The rate of radiation of energy
Now we shall calculate the total energy radiated by an accelerating charge. To keep the discussion general, we shall take the case of a charge accelerating any which way, but nonrelativistically. At a moment when the acceleration is, say, vertical, we know that the electric field that is generated is the charge multiplied by the projection of the retarded acceleration, divided by the distance. So we know the electric field at any point, and we therefore know the square of the electric field and thus the energy $\epsO cE^2$ leaving through a unit area per second. The quantity $\epsO c$ appears quite often in expressions involving radiowave propagation. Its reciprocal is called the impedance of a vacuum, and it is an easy number to remember: it has the value $1/\epsO c = 377$ ohms. So the power in watts per square meter is equal to the average of the field squared, divided by $377$. Using our expression (29.1) for the electric field, we find that \begin{equation} \label{Eq:I:32:2} S=\frac{q^2a'^2\sin^2\theta}{16\pi^2\epsO r^2c^3} \end{equation} is the power per square meter radiated in the direction $\theta$. We notice that it goes inversely as the square of the distance, as we said before. Now suppose we wanted the total energy radiated in all directions: then we must integrate (32.2) over all directions. First we multiply by the area, to find the amount that flows within a little angle $d\theta$ (Fig. 32–1). We need the area of a spherical section. The way to think of it is this: if $r$ is the radius, then the width of the annular segment is $r\,d\theta$, and the circumference is $2\pi r \sin \theta$, because $r\sin\theta$ is the radius of the circle. So the area of the little piece of the sphere is $2\pi r \sin \theta$ times $r\,d\theta$: \begin{equation} \label{Eq:I:32:3} dA=2\pi r^2\sin\theta\,d\theta. \end{equation} By multiplying the flux [(32.2), the power per square meter] by the area in square meters included in the small angle $d\theta$, we find the amount of energy that is liberated in this direction between $\theta$ and $\theta + d\theta$; then we integrate that over all the angles $\theta$ from $0$ to $180^\circ$: \begin{equation} \label{Eq:I:32:4} P=\int S\,dA=\frac{q^2a'^2}{8\pi\epsO c^3} \int_0^\pi\sin^3\theta\,d\theta. \end{equation} By writing $\sin^3 \theta=(1 - \cos^2\theta)\sin\theta$ it is not hard to show that $\int_0^\pi\sin^3\theta\,d\theta=4/3$. Using that fact, we finally get \begin{equation} \label{Eq:I:32:5} P=\frac{q^2a'^2}{6\pi\epsO c^3}. \end{equation} This expression deserves some remarks. First of all, since the vector $\FLPa'$ had a certain direction, the $a'^2$ in (32.5) would be the square of the vector $\FLPa'$, that is, $\FLPa'\cdot\FLPa'$, the length of the vector, squared. Secondly, the flux (32.2) was calculated using the retarded acceleration; that is, the acceleration at the time at which the energy now passing through the sphere was radiated. We might like to say that this energy was in fact liberated at this earlier time. This is not exactly true; it is only an approximate idea. The exact time when the energy is liberated cannot be defined precisely. All we can really calculate precisely is what happens in a complete motion, like an oscillation or something, where the acceleration finally ceases. Then what we find is that the total energy flux per cycle is the average of acceleration squared, for a complete cycle. This is what should really appear in (32.5). Or, if it is a motion with an acceleration that is initially and finally zero, then the total energy that has flown out is the time integral of (32.5). To illustrate the consequences of formula (32.5) when we have an oscillating system, let us see what happens if the displacement $x$ of the charge is oscillating so that the acceleration $a$ is $-\omega^2x_0e^{i\omega t}$. The average of the acceleration squared over a cycle (remember that we have to be very careful when we square things that are written in complex notation—it really is the cosine, and the average of $\cos^2\omega t$ is one-half) thus is \begin{equation} \avg{a'^2} = \tfrac{1}{2}\omega^4x_0^2.\notag \end{equation} Therefore \begin{equation} \label{Eq:I:32:6} P=\frac{q^2\omega^4x_0^2}{12\pi\epsO c^3}. \end{equation} The formulas we are now discussing are relatively advanced and more or less modern; they date from the beginning of the twentieth century, and they are very famous. Because of their historical value, it is important for us to be able to read about them in older books. In fact, the older books also used a system of units different from our present mks system. However, all these complications can be straightened out in the final formulas dealing with electrons by the following rule: The quantity $q_e^2/4\pi\epsO$, where $q_e$ is the electronic charge (in coulombs), has, historically, been written as $e^2$. It is very easy to calculate that $e$ in the mks system is numerically equal to $1.5188\times10^{-14}$, because we know that, numerically, $q_e = 1.60206\times10^{-19}$ and $1/4\pi\epsO = 8.98748\times10^9$. Therefore we shall often use the convenient abbreviation \begin{equation} \label{Eq:I:32:7} e^2=\frac{q_e^2}{4\pi\epsO}. \end{equation} If we use the above numerical value of $e$ in the older formulas and treat them as though they were written in mks units, we will get the right numerical results. For example, the older form of (32.5) is $P = \tfrac{2}{3}e^2a'^2/c^3$. Again, the potential energy of a proton and an electron at distance $r$ is $q_e^2/4\pi\epsO r$ or $e^2/r$, with $e = 1.5188\times10^{-14}$ (mks).