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

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

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 1 2 𝜀 0 𝐸 2 and 𝐵 2 2 𝜇 0 , respectively

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

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

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

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

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!

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

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× 10 8 m/s : the speed of light

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

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°!

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

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?

Energy of the waves Waves transport energy from one place to another Energy is stored in the electric field with density 1 2 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 1 2 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!

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

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

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

Summary Maxwell’s equations predict that the 𝐸 - and 𝐵 -fields satisfy linked wave equations with speed 1 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