Chapter 8ElectromagneticwavesChapter 8IntroductionThe basic laws of electromagnetism were established in thenineteenth century. Initially, the electric and magneticinteraction were considered to be independent, each of themgoverned by its own Gauss-like law. Ampere's law providedthe first hint of a linkage, since it acknowledged the factthat moving charges produce magnetic fields. The linkbetween electric and magnetic fields was further clarified byFaraday's induction law. Finally, James Clerck Maxwell,completed the set of laws that fully characterize the electro-magnetic interaction. Maxwell also went a step further: bycombining the electromagnetic equations, he was able toshow that electric and magnetic fields in free space satisfy awave equation in which the wave speed happens to beexactly equal to the measured speed of light. The obviousconclusion was that light is nothing but an electromagneticwave. In this chapter, we will explore this fascinatingdiscovery.MaxwellÕs equationsMaxwell's equations in integral formWhen Maxwell started his work on electromagnetism, theknown field equations were GaussÕ laws for the electric andmagnetic fields, Amp•reÕs law, and FaradayÕs induction law.You probably know these equations in integral form. GaussÕlaw for the electric field states that E ⋅=∫dSqSε0 (1)where E is the electric field, q the electric charge enclosed bythe surface S and ε0 = 8.854 × 10-12 N-1 m-2 C2 is thevacuum permittivity. Since there are no magnetic charges,GaussÕs law for the magnetic field is B ⋅=∫dSS0 (2)Amp•reÕs law relates any current I with the magnetic field Bit creates: B ⋅=∫dlILµ0 , (3)phy 241 MenŽndez 138Chapter 8where the magnetic field is integrated along any closed paththat encircles the current and µ0 is the magnetic permeabil-ity of vacuum µ0 = 4π × 10-7 m kg C-2. Finally, FaradayÕslaw describes the electric field produced by a changingmagnetic flux: E ⋅=− ⋅∫∫ddddlStBLS (4)where the surface integral is over the any surface S enclosedby the line L along which the line integral is performed.Maxwell noted an inconsistency in the accepted form ofAmp•reÕs law. Suppose you apply this law to the circuitillustrated in Fig. 111111111110000000000 L u N dSFigure 1 The line integral of B along L isproportional to the currentacross any surface S boundedby L. When L shrinks to zero,the surface becomes closed.If the line L shrinks to a point, the line integral of Bbecomes zero. At this point the surface is closed, so thatAmp•reÕs law implies that the net current leaving a closedsurface is always zero. This is perfectly reasonable if weimagine a wire entering and leaving the surface, for thecurrent that enters the volume enclosed by the surface isphy 241 MenŽndez 139Chapter 8exactly equal to the current that leaves the volume. Thus thenet current is zero. Imagine, however, that the above surfaceencloses one of the plates of a capacitor. If the net currentflowing across the surface is zero, this means that the netcharge enclosed by the surface is constant. In other words,the charge on a capacitor plate should remain constant. This,of course, is not true. Maxwell figured out an ad hoc way offixing this problem. He reasoned that for the case of a closedsurface the inconsistency is removed if the right-hand side ofEq. (3) is replaced by µ0 (I + dq/dt), where q is the chargeenclosed by the volume. When the line shrinks to a point,we get I + dq/dt = 0, or I = - dq/dt. This is obviously true,for the net current leaving a volume is by definition the rate ofchange of the charge inside the volume. Using GaussÕ law toexpress the charge in terms of a surface integral of theelectric field, Maxwell proposed that Amp•reÕs law bemodified to B ⋅= + ⋅∫∫dddEdlSµε00ItSL (5)For the case of a closed surface, MaxwellÕs modification ofAmp•reÕs law must be correct. But he also assumed that thenew expression would be valid even in cases when the surfaceis not closed. Of course, this would have to be verified byexperiment. For cases where the electric field does notdepend on time, such as in DC circuits, the new term addedby Maxwell is zero and we recover the standard Amp•reÕslaw. Eqs. (1), (2), (4), and (140) are known, together, asMaxwell equations. They characterize the electromagneticfields completely. They also lead to the wave equation for theelectromagnetic field. However, the wave equation is adifferential equation. To show that it follows from MaxwellÕsequations, we must first rewrite MaxwellÕs equations indifferential form.MaxwellÕs equations in differential formLet us first consider GaussÕ law for the electric field. Supposethat we apply Eq. (1) to the infinitesimal volume element inphy 241 MenŽndez 140Chapter 8Fig. (2).111111000000111111111111000000000000111111000000 E (x + dx) E (x) E (y + dy) E (y) x y x yFigure 2 Infinitesimal volume on which we apply GaussÕ law forthe electric fieldThe surface integral over the six faces of the cube becomes SxxyyzzEx x Ex yzEy y Ey xzEz z Ez xyq∫=+−[]++−[]++−[]()()()()()()ddddddddd=ε0 (where q is the charge inside the cube. This charge can bewritten q = ρ dV = ρ dx dy dz. On the other hand, theterms inside the square brackets can be written in terms ofthe derivatives of the field, so that we obtain ∂∂∂∂∂∂ρεExEyEzxyz++=0 (This is GaussÕ law in differential form. We can now define aspecial vector that simplifies the notation. Our ÒvectorÒ,denoted as ∇∇∇∇, is defined, in terms of its components, asphy 241 MenŽndez 141Chapter 8 ∇=∂∂∂∂∂∂xyz,, (8)This vector follows the usual rules of vector multiplication,except that multiplication means in this case differentiation.For example, for two vectors A and B it is known that thedot product can be written as A¥B = AxBx+AyBy+AzBz.Similarly, the dot product ∇∇∇∇¥E is given by ∇⋅ =⋅()=++E∂∂∂∂∂∂∂∂∂∂∂∂xyzEEEExEyEzxyzxyz,, ,,(9)Hence we can write GaussÕ law for the electric field as ∇⋅ =Eρε0 (10)Similarly, we can write GaussÕs law for the magnetic field as ∇⋅ =B 0 (11)FaradayÕs law can also be written in differential form byconsidering three infinitesimal squares on the planes XY, YZ,and ZX. Computing the line integral of the electric fieldalong the sides of the squares and equating it to the flux ofthe electric field, one can show (see homework