Waves in free space; plane waves
In Chapter 18 we had reached the point where we had the Maxwell equations in complete form. All there is to know about the classical theory of the electric and magnetic fields can be found in the four equations:  \begin{equation}  \begin{alignedat}{6} &\text{I.}&\;\; &\FLPdiv{\FLPE}&&\;=\frac{\rho}{\epsO}\quad&\quad &\text{II.}&\;\; &\FLPcurl{\FLPE}&&\;=-\ddp{\FLPB}{t}\\[1ex] &\text{III.}&\;\; &\FLPdiv{\FLPB}&&\;=0\quad&\quad &\text{IV.}&\;\; c^2&\FLPcurl{\FLPB}&&\;=\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t} \end{alignedat} \label{Eq:II:20:1} \end{equation} 
 \begin{equation}  \begin{alignedat}{3} &\text{I.}&&\FLPdiv{\FLPE}&\;=&\;\frac{\rho}{\epsO}\\[1ex] &\text{II.}&&\FLPcurl{\FLPE}&\;=&\;-\ddp{\FLPB}{t}\\[1ex] &\text{III.}&&\FLPdiv{\FLPB}&\;=&\;0\\[1ex] &\text{IV.}&\quad c^2&\FLPcurl{\FLPB}&\;=&\;\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t} \end{alignedat} \label{Eq:II:20:1} \end{equation}  When we put all these equations together, a remarkable new phenomenon occurs: fields generated by moving charges can leave the sources and travel alone through space. We considered a special example in which an infinite current sheet is suddenly turned on. After the current has been on for the time $t$, there are uniform electric and magnetic fields extending out the distance $ct$ from the source. Suppose that the current sheet lies in the $yz$-plane with a surface current density $J$ going toward positive $y$. The electric field will have only a $y$-component, and the magnetic field, only a $z$-component. The field components are given by \begin{equation} \label{Eq:II:20:2} E_y=cB_z=-\frac{J}{2\epsO c}, \end{equation}  for positive values of $x$ less than $ct$. For larger $x$ the fields are zero. There are, of course, similar fields extending the same distance from the current sheet in the negative $x$-direction. In Fig. 20–1 we show a graph of the magnitude of the fields as a function of $x$ at the instant $t$. As time goes on, the “wavefront” at $ct$ moves outward in $x$ at the constant velocity $c$. Now consider the following sequence of events. We turn on a current of unit strength for a while, then suddenly increase the current strength to three units, and hold it constant at this value. What do the fields look like then? We can see what the fields will look like in the following way. First, we imagine a current of unit strength that is turned on at $t=0$ and left constant forever. The fields for positive $x$ are then given by the graph in part (a) of Fig. 20–2. Next, we ask what would happen if we turn on a steady current of two units at the time $t_1$. The fields in this case will be twice as high as before, but will extend out in $x$ only the distance $c(t-t_1)$, as shown in part (b) of the figure. When we add these two solutions, using the principle of superposition, we find that the sum of the two sources is a current of one unit for the time from zero to $t_1$ and a current of three units for times greater than $t_1$. At the time $t$ the fields will vary with $x$ as shown in part (c) of Fig. 20–2. Now let’s take a more complicated problem. Consider a current which is turned on to one unit for a while, then turned up to three units, and later turned off to zero. What are the fields for such a current? We can find the solution in the same way—by adding the solutions of three separate problems. First, we find the fields for a step current of unit strength. (We have solved that problem already.)  Next, we find the fields produced by a step current of two units. Finally, we solve for the fields of a step current of minus three units. When we add the three solutions, we will have a current which is one unit strong from $t=0$ to some later time, say $t_1$, then three units strong until a still later time $t_2$, and then turned off—that is, to zero. A graph of the current as a function of time is shown in Fig. 20–3(a). When we add the three solutions for the electric field, we find that its variation with $x$, at a given instant $t$, is as shown in Fig. 20–3(b). The field is an exact representation of the current. The field distribution in space is a nice graph of the current variation with time—only drawn backwards. As time goes on the whole picture moves outward at the speed $c$, so there is a little blob of field, travelling toward positive $x$, which contains a completely detailed memory of the history of all the current variations. If we were to stand miles away, we could tell from the variation of the electric or magnetic field exactly how the current had varied at the source. You will also notice that long after all activity at the source has completely stopped and all charges and currents are zero, the block of field continues to travel through space. We have a distribution of electric and magnetic fields that exist independently of any charges or currents. That is the new effect that comes from the complete set of Maxwell’s equations. If we want, we can give a complete mathematical representation of the analysis we have just done by writing that the electric field at a given place and a given time is proportional to the current at the source, only not at the same time, but at the earlier time $t-x/c$. We can write \begin{equation} \label{Eq:II:20:3} E_y(t)=-\frac{J(t-x/c)}{2\epsO c}. \end{equation}  We have, believe it or not, already derived this same equation from another point of view in Vol. I, when we were dealing with the theory of the index of refraction. Then, we had to figure out what fields were produced by a thin layer of oscillating dipoles in a sheet of dielectric material with the dipoles set in motion by the electric field of an incoming electromagnetic wave. Our problem was to calculate the combined fields of the original wave and the waves radiated by the oscillating dipoles. How could we have calculated the fields generated by moving charges when we didn’t have Maxwell’s equations? At that time we took as our starting point (without any derivation) a formula for the radiation fields produced at large distances from an accelerating point charge. If you will look in Chapter 31 of Vol. I, you will see that Eq. (31.9) there is just the same as the Eq. (20.3) that we have just written down. Although our earlier derivation was correct only at large distances from the source, we see now that the same result continues to be correct even right up to the source. We want now to look in a general way at the behavior of electric and magnetic fields in empty space far away from the sources, i.e., from the currents and charges. Very near the sources—near enough so that during the delay in transmission, the source has not had time to change much—the fields are very much the same as we have found in what we called the electrostatic or magnetostatic cases. If we go out to distances large enough so that the delays become important, however, the nature of the fields can be radically different from the solutions we have found. In a sense, the fields begin to take on a character of their own when they have gone a long way from all the sources. So we can begin by discussing the behavior of the fields in a region where there are no currents or charges. Suppose we ask: What kind of fields can there be in regions where $\rho$ and $\FLPj$ are both zero? In Chapter 18 we saw that the physics of Maxwell’s equations could also be expressed in terms of differential equations for the scalar and vector potentials: \begin{align} \label{Eq:II:20:4} \nabla^2\phi- \frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}&= -\frac{\rho}{\epsO},\\[1ex] \label{Eq:II:20:5} \nabla^2\FLPA- \frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}&= -\frac{\FLPj}{\epsO c^2}. \end{align}  If $\rho$ and $\FLPj$ are zero, these equations take on the simpler form \begin{align} \label{Eq:II:20:6} \nabla^2\phi- \frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}&=0,\\[1ex] \label{Eq:II:20:7} \nabla^2\FLPA- \frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}&=\FLPzero. \end{align}  Thus in free space the scalar potential $\phi$ and each component of the vector potential $\FLPA$ all satisfy the same mathematical equation. Suppose we let $\psi$ (psi) stand for any one of the four quantities $\phi$, $A_x$, $A_y$, $A_z$; then we want to investigate the general solutions of the following equation: \begin{equation} \label{Eq:II:20:8} \nabla^2\psi- \frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0. \end{equation}  This equation is called the three-dimensional wave equation—three-dimensional, because the function $\psi$ may depend in general on $x$, $y$, and $z$, and we need to worry about variations in all three coordinates. This is made clear if we write out explicitly the three terms of the Laplacian operator: \begin{equation} \label{Eq:II:20:9} \frac{\partial^2\psi}{\partial x^2}+ \frac{\partial^2\psi}{\partial y^2}+ \frac{\partial^2\psi}{\partial z^2}- \frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0. \end{equation}  In free space, the electric fields $\FLPE$ and $\FLPB$ also satisfy the wave equation. For example, since $\FLPB=\FLPcurl{\FLPA}$, we can get a differential equation for $\FLPB$ by taking the curl of Eq. (20.7). Since the Laplacian is a scalar operator, the order of the Laplacian and curl operations can be interchanged: \begin{equation*} \FLPcurl{(\nabla^2\FLPA)}=\nabla^2(\FLPcurl{\FLPA})=\nabla^2\FLPB. \end{equation*}  Similarly, the order of the operations curl and $\ddpl{}{t}$ can be interchanged: \begin{equation*} \FLPcurl{\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}}= \frac{1}{c^2}\,\frac{\partial^2}{\partial t^2}(\FLPcurl{\FLPA})= \frac{1}{c^2}\,\frac{\partial^2\FLPB}{\partial t^2}. \end{equation*}  Using these results, we get the following differential equation for $\FLPB$: \begin{equation} \label{Eq:II:20:10} \nabla^2\FLPB- \frac{1}{c^2}\,\frac{\partial^2\FLPB}{\partial t^2}=\FLPzero. \end{equation}  So each component of the magnetic field $\FLPB$ satisfies the three-dimensional wave equation. Similarly, using the fact that $\FLPE=-\FLPgrad{\phi}-\ddpl{\FLPA}{t}$, it follows that the electric field $\FLPE$ in free space also satisfies the three-dimensional wave equation: \begin{equation} \label{Eq:II:20:11} \nabla^2\FLPE- \frac{1}{c^2}\,\frac{\partial^2\FLPE}{\partial t^2}=\FLPzero. \end{equation}  All of our electromagnetic fields satisfy the same wave equation, Eq. (20.8). We might well ask: What is the most general solution to this equation? However, rather than tackling that difficult question right away, we will look first at what can be said in general about those solutions in which nothing varies in $y$ and $z$. (Always do an easy case first so that you can see what is going to happen, and then you can go to the more complicated cases.) Let’s suppose that the magnitudes of the fields depend only upon $x$—that there are no variations of the fields with $y$ and $z$. We are, of course, considering plane waves again. We should expect to get results something like those in the previous section. In fact, we will find precisely the same answers. You may ask: “Why do it all over again?” It is important to do it again, first, because we did not show that the waves we found were the most general solutions for plane waves, and second, because we found the fields only from a very particular kind of current source. We would like to ask now: What is the most general kind of one-dimensional wave there can be in free space? We cannot find that by seeing what happens for this or that particular source, but must work with greater generality. Also we are going to work this time with differential equations instead of with integral forms. Although we will get the same results, it is a way of practicing back and forth to show that it doesn’t make any difference which way you go. You should know how to do things every which way, because when you get a hard problem, you will often find that only one of the various ways is tractable. We could consider directly the solution of the wave equation for some electromagnetic quantity. Instead, we want to start right from the beginning with Maxwell’s equations in free space so that you can see their close relationship to the electromagnetic waves. So we start with the equations in (20.1), setting the charges and currents equal to zero. They become \begin{equation} \begin{alignedat}{3} &\text{I.}&&\FLPdiv{\FLPE}&\;=&\;0\\[1ex] &\text{II.}&&\FLPcurl{\FLPE}&\;=&\;-\ddp{\FLPB}{t}\\[1ex] &\text{III.}&&\FLPdiv{\FLPB}&\;=&\;0\\[1ex] &\text{IV.}&\quad c^2&\FLPcurl{\FLPB}&\;=&\;\ddp{\FLPE}{t} \end{alignedat} \label{Eq:II:20:12} \end{equation}  We write the first equation out in components: \begin{equation} \label{Eq:II:20:13} \FLPdiv{\FLPE}=\ddp{E_x}{x}+\ddp{E_y}{y}+\ddp{E_z}{z}=0. \end{equation}  We are assuming that there are no variations with $y$ and $z$, so the last two terms are zero. This equation then tells us that \begin{equation} \label{Eq:II:20:14} \ddp{E_x}{x}=0. \end{equation}  Its solution is that $E_x$, the component of the electric field in the $x$-direction, is a constant in space. If you look at IV in (20.12), supposing no $\FLPB$-variation in $y$ and $z$ either, you can see that $E_x$ is also constant in time. Such a field could be the steady dc field from some charged condenser plates a long distance away. We are not interested now in such an uninteresting static field; we are at the moment interested only in dynamically varying fields. For dynamic fields, $E_x=0$. We have then the important result that for the propagation of plane waves in any direction, the electric field must be at right angles to the direction of propagation. It can, of course, still vary in a complicated way with the coordinate $x$. The transverse $\FLPE$-field can always be resolved into two components, say the $y$-component and the $z$-component. So let’s first work out a case in which the electric field has only one transverse component. We’ll take first an electric field that is always in the $y$-direction, with zero $z$-component. Evidently, if we solve this problem we can also solve for the case where the electric field is always in the $z$-direction. The general solution can always be expressed as the superposition of two such fields.  How easy our equations now get. The only component of the electric field that is not zero is $E_y$, and all derivatives—except those with respect to $x$—are zero. The rest of Maxwell’s equations then become quite simple. Let’s look next at the second of Maxwell’s equations [II of Eq. (20.12)]. Writing out the components of the curl $\FLPE$, we have \begin{alignat*}{4} &(\FLPcurl{\FLPE})_x&&=\ddp{E_z}{y}&&-\ddp{E_y}{z}&&=0,\\[1.5ex] &(\FLPcurl{\FLPE})_y&&=\ddp{E_x}{z}&&-\ddp{E_z}{x}&&=0,\\[1.5ex] &(\FLPcurl{\FLPE})_z&&=\ddp{E_y}{x}&&-\ddp{E_x}{y}&&=\ddp{E_y}{x}. \end{alignat*}  The $x$-component of $\FLPcurl{\FLPE}$ is zero because the derivatives with respect to $y$ and $z$ are zero. The $y$-component is also zero; the first term is zero because the derivative with respect to $z$ is zero, and the second term is zero because $E_z$ is zero. The only components of the curl of $\FLPE$ that is not zero is the $z$-component, which is equal to $\ddpl{E_y}{x}$. Setting the three components of $\FLPcurl{\FLPE}$ equal to the corresponding components of $-\ddpl{\FLPB}{t}$, we can conclude the following: \begin{align} \label{Eq:II:20:15} &\ddp{B_x}{t}=0,\quad\ddp{B_y}{t}=0.\\[1ex] \label{Eq:II:20:16} &\ddp{B_z}{t}=-\ddp{E_y}{x}. \end{align}  Since the $x$-component of the magnetic field and the $y$-component of the magnetic field both have zero time derivatives, these two components are just constant fields and correspond to the magnetostatic solutions we found earlier. Somebody may have left some permanent magnets near where the waves are propagating. We will ignore these constant fields and set $B_x$ and $B_y$ equal to zero.  Incidentally, we would already have concluded that the $x$-component of $\FLPB$ should be zero for a different reason. Since the divergence of $\FLPB$ is zero (from the third Maxwell equation), applying the same arguments we used above for the electric field, we would conclude that the longitudinal component of the magnetic field can have no variation with $x$. Since we are ignoring such uniform fields in our wave solutions, we would have set $B_x$ equal to zero. In plane electromagnetic waves the $\FLPB$-field, as well as the $\FLPE$-field, must be directed at right angles to the direction of propagation. Equation (20.16) gives us the additional proposition that if the electric field has only a $y$-component, the magnetic field will have only a $z$-component. So $\FLPE$ and $\FLPB$ are at right angles to each other. This is exactly what happened in the special wave we have already considered. We are now ready to use the last of Maxwell’s equations for free space [IV of Eq. (20.12)]. Writing out the components, we have \begin{equation} \begin{alignedat}{4} &c^2(\FLPcurl{\FLPB})_x&&= c^2\,\ddp{B_z}{y}&&-c^2\,\ddp{B_y}{z}&&=\ddp{E_x}{t},\\[1ex] &c^2(\FLPcurl{\FLPB})_y&&= c^2\,\ddp{B_x}{z}&&-c^2\,\ddp{B_z}{x}&&=\ddp{E_y}{t},\\[1ex] &c^2(\FLPcurl{\FLPB})_z&&= c^2\,\ddp{B_y}{x}&&-c^2\,\ddp{B_x}{y}&&=\ddp{E_z}{t}. \end{alignedat} \label{Eq:II:20:17} \end{equation}  Of the six derivatives of the components of $\FLPB$, only the term $\ddpl{B_z}{x}$ is not equal to zero. So the three equations give us simply \begin{equation} \label{Eq:II:20:18} -c^2\,\ddp{B_z}{x}=\ddp{E_y}{t}. \end{equation}  The result of all our work is that only one component each of the electric and magnetic fields is not zero, and that these components must satisfy Eqs. (20.16) and (20.18). The two equations can be combined into one if we differentiate the first with respect to $x$ and the second with respect to $t$; the left-hand sides of the two equations will then be the same (except for the factor $c^2$). So we find that $E_y$ satisfies the equation \begin{equation} \label{Eq:II:20:19} \frac{\partial^2E_y}{\partial x^2}-\frac{1}{c^2}\,\frac{\partial^2E_y}{\partial t^2}=0. \end{equation}  We have seen the same differential equation before, when we studied the propagation of sound. It is the wave equation for one-dimensional waves. You should note that in the process of our derivation we have found something more than is contained in Eq. (20.11). Maxwell’s equations have given us the further information that electromagnetic waves have field components only at right angles to the direction of the wave propagation. Let’s review what we know about the solutions of the one-dimensional wave equation. If any quantity $\psi$ satisfies the one-dimensional wave equation \begin{equation} \label{Eq:II:20:20} \frac{\partial^2\psi}{\partial x^2}-\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0, \end{equation}  then one possible solution is a function $\psi(x,t)$ of the form \begin{equation} \label{Eq:II:20:21} \psi(x,t)=f(x-ct), \end{equation}  that is, some function of the single variable $(x-ct)$. The function $f(x-ct)$ represents a “rigid” pattern in $x$ which travels toward positive $x$ at the speed $c$ (see Fig. 20–4). For example, if the function $f$ has a maximum when its argument is zero, then for $t=0$ the maximum of $\psi$, will occur at $x=0$. At some later time, say $t=10$, $\psi$ will have its maximum at $x=10c$. As time goes on, the maximum moves toward positive $x$ at the speed $c$. Sometimes it is more convenient to say that a solution of the one-dimensional wave equation is a function of $(t-x/c)$. However, this is saying the same thing, because any function of $(t-x/c)$ is also a function of $(x-ct)$: \begin{equation*} F(t-x/c)=F\biggl[-\frac{x-ct}{c}\biggr]=f(x-ct). \end{equation*}  Let’s show that $f(x-ct)$ is indeed a solution of the wave equation. Since it is a function of only one variable—the variable $(x-ct)$—we will let $f'$ represent the derivative of $f$ with respect to its variable and $f''$ represent the second derivative of $f$. Differentiating Eq. (20.21) with respect to $x$, we have \begin{equation*} \ddp{\psi}{x}=f'(x-ct), \end{equation*}  since the derivative of $(x-ct)$ with respect to $x$ is $1$. The second derivative of $\psi$, with respect to $x$ is clearly \begin{equation} \label{Eq:II:20:22} \frac{\partial^2\psi}{\partial x^2}=f''(x-ct). \end{equation}  Taking derivatives of $\psi$ with respect to $t$, we find \begin{align} &\ddp{\psi}{t}=f'(x-ct)(-c),\notag\\[1.5ex] \label{Eq:II:20:23} &\frac{\partial^2\psi}{\partial t^2}=+c^2f''(x-ct). \end{align}  We see that $\psi$ does indeed satisfy the one-dimensional wave equation. You may be wondering: “If I have the wave equation, how do I know that I should take $f(x-ct)$ as a solution? I don’t like this backward method. Isn’t there some forward way to find the solution?” Well, one good forward way is to know the solution. It is possible to “cook up” an apparently forward mathematical argument, especially because we know what the solution is supposed to be, but with an equation as simple as this we don’t have to play games. Soon you will get so that when you see Eq. (20.20), you nearly simultaneously see $\psi=f(x-ct)$ as a solution. (Just as now when you see the integral of $x^2\,dx$, you know right away that the answer is $x^3/3$.) Actually you should also see a little more. Not only is any function of $(x-ct)$ a solution, but any function of $(x+ct)$ is also a solution. Since the wave equation contains only $c^2$, changing the sign of $c$ makes no difference. In fact, the most general solution of the one-dimensional wave equation is the sum of two arbitrary functions, one of $(x-ct)$ and the other of $(x+ct)$: \begin{equation} \label{Eq:II:20:24} \psi=f(x-ct)+g(x+ct). \end{equation}  The first term represents a wave travelling toward positive $x$, and the second term an arbitrary wave travelling toward negative $x$. The general solution is the superposition of two such waves both existing at the same time. We will leave the following amusing question for you to think about. Take a function $\psi$ of the following form: \begin{equation*} \psi=\cos kx\cos kct. \end{equation*}  This equation isn’t in the form of a function of $(x-ct)$ or of $(x+ct)$. Yet you can easily show that this function is a solution of the wave equation by direct substitution into Eq. (20.20). How can we then say that the general solution is of the form of Eq. (20.24)? Applying our conclusions about the solution of the wave equation to the $y$-component of the electric field, $E_y$, we conclude that $E_y$ can vary with $x$ in any arbitrary fashion. However, the fields which do exist can always be considered as the sum of two patterns. One wave is sailing through space in one direction with speed $c$, with an associated magnetic field perpendicular to the electric field; another wave is travelling in the opposite direction with the same speed. Such waves correspond to the electromagnetic waves that we know about—light, radiowaves, infrared radiation, ultraviolet radiation, x-rays, and so on. We have already discussed the radiation of light in great detail in Vol. I. Since everything we learned there applies to any electromagnetic wave, we don’t need to consider in great detail here the behavior of these waves. We should perhaps make a few further remarks on the question of the polarization of the electromagnetic waves. In our solution we chose to consider the special case in which the electric field has only a $y$-component. There is clearly another solution for waves travelling in the plus or minus $x$-direction, with an electric field which has only a $z$-component. Since Maxwell’s equations are linear, the general solution for one-dimensional waves propagating in the $x$-direction is the sum of waves of $E_y$ and waves of $E_z$. This general solution is summarized in the following equations: \begin{equation} \begin{aligned} \FLPE&=(0,E_y,E_z)\\[.5ex] E_y&=f(x-ct)+g(x+ct)\\[.5ex] E_z&=F(x-ct)+G(x+ct)\\[1ex] \FLPB&=(0,B_y,B_z)\\[.5ex] cB_z&=f(x-ct)-g(x+ct)\\[.5ex] cB_y&=-F(x-ct)+G(x+ct). \end{aligned} \label{Eq:II:20:25} \end{equation}  Such electromagnetic waves have an $\FLPE$-vector whose direction is not constant but which gyrates around in some arbitrary way in the $yz$-plane. At every point the magnetic field is always perpendicular to the electric field and to the direction of propagation. If there are only waves travelling in one direction, say the positive $x$-direction, there is a simple rule which tells the relative orientation of the electric and magnetic fields. The rule is that the cross product $\FLPE\times\FLPB$—which is, of course, a vector at right angles to both $\FLPE$ and $\FLPB$—points in the direction in which the wave is travelling. If $\FLPE$ is rotated into $\FLPB$ by a right-hand screw, the screw points in the direction of the wave velocity. (We shall see later that the vector $\FLPE\times\FLPB$ has a special physical significance: it is a vector which describes the flow of energy in an electromagnetic field.)