一个公式教你背会 矩形波导或圆波导(或者矩形或圆形谐振腔)以纵向分量为领矢得到全部的场表达式
看不懂的我抽空来补充完整
开心！！之前这个我一直没背下来，现在摸到石头边了
以纵向分量为领向矢量
\[\left[
\begin{array}{cccc}
E_{u}\\
E_{v}\\
H_{u}\\
H_{v}
\end{array}
\right ]=
\frac{1}{K_c^2}
\left[
\begin{array}{cccc}
-\gamma& 0 & 0 & {-j\omega\mu}\\
0& -\gamma &j\omega \mu &0\\
0& j\omega\epsilon &-\gamma&0 \\
-j\omega \epsilon& 0 &0 &-\gamma
\end{array}
\right ]
\left[
\begin{array}{cccc}
\frac{\partial E_z}{h_1\partial u}\\
\frac{\partial E_z}{h_2\partial v}\\
\frac{\partial H_z}{h_1\partial u} \\
\frac{\partial H_z}{h_2\partial v}
\end{array}
\right ]
\]
矩形波导的情况就是
\[u=x,\quad v=y\\
h_1=1,\quad h_2=1
\]
圆波导的情况就是
\[u=r,\quad v=\varphi\\
h_1=1,\quad h_2=r
\]
波导，纵向分量为领向矢量
矩形波导\(TE_{mn}\)
\[H_z=H_0\cos(\frac{m\pi}{a}x)\cos(\frac{n\pi}{b}y)e^{-j\beta z}\\
E_z=0
\]
矩形波导\(TM_{mn}\)
\[E_z=E_0\sin(\frac{m\pi}{a}x)\sin(\frac{n\pi}{b}y)e^{-j\beta z}\\
H_z=0
\]
圆波导\(TE_{mn}\)
\[H_z=H_0J_m(K_cr)\mathop{}\limits_{\sin(m\varphi)}^{cos(m\varphi)}e^{-j\beta z}\\
E_z=0
\]
圆波导\(TM_{mn}\)
\[E_z=E_0J_m(K_cr)\mathop{}\limits_{\sin(m\varphi)}^{cos(m\varphi)}e^{-j\beta z}\\
H_z=0
\]
如果是谐振腔的话公式也很像。
4个\(-\gamma\)都变成 \(\frac{\partial}{\partial z}\)
\[\left[
\begin{array}{cccc}
E_{u}\\
E_{v}\\
H_{u}\\
H_{v}
\end{array}
\right ]=
\frac{1}{K_c^2}
\left[
\begin{array}{cccc}
\frac{\partial}{\partial z}& 0 & 0 & {-j\omega\mu}\\
0& \frac{\partial}{\partial z} &j\omega \mu &0\\
0& j\omega\epsilon &\frac{\partial}{\partial z}&0 \\
-j\omega \epsilon& 0 &0 &\frac{\partial}{\partial z}
\end{array}
\right ]
\left[
\begin{array}{cccc}
\frac{\partial E_z}{h_1\partial u}\\
\frac{\partial E_z}{h_2\partial v}\\
\frac{\partial H_z}{h_1\partial u} \\
\frac{\partial H_z}{h_2\partial v}
\end{array}
\right ]
\]
谐振腔，纵向分量为领向矢量
矩形谐振腔，\(TM_{mnp}\)
\[E_z=2E_0\sin(\frac{m\pi}{a}x)\sin(\frac{n\pi}{b}y)\cos(\frac{p\pi}{l}z)\\
H_z=0
\]
矩形谐振腔，\(TE_{mnp}\)
\[H_z=-2j H_0\cos(\frac{m\pi}{a}x)\cos(\frac{n\pi}{b}y)\sin(\frac{p\pi}{l}z)\\
E_z=0
\]
圆形谐振腔，\(TM_{mnp}\)
\[E_z=2E_0 J_m(K_c r)\mathop{}\limits_{\sin(m\varphi)}^{cos(m\varphi)}\cos(\frac{p\pi}{l}z)\\
H_z=0
\]
圆形谐振腔，\(TE_{mnp}\)
\[H_z=-2j H_0 J_m(K_c r)\mathop{}\limits_{\sin(m\varphi)}^{cos(m\varphi)}\sin(\frac{p\pi}{l}z)\\
E_z=0
\]
其他的
另一个有意思的是Lorentz变换矩阵，虽然用的频率低，但是结构很漂亮
\[\left(\begin{array}{c}
x \\
y \\
z \\
i c t
\end{array}\right)=\left(\begin{array}{cccc}
\gamma & 0 & 0 & i \beta \gamma \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-i \beta \gamma & 0 & 0 & \gamma
\end{array}\right)\left(\begin{array}{c}
x^{\prime} \\
y^{\prime} \\
z^{\prime} \\
i c t^{\prime}
\end{array}\right)
\]
其中，
\[\beta:=\frac{u}{c}, \quad \gamma:=\frac{1}{\sqrt{1-\beta^{2}}}
\]
Lorentz矩阵是个正交矩阵，\(A^{-1}=A^{\mathsf{T}}\)
