Plane electromagnetic waves and Gaussian beams (Lecture 17) February 2, 2016 305/441
Lecture outline In this lecture we will study electromagnetic field propagating in space free of charges and currents. We focus on two types of solutions of Maxwell s equations: plane electromagnetic waves and Gaussian beams. 306/441
Plane electromagnetic waves A plane electromagnetic wave propagates in free space (without charges and currents). All components of the field depend only on the variable ξ = z ct, E(r, t) = F (ξ), B(r, t) = G(ξ). (17.1) From the equation E = 0 if follows that F z / ξ = 0, and hence F z = 0. Similarly, G z = 0 because of B = 0. We see that a plane wave is transverse. We now apply Maxwell s equation B/ t = E to the fields (17.1). We have hence F x = cg y and F y = cg x. F x = cg y, F y = cg x, (17.2) 307/441
Plane electromagnetic waves In vector notation F = cn G or E = cn B, (17.3) where n is a unit vector in the direction of propagation (in our case along the z axis). Multiplying vectorially Eq. (17.3) by n, we also obtain B = 1 c n E. (17.4) If we use potentials φ and A to describe a plane wave, they would also depend on ξ only: φ = φ(ξ), A = A(ξ). 308/441
Plane electromagnetic waves We have B = A = ^xa y + ^ya x = n A = 1 c n A t. (17.5) After the magnetic field is found, we can find the electric field using Eq. (17.3). 309/441
Plane electromagnetic waves Often a plane wave has a sinusoidal time dependence with some frequency ω. In this case it is convenient to use the complex notation: E = Re (E 0 e iωt+i k r+iφ 0 ), B = Re (B 0 e iωt+i k r+iφ 0 ), where E 0 and B 0 are the amplitudes of the wave, and k = nω/c is the wave number. The wave propagates in the direction of k; the amplitude of the electric and magnetic fields are E 0 = cb 0. In general, E 0 and B 0 can be complex vectors orthogonal to k, e.g., E 0 = E (r) 0 + ie (i) 0 with E (r) 0 and E (i) 0 real. Purely real of purely imaginary E 0 corresponds to a linear polarization of the wave; a complex vector E 0 describes an elliptical polarization. 310/441
Plane electromagnetic waves The Poynting vector gives the energy flow in the wave ɛ0 S = E H = E0 2 µ n cos2 (ωt + kr + φ 0 ), (17.6) 0 (in this formula E 0 is assumed real). The energy flows in the direction of the propagation k. Averaged over time energy flow, S, is S = 1 2 ɛ0 µ 0 E 2 0 n = 1 2Z 0 E 2 0 n = c2 2Z 0 B 2 0 n. (17.7) Energy density in the plane wave We have ū = 1 ɛ 0 2 2 (E 0 2 + c 2 B0 2 ) = ɛ 0 2 E 0 2 S = cu. 311/441
Plane electromagnetic waves source of radiation n local plane wave here Electromagnetic field looks like a plane wave locally in some limited region. local plane wave here n An arbitrary solution of Maxwell s equations in free space (without charges) can be represented as a superposition of plane waves with various amplitudes and directions of propagation. 312/441
Gaussian beams We consider another important example of electromagnetic field in vacuum Gaussian beams. They are typically used for description of laser beams. They can be understood as a collection of plane waves with the same frequency propagating at small angles to a given direction a so called paraxial approximation. 313/441
Gaussian beams Start from the wave equation (13.4) for the x component of the electric field (a linear polarization of the laser light) 2 E x x 2 + 2 E x y 2 + 2 E x z 2 1 c 2 2 E x t 2 = 0, (17.8) and assume E x (x, y, z, t) = u(x, y, z)e iωt+ikz, (17.9) where u is a slow function of its arguments, and ω = ck. More specifically, we require 1 u u z k, 1 u u t ω. (17.10) Physical fields are the real part of E x. 314/441
Gaussian beams Main steps in the derivation of Gaussian beams: Neglect 2 u/ z 2 in comparison with k u/ z. Assume axisymmetry u = u(ρ, z) with ρ = x 2 + y 2. Seek solution in the form The result is u = A(z)e Q(z)ρ2 1/w0 2 Q(z) = 1 + 2iz/kw0 2, (17.11) where w 0 is one constant of integration (called the waist) and E 0 A(z) = 1 + 2iz/kw0 2, (17.12) where E 0 is another constant of integration. 315/441
Gaussian beams We now introduce important geometrical parameters: the Rayleigh length Z R and the angle θ: Z R = kw 2 0 2, θ = w 0 Z R = 2 kw 0. (17.13) They can also be written as where λ = k 1 = c/ω. Z R = 2 λ θ 2, w 0 = 2 λ θ, (17.14) 316/441
Gaussian beams At z = 0 the radial dependence of u is e ρ2 /w 2 0 w0 gives the transverse size of the focal spot here. At ρ = 0 u = E 0 /(1 + iz/z R ) Z R is the characteristic length of the focal region along the z axis. 2 Ρ/w 0 1 0-1 Contour lines of constant amplitude u. -2-3 -2-1 0 1 2 3 z/z R 2 Ρ/w 0 1 0-1 Contour lines of constant phase φ, with u = u e iφ. -2-3 -2-1 0 1 2 3 z/z R 317/441
Gaussian beams Condition for the validity of the paraxial approximation: 2 u/ z 2 u/z 2 R k u/ z ku/z R Z R λ. From this condition it follows that λ w 0 Z R and θ 1. This means that the size of the focal spot w 0 is much larger than the reduced wavelength, and θ 1. The magnetic field in a Gaussian beam the lowest approximation can be found, in the lowest order, by using Eq. (17.4), where n is directed along z, B y = 1 c E x. (17.15) 318/441