Easy derivation of Maxwell’s and Wave Equation. This starts from observations due to Faraday and Ampere and a suppostion of Maxwell. Together with a vector identity due to Stokes d~ℓ V~ = d~a V~ , IC · ZS · ³∇ × ´ we will derive wave equation. Faraday summarizes his observations of electric field (emf) being induced by time-variation of magnetic flux. The geometry is a loop C bounding an area A. The electric field in the loop is proportional to the time derivative of the magnetic flux in A. d d~ℓ E~ = d~a B~ IC · −dt ZA · ∂ d~a E~ = d~a B~ ZA · ³∇ × ´ − ZA · ∂t ∂ E~ = B.~ (1) ∇ × −∂t Ampere connected the current J~ thru a loop area with the intergral of magnetic field B~ around the loop: d~ℓ B~ = µ d~a J.~ I · ZA · Maxwell supposed that for time-varying situtation there would be a dis- placement current: J~D = ǫ(∂/∂t)E.~ In free space the only current would the displacement curret. Using Stokes again, we get the remaining needed equation ∂ B~ = ǫµ E.~ (2) ∇ × ∂t Wave equations. Using vector identity ( V~ )= ( V~ ) 2V~ ∇ × ∇ × ∇ ∇ · − ∇ on (2) and lack of a magnetic “charge” B~ = 0, we get the wave equation ∇ · on B~ : ∂ ( B~ )=ǫµ ( E~ ) ∇ × ∇ × ∂t ∇ × 2 2 ∂ B~ = ǫµ B~ (3a) −∇ − ∂t2 Similarly, using E~ = 0 in free space (no charges), we can get wave ∇ · equation for electric field E~ : 2 2 ∂ E~ = ǫµ E.~ (3b) ∇ ∂t2 Making a plane wave. For simplicity we consider a wave propagating in the z-direction. So E,~ B~ eikz−iωt. { }∝ To begin, suppose electric field only has component in x-direction. Then ∂ B~ = xEˆ x −∂t ∇ × xˆ yˆ zˆ ~ ∂ iωBamp = 0 0 ∂z E eikz 0 0 x =ikExy.ˆ With B~ in y-direction, we see E,B,k form a right-hand rule connecting E, B and propagation direction (z) The only thing loose is the relative magnitudes of the fields. From the wave equation we see that k2 = ǫµω2. The velcity of wave is give by ω/k. So plane-wave velocity is 1/√ǫµ. Simpler units. We can discard carrying around MKS units and factors such ǫµ by referring all quantities to their values in free space, where velocity of light is c and we define the index of refraction as ratio of √ǫµ to its value in free space. Thus nfree = 1, and the relative index of refraction for materials is dimensionless and small: water- 4/3; good glass- 3/2; diamond- 2.4; GaP- 3.5 (c.f., page 94 Hecht). Going to equation above and replacing all values by those relative to free-space values, B/E = (k/ω) = √ǫµrelative = n. Henceforth the ampitude of Bamp = nEamp. This will be very useful is reflection, refraction, lens and most optical phenomena. Boundary conditions. At an interface between two materials (e.g., air ~ ~ d ~ and glass) we apply integral Maxwll relatons. E.g., C dℓ E = dt A d~a B. Let area A be a narrow rectangle with long sides parllelH · the− interface.R · As the width of rectangle squeeze the interface, the right-hand-side integral goes to zero. Thus the left-hand side is can be written d~ℓ E~one−side + d~ℓ E~other−side = 0. Zpath to right · Zpath to left · As the paths get shorter, says tangential compenents of E~ are equal. The other maxwell equation becomes d~ℓ B~ = ǫµ d d~a ∂/∂tE~ . As the C · − dt A · area shrinks, the tangential componetsH of B~ are continuous.R Summary of handling fields at interface of two materials. 1. Draw interface. 2. On each side draw components of each field so set obeys right hand rule, with amplitude of each B = n (matching) E. 3. Apply boundary conditions using angle propagation make with per- pendicular to surface to specify tangential components components..