Supongamos que el medio en el que se propaga la onda electromagnética es isótropo, lineal y homogéneo, no conductor. La onda electromagnética se propaga a lo largo del eje Z.

${\displaystyle \begin{array}{l}\overrightarrow{E}=E_x\widehat{i}+E_y\widehat{j}+E_z\widehat{k},\{\begin{array}{l}{E}_x=E_{0x}\exp \left(i\left(kz-\omega t\right)\right)\\ E_y=E_{0y}\exp \left(i\left(kz-\omega t\right)\right)\\ E_z=E_{0z}\exp \left(i\left(kz-\omega t\right)\right)\end{array}\\ \overrightarrow{B}=B_x\widehat{i}+B_y\widehat{j}+B_z\widehat{k},\{\begin{array}{l}{B}_x=B_{0x}\exp \left(i\left(kz-\omega t\right)\right)\\ B_y=B_{0y}\exp \left(i\left(kz-\omega t\right)\right)\\ B_z=B_{0z}\exp \left(i\left(kz-\omega t\right)\right)\end{array}\end{array}}$

Las amplitudes E_0x, E_0y, E_0z, B_0x, B_0y, B_0z son funciones de x e y, que por ahora no conocemos, la dependencia con z y t aparece en la función exponencial. La onda electromagnética no es transversal ya que tiene una compoenete E_z a lo largo de la dirección de propagación

Apicamos la ley de Faraday en forma diferencial

${\displaystyle \begin{array}{l}\overrightarrow{\nabla }\times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}\\ \left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\widehat{i}+\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)\widehat{j}+\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)\widehat{k}=-\frac{\partial }{\partial t}\left(B_x\widehat{i}+B_y\widehat{j}+B_z\widehat{k}\right)\\ \{\begin{array}{l}\frac{\partial E_{0z}}{\partial y}{e}^{i(kz-\omega t)}-ik{E}_{0y}{e}^{i(kz-\omega t)}=i\omega B_{0x}{e}^{i(kz-\omega t)}\\ ik{E}_{0x}{e}^{i(kz-\omega t)}-\frac{\partial E_{0z}}{\partial x}{e}^{i(kz-\omega t)}=i\omega B_{0y}{e}^{i(kz-\omega t)}\\ \frac{\partial E_{0y}}{\partial x}{e}^{i(kz-\omega t)}-\frac{\partial E_{0x}}{\partial y}{e}^{i(kz-\omega t)}=i\omega B_{0z}{e}^{i(kz-\omega t)}\end{array}\\ \{\begin{array}{l}\frac{\partial E_{0z}}{\partial y}-ik{E}_{0y}=i\omega B_{0x}\\ ik{E}_{0x}-\frac{\partial E_{0z}}{\partial x}=i\omega B_{0y}\\ \frac{\partial E_{0y}}{\partial x}-\frac{\partial E_{0x}}{\partial y}=i\omega B_{0z}\end{array}\end{array}}$

La ley de Ampère-Maxwell, sin corrientes

${\displaystyle \begin{array}{l}\overrightarrow{\nabla }\times \overrightarrow{B}-\frac{1}{c^2}\frac{\partial \overrightarrow{E}}{\partial t}=0\\ \left(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\right)\widehat{i}+\left(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\right)\widehat{j}+\left(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}\right)\widehat{k}-\frac{1}{c^2}\frac{\partial }{\partial t}\left(E_x\widehat{i}+E_y\widehat{j}+E_z\widehat{k}\right)=0\\ \{\begin{array}{l}\frac{\partial B_{0z}}{\partial y}-ik{B}_{0y}+\frac{1}{c^2}i\omega E_{0x}=0\\ ik{B}_{0x}-\frac{\partial B_{0z}}{\partial x}+\frac{1}{c^2}i\omega E_{0y}=0\\ \frac{\partial B_{0y}}{\partial x}-\frac{\partial B_{0x}}{\partial y}+\frac{1}{c^2}i\omega E_{0z}=0\end{array}\end{array}}$

Las cuatro componentes transversales E_0x, E_0y, B_0x, B_0y se pueden deducir de las componentes longitudinales, E_0z, B_0z del siguiente modo

• A partir de las ecuaciones

${\displaystyle \{\begin{array}{l}\frac{\partial B_{0z}}{\partial y}-ik{B}_{0y}+\frac{1}{c^2}i\omega E_{0x}=0\\ ik{E}_{0x}-\frac{\partial E_{0z}}{\partial x}=i\omega B_{0y}\end{array}}$

Despejamos E_0x y B_0y

${\displaystyle \begin{array}{l}{E}_{0x}=\frac{i}{{\gamma }^2}\left(\omega \frac{\partial B_{0z}}{\partial y}+k\frac{\partial E_{0z}}{\partial x}\right),{\gamma }^2=\frac{{\omega }^2}{c^2}-k^2\\ B_{0y}=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial y}+\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial x}\right)\end{array}}$

• A partir de las ecuaciones

${\displaystyle \{\begin{array}{l}ik{B}_{0x}-\frac{\partial B_{0z}}{\partial x}+\frac{1}{c^2}i\omega E_{0y}=0\\ \frac{\partial E_{0z}}{\partial y}-ik{E}_{0y}=i\omega B_{0x}\end{array}}$

Despejamos B_0x y E_0y

${\displaystyle \begin{array}{l}{E}_{0y}=\frac{i}{{\gamma }^2}\left(k\frac{\partial E_{0z}}{\partial y}-\omega \frac{\partial B_{0z}}{\partial x}\right)\\ B_{0x}=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial x}-\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial y}\right)\end{array}}$

Las ecuaciones de propagación de las ondas electromagnéticas son

${\displaystyle \{\begin{array}{l}{\nabla }^2\overrightarrow{E}-{\epsilon }_0{\mu }_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}=0\\ {\nabla }^2\overrightarrow{B}-{\epsilon }_0{\mu }_0\frac{{\partial }^2\overrightarrow{B}}{\partial t^2}=0\end{array}}$

La onda electromagnética se propaga a lo largo del eje Z

${\displaystyle \begin{array}{l}\left(\frac{{\partial }^2{E}_{0z}}{\partial x^2}+\frac{{\partial }^2{E}_{0z}}{\partial y^2}\right)e^{i\left(kz-\omega t\right)}-k^2{E}_{0z}{e}^{i\left(kz-\omega t\right)}+\frac{{\omega }^2}{c^2}{E}_{0z}{e}^{i\left(kz-\omega t\right)}=0\\ \frac{{\partial }^2{E}_{0z}}{\partial x^2}+\frac{{\partial }^2{E}_{0z}}{\partial y^2}+\left(\frac{{\omega }^2}{c^2}-k^2\right)E_{0z}=0\\ \frac{{\partial }^2{E}_{0z}}{\partial x^2}+\frac{{\partial }^2{E}_{0z}}{\partial y^2}+{\gamma }^2{E}_{0z}=0\end{array}}$

De modo similar

${\displaystyle \frac{{\partial }^2{B}_{0z}}{\partial x^2}+\frac{{\partial }^2{B}_{0z}}{\partial y^2}+{\gamma }^2{B}_{0z}=0}$

Los valores del número de onda k son discretos y vienen determinados por la frecuencia ω de la onda electromagnética y por la geometría y características del medio en el que se propaga ε, μ (no conductor)

Resolvemos la ecuación diferencial utilizando el procedimiento de variables separadas

${\displaystyle B_z\left(x,y,t\right)=X(x)Y(y)\exp \left(i\left(kz-\omega t\right)\right)}$

Introducimos en la ecuación diferencial

${\displaystyle \begin{array}{l}\frac{{\partial }^2{B}_{0z}}{\partial x^2}+\frac{{\partial }^2{B}_{0z}}{\partial y^2}+{\gamma }^2{B}_{0z}=0\\ \left(Y\frac{d^{2}X}{d{x}^2}+X\frac{d^{2}Y}{d{y}^2}+{\gamma }^{2}XY\right)\exp \left(i\left(kz-\omega t\right)\right)=0\\ \frac{1}{X(x)}\frac{d^{2}X(x)}{d{x}^2}+\frac{1}{Y(y)}\frac{d^{2}Y(y)}{d{y}^2}+{\gamma }^2=0\end{array}}$

El primer término, es solamente función de x el segundo solamente de y. Convertimos así, una ecuación diferencial en un sistema de dos ecuaciones diferenciales cuya solución es sencilla

${\displaystyle \begin{array}{l}\{\begin{array}{l}\frac{1}{X(x)}\frac{d^{2}X(x)}{d{x}^2}=-{\alpha }^2\\ \frac{1}{Y(y)}\frac{d^{2}Y(y)}{d{y}^2}=-{\beta }^2\end{array}\\ {\gamma }^2={\alpha }^2+{\beta }^2\end{array}}$

La solución de las ecuaciones diferenciales son

${\displaystyle \{\begin{array}{l}\frac{d^{2}X}{d{x}^2}+{\alpha }^{2}X=\mathrm{0,}X(x)=C_1\sin \left(\alpha x\right)+C_2\cos \left(\alpha x\right)\\ \frac{d^{2}Y}{d{y}^2}+{\beta }^{2}Y=\mathrm{0,}Y(y)=C_3\sin \left(\beta y\right)+C_4\cos \left(\beta y\right)\end{array}}$

La expresión de la amplitud B_0z del campo magnético es

${\displaystyle B_{0z}=\left(C_1\sin \left(\alpha x\right)+C_2\cos \left(\alpha x\right)\right)\left(C_3\sin \left(\beta y\right)+C_4\cos \left(\beta y\right)\right)}$

Sus derivadas resopecto de x e y son

${\displaystyle \begin{array}{l}\frac{\partial B_{0z}}{\partial x}=\alpha \left(C_1\cos \left(\alpha x\right)-C_2\sin \left(\alpha x\right)\right)\left(C_3\sin \left(\beta y\right)+C_4\cos \left(\beta y\right)\right)\\ \frac{\partial B_{0z}}{\partial y}=\beta \left(C_1\sin \left(\alpha x\right)+C_2\cos \left(\alpha x\right)\right)\left(C_3\cos \left(\beta y\right)-C_4\sin \left(\beta y\right)\right)\end{array}}$

TE, ondas transversales eléctricas

Las ondas transversales eléctricas están caracterizadas por que E_z=0

El caso más sencillo, es una guía rectangular de dimensiones a y b que se extiende a lo largo del eje Z

La condición de contorno es ${\frac{\partial B_z}{\partial n}|}_{\text{pared}}=0$ , donde el símbolo n indica derivada en la dirección normal a la superficie metálica evaluada en la posición de dicha superficie de la guía. Estas condiciones son equivalentes a las siguientes

• Sabiendo que E_0z=0, de la relación

${\displaystyle \begin{array}{l}{E}_{0x}=\frac{i}{{\gamma }^2}\left(\omega \frac{\partial B_{0z}}{\partial y}+k\frac{\partial E_{0z}}{\partial x}\right)=\frac{i}{{\gamma }^2}\omega \frac{\partial B_{0z}}{\partial y}\\ {E_{0x}|}_{y=0}=\frac{i}{{\gamma }^2}\omega {\frac{\partial B_{0z}}{\partial y}|}_{y=0}=0\\ {E_{0x}|}_{y=b}=\frac{i}{{\gamma }^2}\omega {\frac{\partial B_{0z}}{\partial y}|}_{y=b}=0\end{array}}$

El campo eléctrico E_0y es perpendicular a los planos conductores y=0 e y=b

• Por otra parte, sabiendo que E_0z=0, de la relación

${\displaystyle \begin{array}{l}{E}_{0y}=\frac{i}{{\gamma }^2}\left(k\frac{\partial E_{0z}}{\partial y}-\right)=-\frac{i}{{\gamma }^2}\omega \frac{\partial B_{0z}}{\partial x}\\ {E_{0y}|}_{x=0}=-\frac{i}{{\gamma }^2}\omega {\frac{\partial B_{0z}}{\partial x}|}_{x=0}=0\\ {E_{0y}|}_{x=a}=-\frac{i}{{\gamma }^2}\omega {\frac{\partial B_{0z}}{\partial x}|}_{x=a}=0\end{array}}$

El campo eléctrico E_0x es perpendicular a los planos conductores x=0 y x=a

En las proximidades de los planos conductores el campo eléctrico tiene dirección perpendicular a los mismos

Los coeficientes C₁, C₂, C₃ y C₄ se determinan a partir de las condiciones de contorno

• Las condiciones de contorno para los planos paralelos y=0 e y=b son

${\displaystyle \begin{array}{l}{\frac{\partial B_{0z}}{\partial y}|}_{y=0}=\mathrm{0,}{\frac{\partial B_{0z}}{\partial y}|}_{y=b}=0\\ \{\begin{array}{l}\beta \left(C_1\sin \left(\alpha x\right)+C_2\cos \left(\alpha x\right)\right)\left(C_3\right)=\mathrm{0,}{C}_3=0\\ \beta \left(C_1\sin \left(\alpha x\right)+C_2\cos \left(\alpha x\right)\right)\left(-C_4\sin \left(\beta b\right)\right)=\mathrm{0,}\sin \left(\beta b\right)=0\end{array}\end{array}}$

• Las condiciones de contorno para los planos paralelos x=0 y x=a son

${\displaystyle \begin{array}{l}{\frac{\partial B_{0z}}{\partial x}|}_{x=0}=\mathrm{0,}{\frac{\partial B_{0z}}{\partial x}|}_{x=a}=0\\ \{\begin{array}{l}\alpha \left(C_1\right)\left(C_3\sin \left(\beta y\right)+C_4\cos \left(\beta y\right)\right)=\mathrm{0,}{C}_1=0\\ \alpha \left(-C_2\sin \left(\alpha a\right)\right)\left(C_3\sin \left(\beta y\right)+C_4\cos \left(\beta y\right)\right)=\mathrm{0,}\sin \left(\alpha a\right)=0\end{array}\end{array}}$

Obtenemos valores discretos de α y β

${\displaystyle \begin{array}{l}\{\begin{array}{l}\sin \left(\alpha a\right)=\mathrm{0,}\alpha a=m\pi ,m=\mathrm{0,1,2,3...}\\ \sin \left(\beta b\right)=\mathrm{0,}\beta b=n\pi ,n=\mathrm{0,1,2,3...}\end{array}\\ {\gamma }_{m,n}^2={\alpha }^2+{\beta }^2={\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2\end{array}}$

Amplitudes

La amplitud B_0z de la componente B_z del campo magnético vale

${\displaystyle \begin{array}{l}{B}_{0z}=C_2\cos \left(\frac{m\pi }{a}x\right)C_4\cos \left(\frac{n\pi }{b}y\right)\\ B_{0z}=B_0\cos \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\end{array}}$

Donde el nuevo coeficiente B₀=C₂C₄

La componente B_z del campo magnético vale

${\displaystyle B_z=\mathrm{Re}\left\{B_{0z}\exp \left(i\left(kz-\omega t\right)\right)\right\}=B_0\cos \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\cos \left(kz-\omega t\right)}$

Conocido B_0z y E_0z=0, determinamos las otras amplitudes de las componentes del campo eléctrico y magnético

• Componente E_x

${\displaystyle \begin{array}{l}{E}_{0x}=\frac{i}{{\gamma }^2}\left(\omega \frac{\partial B_{0z}}{\partial y}+k\frac{\partial E_{0z}}{\partial x}\right)=\frac{i}{{\gamma }^2}\omega \frac{\partial B_{0z}}{\partial y}\\ E_{0x}=\frac{i}{{\gamma }^2}\omega \frac{n\pi }{b}\left(-B_0\right)\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\\ E_x=\mathrm{Re}\left\{E_{0x}\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{\omega }{{\gamma }^2}\frac{n\pi }{b}{B}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente E_y

${\displaystyle \begin{array}{l}{E}_{0y}=\frac{i}{{\gamma }^2}\left(k\frac{\partial E_{0z}}{\partial y}-\omega \frac{\partial B_{0z}}{\partial x}\right)=-\frac{i}{{\gamma }^2}\omega \frac{\partial B_{0z}}{\partial x}\\ E_{0y}=-\frac{i}{{\gamma }^2}\omega \frac{m\pi }{a}\left(-B_0\right)\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\\ E_y=\mathrm{Re}\left\{E_{0y}\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{\omega }{{\gamma }^2}\frac{m\pi }{a}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente B_x

${\displaystyle \begin{array}{l}{B}_{0x}=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial x}-\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial y}\right)=\frac{i}{{\gamma }^2}k\frac{\partial B_{0z}}{\partial x}\\ B_{0x}=\frac{i}{{\gamma }^2}k\frac{m\pi }{a}\left(-B_0\right)\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\\ B_x=\mathrm{Re}\left\{B_{0x}\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{k}{{\gamma }^2}\frac{m\pi }{a}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente B_y

${\displaystyle \begin{array}{l}{B}_{0y}=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial y}+\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial x}\right)=\frac{i}{{\gamma }^2}k\frac{\partial B_{0z}}{\partial y}\\ B_{0y}=\frac{i}{{\gamma }^2}k\frac{n\pi }{b}\left(-B_0\right)\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\\ B_y=\mathrm{Re}\left\{B_{0y}\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{k}{{\gamma }^2}\frac{n\pi }{b}{B}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

Resumiendo, las componentes del campo eléctrico $\overrightarrow{E}$ y del campo magnético $\overrightarrow{B}$, son

${\displaystyle \begin{array}{l}\overrightarrow{E}=\{\begin{array}{l}{E}_x=\frac{\omega }{{\gamma }^2}\frac{n\pi }{b}{B}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ E_y=-\frac{\omega }{{\gamma }^2}\frac{m\pi }{a}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ E_z=0\end{array}\\ \overrightarrow{B}=\{\begin{array}{l}{B}_x=\frac{k}{{\gamma }^2}\frac{m\pi }{a}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ B_y=\frac{k}{{\gamma }^2}\frac{n\pi }{b}{B}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ B_z=B_0\cos \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\cos \left(kz-\omega t\right)\end{array}\end{array}}$

Energía por unidad de tiempo

El vector de Poynting es el producto vectorial

${\displaystyle \overrightarrow{S}=\overrightarrow{E}\times \overrightarrow{H}=\frac{1}{{\mu }_0}\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ E_x& E_y& 0\\ B_x& B_y& B_z\end{array}\right|=\left(E_y{B}_z\right)\widehat{i}-\left(E_x{B}_z\right)\widehat{j}+\left(E_x{B}_y-E_y{B}_x\right)}$

Teniendo en cuenta que el valor medio de una función periódica

${\displaystyle \begin{array}{l}\langle f(t)\rangle =\frac{1}{P}{\displaystyle \underset{0}{\overset{P}{\int }}f(t)dt}\\ \langle {\cos }^2\left(kz-\omega t\right)\rangle =\langle {\sin }^2\left(kz-\omega t\right)\rangle =\frac{1}{2}\\ \langle \cos \left(kz-\omega t\right)\sin \left(kz-\omega t\right)\rangle =0\end{array}}$

>> syms k z w t;
>> w*int(sin(k*z-w*t)^2,t,0,pi/w)/pi
 ans =1/2
>> w*int(cos(k*z-w*t)^2,t,0,pi/w)/pi
 ans =1/2
>> w*int(cos(k*z-w*t)*sin(k*z-w*t),t,0,pi/w)/pi
 ans =0

El valor medio de <S_x>=<S_y>0, es nulo ya que son productos del seno por el coseno, solamente queda el valor medio a lo largo del eje Z, <S_z>

${\displaystyle \begin{array}{l}{S}_z=\frac{1}{{\mu }_0}\left(\begin{array}{l}\frac{\omega }{{\gamma }^2}\frac{n\pi }{b}{B}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\frac{k}{{\gamma }^2}\frac{n\pi }{b}{B}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)+\\ \frac{\omega }{{\gamma }^2}\frac{m\pi }{a}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\frac{k}{{\gamma }^2}\frac{m\pi }{a}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}\right)\\ \langle S_z\rangle =\frac{1}{{\mu }_0}\frac{k\omega }{{\gamma }^4}{B}_0^2\left({\left(\frac{n\pi }{b}\right)}^2{\cos }^2\left(\frac{m\pi }{a}x\right){\sin }^2\left(\frac{n\pi }{b}y\right)+{\left(\frac{m\pi }{a}\right)}^2{\sin }^2\left(\frac{m\pi }{a}x\right){\cos }^2\left(\frac{n\pi }{b}y\right)\right)\langle {\sin }^2\left(kz-\omega t\right)\rangle \\ \langle S_z\rangle =\frac{k\omega }{2{\mu }_0{\pi }^2}\frac{a^4{b}^4}{{\left(m^2{b}^2+n^2{a}^2\right)}^2}{B}_0^2\left({\left(\frac{n}{b}\right)}^2{\cos }^2\left(\frac{m\pi }{a}x\right){\sin }^2\left(\frac{n\pi }{b}y\right)+{\left(\frac{m}{a}\right)}^2{\sin }^2\left(\frac{m\pi }{a}x\right){\cos }^2\left(\frac{n\pi }{b}y\right)\right)\end{array}}$

se ha tenido en cuenta que γ²=α²+β²

La energía por unidad de tiempo (potencia) es

${\displaystyle \begin{array}{l}P={\displaystyle \underset{0}{\overset{b}{\int }}{\displaystyle \underset{0}{\overset{a}{\int }}\langle S_z\rangle dx·dy}}=\\ \frac{k\omega }{2{\mu }_0{\pi }^2}\frac{a^4{b}^4}{{\left(m^2{b}^2+n^2{a}^2\right)}^2}{B}_0^2\left({\left(\frac{n}{b}\right)}^2{\displaystyle \underset{0}{\overset{b}{\int }}{\displaystyle \underset{0}{\overset{a}{\int }}{\cos }^2\left(\frac{m\pi }{a}x\right){\sin }^2\left(\frac{n\pi }{b}y\right)dx·dy}}+{\left(\frac{m}{a}\right)}^2{\displaystyle \underset{0}{\overset{b}{\int }}{\displaystyle \underset{0}{\overset{a}{\int }}{\sin }^2\left(\frac{m\pi }{a}x\right){\cos }^2\left(\frac{n\pi }{b}y\right)dx·dy}}\right)\end{array}}$

Teniendo en cuenta los resultados de las integrales

${\displaystyle \begin{array}{l}{\displaystyle \int {\cos }^{2}x·dx}={\displaystyle \int \frac{1+\cos \left(2x\right)}{2}dx=\frac{1}{2}\left(x+\frac{1}{2}\sin \left(2x\right)\right)}\\ {\displaystyle \int {\sin }^{2}x·dx}={\displaystyle \int \frac{1-\cos \left(2x\right)}{2}dx=\frac{1}{2}\left(x-\frac{1}{2}\sin \left(2x\right)\right)}\end{array}}$

La potencia es

${\displaystyle \begin{array}{l}P=\frac{k\omega }{2{\mu }_0{\pi }^2}\frac{a^4{b}^4}{{\left(m^2{b}^2+n^2{a}^2\right)}^2}{B}_0^2\left\{\begin{array}{l}{\left(\frac{n}{b}\right)}^2\frac{a}{m\pi }\frac{1}{2}\left(m\pi +\frac{1}{2}\sin \left(2m\pi \right)\right)\frac{b}{n\pi }\frac{1}{2}\left(n\pi -\frac{1}{2}\sin \left(2n\pi \right)\right)+\\ {\left(\frac{m}{a}\right)}^2\frac{a}{m\pi }\frac{1}{2}\left(m\pi -\frac{1}{2}\sin \left(2m\pi \right)\right)\frac{b}{n\pi }\frac{1}{2}\left(n\pi +\frac{1}{2}\sin \left(2n\pi \right)\right)\end{array}\right\}\\ P=\frac{k\omega }{8{\mu }_0{\pi }^2}\frac{a^5{b}^5}{{\left(m^2{b}^2+n^2{a}^2\right)}^2}{B}_0^2\left\{{\left(\frac{n}{b}\right)}^2\left(1+\frac{\sin \left(2m\pi \right)}{2m\pi }\right)\left(1-\frac{\sin \left(2n\pi \right)}{2n\pi }\right)+{\left(\frac{m}{a}\right)}^2\left(1-\frac{\sin \left(2m\pi \right)}{2m\pi }\right)\left(1+\frac{\sin \left(2n\pi \right)}{2n\pi }\right)\right\}\end{array}}$

Cuando n o m son cero, tenemos la indeterminación 0/0 que es 1

${\displaystyle \underset{x\to 0}{\lim }\frac{\sin x}{x}=1}$

>> syms x;
>> limit(sin(x)/x,x,0)
ans =1

Velocidad de fase y de grupo

La velocidad de fase es v=ω/k

La velocidad de grupo es

${\displaystyle \begin{array}{l}{\gamma }^2=\frac{{\omega }^2}{c^2}-k^2\\ \omega =c\sqrt{{\gamma }^2+k^2}\\ v_g=\frac{d\omega }{dk}=c\frac{2k}{2\sqrt{{\gamma }^2+k^2}}=\frac{k{c}^2}{\omega }=c\sqrt{1-\frac{c^2{\gamma }^2}{{\omega }^2}}\end{array}}$

El producto v_g·v=c². La velocidad de fase es mayor que la velocidad de la luz

Ejemplo

Sea una guía rectangular de cobre de lados a=3 cm y b=1 cm,

${\displaystyle {\gamma }^2=\frac{{\omega }_{m,n}^2}{c^2}={\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2}$

m y n no pueden ser cualquier número entero ya que γ²=(ω/c)²-k². Dado ω, k² no puede ser negativo.

Como a>b la frecuencia más baja corresponde a m=1, n=0

${\displaystyle \frac{{\omega }_{\mathrm{1,0}}^2}{c^2}={\left(\frac{\pi }{a}\right)}^2,f_{\mathrm{1,0}}=\frac{{\omega }_{\mathrm{1,0}}}{2\pi }=\frac{c}{2a}=5·{10}^9\text{Hz=}5\text{GHz}}$

El número de onda k₀ de una onda electromagnética de frecuencia 8 GHz en el vacío es

${\displaystyle k_0=\frac{\omega }{c}=\frac{2\pi ·8·{10}^9}{3·{10}^8}=167.55{\text{m}}^{-1},{\lambda }_0=\frac{2\pi }{k_0}=0.0375m}$

El número de onda k en la guía es

${\displaystyle \begin{array}{l}{\gamma }^2=\frac{{\omega }^2}{c^2}-k^2\\ k^2=\frac{{\omega }^2}{c^2}-\frac{{\omega }_{\mathrm{1,0}}^2}{c^2}=\frac{{\left(2\pi ·8·{10}^9\right)}^2-\left(2\pi ·5·{10}^9\right)}{{\left(3·{10}^8\right)}^2},k=130.795{\text{m}}^{-1},\lambda =0.048\text{m}\\ \frac{\lambda }{{\lambda }_0}=1.281\end{array}}$

Velocidad de fase y de grupo son

${\displaystyle \begin{array}{l}v=\frac{\omega }{k}=\frac{2\pi ·8·{10}^9}{130.795}=1.281·c\\ v·v_g=c^2,v_g=0.781·c\end{array}}$

Si el valor máximo del la amplitud del campo eléctrico es 30 kV/cm=3·10⁶ V/m, calcular la amplitud del campo magnético para m=1, n=0

E_x es nulo y la amplitud de E_y es

${\displaystyle \begin{array}{l}\frac{\omega }{{\gamma }^2}\frac{m\pi }{a}{B}_0=3·{10}^6\\ \frac{\omega }{{\left(\frac{\pi }{a}\right)}^2}\frac{1·\pi }{a}{B}_0=3·{10}^6\\ \frac{2\pi ·8·{10}^9·0.03}{\pi }{B}_0=3·{10}^6,B_0=0.0063\text{T}\end{array}}$

Para el modo m=1, n=0, la potencia es

${\displaystyle \begin{array}{l}P=\frac{k\omega }{8{\mu }_0{\pi }^2}\frac{a^5{b}^5}{b^4}{B}_0^2\frac{1}{a^2}\left(1-\frac{\sin \left(2\pi \right)}{2\pi }\right)\left(1+\frac{\sin 0}{0}\right)=\frac{k\omega }{4{\mu }_0{\pi }^2}{a}^{3}b·B_0^2\\ P=\frac{130.795·\left(2\pi ·8·{10}^9\right){0.03}^3·0.01·{0.0063}^2}{4·4\pi ·{10}^{-7}{\pi }^2}=1.4202·{10}^6\text{W}\end{array}}$

TM, ondas transversales magnéticas

Las ondas transversales magnéticas están caracterizadas por que B_z=0.

La solución de la ecuación diferencial

${\displaystyle \frac{{\partial }^2{E}_{0z}}{\partial x^2}+\frac{{\partial }^2{E}_{0z}}{\partial y^2}+{\gamma }^2{E}_{0z}=0}$

es, de modo análogo al primer apartado

${\displaystyle \begin{array}{l}{E}_{0z}\left(x,y\right)=\left(C_1\sin \left(\alpha x\right)+C_2\cos \left(\alpha x\right)\right)\left(C_3\sin \left(\beta y\right)+C_4\cos \left(\beta y\right)\right)\\ {\gamma }^2={\alpha }^2+{\beta }^2\end{array}}$

Las condiciones de contorno son

• $E_{0z}\left(\mathrm{0,}y\right)=0$

${\displaystyle E_{0z}\left(\mathrm{0,}y\right)=C_2\left(C_3\sin \left(\beta y\right)+C_4\cos \left(\beta y\right)\right)=\mathrm{0,}{C}_2=0}$

• $E_{0z}\left(x\mathrm{,0}\right)=0$

${\displaystyle E_{0z}\left(x\mathrm{,0}\right)=\left(C_1\sin \left(\alpha x\right)+C_2\cos \left(\alpha x\right)\right)C_4=\mathrm{0,}{C}_4=0}$

• $E_{0z}\left(x,b\right)=0$

${\displaystyle E_{0z}\left(x,b\right)=C_1\sin \left(\alpha x\right)C_3\sin \left(\beta b\right)=\mathrm{0,}\sin \left(\beta b\right)=0}$

• $E_{0z}\left(a,y\right)=0$

${\displaystyle E_{0z}\left(a,y\right)=C_1\sin \left(\alpha a\right)C_3\sin \left(\beta y\right)=\mathrm{0,}\sin \left(\alpha a\right)=0}$

Obtenemos valores discretos de α y β

${\displaystyle \begin{array}{l}\{\begin{array}{l}\sin \left(\alpha a\right)=\mathrm{0,}\alpha a=m\pi ,m=\mathrm{0,1,2,3...}\\ \sin \left(\beta b\right)=\mathrm{0,}\beta b=n\pi ,n=\mathrm{0,1,2,3...}\end{array}\\ {\gamma }_{m,n}^2={\alpha }^2+{\beta }^2={\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2\end{array}}$

Amplitudes

La amplitud E_0z de la componente E_z del campo magnético vale

${\displaystyle \begin{array}{l}{E}_{0z}=C_1\sin \left(\frac{m\pi }{a}x\right)C_3\sin \left(\frac{n\pi }{b}y\right)\\ E_{0z}=E_0\sin \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\end{array}}$

Donde el nuevo coeficiente E₀=C₁C₃

La componente E_z del campo magnético vale

${\displaystyle E_z=\mathrm{Re}\left\{E_{0z}\exp \left(i\left(kz-\omega t\right)\right)\right\}=E_0\sin \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\cos \left(kz-\omega t\right)}$

Conocido E_0z y B_0z=0, determinamos las otras amplitudes de las componentes del campo eléctrico y magnético

• Componente E_x

${\displaystyle \begin{array}{l}{E}_{0x}=\frac{i}{{\gamma }^2}\left(\omega \frac{\partial B_{0z}}{\partial y}+k\frac{\partial E_{0z}}{\partial x}\right)=\frac{i}{{\gamma }^2}k\frac{\partial E_{0z}}{\partial x}\\ E_{0x}=\frac{i}{{\gamma }^2}k{E}_0\frac{m\pi }{a}\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\\ E_x=\mathrm{Re}\left\{E_{0x}\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{k}{{\gamma }^2}\frac{m\pi }{a}{E}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente E_y

${\displaystyle \begin{array}{l}{E}_{0y}=\frac{i}{{\gamma }^2}\left(k\frac{\partial E_{0z}}{\partial y}-\omega \frac{\partial B_{0z}}{\partial x}\right)=\frac{i}{{\gamma }^2}k\frac{\partial E_{0z}}{\partial y}\\ E_{0y}=\frac{i}{{\gamma }^2}k\frac{n\pi }{b}{E}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\\ E_y=\mathrm{Re}\left\{E_{0y}\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{k}{{\gamma }^2}\frac{n\pi }{b}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente B_x

${\displaystyle \begin{array}{l}{B}_{0x}=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial x}-\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial y}\right)=-\frac{i}{{\gamma }^2}\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial y}\\ B_{0x}=-\frac{i}{{\gamma }^2}\frac{\omega }{c^2}\frac{n\pi }{b}{E}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\\ B_x=\mathrm{Re}\left\{B_{0x}\exp \left(i\left(kz-\omega t\right)\right)\right\}=\frac{\omega }{c^2{\gamma }^2}\frac{n\pi }{b}{E}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

• Componente B_y

${\displaystyle \begin{array}{l}{B}_{0y}=\frac{i}{{\gamma }^2}\left(k\frac{\partial B_{0z}}{\partial y}+\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial x}\right)=\frac{i}{{\gamma }^2}\frac{\omega }{c^2}\frac{\partial E_{0z}}{\partial x}\\ B_{0y}=\frac{i}{{\gamma }^2}\frac{\omega }{c^2}\frac{m\pi }{a}{E}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\\ B_y=\mathrm{Re}\left\{B_{0y}\exp \left(i\left(kz-\omega t\right)\right)\right\}=-\frac{\omega }{c^2{\gamma }^2}\frac{m\pi }{a}{E}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\end{array}}$

Resumiendo, las componentes del campo eléctrico y magnéticos son

${\displaystyle \begin{array}{l}\overrightarrow{E}=\{\begin{array}{l}{E}_x=-\frac{k}{{\gamma }^2}\frac{m\pi }{a}{E}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ E_y=-\frac{k}{{\gamma }^2}\frac{n\pi }{b}{B}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ E_z=E_0\sin \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\cos \left(kz-\omega t\right)\end{array}\\ \overrightarrow{B}=\{\begin{array}{l}{B}_x=\frac{\omega }{c^2{\gamma }^2}\frac{n\pi }{b}{E}_0\sin \left(\frac{m\pi }{a}x\right)\cos \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ B_y=-\frac{\omega }{c^2{\gamma }^2}\frac{m\pi }{a}{E}_0\cos \left(\frac{m\pi }{a}x\right)\sin \left(\frac{n\pi }{b}y\right)\sin \left(kz-\omega t\right)\\ B_z=0\end{array}\end{array}}$