Electromagnetic wave equation in free space

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations.

1. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$
2. $\overrightarrow{\nabla}. \overrightarrow{B}=0$
3. $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$
4. $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}$
   
   
   Modified Form:
   
   
   $\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}+ \frac{\partial \overrightarrow{D}}{\partial t}$

For free space:

Current density ($\overrightarrow{J}$) = 0
Volume charge distribution ($\rho$) = 0
Permittivity $\epsilon = \epsilon_{0}$
Permeability $\mu = \mu_{0}$

Now, Maxwell's equation for free space:

$\overrightarrow{\nabla}. \overrightarrow{E}=0 \qquad(1)$
$\overrightarrow{\nabla}. \overrightarrow{B}=0 \qquad(2)$
$\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} \qquad(3)$
$\overrightarrow{\nabla} \times \overrightarrow{H}= 0$

Modified form for free space:

$\overrightarrow{\nabla} \times \overrightarrow{H}= \frac{\partial \overrightarrow{D}}{\partial t} \qquad(4)$

Now, On solving Maxwell's equation for free space we get the electromagnetic wave equation for free space. The electromagnetic wave equation has both an electric field vector and a magnetic field vector. So Maxwell's equations for free space give two-equation for electromagnetic wave i.e. one is for electric field vector($\overrightarrow{E}$) and the second is for magnetic field vector ($\overrightarrow{H}$).

Electromagnetic wave equation for free space in term of $\overrightarrow{E}$:

Now from equation $(3)$

$\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} $

Now take the curl on both sides of the above equation

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{E})=-\overrightarrow{\nabla} \times \frac{\partial \overrightarrow{B}}{\partial t} $

$(\overrightarrow{\nabla}. \overrightarrow{E}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{E}=-\frac{\partial}{\partial t} (\overrightarrow{\nabla} \times \overrightarrow{B}) \qquad(5)$

But for free space:

$\overrightarrow{\nabla}. \overrightarrow{E}=0 \qquad$   {from equation (1)}

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu_{0} \frac{\partial \overrightarrow{D}}{\partial t}\qquad $   {from equation (4)}

Now substitute these values in equation $(5)$. So

$ -\nabla^{2}.\overrightarrow{E}=-\frac{\partial}{\partial t} (\mu_{0} \frac{\partial \overrightarrow{D}}{\partial t})$

$ -\nabla^{2}.\overrightarrow{E}=-\mu_{0} \frac{\partial^{2} \overrightarrow{D}}{\partial t^{2}} $

$ \nabla^{2}.\overrightarrow{E}=\mu_{0} \frac{\partial^{2}}{\partial t^{2}} (\epsilon_{0} \overrightarrow{E}) \qquad \left( \because \overrightarrow{D}=\epsilon_{0} \overrightarrow{E} \right)$

$\nabla^{2}.\overrightarrow{E}=\mu_{0} \epsilon_{0} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}$

The value of $\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}= c$. Where $c$ is the speed of the wave in free space. So the above equation is often written as

$\nabla^{2}.\overrightarrow{E}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}$

This is an electromagnetic wave equation for free space in terms of electric field vector ($\overrightarrow{E}$).

Electromagnetic wave equation for free space in term of $\overrightarrow{H}$:

Now from equation $(4)$

$\overrightarrow{\nabla} \times \overrightarrow{H}= \frac{\partial \overrightarrow{D}}{\partial t}$

Now take the curl on both sides of the above equation

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{H})=\overrightarrow{\nabla} \times \frac{\partial \overrightarrow{D}}{\partial t} $

$(\overrightarrow{\nabla}. \overrightarrow{H}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{H}=\frac{\partial}{\partial t} (\overrightarrow{\nabla} \times \overrightarrow{D}) \qquad(6)$

But for free space:

$\overrightarrow{\nabla}. \overrightarrow{B}=0 \qquad $  {from equation (2)}

$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$

$\overrightarrow{\nabla} \times \overrightarrow{D}= -\epsilon_{0} \frac{\partial \overrightarrow{B}}{\partial t} \qquad$   {from equation (3)}

Now substitute these values in equation $(6)$. So

$ -\nabla^{2}.\overrightarrow{H}=-\frac{\partial}{\partial t} (\epsilon_{0} \frac{\partial \overrightarrow{B}}{\partial t})$

$ -\nabla^{2}.\overrightarrow{H}=-\epsilon_{0} \frac{\partial^{2} \overrightarrow{B}}{\partial t^{2}} $

$ \nabla^{2}.\overrightarrow{H}=\epsilon_{0} \frac{\partial^{2}}{\partial t^{2}} (\mu_{0} \overrightarrow{H}) \qquad \left(\because \overrightarrow{B}=\mu_{0} \overrightarrow{H} \right)$

$\nabla^{2}.\overrightarrow{H}=\mu_{0} \epsilon_{0} \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}}$

The value of $\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}= c$. Where $c$ is the speed of the wave in free space. So the above equation is often written as

$\nabla^{2}.\overrightarrow{H}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{H}}{\partial t^{2}}$

This is the electromagnetic wave equation for free space in terms of electric field vector ($\overrightarrow{H}$).