The nuclear force field
We would like to make some further remarks about the part of the mass of nuclear particles that is not electromagnetic. Where does this other large fraction come from? There are other forces besides electrodynamics—like nuclear forces—that have their own field theories, although no one knows whether the current theories are right. These theories also predict a field energy which gives the nuclear particles a mass term analogous to electromagnetic mass; we could call it the “$\pi$-mesic-field-mass.” It is presumably very large, because the forces are great, and it is the possible origin of the mass of the heavy particles. But the meson field theories are still in a most rudimentary state. Even with the well-developed theory of electromagnetism, we found it impossible to get beyond first base in explaining the electron mass. With the theory of the mesons, we strike out. We may take a moment to outline the theory of the mesons, because of its interesting connection with electrodynamics. In electrodynamics, the field can be described in terms of a four-potential that satisfies the equation \begin{equation*} \Box^2A_\mu=\text{sources}. \end{equation*}  Now we have seen that pieces of the field can be radiated away so that they exist separated from the sources. These are the photons of light, and they are described by a differential equation without sources: \begin{equation*} \Box^2A_\mu=0. \end{equation*}  People have argued that the field of nuclear forces ought also to have its own “photons”—they would presumably be the $\pi$-mesons—and that they should be described by an analogous differential equation. (Because of the weakness of the human brain, we can’t think of something really new; so we argue by analogy with what we know.) So the meson equation might be \begin{equation*} \Box^2\phi=0, \end{equation*}  where $\phi$ could be a different four-vector or perhaps a scalar. It turns out that the pion has no polarization, so $\phi$ should be a scalar. With the simple equation $\Box^2\phi=0$, the meson field would vary with distance from a source as $1/r^2$, just as the electric field does. But we know that nuclear forces have much shorter distances of action, so the simple equation won’t work. There is one way we can change things without disrupting the relativistic invariance: we can add or subtract from the d’Alembertian a constant, times $\phi$. So Yukawa suggested that the free quanta of the nuclear force field might obey the equation \begin{equation} \label{Eq:II:28:17} -\Box^2\phi-\mu^2\phi=0, \end{equation}  where $\mu^2$ is a constant—that is, an invariant scalar. (Since $\Box^2$ is a scalar differential operator in four dimensions, its invariance is unchanged if we add another scalar to it.) Let’s see what Eq. (28.17) gives for the nuclear force when things are not changing with time. We want a spherically symmetric solution of \begin{equation*} \nabla^2\phi-\mu^2\phi=0 \end{equation*}  around some point source at, say, the origin. If $\phi$ depends only on $r$, we know that \begin{equation*} \nabla^2\phi=\frac{1}{r}\,\frac{\partial^2}{\partial r^2} (r\phi). \end{equation*}  So we have the equation \begin{equation*} \frac{1}{r}\,\frac{\partial^2}{\partial r^2} (r\phi)-\mu^2\phi=0 \end{equation*}  or \begin{equation*} \frac{\partial^2}{\partial r^2}(r\phi)=\mu^2(r\phi). \end{equation*}  Thinking of $(r\phi)$ as our dependent variable, this is an equation we have seen many times. Its solution is \begin{equation*} r\phi=Ke^{\pm\mu r}. \end{equation*}  Clearly, $\phi$ cannot become infinite for large $r$, so the $+$ sign in the exponent is ruled out. The solution is \begin{equation} \label{Eq:II:28:18} \phi=K\,\frac{e^{-\mu r}}{r}. \end{equation}  This function is called the Yukawa potential. For an attractive force, $K$ is a negative number whose magnitude must be adjusted to fit the experimentally observed strength of the forces. The Yukawa potential of the nuclear forces dies off more rapidly than $1/r$ by the exponential factor. The potential—and therefore the force—falls to zero much more rapidly than $1/r$ for distances beyond $1/\mu$, as shown in Fig. 28–6. The “range” of nuclear forces is much less than the “range” of electrostatic forces. It is found experimentally that the nuclear forces do not extend beyond about $10^{-13}$ cm, so $\mu\approx10^{15}$ m$^{-1}$. Finally, let’s look at the free-wave solution of Eq. (28.17). If we substitute \begin{equation*} \phi=\phi_0e^{i(\omega t-kz)} \end{equation*}  into Eq. (28.17), we get that \begin{equation*} \frac{\omega^2}{c^2}-k^2-\mu^2=0. \end{equation*}  Relating frequency to energy and wave number to momentum, as we did at the end of Chapter 34 of Vol. I, we get that \begin{equation*} \frac{E^2}{c^2}-p^2=\mu^2\hbar^2, \end{equation*}  which says that the Yukawa “photon” has a mass equal to $\mu\hbar/c$. If we use for $\mu$ the estimate $10^{15}$ m$^{-1}$, which gives the observed range of the nuclear forces, the mass comes out to $3\times10^{-25}$ g, or $170$ MeV, which is roughly the observed mass of the $\pi$-meson. So, by an analogy with electrodynamics, we would say that the $\pi$-meson is the “photon” of the nuclear force field. But now we have pushed the ideas of electrodynamics into regions where they may not really be valid—we have gone beyond electrodynamics to the problem of the nuclear forces.