Wave Equations and Waves

Question 1

General Function of a Wave in One Dimension

moving to the right

Answer 1

y = f (x-vt+𝛿)

Question 2

General Function of a Wave in One Dimension

moving to the left

Answer 2

y = f(x+vt+𝛿)

Question 3

The Wave Equation in One Dimension

Answer 3

∂²y/∂x² = 1/v² ∂²y/∂t²

• where v is the speed of the propagation of the wave
• any function of the form y=f(x-vt+𝛿) will satisfy the 1D wave equation

Question 4

General Form of a Harmonic Wave

Answer 4

y = A sin(kx - ωt + 𝛿)

-where ω=2πf , k=2π/λ , v=ω/k

Question 5

Proof that a harmonic wave satisfies the 1D wave equation

Answer 5

y = Asin(kx - ωt + 𝛿) = Asin( k(x - ω/k t + 𝛿/k) = f(x-vt+𝛿)

Question 6

The Laplacian

Answer 6

∇²V = |∇.|∇ V = ∂²V/∂x² + ∂²V/∂y² + ∂²V/∂z²

-where V is a scalar

Question 7

Vector Form of the Laplacian

Answer 7

∇²|E = ∇²Ex ^i + ∇²Ey ^j + ∇²Ez ^k

Question 8

The Wave Equation in Three Dimensions

Answer 8

∇² |E = 1/v² * ∂²|E/∂t²

Question 9

General Equation of a Plane Wave

Answer 9

|E = Eo ^i sin ( |k.|r - ωt + 𝛿)

• where |E is a plane wave travelling in the |k direction
• its a plane wave because if you change your position |r by moving perpendicular to |k then |k.|r does not change, so in a plane perpendicular to |k all the values for |E are the same

Question 10

Compare the Wave Equations in 3D and 1D

Answer 10

• the 3D wave equation is similar to the 1D wave equation, but:
• -now a vector whose size oscillates and,
• -to move from one wavelength you have to change |k.|r by 2π as this makes the sine function go through one oscillation, therefore creating one wavelength

Question 11

A Vector Identity

Answer 11

|∇ x (|∇ x |E) = |∇ (|∇ . |E) - ∇² |E

Question 12

Maxwell’s Equations in a Vacuum

Answer 12

-for |E and |B in a vacuum, there can be no conduction current density in a vacuum as there are no charges to conduct with:
|∇ x |B = μoεo * ∂|E/∂t
and
|∇ . |E = 0

Question 13

Electric Field and Waves

Answer 13

-sub in to the R.H.S of the vector identity
|∇ x (|∇ x |E) = |∇ x (-∂|B/∂t) from Faraday’s Law
-switch the order of differentiation:
= -∂/∂t (|∇ x |B)
-sub in Ampere-Maxwell Law:
= - ∂/∂t (μoεo * ∂|E/∂t)
= - μoεo * ∂²|E/∂t²
-using the L.H.S. of the vector identity and Gauss’s Law for electric fields in a vacuum:
|∇ (|∇ . |E) - ∇² |E = |∇ (0) - ∇² |E
-equate L.H.S. and R.H.S.:
- ∇² |E = - ∂/∂t (μoεo * ∂|E/∂t)

∇² |E = μoεo * ∂²|E/∂t²
-this is the same form as the wave equation:
∇² |E = 1/v² * ∂²|E/∂t²
-so we can equate
1/v² = μoεo
-this gives v = 2.9979 x 10^8 m/s , the speed of light
-electric fields propagate through space at the speed of light

Question 14

Magnetic Field and Waves

Answer 14

-sub in to the R.H.S of the vector identity
|∇ x (|∇ x |B) = |∇ x (μoεo * ∂|E/∂t)
-switch the order of differentiation:
= μoεo ∂/∂t (|∇x|E)
-sub in using Faraday’s Law:
= μoεo ∂/∂t (- ∂|B/∂t)
= - μoεo * ∂²|B/∂t²
-using the R.H.S of the vector identity and Gauss’s Law for magnetic fields:
|∇ (|∇ . |B) - ∇² |B = |∇ (0) - ∇² |B = - ∇² |B
-equate the L.H.S and R.H.S of the vector identity:
- μoεo * ∂²|B/∂t² = - ∇² |B

μoεo * ∂²|B/∂t² = ∇² |B
-this is the same form as the wave equation:
∇² |B = 1/v² * ∂²|B/∂t²

-so we can equate
1/v² = μoεo
-this gives v = 2.9979 x 10^8 m/s , the speed of light
-magnetic fields propagate through space at the speed of light

Question 15

Mawell - Light and Magnetism

Answer 15

“the agreement of the results seems to show that light and magnetism are affections of the same substance, and that light is an electromagnetic disturbance propagated through the field according to electromagnetic laws.”

Question 16

What is a wave?

Answer 16

-a disturbance that propagates in time and space

Question 17

What is an electromagnetic wave?

Answer 17

-for electromagnetic waves, a time dependent magnetic field induces a time dependent electric field that then induces a time dependent magnetic field and so on
-electromagnetic waves are therefore self proagating
-in a vacuum there are no sources or sinks of |E and |B so:
|∇.|E = 0 and |∇.|B = 0
-which means that there are no beginnings or ends to our field lines
-these electromagnetic waves therefore propagate indefinitely through free space

Question 18

Representing Plane Waves Using Complex Numbers

Description

Answer 18

-a plane wave is represented by:
|E = |Eo cos (|k . |r - ωt + 𝛿) where c=ω/k
-a useful way to represent a plane wave is using complex numbers, where e^iθ = cosθ + isinθ
|E = |Eo exp (i (|k . |r - ωt + 𝛿))
-the real part of this is what we observe and measure, the imaginary part just makes the maths easier

Question 19

Representing Plane Waves Using Complex Numbers

∂/∂t

Answer 19

-differentiate |E
∂|E/∂t = (-iω) |Eo exp (i (|k . |r - ωt + 𝛿)) = -iω |E
∂²|E/∂t² = (-iω)(-iω) |Eo exp (i (|k . |r - ωt + 𝛿)) = -ω² |E
-therefore for our complex plane wave:
∂/∂t = -iω

Question 20

Representing Plane Waves Using Complex Numbers

|∇

Answer 20

-evaluate the dot product in the plane wave equation in terms of components:
|k . |r = (kx,ky,kz) . (x,y,z) = kxx + kyy + kzz
so,
|E = Eo exp(i (kxx + kyy + kzz - ωt + 𝛿))
-find |∇E for each component of |E:
∂Ex/∂x = i kx |E
∂Ey/∂y = i ky |E
∂Ez/∂z = i kz |E
-therefore del is represented by:
|∇ = i |k , where |k is the wave vector (not z)

Question 21

Representing Plane Waves Using Complex Numbers

Divergence

Answer 21

-since |∇ = i |k

|∇ . |E = i |k . |E

Question 22

Representing Plane Waves Using Complex Numbers

Curl

Answer 22

-since |∇ = i |k

|∇ x |E = i |k x |E

Question 23

Orientation of |E and |B relative to |k

Answer 23

-in free space (vacuum) we know that |∇ . |E = 0
so |∇ . |E = i |k . |E = 0
-this means that |E is perpendicular to the direction of propagation |k
-also, in free space |∇ . |B = 0
so |∇ . |B = i |k . |B = 0
-this means that |B is also perpendicular to the direction of propagation |k
-but we can’t tell from this the relative orientations of |E and |B

Question 24

Relative Orientations of |E and |B

Answer 24

-from Faraday’s Law:
|∇ x |E = - ∂|B/∂t , using complex notation
i |k x |E = iω |B, rewrite |k = k* ^k and cancel i
k* ^k x |E = ω |B , divide by k
^k x |E = ω/k |B = c |B
-from Ampere-Maxwell Law in free space
|∇ x |B = 1/c² ∂|E/∂t , using complex notation:
i |k x |B = -iω|E / c², rewrite |k = k* ^k and cancel i
k* ^k x |B = -ω/c² |E , divide by k
^k x |B = -ω/kc² |E = - 1/c |E
-using both equations:
^k x |B = - 1/c |E
-so |E is perpendicular to both ^k and |B
-also taking the modulus over both sides:
| ^k x |B | = | - 1/c |E |
-as ^k and |B are orthogonal and ^k is a unit vector:
B = E / c

Question 25

Proving that |E and |B in Electromagnetic Waves are in Phase

Answer 25

-using faraday's law:
∫ |E . d|l = - d/dt ∫ |B . ^n dA
-calculate the circulation of E (L.H.S.)
-equate with R.H.S.
∂Ey/∂x = - ∂Bz/∂t
-let Ey = Eyo sin (kx-ωt)
∂Bz/∂t = - ∂Ey/∂x = - Eyo * k cos(kx-ωt)
-integrate w.r.t. t to find Bz:
Bz =  Eyo*k/ω sin (kx-ωt) + C = Eyo/c sin (kx-ωt) + C
-constant of integration, C, is zero since if there was no oscillating field, there would be no induced magnetic field in free space, Eyo=0 => Bz = 0
Bz = Eyo/c sin (kx-ωt)
-B and E have the same k and ω, so must be in phase

Question 26

Poynting Vector

Definition

Answer 26

|S = (|E x |B) / μo

• the vector S is in the direction of propagation of the wave
• its size is the rate of energy flow across a unit area placed perpendicular to the flow direction