1. Lecture 14 : Electromagnetic Waves
Wave equations
Properties of the waves
Energy of the waves

2. Recap (1) 𝜵 × 𝑬 =− 𝝏 𝑩 𝝏𝒕 𝜵 . 𝑬 =𝟎 𝜵 × 𝑩 = 𝝁 𝟎 𝜺 𝟎 𝝏 𝑬 𝝏𝒕 𝜵 . 𝑩 =𝟎
In a region of space containing no free charge ( 𝜌 𝑓 =0) or current ( 𝐽 𝑓 = 0 ), Maxwell’s equations can be written as
𝜵 × 𝑬 =− 𝝏 𝑩 𝝏𝒕
𝜵 . 𝑬 =𝟎
𝜵 × 𝑩 = 𝝁 𝟎 𝜺 𝟎 𝝏 𝑬 𝝏𝒕
𝜵 . 𝑩 =𝟎

3. Recap (2)
Work needs to be performed when creating an 𝐸 -field (e.g. by assembling charges), or a 𝐵 -field (e.g. by establishing a current against an inductance)
This creates a potential energy stored in the electromagnetic field with energy densities 𝜀 0 𝐸 2 and 𝐵 2 2 𝜇 0 , respectively

4. Wave equations
We will now explore how Maxwell’s equations lead to wave motion in the form of light – an electromagnetic wave

5. Wave equations
A wave is a propagating disturbance that carries energy from one point to another
A wave may be characterized by its amplitude, wavelength and speed and can be longitudinal or transverse
speed

6. Wave equations To discover EM waves, consider evaluating 𝛻 × 𝛻 × 𝐸
Substituting in Maxwell’s equations, this equals 𝛻 × − 𝜕 𝐵 𝜕𝑡 =− 𝜕 𝑑𝑡 𝛻 × 𝐵 =− 𝜕 𝜕𝑡 𝜇 0 𝜀 0 𝜕 𝐸 𝜕𝑡 =− 𝜇 0 𝜀 0 𝜕 2 𝐸 𝜕 𝑡 2
Also by vector calculus, 𝛻 × 𝛻 × 𝐸 = 𝛻 𝛻 . 𝐸 − 𝛻 2 𝐸
Using Maxwell’s equation 𝛻 . 𝐸 =0, this just equals − 𝛻 2 𝐸
Putting the two sides together, we find 𝜵 𝟐 𝑬 = 𝝁 𝟎 𝜺 𝟎 𝝏 𝟐 𝑬 𝝏 𝒕 𝟐

7. Wave equations The equation represents a travelling wave
To see why, consider a 1D wave travelling along the 𝑥-axis with displacement 𝑦 𝑥,𝑡 =𝐴 sin (𝑘𝑥−𝜔𝑡) in terms of its amplitude 𝐴, wavelength 𝜆= 2𝜋 𝑘 , time period 𝑇= 2𝜋 𝜔 , velocity 𝑣= 𝜆 𝑇 = 𝜔 𝑘

8. Wave equations
The displacement 𝒚(𝒙,𝒕) of a wave travelling with speed 𝒗 satisfies the wave equation 𝝏 𝟐 𝒚 𝝏 𝒙 𝟐 = 𝟏 𝒗 𝟐 𝝏 𝟐 𝒚 𝝏 𝒕 𝟐
To check this, substitute in 𝑦 𝑥,𝑡 =𝐴 sin (𝑘𝑥−𝜔𝑡) where 𝑣= 𝜔 𝑘
On the left-hand side: 𝜕 2 𝑦 𝜕 𝑥 2 =− 𝑘 2 𝐴 sin (𝑘𝑥−𝜔𝑡)
On the right-hand side: 𝜕 2 𝑦 𝜕 𝑡 2 =− 𝜔 2 𝐴 sin (𝑘𝑥−𝜔𝑡)
Using 1 𝑣 2 = 𝑘 2 𝜔 2 , we recover the wave equation!

9. Wave equations
The equation 𝝏 𝟐 𝒚 𝝏 𝒙 𝟐 = 𝟏 𝒗 𝟐 𝝏 𝟐 𝒚 𝝏 𝒕 𝟐 describes general wave travel

10. Wave equations
The relation 𝛻 2 𝐸 = 𝜇 0 𝜀 0 𝜕 2 𝐸 𝜕 𝑡 2 implies that each component of the electric field is a wave travelling at speed 𝒗= 𝟏 𝝁 𝟎 𝜺 𝟎 =𝒄
Evaluating, we obtain 𝑐=3× m/s : the speed of light

11. Properties of the waves
Let’s find out more about the properties of electromagnetic waves
To simplify the case, suppose that the electric field is along the 𝑦-axis, such that 𝐸 𝑦 𝑥,𝑡 =𝐴 sin (𝑘𝑥−𝜔𝑡) and 𝐸 𝑥 = 𝐸 𝑧 =0. This is known as a linearly polarized wave. 𝑦 𝑥

12. Properties of the waves
We can deduce the 𝐵 -field using the relation 𝛻 × 𝐸 =− 𝜕 𝐵 𝜕𝑡
𝑥-component: − 𝜕 𝐵 𝑥 𝜕𝑡 = 𝜕 𝐸 𝑧 𝜕𝑦 − 𝜕 𝐸 𝑦 𝜕𝑧 =0, hence 𝐵 𝑥 =0
𝑦-component: − 𝜕 𝐵 𝑦 𝜕𝑡 = 𝜕 𝐸 𝑥 𝜕𝑧 − 𝜕 𝐸 𝑧 𝜕𝑥 =0, hence 𝐵 𝑦 =0
𝑧-component: − 𝜕 𝐵 𝑧 𝜕𝑡 = 𝜕 𝐸 𝑦 𝜕𝑥 − 𝜕 𝐸 𝑥 𝜕𝑦 =𝑘𝐴 cos (𝑘𝑥−𝜔𝑡) , hence 𝐵 𝑧 𝑥,𝑡 = 𝐴 𝑣 sin (𝑘𝑥−𝜔𝑡) = 1 𝑣 𝐸 𝑦 (𝑥,𝑡)
The magnetic field is perfectly in phase with the electric field and oriented at 90°!

13. Properties of the waves
The form of a linearly-polarized electromagnetic wave:
The changing 𝐸 -field causes a changing 𝐵 -field, which in its turn causes a changing 𝐸 -field

14. Properties of the waves
For an electromagnetic wave travelling in a medium with relative permittivity 𝜀 𝑟 and relative permeability 𝜇 𝑟 we can use Maxwell’s Equations to show that speed 𝒗= 𝒄 𝜺 𝒓 𝝁 𝒓
The quantity 𝜀 𝑟 𝜇 𝑟 is known as the refractive index
Why does this produce a spectrum of light?

15. Energy of the waves Waves transport energy from one place to another
Energy is stored in the electric field with density 𝜀 0 𝐸 2 and the magnetic field with density 1 2 𝜇 0 𝐵 2
For a linearly polarized wave, 𝐵=𝐸/𝑣, where 𝑣= 1 𝜇 0 𝜀 0 , hence 𝜀 0 𝐸 2 = 1 2 𝜀 0 𝑣 2 𝐵 2 = 1 2 𝜇 0 𝐵 2
The energy contained in the electric and magnetic fields is equal!

16. Energy of the waves What is the rate of energy flow?
The energy flowing past area 𝐴 in time 𝑡 is contained in a volume 𝐴𝑣𝑡
If the energy density is 𝑈, this energy is 𝑈𝐴𝑣𝑡 hence the energy flow per unit area per unit time is equal to 𝑈𝑣
For linearly-polarized electromagnetic waves, 𝑈𝑣= 1 𝜇 0 𝐵 𝑧 2 𝑣= 1 𝜇 0 𝐸 𝑦 𝐵 𝑧
The energy flow can be represented by the Poynting vector 𝟏 𝝁 𝟎 ( 𝑬 × 𝑩 )
𝑣×𝑡
Energy density 𝑈 𝐴
Propagation speed 𝑣

17. Energy of the waves
The Poynting vector, 1 𝜇 0 𝐸 × 𝐵 = 𝐸 × 𝐻 , is oriented in the direction of travel, perpendicular to the 𝐸 - and 𝐵 -fields
𝐸 × 𝐻

18. Energy of the waves
As an example, consider current 𝐼 flowing along a cylindrical conductor driven by a potential difference 𝑉
The energy flux is equal to the Poynting vector 1 𝜇 0 𝐸 × 𝐵
This is non-zero at the surface of the conductor, flowing inward
If the curved surface area is 𝐴, the power flowing in is 𝑃= 1 𝜇 0 𝐸𝐵𝐴= 1 𝜇 0 𝑉 𝐿 𝜇 0 𝐼 2𝜋𝑟 2𝜋𝑟𝐿=𝑉𝐼
We recover Joule heating!
𝐵= 𝜇 0 𝐼 2𝜋𝑟
𝐸= 𝑉 𝐿 𝑟 𝐿

19. Summary
Maxwell’s equations predict that the 𝐸 - and 𝐵 -fields satisfy linked wave equations with speed 𝜇 0 𝜀 0
For a linearly-polarized wave, the 𝐸 - and 𝐵 -fields are in phase and oriented at 𝟗𝟎°
The energy densities stored in the 𝐸 - and 𝐵 -fields is equal, and the energy flow per unit area per unit time is equal to the Poynting vector 𝐸 × 𝐻