Maxwell's Equations and Light Waves Longitudinal vs. transverse waves Derivation of wave equation from Maxwell's Equations Why light waves are transverse waves Why we neglect the magnetic field Photons and photon statistics The equations of optics are Maxwell’s equations. E 0 B 0 B E t E B me t where E is the electric field, B is the magnetic field, e is the permittivity, and m is the permeability of the medium. As written, they assume no charges (or free space). Derivation of the Wave Equation from Maxwell’s Equations Take of: B E t B [ E] [ ] t Change the order of differentiation on the RHS: [ E] [ B] t Derivation of the Wave Equation from Maxwell’s Equations (cont’d) But: E B me t Substituting for B , we have: E [ E] [ B] [ E] [ me ] t t t Or: E [ E] me 2 t 2 assuming that m and e are constant in time. Derivation of the Wave Equation from Maxwell’s Equations (cont’d) Identity: r r r r r r r 2 [ f ] ( f ) f Using the identity, 2E [ E] me 2 t becomes: (E)2 E me E t 2 2 If we now assume zero charge density: r = 0, then E 0 and we’re left with the Wave Equation! 2 E me E t 2 2 where µe 1/c2 Why light waves are transverse Suppose a wave propagates in the x-direction. Then it’s a function of x and t (and not y or z), so all y- and z-derivatives are zero: Ey Ez By Bz 0 y z y z Now, in a charge-free medium, that is, E 0 andΚΚB0 Ex Ey Ez 0 Bx By Bz 0 x y z x y z Substituting the zero values, we have: Ex 0 and Bx 0 x x So the longitudinal fields are at most constant, and not waves. The magnetic-field direction in a light wave Suppose a wave propagates in the x-direction and has its electric field along the y-direction [so Ex = Ez= 0, and Ey = Ey(x,t)]. What is the direction of the magnetic field? Use: x E y Ex Ez E y E E B z E , , t y z z x x y So: In other words: E y B 0,0, t x E y B z t x And the magnetic field points in the z-direction. The magnetic-field strength in a light wave Suppose a wave propagates in the x-direction and has its electric field in the y-direction. What is the strength of the magnetic field? Start with: E y B z t x and E r,t E0 exp i kx w t y t We can integrate: Ey Bz (x,t) Bz (x,0) dt x 0 Take Bz(x,0) = 0 So: Bz (x,t) ik E0 expi(kxwt) iw But w / k = c: Bz (x,t) 1c Ey(x,t) Differentiating Ey with respect to x yields an ik, and integrating with respect to t yields a 1/-iw. An Electromagnetic Wave The electric and magnetic fields are in phase. snapshot of the wave at one time The electric field, the magnetic field, and the k-vector are all perpendicular: E B k The Energy Density of a Light Wave The energy density of an electric field is: The energy density of a magnetic field is: 1 Using B = E/c, and c we have: em 1 2 UE eE 2 11 2 UB B 2m , which together imply that B E em 11 2 1 2 UB E em e E U E 2m 2 Total energy density: U U E U B e E2 So the electrical and magnetic energy densities in light are equal. Why we neglect the magnetic field The force on a charge, q, is: Felectrical Fmagnetic F qE qv B Taking the ratio of the magnitudes of the two forces: Since B = E/c: Fmagnetic Felectrical Fmagnetic Felectrical qvB qE where v is the charge velocity v B vBsin vB v c So as long as a charge’s velocity is much less than the speed of light, we can neglect the light’s magnetic force compared to its electric force. The Poynting Vector: S = c2 e E x B The power per unit area in a beam. U = Energy density A Justification (but not a proof): Energy passing through area A in time Dt: = U V = U A c Dt So the energy per unit time per unit area: c Dt = U V / ( A Dt ) = U A c Dt / ( A Dt ) = U c = c e E2 = c2 e E B And the direction E B k is reasonable. The Irradiance (often called the Intensity) A light wave’s average power per unit area is the irradiance. 1 S(r ,t) T tT /2 S(r ,t ') dt ' tT /2 Substituting a light wave into the expression for the Poynting vector, S c 2 e E B, yields: real amplitudes S (r ,t) c 2 e E0 B0 cos 2 (k r w t ) The average of cos2 is 1/2: I(r ,t) S (r ,t) c 2 e E0 B0 (1 / 2) The Irradiance (continued) Since the electric and magnetic fields are perpendicular and B0 = E0 / c, I 1 2 c e E0 B0 2 I or: becomes: I 1 ce E 0 2 ~ 2 1 2 c e E0 2 because the real amplitude squared is the same as the mag-squared complex one. where: 2 E0 E0 x E0*x E0 y E0*y E0 z E0*z Remember: this formula only works when the wave is of the form: E r ,t Re E0 exp i k r w t that is, when all the fields involved have the same k r wt Sums of fields: Electromagnetism is linear, so the principle of Superposition holds. If E1(x,t) and E2(x,t) are solutions to the wave equation, then E1(x,t) + E2(x,t) is also a solution. Proof: 2 (E1 E2 ) 2 E1 2 E2 2 2 x x x2 and 2 (E1 E2 ) 2 E1 2 E2 2 2 t t t 2 2 (E1 E2 ) 1 2 (E1 E2 ) 2 E1 1 2 E1 2 E2 1 2 E2 2 2 2 2 2 0 2 2 2 2 x c t c t x c t x This means that light beams can pass through each other. It also means that waves can constructively or destructively interfere. The irradiance of the sum of two waves If they’re both proportional to exp i(k r w t) , then the irradiance is: I 12 ce E0 E0* 12 ce E0 x E0 x * E0 y E0 y * E0 z E0 z * Different polarizations (say x and y): I 12 ce E0 x E0 x * E0 y E0 y * I x I y Same polarizations (say E0x Intensities add. E1 E2): I I ce Re E E I I 12 ce E1 E1* 2 Re E1 E2* E2 E2* Therefore: * 1 1 2 2 Note the cross term! The cross term is the origin of interference! Interference only occurs for beams with the same polarization. The irradiance of the sum of two waves of different color We can’t use the formula because the k’s and w’s are different. So we need to go back to the Poynting vector, S (r ,t) c 2 e EB S (r ,t) c 2e E1 E2 B1 B2 c 2e E B 1 1 E1 B2 E2 B1 E2 B2 E10 cos(k1 r w1 t 1 ) B20 cos(k2 r w 2 t 2 ) This product averages to zero, as does Different colors: I I1 I 2 E2 B1 Intensities add. Waves of different color (frequency) do not interfere! Irradiance of a sum of two waves Same polarizations Same colors Different colors I I1 I 2 ce Re E1 E2 I I1 I 2 * Different polarizations I I1 I 2 I I1 I 2 Interference only occurs when the waves have the same color and polarization. Light is not only a wave, but also a particle. Photographs taken in dimmer light look grainier. Very very dim Bright Very dim Very bright Dim Very very bright When we detect very weak light, we find that it’s made up of particles. We call them photons. Photons The energy of a single photon is: hn or w = (h/2p)w where h is Planck's constant, 6.626 x 10-34 Joule-seconds One photon of visible light contains about 10-19 Joules, not much! F is the photon flux, or the number of photons/sec in a beam. F = P / hn where P is the beam power. Counting photons tells us a lot about the light source. Random (incoherent) light sources, such as stars and light bulbs, emit photons with random arrival times and a Bose-Einstein distribution. Laser (coherent) light sources, on the other hand, have a more uniform (but still random) distribution: Poisson. Bose-Einstein Poisson Photons have momentum If an atom emits a photon, it recoils in the opposite direction. If the atoms are excited and then emit light, the atomic beam spreads much more than if the atoms are not excited and do not emit. Photons—Radiation Pressure Photons have no mass and always travel at the speed of light. The momentum of a single photon is: h/l, or Radiation pressure = Energy Density (Force/Area = Energy/Volume) When radiation pressure cannot be neglected: Comet tails (other forces are small) Viking space craft (would've missed Mars by 15,000 km) Stellar interiors (resists gravity) PetaWatt laser (1015 Watts!) k Photons "What is known of [photons] comes from observing the results of their being created or annihilated." Eugene Hecht What is known of nearly everything comes from observing the results of photons being created or annihilated.