Finding the Speed of Electromagnetic Light from Maxwell's Equations
📂Electrodynamics Finding the Speed of Electromagnetic Light from Maxwell's Equations Vacuum Maxwell’s Equations
∇ ⋅ E = 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 ϵ 0 ∂ E ∂ t
\begin{align}
\nabla \cdot \mathbf{E} &= 0 \\[1em]
\nabla \cdot \mathbf{B} &= 0 \\[1em]
\nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \\[1em]
\nabla \times \mathbf{B} &= \mu_{0}\epsilon_{0}\frac{\partial \mathbf{E}}{\partial t}
\end{align}
∇ ⋅ E ∇ ⋅ B ∇ × E ∇ × B = 0 = 0 = − ∂ t ∂ B = μ 0 ϵ 0 ∂ t ∂ E
One-dimensional wave equation
∂ 2 f ∂ x 2 = 1 v 2 ∂ 2 f ∂ t 2
\frac{\partial^2 f}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 f}{\partial t^2}
∂ x 2 ∂ 2 f = v 2 1 ∂ t 2 ∂ 2 f
Three-dimensional wave equation
∇ 2 f = 1 v 2 ∂ 2 f ∂ t 2
\nabla ^2 f = \frac{1}{v^2}\frac{\partial ^2 f}{\partial t^2}
∇ 2 f = v 2 1 ∂ t 2 ∂ 2 f
Derivation The goal is to derive a wave equation form from Maxwell’s equations, regarding E \mathbf{E} E B \mathbf{B} B ( 3 ) (3) ( 3 )
∇ × ( ∇ × E ) = ∇ × ( − ∂ B ∂ t ) = − ∂ ∂ t ( ∇ × B ) = − ∂ ∂ t ( μ 0 ϵ 0 ∂ E ∂ t ) = − μ 0 ϵ 0 ∂ 2 E ∂ t 2
\begin{align*}
\nabla \times (\nabla \times \mathbf{E}) &= \nabla \times \left( -\frac{\partial \mathbf{B}}{\partial t} \right)
\\ &= -\frac{\partial }{\partial t} (\nabla \times \mathbf{B})
\\ &= -\frac{\partial}{\partial t} \left( \mu_{0}\epsilon_{0} \frac{\partial \mathbf{E}}{\partial t} \right)
\\ &= -\mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{E}}{\partial t^2}
\end{align*}
∇ × ( ∇ × E ) = ∇ × ( − ∂ t ∂ B ) = − ∂ t ∂ ( ∇ × B ) = − ∂ t ∂ ( μ 0 ϵ 0 ∂ t ∂ E ) = − μ 0 ϵ 0 ∂ t 2 ∂ 2 E
The third equality holds by ( 4 ) (4) ( 4 ) ( 4 ) (4) ( 4 )
∇ × ( ∇ × B ) = ∇ × ( μ 0 ϵ 0 ∂ E ∂ t ) = μ 0 ϵ 0 ∂ ∂ t ( ∇ × E ) = − μ 0 ϵ 0 ∂ 2 B ∂ t 2
\begin{align*}
\nabla \times (\nabla \times \mathbf{B}) &= \nabla \times \left( \mu_{0} \epsilon_{0} \frac{\partial \mathbf{E}}{\partial t} \right)
\\ &= \mu_{0}\epsilon_{0} \frac{\partial }{\partial t}(\nabla \times \mathbf{E})
\\ &= -\mu_{0}\epsilon_{0} \frac {\partial ^2 \mathbf{B} }{\partial t^2}
\end{align*}
∇ × ( ∇ × B ) = ∇ × ( μ 0 ϵ 0 ∂ t ∂ E ) = μ 0 ϵ 0 ∂ t ∂ ( ∇ × E ) = − μ 0 ϵ 0 ∂ t 2 ∂ 2 B
And since ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A \nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A})-\nabla^2\mathbf{A} ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A
∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E = − μ 0 ϵ 0 ∂ 2 E ∂ t 2
\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E})-\nabla^2\mathbf{E}=-\mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{E}}{\partial t^2}
∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E = − μ 0 ϵ 0 ∂ t 2 ∂ 2 E
∇ × ( ∇ × B ) = ∇ ( ∇ ⋅ B ) − ∇ 2 B = − μ 0 ϵ 0 ∂ 2 B ∂ t 2
\nabla \times (\nabla \times \mathbf{B}) = \nabla(\nabla \cdot \mathbf{B})-\nabla^2\mathbf{B}=-\mu_{0}\epsilon_{0}\frac{\partial ^2 \mathbf{B}}{\partial t^2}
∇ × ( ∇ × B ) = ∇ ( ∇ ⋅ B ) − ∇ 2 B = − μ 0 ϵ 0 ∂ t 2 ∂ 2 B
Finally, because of ∇ ⋅ E = 0 \nabla \cdot \mathbf{E}=0 ∇ ⋅ E = 0 ∇ ⋅ B = 0 \nabla \cdot \mathbf{B}=0 ∇ ⋅ B = 0
∇ 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}
∇ 2 E = μ 0 ϵ 0 ∂ t 2 ∂ 2 E
∇ 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}
∇ 2 B = μ 0 ϵ 0 ∂ t 2 ∂ 2 B
Now, from Maxwell’s equations, E \mathbf{E} E B \mathbf{B} B 1 v 2 = μ 0 ϵ 0 \dfrac{1}{v^2}=\mu_{0}\epsilon_{0} v 2 1 = μ 0 ϵ 0 v = 1 μ 0 ϵ 0 = 3.00 × 1 0 8 m / s v=\dfrac{1}{\sqrt{\mu_{0}\epsilon_{0}}}=3.00\times 10^8 m/s v = μ 0 ϵ 0 1 = 3.00 × 1 0 8 m / s
