Polarisation of electromagnetic waves

Polarisation of electromagnetic waves

As we have seen, the electric and magnetic fields of a plane EM wave oscillate in
a direction perpendicular to the direction of propagation defined by the wavevector
k, and therefore the plane EM waves are transversal. A property associated with a
transversal wave is its polarisation character, related to the closed curve described by
the tip of the electric (or magnetic) field vector at a fixed point r = r0 in the space.
In order to analyse the polarisation character of an EM plane wave, let us assume,
without loss of generality, that the EM wave propagates along the z-axis. In this case
we have:
k = kuz (2.54)
where we will use ux, uy and uz as the unitary vectors along the x, y and z-axis
respectively.
First, let us assume the simplest situation of an EM wave in which its associated
electric field is along the x-axis:
E = E0 cos(ωt − kz)ux (2.55)
The magnetic field of this EM wave is obtained using the relation (2.51):
H = H0 cos(ωt − kz)uy (2.56)
where the amplitude H0 is related to the amplitude E0 by:
H0 = (k/ωμ0)E0 = (ε/μ0)1/2E0

Note that the electric and magnetic fields are in phase, that is, if at a fixed time and
at a particular plane z = z0 (z being arbitrary) the electric field E reaches its maximum

value, the magnetic field H will also be at its maximum value. The wave described
by equations (2.55) and (2.56) is said to be linearly polarised (or more specifically,
linearly x-polarised) because the electric field vector E (or H) is always along a
particular direction (x direction in this case) (see Figure 2.2).
Let consider now a linearly y-polarised wave, with an addition phase of +π/2
described by:
E = E0 cos(ωt − kz + π/2)uy = −E0 sin(ωt − kz)uy (2.58)
and
H = −H0 cos(ωt − kz + π/2)ux = H0 sin(ωt − kz)ux (2.59)
with H0 = (ε/μ0)1/2E0. Because Maxwell’s equations are linear, a linear combination
of several solutions will also be a solution. In particular, the sum of plane waves
described in (2.55) and (2.56), and those described by (2.58) and (2.59) will give us a
different, but valid, solution of the wave equation:
E = E0[cos(ωt − kz)ux − sin(ωt − kz)uy] (2.60)
and
H = H0[cos(ωt − kz)uy + sin(ωt − kz)ux] (2.61)
In order to examine the polarisation character of this new wave, let us study the
curve described by the tip of the electric field vector at a fixed plane, for instance, the
plane defined by z = 0. At this position, the time dependence of the fields is:
Ex = E0 cos ωt and Ey = −E0 sin ωt (2.62)
Hx = H0 sin ωt and Hy = H0 cos ωt (2.63)
The modulus of the electric field vector is therefore:
E2 = E2x
+ E2y
= E2
0 (2.64)
and for the magnetic field:
H2 = H2x
+H2y
= H2


which indicates that, at a fixed plane, the tip of the electric field vector (and the
magnetic field vector) describe a circle. For that reason, this wave is said to be circularly
polarised. Moreover, looking at the wave along the propagation direction, one can
observe that the electric field vector rotates contra-clockwise, and thus we are dealing
with a left-hand circularly polarised wave.
On a general form, if two linearly polarised waves, mutually perpendicular, are
superposed, having the same propagation direction and frequency, but with different
amplitudes and relative phases, at a generic plane (for instance, at z = 0), we will have:

Ex = E01 cos(ωt − θ1) (2.66)
Ey = −E02 cos(ωt − θ2)


For such a wave, the relation between the Cartesian components of the electric field is:
(Ex/E01)2 + (Ey/E02)2 − 2(Ex/E01)(Ey/E02) cos θ = sin2θ (2.68)
where we have defined θ as the relative phase between Ex and Ey (θ ≡ θ2 − θ1). This
equation represents an ellipse, being the curve drawn by the electric field, and describes
an elliptically polarised wave.
In general, the principal axis of the ellipse will be tilted with respect to the x and y
axis. In particular, for θ = π/2, 3π/2, . . ., the major and minor axis of the ellipse will
lie along the x and y axis. In this case, if in addition the amplitude of the components
are equal (E01 = E02), then the ellipse will degenerate into a circle. For relative phase
of θ = 0, π, 2π, . . ., the ellipse will become a straight line, with:
Ex = ±(E02/E01)Ey (2.69)
which represents once again a linearly polarised wave.