从 麦克斯韦方程组 到 波动方程 再到 二维波动方程

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从麦克斯韦方程组,推导出了电场和磁场的波动方程。这两个波动方程描述了电磁场的传播,并且它们的形式与标准的波动方程一致。波动方程表明电场和磁场都以波的形式传播,传播速度 c c c 由以下关系确定:
c = 1 μ 0 ϵ 0 c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} c=μ0ϵ0 1这个速度 c c c 恰好是光速,这证明了光是一种电磁波。

麦克斯韦方程组在真空中的形式

  1. 高斯定律(电场)
    ∇ ⋅ E = ρ ϵ 0 \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} E=ϵ0ρ

  2. 高斯定律(磁场)
    ∇ ⋅ B = 0 \nabla \cdot \mathbf{B} = 0 B=0

  3. 法拉第电磁感应定律
    ∇ × E = − ∂ B ∂ t \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} ×E=tB

  4. 安培-麦克斯韦方程
    ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} ×B=μ0J+μ0ϵ0tE
    在真空中,电荷密度 ρ \rho ρ 和电流密度 J \mathbf{J} J 都为零,因此方程简化为:

  5. ∇ ⋅ E = 0 \nabla \cdot \mathbf{E} = 0 E=0

  6. ∇ ⋅ B = 0 \nabla \cdot \mathbf{B} = 0 B=0

  7. ∇ × E = − ∂ B ∂ t \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} ×E=tB

  8. ∇ × B = μ 0 ϵ 0 ∂ E ∂ t \nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} ×B=μ0ϵ0tE

推导电场的波动方程

首先对法拉第电磁感应定律取旋度:
∇ × ( ∇ × E ) = ∇ × ( − ∂ B ∂ t ) \nabla \times (\nabla \times \mathbf{E}) = \nabla \times \left( -\frac{\partial \mathbf{B}}{\partial t} \right) ×(×E)=×(tB)利用向量分析中的恒等式 ∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E \nabla \times (\nabla \times \mathbf{E}) = \nabla (\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} ×(×E)=(E)2E,并结合高斯定律(电场) ∇ ⋅ E = 0 \nabla \cdot \mathbf{E} = 0 E=0,得到:
∇ × ( ∇ × E ) = − ∇ 2 E \nabla \times (\nabla \times \mathbf{E}) = -\nabla^2 \mathbf{E} ×(×E)=2E所以:
− ∇ 2 E = ∇ × ( − ∂ B ∂ t ) -\nabla^2 \mathbf{E} = \nabla \times \left( -\frac{\partial \mathbf{B}}{\partial t} \right) 2E=×(tB)
∇ 2 E = ∂ ∂ t ( ∇ × B ) \nabla^2 \mathbf{E} = \frac{\partial}{\partial t} (\nabla \times \mathbf{B}) 2E=t(×B)再利用安培-麦克斯韦方程中的 ∇ × B = μ 0 ϵ 0 ∂ E ∂ t \nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} ×B=μ0ϵ0tE,得到:
∇ 2 E = ∂ ∂ t ( μ 0 ϵ 0 ∂ E ∂ t ) \nabla^2 \mathbf{E} = \frac{\partial}{\partial t} \left( \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right) 2E=t(μ0ϵ0tE)
∇ 2 E = μ 0 ϵ 0 ∂ 2 E ∂ t 2 \nabla^2 \mathbf{E} = \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} 2E=μ0ϵ0t22E这就是电场的波动方程:
∇ 2 E − μ 0 ϵ 0 ∂ 2 E ∂ t 2 = 0 \nabla^2 \mathbf{E} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} = 0 2Eμ0ϵ0t22E=0

推导磁场的波动方程

对安培-麦克斯韦方程取旋度:
∇ × ( ∇ × B ) = ∇ × ( μ 0 ϵ 0 ∂ E ∂ t ) \nabla \times (\nabla \times \mathbf{B}) = \nabla \times \left( \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right) ×(×B)=×(μ0ϵ0tE)同样利用向量恒等式 ∇ × ( ∇ × B ) = ∇ ( ∇ ⋅ B ) − ∇ 2 B \nabla \times (\nabla \times \mathbf{B}) = \nabla (\nabla \cdot \mathbf{B}) - \nabla^2 \mathbf{B} ×(×B)=(B)2B 和高斯定律(磁场) ∇ ⋅ B = 0 \nabla \cdot \mathbf{B} = 0 B=0,得到:
∇ × ( ∇ × B ) = − ∇ 2 B \nabla \times (\nabla \times \mathbf{B}) = -\nabla^2 \mathbf{B} ×(×B)=2B所以:
− ∇ 2 B = ∇ × ( μ 0 ϵ 0 ∂ E ∂ t ) -\nabla^2 \mathbf{B} = \nabla \times \left( \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right) 2B=×(μ0ϵ0tE)
∇ 2 B = μ 0 ϵ 0 ∂ ∂ t ( ∇ × E ) \nabla^2 \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial}{\partial t} (\nabla \times \mathbf{E}) 2B=μ0ϵ0t(×E)再利用法拉第电磁感应定律 ∇ × E = − ∂ B ∂ t \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} ×E=tB,得到:
∇ 2 B = μ 0 ϵ 0 ∂ ∂ t ( − ∂ B ∂ t ) \nabla^2 \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial}{\partial t} \left( -\frac{\partial \mathbf{B}}{\partial t} \right) 2B=μ0ϵ0t(tB)
∇ 2 B = μ 0 ϵ 0 ∂ 2 B ∂ t 2 \nabla^2 \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2} 2B=μ0ϵ0t22B这就是磁场的波动方程:
∇ 2 B − μ 0 ϵ 0 ∂ 2 B ∂ t 2 = 0 \nabla^2 \mathbf{B} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2} = 0 2Bμ0ϵ0t22B=0

从标准波动方程到二维波动方程

从标准波动方程推导出二维波动方程。先从三维标准波动方程出发,逐步简化到二维波动方程。
标准波动方程在三维空间中的一般形式为:
1 v 2 ∂ 2 ψ ∂ t 2 = ∇ 2 ψ \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} = \nabla^2 \psi v21t22ψ=2ψ
其中:

  • ψ ( x , y , z , t ) \psi(x, y, z, t) ψ(x,y,z,t) 是描述波动的函数。

  • v v v 是波速。

  • ∇ 2 ψ \nabla^2 \psi 2ψ 是拉普拉斯算子,定义为:
    ∇ 2 ψ = ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 + ∂ 2 ψ ∂ z 2 \nabla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} 2ψ=x22ψ+y22ψ+z22ψ

推导二维波动方程

为了得到二维波动方程,需要假设波动在 z z z 方向上没有变化。这意味着:
∂ ψ ∂ z = 0 \frac{\partial \psi}{\partial z} = 0 zψ=0
从而: ∂ 2 ψ ∂ z 2 = 0 \frac{\partial^2 \psi}{\partial z^2} = 0 z22ψ=0因此,拉普拉斯算子简化为:
∇ 2 ψ = ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 \nabla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} 2ψ=x22ψ+y22ψ将这个简化的拉普拉斯算子代入标准波动方程中,得到:
1 v 2 ∂ 2 ψ ∂ t 2 = ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} v21t22ψ=x22ψ+y22ψ通常会把 ψ \psi ψ 改写为 u u u 以表示二维波动函数,且将波速 v v v 改写为 c c c ,于是方程变为:
1 c 2 ∂ 2 u ∂ t 2 = ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} c21t22u=x22u+y22u为了使方程更加简洁,将 c 2 c^2 c2 移到右边:
∂ 2 u ∂ t 2 = c 2 ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ) \frac{\partial^2 u}{\partial t^2} = c^2 \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) t22u=c2(x22u+y22u)
这就是二维波动方程。

把步骤简化就是

  1. 从标准波动方程出发:
    1 v 2 ∂ 2 ψ ∂ t 2 = ∇ 2 ψ \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} = \nabla^2 \psi v21t22ψ=2ψ

  2. 将拉普拉斯算子展开为三维形式:
    ∇ 2 ψ = ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 + ∂ 2 ψ ∂ z 2 \nabla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} 2ψ=x22ψ+y22ψ+z22ψ

  3. 假设波动在 z z z 方向上没有变化,即 ∂ ψ ∂ z = 0 \frac{\partial \psi}{\partial z} = 0 zψ=0 ,于是 ∂ 2 ψ ∂ z 2 = 0 \frac{\partial^2 \psi}{\partial z^2} = 0 z22ψ=0

  4. 简化拉普拉斯算子:
    ∇ 2 ψ = ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 \nabla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} 2ψ=x22ψ+y22ψ

  5. 将简化后的拉普拉斯算子代入标准波动方程:
    1 v 2 ∂ 2 ψ ∂ t 2 = ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} v21t22ψ=x22ψ+y22ψ

  6. 改写为二维波动函数 u u u 和波速 c c c ,得到:
    1 c 2 ∂ 2 u ∂ t 2 = ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} c21t22u=x22u+y22u

  7. 最终整理得到二维波动方程:
    ∂ 2 u ∂ t 2 = c 2 ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ) \frac{\partial^2 u}{\partial t^2} = c^2 \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) t22u=c2(x22u+y22u)

这样就从标准波动方程推导出了二维波动方程。

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