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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 Z I ³ ´ ~ ~ ~ d~a · ∇ × V , dℓ · V = C S 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. I Z d ~ =− ~ d~ℓ · E d~a · B dt C Z A Z ³ ´ ∂ ~ ~ =− d~a · B d~a · ∇ × E ∂t A A ~ ~ = − ∂ B. (1) ∇×E ∂t Ampere connected the current J~ thru a loop area with the intergral of ~ around the loop: magnetic field B I Z ~ ~ =µ d~a · J. d~ℓ · B A Maxwell supposed that for time-varying situtation there would be a displacement 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 ~ = ǫµ ∂ E. ~ ∇×B ∂t (2) ~ ) = ∇(∇ · V ~ ) − ∇2 V ~ Wave equations. Using vector identity ∇ × (∇ × V ~ = 0, we get the wave equation on (2) and lack of a magnetic “charge” ∇ · B ~ on B: ~ ~ =ǫµ ∂ (∇ × E) ∇ × (∇ × B) ∂t ∂2 ~ 2~ −∇ B = − ǫµ 2 B ∂t (3a) ~ = 0 in free space (no charges), we can get wave Similarly, using ∇ · E ~ equation for electric field E: ∂2 ~ ∇ E = ǫµ 2 E. ∂t 2~ (3b) Making a plane wave. For simplicity we consider a wave propagating in ~ B} ~ ∝ eikz−iωt . the z-direction. So {E, To begin, suppose electric field only has component in x-direction. Then ∂ ~ B =∇ × x̂Ex ∂t x̂ ~ amp = 0 iω B Ex eikz − ŷ 0 0 ẑ ∂ ∂z 0 =ikEx ŷ. ~ in y-direction, we see E, B, k form a right-hand rule connecting With B E, B and propagation direction (z) The only thing loose is the relative magnitudes of the fields. From the wave equation we see that k 2 = ǫµω 2 . √ 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 nf ree = 1, and the relative index of refraction for materials is dimensionless and small: water- 4/3; good glass- 3/2; diamond- 2.4; GaP3.5 (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 H R ~ =−d ~ and glass) we apply integral Maxwll relatons. E.g., C d~ℓ· E a · B. dt A d~ Let area A be a narrow rectangle with long sides parllel the interface. 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 Z Z ~ other−side = 0. ~ one−side + d~ℓ · E d~ℓ · E path to left path to right ~ are equal. The As the paths get shorter, says tangential compenentsR of E H d ~ As the ~ = −ǫµ a · ∂/∂tE. other maxwell equation becomes C d~ℓ · B dt A d~ ~ are continuous. area shrinks, the tangential componets of B 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 perpendicular to surface to specify tangential components components.