## Physics Friday 134

Part 14: Electromagnetic Waves in Vacuum

Writing Maxwell’s Laws so that all fields are on the left hand sides of the equations, we have



.
The second and third equations, Gauss’ Law of Magnetism and Faraday’s Law, are homogeneous equations; while the first and fourth, Gauss’ Law and Ampère’s Law, are inhomogeneous; they depend on the charge and current densities.
However, let us consider fields in empty space, so that ρ and J are zero as well; all four equations are then homogeneous, and so solutions may be superimposed (added).
Let us first examine solutions to these with harmonic time dependence, since Fourier superposition allows us to construct arbitrary solutions from these. So, we start with

,
where  and  are complex numbers, and we take real components of the results to get the physical fields.
Then our vacuum Maxwell’s Laws equations



,
become



,
or, eliminating the common complex exponentials (which are always nonzero), we obtain equations for the (complex) amplitudes



.
Note that if we take the divergence of the third equation, we obtain the second, and if we take the divergence of the fourth equation, we obtain the first, as the divergence of a curl is always zero. Thus, the first two equations are not independent, but are implicit in the latter two, so we just need

and
.
Taking the curl of the first equation, we have
.
Now,
,
since E has zero divergence. Thus
.
but our second equation tells us that
,
so we can plug this in to get
.

(a Helmholtz equation).
Similarly, taking the curl of the second equation
,
so the zero divergence of the magnetic field gives
,
and substituting the first equation into this gives
, the same differential equation which E must obey.
Letting our fields vary in one direction only, say z, the field equation reduces from

to
,
which has solutions of the form , where . We see then that k is the angular wavenumber of a wave with frequency ω and phase velocity c. Adding back the time dependence, we see the basic one-dimensional solution is
,
or using the dispersion relation ,
,
plane waves propagating in ±z.
Now, solving the Helmholtz equations for E and B and Maxwell’s equations, as well, we begin with plane waves with vector (angular) wavenumber k

,
where E0 and B0 are constant (complex) vectors, and, as before, physical quantities are obtained by taking the real part of complex quantities.
Plugging these into the Helmholtz equation, we use the fact that

to find that

and so

remains our dispersion relation.
Now, we need only find what the constant vectors E0, B0, and k must obey to satisfy Maxwell’s laws. Given a vector field that is the product of a constant vector and a scalar function, the divergence is thus the dot product of the constant vector and the gradient of the scalar ( for v constant), so the zero divergence equation for the electric field gives





.
Similarly, the divergence equation for B gives
.
Thus, both the electric and magnetic fields are perpendicular to the direction of propagation: electromagnetic waves are transverse waves.
For a scalar field φ and a constant vector v, one has
,
so the first curl equation

becomes




,
or, since , this is

,
where  is the unit vector in the direction of propagation; hence, E0 and B0are perpendicular; the electric field, magnetic field, and direction of propagation are all mutually orthogonal for an electromagnetic wave, and for a wave in vacuum, the electric and magnetic field amplitudes are proportional with .

### 2 Responses to “Physics Friday 134”

1. Physics Friday 135 « Twisted One 151's Weblog Says:

[…] Friday 135 By twistedone151 Part 15: Polarization of Electromagnetic Waves Last week, we began discussing electromagnetic waves in a vacuum, obtaining the fact that light and other […]

2. Physics Friday 136 « Twisted One 151's Weblog Says:

[…] of cross-sectional area. Now, let us consider electromagnetic plane waves in vacuum. As we showed here, they are transverse waves with the electric and magnetic fields both perpendicular to the […]