Physics 504, Physics 504, Electromagnetic Waves Spring 2011 Plane Waves Spring 2011 Electricity i(~k ~x ωt) Electricity We begin with waves in a non-conducting uniform linear and If we have a plane wave, E~ (~x, t) e · − , this will and Magnetism ∝ Magnetism medium, so we are discussing solutions of Maxwell’s satisfy the equation provided the wave number equations without sources. As we are assuming no Shapiro Shapiro k := ~k 2 = √µ ω. A fixed phase of this wave moves at time-dependence of the properties of the medium, we will Electromagneti ~v = ~kω/k2 so v =1/√µ, which is called the phase Electromagneti fourier transform in time and consider the “harmonic” Waves p Waves Reflection µ Reflection fields, so and velocity. The index of refraction is defined as n = , and Refraction µ00 Refraction ~ B~ =0 ~ E~ iωB~ =0 B~ = µH~ r ∇· ∇× − and so v = c/n. ~ D~ =0 ~ H~ + iωD~ =0 D~ = E~ If we consider plane waves in the x direction, uniform in y ∇· ∇× ik(x vt) ik(x+vt) where the permittivity and permeability µ are constant and z,wehaveuk(x, t)=ae − + be− , in space. corresponding to right and left moving sinusoidal waves respectively. If the medium is nondispersive, so n is constant, we may So 2E~ = ~ ~ E~ = ~ (iωB)= iωµ~ H~ superimpose these waves with different k to have vaves of ∇ −∇× ∇× −∇× − ∇× arbitrary shape, u(x, t)=f(x vt)+g(x + vt), but if = ω2µD~= ω2µE,~ − − − there is dispersion, having created such a wave packet at which tells us 2 + µω2 E~ = 0, and by taking the curl t = 0 will not produce pulses of unchanged shape at later ∇ ~ times, because the vt terms in the phase will vary with k. of this, the same! equation holds for B.
Thus a general solution of Maxwell’s sourceless equations Physics 504, Physics 504, Spring 2011 Polarization Spring 2011 will be a linear superposition of (the real parts of) Electricity Electricity and and Magnetism Magnetism ~ ~ i~k ~x iωt E(~x, t)= e · − 2 2 From these constraints we see that ~ is a vector E with k = µω , Shapiro Shapiro i~k ~x iωt ~E B~ (~x, t)= ~e · − ) perpendicular to the wavenumber k, and if we set up B Electromagneti Electromagneti Waves orthonormal basis vectors ~1 and ~2 for that plane, with ~1 Waves ˆ but with constraints on ~ and ~ coming from the rest of Reflection and ~2 for that plane, with ~1 ~2 = k,wehave Reflection E B and ~ ~ × and Maxwell’s equations. Refraction = E1~1 + E2~2, and then = √µ(E1~2 E2~1). Refraction ~ ~ ~ ~ E B − The divergence equations require k = 0 and k =0 The amplitudes E and E may be complex. If only one of ~ ~· E ~ · B 1 2 while one curl equation gives ik = iω or i~k ~x iωt × E B them is nonzero, say E1, E~ (~x, t)=Re E1e ~1 = √µkˆ ~ = ~. The magnetic field H~ = B/µ~ so · − × E B ~ ~ = kˆ ~/Z where Z = µ/ is an impediance. E1 cos(k ~x ωt + arg E1)~1 so the argument of E1 is H × E | | · − The impediance of free space is µ / = 376.7Ω. just a phase shift, pretty much irrelevant. In this case, the p 0 0 We have not specified that k and ω are real, which one or electric field is linearly polarized, oscillating but always in p the other might not be, as the permittivity and the direction ~1. The same is true if E1 and E2 are not ~ permeability are in general complex. Still, in many zero but have the same phase, with the resulting E ~ oscillating in the direction E1 ~1 + E2 ~2. contexts they are close to real and if we take k to be real, | | | | and will be in phase, with vB~ and E~ equal in E B magnitude.
iφ Physics 504, Physics 504, But if E2/E1 = Ae is not real, the electric field Spring 2011 Reflection and Refraction Spring 2011 Electricity Electricity components in the two directions are out of phase, and at and Consider a planar interface between two uniform linear and Magnetism Magnetism a given ~x the field sweeps out an ellipse in time. media. In each medium, z If A = 1 and φ = π/2, this is a circle, we have the Shapiro the fields must be a com- r Shapiro | | ~ k’ ~ ik ~x iωt bination of plane waves. n complex E(~x, t)=E1(1 + i2)e · − . The real field Electromagneti µ’ , ε’ Electromagneti cos~k ~x sin~k ~x then spirals clockwise as ~x moves Waves Suppose a wave Waves ∝ 1 · − 2 · µ , ε x ~ Reflection ~ Reflection along k, if we are looking into the wave. It spirals ik ~x iωt r’ and E~ = E~0 e · − , k i k’’ and counterclockwise as time progresses. This is called a left Refraction Refraction B~ = √µ kˆ E~ circularly polarized wave, or a wave of positive helicity.Of × course the opposite phase, with E1/E2 = i, is a right (kˆ is a unit vector) is incident from below, inducing a − circularly polarized wave of negative helicity. refracted wave in the upper medium: 1 Define ~ = (~1 i~2). Then the electric field can be ~ √2 ik 0 ~x iωt ± ± E~ 0 = E~ 0 e · − , B~ 0 = µ00 kˆ0 E~ 0 decomposed into ~ components rather than ~j (j =1, 2) × ± components, and each of these is complex. In either case and a reflected wave in the lower mediump ~ there are four real parameters giving the amplitude of the ik 00 ~x iωt E~ 00 = E~ 000 e · − , B~ 00 = √µ kˆ00 E~ 00. wave. I think we will not need to discuss the Stokes × parameters that give these in terms of measurable As all the equations are linear with time-independent quantities. parameters, only the one fourier component is involved. Physics 504, Physics 504, Kinematics Spring 2011 The Boundary Conditions Spring 2011 Electricity Electricity That is, all the waves have the same frequency. This can and Assume we are considering non-conducting materials with and Magnetism Magnetism also be viewed as saying the boundary values must no free charges, so within each material the fields must be Shapiro Shapiro oscillate together in time. differentiable, and at the interface, Gauss’ law tells us the normal components of B~ and D~ are continuous, while the The x and y dependence of the fields at the boundary will Electromagneti Electromagneti also need to match whatever the boundary conditions. Waves other two, integrated on a path just below and just above Waves Reflection the interface, tells us the components of E~ and vecH Reflection This tells us kx = kx0 = kx00 and ky = ky0 = ky00. The and and magnitudes of the three k’s are determined by ω and the Refraction perpendicular to ~n are continuous. Of course below the Refraction material parameters, k = ~k = ~k 00 = ω√µ, interface we need to add the incident and reflected fields, | | | | k = ~k = ω√µ . The x, y matching means so we have 0 | 0| 0 0 k sin i = k sin r = k sin r ,sok = k implies i = r ,or D~ ~n continuous: (E~0 + E~ 00) 0E~ 0 ~n =0 0 00 0 00 0 · 0 − 0 · the angle of reflection is equal to the angle of h i ωB~ ~n continuous: ~k E~ + ~k 00 E~ 00 ~k 0 E~ 0 ~n =0 incidence. · × 0 × 0 − × 0 · sin i k0 µ00 n0 ~ ~ ~ ~ But we also have = = = , where n and E ~n continuous: E0 + E 000 E 0 ~n =0 sin r k s µ n × − × H~ ~n continuous: n0 are the indices of refraction above and below the × interface. This is Snell’s Law. 1 1 ~k E~0 + ~k 00 E~ 000 ~k 0 E~ 0 ~n =0 µ × × − µ0 × ×
Physics 504, ~ Physics 504, Plane of Incidence Spring 2011 E perpendicular (continued) Spring 2011 Electricity Electricity The transverse conditions on the k’s means that they all and Thus E0 , E0 , B0 , E00, E00 and B00 must all vanish, and the and Magnetism x z y x z y Magnetism lie in the plane of incidence, defined by the incident reflected and transmitted waves are linearly polarized as direction and the interface normal ~n (assuming i = 0). Shapiro ~ Shapiro 6 shown, with all E’s the plane of incidence. The solution of the interface equations in general is ⊥ ~ Electromagneti Thus E0 + E000 E0 = 0 from the continuity of E ~n, and Electromagneti messy, but we can consider separately the two linear Waves − × Waves /µ cos i (E0 E000)= 0/µ0 cos rE0 from the polarizations of the incident electric field, in that plane Reflection − Reflection and continuity of H~ ~n. The two equations enable solving for and and perpendicular to that plane. Take that plane to Refraction p × p Refraction the ratios E0 /E0 and E000/E0 whose squares give the be the xz plane. transmission and reflection coefficients. z First consider the incident B’ ~ ~ r k’ E0 2nµ0 cos i E000 nµ0 cos i n0µ cos r into the plane, so is n − E B E’ = , = in the direction shown, and µ’ , ε’ E0 nµ0 cos i + n0µ cos r E0 nµ0 cos i + n0µ cos r µ , ε x E0 x = B0 y = E0 z =0. We see they depend on the ratios n /n and µ /µ of the B B’’ 0 0 Consider a reflection in the i i indices of refraction and the permeabilities, but for plane of incidence, where k E E’’ k’’ optical frequencies we may usually take µ0/µ = 1. The all E , B and B should change sign, but not the other y x z expression is still somewhat complicated. Jackson ~ ~ components, because E is a vector and B a pseudovector. eliminates the r dependence using But this simply reverses the incident fields, and therefore 2 2 2 2 2 2 n0 cos r = n n sin r = n n sin i. all components must change sign. 0 − 0 0 − p p
~ Physics 504, Physics 504, E in the plane of incidence Spring 2011 Normal Incidence Spring 2011 Electricity Electricity The argument for the op- z and and E’ Magnetism Magnetism posite linear polarization, r k’ with E~ in the plane of in- n B’ Shapiro Shapiro µ’ , ε’ cidence and B~ perpendicu- If the angle of incidence goes to zero, the two results must µ , ε x Electromagneti Electromagneti lar to it proceeds similarly. E Waves converge, as the plane of incidence is not well defined. As Waves E’’ The E~ ~n equation gives i i Reflection i = r = 0 in that limit, the E~ results give Reflection × k B B’’ k’’ and ⊥ and (E E00) cos i = E0 cos r Refraction Refraction − E0 2 E000 1 and the H~ ~n equation gives us = , =1 2 , E0 n0 µ E0 − n µ0 × 1+ 1+ 1 1 k0 1 n0 n µ0 n0 µ (E0 + E000)= E0 = E0 µ µ0 k µ0 n and so does the E~ in the plane, except that the sign of Again the results are not simple. We get E000/E0 is reversed, but that is because the directions defining E000 are reversed in the two cases. E0 2nµ0 cos i 0 = E0 n0µ cos i + nµ0 cos r E00 n0µ cos i nµ0 cos r 0 = − E0 n0µ cos i + nµ0 cos r Physics 504, Physics 504, Simplification if µ = µ0 Spring 2011 Normal incidence, again, with µ = µ0 Spring 2011 Electricity Electricity and and We are often interested in visible light in dielectric Magnetism Magnetism materials which have little magnetic susceptibility, so let Shapiro Shapiro us assume µ = µ0. Then for E~ , For normal incidence, we need to not use the sines, which ⊥ Electromagneti Electromagneti Waves vanish. Using the E~ definition where positive values are Waves ⊥ E0 2n cos i 2 cos i sin r Reflection all in the same direction, we see Reflection = = and and E0 n cos i + n0 cos r sin(i + r) Refraction Refraction E n cos i n cos r sin(r i) E0 2n E000 n n0 000 − 0 − = , = − . = = . E0 n + n0 E0 n + n0 E0 n cos i + n0 cos r sin(r + i) Note that if the reflection is off a more dense medium and for E~ , ~ k (n0 >n), the sign of E is reversed on reflection, which is the origin of the rule we use in elementary discussions of E0 2n cos i 4 sin r cos i 0 = = interference, that there is an extra phase shift by π under E n cos i + n cos r sin(2i) + sin(2r) 0 0 reflection off a more-dense medium. E00 sin(2i) sin(2r) tan(i r) 0 = − = − . E0 sin(2i) + sin(2r) tan(i + r)
Physics 504, Physics 504, Brewster angle Spring 2011 Total Internal Reflection Spring 2011 Electricity Electricity and and For the electric field in the plane of incidence, there is an Magnetism Magnetism Shapiro Earlier we derived Snell’s law, n sin i = n0 sin r, which Shapiro angle for which E000 vanishes, called Brewster’s angle. shows that, if n>n0, for an angle of incidence i greater Assuming µ0 = µ for simplicity (and it is nearly true for 1 Electromagneti than sin− (n0/n) we need a refraction angle with a sine Electromagneti optics in dielectric materials), this is when Waves Waves Reflection greater than 1. How do we interpret that? Really we got Reflection n0 cos i = n cos r, and as n sin i = n0 sin r, multiplying and and Refraction this by requiring kx = kx0 with k0 = n0k/n, from which Refraction them, we see sin 2i = sin 2r,orr = π/2 i, and the 2 2 − k0 = k k which is imaginary (say iκ under these reflected and refracted waves are perpendicular to each z 0 − x circumstances. Then the fields other. Then n0 = n sin i/ sin r = n tan i,so p
ikxx κz iωt 1 E~ 0 = E~ 0 e − − , H~ 0 = ~k0 E~ 0. 1 n0 0 × Brewster’s angle: iB = tan− . µ0ω n Note the fields fall off exponentially with depth z into the This is a way to get complete linear polarization of light. less dense medium for angles beyond that of total internal It also indicates that even at other angles, the reflection, so there is no continuing beam of refracted transmission of one polarization will be greater than the light, though there is field close to the interface. other.
There is no flow of energy into the n0 medium, because Physics 504, Spring 2011 the Poynting vector’s z component is Electricity and 1 Magnetism S~ ~n = Re ~n E~ 0 H~ 0∗ · 2 · × Shapiro 1 h i = Re ~n E~ 0 ~k 0 E~ 0∗ Electromagneti 2ωµ0 · × × Waves 1 h 2 i Reflection ~ 0 ~ 0 and = Re ~n k E 0 =0, Refraction 2ωµ0 ·
as ~n ~k is pure imaginary. · 0 The naive geometrical picture of well defined beams that are reflected exactly at the boundary should be doubted. Perhaps it is effectively off a plane somewhat inside the boundary, and the reflected ray would intersect the incident ray inside the second medium. This is called the Goos-H¨anchen effect. To really describe how much the reflected beam is moved from the naive path, we would need wave packets in x rather than pure wavenumber. We are not going to pursue this.