Radiacion: Guias de onda

Guías de onda

Las guías de onda se analizan resolviendo las ecuaciones de Maxwell.

Comencemos escribiendolas:

$\nabla \cdot {\vec {E}}=0\quad \quad \quad (1)$

$\nabla \cdot {\vec {B}}=0\quad \quad \quad (2)$

$\nabla \times {\vec {E}}=-{\frac {\partial {\vec {B}}}{\partial \mathbf {t} }}\quad \quad \quad (3)$

$\nabla \times {\vec {E}}={\frac {1}{\mathrm {c} ^{2}}}{\frac {\partial {\vec {E}}}{\partial \mathbf {t} }}\quad \quad \quad (4)$

Ahora suponemos un conductor perfecto

esto es que tanto el campo eléctrico, como el magnético son nulos dentro del conductor.

${\vec {E}}=0$ y ${\vec {B}}=0$

luego las condiciones de frontera en el interior del conductor serán :

${\vec {E}}_{paralelo}=0\quad \quad \quad (i)$

${\vec {B}}_{perpendicular}=0\quad \quad \quad (ii)$

Entonces estamos buscando expresiones del tipo

${\vec {E}}{(\mathbf {r,t} )=E_{0}(\mathbf {x,y} )}e^{\mathbf {i(k\ z-\omega t)} }\quad \quad \quad (I)$

${\vec {B}}{(\mathbf {r,t} )=B_{0}(\mathbf {x,y} )}e^{\mathbf {i(k\ z-\omega t)} }\quad \quad \quad (II)$

donde consideramos $\mathbf {k} \in \ Re$.

Tanto I como II deben satisfacer las ecuaciones de maxwell, asi pues debemos encontrar $E_{0(r)}$ y $B_{0(r)}$ tal que satisfagan las ecuaciones (1-4),sujetas a las condiciones de fronteras i) y ii).

Ahora re-escribimos $E_{0(r)}$ y $B_{0(r)}$ de la siguiente manera:

Error al representar (error de sintaxis): \vec{E_0} = E_x(\mathbf{x,y})x + E_y(\mathbf{x,y})y +E_z(\mathbf{x,y})z \quad\quad \quad (1´)

Error al representar (error de sintaxis): \vec{B_0} = B_x(\mathbf{x,y})x + B_y(\mathbf{x,y})y +B_z(\mathbf{x,y})z \quad\quad \quad (2´)

Comencemos demostrando que I satisface a 3, esto es que .

$\nabla \times {\vec {E}}=-{\frac {\partial {\vec {B}}}{\partial \mathbf {t} }}$ ⇒

$\nabla \times {\vec {E_{x}}}={\frac {\partial {\vec {E_{z}}}}{\partial \mathbf {y} }}-{\frac {\partial {\vec {E_{y}}}}{\partial \mathbf {z} }}=({\frac {\partial {\vec {E_{0}z}}}{\partial \mathbf {y} }}-ik{\vec {E_{0}y}})e^{\mathbf {i(k\ z-\omega t)} }$

⇒

$({\frac {\partial {\vec {E_{z}}}}{\partial \mathbf {y} }}-ik{\vec {E_{y}}})=iw\mathbf {B_{z}}$

.

$\nabla \times {\vec {E_{y}}}={\frac {\partial {\vec {E_{x}}}}{\partial \mathbf {z} }}-{\frac {\partial {\vec {E_{z}}}}{\partial \mathbf {x} }}=(ik{\vec {E_{0}x}}-{\frac {\partial {\vec {E_{0}z}}}{\partial \mathbf {x} }})e^{\mathbf {i(k\ z-\omega t)} }$

⇒

$(ik{\vec {E_{x}}}-{\frac {\partial {\vec {E_{z}}}}{\partial \mathbf {x} }})=iw{\vec {B_{y}}}$

.

De manera que

$\nabla \times {\vec {E}}=-{\frac {\partial {\vec {B}}}{\partial \mathbf {t} }}=iw{\vec {B_{y}}}e^{\mathbf {i(k\ z-\omega t)} }$

y el mismo procedimiento se le aplica a 

$\nabla \times {\vec {E}}={\frac {1}{\mathrm {c} ^{2}}}{\frac {\partial {\vec {E}}}{\partial \mathbf {t} }}$

⇒

$\nabla \times {\vec {B_{x}}}={\frac {\partial {\vec {B_{z}}}}{\partial \mathbf {y} }}-{\frac {\partial {\vec {B_{y}}}}{\partial \mathbf {z} }}=({\frac {\partial {\vec {B_{0}z}}}{\partial \mathbf {y} }}-ik{\vec {B_{0}y}})e^{\mathbf {i(k\ z-\omega t)} }$

⇒

$({\frac {\partial {\vec {B_{z}}}}{\partial \mathbf {y} }}-ik{\vec {B_{y}}})=-iw{\frac {1}{\mathrm {c} ^{2}}}{\vec {E_{z}}}$

.

Continuando con este mismo proceso , obtenemos lo siguiente :

1)${\partial _{x}{\vec {E_{y}}}}-{\partial _{y}{\vec {E_{x}}}}=iw{\vec {B_{z}}}$

2)${\partial _{y}{\vec {E_{z}}}}-ik{\vec {E_{y}}}=iw{\vec {B_{x}}}$

3)$ik{\vec {E_{x}}}-{\partial _{x}{\vec {E_{z}}}}=iw{\vec {B_{y}}}$

---

4)${\partial _{x}{\vec {B_{y}}}}-{\partial _{y}{\vec {B_{x}}}}=-iw{\frac {1}{\mathrm {c} ^{2}}}{\vec {E_{z}}}$

5)${\partial _{y}{\vec {B_{z}}}}-ik{\vec {B_{y}}}=-iw{\frac {1}{\mathrm {c} ^{2}}}{\vec {E_{x}}}$

6)$ik{\vec {B_{x}}}-{\partial _{x}{\vec {B_{z}}}}=-iw{\frac {1}{\mathrm {c} ^{2}}}{\vec {E_{y}}}$

Ya con estas ecuaciones, queremos encontrar ${\vec {E_{x}}},{\vec {E_{y}}},{\vec {B_{x}}},{\vec {B_{y}}}$, en términos de ${\vec {E_{z}}},{\vec {B_{z}}}$.

Resolviendo el conjunto de ecuaciones de la 1-6. tenemos:

${\vec {E_{x}}}={\frac {i}{\mathrm {(} {w/c})^{2}-k^{2}}}(k{\partial _{x}{\vec {E_{z}}}}+w{\partial _{y}{\vec {B_{z}}}})$

${\vec {E_{y}}}={\frac {i}{\mathrm {(} {w/c})^{2}-k^{2}}}(k{\partial _{y}{\vec {E_{z}}}}-w{\partial _{x}{\vec {B_{z}}}})$

${\vec {B_{x}}}={\frac {i}{\mathrm {(} {w/c})^{2}-k^{2}}}(k{\partial _{x}{\vec {B_{z}}}}-{\frac {w}{\mathrm {c} ^{2}}}{\partial _{y}{\vec {E_{z}}}})$

${\vec {B_{x}}}={\frac {i}{\mathrm {(} {w/c})^{2}-k^{2}}}(k{\partial _{y}{\vec {B_{z}}}}+{\frac {w}{\mathrm {c} ^{2}}}{\partial _{x}{\vec {E_{z}}}})$

.

Sustituyendo estos resultados en $\nabla \cdot {\vec {E}}=0=\nabla \cdot {\vec {B}}$ , tenemos.

$\nabla \cdot {\vec {E}}={\partial _{x}{\vec {E_{x}}}}+{\partial _{y}{\vec {E_{y}}}}+{\partial _{z}{\vec {E_{z}}}}=({\partial _{x}{\vec {E_{x}}}}+{\partial _{y}{\vec {E_{y}}}}+ik{\vec {E_{0}z}})e^{\mathbf {i(k\ z-\omega t)} }=0$

            ⇒

$({\partial _{x}{\vec {E_{x}}}}+{\partial _{y}{\vec {E_{y}}}}+ik{\vec {E_{0}z}})=0\quad \quad \quad (\diamondsuit )$

Usando las expresiones para ${\vec {E_{x}}},{\vec {E_{y}}}$ , y sustituyendo en $\diamondsuit$ tenemos.

${\frac {i}{\mathrm {(} {w/c})^{2}-k^{2}}}(k{\partial _{x}^{2}{\vec {E_{z}}}}+w{\partial _{x}^{2}y{\vec {B_{z}}}})+{\frac {i}{\mathrm {(} {w/c})^{2}-k^{2}}}(k{partial_{y}^{2}{\vec {E_{z}}}}-w{\partial _{x}^{2}y{\vec {B_{z}}}}+ik{\vec {E_{z}}})=0$

ó

${\partial _{y}^{2}{\vec {E_{z}}}}+{\partial _{x}^{2}{\vec {E_{z}}}}+[{\mathrm {(} {w/c})^{2}-k^{2}}]{\vec {E_{z}}}=0$

⇒

${{\partial _{y}^{2}}+{\partial _{x}^{2}}+[{\mathrm {(} {w/c})^{2}-k^{2}}]}{\vec {E_{z}}}=0\quad \quad \quad (a)$

.

y al hacerlo para

$\nabla \cdot {\vec {B}}=0$ , obtenemos algo similar:

${{\partial _{x}^{2}}+{\partial _{y}^{2}}+[{\mathrm {(} {w/c})^{2}-k^{2}}]}{\vec {B_{z}}}=0\quad \quad \quad (b)$

.

De (a) y (b) , podemos decir lo siguiente:

Si ${\vec {E_{z}}}=0$ , llamamos TE (onda transversal eléctrica)

Si ${\vec {B_{z}}}=0$ , llamamos TM (onda transversal magnética)

Si ${\vec {E_{z}}}={\vec {B_{z}}}=0$ , llamamos TEM (onda transversal electromagnética) , sin embargo se puede ver que el tipo de onda TEM , no puede existir en una guía de onda.