E&M Lecture 15Topics:(1) Maxwell’s Equations(2) 1st, 2nd and 3rd forms(3) Electromagnetic waves in vacuum(4) Speed of light in vacuum(5) EM waves in non-conducting media(6) EM waves in conducting media(7) Other properties of EM waves

Tuesday @ 16.00 No lecture (Dr Schvets)Thursday @ 11.00 Tutorial (E&M)Thursday @ 15.00 Prof Ray (not AM)

Maxwell’s Equations

This “form” reflects the initial experimental basis of “Laws”;

3 other forms possible……field vs source?€

(1) ∇.E =ρεo

(ρ = ρ f + ρb )

(2) ∇.B = 0

(3) ∇ × E = −∂B∂t

(4) ∇ × B = µo J + εoµo∂E∂t

(J = J f + JM + JP )

1st form of Maxwell’s Equations

“all field terms on LHS and all source terms on RHS”The sources (ρ and J) are multiple (free, bound, mag, pol)

but special status of free sources…suggests 2nd Form?€

(1) ∇.E =ρεo

(ρ = ρ f + ρb )

(2) ∇.B = 0

(3) ∇ × E +∂B∂t

= 0

(4) ∇ × B −εoµo∂E∂t

= µo J (J = J f + JM + JP )

2nd form of Maxwell’s EquationsApplies only to “well behaved” LIH media:

Focus on “sources” means equations (2) and (3) unchanged!Recall Gauss’ Law for D:

In this version of (1), ρ → ρf and εo → εRecall H-version of (4):

In this version of (4), J → Jf , also µo → µ and εo → ε

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D = εrεo E = εE and B = µrµoH = µH

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∇.D = ρ f ⇒∇. εE( ) = ρ f ⇒∇.E =ρ f

ε

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∇ ×H = J f +∂D∂t

⇒∇×H −∂D∂t

= J f

µ∇ ×H −µ∂D∂t

= µJ f ⇒∇× B −εµ∂E∂t

= µJ f

2nd and 3rd forms

LHS: 2nd form, free sources only, other sources hidden inpermittivity and permeability constants

RHS: 3rd form (Minkowsky) free sources only, mixedfields, no constants

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(1) ∇.E =ρ f

ε∇.D = ρ f

(2) ∇.B = 0 ∇.B = 0

(3) ∇ × E +∂B∂t

= 0 ∇ × E +∂B∂t

= 0

(4) ∇ × B −εµ∂E∂t

= µJ f ∇ ×H −∂D∂t

= J f

Electromagnetic Wave Equation

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(1) ∇.E =ρεo

(2) ∇.B = 0

(3) ∇ × E +∂B∂t

= 0 (4) ∇ × B −εoµo∂E∂t

= µo J

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(3)⇒ ∇×∇ × E +∂∂t

∇ × B( ) = 0

(4)⇒ ∇×∇ × E +∂∂t

µo J + εoµo∂E∂t

= 0

identity⇒∇ ∇.E( ) −∇2E +∂∂t

µoJ + εoµo∂E∂t

= 0

(1)⇒∇2E −εoµo∂ 2E∂t 2

=∇ρεo

+ µo∂J∂t

Firstform

Electromagnetic Waves in Vacuum

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∇2E −εoµo∂ 2E∂t 2

=∇ρεo

+ µo∂J∂t

in vacuum⇒∇2E −εoµo∂ 2E∂t 2

= 0

possible solution : E = E o expi ωt − kz( )

i.e. a plane wave, z-dependence ⇒ direction of propagationEo : arb. constant vectorω : angular frequency = 2πfk : wave number = 2π/λ vp = ω/k : phase velocity

Electromagnetic Waves

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possible solution : E = E o expi ωt − kz( )

∇2 expi ωt − kz( )( ) ≡ ∂ 2

∂z2exp− ikz( ) = −k 2

∂ 2

∂t2expi ωt − kz( )( ) ≡ ∂ 2

∂t 2expiωt( ) = −ω 2

∇2E −εoµo∂ 2E∂t 2

= 0⇒−k 2E +ω 2εoµo E = 0

⇒ k 2 =ω 2εoµo ⇒ vp =ωk

=1εoµo

= c speed of lightin vacuum! - unpick?

EM Waves in non-conducting LIH medium

Less than speed of light in vacuum!

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2nd form⇒∇2E −εµ∂2E∂t 2

=∇ρ f

ε+ µ

∂J f

∂t= 0

can always avoid free sources( ) ρ f = J f = 0

same possible solution : E = E o expi ωt − kz( )

∇2E −εµ∂2E∂t 2

= 0⇒−k 2E +ω 2εµE = 0

⇒ k 2 =ω 2εµ ⇒ vp =ωk

=1εµ

=cεrµr

EM Waves in conducting LIH medium(cannot avoid the free sources now!)

Attenuated wave; α is absorption coefficient Next:individual Maxwell versus combined Maxwell?

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2nd form⇒∇2E −εµ∂2E∂t 2

=∇ρ f

ε+ µ

∂J f

∂t= 0

usually ρ f = 0 and write J f =σ E ⇒ µ∂J f

∂t=σµ

∂E∂t

∇2E −εµ∂2E∂t 2

−σµ∂E∂t

= 0⇒ k 2 =ω 2εµ − iωσµ

⇒ k is complex! write k = β − iα⇒ E = E o expi ωt − kz( ) = E o exp −αz( )expi ωt −βz( )

Additional properties of EM Waves?

Eo perp to dir of motion! (transverse)

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write E = Ex ˆ x + Ey ˆ y + Ez ˆ z where Ex = EO x exp i(ωt − kz) etc

(1)⇒∇.E =∂Ex

∂x+∂Ey

∂y+∂Ez

∂z= 0 (vacuum)

⇒∂Ez

∂z= 0⇒ ∂

∂zEO z expi(ωt − kz)[ ] = 0⇒ EO z ≡ 0

∴E = Ex ˆ x + Ey ˆ y

when solve individual Maxwell Equation(s):(a) implication of z-only spatial dependence

Exercise: show solution B=Boexpi(ωt-kz) exists + is transverse

Additional properties of EM Waves

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write E = ˆ x Eo expi(ωt − kz)

(3)⇒∇× E = −∂B∂t

=

ˆ x ˆ y ˆ z ∂∂x

∂∂y

∂∂z

Eo expi() 0 0

−∂B∂t

= − ˆ y ikEo expi(ωt − kz)⇒ B = ˆ y kω

Eo expi(ωt − kz)

⇒ B perp to E and Bo =kω

Eo

(b) connection between E and B solutionsChoose Eo along x-axis (also in vacuum):

Mutually perpendicular?

Picturing EM Waves

E and B are in phase for vacuum and non-conducting media,and out of phase for conducting medium!

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exercise : write B = ˆ x Bo expi(ωt − kz)

show E = − ˆ y kωεoµo

Bo expi(ωt − kz)

⇒ E perp to B and Eo =k

ωεoµo

Bo

but Bo =kω

Eo ⇒kω

=ωεoµo

k⇒ k 2 =ω 2εoµo prev ?( )