Deriving the Wave Equation from Maxwell’s Equations | Quantum Science Philippines

# Derivation of the Wave Equation from Maxwell’s Equations

Our aim is to start from the Maxwell’s Equations in order for us to obtain the Wave Equation for the field vectors $\vec{E}$ and $\vec{H}$. Hence, we recall that the Maxwell’s Equations in free space is given by

(i) $\vec{\nabla}\cdot\vec{E}=0$     (ii) $\vec{\nabla}\cdot\vec{B}=0$      (iii) $\vec{\nabla}\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}$      (iv) $\vec{\nabla}\times\vec{H}=\frac{\partial \vec{D}}{\partial t}$

We start from equation (ii) $\vec{\nabla}\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}$

If we take the curl on both sides of the equation, we get

$\vec{\nabla}\times ( \vec{\nabla}\times\vec{E})=-\vec{\nabla}\times\frac{\partial \vec{B}}{\partial t}$                              [1]

But we recall the vector identity: $\vec{\nabla} \times ( \vec{\nabla}\times\vec{A}) = \vec{\nabla} (\vec{\nabla} \cdot \vec{A}) - \vec{\nabla}^2 \vec{A}$

Applying this identity,

$\vec{\nabla}\times ( \vec{\nabla}\times\vec{E})=\vec{\nabla} (\vec{\nabla} \cdot \vec{E}) - \vec{\nabla}^2 \vec{E}$                  [2]

$\vec{\nabla} (\vec{\nabla} \cdot \vec{E}) - \vec{\nabla}^2 \vec{E} =-\vec{\nabla}\times\frac{\partial \vec{B}}{\partial t}$                         [3]

$= - \frac{\partial}{\partial t} (\vec{\nabla} \times \vec{B})$                 [4]

We remember also that equation (iii) can also be written as

$\vec{\nabla}\times\vec{H}=\vec{\nabla}\times(\frac{1}{\mu_0} \vec{B})=\frac{\partial\vec{D}}{\partial t}$                             [5]

$\vec{\nabla} \times \vec{B} = \mu_0 \frac{\vec{\partial D}}{\partial t}$                                                   [6]

And using expression [6] to substitute in [4], we have

$\vec{\nabla} (\vec{\nabla} \cdot \vec{E}) - \vec{\nabla}^2 \vec{E} = - \frac{\partial}{\partial t} (\mu_0 \frac{\vec{\partial D}}{\partial t})$                      [7]

Note that in free space, there is no charge density. As a result, we have equation (i) $\vec{\nabla}\cdot\vec{E}=0$.

Using this, expression [7] reduces to

$- \vec{\nabla}^2 \vec{E} = - \mu_0\frac{\vec{\partial^2 D}}{\partial t^2}$                                              [8]

Since we want the wave equation for $\vec{E}$, we  use that

$\vec{D}=\varepsilon_0\vec{E} \rightarrow$ $\vec{\nabla}^2 \vec{E} = \mu_0\frac{\partial^2}{\partial t^2}(\varepsilon_0\vec{E})$

$\vec{\nabla}^2 \vec{E} = \mu_0 \varepsilon_0\frac{\partial^2\vec{ E}}{\partial t^2}$

Let $\frac{1}{c^2}=\mu_0 \varepsilon_0$, and finally we obtained the Wave Equation for the field vector $\vec{E}$.

$\mathbf{\vec{\nabla}^2 \vec{E} = \frac{1}{c^2}\frac{\partial^2\vec{ E}}{\partial t^2}}$

In a similar manner, we can obtain an expression for $\vec{H}$ using (iii).

Following the same process, we will obtain the Wave Equation for the field vector $\vec{H}$.

$\mathbf{\vec{\nabla}^2 \vec{H} = \frac{1}{c^2}\frac{\partial^2\vec{ H}}{\partial t^2}}$