Lecture 13
Electromagnetic Waves in Free Space
In this lecture you will learn:
• Electromagnetic wave equation in free space
• Uniform plane wave solutions of the wave equation
• Energy and power of electromagnetic waves
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Basic Wave Motion
Consider a wave moving in the +x-direction:
The wave travels a distance equal to one wavelength
in one time period
a( x , t )
v
x
λ
v = velocity of wave propagation
λ = wavelength of the wave
f = frequency of the wave
T = period = 1/f
∂ 2a ( x , t )
∂x 2
distance λ
= = fλ
time
T
fλ =v
Basic relation for wave motion:
1-D wave equation:
v = velocity =
=
1 ∂ 2 a( x , t )
v2
∂t 2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
1
Electromagnetic Wave Motion - I
r
r
∂ µoH
∇×E = −
∂t
r
∇. ε o E = ρ
r
∇ . µo H = 0
r
r r ∂ε E
∇×H = J + o
∂t
Time varying electric and magnetic fields are coupled - this coupling is
responsible for the propagation of electromagnetic waves
Electromagnetic Wave Equation:
Assume free space:
r
ρ =J=0
⇒
r
r
r
r
⎛∂µ H⎞
∂ µo ∇ × H
∂2 E
= − µo ε o 2
∇ × ∇ × E = −∇ × ⎜⎜ o ⎟⎟ = −
∂t
∂t
⎝ ∂t ⎠
r
2
r
1
∂ E
c=
≈ 3 × 108 m/s
⇒ ∇ × ∇ × E = − µo ε o 2
ε o µo
∂t
r
r
1 ∂2 E
⇒ ∇× ∇×E = − 2
c ∂t 2
(
)
(
)
(
)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Electromagnetic Wave Motion - II
⇒
⇒
⇒
(1)
(2)
r
r
1 ∂2 E
∇× ∇×E = − 2
c ∂t 2
(
)
(
0
r
r
r
1 ∂2 E
∇ ∇ . E − ∇ 2E = − 2
c ∂t 2
r
r
1 ∂2 E
∇ 2E = 2
c ∂t 2
(
r
)
(
r
)
r
Use the vector Identity: ∇ × ∇ × F = ∇ ∇ . F − ∇ 2F
)
r
∇. ε o E = ρ = 0
Equation for a wave traveling at speed c in
free space
∂ 2E x ∂ 2E x ∂ 2E x
1 ∂2 Ex
+
+
=
∂x 2
∂y 2
∂z 2
c 2 ∂t 2
∂ 2E y
∂x 2
+
∂ 2E y
∂y 2
+
∂ 2E y
∂z 2
=
Wave equation is essentially
three equations stacked
together – one for each
component of the E-field
2
1 ∂ Ey
c 2 ∂t 2
2
2
2
2
(3) ∂ E2z + ∂ E2z + ∂ E2z = 12 ∂ E2 z
∂x
∂y
∂z
c
∂t
r
Wave must also satisfy: ∇. ε E = 0
o
⇒
∂E x ∂E y ∂E z
+
+
=0
∂x
∂y
∂z
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
2
Electromagnetic Wave Motion - III
The H-field also satisfies a similar wave equation
r
Start from Maxwell’s equations: ∇ × E = −
r
∂ µoH
∂t
r
⇒ ρ =J=0
r
r
r
r
⎛ ∂ ε oE ⎞
∂ εo ∇ × E
∂2 H
⎜
⎟
∇× ∇×H = ∇×⎜
= − µo ε o
⎟=−
∂t
∂t 2
⎝ ∂t ⎠
r
r
∂2 H
⇒ ∇ × ∇ × H = − µo ε o
∂t 2
r
r
1 ∂2 H
⇒ ∇× ∇×H = − 2
c ∂t 2
r
r r ∂ε E
∇×H = J + o
∂t
Assume free space:
(
)
(
)
(
)
0
r
⇒
1 ∂ H
∇ (∇ . H ) − ∇ 2H = − 2
c ∂t 2
⇒
r
r
1 ∂2 H
∇ H= 2
c ∂t 2
r
r
2
r
∇ . µo H = 0
2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
General Solutions of Electromagnetic Wave Equation
Assume only x-component of the E-field:
r
∇. ε o E = 0
⇒
r
E = xˆ E x
∂E x
=0
∂x
The x-component cannot
have x-dependence
r
So assume: E = xˆ E x (z , t )
And plug it into the wave equation:
To obtain:
r
r
1 ∂2 E
∇ 2E = 2
c ∂t 2
∂ 2E x ( z , t ) 1 ∂ 2 E x ( z , t )
= 2
∂z 2
c
∂t 2
Solution is: E x (z , t ) = g (z ± c t )
c
Any function g whose dependence on co-ordinate
z and time t is in the form of (z±ct ) will satisfy the
above equation
Example: E x (z , t ) = Eo exp (− α z − c t )
z
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
3
Sinusoidal Solutions of Electromagnetic Wave Equation - I
∂ 2E x ( z , t ) 1 ∂ 2 E x ( z , t )
= 2
∂z 2
c
∂t 2
r
E = xˆ E x (z − ct )
⇒
The most commonly used solutions are sinusoids, for example:
r
⎛ 2π
(z − c t ) ⎞⎟
E = xˆ E x (z − ct ) = xˆ Eo cos⎜
⎝ λ
⎠
This solution represents a wave that:
i)
ii)
iii)
iv)
Has electric field pointing in x-direction
Has wavelength λ
Has frequency f = c/λ
Is moving in the +z-direction
c
Eo
z
λ
In space the electric field looks x
as shown here (remember that
the density of field lines
correspond to the strength of
the E-field)
c
z
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Sinusoidal Solutions of Electromagnetic Wave Equation - II
The sinusoidal solution,
r
⎛ 2π
(z − c t ) ⎞⎟
E = xˆ E x (z − ct ) = xˆ Eo cos⎜
⎝ λ
⎠
can also be written as:
r
2π ⎞
z ⎞⎞
⎛ 2π c ⎛
⎛ 2π c
E = xˆ Eo cos⎜
t−
z⎟
⎜ t − ⎟ ⎟ = xˆ Eo cos⎜
λ
λ
λ ⎠
c
⎝
⎝
⎠⎠
⎝
Define:
ω=
2π c
λ
= 2π f
and
To get:
r
E = xˆ Eo cos(ω t − k z )
k=
2π
λ
ω = angular frequency (units: radians/sec)
k = wave-vector (units: 1/m)
Note that:
fλ =v
⇒
ω = kc
A dispersion relation
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
4
Sinusoidal Solutions of Electromagnetic Wave Equation - III
r
r
∂ µoH
∇×E = −
∂t
What about the magnetic field?
Recall the E and H-fields are coupled
r
r ∂ εo E
∇×H =
∂t
r
r
r
r ∂ εo E ⎞
⎛
∂ µoH
⎜⎜ or even from ∇ × H =
⎟
∇×E = −
∂t
∂t ⎟⎠
⎝
r
Plug in the following solution for the E-field: E = xˆ Eo cos(ω t − k z )
r
r
k
∂H
1
E sin(ω t − k z )
=−
∇ × E = − yˆ
µo
µo o
∂t
r
k
µoω
µo
⇒ H = yˆ
E cos(ω t − k z )
= µo c =
= ηo
µo ω o
εo
k
r
E
⇒ H = yˆ o cos(ω t − k z )
ηo = impedance of free space ≈ 377 Ω
So an E-field must be accompanied by an H-field
which can be calculated from the equation:
ηo
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Sinusoidal Solutions of Electromagnetic Wave Equation - IV
x
E
c
r
E = xˆ Eo cos(ω t − k z )
r
E
H = yˆ o cos(ω t − k z )
z
ηo
y
H
E x̂
Cartoon depiction:
ẑ
H ŷ
• These solutions of the wave equation are called uniform plane waves
• The E-field (and the H-field as well) is constant over any infinite plane that is
parallel to the x-y plane – in more technical terms, the surfaces of constant phase
are infinite planes
• The pattern shown above moves with a velocity equal to
ω
k
=c
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
5
Sinusoidal Solutions of Electromagnetic Wave Equation - V
Consider the plane wave:
r
E = xˆ Eo cos(ω t − k z )
r
E
H = yˆ o cos(ω t − k z )
ηo
If a person takes a snapshot of the wave in space at any time, say at t = 0, he will
see E-field look like:
z
λ
r
⎛ 2π ⎞
E = xˆ Eo cos(k z ) = xˆ Eo cos⎜
z⎟
⎝ λ ⎠
If a person sits at one location, say z = 0, he will see an oscillating E-field in time that
looks like:
t
r
⎛ 2π ⎞
E = xˆ Eo cos(ω t ) = xˆ Eo cos⎜
t⎟
⎝T ⎠
T
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Plane Waves in 3D - I
E x̂
So far we have found one solution:
ẑ
H ŷ
What about plane waves with E-field pointing in direction n̂ and traveling in
some arbitrary direction in 3D ?
E n̂
H
The answer is:
(
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
)
r
k = k x xˆ + k y yˆ + k z zˆ
r r
⇒ k . k = k 2 = k x2 + k y2 + k z2
r
r = x xˆ + y yˆ + z zˆ
Lets see if this solution satisfies the wave equation:
r
r
1 ∂2 E
∇ 2E = 2
c ∂t 2
r r
rr
rr
ω2
⇒ − k . k nˆ Eo cos ωt − k .r = − 2 nˆ Eo cos ωt − k .r
c
r r
⇒ ω 2 = k . k c 2 = k 2c 2
The solution can only be correct if: ω = k c
⇒ ω = kc
[
(
(
)]
[
(
)]
)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
6
Plane Waves in 3D - II
r
k = k x xˆ + k y yˆ + k z zˆ
r r
⇒ k . k = k 2 = k x2 + k y2 + k z2
r
r = x xˆ + y yˆ + z zˆ
ω = kc
The solution for a plane wave in 3D is:
(
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
)
The solution must also satisfy:
r
∇. ε o E = 0
Plug in the solution to check:
[
(
r
r r
∇. ε o E = ε o ∇. nˆ Eo cos ω t − k . r
r
∇. ε o E = 0
provided
) ] = ε o (k . nˆ )Eo cos(ω t − k . rr )
r
r
r
k . nˆ = 0
r
The solution can only be correct if the unit vector n̂ is perpendicular to k
r
The solution in fact corresponds to a plane wave traveling in the direction: k
r
With E-field pointing in the direction: n̂
k
r
k . nˆ = 0
⇒ E-field is perpendicular
E
n̂
to the direction of travel
H
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Plane Waves in 3D - III
The solution for a plane wave in 3D is:
(
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
)
r
k = k x xˆ + k y yˆ + k z zˆ
r r
⇒ k . k = k 2 = k x2 + k y2 + k z2
ω = kc
r
k . nˆ = 0
r
k
E
λ=
n̂
x
H
2π
=
k
2π
k x2 + k y2 + k z2
λ
z
y
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
7
Plane Waves in 3D - IVr
If the E-field for a plane wave solution in 3D is:
(
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
)
k = k x xˆ + k y yˆ + k z zˆ
r r
⇒ k . k = k 2 = k x2 + k y2 + k z2
ω = kc
Then what is the magnetic field?
r
k . nˆ = 0
Use the same old equation:
r
r ∂ εo E ⎞
⎛
⎟
⎜⎜ or even from ∇ × H =
∂t ⎟⎠
⎝
r r
r r
Plug in the following solution for the E-field: E (r , t ) = nˆ Eo cos ω t − k .r
r
r
r
r r
k × nˆ
∂H
1
Eo sin ω t − k . r
=−
∇×E = −
µo
µo
∂t
v
r (k × nˆ )
r r
r
Eo cos ω t − k . r
⇒ H=
k = k kˆ
r
r
∂ µoH
∇×E = −
∂t
(
(
µo ω
⇒
(
)η
)
(
(
)
(
)
r
r r
E
H = kˆ × nˆ o cos ω t − k . r
)
)
Direction of H-field given by kˆ × nˆ
is also perpendicular to the
direction of travel
o
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Plane Waves in 3D - V
r
k = k x xˆ + k y yˆ + k z zˆ
r r
⇒ k . k = k 2 = k x2 + k y2 + k z2
E-field for a plane wave solution in 3D:
(
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
)
ω = kc
H-field for a plane wave solution in 3D:
(
)η
(
r r
r r
E
H (r , t ) = kˆ × nˆ o cos ω t − k . r
)
r
k . nˆ = 0
o
E
H
n̂
k̂
kˆ × nˆ
Questions:
i) What about energy and power?
ii) How much power per unit area does a plane wave carry?
iii) What is the energy density of a plane wave?
r r
r r
r r
Try using the Poynting vector: S (r , t ) = E (r , t ) × H (r , t )
(units: Joules/(m2-sec)
or: Watts/m2 )
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
8
Power Carried By Plane Waves
E-field and H-field for a plane wave solution in 3D:
(
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
(
)η
(
r r
r r
E
H (r , t ) = kˆ × nˆ o cos ω t − k . r
)
)
o
Poynting vector for a plane wave:
r r
r r
r r
S (r , t ) = E (r , t ) × H (r , t )
(
= nˆ × kˆ × nˆ
= kˆ
Eo2
ηo
)η
Eo2
Use the vector identity:
(
r r
cos2 ω t − k . r
(
)
(
)
r r r
r r r
r r r
A × B × C = B A .C − C A . B
)
(
)
o
(
r r
cos2 ω t − k . r
E
)
H
Power flows in the direction of the wave-vector
Time average power per unit area of a plane wave:
k̂
kˆ × nˆ
cos2 (ω t + θ )
Usually one is interested in the time-average
power per unit area:
=
2π ω
1
2π ω
1
=
2
r r
r r
E2
Eo2
S (r , t ) = kˆ o cos2 ω t − k . r = kˆ
ηo
2 ηo
(
n̂
)
2
∫ cos (ω t + θ ) dt
0
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Power and Energy Density of Plane Waves
E-field and H-field for a plane wave solution in 3D:
(
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
(
)η
(
r r
r r
E
H (r , t ) = kˆ × nˆ o cos ω t − k . r
)
)
o
r
Electric field energy density = We (r , t ) =
r
(
r r
r r
r r
1
1
ε o E (r , t ) . E (r , t ) = ε o Eo2 cos2 ω t − k . r
2
2
Magnetic field energy density = Wm (r , t ) =
Notice that:
)
(
r r
r r
r r
1
1 E2
µo H (r , t ) . H (r , t ) = µo o2 cos2 ω t − k . r
2
2 ηo
)
x
r
r
We (r , t ) = Wm (r , t )
E
c
r r
r
r
S (r , t ) = c (We (r , t ) + Wm (r , t ))
The electric and magnetic
energy densities are moving at
the speed c and this constitutes
y
the power of an electromagnetic
plane wave
z
H
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
9
Example: Plane Wavesr
If the fields for a plane wave solution in 3D is:
(
)
r r
r r
E (r , t ) = nˆ Eo cos ω t − k .r
r r
r r
E
H (r , t ) = kˆ × nˆ o cos ω t − k . r
(
)η
(
)
k = k x xˆ + k y yˆ + k z zˆ
r r
⇒ k . k = k 2 = k x2 + k y2 + k z2
r
k . nˆ = 0
ω = kc
o
Lets try to write solution for a plane wave that:
i)
ii)
iii)
iv)
Is moving in the -x-direction
Has E-field pointing in the y-direction
Has E-field amplitude Ro
Has wavelength λo
r
(i) And (iv) imply: k = − k xˆ
and
k=
2π
λo
r r
Answer is: E (r , t ) = yˆ Ro cos(ω t + k x )
Wavevector is in the -x-direction
− ẑ H
r r
R
H (r , t ) = − zˆ o cos(ω t + k x )
ηo
E
ŷ
− x̂
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
10