aB. a(aB. (13.4.14) ax at Then substitute Eq.(13.4.12)into Eq.(13.4.13)yielding E. E, Hoo dt2 (13.4.15) Ox2 Eq.(13.4.15)is called the one-dimensional wave equation,which refers to the fact that the only spatial variation of our plane wave is in the x-direction.This equation involves second partial derivatives in space and in time.By a dimensional analysis the quantity 1/has the same dimensions as speed squared. We can repeat the argument to find a one-dimensional wave equation satisfied by the z- component of the magnetic field.We start by taking a/ax of Eq.(13.4.12) ad止y=ueo -=Hoo dx dt (13.4.16) dx Now we substitute Eq.(13.4.6)into Eq.(13.4.16),yielding a one-dimensional wave equation satisfied by the z-component of the magnetic field, B B. =00t dr2 (13.4.17) The general form of a one-dimensional wave equation is given by w(x,t)1 'w(x,t) (13.4.18) where v is the speed of propagation and y(x,t)is the wave function.Both E,and B. satisfy the wave equation and propagate with speed V= (13.4.19) uEo We can rewrite Eq.(13.4.19)as 1 Eo=- %2 (13.4.20) Recall from(Eq.2.2.3),that is defined to be exactly equal to 13-11
13-11 Bz Bz x t t x ! " ! # ! " ! # $ % = $ % ! & ! ' ! & ! ' (13.4.14) Then substitute Eq. (13.4.12) into Eq. (13.4.13) yielding !2 E y !x 2 = µ0 " 0 !2 E y !t 2 . (13.4.15) Eq. (13.4.15) is called the one-dimensional wave equation, which refers to the fact that the only spatial variation of our plane wave is in the x -direction. This equation involves second partial derivatives in space and in time. By a dimensional analysis the quantity 1/ µ0 ! 0 has the same dimensions as speed squared. We can repeat the argument to find a one-dimensional wave equation satisfied by the zcomponent of the magnetic field. We start by taking ! / !x of Eq. (13.4.12). ! "2 Bz "x 2 = µ0 # 0 " "x "E y "t = µ0 # 0 " "t "E y "x $ % & ' ( ) . (13.4.16) Now we substitute Eq. (13.4.6) into Eq. (13.4.16), yielding a one-dimensional wave equation satisfied by the z-component of the magnetic field, !2 Bz !x 2 = µ0 " 0 !2 Bz !t 2 . (13.4.17) The general form of a one-dimensional wave equation is given by !2 "(x,t) !x 2 = 1 v 2 !2 "(x,t) !t 2 , (13.4.18) where v is the speed of propagation and ! (x,t) is the wave function. Both E y and Bz satisfy the wave equation and propagate with speed v = 1 µ0 ! 0 . (13.4.19) We can rewrite Eq. (13.4.19) as ! 0 = 1 µ0 v 2 . (13.4.20) Recall from (Eq. 2.2.3), that ! 0 is defined to be exactly equal to
6、1 %c3, (13.4.21) Comparing Eqs.(13.4.21)and (13.4.20),we conclude that the speed of propagation is exactly equal to the speed of light, v=c=299792458m.s (13.4.22) Thus,we conclude that Maxwell's Equations predict that electric and magnetic field can propagate through space at the speed of light.The spectrum of electromagnetic waves is shown in Figure 13.4.4. 1022 Gamma rays 10 108 X-ravs Inm Ultraviolet 1016 visble light入 m 101 Infrared 1012 1cm 1010 Microwaves 10 TV FM Radiowaves 10 AM 1km 104 Long wave Frequency.Hz Wavelength Figure 13.4.4 Electromagnetic spectrum 13.4.1 One-Dimensional Wave Equation We shall now explore the properties of wave functions u(x,t)that are solutions to the one-dimensional wave equation,Eq.(13.4.18).Many types of physical phenomena can be described by wave functions.We have already seen in the previous section that the plane transverse electric and magnetic fields,E(x,t)and B.(x,t),propagating at the speed of light satisfy a one-dimensional wave equations.The transverse displacement of a stretched string,y(x,t)will propagate along a string oriented along the x-direction at s speed dependent on the material properties and tension of the string.In each of these 13-12
13-12 ! 0 = 1 µ0 c 2 , (13.4.21) Comparing Eqs. (13.4.21) and (13.4.20), we conclude that the speed of propagation is exactly equal to the speed of light, v = c = 299792458 m1 !s -1 . (13.4.22) Thus, we conclude that Maxwell’s Equations predict that electric and magnetic field can propagate through space at the speed of light. The spectrum of electromagnetic waves is shown in Figure 13.4.4. Figure 13.4.4 Electromagnetic spectrum 13.4.1 One-Dimensional Wave Equation We shall now explore the properties of wave functions ! (x,t) that are solutions to the one-dimensional wave equation, Eq. (13.4.18). Many types of physical phenomena can be described by wave functions. We have already seen in the previous section that the plane transverse electric and magnetic fields, E y (x,t) and Bz (x,t), propagating at the speed of light satisfy a one-dimensional wave equations. The transverse displacement of a stretched string, y(x,t) will propagate along a string oriented along the x -direction at s speed dependent on the material properties and tension of the string. In each of these
instances,the wave function is a function of both position and time.The propagation of the wave a constant speed v is a means of translating the wave function from a position and time,(x,),to a position and time (x2).We shall now explore wave functions that solve the one-dimensional wave equation,Eq.(13.4.18).We shall first begin with an example. 13.4.2 Gaussian One Dimensional Wave Pulse Consider the function y(x,t)=yoe-(-l (13.4.23) The variables (x,t)appear together as x-vt.At =0, y(x,0)=ye- (13.4.24) In Figure 13.4.5(a)we show the position of the wave function given in Eq.(13.4.23)for three times,=0,=(2 m)/v,and=(4 m)/v.The wave function is propagating in the positive x-direction.Figure 13.4.5(b)shows the propagation of a wave function given by y(x,t)=yoe-(l (13.4.25) When variables (x,t)appear together as x+vt,as in y(x,t)=yoe-(xtwl (13.4.26) The wave function is propagating in the negative x-direction.Figure 13.4.5(b)shows the propagation of the wave function given in Eq.(13.4.26). yx） y(x,t)=ye-(x-mi x+m)）=%e*w广 vt=4 m vt=2m vt vt=2m vt=4 m (a) (b) Figure 13.4.5 (a)Wave function propagating in the (a)positive x-direction,(b)negative x-direction. Let's show by direct computation that Eq.(13.4.23)satisfies the wave equation, 13-13
13-13 instances, the wave function is a function of both position and time. The propagation of the wave a constant speed v is a means of translating the wave function from a position and time, (x1,t 1), to a position and time (x2 ,t 2 ). We shall now explore wave functions that solve the one-dimensional wave equation, Eq. (13.4.18). We shall first begin with an example. 13.4.2 Gaussian One Dimensional Wave Pulse Consider the function y(x,t) = y0e!(x!vt) 2 /a2 . (13.4.23) The variables (x,t) appear together as x ! vt . At t 1 = 0 , y(x,0) = y0e! x2 /a2 . (13.4.24) In Figure 13.4.5(a) we show the position of the wave function given in Eq. (13.4.23) for three times, t 1 = 0 , t 2 = (2 m) / v , and t 3 = (4 m) / v . The wave function is propagating in the positive x -direction. Figure 13.4.5(b) shows the propagation of a wave function given by y(x,t) = y0e!(x+vt) 2 /a2 . (13.4.25) When variables (x,t) appear together as x + vt , as in y(x,t) = y0e!(x+vt) 2 /a2 . (13.4.26) The wave function is propagating in the negative x -direction. Figure 13.4.5(b) shows the propagation of the wave function given in Eq. (13.4.26). (a) (b) Figure 13.4.5 (a) Wave function propagating in the (a) positive x -direction, (b) negative x -direction. Let’s show by direct computation that Eq. (13.4.23) satisfies the wave equation
x,=1x, (13.4.27) dx2 v20t2 The left-hand-side of Eq.(13.4.27)is -品e）新-2a-mo (13.4.28) =(-2Ia')yoe--myn+(-2(x-v)Ia')yoe--wyl The right-hand-side of Eq.(13.4.27)is -(品x）2-w1dwe心） 1'x,0=1a(ag v2 -(-2vla)ye+(xx-v)layy.ww) (13.4.29) =(-21a2%e-m1+(20x-0）1a2ye--mr1d A comparison of Eqs.(13.4.28)and(13.4.29)shows that our wave function (Eq.(13.4.23) is a solution to the one-dimensional wave equation(Eq.(13.4.27)). Example 13.4.2 illustrates the general property that any function of the form (x+vt)or w(x-vt)satisfies the one-dimensional wave equation,Eq.(13.4.18).We shall now demonstrate this property.Let x'=x+vt.The space and time partial derivatives are dx'/dx =1, (134.30) dx'/dt=tv. (13.4.31) Using the chain rule,the first partial derivatives with respect to x is du(x')du dx' (13.4.32) dx dx'dx Substitute Eq.(13.4.30)into Eq.(13.4.32)yielding dv(x)ov (13.4.33) ax dx' We now repeat this calculation for the second partial derivative, 13-14
13-14 !2 y(x,t) !x 2 = 1 v 2 !2 y(x,t) !t 2 . (13.4.27) The left-hand-side of Eq. (13.4.27) is !2 y(x,t) !x 2 = ! !x ! !x y0e"(x"vt) 2 /a # 2 $ % & ' ( = ! !x ("2(x " vt) / a 2 )y0e"(x"vt) 2 /a2 ( ) = ("2 / a 2 )y0e"(x"vt) 2 /a2 + ("2(x " vt) / a 2 ) 2 y0e"(x"vt) 2 /a2 . (13.4.28) The right-hand-side of Eq. (13.4.27) is 1 v 2 !2 y(x,t) !t 2 = 1 v 2 ! !t ! !t y0e"(x"vt) 2 /a # 2 $ % & ' ( = 1 v 2 ! !t (2v(x " vt) / a 2 )y0e"(x"vt) 2 /a2 ( ) = 1 v 2 ("2v 2 / a 2 )y0e"(x"vt) 2 /a2 + (2v(x " vt) / a 2 ) 2 y0e"(x"vt) 2 /a2 ( ) = ("2 / a 2 )y0e"(x"vt) 2 /a2 + (2(x " vt) / a 2 ) 2 y0e"(x"vt) 2 /a2 .(13.4.29) A comparison of Eqs. (13.4.28) and (13.4.29) shows that our wave function (Eq. (13.4.23) is a solution to the one-dimensional wave equation (Eq. (13.4.27)). Example 13.4.2 illustrates the general property that any function of the form !(x + vt) or !(x " vt) satisfies the one-dimensional wave equation, Eq. (13.4.18). We shall now demonstrate this property. Let x! = x ± vt . The space and time partial derivatives are !x" / !x = 1, (13.4.30) !x" / !t = ±v . (13.4.31) Using the chain rule, the first partial derivatives with respect to x is !"(x#) !x = !" !x# !x# !x . (13.4.32) Substitute Eq. (13.4.30) into Eq. (13.4.32) yielding !"(x#) !x = !" !x# . (13.4.33) We now repeat this calculation for the second partial derivative
o'v= a∂w d'w dx'o'w dx2axax' dx'2 dx (13.4.34) dx2 Similarly,the partial derivatives in t are given by ow owax' (13.4.35) at ax'at st⊙w dr' a2y∂年 aΨ 土y 'wa dr2 (13.4.36) at ax' dx"2 dt Comparing Eq.(13.4.34)with Eq.(13.4.36)results in d'wd'w 1d'w dx2 dx2v2012 (13.4.37) which shows that y(x+vt)satisfies the one-dimensional wave equation. The wave equation is an example of a linear differential equation,which means that if w (x,t)and w2 (x,t)are solutions to the wave equation,then w(x,t)+(x,t)is also a solution. Let's now return to our electromagnetic fields.One possible solution to the one- dimensional wave equations for the electric and magnetic field is E=Exi=6,sx-a月 (13.4.38) B=Rc诚=Rs-aj水 where the fields are sinusoidal,with amplitudes Eand B The electric field,E (x,t=0),at time t=0,as a function of position x is given by the expression E,(x,1=0)=E。sin 2元 (13.4.39) In Figure 13.4.6,we show a plot E(x,t=0)as a function of x.On the plane x=0,the electric field is zero,E(0,0)=0.On the planex=,the electric field has completed 13-15
13-15 2 2 2 2 2 2 x x x x x x x # ! # $ #! % # ! # " # ! = & ' = = # # ( # " ) # " # # " . (13.4.34) Similarly, the partial derivatives in t are given by x v t x t x #! #! # " #! = = ± # # " # # " , (13.4.35) 2 2 2 2 2 2 2 x v v v t t x x t x # ! # $ #! % # ! # " # ! = &± ' = ± = # # ( # " ) # " # # " . (13.4.36) Comparing Eq. (13.4.34) with Eq. (13.4.36) results in 2 2 2 2 2 2 2 1 x ' x v t " ! " ! " ! = = " " " , (13.4.37) which shows that ! (x ± vt) satisfies the one-dimensional wave equation. The wave equation is an example of a linear differential equation, which means that if 1 ! (x,t) and 2 ! (x,t) are solutions to the wave equation, then 1 2 ! (x,t) ±! (x,t) is also a solution. Let’s now return to our electromagnetic fields. One possible solution to the onedimensional wave equations for the electric and magnetic field is ! E = E y (x,t)ˆ j = E0 sin 2! " (x # ct) $ % & ' ( ) ˆ j, ! B = Bz (x,t)kˆ = B0 sin 2! " (x # ct) $ % & ' ( ) kˆ, (13.4.38) where the fields are sinusoidal, with amplitudes E0 and B0 . The electric field, E y (x,t = 0) , at time t = 0 , as a function of position x is given by the expression E y (x,t = 0) = E0 sin 2! " x # $ % & ' ( . (13.4.39) In Figure 13.4.6, we show a plot E y (x,t = 0) as a function of x . On the plane x = 0 , the electric field is zero, E y (0,0) = 0 . On the plane x = ! , the electric field has completed