The fields of an oscillating dipole
We have still not lived up to our promise to derive Eq. (21.1) for the electric field of a point charge in motion. Even with the results we already have, it is a relatively complicated thing to derive. We have not found Eq. (21.1) anywhere in the published literature except in Vol. I of these lectures.1 So you can see that it is not easy to derive. (The fields of a moving charge have been written in many other forms that are equivalent, of course.) We will have to limit ourselves here just to showing that, in a few examples, Eqs. (21.15) and (21.16) give the same results as Eq. (21.1). First, we will show that Eq. (21.1) gives the correct fields with only the restriction that the motion of the charged particle is nonrelativistic. (Just this special case will take care of $90$ percent, or more, of what we said about light.)  We consider a situation in which we have a blob of charge that is moving about in some way, in a small region, and we will find the fields far away. To put it another way, we are finding the field at any distance from a point charge that is shaking up and down in very small motion. Since light is usually emitted from neutral objects such as atoms, we will consider that our wiggling charge $q$ is located near an equal and opposite charge at rest. If the separation between the centers of the charges is $\FLPd$, the charges will have a dipole moment $\FLPp=q\FLPd$, which we take to be a function of time. Now we should expect that if we look at the fields close to the charges, we won’t have to worry about the delay; the electric field will be exactly the same as the one we have calculated earlier for an electrostatic dipole—using, of course, the instantaneous dipole moment $\FLPp(t)$. But if we go very far out, we ought to find a term in the field that goes as $1/r$ and depends on the acceleration of the charge perpendicular to the line of sight. Let’s see if we get such a result.  We begin by calculating the vector potential $\FLPA$, using Eq. (21.16). Suppose that our moving charge is in a small blob whose charge density is given by $\rho(x,y,z)$, and the whole thing is moving at any instant with the velocity $\FLPv$. Then the current density $\FLPj(x,y,z)$ will be equal to $\FLPv\rho(x,y,z)$. It will be convenient to take our coordinate system so that the $z$-axis is in the direction of $\FLPv$; then the geometry of our problem is as shown in Fig. 21-2. We want the integral \begin{equation} \label{Eq:II:21:17} \int\frac{\FLPj(2,t-r_{12}/c)}{r_{12}}\,dV_2. \end{equation}  Now if the size of the charge-blob is really very small compared with $r_{12}$, we can set the $r_{12}$ term in the denominator equal to $r$, the distance to the center of the blob, and take $r$ outside the integral. Next, we are also going to set $r_{12}=r$ in the numerator, although that is not really quite right. It is not right because we should take $\FLPj$ at, say, the top of the blob at a slightly different time than we used for $\FLPj$ at the bottom of the blob. When we set $r_{12}=r$ in $\FLPj(t-r_{12}/c)$, we are taking the current density for the whole blob at the same time $(t-r/c)$. That is an approximation that will be good only if the velocity $v$ of the charge is much less than $c$. So we are making a nonrelativistic calculation. Replacing $\FLPj$ by $\rho\FLPv$, the integral (21.17) becomes \begin{equation*} \frac{1}{r}\int\FLPv\rho(2,t-r/c)\,dV_2. \end{equation*} Since all the charge has the same velocity, this integral is just $\FLPv/r$ times the total charge $q$. But $q\FLPv$ is just $\ddpl{\FLPp}{t}$, the rate of change of the dipole moment—which is, of course, to be evaluated at the retarded time $(t-r/c)$. We will write it as $\dot{\FLPp}(t-r/c)$. So we get for the vector potential \begin{equation} \label{Eq:II:21:18} \FLPA(1,t)=\frac{1}{4\pi\epsO c^2}\,\frac{\dot{\FLPp}(t-r/c)}{r}. \end{equation}  Our result says that the current in a varying dipole produces a vector potential in the form of spherical waves whose source strength is $\dot{\FLPp}/\epsO c^2$.  We can now get the magnetic field from $\FLPB=\FLPcurl{\FLPA}$. Since $\dot{\FLPp}$ is totally in the $z$-direction, $\FLPA$ has only a $z$-component; there are only two nonzero derivatives in the curl. So $B_x=\ddpl{A_z}{y}$ and $B_y=-\ddpl{A_z}{x}$. Let’s first look at $B_x$: \begin{equation} \label{Eq:II:21:19} B_x=\ddp{A_z}{y}=\frac{1}{4\pi\epsO c^2}\,\ddp{}{y}\, \frac{\dot{p}(t-r/c)}{r}. \end{equation} To carry out the differentiation, we must remember that $r=\sqrt{x^2+y^2+z^2}$, so  \begin{equation} \label{Eq:II:21:20} B_x=\frac{1}{4\pi\epsO c^2}\,\dot{p}(t-r/c)\,\ddp{}{y}\, \biggl(\frac{1}{r}\biggr)+\frac{1}{4\pi\epsO c^2}\,\frac{1}{r}\, \ddp{}{y}\,\dot{p}(t-r/c). \end{equation} 
 \begin{align} B_x=\frac{1}{4\pi\epsO c^2}\,\dot{p}(t-r/c)\,&\ddp{}{y}\, \biggl(\frac{1}{r}\biggr)\,+\notag\\[1ex] \label{Eq:II:21:20} \frac{1}{4\pi\epsO c^2}\,\frac{1}{r}\, &\ddp{}{y}\,\dot{p}(t-r/c). \end{align}  Remembering that $\ddpl{r}{y}=y/r$, the first term gives \begin{equation} \label{Eq:II:21:21} -\frac{1}{4\pi\epsO c^2}\,\frac{y\dot{p}(t-r/c)}{r^3}, \end{equation} which drops off as $1/r^2$ like the potential of a static dipole (because $y/r$ is constant for a given direction).  The second term in Eq. (21.20) gives us the new effects. Carrying out the differentiation, we get \begin{equation} \label{Eq:II:21:22} -\frac{1}{4\pi\epsO c^2}\,\frac{y}{cr^2}\,\ddot{p}(t-r/c), \end{equation} where $\ddot{p}$ means, of course, the second derivative of $p$ with respect to $t$. This term, which comes from differentiating the numerator, is responsible for radiation. First, it describes a field which decreases with distance only as $1/r$. Second, it depends on the acceleration of the charge. You can begin to see how we are going to get a result like Eq. (21.1´), which describes the radiation of light.  Let’s examine in a little more detail how this radiation term comes about—it is such an interesting and important result. We start with the expression (21.18), which has a $1/r$ dependence and is therefore like a Coulomb potential, except for the delay term in the numerator. Why is it then that when we differentiate with respect to space coordinates to get the fields, we don’t just get a $1/r^2$ field—with, of course, the corresponding time delays?  We can see why in the following way: Suppose that we let our dipole oscillate up and down in a sinusoidal motion. Then we would have \begin{equation*} p=p_z=p_0\sin\omega t \end{equation*} and \begin{equation*} A_z=\frac{1}{4\pi\epsO c^2}\,\frac{\omega p_0\cos\omega(t-r/c)}{r}. \end{equation*} If we plot a graph of $A_z$ as a function of $r$ at a given instant, we get the curve shown in Fig. 21-3. The peak amplitude decreases as $1/r$, but there is, in addition, an oscillation in space, bounded by the $1/r$ envelope. When we take the spatial derivatives, they will be proportional to the slope of the curve. From the figure we see that there are slopes much steeper than the slope of the $1/r$ curve itself. It is, in fact, evident that for a given frequency the peak slopes are proportional to the amplitude of the wave, which varies as $1/r$. So that explains the drop-off rate of the radiation term.  It all comes about because the variations with time at the source are translated into variations in space as the waves are propagated outward, and the magnetic fields depend on the spatial derivatives of the potential.  Let’s go back and finish our calculation of the magnetic field. We have for $B_x$ the two terms (21.21) and (21.22), so \begin{equation*} B_x=\frac{1}{4\pi\epsO c^2}\biggl[ -\frac{y\dot{p}(t-r/c)}{r^3}-\frac{y\ddot{p}(t-r/c)}{cr^2} \biggr]. \end{equation*} With the same kind of mathematics, we get \begin{equation*} B_y=\frac{1}{4\pi\epsO c^2}\biggl[ \frac{x\dot{p}(t-r/c)}{r^3}+\frac{x\ddot{p}(t-r/c)}{cr^2} \biggr]. \end{equation*} Or we can put it all together in a nice vector formula: \begin{equation} \label{Eq:II:21:23} \FLPB=\frac{1}{4\pi\epsO c^2}\, \frac{[\dot{\FLPp}+(r/c)\ddot{\FLPp}]_{t-r/c}\times\FLPr}{r^3}. \end{equation}  Now let’s look at this formula. First of all, if we go very far out in $r$, only the $\ddot{\FLPp}$ term counts. The direction of $\FLPB$ is given by $\ddot{\FLPp}\times\FLPr$, which is at right angles to the radius $\FLPr$ and also at right angles to the acceleration, as in Fig. 21-4. Everything is coming out right; that is also the result we get from Eq. (21.1´).  Now let’s look at what we are not used to—at what happens closer in. In Section 14-7 we worked out the law of Biot and Savart for the magnetic field of an element of current. We found that a current element $\FLPj\,dV$ contributes to the magnetic field the amount \begin{equation} \label{Eq:II:21:24} d\FLPB=\frac{1}{4\pi\epsO c^2}\, \frac{\FLPj\times\FLPr}{r^3}\,dV. \end{equation} You see that this formula looks very much like the first term of Eq. (21.23), if we remember that $\dot{\FLPp}$ is the current. But there is one difference. In Eq. (21.23), the current is to be evaluated at the time $(t-r/c)$, which doesn’t appear in Eq. (21.24). Actually, however, Eq. (21.24) is still very good for small $r$, because the second term of Eq. (21.23) tends to cancel out the effect of the retardation in the first term. The two together give a result very near to Eq. (21.24) when $r$ is small.  We can see that this way: When $r$ is small, $(t-r/c)$ is not very different from $t$, so we can expand the bracket in Eq. (21.23) in a Taylor series. For the first term, \begin{equation*} \dot{\FLPp}(t-r/c)=\dot{\FLPp}(t)-\frac{r}{c}\,\ddot{\FLPp}(t)+\text{etc.}, \end{equation*} and to the same order in $r/c$, \begin{equation*} \frac{r}{c}\,\ddot{\FLPp}(t-r/c)=\frac{r}{c}\,\ddot{\FLPp}(t)+\text{etc.} \end{equation*} When we take the sum, the two terms in $\ddot{\FLPp}$ cancel, and we are left with the unretarded current $\dot{\FLPp}$: that is, $\dot{\FLPp}(t)$—plus terms of order $(r/c)^2$ or higher [e.g., $\tfrac{1}{2}(r/c)^2\dddot{\FLPp}\,$] which will be very small for $r$ small enough that $\dot{\FLPp}$ does not alter markedly in the time $r/c$.  So Eq. (21.23) gives fields very much like the instantaneous theory—much closer than the instantaneous theory with a delay; the first-order effects of the delay are taken out by the second term. The static formulas are very accurate, much more accurate than you might think. Of course, the compensation only works for points close in. For points far out the correction becomes very bad, because the time delays produce a very large effect, and we get the important $1/r$ term of the radiation.  We still have the problem of computing the electric field and demonstrating that it is the same as Eq. (21.1´). For large distances we can see that the answer is going to come out all right. We know that far from the sources, where we have a propagating wave, $\FLPE$ is perpendicular to $\FLPB$ (and also to $\FLPr$), as in Fig. 21-4, and that $cB=E$. So $\FLPE$ is proportional to the acceleration $\ddot{\FLPp}$, as expected from Eq. (21.1´).  To get the electric field completely for all distances, we need to solve for the electrostatic potential. When we computed the current integral for $\FLPA$ to get Eq. (21.18), we made an approximation by disregarding the slight variation of $r$ in the delay terms. This will not work for the electrostatic potential, because we would then get $1/r$ times the integral of the charge density, which is a constant. This approximation is too rough. We need to go to one higher order. Instead of getting involved in that higher-order computation directly, we can do something else—we can determine the scalar potential from Eq. (21.6), using the vector potential we have already found. The divergence of $\FLPA$, in our case, is just $\ddpl{A_z}{z}$—since $A_x$ and $A_y$ are identically zero. Differentiating in the same way that we did above to find $\FLPB$, \begin{align*} \FLPdiv{\FLPA}&=\frac{1}{4\pi\epsO c^2}\biggl[ \dot{p}(t-r/c)\,\ddp{}{z}\,\biggl(\frac{1}{r}\biggr)+ \frac{1}{r}\,\ddp{}{z}\,\dot{p}(t-r/c) \biggr]\\[2ex] &=\frac{1}{4\pi\epsO c^2}\biggl[ -\frac{z\dot{p}(t-r/c)}{r^3}-\frac{z\ddot{p}(t-r/c)}{cr^2} \biggr]. \end{align*} Or, in vector notation, \begin{equation*} \FLPdiv{\FLPA}=-\frac{1}{4\pi\epsO c^2}\, \frac{[\dot{\FLPp}+(r/c)\ddot{\FLPp}]_{t-r/c}\cdot\FLPr}{r^3}. \end{equation*} Using Eq. (21.6), we have an equation for $\phi$: \begin{equation*} \ddp{\phi}{t}=\frac{1}{4\pi\epsO}\, \frac{[\dot{\FLPp}+(r/c)\ddot{\FLPp}]_{t-r/c}\cdot\FLPr}{r^3}. \end{equation*}  Integrating with respect to $t$ just removes one dot from each of the $\FLPp$'s, so \begin{equation} \label{Eq:II:21:25} \phi(\FLPr,t)=\frac{1}{4\pi\epsO}\, \frac{[\FLPp+(r/c)\dot{\FLPp}]_{t-r/c}\cdot\FLPr}{r^3}. \end{equation} (The constant of integration would correspond to some superposed static field which could, of course, exist. For the oscillating dipole we have taken, there is no static field.)  We are now able to find the electric field $\FLPE$ from \begin{equation*} \FLPE=-\FLPgrad{\phi}-\ddp{\FLPA}{t}. \end{equation*} Since the steps are tedious but straightforward [providing you remember that $\FLPp(t-r/c)$ and its time derivatives depend on $x$, $y$, and $z$ through the retardation $r/c$], we will just give the result:  \begin{equation} \label{Eq:II:21:26} \FLPE(\FLPr,t)=\frac{1}{4\pi\epsO r^3}\biggl[ \frac{3(\FLPp\stared\cdot\FLPr)\FLPr}{r^2}- \FLPp\stared+\frac{1}{c^2} \{\ddot{\FLPp}(t-r/c)\times\FLPr\}\times\FLPr \biggr] \end{equation} 
  \begin{align} \FLPE(\FLPr,t)=\frac{1}{4\pi\epsO r^3}\biggl[ &\frac{3(\FLPp\stared\cdot\FLPr)\FLPr}{r^2}- \FLPp\stared+\notag\\ \label{Eq:II:21:26} &\frac{1}{c^2} \{\ddot{\FLPp}(t-r/c)\times\FLPr\}\times\FLPr \biggr] \end{align}   with \begin{equation} \label{Eq:II:21:27} \FLPp\stared=\FLPp(t-r/c)+\frac{r}{c}\,\dot{\FLPp}(t-r/c). \end{equation}  Although it looks rather complicated, the result is easily interpreted. The vector $\FLPp\stared$ is the dipole moment retarded and then “corrected” for the retardation, so the two terms with $\FLPp\stared$ give just the static dipole field when $r$ is small. [See Chapter 6, Eq. (6.14).]  When $r$ is large, the term in $\ddot{\FLPp}$ dominates, and the electric field is proportional to the acceleration of the charges, at right angles to $\FLPr$, and, in fact, directed along the projection of $\ddot{\FLPp}$ in a plane perpendicular to $\FLPr$.  This result agrees with what we would have gotten using Eq. (21.1). Of course, Eq. (21.1) is more general; it works with any motion, while Eq. (21.26) is valid only for small motions for which we can take the retardation $r/c$ as constant over the source. At any rate, we have now provided the underpinnings for our entire previous discussion of light (excepting some matters discussed in Chapter 34 of Vol. I), for it all hinged on the last term of Eq. (21.26). We will discuss next how the fields can be obtained for more rapidly moving charges (leading to the relativistic effects of Chapter 34 of Vol. I). 