Electromagnetic Wave Equation and Waves in Free Space by using Maxwell’s equations

Electromagnetic Wave Equation by using Maxwell’s Equations

Maxwell’s equations are applicable to find a wave equation for electromagnetic waves can be derived by using them. Let us assume an isotropic medium having permittivity, $\varepsilon$, permeability $\mu$ and conductivity $\sigma$ but free from any other charge and current other than determined by Ohm’s law. The properties of the medium has been summarized as

$$\begin{align*}\left. \begin{matrix}\vec D &= \varepsilon \: \vec E \\\vec B &= \mu \: \vec H \\ \vec J &= \sigma\vec E \\\rho &= 0 \end{matrix}\right )\dots (1)\end{align*}$$

The Maxwell’s equations are,

$$\begin{align*}\left. \begin{matrix}\nabla . \vec {D} = \rho \\\nabla . \vec B = 0 \\ \nabla \times \vec E = -\frac {\partial \vec {B}}{\partial t}\\\nabla \times \vec H = \vec j + \frac {\partial \vec {D}}{\partial t}\\ \end{matrix}\right )\dots (2)\end{align*}$$

The Maxwell’s equations can be re-written using parameters of medium as explained in equation (1)

$$\begin{align*}\left. \begin{matrix}\nabla . \vec {D} = \rho \\\nabla . \vec B = 0 \\ \nabla \times \vec E = -\frac {\partial \vec {B}}{\partial t}= -\mu \frac {\partial \vec H}{\partial t}\\\nabla \times \vec H = \vec j + \frac {\partial \vec {D}}{\partial t}= \sigma \vec E+ \frac {\partial \vec E}{\partial t}\\ \end{matrix}\right )\dots (3)\end{align*}$$

Using 3^rd Maxwell’s equation, it can be written as

$$\begin{align*}\vec {\nabla} \times (\vec {\nabla}\ \times \vec E) &= \vec {\nabla}\times \left (-\mu \frac {\partial \vec H}{\partial t} \right ) \\ &= \mu\frac {\partial}{\partial t}(\vec {\nabla} \times H) \\&=-\mu \frac {\partial}{\partial t}\left (\sigma \vec E+ \varepsilon \frac {\partial \vec E}{\partial t} \right ) \\ &= -\mu\sigma \frac {\partial \vec E}{\partial t} - \mu\varepsilon \frac {\partial^2 \vec E}{\partial t^2}\end{align*}$$

$$\begin{align*}\vec {\nabla}\times (\vec {\nabla}\times \vec H)&= \vec{\nabla}\times\left (\sigma\vec E + \varepsilon \frac {\partial \vec E}{\partial t} \right ) \\&= \sigma (\vec {\nabla}\times \vec E)+ \varepsilon\frac {\partial}{\partial t}(\vec {\nabla}\times \vec E) \\\end{align*}$$

$$\begin{align*}\text {or,}\:\vec {\nabla} \times (\vec {\nabla} \times \vec H) &= \sigma \left (-\mu \frac {\partial \vec H}{\partial t} \right ) +\varepsilon \frac {\partial}{\partial t}\left (-\mu \frac {\partial \vec H}{\partial t} \right ) \\ &=-\mu\sigma \frac {\partial \vec H}{\partial t} - \mu\varepsilon \frac {\partial^2 \vec H}{\partial t^2}\dots (5)\end{align*}$$

One of the usual vector identities is

$$\begin{align*}\vec {\nabla}\times (\vec {\nabla}\times \vec A)= \vec {\nabla}(\vec {\nabla}.\vec A)-(\vec {\nabla}.\vec {\nabla})\vec A &= \vec {\nabla}(\vec {\nabla}.\vec A)-\vec {\nabla}.^2\vec A \\ \text {Similary},\\ \vec {\nabla}\times (\vec {\nabla}\times \vec E)&= \vec {\nabla}(\vec {\nabla}.\vec E)-\vec {\nabla}.^2\vec E \\ &= 0 -\nabla^2.\vec E\:(\because\text {For linear medium} \: \vec {\nabla}.\vec E = 0)\\&= -\nabla ^2.\vec E\end{align*}$$

$$\begin{align*} \vec {\nabla}\times (\vec {\nabla}\times \vec H)&= \vec {\nabla}(\vec {\nabla}.\vec H\nabla .^2\vec H \\ &= 0 -\vec {\nabla}.^2\vec H\:(\because \vec {\nabla}.\vec H = 0)\\&= \nabla ^2\vec H\end{align*}$$

Equations (3) and (4) becomes

$$\begin{align*} \nabla ^2\vec E &= -\mu\sigma \frac {\partial \vec E}{\partial t} - \mu\varepsilon \frac {\partial^2 \vec E}{\partial t^2}\\ \nabla ^2\vec E -\mu\sigma \frac {\partial \vec E}{\partial t} - \mu\varepsilon \frac {\partial^2 \vec E}{\partial t^2}&= 0 \dots (6)\\ \text {Similarly}\\\nabla ^2\vec H-\mu\sigma \frac {\partial \vec H}{\partial t} - \mu\varepsilon \frac {\partial^2 \vec H}{\partial t^2}&= 0 \dots (7) \end{align*}$$

Equation (6) and (7) represent the wave equations govern the electromagnetic field in isotropic medium where charge density is zero.

Plane Electromagnetic Waves in Free Space

The Maxwell’s equations are

$$\begin{align*}\left. \begin{matrix}\nabla . \vec {D} = \rho \\\nabla . \vec B = 0 \\ \nabla \times \vec E = -\frac {\partial \vec {B}}{\partial t}\\\nabla \times \vec H = \vec J+ \frac {\partial \vec {D}}{\partial t}\\ \vec B = \mu \vec H,\:\vec D= \varepsilon\vec E,\: \vec J = \sigma \vec E\end{matrix}\right )\dots (1)\end{align*}$$

$$\begin{align*}\text {In free space,}\left. \begin{matrix}\rho = 0 \\ \sigma = 0 \\ \mu = \mu_0 \\ \varepsilon =\varepsilon_0\end{matrix}\right )\dots (2)\end{align*}$$

On substituting these values , Maxwell’s equation become

$$\begin{align*}\left. \begin{matrix}\nabla . \vec E = 0\\ \nabla . \vec H = 0 \\ \nabla \times \vec E =- \mu_0 \frac {\partial \vec H}{\partial t}\\ \nabla \times \vec H =\varepsilon_0 \frac {\partial \vec E}{\partial t}\\\end{matrix}\right )\dots (3)\end{align*}$$

$$\begin{align*}\text {Similary}\\ \vec {\nabla}\times (\vec {\nabla}\times E) = \vec {\nabla}(\nabla .\vec E) - \nabla ^2 \vec E\: \dots (4)\\ \vec {\nabla}\times (\vec {\nabla}\times E) = 0 - \mu_0 \frac {\partial}{\partial}(\vec {nabla } \times \vec H)\\ &= - \mu_0\varepsilon_0\frac {\partial ^2\vec E}{\partial t^2}\: \text {using equation}\: (1) \: \dots (5)\end{align*}$$

$$\begin{align*} \text{From equations}\:(4)\:\text {and}\:(5)\\\nabla ^2 \vec E - \mu_0\varepsilon_0\frac {\partial ^2\vec E}{\partial t^2} &= 0\: \dots (6) \:(\because\vec {\nabla}\times (\vec {\nabla}\times E &= \nabla ^2 \vec E) \\ \nabla ^2 \vec H - \mu_0\varepsilon_0\frac {\partial ^2\vec H}{\partial t^2} &= 0\: \dots (7)\\ \end{align*}$$

These equations (6) and (7) represent wave equations governing electromagnetic fields $\vec E$ and $\vec H$ free space. On combination,

$$\begin{align*}\text {From equations}\: (6)\: \text {and}\:(7)\\ \nabla ^2 \vec A - \mu_0\varepsilon_0\frac {\partial ^2\vec A}{\partial t^2} = 0\end{align*}$$

General wave equation

$$\begin{align*} \nabla ^2 \vec A - \frac {1}{v^2}\frac {\partial ^2\vec A}{\partial t^2} = 0\end{align*}$$

where v is velocity of wave.

$$\begin{align*} v^2 &= \frac {1}{\mu_0\varepsilon_0} \\ \therefore v &= \frac {1}{\sqrt {\mu_0\varepsilon_0} } \dots (8)\end{align*}$$

Equation (8) conforms that light is electromagnetic wave.

Relation between $\vec K,\: \vec E\: \text {and} \vec H$ Vectors

A plane wave is defined as a wave whose amplitude is the same at any point in a plane perpendicular to a specified direction. The plane solutions of above equations in well known form may be written as

$$\begin{align*} \vec E (r, t) &= E_0e^{i(\vec k. \vec r - \omega t)} \: \dots (8) \\ \vec H (r, t) &= H_0e^{i(\vec k. \vec r - \omega t)} \: \dots (9) \\ \vec A (r, t) &= A_0e^{i(\vec k. \vec r - \omega t)} \: \dots (10) \end{align*}$$

From this, we can write

$$\begin{align*}\vec {\nabla}.\vec E &=\left (\hat i \frac {\partial }{\partial x} + \hat j \frac {\partial }{\partial y} + \hat k \frac {\partial}{\partial z} \right ). E_0e^{i(\vec K. \vec r - \omega t)} \\ &=\left (\hat i \frac {\partial }{\partial x} + \hat j \frac {\partial }{\partial y} + \hat k \frac {\partial}{\partial z} \right ) (\hat iE_{ox} + \hat jE_{oy}+ \hat zE_{oz})e^{i(k_x x+ K_yy +k_zz - \omega t)} \\ &= i. \vec K. \vec E_0 e^{i(\vec K. \vec r - \omega t)} = i\vec K. \vec E \\ \text {Similarly}\\ (\vec {\nabla } \times \vec E) &= i\vec K. \vec H \end{align*}$$

Thus the requirements $\vec {\nabla } .\vec E = 0$ and $\vec K .\vec H = 0$

$$i \vec k. \vec E = 0 \: \text {and} i\vec k. \vec H = 0$$

This shows that electromagnetic field vectors $\vec E$ and (\vec H\) are perpendicular to the direction of propagation vector $\vec k$. This implies that electromagnetic waves are transverse in character. $$\begin{align*} i \vec K \times \vec E = -\mu_0(-i\omega \vec H) \rightarrow \vec K \times \vec E = \mu_0i\omega \vec H \\ i \vec k\times \vec E = \varepsilon_0.(-i\omega \vec E)\rightarrow \vec K \times \vec H = -\varepsilon_0 \omega \vec E \\\end{align*}$$

We know

$$\begin{align*} \vec {\nabla} \times \vec E =- \mu_0 \frac {\partial \vec H}{\partial t}\:\:\:\:\: \nabla \approx ik\\ \vec {\nabla } \times \vec H =\varepsilon_0 \frac {\partial \vec E}{\partial t}\:\:\: \:\:\frac {\partial }{\partial t} \approx -i\omega\end{align*}$$

$$\begin{align*} \vec {\nabla} \times \vec E &=- \mu_0 \frac {\partial \vec H}{\partial t}\: \:\:\:\:\: \text {and} \:\vec {\nabla } \times \vec H =\varepsilon_0 \frac {\partial \vec E}{\partial t}\\ i\vec k \times \vec E &= -\mu_0(-i\omega\vec H)\:\:\: \:\:\: i\vec k \times \vec H = \varepsilon_0(-i\omega\vec E)\\ \vec k \times \vec E &= \mu_0\omega\vec H \:\:\:\:\:\: \vec k \times \vec H = -\varepsilon_0\omega\vec E \end{align*}$$

These two equations shows that $\vec E,\: \vec H, \: \vec K$ are perpendicular vectors.

Bibliography

P.B. Adhikari, Bhoj Raj Gautam, Lekha Nath Adhikari. A Textbook of Physics. kathmandu: Sukunda Pustak Bhawan, 2011.

Jha, V. K.; 'Lecture title'; Maxwell's Equation; St. Xavier's College, Kathmandu; 2016.