Wave Equation for Electromagnetic Radiation

The wave equation for electromagnetic radiation in free space can be derived from Maxwell's equations. It describes how electric and magnetic fields propagate as waves through space. The equations are given by:

$$\nabla^2 \mathbf{E} - \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} = 0$$ $$\nabla^2 \mathbf{B} - \frac{1}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} = 0$$

Here:

• $\mathbf{E}$ is the electric field vector.
• $\mathbf{B}$ is the magnetic field vector.
• $\nabla^2$ is the Laplacian operator, which represents the spatial part of the wave equation.
• $c$ is the speed of light in a vacuum.
• $\frac{\partial^2}{\partial t^2}$ is the second derivative with respect to time.

These equations indicate that both the electric and magnetic fields satisfy the same form of wave equation and propagate at the speed of light $c$.

In terms of a scalar field $\psi$, the wave equation in three-dimensional space is often written as:

$$\nabla^2 \psi - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} = 0$$

This general form can represent various types of waves, including electromagnetic waves, when $\psi$ is replaced with the corresponding field components.

Electromagnetic Wave Propagation in Free Space

In free space, electromagnetic waves propagate according to Maxwell's equations without any interaction with a material medium. This scenario is fundamental to understanding the behavior of electromagnetic radiation, including light and radio waves.

EM Wave Equations for Free Space in Terms of Electric Field (E)

Gauss's Law for Electricity

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

For Free space $\rho = 0$. Therefore:

$$\nabla \cdot \mathbf{E} = 0 \quad \text{(i)}$$

Gauss's Law for Magnetism

$$\nabla \cdot \mathbf{B} = 0 \quad \text{(ii)}$$

Faraday's Law of Electromagnetic Induction

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \quad \text{(iii)}$$

Ampère's Law with Maxwell's Correction

$$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$

or

$$\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}}{\partial t} \quad \text{(iv)}$$

Now, taking curl of Eq. (iii)

$$\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B} ) \quad \text{(v)}$$

From vector identity, we know:

$$\nabla \times (\nabla \times \mathbf{E}) = \nabla (\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$$

From Eq (v):

$$\nabla (\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B} )$$

Therefore, from Eq. (i) and using $\mathbf{B} = \mu \mathbf{H}$

$$0 - \nabla^2 \mathbf{E} = -\mu_0 \frac{\partial}{\partial t} (\nabla \times \mathbf{H})$$

From Eq. (iv):

$$\nabla^2 \mathbf{E} = \mu_0 \frac{\partial}{\partial t} \frac{\partial \mathbf{D}}{\partial t}$$

As $\mathbf{D} = \epsilon_0 \mathbf{E}$, therefore:

$$\nabla^2 \mathbf{E} = \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{D}}{\partial t^2} \quad \text{(vi)}$$

General Wave Expression,

$$\nabla^2 \psi = \frac{1}{V^2} \frac{\partial^2 \psi}{\partial t^2} \quad \text{(vii)}$$

Comparing Eq. (vi) and (vii)

$$\frac{1}{V^2} = \mu_0 \epsilon_0$$

or

$$V = \frac{1}{\sqrt {\mu_0 \epsilon_0}} \quad \text{(viii)}$$

Multiplying and dividing Eq. (viii) by $\sqrt {4 \pi}$ and we know $\frac{1}{\sqrt {4 \pi \epsilon_0}} = \sqrt {9 \times 10^9}$, $\mu_0 = 10^{-7}$:

$$V = 3 \times 10^8 m/s$$

EM Wave Equations for Free Space in Terms of Magnetic Field (H)

Gauss's Law for Electricity

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

For Free space $\rho = 0$. Therefore:

$$\nabla \cdot \mathbf{E} = 0 \quad \text{(i)}$$

Gauss's Law for Magnetism

$$\nabla \cdot \mathbf{B} = 0 \quad \text{(ii)}$$

Faraday's Law of Electromagnetic Induction

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \quad \text{(iii)}$$

Ampère's Law with Maxwell's Correction

$$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$

or

$$\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}}{\partial t} \quad \text{(iv)}$$

Now, taking curl of Eq. (iv)

$$\nabla \times (\nabla \times \mathbf{H}) = \frac{\partial}{\partial t} (\nabla \times \mathbf{D} ) \quad \text{(v)}$$

From vector identity we know:

$$\nabla \times (\nabla \times \mathbf{H}) = \nabla (\nabla \cdot \mathbf{H}) - \nabla^2 \mathbf{H}$$

From Eq (v):

$$\nabla (\nabla \cdot \mathbf{H}) - \nabla^2 \mathbf{H} = \frac{\partial}{\partial t} (\nabla \times \mathbf{D} )$$

Therefore, from Eq. (ii) and using $\mathbf{B} = \mu_0 \mathbf{H}$ and $\mathbf{D} = \epsilon_0 \mathbf{E}$

$$0 - \nabla^2 \mathbf{H} = \epsilon_0 \frac{\partial}{\partial t} (\nabla \times \mathbf{E})$$

From Eq. (iii),

$$-\nabla^2 \mathbf{H} = \epsilon_0 \frac{\partial}{\partial t} (-\frac{\partial \mathbf{B}}{\partial t})$$

As $\mathbf{B} = \mu_0 \mathbf{H}$ therefore:

$$\nabla^2 \mathbf{H} = \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{H}}{\partial t^2} \quad \text{(vi)}$$

General Wave Expression,

$$\nabla^2 \psi = \frac{1}{V^2} \frac{\partial^2 \psi}{\partial t^2} \quad \text{(vii)}$$

Comparing Eq. (vi) and (vii)

$$\frac{1}{V^2} = \mu_0 \epsilon_0$$

or

$$V = \frac{1}{\sqrt {\mu_0 \epsilon_0}} = c \quad \text{(viii)}$$

Hence velocity of EM wave in free space is equal to speed of light.

Multiplying and dividing Eq. (viii) by $\sqrt {4 \pi}$ and we know $\frac{1}{\sqrt {4 \pi \epsilon_0}} = \sqrt {9 \times 10^9}$, $\mu_0 = 10^{-7}$

$$V = 3 \times 10^8 m/s$$

Characteristics of Electromagnetic Waves in Free Space

• Speed of Light: From Maxwell's equations, the speed $v$ of electromagnetic waves in free space is:$v = c \approx 3 \times 10^8 \, \text{m/s}$

• Transverse Nature: Electric and magnetic fields oscillate perpendicular to the direction of propagation.

• Wave Vector: The wave vector $k$ indicates the direction of propagation and is related to the wavelength $λ$ by $\mathbf{k} = \frac{2\pi}{\lambda} \hat{\mathbf{n}}$ where $\hat{\mathbf{n}}$ is the unit vector in the direction of propagation.

Electromagnetic Wave Propagation in Isotropic Dielectric Medium

In an isotropic dielectric medium, electromagnetic waves interact with the medium's atoms and molecules, influencing their propagation characteristics compared to free space. Understanding wave propagation in such mediums is crucial for various applications in optics, telecommunications, and materials science.

Maxwell's Equations in Dielectric Medium

Maxwell's equations describe electromagnetic wave behavior in a dielectric medium:

Gauss's Law for Electricity

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon}$$

For Free space $\rho = 0$. Therefore:

$$\nabla \cdot \mathbf{E} = 0 \quad \text{(i)}$$

Gauss's Law for Magnetism

$$\nabla \cdot \mathbf{B} = 0 \quad \text{(ii)}$$

Faraday's Law of Electromagnetic Induction

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \quad \text{(iii)}$$

Ampère's Law with Maxwell's Correction

$$\nabla \times \mathbf{B} = \mu \epsilon \frac{\partial \mathbf{E}}{\partial t}$$

or

$$\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}}{\partial t} \quad \text{(iv)}$$

Now, taking curl of Eq. (iii)

$$\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B} ) \quad \text{(v)}$$

From vector identity we know:

$$\nabla \times (\nabla \times \mathbf{E}) = \nabla (\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E}$$

From Eq (v):

$$\nabla (\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B} )$$

Therefore, from Eq. (i) and using $\mathbf{B} = \mu \mathbf{H}$

$$0 - \nabla^2 \mathbf{E} = -\mu \frac{\partial}{\partial t} (\nabla \times \mathbf{H})$$

From Eq. (iv):

$$\nabla^2 \mathbf{E} = \mu \frac{\partial}{\partial t} \frac{\partial \mathbf{D}}{\partial t}$$

As $\mathbf{D} = \epsilon \mathbf{E}$, therefore:

$$\nabla^2 \mathbf{E} = \mu \epsilon \frac{\partial^2 \mathbf{D}}{\partial t^2} \quad \text{(vi)}$$

General Wave Expression:

$$\nabla^2 \psi = \frac{1}{V^2} \frac{\partial^2 \psi}{\partial t^2} \quad \text{(vii)}$$

Comparing Eq. (vi) and (vii)

$$\frac{1}{V^2} = \mu \epsilon$$

or

$$V = \frac{1}{\sqrt {\mu \epsilon}}$$

Eq. (viii) shows that the propagation velocity of an electromagnetic wave in a dielectric medium is less than that in free space.

For a dielectric medium with relative permittivity $\epsilon_r$ and relative permeability $\mu_r$:

$$V = \frac{c}{\sqrt {\mu_r \epsilon_r}}$$

where $c = \frac{1}{\sqrt {\mu_0 \epsilon_0}}$

In non-magnetic dielectric media, $\mu_r \approx 1$, so:

$$V = \frac{c}{\sqrt {\mu_r}}$$

Hence,

Thus, the refractive index $n$ = $\sqrt {\text{Relative permittivity}} \; \;$ or $\; \; n = \sqrt {\epsilon_r}$

Characteristics of Electromagnetic Waves in Dielectric Medium

• Speed of Light: In a dielectric medium, the speed $v$ of electromagnetic waves is: $v = \frac{c}{\sqrt{\epsilon_r}}$, where $\epsilon_r$ is the relative permittivity of the medium.

• Wave Vector: The wavelength $\lambda$ in the medium is related to the wavelength in a vacuum $\lambda_0$ by: $\lambda = \frac{\lambda_0}{n}$ where $n$ is the refractive index of the medium.

• Wave Polarization: The interaction of electromagnetic waves with the dielectric medium can lead to effects like refraction, reflection, and absorption. Polarization of the wave may affect these interactions.

Poynting Vector and Poynting Theorem

Electromagnetic waves carry energy as they propagate through space. This energy is associated with both the electric and magnetic fields. The amount of energy flowing through a unit area, perpendicular to the direction of energy propagation per unit time, is called the Poynting vector, denoted by $\mathbf{S}$ . Mathematically it is defined:

$$\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}$$

In free space, where $\mathbf{D} = \epsilon_0 \mathbf{E}$ and $\mathbf{B} = \mu_0 \mathbf{H}$, the Poynting vector can be expressed as:

$$\mathbf{S} = \mathbf{E} \times \mathbf{H}$$

where $\mathbf{E}$ and $\mathbf{H}$ represent the instantaneous values of the electric and magnetic field vectors. The Poynting vector $\mathbf{S}$ is perpendicular to both $\mathbf{E}$ and $\mathbf{H}$, and points in the direction of wave propagation. Its units are watts per square meter $W/m^2$, representing the rate of energy transport per unit area.

The Poynting theorem is a fundamental principle in electromagnetism that describes the conservation of energy for electromagnetic fields. It provides a relationship between the power flowing through a surface and the rate of change of electromagnetic energy within a volume. This theorem is named after the physicist John Henry Poynting.

The Poynting theorem states that the rate at which electromagnetic energy is transferred through a surface is equal to the negative rate of change of electromagnetic energy stored in the volume enclosed by that surface, plus the work done by external forces (if any).

Mathematically, the theorem can be expressed as:

where:

• $E$ is the electric field vector.
• $H$ is the magnetic field vector.
• $J$ is the current density vector.
• $D$ is the electric displacement field vector, where $D = \epsilon E$.
• $B$ is the magnetic flux density vector, where $B = \mu H$.
• $\frac{\partial}{\partial t}$ represents the time derivative.
• $\oint_S (E \times H ) \cdot dS$ is the integral of the Poynting vector across a closed surface $S$.

Derivation

We can calculate the energy density carried by electromagnetic waves with the help of Maxwell's equations given below.

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

Take scalar (dot) product of Eq. (iii) and Eq. (iv) with $\vec{H}$ and $\vec{E}$ respectively, i.e.,

$$\vec{H} \cdot (\nabla \times \vec{E}) = -\vec{H} \cdot \frac{\partial \vec{B}}{\partial t} \quad \text{(v)}$$ $$\vec{E} \cdot (\nabla \times \vec{H}) = \vec{E} \cdot \vec{J} + \vec{E} \cdot \frac{\partial \vec{D}}{\partial t} \quad \text{(vi)}$$

Subtract Eq. (v) from Eq. (vi), i.e.,

or

$$[ \text{div} (\vec{A} \times \vec{B}) = \vec{B} \cdot \text{curl} \vec{A} - \vec{A} \cdot \text{curl} \vec{B} ]$$

Using the relations $\vec{B} = \mu \vec{H}$ and $\vec{D} = \epsilon \vec{E}$, we can get

$$\vec{E} \cdot \frac{\partial \vec{D}}{\partial t} = \vec{E} \cdot \frac{\partial}{\partial t} (\epsilon \vec{E}) = \frac{1}{2} \frac{\partial}{\partial t} (E^2) =$$ $$\frac{\partial}{\partial t} \left( \frac{1}{2} \vec{E} \cdot \vec{D} \right) \quad [E^2 = \vec{E} \cdot \vec{E}]$$

$\vec{H} \cdot \frac{\partial \vec{B}}{\partial t} = \vec{H} \cdot \frac{\partial}{\partial t} (\mu \vec{H}) = \frac{1}{2} \mu \frac{\partial}{\partial t} (H^2) =$

$$\frac{\partial}{\partial t} \left( \frac{1}{2} \vec{H} \cdot \vec{B} \right) \quad [H^2 = \vec{H} \cdot \vec{H}]$$

Now Eq. (vii) can be written as

or

$$\vec{E} \cdot \vec{J} = \frac{\partial}{\partial t} \left[ \frac{1}{2} (\vec{H} \cdot \vec{B} + \vec{E} \cdot \vec{D}) \right]$$ $$- \text{div} (\vec{E} \times \vec{H}) \quad \text{(viii)}$$

Integrating Eq. (viii) over a volume $V$ enclosed by a surface $S$, we get

$- \int_V \text{div} (\vec{E} \times \vec{H}) \, dV$ or

$\int_V \vec{E} \cdot \vec{J} \, dV = \frac{\partial}{\partial t} \left[ \int_V \left( \frac{1}{2} \mu H^2 + \frac{1}{2} \epsilon E^2 \right) dV \right]$

$- \oint_S (\vec{E} \times \vec{H}) \cdot d \vec{S} \quad \text{(ix)}$

$$[\vec{B} = \mu \vec{H}, \, \vec{D} = \epsilon \vec{E}]$$

and

$\int_V \text{div} (\vec{E} \times \vec{H}) \, dV = \oint_S (\vec{E} \times \vec{H}) \cdot d \vec{S} \quad \text{(x)}$

Eq. (ix) can also be written as

$\int_V \vec{E} \cdot \vec{J} \, dV = \frac{\partial}{\partial t} \left[ \int_V \left( \frac{1}{2} \mu H^2 + \frac{1}{2} \epsilon E^2 \right) dV \right]$

$- \oint_S (\vec{E} \times \vec{H}) \cdot d \vec{S}$

Interpretation

(a) $\int_v (\mathbf{E} \cdot \mathbf{J}) dV$ : Represents the total power dissipated in the volume $V$ due to the motion of charges.

(b) $\frac{\partial}{\partial t} \int_v \left[ \frac{1}{2} \mu H^2 + \frac{1}{2} \epsilon E^2 \right] dV$ : Represents the rate of change of energy stored in the electric and magnetic fields within volume volume $V$.

(c) $\int_s (\mathbf{E} \times \mathbf{H}) \cdot d\mathbf{S}$ : Represents the rate of energy flow (power) through the surface $S$, i.e., the net power flowing out of volume $V$. The vector $S = E \times H$ is known as the Poynting vector.

The equation (x) derived is known as the Poynting theorem, which describes the conservation of energy for electromagnetic fields. It states that the power transferred into the field is equal to the sum of the rate of change of electromagnetic energy within a volume and the rate of energy flow out through the boundary surface. This represents the energy conservation law in electromagnetism.