Physics 4D ELECTROMAGNETIC WAVE Hans P. Paar 26 January 2006 i
Chapter 1 Vector Calculus 1.1 Introduction Vector calculus is a branch of mathematics that allows differentiation and integration of (scalar) functions and vector function in several variables at once. It is based upon multivariable calculus. A (scalar) function is a scalar whose value depends upon several variables. Examples are the temperature and pressure of the atmosphere, the density of an inhomogeneous solid and so on. A vector function (also called a vector field) is a vector whose components depend upon several variables. Examples of these are the electric and magnetic field, the velocity of a fluid, and so on. In the present case we are dealing with vectors in three-dimensional space so they have three components. The number of variables that functions and vector field can depend on is also three. In these notes we review the minimum amount of vector calculus needed to understand electromagnetic waves. We will often follow the notation of H.M. chey, Div, Grad, Curl, and All That (fourth edition). Wherever possible, a reference will be given to this book. 1.2 The Operator The fundamental operator we deal with in vector calculus is the operator. It is defined as = e x x + e y y + e z z (1.1) ii
see chey, p.44. The vectors e x, e y, e z are three mutually perpendicular unit vectors that form a right-handed triplet with e x along the positive x-axis and similarly e y along the positive y-axis and e z along the positive z-axis. We do not use chey s notation i, j, k for these. The partial differentiations are familiar from multivariable calculus. It is seen that the operator is a vector because it is the sum of three terms that are each a vector. The order of the factors in each term does not matter because the unit vectors are constant vectors, independent of position x, y, z so the partial differentiations will do nothing to them. This is all you need to know (besides some nomenclature) because everything in vector calculus we need follows from the definition (1.1) of the operator. 1.3 The Gradient of a calar Function The gradient of a (scalar) function f(x,y,z), defined as F, is evaluated as f = ( e x x + e y y + e ) z f(x, y, z) (1.2) z Working out the parentheses in (1.2) we get f = ( f e x x + e f y y + e f ) z z (1.3) see chey, p.121. chey muddles the gradient discussion a bit. It is seen that the gradient of a (scalar) function is a vector. 1.4 The Divergence of a Vector Function The divergence of a vector function v(x, y, z), defined as v, is evaluated as v = ( e x x + e y y + e ) z (vx e x + v y e y + v z e z ) (1.4) z Working out the parentheses in (1.4) we get nine terms. Using the fact that the three unit vectors e x, e y, e z form a triplet of mutually perpendicular unit vectors we get v = v x x + v y y + v z z (1.5) iii
see chey p. 41. We have used for example that e x e y = 0 and similar relations for the five other combinations of dot products of two different unit vectors. We also used for example that e x e x = 1 and similar relations for the other two combinations of dot products of a unit vector with itself. It is seen that the divergence of a vector function is a scalar. 1.5 The Curl of a Vector Function The curl of a vector function v(x, y, z), defined as v, is evaluated as v = ( e x x + e y y + e ) z (vx e x + v y e y + v z e z ) (1.6) z uch a cross product can be evaluated using a determinant e x e y e z v = x y z v x v y v z (1.7) to get v = ( v z y v y ) ex ( v z z x v x ) ey + ( v y z x v x ) ez (1.8) y see chey p. 80-83. It is seen that the curl of a vector function is a vector. 1.6 Examples The simplest example of a gradient is r where r = x 2 + y 2 + z 2 is the length of the vector r = xe x + ye y + ze z. We get r = ( e x x + e y y + e ) z (x 2 + y 2 + z 2 ) 1 2 (1.9) z Working out the parentheses in the first factor we get terms proportional to e x, e y, and e z. The first of these can be evaluated as e x x (x2 + y 2 + z 2 ) 1 2 1 = e x 2 (x2 + y 2 + z 2 ) 1 x 2 2x = e x r (1.10) imilar expressions hold for the other two terms. This can be seen by an explicit calculation or by replacing x by y and replacing x by z respectively iv
in (1.10). It is a good idea to pick up these substitution tricks and save time. They also serve as a check if you did the explicit calculation after all. All three expressions have a factor 1/r in common which we take outside parentheses to get r = e xx + e y y + e z z r = r r We expect r to be a vector and it is. The units work out too. If f(x, y, z) = r n then the gradient of f is (1.11) f = ( e x x + e y y + e ) z (x 2 + y 2 + z 2 ) n 2 (1.12) z Working out the parentheses in the first factor we get terms proportional to e x, e y, and e z. The first of these can be evaluated as e x x (x2 + y 2 + z 2 ) n 2 = ex n 2 (x2 + y 2 + z 2 ) n 2 1 2x = e x nx r n 2 (1.13) imilar expressions hold for the other two terms. All three expressions have a factor nr n 2 in common which we take outside parentheses to get f = (e x x + e y y + e z z) nr n 2 = rnr n 2 (1.14) We expect f to be a vector and indeed it is. We could have guessed the answer by using f(r) = df r (1.15) dr which results in (1.14) as well. It is often true in vector calculus that when you guess a relation it is often correct but it bears checking. This feature is what makes vector calculus useful. The special case n = 1 corresponds to a 1/r potential such as we encounter in electrostatics and gravity. The force is proportional to the negative of the gradient of f. etting n = 1 in (1.14) we find that the force is proportional to r/r 3. The force is along r and has a magnitude proportional to 1/r 2 as expected. The proportionality constant can be positive or negative so the direction of the force can be in the direction of r or in the direction of r. Next consider a vector function v(x, y, z) = r/r n. The divergence of this vector function is v = ( e x x + e y y + e ) x z ( z r n e x + y r n e y + z r n e z) (1.16) v
Working out the parentheses we get nine terms. As in ec. 1.4, only three terms survive because e x, e y, e z are three mutually perpendicular vectors. One of the three terms is proportional to e x e x (which equals 1) and is of the form x ( x r n ) = rn xnr n 1 r x r 2n (1.17) This a good point to verify that the units work out (they do). The partial derivative of r with respect to x can be evaluated as r x = x (x2 + y 2 + z 2 ) 1 2 = x r compare (1.13), and (1.17) becomes ( x ) r n x 2 nr n 2 = x r n r 2n (1.18) = 1 n x2 r 2 r n (1.19) This is again a good point to verify that the units work out (they do). The other two terms from (1.16) are of the form 1 n y2 r 2 r n and 1 n z2 r 2 r n (1.20) This can be seen by an explicit calculation or by replacing x by y and replacing x by z respectively in (1.20), see the discussion below (1.10). To complete the calculation of v of (1.16) we must sum the term in (1.19) and the two terms in (1.20). They have a common denominator r n. The numerators neatly sum to 3 nr 2 /r 2 = 3 n so finally we get and the units work out. function is v = r we get r r n = 3 n r n (1.21) In the special case that n = 0 and the vector r = 3 (1.22) Next consider the curl of v(x, y, z) = r/r n. It can be evaluated as in (1.7) as e x e y e z v = x y z (1.23) x r n vi y r n z r n
The first term is proportional to e x and is [ ( z ) ( y )] ( n)z r e x = ex y r n z r n r n+1 y ( n)y r r n+1 z (1.24) We know from a calculation similar to the one leading to (1.10) that r x = x r (1.25) imilar relations hold for r/y and r/z. Again, all that is required is to replace x by y and x by z in (1.25). Thus (1.24) becomes [ ( z ) ( y )] ( n) e x = ex y r n z r n (yz zy) = 0 (1.26) rn+2 The other two terms proportional to e y and e z can be obtained in a similar fashion and are found to be zero as well. They are equal to ( n) e y r n+2 (zx xz) = 0 and e ( n) z (xy yx) = 0 (1.27) rn+2 To see this without doing an explicit calculation is somewhat harder than in the cases discussed earlier. Thus we find that r r n = 0 (1.28) This is true for all n so we have in particular that r = 0. vii
Chapter 2 Waves and their Properties 2.1 Plane Waves Waves were discussed in physics 4B. We review the discussion here, using a slightly different notation. In 4B the expression for a plane wave traveling in the +x-direction was written for example as D(x, t) = D M sin (kx ωt) (2.1) where the wave number k and the angular frequency ω are given by k = 2π λ and ω = 2π T with λ the wavelength and T the period. It was also shown there that (2.2) λ = vt (2.3) where v is the propagation velocity. The expression φ = kx ωt, the argument of the sine in (2.1), is called the phase of the wave. The expression (2.1) represents a plane wave (plane along the y and z direction) because (2.1) is independent of y and z and its amplitude for given values of x and t has the same value at all y and z. One could as well have used to cosine instead of the sine in (2.1). Instead of choosing one or the other, we employ a complex number notation using e iφ = cos φ + i sin φ (2.4) Using (2.4) we can write for a wave traveling in the +x direction D(x, t) = D M e i(kx ωt) (2.5) viii
where it is optional to take either the real or the imaginary part of (2.5), compare (2.1). We generalize the notation to represent a wave traveling in a direction along the vector k as follows u(r, t) = u 0 e i(k r ωt) (2.6) where u 0 is a constant and represents the amplitude of the wave. Here r is the position at which the amplitude of the wave is to be evaluated and φ = k r ωt is called the phase. The vector k is now called the wave vector. The wave is a plane wave traveling in the direction of k while the plane of the wave is perpendicular to k as can be seen as follows. We may choose the orientation of our coordinate system as we like without loss of generality. Therefore, let us choose a coordinate system with its z-axis along k so k = (0, 0, k) with k = k. The phase φ at a fixed time t 0 is given by φ = k r ωt 0 = kz ωt 0. The endpoints of all vectors r that satisfy the condition of constant phase φ = kz ωt 0 have a constant z = (φ + ωt 0 )/k with x and y unconstrained. This relation represents a plane perpendicular to the z-axis, parallel to the x, y plane. Thus the condition of constant phase φ forces the endpoints of all vectors r to lay in a plane that is perpendicular to the vector k. Keeping k parallel to the z-axis and keeping the phase φ constant (the observer rides with the traveling wave at a fixed position on the wave) we find that as z and t change, their change is such that φ = k(z + dz) ω(t + dt) = kz ωt. Canceling terms we find we find kdz ωdt = 0 or v = dz dt = ω k (2.7) where v is the propagation velocity of the traveling wave. To keep the phase φ constant, z and t must both increase (decreasing time is not possible of course) implying a wave traveling in the +z-direction. Had we written (2.6) with a plus sign as in u(r, t) = u 0 e i(k r+ωt) (2.8) we would have found that v would have had the opposite sign, corresponding to a wave traveling in the negative z-direction with our special choice of coordinate system. Of course it is the relative sign of k r and ωt that determines the direction of propagation, not their individual signs. If we keep t fixed (so we take a snapshot of the traveling wave) we can compare the amplitude of the wave at z and z + λ. These two amplitudes must be the same by the definition of the wavelength λ. Increasing z by an ix
amount λ at fixed t changes the phase but the phase change should be such that the amplitude of the wave is unchanged. Thus we must have Canceling terms we find k(z + λ) = kz + 2π (2.9) kλ = 2π or k = 2π λ = k (2.10) This is the first relation in (2.2) for k, the magnitude of the wave vector k. Instead of keeping t fixed, we can keep z fixed and compare the amplitude of the wave at t and t + T. These two amplitudes must be the same by the definition of the period T. Increasing t by an amount T at fixed z changes the phase but the phase change should be such that the amplitude of the wave is unchanged. Thus we must have Canceling terms we find ω(t + T ) = ωt + 2π (2.11) ωt = 2π or ω = 2π T (2.12) This is the second relation in (2.2). Using (2.10) and (2.12) in the second relation of (2.7) we find λ = vt, compare (2.3). Note that v is parallel to k. The waves considered in the previous section could be called scalar waves because the amplitude of the waves is a scalar. This is useful to distinguish those from vector waves where the amplitude of the wave is a vector. We write such a vector wave in general as v(r, t) = v 0 e i(k r ωt) (2.13) with v 0 a constant vector representing the vector amplitude of the vector wave. The meaning of the other quantities in (2.13) are the same as those for the scalar waves discussed above and the discussion following (2.6) applies here as well. The expressions (2.1), (2.5), (2.6), (2.8), and (2.13) satisfy wave equations that we discuss next. x
2.2 Wave Equations for calar Waves We start with a calculation of the gradient of a (scalar) wave (2.6) u(r, t) = u 0 e i(k r ωt) (2.14) = u 0 e i(k r ωt) (2.15) ( = u 0 ex x + e y y + e ) z e i(k xx+k yy+k zz ωt) (2.16) z where in (2.15) we exchanged the operator with the constant u 0. Working out the parentheses we get terms proportional to e x, e y and e z, compare (1.9). The first of these can be evaluated as u 0 e x x = u 0 e x (ik x ) e i(kxx+kyy+kzz ωt) (2.17) = e x (ik x ) u(r, t) (2.18) imilar expressions hold for the other two terms. This can be seen by an explicit calculation or by replacing x by y and replacing x by z respectively in (2.18). All three expressions have a factor iu(r, t) in common which we take outside parentheses to get u(r, t) = ( e x k x + e y k y + e z k z ) iu(r, t) (2.19) = ( ik) u(r, t) (2.20) We expect u(r, t) to be a vector and it is. Note that the result is exactly what you would expect intuitively: the operator leaves the exponential intact but the chain rule brings down a factor ik, something we are used to in single variable calculus. We can now evaluate 2 u(r, t) as 2 u(r, t) = u(r, t) (2.21) = (ik) u(r, t) (2.22) = (ik) u(r, t) (2.23) = (ik) 2 u(r, t) (2.24) = k 2 u(r, t) (2.25) = k 2 u(r, t) (2.26) xi
where we used (2.20) in (2.21) and (2.23). The 2 is a scalar. It multiplies the scalar u(r, t) so the result 2 u(r, t) should be a scalar and (2.26) shows that it is a scalar. It is much easier to differentiate the scalar wave (2.6) with respect to time. We get u(r, t) = u 0e i(k r ωt) (2.27) = u 0 (2.28) = u 0 ( iω) e i(k r ωt) (2.29) = ( iω)u 0 e i(k r ωt) (2.30) = ( iω) u(r, t) (2.31) We can now evaluate the second derivative with respect to time of the scalar wave (2.6) as 2 u(r, t) = u(r, t) (2.32) 2 = ( iω)u(r, t) (2.33) = ( iω) u(r, t) (2.34) = ( iω) 2 u(r, t) (2.35) = ω 2 u(r, t) (2.36) where we used (2.31) in (2.32) and (2.34). Using (2.26) and (2.36) we see that the scalar wave u(r, t) satisfies 2 u(r, t) k2 2 ) u(r, t) ω 2 2 = ( 2 k2 ω 2 u(r, t) = 0 (2.37) or with (2.7) 2 u(r, t) 1 v 2 2 u(r, t) 2 = ( 2 1v 2 ) u(r, t) = 0 (2.38) The expressions (2.37) and (2.38) are called wave equations because their solutions describe (traveling) waves. These wave equations are pretty complicated expressions: they are second order partial differential equations. xii
They are also linear and homogeneous. Linear because all u(r, t) and its derivatives appear to the first power, homogeneous because all terms involve u(r, t). uch linear homogeneous differential equations have an extremely important property: if we have two linearly independent solutions u 1 (r, t) and u 2 (r, t) (never mind where we got them) then the linear superposition u(r, t) = a 1 u 1 (r, t) + a 2 u 2 (r, t) is also a solution. The proof is easy and you should try to prove this property. Verify that the proof fails when the equation is non-linear or inhomogeneous (or both). Because a second order differential equation has at most two linearly independent solutions, this linear superposition of two linearly independent solutions is also the most general solution of the wave equation. The two linearly independent solutions of (2.37) and (2.38) are (2.6) and (2.8) that is traveling waves in the +r and r directions respectively. Under certain circumstances their linear superposition can form standing waves as discussed in Physics 4B. 2.3 Wave Equations for Vector Waves We follow the development of the previous section for scalar waves. calculate the divergence of the vector wave (2.13) We v(r, t) = v 0 e i(k r ωt) (2.39) = v 0 e i(k r ωt) (2.40) = v 0 (ik)e i(k r ωt) (2.41) = i k v(r, t) (2.42) where we exchanged the operator with the constant vector v 0 and used (2.20) in (2.40). We can now evaluate 2 v(r, t) as 2 v(r, t) = v(r, t) (2.43) = [ (ik)] v(r, t) (2.44) = (ik) v(r, t) (2.45) = (ik) 2 v(r, t) (2.46) = k 2 v(r, t) (2.47) = k 2 v(r, t) (2.48) xiii
where we used (2.42) in (2.43) and (2.45). The 2 is a scalar. It multiplies the vector v(r, t) so the result 2 v(br, t) should be a vector and (2.48) shows that it is a vector. This result can be compared with (2.26). We now differentiate the vector wave (2.13) with respect to time. We get v(r, t) = v 0e i(k r ωt) (2.49) = v 0 (2.50) = v 0 ( iω) e i(k r ωt) (2.51) = ( iω)v 0 e i(k r ωt) (2.52) = ( iω) v(r, t) (2.53) We can now evaluate the second derivative with respect to time of the vector wave (2.13) as 2 v(r, t) = v(r, t) (2.54) 2 = ( iω)v(r, t) (2.55) = ( iω) v(r, t) (2.56) = ( iω) 2 v(r, t) (2.57) = ω 2 v(r, t) (2.58) This result can be compared with (2.36). Using (2.48) and (2.58) we see that the vector wave v(r, t) satisfies 2 v(r, t) k2 2 ) v(r, t) ω 2 2 = ( 2 k2 ω 2 v(r, t) = 0 (2.59) or with (2.7) 2 v(r, t) 1 v 2 2 v(r, t) 2 = ( 2 1v 2 ) v(r, t) = 0 (2.60) The expressions (2.59) and (2.60) are called vector wave equations because their solutions describe (traveling) vector waves. The same comments apply here that were made below (2.38). xiv
Chapter 3 Gauss and tokes Theorems 3.1 Gauss Theorem Consider a volume V that is enclosed by a surface. Unit vectors ˆn are normal to the surface and are directed outward. Gauss Theorem states v dv = v ˆn d (3.1) V where v(r) is a vector function, also called a vector field. ometimes the notation d = ˆnd is used, see for example Giancoli s formulation of Maxwell s equations. The vector v may also depend upon other variables such as time but those are irrelevant for Gauss Theorem. Gauss Theorem is also called the Divergence Theorem. As an example we consider the case where the surface is a sphere with radius R and the vector field is the radius vector r. For the left hand side of (3.1) we must evaluate v = r. We find r = 3 according to (1.22). Thus the integrand of the volume integral has the constant value 3 and can be taken outside the volume integral. The integral over the volume is 4πR 3 /3 so the left hand side of (3.1) is 4πR 3. For the right hand side of (3.1) we use that ˆn = r/r so that v ˆn = r r/r = r. Thus the integrand of the surface integral has the constant value R and can be taken outside the surface integral. The integral over is 4πR 2 so the right hand side of (3.1) is 4πR 3. The two sides of (3.1) are seen to be equal. xv
3.2 tokes Theorem Consider a surface that is bounded by a line s. The direction for the line integral is defined by the direction of unit vectors ˆτ tangent to the curve. Unit vectors ˆn are normal to the surface in a direction relative to the direction of the line integral defined by the right-hand rule. This definition is different from the one for Gauss Theorem because here there is no definition possible for inside or outside the surface. tokes Theorem states v ˆn d = v ˆτ ds (3.2) where v(r) is a vector function as above. ometimes the notation ds = ˆτ ds is used, see for example Giancoli s formulation of Maxwell s equations.. The vector v may also depend upon other variables such as time but those are irrelevant for tokes Theorem. tokes Theorem is also called the Curl Theorem. As an example we consider the case where the line s is a circle of radius R. The vector r has its endpoint laying on the circle so r = x e x + y e y with x 2 + y 2 = R 2. Here e x and e y are the unit vectors along the positive x and y axes respectively. The vector s = y e x x e y is tangential to the circle as can be seen by evaluating s r, which is zero so s and r are perpendicular. For the left hand side of (3.2) we must evaluate v = s. We find e x e y e z s = x y z y x 0 [ = e x y 0 z ( x)] [ e y x 0 z y] [ + e z x ( x) y y] = 2e z (3.3) At the point (x, y) = (0, R) the vector s = Re y. We choose ˆτ parallel to s. Using the right-hand rule we find that the normal on the plane of the circle is given by ˆn = e z. Thus s ˆn = 2e z ( e z ) = +2 and the integrand of the surface integral has the constant value 2 and can be taken outside the surface integral. The integral over the surface is πr 2 so the left hand side of (3.2) is 2πR 2. For the right hand side of (3.2) we use that s and ˆτ are parallel and that the length of the vector s = R so that s ˆτ = R. Thus the integrand of the line integral has the constant value R and can be taken outside the line integral. The integral along s is 2πR so the right hand side of (3.2) is 2πR 2. The two sides of (3.2) are seen to be equal. s xvi
Chapter 4 Differential Form of Maxwell s Equations Maxwell s equations are ɛ 0 E ˆn d = ρ dv (4.1) B ˆn d = 0 (4.2) E τ ds = B ˆn d (4.3) s B τ ds = µ 0 j ˆn d + µ 0 ɛ 0 E ˆn d (4.4) s We will not discuss the Maxwell equations but instead refer to Giancoli. We use Gauss Theorem to get E ˆn d = E dv (4.5) B ˆn d = B dv (4.6) and tokes Theorem to get E τ ds = s B τ ds = s xvii E ˆn d (4.7) B ˆn d (4.8) (4.9)
Combining (4.1) with (4.5), (4.2) with (4.6), (4.3) with (4.7), and (4.5) with (4.8) we get ɛ 0 E dv = ρ dv (4.10) B dv = 0 (4.11) E ˆn d = B ˆn d (4.12) B ˆn d = µ 0 j ˆn d + µ 0 ɛ 0 E ˆn d (4.13) The relations (4.10) through (4.13) must hold for arbitrary volumes, surfaces, and lines so in each equation, the integrands must be equal. Thus we get E = ρ ɛ 0 (4.14) B = 0 (4.15) E = B (4.16) E B = µ 0 j + µ 0 ɛ 0 (4.17) In vacuum ρ = j = 0 and the Maxwell equations become E = 0 (4.18) B = 0 (4.19) E = B (4.20) E B = µ 0 ɛ 0 (4.21) There is a certain symmetry in these equations. Of particular note is the minus sign in (4.20), a manifestation of Lenz law. If this minus sign was not there, or more accurately if (4.20) and (4.21) did not have opposite signs, electromagnetic waves would not exist! We will show next the existence of traveling electromagnetic waves. xviii
Chapter 5 Electromagnetic Waves 5.1 Wave Equations for E and B The relations (4.20) and (4.21) are two coupled equations because each involves E and B. We can separate E and B, at the cost of obtaining higher derivatives, by taking the curl of either equation then substitute the other equation. Let s start with taking the curl of both sides of (4.21) to get ( B) = µ 0 ɛ 0 E (5.1) = µ 0 ɛ 0 ( E) (5.2) = µ 0 ɛ 0 ( B ) (5.3) = µ 0 ɛ 0 2 B 2 (5.4) where we have used (4.20) in (5.2). We now use the general relation for a vector triple product a (b c) = (a c) b (a b) c (5.5) Using this relation to work out the triple product on the left hand side of (5.1), we must take care with the order of the operators. The operator works only on vectors to its right. Therefore v(r) is not the same as v(r). The two operators in (5.1) need to be kept to the left of B xix
when using (5.5). Thus we get ( B) = ( B) 2 B (5.6) = 2 B (5.7) where we have used (4.19) in (5.6). ubstitution of (5.7) in (5.1) gives 2 B µ 0 ɛ 0 2 B 2 = 0 (5.8) We have succeeded in decoupling E from B and found that B satisfies a vector wave equation, compare (2.59) and (2.60). The solutions of this wave equation are traveling waves of the form (2.13) with v(r, t) = B(r, t) and with either a plus or a minus sign in the exponent, see the discussion near (2.8). Comparing (5.8) with (2.60) we see immediately that the propagation velocity v of the vector wave B(r, t) is given by v = 1 ɛ0 µ 0 (5.9) With 1/(4πɛ 0 ) = 9 10 9 and µ 0 /(4π) = 10 7 we find that v = 3 10 8 m/s, the speed of light. This is a strong indication that light is nothing else but an electromagnetic wave phenomenon. We use the letter c for the velocity of light so (5.9) becomes c = 1 ɛ0 µ 0 (5.10) Instead of starting with taking the curl of both sides of (4.17) we could have started with (4.16). We find ( E) = ( B ) (5.11) = E ( B) (5.12) = ( E µ0 ɛ 0 ) (5.13) = µ 0 ɛ 0 2 E 2 (5.14) xx
where we have used (4.21) in (5.12). As in (5.7) we have that ( E) = ( E) 2 E (5.15) = 2 E (5.16) where we have used (4.18) in (5.15). ubstitution of (5.16) in (5.15) gives 2 E µ 0 ɛ 0 2 E 2 = 0 (5.17) We find that in decoupling E and B that E satisfies the same vector wave equation as B, see (5.8) and thus the solution of (5.17) are traveling waves of the form (2.13) with a propagation velocity given by (5.9) and (5.10). 5.2 Plane Wave olutions We try solutions of the form (2.13) except that we factor the vector amplitude v 0 in (2.13) into a unit vector ɛ and a scalar that represents the amplitude of the vector wave. Because E and B may point in different directions and have different amplitudes we introduce two units vectors ɛ 1 and ɛ 2 and two amplitudes E m and B m. Thus we write E(r, t) = ɛ 1 E m e i(k r ωt) (5.18) B(r, t) = ɛ 2 B m e i(k r ωt) (5.19) We now want to find expressions for ɛ 1 and ɛ 2 and for E m and B m. We obtain these by substitution of the vector waves (5.18) and (5.19) in Maxwell s equations (4.18) through (4.21). ubstitution of (5.18) and (5.19) in (4.18) and (4.19) respectively we find E = [ɛ 1 E m e i(k r ωt)] = ik E = 0 (5.20) E = [ɛ 2 B m e i(k r ωt)] = ik B = 0 (5.21) where we have used (2.42). We find therefore that k is perpendicular to ɛ 1 and perpendicular to ɛ 2 and thus thta k is perpendicular to E and perpendicular to B. Because k point in the direction of propagation of the xxi
traveling vector wave, we find that E and B are perpendicular to the direction of propagation. We call such waves transverse (plane) waves and electromagnetic waves are thus transverse waves. Next we want to find the relative orientation of ɛ 1 and ɛ 2 and thus of E and B. To simplify the calculation we use the freedom we have to choose our coordinate system. Without loss of generality we choose the z-axis parallel to k so k = (0, 0, k) where k = k. The dot product k r = kz. We can still choose the orientation of for example the x-axis without affecting the orientation of the z-axis. We choose the x-axis parallel to ɛ 1 so ɛ 1 = (1, 0, 0) and therefore E = (E m e i(kz ωt), 0, 0) (5.22) We now use (4.20). We need E which can be evaluated as E = e x e y e z x y z E m e i(k z ωt) 0 0 = e y E m (ik)e i(k z ωt) (5.23) We also need the derivative of B with respect to time. This can be evaluated as B = ɛ 2B m ( iω)e i(k z ωt) = iωɛ 2 B m e i(k z ωt) (5.24) compare (2.53). ubstitution of (5.23) and (5.24) in (4.20) gives e y E m (ik)e i(k z ωt) = iωɛ 2 B m e i(k z ωt) (5.25) Canceling common factors in (5.25) we get or e y E m k = ωɛ 2 B m (5.26) ɛ 2 = e y ke m ωb m (5.27) We find that ɛ 2 is parallel to the y-axis. We already have that ɛ 1 is parallel to the x-axis so we find that the vectors (ɛ 1, ɛ 2, k) form a right handed triplet xxii
and therefore also (E, B, k) form a right handed triplet. We now require that ɛ 2 = 1. With (5.27) we find that or using (2.7) with v = c we find B m = E m c ke m = ωb m (5.28) = ɛ 0 µ 0 E m (5.29) The electric and magnetic fields are in phase, that at each position they oscillate in phase. The amplitude of the magnetic field is much smaller than the amplitude of the electric field. However, the stored energies in each are equal, see Giancoli. xxiii