`\overrightarrow{E}`$ · *d***l** = $`\overrightarrow{E}`$~1~ · *A_B* + $`\overrightarrow{E}`$~2~ · *C_D* = −
-[d]{.underline} { $`\overrightarrow{B}`$ · *d*$`\overrightarrow{S}`$ → 0 (3.13)
->
+$`\overrightarrow{n_a}`$. By letting $`\delta\right 0`$, we get
+
+@@@@@@@@@@@
+
 as the line integral along the sides goes to zero and the flux of the
 induction field $`\overrightarrow{B}`$, which is a finite quantity, approaches 0.
-Considering that *C_D* =
->
-−*A_B* = d**l**, we get:
->
-($`\overrightarrow{E}`$~1~ − $`\overrightarrow{E}`$~2~) · d**l** = 0 ∀d**l** /I to the separation
-surface (3.14)
->
+Considering that $`\overrightarrow{CD}-\overrightarrow{AB}=d\overrightarrow{l}`$, we get:
+
+@@@@@@@@@@
+
 or again
 
-*E*~1*T*~ − *E*~2*T*~ = 0 (3.15)
+@@@@@@@@@@@@
 
 i.e. the tangential components of the electric field are always
 conserved at the interface. This condition can also be written:
->
-**nˆ**~2→1~ ***×*** ($`\overrightarrow{E}`$~1~ − $`\overrightarrow{E}`$~2~) = 0*.* (3.16)
 
-__chap5 H vector__
+@@@@@@@@@@ $`\quad equ. 3.16)`$
 
-Following the same reasoning as for the $`\overrightarrow{E}`$ vector we write for the
-left hand
->
-side when *δ* → 0: $`\overrightarrow{H}`$
+__H vector__
 
-C
+Following the same reasoning as for the $`\overrightarrow{E}`$ vector we write for the
+left hand side when $`\delta\right 0`$ :
 
-· *d***l** = $`\overrightarrow{H}`$
+$`\overrightarrow{H}`$
 
-~1~ · *A_B*
+@@@@@@@@@@@@@ $`\quad equ. 3.17)`$
 
-\+ $`\overrightarrow{H}`$~2~ · *C_D.*
->
-(3.17)
->
 The right hand side needs more attention. The flux of the vector $`\overrightarrow{D}`$
-approaches
->
-0\. However the flux of the vector $`\overrightarrow{j}`$ over an infinitesimal surface
+approaches 0. However the flux of the vector $`\overrightarrow{J}`$ over an infinitesimal surface
 can give a finite value: the surface integral reduces to an integral
 over the line of width W equal to the side of the rectangle. No volume
-current can contribute, however a surface current $`\overrightarrow{j}`$*~s~* (i.e. a
-current flowing on the separation surface with dimensions \[*A/m*\]
-not \[*A/m*^2^\] as $`\overrightarrow{j}`$) can. This is typically the case of a
+current can contribute, however a surface current $`\overrightarrow{J}_S`$ (i.e. a
+current flowing on the separation surface with dimensions $`[A/m]`$
+not $`[A/m*^2]`$ as $`\overrightarrow{J}`$) can. This is typically the case of a
 perfect conductor where a finite current can flow through a
 infinitesimally small area. We get:
->
+
+@@@@@@@@ $`\quad equ. 3.18)`$
+
 or again
->
-($`\overrightarrow{H}`$~1~ − $`\overrightarrow{H}`$~2~) · d**l** = *J~s~* (3.18)
->
-*H*~1*T*~ − *H*~2*T*~ = *J~s~* (3.19)
->
+
+@@@@@@@@ $`\quad equ. 3.19)`$
+
 i.e. the tangential components of the magnetic field $`\overrightarrow{H}`$ is
 discontinuous unless no surface currents exist. As for the electric
 field, this condition can also be written:
->
-**nˆ**~2→1~ ***×*** ($`\overrightarrow{H}`$~1~ − $`\overrightarrow{H}`$~2~) = $`\overrightarrow{j}`$*~s~* (3.20)
+
+@@@@@@@@ $`\quad equ. 3.20)`$
+
+!! *Summary*   
+!! 
+
+! *Remarks*   
+!
+! * If medium 1 & 2 are perfect dielectrics then there are no charges
+! nor surface currents at the interface, and so the tangential component
+! of $`\overrightarrow{H}`$ and the normal component of $`\overrightarrow{D}`$ are both continuous.   
+!
+! * If medium 1 is a perfect dielectric and medium 2 is a perfect metal,
+! there are charges and surface currents at the interface, and so the
+! tangential component of $`\overrightarrow{H}`$ and the normal component of $`\overrightarrow{D}`$ are
+! not continuous.   
+!
+! * In case of linear media, the 4 relations can be expressed in terms
+! of $`\overrightarrow{E}`$ and $`\overrightarrow{B}`$ alone using the constitutive relations
+! $`\overrightarrow{D}=\epsilon\,$`\overrightarrow{E}`$ and $`\overrightarrow{B}=\mu\,$`\overrightarrow{H}`$.
+
+&&&&&&&&&&&&&&&&&&&&&&&&&
 
 #### chap2 Reflection and transmission at normal incidence