##### Contents

A sigle wave

Superposition of waves

# Electromagnetic equations revisited

### 1. Maxwell equations

The Maxwell's equations in their differential forms related to the free space are:

• . E = ρ/ε0 = 0       (1)
• x E = - ∂B/∂t       (2)
• . B = 0       (3)
• x B = μ0 (J + ε0 ∂E/∂t) = μ0 ε0 ∂E/∂t       (4)

### 2. ∇ x∇ xA = ?

Let's take the curl of the vector A (Ax,Ay,Az):
x A = ( ∂Az/∂y - ∂Ay/∂z, ∂Ax/∂z - ∂Az/∂x, ∂Ay/∂x - ∂Ax/∂y)

Now let's take the curl of its curl:
x x A = (Component-1, Component-2, Component-3)
Component-1 = ∂(∂Ay/∂x - ∂Ax/∂y)∂y - ∂(∂Ax/∂z - ∂Az/∂x)∂z
Component-2 = ∂(∂Az/∂y - ∂Ay/∂z)∂z - ∂(∂Ay/∂x - ∂Ax/∂y)∂x
Component-3 = ∂(∂Ax/∂z - ∂Az/∂x)∂x - ∂(∂Az/∂y - ∂Ay/∂z)∂y

That is :
Component-1 = ∂2Ay/∂x∂y - ∂2Ax/∂y2 - ∂2Ax/∂z2 + ∂2Az/∂x∂z
Component-2 = ∂2Az/∂y∂z - ∂2Ay/∂z2 - ∂2Ay/∂x2 + ∂2Ax/∂y∂x
Component-3 = ∂2Ax/∂z∂x - ∂2Az/∂x22Az/∂y2 + ∂2Ay/∂z∂y

Adding and subtracting: ∂2Ax/∂x2, ∂2Ay/∂y2, and ∂2Az/∂z2 in the first, second and third components of the vector A respectively, we get:
Component-1 = ∂2Ay/∂x∂y - ∂2Ax/∂y2 - ∂2Ax/∂z2 + ∂2Az/∂x∂z + ∂2Ax/∂x2 - ∂2Ax/∂x2
Component-2 = ∂2Az/∂y∂z - ∂2Ay/∂z2 - ∂2Ay/∂x2 + ∂2Ax/∂y∂x + ∂2Ay/∂y2 - ∂2Ay/∂y2
Component-3 = ∂2Ax/∂z∂x - ∂2Az/∂x2 - ∂2Az/∂y2 + ∂2Ay/∂z∂y + ∂2Az/∂z2 - ∂2Az/∂z2

Hence:
Component-1 = ∂( . A)∂x - . ( Ax)
Component-2 = ∂( . A)∂y - . ( Ay)
Component-3 = ∂( . A)∂z - . ( Az)

That is the scalar: ( . A) - . ( A)

x xA = ( .A) - ( . A)       (5)

Since . = 2 = Δ , that is the Laplacian operator, we have:

x xA = ( .A) - ΔA       (6)

### 3. The two electromagnetic equations

x x E = - ∂/∂( x B) = - μ0 ε02E/∂t2

Since .E = 0, it remains, according to the relationship (6): - Δ E = - μ0 ε02E/∂t2
That is: Δ E - μ0 ε02E/∂t2 = 0. Or:
Δ E - (1/c2) ∂2E/∂t2 = 0       (7)
with: c = 1/[μ0 ε0]1/2. c = 3.00 x 108 m/s, the speed of light in the vacuum.

Now, let's take the curl of the curl of B, we have:
x x B = μ0 ε0 ∂ ( x E) /∂t, according to the relationship (4). According to the relationship (3), we have then:
∂ ( x E) /∂t = - ∂2 B /∂t2. From the relationship (6), we have then: x x B = - Δ B = - μ0 ε02 B /∂t2 Or:
Δ B - (1/c2) ∂2B /∂t2 = 0       (8)

### 4. Example: simple case

We will show that the plane waves are the solutions for the Maxwell's equations:

Let's take the simple case where E = Ex = Ex0 cos(ωt - kz), Ey = 0 and Ez = 0. We have:
x E = (0, ∂Ex/∂z,0) = k Ex0 sin(ωt - kz) (in the y direction)
Using the equation (2), we have:
x E = k Ex0 sin(ωt - kz) = - ∂B/∂t. We integrate to obtain:
B = (k/ω) Ex0 cos(ωt - kz)
Since: ω/k = c, then:
B = (Ex0/c) cos(ωt - kz) (in the y direction)
Therefore:

if:
E(z,t) = Ex = Ex0 cos(ωt - kz) (in the x direction)
Then:
B(z,t) = (Ex0/c) cos(ωt - kz) (in the y direction)

The above expression of E(z,t) and B(z,t) are solutions of the Maxwell equations. E(-z,t) and B(-z,t) are also solutions of these equations. Therefore, all the linear combination of these expressions:

E(z,t) = ∑ [αiE(z,t) + βi E(-z,t)]
B(z,t) = ∑ [αiB(z,t) + βi B(-z,t)]
are also solutions.