The boundary conditions
All we have done so far is to describe the three waves; our problem now is to work out the parameters of the reflected and transmitted waves in terms of those of the incident wave. How can we do that? The three waves we have described satisfy Maxwell’s equations in the uniform material, but Maxwell’s equations must also be satisfied at the boundary between the two different materials. So we must now look at what happens right at the boundary. We will find that Maxwell’s equations demand that the three waves fit together in a certain way.  As an example of what we mean, the $y$-component of the electric field $\FLPE$ must be the same on both sides of the boundary. This is required by Faraday’s law, \begin{equation} \label{Eq:II:33:19} \FLPcurl{\FLPE}=-\ddp{\FLPB}{t}, \end{equation}  as we can see in the following way. Consider a little rectangular loop $\Gamma$ which straddles the boundary, as shown in Fig. 33–4. Equation (33.19) says that the line integral of $\FLPE$ around $\Gamma$ is equal to the rate of change of the flux of $\FLPB$ through the loop: \begin{equation*} \oint_\Gamma\FLPE\cdot d\FLPs=-\ddp{}{t}\int\FLPB\cdot\FLPn\,da. \end{equation*}  Now imagine that the rectangle is very narrow, so that the loop encloses an infinitesimal area. If $\FLPB$ remains finite (and there’s no reason it should be infinite at the boundary!) the flux through the area is zero. So the line integral of $\FLPE$ must be zero. If $E_{y1}$ and $E_{y2}$ are the components of the field on the two sides of the boundary and if the length of the rectangle is $l$, we have \begin{equation} E_{y1}l-E_{y2}l=0\notag \end{equation}  or \begin{equation} \label{Eq:II:33:20} E_{y1}=E_{y2}, \end{equation}  as we have said. This gives us one relation among the fields of the three waves. The procedure of working out the consequences of Maxwell’s equations at the boundary is called “determining the boundary conditions.” Ordinarily, it is done by finding as many equations like Eq. (33.20) as one can, by making arguments about little rectangles like $\Gamma$ in Fig. 33–4, or by using little Gaussian surfaces that straddle the boundary. Although that is a perfectly good way of proceeding, it gives the impression that the problem of dealing with a boundary is different for every different physical problem. For example, in a problem of heat flow across a boundary, how are the temperatures on the two sides related? Well, you could argue, for one thing, that the heat flow to the boundary from one side would have to equal the flow away from the other side. It is usually possible, and generally quite useful, to work out the boundary conditions by making such physical arguments. There may be times, however, when in working on some problem you have only some equations, and you may not see right away what physical arguments to use. So although we are at the moment interested only in an electromagnetic problem, where we can make the physical arguments, we want to show you a method that can be used for any problem—a general way of finding what happens at a boundary directly from the differential equations. We begin by writing all the Maxwell equations for a dielectric—and this time we are very specific and write out explicitly all the components: \begin{align} \label{Eq:II:33:21} &\FLPdiv{\FLPE}=-\frac{\FLPdiv{\FLPP}}{\epsO}\\[1ex] &\quad \epsO\biggl(\ddp{E_x}{x}\!+\!\ddp{E_y}{y}\!+\!\ddp{E_z}{z}\biggr)= -\biggl(\ddp{P_x}{x}\!+\!\ddp{P_y}{y}\!+\!\ddp{P_z}{z}\biggr)\notag\\[2ex] % ebook break % ebook insert:   \label{Eq:II:0:0} &\FLPcurl{\FLPE}=-\ddp{\FLPB}{t}\notag\\[1ex] % ebook break % ebook indent \label{Eq:II:33:22a} &\quad \ddp{E_z}{y}-\ddp{E_y}{z}=-\ddp{B_x}{t}\tag{33.22a}\\[.75ex] \label{Eq:II:33:22b} &\quad \ddp{E_x}{z}-\ddp{E_z}{x}=-\ddp{B_y}{t}\tag{33.22b}\\[.75ex] \label{Eq:II:33:22c} &\quad \ddp{E_y}{x}-\ddp{E_x}{y}=-\ddp{B_z}{t}\tag{33.22c}\\[2ex] % ebook break \label{Eq:II:33:23} &\FLPdiv{\FLPB}=0\tag{33.23}\\[1.75ex] &\quad \ddp{B_x}{x}+\ddp{B_y}{y}+\ddp{B_z}{z}=0\notag\\[2ex] % ebook break % ebook insert: \label{Eq:II:0:0} &c^2\FLPcurl{\FLPB}=\frac{1}{\epsO}\,\ddp{\FLPP}{t}+\ddp{\FLPE}{t}\notag\\[1ex] % ebook break % ebook indent \label{Eq:II:33:24a} &\quad c^2\biggl(\ddp{B_z}{y}-\ddp{B_y}{z}\biggr)= \frac{1}{\epsO}\,\ddp{P_x}{t}+\ddp{E_x}{t}\tag{33.24a}\\[.75ex] \label{Eq:II:33:24b} &\quad c^2\biggl(\ddp{B_x}{z}-\ddp{B_z}{x}\biggr)= \frac{1}{\epsO}\,\ddp{P_y}{t}+\ddp{E_y}{t}\tag{33.24b}\\[.75ex] \label{Eq:II:33:24c} &\quad c^2\biggl(\ddp{B_y}{x}-\ddp{B_x}{y}\biggr)= \frac{1}{\epsO}\,\ddp{P_z}{t}+\ddp{E_z}{t}\tag{33.24c} \end{align}  Now these equations must all hold in region $1$ (to the left of the boundary) and in region $2$ (to the right of the boundary). We have already written the solutions in regions $1$ and $2$. Finally, they must also be satisfied in the boundary, which we can call region $3$. Although we usually think of the boundary as being sharply discontinuous, in reality it is not. The physical properties change very rapidly but not infinitely fast. In any case, we can imagine that there is a very rapid, but continuous, transition of the index between region $1$ and $2$, in a short distance we can call region $3$. Also, any field quantity like $P_x$, or $E_y$, etc., will make a similar kind of transition in region $3$. In this region, the differential equations must still be satisfied, and it is by following the differential equations in this region that we can arrive at the needed “boundary conditions.” For instance, suppose that we have a boundary between vacuum (region $1$) and glass (region $2$). There is nothing to polarize in the vacuum, so $\FLPP_1=0$. Let’s say there is some polarization $\FLPP_2$ in the glass. Between the vacuum and the glass there is a smooth, but rapid, transition. If we look at any component of $\FLPP$, say $P_x$, it might vary as drawn in Fig. 33–5(a). Suppose now we take the first of our equations, Eq. (33.21). It involves derivatives of the components of $\FLPP$ with respect to $x$, $y$, and $z$. The $y$- and $z$-derivatives are not interesting; nothing spectacular is happening in those directions. But the $x$-derivative of $P_x$ will have some very large values in region $3$, because of the tremendous slope of $P_x$. The derivative $\ddpl{P_x}{x}$ will have a sharp spike at the boundary, as shown in Fig. 33–5(b). If we imagine squashing the boundary to an even thinner layer, the spike would get much higher. If the boundary is really sharp for the waves we are interested in, the magnitude of $\ddpl{P_x}{x}$ in region $3$ will be much, much greater than any contributions we might have from the variation of $\FLPP$ in the wave away from the boundary—so we ignore any variations other than those due to the boundary. Now how can Eq. (33.21) be satisfied if there is a whopping big spike on the right-hand side? Only if there is an equally whopping big spike on the other side. Something on the left-hand side must also be big. The only candidate is $\ddpl{E_x}{x}$, because the variations with $y$ and $z$ are only those small effects in the wave we just mentioned. So $-\epsO(\ddpl{E_x}{x})$ must be as drawn in Fig. 33–5(c)—just a copy of $\ddpl{P_x}{x}$. We have that \begin{equation*} \epsO\,\ddp{E_x}{x}=-\ddp{P_x}{x}. \end{equation*}  If we integrate this equation with respect to $x$ across region $3$, we conclude that \begin{equation} \label{Eq:II:33:25} \epsO(E_{x2}-E_{x1})=-(P_{x2}-P_{x1}). \end{equation}  In other words, the jump in $\epsO E_x$ in going from region $1$ to region $2$ must be equal to the jump in $-P_x$. We can rewrite Eq. (33.25) as \begin{equation} \label{Eq:II:33:26} \epsO E_{x2}+P_{x2}=\epsO E_{x1}+P_{x1}, \end{equation}  which says that the quantity $(\epsO E_x+P_x)$ has equal values in region $2$ and region $1$. People say: the quantity $(\epsO E_x+P_x)$ is continuous across the boundary. We have, in this way, one of our boundary conditions. Although we took as an illustration the case in which $\FLPP_1$ was zero because region $1$ was a vacuum, it is clear that the same argument applies for any two materials in the two regions, so Eq. (33.26) is true in general. Let’s now go through the rest of Maxwell’s equations and see what each of them tells us. We take next Eq. (33.22a). There are no $x$-derivatives, so it doesn’t tell us anything. (Remember that the fields themselves do not get especially large at the boundary; only the derivatives with respect to $x$ can become so huge that they dominate the equation.) Next, we look at Eq. (33.22b). Ah! There is an $x$-derivative! We have $\ddpl{E_z}{x}$ on the left-hand side. Suppose it has a huge derivative. But wait a moment! There is nothing on the right-hand side to match it with; therefore $E_z$ cannot have any jump in going from region $1$ to region $2$. [If it did, there would be a spike on the left of Eq. (33.22b) but none on the right, and the equation would be false.] So we have a new condition: \begin{equation} \label{Eq:II:33:27} E_{z2}=E_{z1}. \end{equation}  By the same argument, Eq. (33.22c) gives \begin{equation} \label{Eq:II:33:28} E_{y2}=E_{y1}. \end{equation}  This last result is just what we got in Eq. (33.20) by a line integral argument. We go on to Eq. (33.23). The only term that could have a spike is $\ddpl{B_x}{x}$. But there’s nothing on the right to match it, so we conclude that \begin{equation} \label{Eq:II:33:29} B_{x2}=B_{x1}. \end{equation}  On to the last of Maxwell’s equations! Equation (33.24a) gives nothing, because there are no $x$-derivatives. Equation (33.24b) has one, $-c^2\,\ddpl{B_z}{x}$, but again, there is nothing to match it with. We get \begin{equation} \label{Eq:II:33:30} B_{z2}=B_{z1}. \end{equation}  The last equation is quite similar, and gives \begin{equation} \label{Eq:II:33:31} B_{y2}=B_{y1}. \end{equation}  The last three equations gives us that $\FLPB_2=\FLPB_1$. We want to emphasize, however, that we get this result only when the materials on both sides of the boundary are nonmagnetic—or rather, when we can neglect any magnetic effects of the materials. This can usually be done for most materials, except ferromagnetic ones. (We will treat the magnetic properties of materials in some later chapters.) Our program has netted us the six relations between the fields in region $1$ and those in region $2$. We have put them all together in Table 33–1. We can now use them to match the waves in the two regions. We want to emphasize, however, that the idea we have just used will work in any physical situation in which you have differential equations and you want a solution that crosses a sharp boundary between two regions where some property changes. For our present purposes, we could have easily derived the same equations by using arguments about the fluxes and circulations at the boundary. (You might see whether you can get the same result that way.) But now you have seen a method that will work in case you ever get stuck and don’t see any easy argument about the physics of what is happening at the boundary—you can just work with the equations.