Maxwell’s equations describe all classical electromagnetic phenomena, including electromagnetic waves.

In SI units Maxwell’s equations in differential form are

,

.

In a macroscopic medium the charge density r is the sum of the free charge density r_f and the polarization charge density r_p.

r = r_f + r_p.

The polarization charge density is given by , so .

In a macroscopic medium the current density j is the sum of the free current density j_f , the magnetization current density j_M, and the polarization current density j_p.

.

The magnetization and polarization current densities are given by

,  so  .

We define the electric displacement D and the magnetic intensity H as

D = e₀E + P,    H = B/m₀ – M.

Then in terms of D and H we may write Maxwell’s equations as

,

.

In a linear, isotropic material P is proportional to E, and M is proportional to H.  We write

P = e₀c_eE,    M = c_mH.

We then have

D = e₀(1+c_e)E = eE,     B = m₀(1+c_m)H = mH.

We now can write Maxwell’s equations entirely in terms of E and H.

,

.

In a source free medium (r_f = 0, j_f = 0) this becomes

.

In a homogeneous material e and m are independent of position.  Then

.

Now we can derive the homogeneous wave equation for E.

.

.

Similarly we can derive the homogeneous wave equation for H.

[Note that if e = e(r), i.e. we have a spatially non-uniform dielectric material, then the divergence of E is not zero and we obtain a wave equation with a source term.]

Each Cartesian component of E and H satisfies the three-dimensional homogeneous wave equation with wave speed v = (1/(me))^½_.

Plane wave solutions to this equation exist.  For a plane wave propagating in the direction we have

.

For electromagnetic waves in lih materials E and H are always perpendicular to each other and perpendicular to the direction of propagation.  The direction of propagation is the direction of E ´ H.  The magnitudes of E and H are related through |E|/|B| = |E|/|mH| = v.

Sinusoidal plane waves are a particularly important type of solution to the wave equation.

E(r,t) = Re[E₀exp(i(k×r-wt))] where E₀ is a complex amplitude.

We write E(r,t) = E₀ exp(i(k×r-wt) and understand that the physics is contained in the real part of the solution.  This works because Maxwell’s equations are linear equations.

For linear equations we have:

Given A, B are complex numbers and f(A) = B.

Then f(Re(A)) = Re(B), f(Im(A)) = Im(B).

Let E(r,t) = E₀exp(i(k×r-wt).

Then:

Similarly:

Thus for monochromatic plane waves we have

,

and Maxwell’s equations in a source-free, lih medium become

.

Very few simple macroscopic media for which m and e are constant exist.  For monochromatic, sinusoidal plane waves we often find that m(w) and e(w) depend on w, but not on other variables such as the magnitude and direction of the electric field.  The medium is then dispersive, the speed v is a function of w.

A basic plane wave may be written as a linear superposition of monochromatic, sinusoidal plane waves, i.e. in terms of its Fourier integral expansion.

For each Cartesian component of E and H (here denoted by u) we have

.

We may write u(r,t) in terms of its Fourier integral expansion,

.

Then 

,

is the Fourier transform of zero, and u(r,w) satisfies the scalar wave equation

,

with k = w/v(w).

Each Fourier component of a basic plane wave satisfies the scalar wave equation.  Only when we reconstitute a wave as a linear superposition of plane waves with different frequencies does the dispersion produce modifications.  The shape of the wave changes as it propagates.

The Poynting vector S represents the energy flux in an electromagnetic wave.

.

For a monochromatic plane wave we define the intensity I as I = |<S>|.

If we use complex notation then

.

Using Re(A)= ½ (A+A*) for any complex number A we obtain

.

Since E = E₀exp(iwt) and H = H₀exp(iwt) at any fixed point in space we find that averaged over one period t = 2p/w

,

Therefore

.

The average energy density in a monochromatic plane wave is given by

.

So <S> =  v<u>, with v = (me)^-1/2 .

Problem:

The Shiva laser fusion system was capable of delivering 10¹³ W of optical power at 1.06 mm wavelength to a pellet of 100 mm diameter.
(a)  Find the rms E field of linearly polarized light in V/m under the assumption of uniform illumination over the pellet surface.
(b)  Compare the result with the electric field experienced by the electron in a Hydrogen atom.

Solution:

The magnitude of the Poynting vector S = (1/m₀) E´B relates the intensity of an electromagnetic wave to the field strength.  For a sinusoidal electromagnetic wave we have <S> = (|E₀²|/2)(e₀/m₀)^1/2.
The electron in hydrogen atom experiences the electric field of a point charge q = 1.6*10^-19 C a distance a₀ = 5.29*10^-11 m from the point charge, E_q = q/(4pe₀a₀²) .
10¹³W/(p*10^-8/4)m² = (|E₀²|/2)(e₀/m₀)^1/2,  E₀ = 9.8*10¹¹ N/C is the electric field amplitude of the light wave.  The rms E field is E₀/Ö2 = 6.9*10¹¹ N/C.
E_q = q/(4pe₀a₀²) = 5.15*10¹¹ N/C is the electric field experienced by the electron in a Hydrogen atom.
E₀/Ö2 is on the order of  E_q.