动力学习题解答参考 第五章电磁波的辐射 V2A(x,1)=-k2A(x,1) (2)式右边的积分式中,被积函数为0,积分为0。 d-a(o) d2+k2c2a4(1)=0,亦即a满足谐振子方程 2)选取规范V·A=0,q=0,于是有 V.A=v∑[a2)ex+a(0)=∑la(Vex+a(ov.el ∑[k·a()ie3-k·a(0)e-5=0 ∵a4(1),a(1)是线性无关的正交组 要使上式成立,仅当k·a=k·a=0时 故,证得当取V·A=0,9=0时,k·a4=0 3)已知A(x1)=∑[a(l+a()e-s B=V×A=∑[ikaA()e“-a()e- E=-V aA da(①2“k<a((取规范vA=0.=0) at dt 5.设A和φ是满足洛伦兹规范的矢势和标势 (1)引入一矢量函数z(元t)(赫兹矢量),若令g=V2,证明12 (2)若令P=-V·F证明2满足方程V221a1-C40P,写出在真空中的推 迟解。 3)证明E和B可通过Z用下列公式表出,E=Vx(×2)-c2HP,B=V×2 at 解:1)证明:A与φ满足洛仑兹规范,故有V·A+ ∵φ=-V·Z代入洛仑兹规范,有: V·A+-2(-V.Z)=0,即V.A=V(电动力学习题解答参考 第五章 电磁波的辐射 - 5 - ( , ) ( , ) 2 2 A x t k A x t v v v v ∴∇ = − ∴ 2 式右边的积分式中 被积函数为 0 积分为 0 ∴ ( ) 0 ( ) 2 2 2 2 + k c a t = dt d a t k k v v 亦即 k a v 满足谐振子方程 2 选取规范∇ ⋅ A = 0,ϕ = 0 v 于是有 ∑ ∑ ⋅ − ⋅ ⋅ − ⋅ ∇ ⋅ = ∇ ⋅ + = ∇ ⋅ + ∇ ⋅ k ik x k ik x k k ik x k ik x k A [a (t)e a (t)e ] [a (t) e a (t) e ] * v v v v v v v v v v v v v [ ( ) ( ) ] 0 * = ∑ ⋅ ⋅ − ⋅ ⋅ = ⋅ − ⋅ k ik x k ik x k k a t ie k a t ie v v v v v v v v ( ), ( ) * a t a t k k v v Q 是线性无关的正交组 ∴要使上式成立 仅当 0 * k ⋅ ak = k ⋅ ak = v v v v 时 ∴故 证得当取∇ ⋅ A = 0,ϕ = 0 v 时 ⋅ = 0 k k a v v 3 已知 ∑ ⋅ − ⋅ = + k ik x k ik x k A(x,t) [a (t)e a (t)e ] * v v v v v v v v ∑ ⋅ − ⋅ ∴ = ∇ × = − k ik x k ik x k B A [ika (t)e ika (t)e ] * v v v v v v v v v ∑ ⋅ − ⋅ = − + ∂ ∂ = −∇ − k k ik x k ik x e dt da t e dt da t t A E ] ( ) ( ) [ * v v v v v v v v ϕ 取规范∇ ⋅ A = 0,ϕ = 0) v 5. 设 A v 和ϕ 是满足洛伦兹规范的矢势和标势 1 引入一矢量函数 Z(x,t) v v 赫兹矢量 若令 Z v ϕ = ∇ ⋅ 证明 t Z c A ∂ ∂ = v v 2 1 2 若令 P v ρ = −∇ ⋅ 证明 Z v 满足方程 c P t Z c Z v v v 0 2 2 2 2 2 1 = − µ ∂ ∂ ∇ − 写出在真空中的推 迟解 3 证明 E B v v 和 可通过 Z v 用下列公式表出 E Z c P v v v 0 2 = ∇ × (∇ × ) − µ , Z c t B v v ∇ × ∂ ∂ = 2 1 解 1 证明 A v 与ϕ 满足洛仑兹规范 故有 0 1 2 = ∂ ∂ ∇ ⋅ + c t A v ϕ = −∇ ⋅ Ζ v Qϕ 代入洛仑兹规范 有 ( ) 0 1 2 −∇ ⋅ Ζ = ∂ ∂ ∇ ⋅ + ⋅ v v c t A 即 ) 1 ( 2 c t A ∂ ∂Ζ ∇ ⋅ = ∇ ⋅ v v