Electromagnetic Waves

Transcription

1 May 7, J.D.Jackson, Classical Electrodynamics, 2nd Edition, Section 7

2 Maxwell Equations In a region of space where there are no free sources (ρ = 0, J = 0), Maxwell s equations reduce to a simple form given E = 0, E 1 B + c t = 0 B = 0, B µɛ c where D and H are given by relations E t = 0 (1) D = ɛ E and H = 1 µ B (2) where ɛ is the electric permittivity and µ the magnetic permeability which assumed to be independent of the frequency.

3 Plane Maxwell s equations can be written as 2 B µɛ c 2 2 B t 2 = 0 and 2 E µɛ c 2 2 E t 2 = 0 (3) In other words each component of B and E obeys a wave equation of the form 2 u 1 2 u v 2 t 2 = 0 where v = c (4) µɛ is a constant with dimensions of velocity characteristic of the medium. The wave equation admits admits plane-wave solutions: u = e i k x iωt E( x, t) = Ee ik n x iωt and B( x, t) = Be ik n x iωt (5) (6) where the relation between the frequency ω and the wave vector k is k = ω v = µɛ ω c or k k = ( ω v ) 2 (7) also the vectors n, E and B are constant in time and space.

4 If we consider waves propagating in one direction, say x-direction then the fundamental solution is u(x, t) = Ae ik(x vt) + Be ik(x+vt) (8) which represents waves traveling to the right and to the left with propagation velocities v which is called phase velocity of the wave. From the divergence relations of (1) by applying (6) we get n E = 0 and n B = 0 (9) This means that E (or E) and B (or B) are both perpendicular to the direction of propagation n. Such a wave is called transverse wave. The curl equations provide a further restriction B = µɛ n E and E = 1 µɛ n B (10) The combination of equations (9) and (10) suggests that the vectors n, E and B form an orthonormal set. Also, if n is real, then (10) implies that that E and B have the same phase.

5 It is then useful to introduce a set of real mutually orthogonal unit vectors ( ɛ 1, ɛ 2, n). In terms of these unit vectors the field strengths E and B are or E = ɛ 1 E 0, B = ɛ2 µɛe0 (11) E = ɛ 2 E 0, B = ɛ1 µɛe 0 (12) E 0 and E 0 are constants, possibly complex. Thus the wave described by (6) and (11) or (12) is a transverse wave propagating in the direction n. Or that E and B are oscillating in a plane perpendicular to the wave vector k, determining the direction of propagation of the wave.

6 The energy flux of EM waves is described by the real part of the complex Poynting vector 1 c S = E 2 4π H = 1 c [ ER H 2 4π R + E I H I + i( EI H R E R H )] I the time averaged flux of energy is: c ɛ S = 8π µ E 0 2 n (13) The total time averaged density is: u = 1 (ɛ E 16π E + 1µ ) B B = ɛ 8π E 0 2 (14) The ratio of the magnitude of (13) to (14) is the speed of energy flow i.e. v = c/ µɛ. 2 3 (Prove the above relations) 2 Note : we use the magnetic induction H because although B is the applied induction, the actual field that carries the energy and momentum in media is H 3 Note: To prove the above relations use cos 2 x = 1/2 and since E R = ( E + E )/2 we get E 2 R = E E /2.

7 Linear and Circular Polarization of EM Waves The plane wave (6) and (11) is a wave with its electric field vector always in the direction ɛ 1. Such a wave is said to be linearly polarized with polarization vector ɛ 1. The wave described by (12) is linearly polarized with polarization vector ɛ 2 and is linearly independent of the first. The two waves : E 1 = ɛ 1 E 1 e i k x iωt, E2 = ɛ 2 E 2 e i k x iωt with (15) B i = µɛ k E i, i = 1, 2 k Can be combined to give the most general homogeneous plane waves propagating in the direction k = k n, E( x, t) = ( ɛ 1 E 1 + ɛ 2 E 2 ) e i k x iωt (16) [ E( x, t) = ɛ 1 E 1 + ɛ 2 E 2 e i(φ2 φ1)] e i k x iωt+iφ 1 (17) The amplitudes E 1 = E 1 e iφ1 and E 2 = E 2 e iφ2 are complex numbers in order to allow the possibility of a phase difference between waves of different polarization.

8 LINEARLY POLARIZED If the amplitudes E 1 = E 1 e iφ1 and E 2 = E 2 e iφ2 have the same phase (16) represents a linearly polarized wave with the polarization vector making an angle θ = tan 1 (R(E 2 )/I(E 1 )) (which remains constant as the field evolves in space and time) with ɛ 1 and magnitude E = E E 2 2. ELLIPTICALLY POLARIZED If E 1 and E 2 have the different phase the wave (16) is elliptically polarized and the electric vector rotates around k.

9 Circular Polarization E 1 = E 2 = E 0 φ 1 φ 2 = ±π/2 and the wave becomes E( x, t) = E 0 ( ɛ 1 ± i ɛ 2 ) e i k x iωt (18) At a fixed point in space, the fields are such that the electric vector is constant in magnitude, but sweeps around in a circle at a frequency ω. The components of the electric field, obtained by taking the real part of (18) E x ( x, t) = E 0 cos(kz ωt), E y ( x, t) = E 0 cos(kz ωt) (19) For the upper sign ( ɛ 1 + i ɛ 2 ) the rotation is counter-clockwise when the observer is facing into the oncoming wave. The wave is called left circularly polarized in optics while in modern physics such a wave is said to have positive helicity. For the lower sign ( ɛ 1 i ɛ 2 ) the wave is right circularly polarized or it has negative helicity.

10 Elliptically Polarized EM Waves An alternative general expression for E can be given in terms of the complex orthogonal vectors with properties ɛ ± = 1 2 ( ɛ 1 ± i ɛ 2 ) (20) ɛ ± ɛ = 0, ɛ ± ɛ 3 = 0, ɛ ± ɛ ± = 1. (21) Then the general representation of the electric vector E( x, t) = (E + ɛ + + E ɛ ) e i k x iωt (22) where E and E + are complex amplitudes If E and E + have different amplitudes but the same phase eqn (22) represents an elliptically polarized wave with principle axes of the ellipse in the directions of ɛ 1 and ɛ 2. The ratio of the semimajor to semiminor axis is (1 + r)/(1 r), where E /E + = r.

11 The ratio of the semimajor to semiminor axis is (1 + r)/(1 r), where E /E + = r. If the amplitudes have a phase difference between them E /E + = re iα, then the ellipse traced out by the E vector has its axes rotated by an angle α/2. Figure: The figure shows the general case of elliptical polarization and the ellipses traced out by both E and B at a given point in space. Note : For r = ±1 we get back to a linearly polarized wave.

12 Polarization Figure: The figure shows the linear, circular and elliptical polarization

13 Stokes Parameters The polarization content of an EM wave is known if it can be written in the form of either (16) or (22) with known coefficients (E 1, E 2 ) or (E, E + ). In practice, the converse problem arises i.e. given a wave of the form (6), how can we determine from observations on the beam the state of polarization? A useful tool for this are the four Stokes parameters. These are quadratic in the field strength and can be determined through intensity measurements only. Their measurements determines completely the state of polarization of the wave. For a wave propagating in the z-direction the scalar products ɛ 1 E, ɛ 2 E, ɛ + E, ɛ E (23) are the amplitudes of radiation respectively, with linear polarization in the x-direction, linear polarization in the y-direction, positive helicity and negative helicity. The squares of these amplitudes give a measure of the intensity of each type of polarization. The phase information can be taken by using cross products

14 Stokes Parameters In terms of the linear polarization bases ( ɛ 1, ɛ 2 ), the Stokes parameters are: s 0 = ɛ 1 E 2 + ɛ 2 E 2 = a a 2 2 s 1 = ɛ 1 E 2 ɛ 2 E 2 = a1 2 a2 2 [ s 2 = 2R ( ɛ 1 E) ( ɛ 1 E) ] = 2a 1 a 2 cos(δ 1 δ 2 ) (24) [ s 3 = 2I ( ɛ 1 E) ( ɛ 1 E) ] = 2a 1 a 2 sin(δ 1 δ 2 ) where we defined the coefficients of (16) or (22) as magnitude times a phase factor: E 1 = a 1 e iδ1, E 2 = a 2 e iδ2, E + = a + e iδ+, E = a e iδ (25) Here s 0 and s 1 contain information regarding the amplitudes of linear polarization, whereas s 2 and s 3 say something about the phases. Knowing these parameters (e.g by passing a wave through perpendicular polarization filters) is sufficient for us to determine the amplitudes and relative phases of the field components.

15 Stokes Parameters In terms of the linear polarization bases ( ɛ +, ɛ ), the Stokes parameters are: s 0 = ɛ + E 2 + ɛ E 2 = a+ 2 + a 2 [ s 1 = 2R ( ɛ + E) ( ɛ E) ] = 2a + a cos(δ δ + ) (26) [ s 2 = 2I ( ɛ + E) ( ɛ E) ] = 2a + a sin(δ δ + ) s 3 = ɛ + E 2 ɛ E 2 = a 2 + a 2 Notice an interesting rearrangement of roles of the Stokes parameters with respect to the two bases. The four Stokes parameters are not independent since they depend on only 3 quantities a 1, a 2 and δ 1 δ 2. They satisfy the relation s 2 0 = s s s 2 3. (27)