Electromagnetic Waves in Space: Some Radiation Physics
The oscillating electric field established between the plates of a capacitor by a harmonic voltage
generator does not stay between those plates. It leaks out through the sides of the capacitor and
moves away into the surrounding volume. The field is said to be radiated and the capacitor acts
like a broadcasting antenna. In vacuum, the field moves at the speed of light, 2.99792458×108
m/s1, a quantity denoted by c. A cross-sectional view of the electric field at a particular instant of
time is shown below.
fig. 1 A cross-sectional
view of the electric field,
part of the
electromagnetic wave
radiated into space.
It is useful to describe this wave mathematically. The field between the capacitor plates can be
represented as E0 cos( t) or as the real part of E0e-i t . Consider any mathematical function, like
the function y = A(x) shown in figure 2. This function is peaked at x = 0. The function shown in
figure 3 is y = A(x - x0) and this represents the same figure displaced a distance x0 to the right
(along the positive x-axis). In the same way, the function A(x-vt) where v is a constant and t is
time represents a figure moving to the right at speed v.
Left: fig. 2 Graph
of a function A(x)
peaked at x=0.
Right: fig. 3
Graph of
A(x - x0).
We can therefore represent our traveling electric field by E0 cos(k(x - ct)) or as the real part of E0
exp( j k(x - ct)), where k is some constant that makes the argument of the trig function
dimensionless. The electric field is now a function of position x and time t. For this to hold at all
1
This value is exact since the meter is now defined in terms of the speed of light.
5.1
locations and times, E(x,t) must represent the field inside the capacitor when x = 0. With E(x,t)
=E0 cos(k(x - ct)), the electric field at x = 0 is E(0,t) = E0 cos(- kct) = E0 cos(kct). But the field
between the plates of the capacitor is E0 cos( t). Therefore,
(1)
kc =
Equation (1) implies that we can write our electric wave as E(x,t) = E0 cos(kx - t).
The quantity k is called the wave number. To see its physical significance, consider a snapshot or
freeze-frame of the wave. The mathematical representation of this will be the function E(x,t) at a
fixed time t. For simplicity, take t = 0. The distribution of electric field in space at time t = 0 is
E(x,0) = E0 cos(kx) or the real part of E0eikx . This is a harmonic function of position and repeats
after a distance x = 2 /k. This distance is called the wavelength .
(2)
= 2 /k
Problem 1: A. Microwave radiation, like that produced in a microwave oven, has a wavelength
of 12.2 cm in vacuum. Find the frequency of this wave.
B. A wave of yellow-green light has a wavelength of 550nm in vacuum. Find the frequency of
this wave.
The electric field is a vector and points along some direction, although it flips directions every
half-period. If we call this the z-direction, then we can write formally
(3) E(x,t) = E0 cos(kx - t) z or E(x,t) = Re{E0 exp[i(kx - t)] z}
where boldface z is a unit vector along the z-direction. Note that E points in the ±z-direction but
varies along the x-direction. It is said to be a transverse wave, since the electric field points in a
direction perpendicular to the wave motion. The electric field is said to be polarized along the zaxis. The Maxwell-Ampere relation
(4)
∫ B ⋅ ds = ε
0
µ0
d
dt
∫ E ⋅ n dA
implies that a changing electric field produces a magnetic field. A detailed analysis, using the
entire set of Maxwell equations, shows that the electric field of equation (3) is accompanied by a
magnetic field B(x,t) given by
(5) B(x,t) = - B0 cos(kx - t) y or B(x,t) = Re{- B0 exp[i(kx - t)] y}
where
(6) B0 = E0 /c.
In general, the electric and magnetic fields of an electromagnetic wave (EM wave) satisfy
(7) E × B direction of wave motion
Just as energy can be stored in an electric field, it can be stored in a magnetic field. The energy
per unit volume in a magnetic field is given by uB = B2/2µ 0 where µ 0 is the permeability constant
= 4 × 10-7 T2m3/J. Maxwell’s equations relate the speed of light to the electric and magnetic
constants and they imply
(8) c2 = 1/( 0µ 0)
The total energy density of an electromagnetic wave is therefore
(9) uTOT = ½ ( 0E2 + B2/µ 0)
(total energy density of an EM wave)
5.2
Problem 2: Use equations 3, 5, 6, 8, and 9 to show that, for the particular EM wave we have
been considering, uTOT (x, t) = 2 uE (x,t) = 0 E02 cos2(kx - t).
Problem 2 shows that, in any region of space through which our EM wave is moving, the timeaveraged energy density is
(10) uTOT (x, t) = 0 E02 cos2(kx - t) = ½ 0 E02
and is therefore twice as big as the energy density due to the electric field alone.
In terms of complex amplitudes, we can write a general wave that is moving along the +z
direction and is polarized along the x-axis as
E(x,t) = Re{E0 exp[i(kx - t - )]} z = Re{(0 exp[i(kx - t)]} z where (0 = E0 e-j
B(x,t) = - Re{B0 exp[i(kx - t - )]} y = Re{%0 exp[i(kx - t)]}(- y) where %0 = B0 e-j
and %0 = (0 /c.
The time-averaged energy density is uTOT (x, t) = ¼ ( 0 (02 +%02/µ 0) = ½ 0 (02
An important notational change: From now on we will follow the standard practice of writing
fields like the above in the form
(11)
E(x,t) = (0 exp[i(kx - t)] z ,
B(x,t) = %0 exp[i(kx - t)]}(- y)
This makes E and B into complex numbers (actually, complex vectors). It is understood that the
physical E and B fields are the real parts of these complex vectors.
When considering wave motion, we are usually less concerned with an energy density than with
the transport of energy through some surface (such as the active surface of a light detector).
Consider the wave of equation (11). This wave is moving in the +x direction. If we place a
surface perpendicular to this direction as shown in figure 3, the wave will fall on the surface
normally.
fig. 3. The electric part of
the electromagnetic wave of
eq. 11. It is normally
incident on a surface that
lies in the yz plane.
5.3
How much energy gets
through the shaded surface in
figure 11 during a time t?
The wave is traveling in free
space with a speed c, so in
time t it has traveled a
distance of c t.
All the energy that passes the
shaded surface during a time
t was contained in the
volume of a box of edge
length c t at the beginning of
the timing interval. If we take
the area of the shaded surface
to be A, the volume of the box
is A c t.
The average energy density in the box is ½ 0 (02 so that the total energy passing through the
surface in time t is U = (½ 0 (02)(A c t.). The rate at which energy flows through the surface
is the transmitted power P
(12)
P = U/ t = ½ A c
0
(02
The transmitted power is proportional to the surface area A. Light detectors (such as the eye or
the surface of a photocell) are sensitive to the power per unit area, a quantity called the intensity,
I.
(13) I = P / A = ½ c
0
(02 = (02 / (2 µ 0 c)
The quantity µ 0 c has the convenient approximate value 120 V2/W.
Solved Problem
A standard class IIIa helium-neon laser produces 4.mW of beam power. The beam diameter is
approximately 1mm.
A. Compute the intensity of this beam.
B. Compute the electric field amplitude of the beam
C. Compare the value obtained in part B with the electrostatic field that holds an electron in an
atom.
5.4
Solution: A.
B.
(
4 ×10−3 W
3 W
.
I=
−3
2 = 5.03 ×10
π ( 0.5 ×10 m)
m2
= 2 µ0 c I = 2 (120π
V2
W
V
) (5.03 × 103 2 ) = 196
. × 103
W
m
m
C.. The electric field experienced by an electron in an atom is due to the atomic nucleus and has
the value E =
q
4πε0r 2
where q is the nuclear charge and r the radius of the atom. A minimum
value is given by q = e = 1.602×10-19C, appropriate for hydrogen. The radius of an atom is
roughly 10-10 m, so that E (9×109 Vm/C)×(1.602×10-19C) / (10-10 m)2 = 1.4 × 1011 V/m.
This is many orders of magnitude larger than the laser field amplitude of part B. A bound electron
that is immersed in this beam will be only very slightly perturbed by it.
You should not conclude from the above problem that all lasers produce electric fields which are
small compared to the field that binds an electron to an atom. In fact, some laser beams are so
intense they can ionize atoms!
Problem 3:
Light travels to the right in a fiber-optic
cable as shown. At P the cable is coupled to
a pair of branching cables which are joined
together again at Q. Let L1 be the length of
the shorter branch and L2 the length of the
longer one. Assume monochromatic light of
wavelength in the cable material and
suppose that the light intensities in the two
branches are equal.
A. What is the relative phase of the two interfering waves at Q?
B. Suppose the cables are adjusted so that this relative phase is radians. Will the light intensity
in the cable to the right of Q be zero? If so, what happens to the energy of the two interfering
waves?
5.5
Problem 4:
Consider two electromagnetic waves, both described by ( ej (kx - t) in vacuum. The waves have the
same wavelength and phase. One of these
waves moves through a length L of nonabsorbing transparent material of refractive
index n. The two waves are shown at one
particular instant, taken to be t = 0 for
convenience. After having traveled through
the material, the two waves will have a phase
difference . Show that the complex
amplitude of the wave emerging from the
material is given by ( 1 = ( ej (n -1) kL .
(HINT: Compute the phase difference )
Problem 5:
If an electric field is described by a wave E1 = ( ej (kx - t) , what is the meaning of a wave described
by E2 = j ( ej (kx - t) ? How could you produce E2 using E1 as input?
Wavefronts
There are many sorts of three dimensional waves. Three of the simplest geometries are illustrated
below. They are called plane waves, cylindrical waves, and spherical waves. A plane wave is
described by a wave amplitude of the form E(x,t) = E0 e j (kx - t) where x is one of the three
Cartesian coordinates. Notice that the phase kx - t depends only on x and t but not on y or z. The
phase of the wave at the location (x, y, z) at time t is the same as the phase of the wave at the
location (x, y1,z1) at time t. At any given time the phase of the wave is constant on any plane
parallel to the yz plane. Whenever the phase is constant at all points on a surface the surface is
called a wavefront. In this example, the wavefronts are all planes, hence the name plane wave.
In this example, the wavefront is perpendicular to the direction of wave motion. This will usually
be the case.
Plane wavefronts
spherical wavefronts
5.6
An outgoing spherical wave which has the coordinate origin as center has the wave amplitude
E(x,t) = A e j (kx - t) / r where r = distance to the origin = (x2 + y2 + z2) and A is a vector that
depends on the polar angles and . The wavefronts are expanding spheres. A wave like this
describes the motion of light moving in all three dimensions away from a point source. Notice the
factor 1/r, which implies that the intensity falls off like 1/r2.
5.7
Experiment 5.1: Intensity of a light bulb
The purpose of the experiment is to determine how the intensity of a point source of light
varies with distance from the source.
apparatus: Incandescent light source, photometer
Procedure:
A. Cover a lightbulb with a mask that allows only a small portion of the light to escape.
Use a photodiode light meter to measure the intensity approximately 1m, 1.5m, 2m,
2.5m, and 3m along a straight line from the light.
B. Now measure the so-called “dark reading”, the intensity reported by the light meter
when exposed to no light at all. You can find this by turning on the device with the cover
in place. Subtract this reading from the measurements in part A.
C. Graph your results. Are they compatible with I 1/r2 ?
5.8
Principle of Superposition
When two electric fields, E1 and E2 , exist in the same region of space, the net electric field is the
vector sum
Enet = E1 +E2
This is called “the principle of superposition”. It is always true in vacuum and holds to a high
degree of accuracy in transparent media for light that is not too intense. If two light waves
combine in a region of a space the intensity produced by the combinations is proportional to the
squared magnitude of Enet . That is, I Enet 2 = E1 +E2 2
This is one of the most important principles in optics. When two waves combine, the wave
amplitudes add, the intensities generally do not.
Huygens’ Wavelets
Christian Huygens formulated an important principle of optics in a work published in 1690.
Below is a picture of a plane wave on a water surface, incident on a barrier in which there is a
small hole.
The wave emerging from the hole appears to be part of a circular wave. Huygens reasoned that the
circular wave (it would have been a spherical wave for sound waves or light waves in three
dimensions encountering a pinhole in a screen) is produced by the point on the incident wavefront
at the location of the hole and would have been produced whether the barrier were there or not.
These outgoing waves are usually called Huygens wavelets. If the barrier is removed, Huygens
thought, all points on the incident wavefront will produce similar spherical outgoing waves. In
this manner of thinking, the slit does not produce the wavelet but rather allows it to propagate
undisturbed by the effect of other wavelets. Huygens’ principle says that the net wave amplitude
at any point in space will be the superposition of the wave amplitudes due to each spherical
outgoing wave.
Huygens’ principle can be shown to hold generally and is a consequence of the wave equation, a
differential equation satisfied by all waves.
5.9
5.10