EM waves: energy, resonators. Scalar wave equation Maxwell equations to the EM wave equation A simple linear resonator Energy in EM waves 3D waves

Transcription

1 EM waves: energy, resonators Scalar wave equation Maxwell equations to the EM wave equation A simple linear resonator Energy in EM waves 3D waves

2 Simple scalar wave equation 2 nd order PDE 2 z 2 ψ (z,t) 1 c 2 2 t 2 ψ (z,t) = 0 Assume separable solution 1 ( ) 2 f z z f (z) c 2 g t ( ) 2 t 2 g t ( ) = 0 ψ (z,t) = f ( z)g( t) Each part is equal to a constant A 1 ( ) 2 f z z f (z) = A, c 2 g t ( ) 2 t 2 g t ( ) = A f (z) = cos kz ( ) k 2 = A, g t ( ) = cos ω t ( ) ω 2 1 c 2 = A ω = ±k c Sin( ) also works as a second solution

3 Full solution of wave equation Full solution is a linear combination of both ψ (z,t) = f ( z)g( t) = ( A 1 coskz + A 2 sin kz) ( B 1 cosωt + B 2 sinωt) Too messy: use complex solution instead: ψ (z,t) = f ( z)g( t) = A 1 e ikz + A 2 e ikz ( )( B 1 e iωt + B 2 e iωt ) ψ (z,t) = A 1 B 1 e i ( kz+ωt ) i + ( kz+ωt) A2 B 2 e + A1 B 2 e i ( kz ωt ) i + ( kz ωt) A2 B 1 e Constants are arbitrary: rewrite ψ (z,t) = A 1 cos( kz + ωt +φ ) 1 + A 2 cos( kz ωt +φ ) 2

4 Interpretation of solutions Wave vector k = 2π λ Angular frequency ω = 2πν Wave total phase: absolute phase : Phase velocity: c φ Φ = kz ωt +φ Φ = kz k ct +φ = k ( z ct) +φ Φ = constant when z = ct ψ (z,t) = A 1 cos( kz + ωt +φ ) 1 + A 2 cos( kz ωt +φ ) 2 Reverse (to -z) Forward (to +z)

5 Maxwell's Equations to wave eqn The induced polarization, P, contains the effect of the medium: E = 0 E = B B = 0 Take the curl:" E ( ) = t Use the vector ID:" t B = 1 E c 2 t + µ 0 B = t ( ) = B( A C) C( A B) A B C P t 1 E c 2 t + µ 0 E ( ) = E ( ) ( )E = 2 E 2 E 1 c 2 2 E t 2 = µ 0 2 P t 2 P t Inhomogeneous Wave Equation "

6 Maxwell's Equations in a Medium The induced polarization, P, contains the effect of the medium: 2 E 1 c 2 2 E t 2 = µ 0 2 P t 2 Sinusoidal waves of all frequencies are solutions to the wave equation The polarization (P) can be thought of as the driving term for the solution to this equation, so the polarization determines which frequencies will occur. For linear response, P will oscillate at the same frequency as the input. P( E) = ε 0 χe In nonlinear optics, the induced polarization is more complicated: ( ) P( E) = ε 0 χ (1) E + χ (2) E 2 + χ (3) E The extra nonlinear terms can lead to new frequencies.

7 Solving the wave equation: linear induced polarization For low irradiances, the polarization is proportional to the incident field:" P( E) = ε 0 χe, D = ε 0 E + P = ε ( 0 1+ χ )E = ε E = n 2 E In this simple (and most common) case, the wave equation becomes:" 2 E 1 c 2 2 E t 2 = 1 c 2 χ 2 E t 2 Using:" ε 0 µ 0 = 1/ c 2 The electric field is a vector function in 3D, so this is actually 3 equations:" " 2 E n2 c 2 2 E t 2 = 0 ε 0 2 E x 2 E y 2 E z ( 1+ χ ) = ε = n 2 ( r,t) n2 2 c 2 t E 2 x ( r,t) n2 2 c 2 t E 2 y ( r,t) n2 2 c 2 t E 2 z ( r,t) = 0 ( r,t) = 0 ( r,t) = 0

8 Plane wave solutions for the wave equation If we assume the solution has no dependence on x or y:" 2 E( z,t) = 2 x E ( z,t ) y E ( z,t ) z E ( z,t ) = 2 2 z E ( z,t ) 2 2 E z 2 n2 c 2 2 E t 2 = 0 The solutions are oscillating functions, for example" ( ) = ˆx E x cos k z z ωt E z,t ( ) Where" ω = k c, k = 2π n / λ, v ph = c / n " This is a linearly polarized wave."

9 Complex notation for waves Write cosine in terms of exponential 1 E( z,t) = ˆx E x cos( kz ωt +φ) = ˆx E x Note E-field is a real quantity. It is convenient to work with just one part +i kz ωt E 0 e ( ) E 0 = 1 ( ) 2 E x e iφ We will use Svelto: Then take the real part. No factor of 2 i kz ωt e ( ( ) + e i( kz ωt+φ ) ) 2 ei kz ωt+φ In nonlinear optics, we have to explicitly include conjugate term

10 Example: linear resonator (1D) Boundary conditions: conducting ends (mirrors) Field is a superposition of + ve and ve waves: E x ( ) = 0 E x z = L z,t E x z = 0,t ( ) = A + e i k zz ωt+φ + ( ) = 0 ( ) z,t + A e i ( k zz ωt+φ ) Absorb phase into complex amplitude ( z,t) = ( A + e +ikzz + A e ik zz )e iωt Apply b.c. at z = 0 E x ( ) = 0 = A + + A ( z,t) = Asin k z z e iωt E x 0,t E x ( )e iωt A + = A

11 Quantization of frequency: longitudinal modes Apply b.c. at far end ( ) = 0 = Asin k z L z e iωt k z L z = lπ l = 1,2,3, E x L z,t Relate to wavelength: k z = 2π λ = lπ L z L z = l λ 2 Integer number of half-wavelengths Relate to allowed frequencies: ω l c = lπ c ν l = l L z 2L z Equally spaced frequencies: Δν = c = 1 2L z T RT Frequency spacing = 1/ round trip time

12 Wave energy and intensity Both E and H fields have a corresponding energy density (J/m 3 ) For static fields (e.g. in capacitors) the energy density can be calculated through the work done to set up the field ρ = 1 2 εe µh 2 Some work is required to polarize the medium Energy is contained in both fields, but H field can be calculated from E field

13 Calculating H from E in a plane wave Assume a non-magnetic medium µ 0 H t ( ) = ˆx E x cos( kz ωt) E z,t E = B t = µ 0 H t Can see H is perpendicular to E = E = ˆx ŷ ẑ x y z E x 0 0 Integrate to get H-field: = ŷ z E x = ŷk z E 0 sin( k z z ωt) ( ) H = ŷ k E z 0 sin( k µ z z ωt)dt = ŷ k E cos k z 0 z z ωt 0 µ 0 ω

14 H field from E field H field for a propagating wave is in phase with E-field H = ŷh 0 cos( k z z ωt) = ŷ k z E ωµ 0 cos( k z z ωt) 0 Amplitudes are not independent H 0 = k z ωµ 0 E 0 k z = n ω c H 0 = n cµ 0 E 0 = nε 0 ce 0 c 2 = 1 µ 0 ε 0 1 µ 0 c = ε 0 c

15 Energy density in an EM wave Back to energy density, non-magnetic ρ = 1 2 εe µ 0 H 2 ε = ε 0 n 2 ρ = 1 2 ε 0 n 2 E µ 0 n 2 ε 0 2 c 2 E 2 µ 0 ε 0 c 2 = 1 ρ = ε 0 n 2 E 2 = ε 0 n 2 E 2 cos 2 ( k z z ωt) Equal energy in both components of wave H = nε 0 ce

16 Cycle-averaged energy density Optical oscillations are faster than detectors Average over one cycle: ρ = ε 0 n T E 0 cos 2 ( k z z ωt)dt T 0 Graphically, we can see this should = ½ k z = k z = π/ Regardless of position z t/t ρ = 1 2 ε 0 n2 E 0 2

17 Intensity and the Poynting vector Intensity is an energy flux (J/s/cm 2 ) In EM the Poynting vector give energy flux S = E H For our plane wave, S = E H = E 0 cos( k z z ωt)nε 0 ce 0 cos k z z ωt S = nε 0 ce 0 2 cos 2 S is along k Time average: ( k z z ωt)ẑ S = 1 2 nε 0 ce 02ẑ Intensity is the magnitude of S ( ) ˆx ŷ I = 1 2 nε ce 2 = c 0 0 n ρ = V ρ Photon flux: F = I phase hν

18 General plane wave solution Assume separable function E(x, y,z,t) ~ f ( 1 x) f 2 y ( ) f 3 z ( )g t 2 E( z,t) = 2 x E ( z,t ) y E ( z,t ) z E ( z,t ) = n2 2 2 c 2 t E ( z,t ) 2 Solution takes the form: E(x, y,z,t) = E 0 e ikxx e ikyy e ikzz e iωt = E 0 e i ( k xx+k y y+k z z) e iωt E(x, y,z,t) = E 0 e i( k r ωt ) Now k-vector can point in arbitrary direction ( )