Spherical waves
We have seen that there are solutions of the wave equation which correspond to plane waves, and that any electromagnetic wave can be described as a superposition of many plane waves. In certain special cases, however, it is more convenient to describe the wave field in a different mathematical form. We would like to discuss now the theory of spherical waves—waves which correspond to spherical surfaces that are spreading out from some center. When you drop a stone into a lake, the ripples spread out in circular waves on the surface—they are two-dimensional waves. A spherical wave is a similar thing except that it spreads out in three dimensions. Before we start describing spherical waves, we need a little mathematics. Suppose we have a function that depends only on the radial distance $r$ from a certain origin—in other words, a function that is spherically symmetric. Let’s call the function $\psi(r)$, where by $r$ we mean \begin{equation*} r=\sqrt{x^2+y^2+z^2}, \end{equation*}  the radial distance from the origin. In order to find out what functions $\psi(r)$ satisfy the wave equation, we will need an expression for the Laplacian of $\psi$. So we want to find the sum of the second derivatives of $\psi$ with respect to $x$, $y$, and $z$. We will use the notation that $\psi'(r)$ represents the derivative of $\psi$ with respect to $r$ and $\psi''(r)$ represents the second derivative of $\psi$ with respect to $r$. First, we find the derivatives with respect to $x$. The first derivative is \begin{equation*} \ddp{\psi(r)}{x}=\psi'(r)\,\ddp{r}{x}. \end{equation*}  The second derivative of $\psi$ with respect to $x$ is \begin{equation*} \frac{\partial^2\psi}{\partial x^2}= \psi''\biggl(\ddp{r}{x}\biggr)^2+ \psi'\,\frac{\partial^2r}{\partial x^2}. \end{equation*}  We can evaluate the partial derivatives of $r$ with respect to $x$ from \begin{equation*} \ddp{r}{x}=\frac{x}{r},\quad \frac{\partial^2r}{\partial x^2}= \frac{1}{r}\biggl(1-\frac{x^2}{r^2}\biggr). \end{equation*}  So the second derivative of $\psi$ with respect to $x$ is \begin{equation} \label{Eq:II:20:28} \frac{\partial^2\psi}{\partial x^2}=\frac{x^2}{r^2}\psi''+\frac{1}{r}\biggl(1-\frac{x^2}{r^2}\biggr)\psi'. \end{equation} Likewise, \begin{alignat}{3} \label{Eq:II:20:29} \frac{\partial^2\psi}{\partial y^2}&= \frac{y^2}{r^2}\,&&\psi''+ \frac{1}{r}\biggl(1-\frac{y^2}{r^2}&&\biggr)\psi',\\[1.5ex] \label{Eq:II:20:30} \frac{\partial^2\psi}{\partial z^2}&= \frac{z^2}{r^2}\,&&\psi''+ \frac{1}{r}\biggl(1-\frac{z^2}{r^2}&&\biggr)\psi'. \end{alignat}  The Laplacian is the sum of these three derivatives. Remembering that $x^2+y^2+z^2=r^2$, we get \begin{equation} \label{Eq:II:20:31} \nabla^2\psi(r)=\psi''(r)+\frac{2}{r}\,\psi'(r). \end{equation}  It is often more convenient to write this equation in the following form: \begin{equation} \label{Eq:II:20:32} \nabla^2\psi(r)=\frac{1}{r}\,\frac{d^2}{dr^2}(r\psi). \end{equation}  If you carry out the differentiation indicated in Eq. (20.32), you will see that the right-hand side is the same as in Eq. (20.31). If we wish to consider spherically symmetric fields which can propagate as spherical waves, our field quantity must be a function of both $r$ and $t$. Suppose we ask, then, what functions $\psi(r,t)$ are solutions of the three-dimensional wave equation \begin{equation} \label{Eq:II:20:33} \nabla^2\psi(r,t)-\frac{1}{c^2}\, \frac{\partial^2}{\partial t^2}\,\psi(r,t)=0. \end{equation}  Since $\psi(r,t)$ depends only on the spatial coordinates through $r$, we can use the equation for the Laplacian we found above, Eq. (20.32). To be precise, however, since $\psi$ is also a function of $t$, we should write the derivatives with respect to $r$ as partial derivatives. Then the wave equation becomes \begin{equation*} \frac{1}{r}\,\frac{\partial^2}{\partial r^2}\,(r\psi)- \frac{1}{c^2}\,\frac{\partial^2}{\partial t^2}\,\psi=0. \end{equation*}   We must now solve this equation, which appears to be much more complicated than the plane wave case. But notice that if we multiply this equation by $r$, we get \begin{equation} \label{Eq:II:20:34} \frac{\partial^2}{\partial r^2}\,(r\psi)- \frac{1}{c^2}\,\frac{\partial^2}{\partial t^2}\,(r\psi)=0. \end{equation}  This equation tells us that the function $r\psi$ satisfies the one-dimensional wave equation in the variable $r$. Using the general principle which we have emphasized so often, that the same equations always have the same solutions, we know that if $r\psi$ is a function only of $(r-ct)$ then it will be a solution of Eq. (20.34). So we know that spherical waves must have the form \begin{equation*} r\psi(r,t)=f(r-ct). \end{equation*}  Or, as we have seen before, we can equally well say that $r\psi$ can have the form \begin{equation*} r\psi=f(t-r/c). \end{equation*}  Dividing by $r$, we find that the field quantity $\psi$ (whatever it may be) has the following form: \begin{equation} \label{Eq:II:20:35} \psi=\frac{f(t-r/c)}{r}. \end{equation}  Such a function represents a general spherical wave travelling outward from the origin at the speed $c$. If we forget about the $r$ in the denominator for a moment, the amplitude of the wave as a function of the distance from the origin at a given time has a certain shape that travels outward at the speed $c$. The factor $r$ in the denominator, however, says that the amplitude of the wave decreases in proportion to $1/r$ as the wave propagates. In other words, unlike a plane wave in which the amplitude remains constant as the wave runs along, in a spherical wave the amplitude steadily decreases, as shown in Fig. 20–6. This effect is easy to understand from a simple physical argument. We know that the energy density in a wave depends on the square of the wave amplitude. As the wave spreads, its energy is spread over larger and larger areas proportional to the radial distance squared. If the total energy is conserved, the energy density must fall as $1/r^2$, and the amplitude of the wave must decrease as $1/r$. So Eq. (20.35) is the “reasonable” form for a spherical wave. We have disregarded the second possible solution to the one-dimensional wave equation: \begin{equation*} r\psi=g(t+r/c), \end{equation*}  or \begin{equation*} \psi=\frac{g(t+r/c)}{r}. \end{equation*}  This also represents a spherical wave, but one which travels inward from large $r$ toward the origin. We are now going to make a special assumption. We say, without any demonstration whatever, that the waves generated by a source are only the waves which go outward. Since we know that waves are caused by the motion of charges, we want to think that the waves proceed outward from the charges. It would be rather strange to imagine that before charges were set in motion, a spherical wave started out from infinity and arrived at the charges just at the time they began to move. That is a possible solution, but experience shows that when charges are accelerated the waves travel outward from the charges. Although Maxwell’s equations would allow either possibility, we will put in an additional fact—based on experience—that only the outgoing wave solution makes “physical sense.” We should remark, however, that there is an interesting consequence to this additional assumption: we are removing the symmetry with respect to time that exists in Maxwell’s equations. The original equations for $\FLPE$ and $\FLPB$, and also the wave equations we derived from them, have the property that if we change the sign of $t$, the equation is unchanged. These equations say that for every solution corresponding to a wave going in one direction there is an equally valid solution for a wave travelling in the opposite direction. Our statement that we will consider only the outgoing spherical waves is an important additional assumption. (A formulation of electrodynamics in which this additional assumption is avoided has been carefully studied. Surprisingly, in many circumstances it does not lead to physically absurd conclusions, but it would take us too far astray to discuss these ideas just now. We will talk about them a little more in Chapter 28.) We must mention another important point. In our solution for an outgoing wave, Eq. (20.35), the function $\psi$ is infinite at the origin. That is somewhat peculiar. We would like to have a wave solution which is smooth everywhere. Our solution must represent physically a situation in which there is some source at the origin. In other words, we have inadvertently made a mistake. We have not solved the free wave equation (20.33) everywhere; we have solved Eq. (20.33) with zero on the right everywhere, except at the origin. Our mistake crept in because some of the steps in our derivation are not “legal” when $r=0$. Let’s show that it is easy to make the same kind of mistake in an electrostatic problem. Suppose we want a solution of the equation for an electrostatic potential in free space, $\nabla^2\phi=0$. The Laplacian is equal to zero, because we are assuming that there are no charges anywhere. But what about a spherically symmetric solution to this equation—that is, some function $\phi$ that depends only on $r$. Using the formula of Eq. (20.32) for the Laplacian, we have \begin{equation*} \frac{1}{r}\,\frac{d^2}{dr^2}\,(r\phi)=0. \end{equation*}  Multiplying this equation by $r$, we have an equation which is readily integrated: \begin{equation*} \frac{d^2}{dr^2}\,(r\phi)=0. \end{equation*}  If we integrate once with respect to $r$, we find that the first derivative of $r\phi$ is a constant, which we may call $a$: \begin{equation*} \ddt{}{r}\,(r\phi)=a. \end{equation*}  Integrating again, we find that $r\phi$ is of the form \begin{equation*} r\phi=ar+b, \end{equation*}  where $b$ is another constant of integration. So we have found that the following $\phi$ is a solution for the electrostatic potential in free space: \begin{equation*} \phi=a+\frac{b}{r}. \end{equation*}  Something is evidently wrong. In the region where there are no electric charges, we know the solution for the electrostatic potential: the potential is everywhere a constant. That corresponds to the first term in our solution. But we also have the second term, which says that there is a contribution to the potential that varies as one over the distance from the origin. We know, however, that such a potential corresponds to a point charge at the origin. So, although we thought we were solving for the potential in free space, our solution also gives the field for a point source at the origin. Do you see the similarity between what happened now and what happened when we solved for a spherically symmetric solution to the wave equation? If there were really no charges or currents at the origin, there would not be spherical outgoing waves. The spherical waves must, of course, be produced by sources at the origin. In the next chapter we will investigate the connection between the outgoing electromagnetic waves and the currents and voltages which produce them.