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[
egin{array}{cccc}
E_{u}\
E_{v}\
H_{u}\
H_{v}
end{array}
ight ]=
frac{1}{K_c^2}
left[
egin{array}{cccc}
-gamma& 0 & 0 & {-jomegamu}\
0& -gamma &jomega mu &0\
0& jomegaepsilon &-gamma&0 \
-jomega epsilon& 0 &0 &-gamma
end{array}
ight ]
left[
egin{array}{cccc}
frac{partial E_z}{h_1partial u}\
frac{partial E_z}{h_2partial v}\
frac{partial H_z}{h_1partial u} \
frac{partial H_z}{h_2partial v}
end{array}
ight ]
]
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_0cos(frac{mpi}{a}x)cos(frac{npi}{b}y)e^{-jeta z}\
E_z=0
]
矩形波导(TM_{mn})
[E_z=E_0sin(frac{mpi}{a}x)sin(frac{npi}{b}y)e^{-jeta z}\
H_z=0
]
圆波导(TE_{mn})
[H_z=H_0J_m(K_cr)mathop{}limits_{sin(mvarphi)}^{cos(mvarphi)}e^{-jeta z}\
E_z=0
]
圆波导(TM_{mn})
[E_z=E_0J_m(K_cr)mathop{}limits_{sin(mvarphi)}^{cos(mvarphi)}e^{-jeta 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[
egin{array}{cccc}
E_{u}\
E_{v}\
H_{u}\
H_{v}
end{array}
ight ]=
frac{1}{K_c^2}
left[
egin{array}{cccc}
frac{partial}{partial z}& 0 & 0 & {-jomegamu}\
0& frac{partial}{partial z} &jomega mu &0\
0& jomegaepsilon &frac{partial}{partial z}&0 \
-jomega epsilon& 0 &0 &frac{partial}{partial z}
end{array}
ight ]
left[
egin{array}{cccc}
frac{partial E_z}{h_1partial u}\
frac{partial E_z}{h_2partial v}\
frac{partial H_z}{h_1partial u} \
frac{partial H_z}{h_2partial v}
end{array}
ight ]
]
谐振腔，纵向分量为领向矢量
矩形谐振腔，(TM_{mnp})
[E_z=2E_0sin(frac{mpi}{a}x)sin(frac{npi}{b}y)cos(frac{ppi}{l}z)\
H_z=0
]
矩形谐振腔，(TE_{mnp})
[H_z=-2j H_0cos(frac{mpi}{a}x)cos(frac{npi}{b}y)sin(frac{ppi}{l}z)\
E_z=0
]
圆形谐振腔，(TM_{mnp})
[E_z=2E_0 J_m(K_c r)mathop{}limits_{sin(mvarphi)}^{cos(mvarphi)}cos(frac{ppi}{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(mvarphi)}^{cos(mvarphi)}sin(frac{ppi}{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(egin{array}{c}
x \
y \
z \
i c t
end{array}
ight)=left(egin{array}{cccc}
gamma & 0 & 0 & i eta gamma \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
-i eta gamma & 0 & 0 & gamma
end{array}
ight)left(egin{array}{c}
x^{prime} \
y^{prime} \
z^{prime} \
i c t^{prime}
end{array}
ight)
]
其中，
[eta:=frac{u}{c}, quad gamma:=frac{1}{sqrt{1-eta^{2}}}
]
Lorentz矩阵是个正交矩阵，(A^{-1}=A^T)