The Wave Equation and the Speed of Light Chapter 1 Physics 208, Electro-optics Peter Beyersdorf Document info Ch 1 Class Outline Maxwell’s equations Boundary conditions Poynting’s theorem and conservation laws Complex function formalism time average of sinusoidal products Wave equation Ch 1, 2 Maxwell’s Equations → → Electric field (E) and magnetic field (H) in free- space can be generalized to the electric → → displacement (D) and the magnetic induction (B) that include the effects of matter. Maxwell’s equations relate these vectors ∂B! ! E! + = 0 ∇ × ∂t Faraday’s law ∂D! ! H! = J! Ampere’s law ∇ × − ∂t ! D! = ρ Gauss’ law (for magnetism) ∇ · ! B! = 0 Gauss’ law (for electricity) ∇ · What do each of these mean? Ch 1, 3 Faraday’s Law ∂B! ! E! + = 0 ∇ × ∂t The Curl of the electric field is caused by changing magnetic fields A changing magnetic field can produce electric fields with field lines that close on themselves Ch 1, 4 Ampere’s Law ∂D! ! H! = J! ∇ × − ∂t The Curl of the magnetic field is caused by current of charged particles (J) or of the field they produce (dD/dt) A changing electric field can produce magnetic fields (with field lines that close on themselves) For all cases considered in this class, J=0 Ch 1, 5 Gauss’ Law ! D! = ρ ∇ · Electrical charges are the source of the electric field D! = "E! = "0E! + P! For all cases considered in this class, ρ=0 ε is a 3x3 tensor not a scalar (unless the material is isotropic)! ε may be a function of E and H! (giving rise to non- linear optics) ε can be determined via measurements on a parallel plate capacitor filled with a given material using the 6 equation C=εA/d Ch 1, Gauss’ Law for Magnetism There are no source of magnetic fields No magnetic monopoles Magnetic field lines can only circulate B! = µH! = µ0H! + M! μ is a 3x3 tensor not a scalar (unless the material is isotropic)! μ may be a function of E and H! (giving rise to non- linear optics) μ can be measured using the Biot-Savart law Ch 1, 7 Waves and Maxwell’s Equations A charged particle is a source of an electric field When that particle moves it changes the (spatial distribution of) the electric field When the electric field changes it produces a circulating magnetic field If the particle accelerates this circulating magnetic field will change A changing magnetic field produces a circulating electric field The circulating electric field becomes the source of a circulating magnetic field Ch 1, 8 Boundary Conditions When an EM wave propagates across an interface, Maxwell’s equations must be satisfied at the interface as well as in the bulk materials. The constraints necessary for this to occur are called the “boundary conditions” !1, µ1 !2, µ2 D! dA! = σdA · ! " d E! d!s = B! dA! · −dt · ! " B! dA! = 0 · ! d H! d!s = J! dA! + D! dA! · · dt · ! " " Ch 1, 9 Boundary Conditions Gauss’ law can be used to find the boundary conditions on the component of the electric field that is perpendicular to the interface. If the materials are dielectrics there will be no free charge on the surface (σ=0) !1, µ1 !2, µ2 D! dA! = σdA · ! " d E! d!s = B! dA! · −dt · ! " B! dA! = 0 · ! d H! d!s = J! dA! + D! dA! · · dt · ! " 0" D = D D1 D2 = σdA ∴ 1 2 Ch 1, 10 ⊥ − ⊥ ⊥ ⊥ ! → Boundary Conditions Faraday’s law can be applied at the interface. If the loop around which the electric field is computed is made to have an infintesimal area the right side will go to zero giving a relationship between the parallel components of the electric field !1, µ1 !2, µ2 D! dA! = σdA · ! " d E! d!s = B! dA! · −dt · ! " B! dA! = 0 · ! d H! d!s = J! dA! + D! dA! · · dt · ! " d " 0 E2 E1 = B dA E1 = E2 ! − ! −dt · ∴ ! ! Ch 1, 11 ! → Boundary Conditions Gauss’ law for magnetism gives a relationship between the perpendicular components of the magnetic field at the interface !1, µ1 !2, µ2 D! dA! = σdA · ! " d E! d!s = B! dA! · −dt · ! " B! dA! = 0 · ! d H! d!s = J! dA! + D! dA! · · dt · ! " " B1 A B2 A = 0 B1 = B2 ⊥ − ⊥ ∴ ⊥ ⊥ Ch 1, 12 Boundary Conditions Ampere’s law applied to a loop at the interface that has an infintesimal area gives a relationship between the parallel components of the magnetic field. (Note that in most common materials μ=μo) In the absence of currents J=0 so !1, µ1 !2, µ2 D! dA! = σdA · ! " d E! d!s = B! dA! · −dt · ! " B! dA! = 0 · ! d H! d!s = J! dA! + D! dA! · · dt · ! " 0 "d 0 ! ! ! ! H1 = H2 H1 L H2 L = J dA + D∴dA Ch 1, 13 ! − ! · dt · ! ! ! → ! Poynting’s Theorem The flow of electromagnetic energy is given by the Poynting vector S! = E! H! × which has a magnitude that is the power per unit area carried by an electromagnetic wave in the direction of S. S [W/m2]⇔E [v/m] H [A/m] Ch 1, 14 Complex-Function Formalism Steady-state sinusoidal functions of the form a(t) = A cos(ωt + α) can be treated as having a complex amplitude A = Aeiα such that the! function can be written as a(t) = Re Aeiωt ! # or in shorthand!!Ae!iω!t" where it is understood that the real part of this complex expression represents the or!iginal sinusoidal function Ch 1, 15 Phasors The complex amplitude of a sinusoidal function can be represented graphically by a point (often an arrow from the origin to a point) in the complex plane Im Im Im Re Re Re a(t) = cos ωt a(t) = sin ωt a(t) = cos (ωt) + sin (ωt) iπ/4 A = 1 A = i A = √2e− − Ch 1, 16 ! ! ! Phasors Addition of same-frequency sinusoidal functions involves factoring out the time dependance and simply adding the phasor amplitudes. Addition of difference frequency sinusoidal function is often simplified by factoring out a sinusoidal component at the average frequency. Multiplication of sinusoidal functions can not be done by multiplying phasors since Re [x] Re [y] = Re [xy] ! Ch 1, 17 Phasor Example For electric field amplitudes described by E1 = E10 cos (ω1t) and E2 = E20 cos (ω2t) Use the phasor representation to find a representation of E1+E2 as a slow modulation of a _ field at the average frequency ω=(ω1+ω2)/2 Ch 1, 18 Phasor Example 5 iω1t iω2t E1 + E2 = E10e + E20e 2.5 ∆E ∆E = E + eiω1t + E eiω2t avg 2 avg − 2 ! " ! " -4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8 ∆E i(ω¯+ ∆ω )t ∆E i(ω¯ ∆ω )t = E + e 2 + E e − 2 avg 2 avg − 2 Example ! " !-2.5 " ∆ωt ∆ωt E=cos(2=πt)+3c2E os(2cos.5πt) + i∆E sin eiω¯t avg 2 2 -5 # ! " ! "$ 1 ∆ωt ∆ωt 2 = 4E2 cos2 + (∆E)2 sin2 eiω¯t+iα avg 2 2 # ! " ! "$ E + E ω1 + ω2 ∆E ∆ωt 10 20 ω¯ α arctan tan Eavg ≡ 2E 2 ≡ 2 ≡ 2 ! avg " #$ ∆E E E ∆ω ω ω Ch 1, 19 ≡ 10 − 20 ≡ 1 − 2 Time Averages Optical fields vary too fast to be directly detected, instead it is the irradiance averaged over many cycles that is detected as light. 1 T a(t)b(t) = A cos(ωt + α)B cos(ωt + β)dt ! " T !0 In terms of the phasor amplitudes this is 1 a(t)b(t) = Re AB∗ ! " 2 ! # " " Ch 1, 20 Poynting Vector Example For electric and magnetic fields given by iωt+φ E = E0e E H = 0 eiωt+φ where Z0 µ Z = η 0 377Ω 0 0 ≡ " ≈ ! 0 is the impedance of free space, what is the irradiance of the wave? How much power is measured by a detector of area A? Ch 1, 21 Poynting Vector Example For electric and magnetic fields given by iωt+φ E = E0e E H = 0 eiωt+φ where η0 µ η 0 377Ω 0 ≡ " ≈ ! 0 is the impedance of free space, what is the irradiance of the wave? 2 1 1 iφ E0 iφ E0 Savg = E! H! = Re EH∗ = Re E0e e− = × 2 2 η0 2η0 ! " # % & ' 2 This is analogous to Pavg$=V$ /2R for AC circuits 2 E0 Pavg = S!avg A! A · ≤ 2η0 Ch 1, 22 Wave Equation in Isotropic Materials Starting with Faraday’s law take the curl of both sides use vector calculus relationship to get Use Ampere’s law (in free space where J=0) and Gauss’ law (in free space where ρ=0) ! D! = 0 ∇ · in an isotropic medium Ch 1,23 Wave Equation in Crystals In an anisotropic medium does not simplify as much since ! ! ! ! ! ! ! ! does not imply but rather ! D! = ! "E! = "! E! + E! " ∇ · ∇ · ∇ · · ∇ where ∇ε$0. In this case it is usually easiest to write the wave equation as ∂2E! ^ ^ ! ! E! + µ" = 0 D=εE E^ k^ ∇ × ∇ × ∂t2 ! E ! k^ ^^ ^^ D = 0 or kxEkxE∇ · 2 kx^^ ^kxE^ ^^ !k !k E + µ"ω E = 0 kxkxE Ch 1,2 4 × × ! ! Spherical Solutions to the Wave Equation Consider solutions for E such that ∇2E and δ2E/dt2 are both proportional to E - allowing the two sides to differ only by a constant term. is one such solution in spherical coordinates. Using the relationship for the Laplacian of a spherically symmetric function: Show that !! ! given above is a solution to the wave equation Ch 1, 26 Solutions to the Wave Equation Recall the meaning of k and ω (k=2"/λ, ω=2"/T) we can express this as Since λ is the distance travelled by the wave in one cycle, and T is the time to travel one cycle, λ/T is the velocity of the wave, which can be determined from electrostatics and magnetostatics! Ch 1, 27 Solutions to the Wave Equation From our solution in free space and Gauss’ law in free space (ρ=0) We find that since!! ! ! only has a spatial dependance on r its divergence, given by must be implying !! ! ! ! ! ! meaning this must be a transverse wave (and isn’t a solution in anisotropic media) Ch 1, 28 Speed of Light In free-space where ε=ε0 and μ=μ0 the speed of light is defined to be c≡299792458 m/s.