动力学习题解答 四章电磁波的传播 (2)证明D=-2[k2E-(k,E)k (3)证明能流S与波矢k一般不在同方向上。 证明:1)由麦氏方程组 V×H V·B=0 V.B=B0·Ve5xm)=i,Bex)=i·B=0 同理k·D=0 日=[Vex]xF0=ix厅 i×B D B·B B·(k×B)=0 E=Ve-m)]×E0=ik×E=iDB E=-(kxE).E=0, VE=ik E V·E一般≠0,即k.E一般≠0电动力学习题解答 第四章 电磁波的传播 - 3 - 2 证明 [ ( ) ] 1 2 2 D k E k E k v v v v v = − ⋅ ω µ 3 证明能流 S v 与波矢 k v 一般不在同方向上 证明 1 由麦氏方程组          ∇ ⋅ = ∇ ⋅ = ∂ ∂ ∇ × = ∂ ∂ ∇ × = − 0 0 B D t D H t B E v v v v v v 得 0 ( ) 0 ( ) ∇ ⋅ = 0 ⋅∇ = ⋅ = ⋅ = ⋅ − ⋅ − B B e ik B e ik B i k x t i k x t v v v v v v v v v v ω ω ∴k ⋅ B = 0 r v 同理 k ⋅ D = 0 v v H e H ik H i D i k x t v v v v v v v ω ω ∇ × = ∇ × = × = − ⋅ − 0 ( ) [ ] ik B i D v v v ∴ × = − µω ( ) 0 1 ∴ B ⋅ D = − B ⋅ k × B = v v v v v µω E e E ik E i B i k x t v v v v v v v ω ω ∇ × = ∇ × = × = ⋅ − 0 ( ) [ ] ( ) 0 1 ∴ B ⋅ E = k × E ⋅ E = v v v v v ω E ik E v v v ∇ ⋅ = ⋅ D E v v Q ≠ ε E v ∴∇ ⋅ 一般 ≠ 0 即 k E v v ⋅ 一般 ≠ 0