My name is Alexander FufaeV and here I will explain the following topic:

Derivation of The Wave Equation for E-field and B-field

Table of contents

Wave equation for the electric field Here we derive the wave equation for the E-field from Maxwell's third equation in vacuum.
Wave equation for the magnetic field Here we derive the wave equation for the B-field from Maxwell's fourth equation in vacuum.

Electromagnetic wave with E-field and B-field component.

The goal is to derive the wave equation for the electric field $\boldsymbol{E}$ and the magnetic field $\boldsymbol{B}$ using Maxwell's equations in free space.

We use the four Maxwell equations of electrodynamics in charge-free ($\rho = 0$) and current-free ($\boldsymbol{j} = 0$) space:

$$ \begin{align} \nabla ~\cdot~ \boldsymbol{E} ~&=~ 0 \\\\
\nabla ~\cdot~ \boldsymbol{B} ~&=~ 0 \\\\
\nabla ~\times~ \boldsymbol{E} ~&=~ -\frac{\partial \boldsymbol{B}}{\partial t} \\\\
\nabla ~\times~ \boldsymbol{B} ~&=~ \mu_0 \, \varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t} \end{align} $$

Another ingredient necessary for the derivation of the two wave equations is the following relation for the curl of the curl of the vector field $\boldsymbol{F}$ (double cross product):

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{F} ~=~ \nabla \, \left(\nabla \cdot \boldsymbol{F}\right) ~-~ \nabla^2 \, \boldsymbol{F} \end{align} $$

The first summand here is the gradient of the divergence of $\boldsymbol{F}$ and the second summand is the divergence of the gradient of $\boldsymbol{F}$. This is just a mathematical relation that can be derived.

Wave equation for the electric field

Maxwell's equations 1 in vacuum are coupled differential equations. To get the wave equation for the E-field, we have to decouple the third Maxwell equation. Apply the curl "$\nabla \times $" on both sides of the third Maxwell equation:

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{E} ~=~ \nabla \times \left( -\frac{\partial \boldsymbol{B}}{\partial t} \right) \end{align} $$

The time derivative together with the minus sign may be placed before the nabla operator, because the nabla operator ( spatial derivative) does not depend on the time $t$:

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{E} ~=~ - \frac{\partial}{\partial t} \, \nabla \times \boldsymbol{B} \end{align} $$

Now we can replace the curl $ \nabla \times \boldsymbol{B} $ of the magnetic field with the help of the fourth Maxwell equation:

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{E} ~=~ -\frac{\partial}{\partial t} \, \left( \mu_0 \, \varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t} \right) \end{align} $$

The time derivative outside the parenthesis can be placed after the magnetic and electric field constants. Two time derivatives can be compactly combined to the second time derivative:

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{E} ~=~ - \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \boldsymbol{E}}{\partial t^2} \end{align} $$

One side of the wave equation is derived, namely the second time derivative of the electric field. Now all that remains is to rewrite the left-hand side into the correct form, as in a wave equation. Use the double cross product 2:

$$ \begin{align} \nabla \, \left(\nabla \cdot \boldsymbol{E}\right) ~-~ \nabla^2 \boldsymbol{E} ~=~ - \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \boldsymbol{E}}{\partial t^2} \end{align} $$

On the left hand side of Eq. 7 contains the divergence $\nabla \cdot \boldsymbol{E}$ of the electric field. According to the first Maxwell equation, the divergence of the electric field in charge-free space is always zero. Thus Eq. 7 simplifies to:

$$ \begin{align} \nabla^2 \, \class{purple}{\boldsymbol{E}} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \class{purple}{\boldsymbol{E}}}{\partial t^2} \end{align} $$

Wave equation for the magnetic field

To derive the wave equation for the magnetic field $\boldsymbol{B}$, we need to decouple Maxwell's fourth equation in Eq. 1. This is done analogously to the E-field. Apply the curl "$\nabla \times$" on both sides of the fourth Maxwell equation:

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{B} ~=~ \nabla \times \left(\mu_0 \, \varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t}\right) \end{align} $$

Now place the time derivative and the two constants on the right side in front of the Nabla operator:

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{B} ~=~ \mu_0 \, \varepsilon_0 \frac{\partial}{\partial t} \, \nabla \times \boldsymbol{E} \end{align} $$

Use Maxwell's third equation to substitute the curl $ \nabla \times \boldsymbol{E} $ of the electric field on the right-hand side:

$$ \begin{align} \nabla \times \nabla \times \boldsymbol{B} ~=~ \mu_0 \, \varepsilon_0 \frac{\partial}{\partial t} \, \left(-\frac{\partial \boldsymbol{B}}{\partial t}\right) \end{align} $$

The time derivative is combined and the double cross product on the left side is replaced using Eq. 2:

$$ \begin{align} \nabla \, \left(\nabla \cdot \boldsymbol{B}\right) ~-~ \nabla^2 \boldsymbol{B} ~=~ - \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \boldsymbol{B}}{\partial t^2} \end{align} $$

The divergence $\nabla \cdot \boldsymbol{B}$ is zero according to the second Maxwell equation. Thus you get:

$$ \begin{align} \nabla^2 \, \class{violet}{\boldsymbol{B}} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \class{violet}{\boldsymbol{B}}}{\partial t^2} \end{align} $$