Spontaneous magnetization
We now turn to the question of why it is that in ferromagnetic materials a small magnetic field produces such a large magnetization. The magnetization of ferromagnetic materials like iron and nickel comes from the magnetic moment of the electrons in the inner shell of the atom. Each electron has a magnetic moment $\FLPmu$ equal to $q/2m$ times its $g$-factor, times its angular momentum $\FLPJ$. For a single electron with no net orbital motion, $g=2$, and the component of $\FLPJ$ in any direction—say the $z$-direction—is $\pm\hbar/2$, so the component of $\FLPmu$ along the $z$-axis is \begin{equation} \label{Eq:II:36:28} \mu_z=\frac{q\hbar}{2m}=0.928\times10^{-23}\text{ amp$\cdot$m$^2$}. \end{equation}  In an iron atom, there are actually two electrons that contribute to the ferromagnetism, so to keep the discussion simpler we will talk about nickel, which is ferromagnetic like iron but which has only one electron in the inner shell. (It is easy to extend the arguments to iron.)  Now the point is that in the presence of an external field $\FLPB$, the atomic magnets tend to line up with the field, but are knocked about by thermal motions just as we described for paramagnetic materials. In the last chapter we found out that the balance between a magnetic field trying to line up the atomic magnets and the thermal motions trying to derange them produced the result that the mean magnetic moment per unit volume will end up as \begin{equation} \label{Eq:II:36:29} M=N\mu\tanh\frac{\mu B_a}{kT}. \end{equation}  By $\FLPB_a$ we mean the field acting at the atom, and $kT$ is the Boltzmann energy. In the theory of paramagnetism we used for $B_a$ just $B$ itself, neglecting the part of the field at any given atom contributed by the atoms nearby. In the ferromagnetic case, there is a complication. We shouldn’t use the average field in the iron for the $\FLPB_a$ acting on an individual atom. Instead, we must do as we did in the case of dielectrics—we have to find the local field acting at a single atom. For an exact calculation we should add up the fields at the atom in question contributed by all of the other atoms in the crystal lattice. But as we did for dielectrics, we will make the approximation that the field at an atom is the same as we would find in a small spherical hole in the material—assuming that the moments of the atoms in the neighborhood are not changed by the presence of the hole. Following the arguments we made in Chapter 11, we might think that we could write \begin{equation*} \FLPB_{\text{hole}}=\FLPB+\frac{1}{3}\,\frac{\FLPM}{\epsO c^2}\quad (\text{wrong!}). \end{equation*}  But that is not right. We can, however, make use of the results of Chapter 11 if we make a careful comparison of the equations of Chapter 11 with the equations for ferromagnetism in this chapter. Let’s put together the corresponding equations. For regions where there are no conduction currents or charges we have:  \begin{equation} \begin{alignedat}{2} &\text{Electrostatics}&\qquad &\text{Static ferromagnetism}\\ &\FLPdiv{\biggl(\FLPE+\frac{\FLPP}{\epsO}\biggr)}=0&\qquad &\FLPdiv{\FLPB}=0\\ &\FLPcurl{\FLPE}=\FLPzero&\qquad &\FLPcurl{\biggl(\FLPB-\frac{\FLPM}{\epsO c^2}\biggr)}=\FLPzero \end{alignedat} \label{Eq:II:36:30} \end{equation} 
 \begin{equation} \begin{aligned} &\text{Electrostatics}\\[.5ex] &\FLPdiv{\biggl(\FLPE+\frac{\FLPP}{\epsO}\biggr)}=0\\[1ex] &\FLPcurl{\FLPE}=\FLPzero\\[2ex] &\text{Static ferromagnetism}\\[1ex] &\FLPdiv{\FLPB}=0\\[1ex] &\FLPcurl{\biggl(\FLPB-\frac{\FLPM}{\epsO c^2}\biggr)}=\FLPzero \end{aligned} \label{Eq:II:36:30} \end{equation}   These two sets of equations can be thought of as analogous if we make the following purely mathematical correspondences: \begin{equation*} \FLPE\to\FLPB-\frac{\FLPM}{\epsO c^2},\quad \FLPE+\frac{\FLPP}{\epsO}\to\FLPB. \end{equation*}  This is the same as making the analogy \begin{equation} \label{Eq:II:36:31} \FLPE\to\FLPH,\quad\FLPP\to\FLPM/c^2. \end{equation}  In other words, if we write the equations of ferromagnetism as \begin{equation} \begin{aligned} &\FLPdiv{\biggl(\FLPH+\frac{\FLPM}{\epsO c^2}\biggr)}=0,\\[1ex] &\FLPcurl{\FLPH}=\FLPzero, \end{aligned} \label{Eq:II:36:32} \end{equation}  they look like the equations of electrostatics. This purely algebraic correspondence has led to some confusion in the past. People tended to think that $\FLPH$ was “the magnetic field.” But, as we have seen, $\FLPB$ and $\FLPE$ are physically the fundamental fields, and $\FLPH$ is a derived idea. So although the equations are analogous, the physics is not analogous. However, that doesn’t need to stop us from using the principle that the same equations have the same solutions.  We can use our earlier results for the electric field inside of holes of various shapes in dielectrics—summarized in Fig. 36–1—to find the field $\FLPH$ inside of corresponding holes. Knowing $\FLPH$, we can determine $\FLPB$. For instance (using the results we summarized in Section 36–1), the field $\FLPH$ in a needle-shaped hole parallel to $\FLPM$ is the same as the $\FLPH$ in the material, \begin{equation*} \FLPH_{\text{hole}}=\FLPH_{\text{material}}. \end{equation*}  But since $\FLPM$ in the hole is zero, we have \begin{equation} \label{Eq:II:36:33} \FLPB_{\text{hole}}=\FLPB_{\text{material}}-\frac{\FLPM}{\epsO c^2}. \end{equation}  On the other hand, for a disc-shaped hole, perpendicular to $\FLPM$, we have \begin{equation} \FLPE_{\text{hole}}=\FLPE_{\text{dielectric}}+\frac{\FLPP}{\epsO},\notag \end{equation}  which translates into \begin{equation} \FLPH_{\text{hole}}=\FLPH_{\text{material}}+\frac{\FLPM}{\epsO c^2}.\notag \end{equation}  Or, in terms of $\FLPB$, \begin{equation} \label{Eq:II:36:34} \FLPB_{\text{hole}}=\FLPB_{\text{material}}. \end{equation}  Finally, for a spherical hole, by making our analogy with Eq. (36.3) we would have \begin{equation} \FLPH_{\text{hole}}=\FLPH_{\text{material}}+\frac{\FLPM}{3\epsO c^2}\notag \end{equation}  or \begin{equation} \label{Eq:II:36:35} \FLPB_{\text{hole}}=\FLPB_{\text{material}}-\frac{2}{3}\, \frac{\FLPM}{\epsO c^2}. \end{equation}  This result is quite different from what we got for $\FLPE$. It is, of course, possible to get these results in a more physical way, by using the Maxwell equations directly. For example, Eq. (36.34) follows directly from $\FLPdiv{\FLPB}=0$. (You use a Gaussian surface that is half in the material and half out.) Similarly, you can get Eq. (36.33) by using a line integral along a curve that goes up inside the hole and returns through the material. Physically, the field in the hole is reduced because of the surface currents—which are given by $\FLPcurl{\FLPM}$. We will leave it for you to show that Eq. (36.35) can also be obtained by considering the effects of the surface currents on the boundary of the spherical cavity.  In finding the equilibrium magnetization from Eq. (36.29), it turns out to be most convenient to deal with $\FLPH$; so write \begin{equation} \label{Eq:II:36:36} \FLPB_a=\FLPH+\lambda\,\frac{\FLPM}{\epsO c^2}. \end{equation}  In the spherical hole approximation, we would have $\lambda=\tfrac{1}{3}$, but, as you will see, we will want later to use some other value, so we leave it as an adjustable parameter. Also, we will take all the fields in the same direction so that we won’t need to worry about the vector directions. If we were now to substitute Eq. (36.36) into Eq. (36.29), we would have one equation that relates the magnetization $M$ to the magnetizing field $H$: \begin{equation*} M=N\mu\tanh\biggl(\mu\,\frac{H+\lambda M/\epsO c^2}{kT}\biggr). \end{equation*}  It is however, an equation that cannot be solved explicitly, so we will do it graphically. Let’s put the problem in a generalized form by writing Eq. (36.29) as \begin{equation} \label{Eq:II:36:37} \frac{M}{M_{\text{sat}}}=\tanh x, \end{equation}  where $M_{\text{sat}}$ is the saturation value of the magnetization, namely, $N\mu$, and $x$ represents $\mu B_a/kT$. The dependence of $M/M_{\text{sat}}$ on $x$ is shown by curve $a$ in Fig. 36–13. We can also write $x$ as a function of $M$—using Eq. (36.36) for $B_a$—as \begin{equation} \label{Eq:II:36:38} x=\frac{\mu B_a}{kT}=\frac{\mu H}{kT}+\biggl(\frac{\mu\lambda M_{\text{sat}}} {\epsO c^2kT}\biggr)\frac{M}{M_{\text{sat}}}. \end{equation}  For any given value of $H$, this is a straight-line relationship between $M/M_{\text{sat}}$ and $x$. The $x$ intercept is at $x=\mu H/kT$, and the slope is $\epsO c^2kT/\mu\lambda M_{\text{sat}}$. For any particular $H$, we would have a line like the one marked $b$ in Fig. 36–13. The intersection of curves $a$ and $b$ gives us the solution for $M/M_{\text{sat}}$. We have solved the problem. Let’s look at how the solutions will go for various circumstances. We start with $H=0$. There are two possible situations, shown by the lines $b_1$ and $b_2$ in Fig. 36–14. You will notice from Eq. (36.38) that the slope of the line is proportional to the absolute temperature $T$. So, at high temperatures we would have a line like $b_1$. The solution is $M/M_{\text{sat}}=0$. When the magnetizing field $H$ is zero, the magnetization is also zero. But at low temperatures, we would have a line like $b_2$, and there are two solutions for $M/M_{\text{sat}}$—one with $M/M_{\text{sat}}=0$ and one with $M/M_{\text{sat}}$ near one. It turns out that only the upper solution is stable—as you can see by considering small variations about these solutions. According to these ideas, then, a magnetic material should magnetize itself spontaneously at sufficiently low temperatures. In short, when the thermal motions are small enough, the coupling between the atomic magnets causes them all to line up parallel to each other—we have a permanently magnetized material analogous to the ferroelectrics we discussed in Chapter 11. If we start at high temperatures and come down, there is a critical temperature, called the Curie temperature $T_c$, where the ferromagnetic behavior suddenly sets in. This temperature corresponds to the line $b_3$ of Fig. 36–14, which is tangent to the curve $a$, and has, therefore, a slope of $1$. The Curie temperature is given by \begin{equation} \label{Eq:II:36:39} \frac{\epsO c^2kT_c}{\mu\lambda M_{\text{sat}}}=1. \end{equation}  We can, if we wish, write Eq. (36.38) more simply in terms of $T_c$ as \begin{equation} \label{Eq:II:36:40} x=\frac{\mu H}{kT}+\frac{T_c}{T}\biggl(\frac{M}{M_{\text{sat}}}\biggr). \end{equation}  Now we want to see what happens for small magnetizing fields $H$. We can see from Fig. 36–14 how things will go if we shift our straight lines a little to the right. For the low-temperature case, the intersection point will move out a little bit along the low-slope part of curve $a$, and $M$ will change relatively little. For the high-temperature case, however, the intersection point runs up the steep part of curve $a$, and $M$ will change relatively rapidly. In fact, we can approximate this part of curve $a$ by a straight line of unit slope, and write: \begin{equation*} \frac{M}{M_{\text{sat}}}=x=\frac{\mu H}{kT}+ \frac{T_c}{T}\biggl(\frac{M}{M_{\text{sat}}}\biggr). \end{equation*}  Now we can solve for $M/M_{\text{sat}}$: \begin{equation} \label{Eq:II:36:41} \frac{M}{M_{\text{sat}}}=\frac{\mu H}{k(T-T_c)}. \end{equation}  We have a law that is something like the one we had for paramagnetism. For paramagnetism, we had \begin{equation} \label{Eq:II:36:42} \frac{M}{M_{\text{sat}}}=\frac{\mu B}{kT}. \end{equation}  One difference now is that we have the magnetization in terms of $H$, which includes some of the effects of the interaction of the atomic magnets, but the main difference is that the magnetization is inversely proportional to the difference between $T$ and $T_c$, instead of to the absolute temperate $T$, alone. Neglecting the interactions between neighboring atoms corresponds to taking $\lambda=0$, which from Eq. (36.39) means taking $T_c=0$. Then the results are just what we had in Chapter 35. We can check our theoretical picture with the experimental data for nickel. It is observed experimentally that the ferromagnetic behavior of nickel disappears when its temperature is raised above $631^\circ$K. We can compare this with $T_c$ calculated from Eq. (36.39). Remembering that $M_{\text{sat}}=\mu N$, we have \begin{equation*} T_c=\lambda\,\frac{N\mu^2}{k\epsO c^2}. \end{equation*}  From the density and atomic weight of nickel, we get \begin{equation*} N=9.1\times10^{28}\:\text{m}^{-3}. \end{equation*}  Taking $\mu$ from Eq. (36.28), and setting $\lambda=\tfrac{1}{3}$, we get \begin{equation*} T_c=0.24^\circ\text{K}. \end{equation*}  There is a discrepancy of a factor of about $2600$! Our theory of ferromagnetism fails completely. We can try to “patch up” the theory as Weiss did by saying that for some unknown reason $\lambda$ is not one-third, but $(2600)\times\tfrac{1}{3}$—or about $900$. It turns out that one gets similar values for other ferromagnetic materials like iron. To see what this means, let’s go back to Eq. (36.36). We see that a large $\lambda$ means that $B_a$, the local field on the atom, appears to be much, much larger than we would think. In fact, writing $H=B-M/\epsO c^2$, we have \begin{equation*} B_a=B+\frac{(\lambda-1)M}{\epsO c^2}. \end{equation*}  According to our original idea—with $\lambda=\tfrac{1}{3}$ the local magnetization $M$ reduces the effective field $B_a$ by the amount $-\tfrac{2}{3}M/\epsO c^2$. Even if our model of a spherical hole were not very good, we would still expect some reduction. Instead, to explain the phenomenon of ferromagnetism, we have to imagine that the magnetization of the field enhances the local field by some large factor—like one thousand or more. There doesn’t seem to be any reasonable way to manufacture such tremendous fields at an atom—nor even fields of the proper sign! Clearly, our “magnetic” theory of ferromagnetism is a dismal failure. We must conclude, then, that ferromagnetism has to do with some nonmagnetic interaction between the spinning electrons in neighboring atoms. This interaction must generate a strong tendency for all of the nearby spins to line up in one direction. We will see later that it has to do with quantum mechanics and the Pauli exclusion principle.  Finally, we look at what happens at low temperatures—for $T<T_c$. We have seen that there will then be a spontaneous magnetization—even with $H=0$—given by the intersection of the curves $a$ and $b_2$ of Fig. 36–14. If we solve for $M$ for various temperatures—by varying the slope of the line $b_2$—we get the theoretical curve shown in Fig. 36–15. This curve should be the same for all ferromagnetic materials for which the atomic moment comes from a single electron. The curves for other materials are only slightly different. In the limit, as $T$ goes to absolute zero, $M$ goes to $M_{\text{sat}}$. As the temperature is increased, the magnetization decreases, falling to zero at the Curie temperature. The points in Fig. 36–15 are the experimental observations for nickel. They fit the theoretical curve fairly well. Even though we don’t understand the basic mechanism, the general features of the theory seem to be correct. Finally, there is one more disturbing discrepancy in our attempt to understand ferromagnetism. We have found that above some temperature the material should behave like a paramagnetic substance with a magnetization $M$ proportional to $H$ (or $B$), and that below that temperature it should become spontaneously magnetized. But that’s not what we found when we measured the magnetization curve for iron. It only became permanently magnetized after we had “magnetized” it. According to the ideas just discussed, it would magnetize itself! What is wrong? Well, it turns out that if you look at a small enough crystal of iron or nickel, it is indeed completely magnetized! But in large pieces of iron, there are many small regions or “domains” that are magnetized in different directions, so that on a large scale the average magnetization appears to be zero. In each small domain, however, the iron has a locked-in magnetization with $M$ nearly equal to $M_{\text{sat}}$. The consequences of this domain structure are that gross properties of large pieces of material are quite different from the microscopic properties that we have really been treating. We will take up in the next lecture the story of the practical behavior of bulk magnetic materials.