GROWTH OF MAGNETIZATION REVERSAL NUCLEI DURING LARGE BARKHAUSEN JUMPS

V. Dering

For a complete theoretical understanding of many details of the experiments of Sixtus and Tonks1, it is necessary to obtain a more precise picture of the behavior of the boundary between two oppositely magnetized regions. This boundary is not a mathematical surface on which the direction of spontaneous magnetization suddenly changes to the exactly opposite direction, but is instead a certain transition layer of finite thickness, in which there are many intermediate orientations of the magnetization. In what follows we shall be interested primarily in the following two quantities characterizing the properties of the transition zone, namely: the thickness of the transition layer $\delta$ and the energy $\gamma$ per $1\ \mathrm{cm}^2$ of boundary. The theoretical calculation of these quantities was first given by Bloch[^2] in 1932. Here his arguments are presented in a somewhat modified and simplified form. Until now there had been no possibility of experimental verification of his data. However, the theoretical explanation of the experiments of Sixtus[^3] on the growth of magnetization-reversal nuclei during large Barkhausen jumps has recently opened a way to the experimental determination of the boundary energy $\gamma$[^4]. Therefore, in this article mainly the theoretical—

…a theoretical explanation of these experiments, and those experiments from which \(\gamma\) can be obtained are especially distinguished.

We shall make a theoretical estimate of the quantities \(\gamma\) and \(\delta\), relying on Bloch’s work and taking as a basis the Heisenberg model for a ferromagnet, according to which each atom of the lattice possesses one electronic spin and exchange interaction is taken into account only between nearest neighbors in the lattice. We shall neglect this interaction between more distant atoms. The exchange integral is assumed to be positive, so that the parallel arrangement of neighboring spins corresponds to the state of lowest energy. We shall consider a simple cubic lattice. It is easy to understand that the transition of the spin from one direction of easy magnetization to the opposite one does not occur quite abruptly; otherwise, at the boundary, oppositely directed spins would turn out to be nearest neighbors, which is energetically very unfavorable. Owing to the positive exchange integral, the expenditure of energy for creating the boundary will be considerably smaller if, in the transition zone, the angle between the directions of neighboring spins is as small as possible, if, consequently, the transition takes place sufficiently slowly in a very thick transition layer. If exchange interaction were the only cause influencing the direction of the spins, then the energetically most favorable width of the transition layer \(\delta\) would be extremely large (theoretically infinitely large).

However, in addition to exchange interaction, it is necessary also to take into account stresses, which have the opposite tendency, namely: they strive to make the transition abrupt, i.e. \(\delta\) small. In materials with positive magnetostriction, as shown in the experiments of Sixtus and Tonks, stress creates a direction of easiest magnetization parallel to the load. The broader the transition zone, the more spins are oriented nonparallel to this preferred direction, i.e. are in energetically less favorable directions. Because of the competition between exchange interaction and stresses, such a distribution of spins in the boundary is established that the sum of both kinds of energy leads to the smallest possible value of the total energy of the boundary.

On the basis of these considerations it is not difficult to make a rough estimate of the order of magnitude of \(\delta\) and \(\gamma\). For simplicity, in these arguments we shall suppose that the direction of the load coincides with the \(z\)-axis of the cubic lattice. The boundary is considered plane and parallel to the \(y—z\) plane, so that the direction of the spins in the transition zone depends only on their coordinate \(x\). For brevity we shall denote by \(C\) the work needed to rotate the resultant magnetization in \(1\ \text{cm}^3\) of a spontaneously magnetized substance from the direction of easiest magnetization, parallel to the load, into the perpendicular direction. Thus \(C=\frac{3}{2}\lambda\sigma\), where \(-\lambda\) is the magnetostriction of saturation, and \(\sigma\) is the stress. The work \(\gamma_s\), performed…

... against the stresses in creating \(1\ \mathrm{cm}^2\) of boundary, is proportional to \(C\cdot\delta\), since in order to create each square centimeter of boundary it is necessary, in a volume \(\delta\), to rotate the magnetization from the direction of easiest magnetization. This part of the boundary energy increases in proportion to the boundary thickness \(\delta\); therefore it will be the smaller the sharper the transition, as was already indicated above. The numerical factor, whose magnitude is of the order of unity, will of course also depend on the distribution of spin directions in the transition zone. In this first rough estimate we shall simply put it equal to unity: \(\gamma_\sigma=C\cdot\delta\). Quantum theory leads to the following expression for the work of rotating two spins against the forces of exchange interaction from a parallel position into one in which there is an angle \(\varepsilon\) between them: \(I\cdot\dfrac{1-\cos\varepsilon}{2}\), where \(I\) is the exchange integral. In our case the angles between the directions of neighboring spins are small; therefore, approximately, we may replace this expression by \(I\cdot\dfrac{\varepsilon^2}{4}\). On the basis of our simplified assumptions concerning the position of the boundary, for two neighboring atoms with identical coordinates \(x\) the angle between the spin directions is equal to zero. For two neighboring atoms in the \(x\) direction, whose coordinates differ by the lattice constant \(a\), the angle in the transition layer is on the average equal to \(\pi\dfrac{a}{\delta}\) (because over the segment \(\delta\) this angle must increase, relative to the direction of easiest magnetization, from zero to \(\pi\)). This change of angle over the whole thickness of the boundary layer is divided into \(\dfrac{\delta}{a}\) steps. Since for \(1\ \mathrm{cm}^2\) of boundary, i.e. for the volume \(\delta\), there are \(\dfrac{\delta}{a^3}\) such neighboring spin pairs in the \(x\) direction, then, neglecting a numerical factor of the order of unity, we obtain the share of the exchange interaction in the boundary energy equal to

\[ \gamma_a=I\cdot\left(\frac{a}{\delta}\right)^2\frac{\delta}{a^3} =\frac{I}{a\cdot\delta}. \]

\(\gamma_a\) decreases with increasing boundary thickness and therefore tends to make this transition as little sharp as possible. The total boundary energy, up to numerical factors, is equal to

\[ \gamma=\gamma_a+\gamma_\sigma=C\cdot\delta+\frac{I}{a\cdot\delta}. \tag{1} \]

For a visual representation of the dependence of this quantity on \(\delta\), Fig. 1 shows the stress energy \(\gamma_\sigma\), the exchange energy \(\gamma_a\), and their sum \(\gamma\) as functions of \(\delta\). The boundary thickness actually realized corresponds to the minimum of \(\gamma\). Hence one obtains

\[ \delta\sim\sqrt{\frac{I}{a\cdot C}}, \tag{2} \]

where \(I\) is the exchange integral calculated for a pair of atoms. If, instead, one introduces the exchange integral per \(1\ \mathrm{cm}^3\), \(A = \dfrac{I}{a^3}\), then we obtain

\[ \frac{\delta}{a} \sim \sqrt{\frac{A}{C}} . \tag{2a} \]

Thus, the thickness of the transition layer, measured in lattice constants, is approximately equal to the square root of the ratio of the exchange energy \(A\), referred to \(1\ \mathrm{cm}^3\), to the strain energy \(C = \dfrac{3}{2}\lambda_s\).

Since these energies are related approximately as the Weiss molecular field is to that field strength which is necessary for saturation perpendicular to the direction of easiest magnetization (in the direction of the most difficult magnetization), the orders of magnitude are related as \(10^6\) to 100; for \(\delta\) one obtains a value of the order of 100 lattice constants. At small loads, i.e. at small \(C\), \(\delta\) may, however, become considerably larger.

With the aid of the expression for \(\delta\) given above, we obtain for the boundary energy

\[ \gamma \sim 2\sqrt{\frac{CI}{a}} = 2\frac{I}{a^2}\cdot\frac{a}{\delta}. \tag{3} \]

Fig. 1. Dependence of the magnetoelastic part of the boundary energy \(\gamma_\sigma\), the exchange part \(\gamma_a\), and the total energy \(\gamma = \gamma_a + \gamma_\sigma\) on the thickness of the boundary layer \(\delta\).

From the second form it is clearly seen that a gradual change in the direction of the spins in a transition layer of thickness \(\delta\) is energetically much more favorable than an abrupt, discontinuous change. If, for example, at the surface \(x = 0\) the direction of the spins is suddenly changed from the positive \(z\)-axis by \(180^\circ\), then on every square centimeter of this surface there arise \(\dfrac{1}{a^2}\) antiparallel spin pairs. For this an energy \(\dfrac{I}{a^2}\) is required. In reality, however, the boundary energy is smaller by a factor \(2\cdot\dfrac{a}{\delta}\).

This calculation can without any difficulty be carried out quite rigorously. Let \(\vartheta\) be the angle between the spin direction and the direction of easiest magnetization created by the stresses. In the transition zone the angle \(\vartheta\) depends on the coordinate \(x\): \(\vartheta = \vartheta(x)\). For large negative \(x\), \(\vartheta = 0\); for large positive \(x\), \(\vartheta = \pi\), since in the transition layer we have a rotation of the spins by \(180^\circ\). With the aid of this, initially unknown, function \(\vartheta = \vartheta(x)\), one can write the exchange and magnetoelastic parts of the boundary ener-

... in the form of two integrals of functions \(\vartheta\) and \(\dfrac{d\vartheta(x)}{dx}\). The function \(\vartheta(x)\) is determined by the variational method, by finding the minimum of the sum of these integrals. Taking the boundary conditions into account, we obtain the solution in the following form:

\[ \vartheta(x)=2\operatorname{arc\,tg} e^{\frac{2x}{\delta}} \quad \text{or} \quad \cos\vartheta(x)=-\operatorname{tgh}\frac{2x}{\delta}, \tag{4} \]

where \(\delta\) is the value already given above for the thickness of the boundary layer:

\[ \delta=\sqrt{\frac{I}{a\cdot C}} \]

(see Appendix II). A graphical representation of this function is shown in Fig. 2. In the middle part \(\vartheta\) varies approximately linearly with \(x\), while for large values of \(x\) it approaches the limiting values \(0\) and \(\pi\) exponentially.

Fig. 2. The angle \(\vartheta\) between the direction of the spins and the direction of easiest magnetization in the transition zone as a function of the distance \(x\)

With the aid of this function \(\vartheta(x)\) one can now rigorously determine the energy of the boundary. It happens incidentally that for the simple cubic lattice we find exactly the same expression \(\gamma\) as was obtained above by an approximate method, i.e.

\[ \gamma=2\sqrt{\frac{IC}{a}}. \tag{3} \]

This calculation can be carried out in an analogous manner for other orientations of the direction of easiest magnetization and of the boundary layer. It then turns out that the boundary energy is completely independent of the orientation of the direction of easiest magnetization and of the boundary layer relative to one another and relative to the crystallographic axes.

Thus this energy behaves in exactly the same way as ordinary surface tension. But this will be so only as long as we do not take the magnetic field into account. In the general case the normal to the boundary layer is no longer perpendicular to the direction of easiest magnetization, as was the case in our example under the special assumptions that, with the appearance of the boundary layer, there always arises an associated magnetic field. If, however, in the transition zone the divergence of the magnetization is no longer equal to zero, then a magnetic field arises, the sources of which lie in the boundary layer, and the induction \(\mathbf{B}=\mathbf{H}+4\pi\mathbf{I}\), as required by Maxwell’s equations, is source-free \([(\operatorname{div}\mathbf{B}=0)\) (translator’s note)]. Thus, in order to create \(1\ \mathrm{cm}^2\) of boundary field, generally speaking, one must expend not only the energy \(\gamma\) calculated above, but also a certain amount of energy to increase the energy of the magnetic field created by the boundary layer. However

GROWTH OF MAGNETIZATION-REVERSAL NUCLEI

this energy cannot be written in the form of an addition to the boundary energy, since its magnitude depends not only on the size and orientation of the section of the boundary layer that created it, but also on the arrangement of neighboring boundary layers. We shall consider this conclusion rigorously in analyzing a special case of experiments with magnetization-reversal nuclei.

For lattices of other types one obtains the same result as in the case of the simple cubic lattice considered. Apart from the factor \(\sqrt{2}\) in the case of a body-centered cubic lattice and the factor 2 in the case of a face-centered cubic lattice, exactly the same result is obtained for \(\gamma\), if by \(I\) we continue to mean the exchange integral between nearest neighbors. These numerical factors, however, must not be assigned very great importance, since our model is a very rough representation of the actual state of affairs.

In view of this uncertainty of the theory it is extremely important that Sixtus’s experiments with magnetization-reversal nuclei at large Barkhausen jumps make possible an experimental determination of \(\gamma\), and consequently also a check of the conclusions of the theory. For the purposes of the subsequent theoretical calculations we shall idealize the picture and assume that in the wire with which the experiment is carried out, the direction of easiest magnetization everywhere coincides with the axis of the wire. Therefore, on magnetization reversal, the boundary surface passes through the whole wire, and the magnetization curve has an exact rectangular form. In fact, of course, in the Sixtus specimens the tension is not so great as to overcome completely the influence of internal stresses and crystallographic magnetic anisotropy and to attain the ideal state. The latter is realized only approximately. Initially the magnetization throughout the wire is directed to one side, which we shall call negative. A small field in the positive direction is not yet capable of causing the process of magnetization reversal. It begins only when, at some point of the wire, with the aid of an additional field, the starting field \(H_s\) is reached. Then there a small region of positive magnetization forms within the negatively magnetized surroundings. This nucleus grows by motion of the boundaries as long as the additional field is applied. And the boundary can pass through the entire wire if in the remaining part of it the field is greater than some critical value \(H_0\). Sixtus found that, upon a brief exceeding of the starting field \(H_s\) in the additional coil, a magnetization-reversal nucleus is also formed, but only one such that it cannot grow when the additional field is switched off and preserves its dimensions even in the case when the field, after the additional field has been switched off, is greater than the critical field \(H_0\), which is necessary for motion of the boundary. Sixtus succeeded in measuring the dimensions of these “frozen” nuclei by measuring their stray fields. He found that these nuclei have the form of elongated thin regions.

The largest of them are 13 cm in length and only 0.1 mm in maximum cross-section. The reason for the inhibition of the growth of these frozen nuclei is, first, the influence of surface energy and, second, the action of the nucleus’s own field. This demagnetizing field, created by the “free charges” on the surface of the nucleus, inside it and on its lateral surfaces, is antiparallel to the external field and therefore retards the motion of the lateral boundaries. By contrast, the demagnetizing field at the ends of the nucleus cannot retard growth. Nevertheless, growth is still impossible, since it requires the expenditure of energy to increase the surface of the boundaries, i.e. to increase the boundary energy \(\gamma\).

Let us now consider by how much the field must be increased in order for such a “frozen” nucleus to begin to grow and for the remagnetization process to occur throughout the whole wire. The field required for this, \(H_s'\), is, of course, smaller than the field \(H_s\) necessary to create the nucleus, and, moreover, it depends on the dimensions of the nucleus. We wish to derive this relation theoretically. To this end let us consider the energy conditions which must be fulfilled in order for motion of the boundaries to be possible with an increase in the volume of the nucleus. If the volume of the nucleus \(V\) has increased by \(dV\), then the material receives an energy \(2HI_s dV\), where \(H\) is the field created in the wire by the external coil, since in the volume \(dV\) the magnetization changes from \(-I_s\) to \(+I_s\) (\(I_s\) is the saturation magnetization). This quantity of energy is expended in various ways. First, the motion of the boundary is associated with an increase in the surface \(dF\), which requires the energy \(\gamma dF\). Further, the growth of the nucleus is associated with an increase in the energy of the demagnetizing field. Let the energy of this field be denoted by \(W\), and its change by \(dW\). We shall denote the excess of the energy obtained, \(2HI_s dV\), over the sum of these two other energies by

\[ dA = 2HI_s dV - \gamma dF - dW . \tag{5} \]

This excess energy \(dA\) disappears irreversibly during the motion of the boundary. If the motion of the boundary could occur by itself without any expenditure of energy, then growth of the nucleus would be possible, since for positive \(dV\), \(dA\) is also positive. But in reality the very existence of the critical field \(H_0\) indicates that even for an infinitely slow displacement of the boundary there is required at least the energy \(2H_0 I_s\) per \(1\text{ cm}^2\), which is irreversibly expended. The boundary experiences a kind of “friction”; \(2H_0 I_s\) is the smallest value of the energy required to overcome this “friction.” Of course, here we are by no means speaking of friction in the mechanical sense. The cause of \(H_0\) is small spatial fluctuations of \(\gamma\). The conversion of the energy of this “friction” into heat is carried out by the microscopic eddy currents that arise in this process. The question of \(H_0\) has been considered in detail, for example, in Kersten’s paper\(^5\). For the present investigation it is sufficient to indicate that the boundaries of the nucleus can be displaced only in the case where \(dA\), during growth of the volume of the nucleus, exceeds, at least

at least the smallest value of the “friction” energy \(2H_0 I_s dV\), i.e. when the inequality holds

\[ dA > 2H_0 I_s dV. \tag{6} \]

From this inequality one can obtain the conditions for the growth of the nucleus. The demagnetizing field requires that the nucleus being formed be very long and thin. Let us denote its length by \(l\) and its greatest diameter by \(d\). In Sixtus’ experiments, for example, \(l/d > 500\). Growth of nuclei of such a form can proceed only in two quite different ways: either by increasing the length, or by increasing the thickness. Of course, both may also occur simultaneously. But since in these two modes of growth the motion of the boundaries takes place in entirely different places (in one, only at the ends; in the other, only on the lateral surface), simultaneous growth of length and thickness can occur only in the case when, for increasing the length at a given thickness and for increasing the thickness at a given length, one and the same inequality (6) is satisfied. Therefore both modes of growth may be considered independently of one another.

In the case of a long and thin nucleus, its volume is proportional to the length and to the square of the linear dimensions of the cross section, i.e.

\[ V = C_1 \cdot l \cdot d^2. \]

All numerical calculations will be carried out on the assumption that the nuclei have the form of an ellipsoid of revolution. Only in this particular case can the energy of the demagnetizing field \(W\) be calculated exactly. The nuclei actually observed in Sixtus’ experiments have approximately this form. In this case \(C_1 = \frac{\pi}{6}\). For a very elongated nucleus the surface may approximately be written in the form:

\[ F = C_2 \cdot l \cdot d. \]

For an ellipsoid of revolution \(C_2 = \frac{\pi^2}{4}\).

We shall first consider the limiting case of so long and thin a nucleus that the energy of the demagnetizing field \(W\) may be freely neglected. Then for the quantity whose change is given by (5), we find

\[ A = 2C_1 H I_s \cdot l \cdot d^2 - C_2 \gamma \cdot l \cdot d. \]

The condition for the possibility of growth in length according to (6) is

\[ \frac{\partial A}{\partial l} > 2H_0 I_s \frac{\partial V}{\partial l}; \tag{7} \]

neglecting \(W\) and using the above expression for \(A\), we obtain

\[ d>\frac{C_2\gamma}{2C_1 I_s(H-H_o)} . \tag{9} \]

Thus there exists a definite critical thickness

\[ d_k=\frac{C_2\gamma}{2C_1 I_s(H-H_o)}, \]

which is inversely proportional to \((H-H_o)\). In the case of a very long thin nucleus, growth in length is possible if \(d>d_k\). In the case of an ellipsoid of revolution

\[ d_k=\frac{3\pi\gamma}{4I_s(H-H_o)} . \tag{10} \]

The condition for the possibility of thickness growth can be obtained in exactly the same way, and it reads

\[ \frac{\partial A}{\partial d}>2H I_s \frac{\partial V}{\partial d}, \tag{8} \]

whence, again neglecting \(W\), it follows that

\[ d>\frac{C_2\gamma}{4C_1 I_s(H-H_o)}=\frac{1}{2}d_k . \tag{11} \]

Consequently, an increase in the thickness of very long thin nuclei becomes possible already when the thickness has exceeded only one half of the critical value \(d_k\). The fact that thickness growth begins earlier than length growth is explained by the circumstance that the volume depends quadratically on the thickness and linearly on the length. For a relatively identical change of surface, the growth of volume when the thickness is increased is greater than when the length is increased.

But if we pass to short nuclei, then the energy of the demagnetizing field \(W\) becomes appreciable. However, this field acts differently in the case of growth in length or in thickness. On the lateral surface of the nucleus, where the motion of the boundary mainly occurs during thickness growth, the demagnetizing field is antiparallel to the external field. Consequently, it diminishes the external field. Therefore the thickness above which thickness growth can occur increases with decreasing \(\frac{l}{d}\) and will become greater than \(\frac{d_k}{2}\).

At some definite value of the size ratio \(\frac{l}{d}\), the demagnetizing field at the boundary surface becomes equal to \(H-H_o\). Therefore the total field on the lateral surface cannot exceed the value \(H_o\) for any increase of the nucleus. On the contrary, quite a different situation occurs during length growth. At the ends of the nucleus the demagnetizing field outside the nucleus is parallel to the external field. Consequently, it promotes the growth of the nucleus.

Therefore, with decreasing size ratio \(\frac{l}{d}\), it must become-

GROWTH OF REMAGNETIZATION NUCLEI

becomes smaller than \(d_k\), i.e., the thickness above which growth in length is possible. These relations are best represented graphically in the \(l—d\) plane. Each point in this plane\(^1\) corresponds to a nucleus of definite dimensions. The phase \(l—d\) plane is divided into regions whose phase points satisfy definite conditions of growth, either of length or of thickness. The boundary curve of the region where growth in length is possible is given by the equation

\[ \frac{\partial A}{\partial l}=2H_o I_s\frac{\partial V}{\partial l}. \tag{7a} \]

For large \(l\), this curve must asymptotically approach the straight line \(d=d_k\), and, as the length decreases, must fall. The boundary curve of the region where growth in thickness is possible is obtained from the equation

\[ \frac{\partial A}{\partial d}=2H_o I_s\frac{\partial V}{\partial d}. \tag{8a} \]

It must have as its asymptote \(d=\dfrac{d_k}{2}\) as the ratio of the dimensions \(\dfrac{l}{d}\) increases, and must approach the straight line passing through the origin and having a slope equal to \(H-H_o=H_d\), where \(H_d\) denotes the magnitude of the demagnetizing field inside the nucleus. This picture, qualitatively understandable on the basis of the above considerations, is shown in Fig. 3. The boundary curves in this figure are calculated on the assumption that the nucleus has the form of an ellipsoid of revolution. In this case one can compute exactly the energy of the demagnetizing field \(W\). Namely, it is easy to show that

\[ W=2NI_s^2\cdot V, \tag{12} \]

\(N\) is the demagnetizing factor, which for a long and thin nucleus is equal to

\[ N=4\pi \frac{d^2}{l^2}\left(\ln 2\frac{l}{d}-1\right). \]

Fig. 3. The \((l—d)\) plane with boundary curves for \(H-H_o=0.5\) oersted.

Fig. 3 is made for \(H-H_o=0.5\) oersted and \(I_s=1650\) gauss. This latter number is equal to the saturation of the substance \((15\%\,\mathrm{Ni}—85\%\,\mathrm{Fe})\) on which Sixtus carried out his experiments. The unknown quantity \(\gamma\) changes only the scale of the drawing, without changing the form of the curves. As the unit of length \(\beta\), the critical length \(d_k\) was chosen at

\(^1\) We shall hereafter call them phase points and the phase plane. Translator’s note.

the field \(H-H_o = 1\) oersted, since \(d_k=\dfrac{\beta}{H-H_o}\). For an ellipsoid of revolution \(\beta=\dfrac{3\pi\gamma}{4I_s}\).

The limiting curve \(AOB\) separates the region where growth of thickness is possible, and the curve \(COD\) the region where growth of length is possible. Owing to the different action of the demagnetizing field on the growth of length and thickness, the two limiting curves intersect and thereby divide the \(l-d\) plane into 4 regions. In region I there is no inhibition of growth at all. A nucleus whose phase point has fallen into this region is not energetically limited in its growth. The phase point of the nucleus will move in the drawing to the right and upward. The nucleus may become arbitrarily large, and, consequently, reversal of magnetization may occur throughout the whole wire. In region IV neither growth of length nor growth of thickness can occur. Consequently, there lie the points corresponding to the frozen state of the nuclei. In region II growth of length is possible, but growth of thickness is impossible. The phase point moves in this region parallel to the abscissa axis to the right. Its further behavior will depend on whether the thickness of the corresponding nucleus is greater or less than the thickness \(d_o\) of the nucleus corresponding to the point of intersection of the limiting curves. If \(d>d_o\), then the phase point ultimately reaches the limiting curve \(OB\), crosses it, and enters the region of “free” growth. Consequently, such a nucleus causes reversal of magnetization throughout the whole wire. If, on the contrary, \(d<d_o\), then the phase point, as the length grows, reaches the limiting curve \(OD\), and then growth ceases, since the point falls into the region of complete inhibition of growth. In region III growth of length is forbidden, but an increase of thickness is possible. The phase points of this region move parallel to the ordinate axis upward. Crossing the boundary \(CO\), they enter the region of free growth. Only those phase points which are in the small lens-shaped part of region III, and whose length is less than the length \(l_o\) corresponding to the nucleus represented by the point \(O\), are delayed on the limiting curve \(AO\). The entire region in Fig. 3 in which nuclei cannot cause reversal of magnetization throughout the whole wire is hatched.

The arrangement of the limiting curves in Fig. 3 also depends on the magnitude of the field \(H\). But it can be shown that the form of these curves changes qualitatively little when the field is varied. Mainly only the scale changes. Therefore it is sufficient to indicate only the position of the point \(O\) for different fields. In Fig. 3 it is shown by a dashed curve. The indicated numerical values on this curve correspond to the magnitude \(H-H_o\).

We still cannot explain the possibility that the dimensions of a nucleus may also decrease by displacement of the boundaries. For this it would be necessary that the condition be satisfied

\[ dA< -2H_o I_s dV,\quad (dV>0) \tag{13} \]

which is the reverse of (6).

GROWTH OF REMAGNETIZATION NUCLEI

The equations of the boundary curves corresponding to this process of reduction of nuclei can be obtained from those just considered if in them \(H-H_0\) is replaced by \(H+H_0\). The corresponding regions lie in Fig. 3 in the lower left corner. But we shall not dwell on this, since it is of no interest for what follows.

We can easily answer the question at what field \(H'_s\) a given frozen nucleus will lead to the process of remagnetization. Obviously, this will be the field at which the shaded region of Fig. 3 becomes so small that the phase point corresponding to this nucleus falls exactly on its boundary. In doing so one must distinguish two cases.

1. The phase point lies above the (dotted) curve of the points \(O\). As the field is increased, this phase point, initially lying in region IV, where growth is completely frozen, reaches the boundary curve \(OD\) and passes into the region where growth of the length is possible, but not of the thickness. Its length thus begins to grow, but not without limit, only until the phase point falls on the new curve \(OD\). The process of remagnetization will begin only when the thickness \(d_o\) of the nucleus corresponding to the point \(O\) is smaller than the thickness of the frozen nucleus. In this case \(H'_s\) depends only on the thickness of the nucleus.

2. If the phase point lies below the line of the points \(O\), then, if we exclude from consideration the small lens-shaped part of region II near the point \(O\), where \(l<l_0\), the nucleus will begin the process of remagnetization at that field \(H'_s\) for which the boundary curve \(AO\) passes through its phase point. In this case \(H'_s\) depends on the length and thickness of the nucleus. Only in the case when the length is so large that \(AO\) can practically be replaced by its asymptote does \(H'_s\) again depend only on the thickness.

In Sixtus’s experiments the observed nuclei have phase points above the line of the points \(O\). We shall see below that for the largest of the nuclei obtained by him this indeed must be so by virtue of the method by which they were obtained. According to the theory, these nuclei should undergo a limited growth in length before causing the process of remagnetization throughout the whole wire. This, however, was not observed by Sixtus. The starting field \(H'_s\) of such a nucleus should depend only on the thickness. In this point the theory is confirmed by experiment. Assuming that the nucleus has the form of a spheroid of revolution, for large values of \(\dfrac{l}{d}\) we obtain, to a good approximation, that the thickness \(d_o\) of the nucleus with phase point at \(O\) is equal to \(5/6\,d_k\), so that we obtain

\[ d_o=\frac{5\pi}{8}\cdot \frac{\gamma}{I_s(H-H_0)} . \tag{14} \]

The difference between the starting field \(H'_s\) and the boundary field \(H_o\) in these nuclei is inversely proportional to the thickness of the nucleus, i.e.

\[ H'_s-H_o=\frac{K}{d}, \tag{15} \]

where the proportionality constant \(K\), besides known quantities, also contains the boundary energy

\[ K=\frac{5\pi}{8}\cdot \frac{\gamma}{I_s}. \tag{16} \]

A comparison of the theoretical formula (15) with experiment is presented in Fig. 4. The points in the drawing show the relation between \(H_s' - H_0\) and the thickness of the nucleus, obtained from Siksus’s experiments. The curve gives the theoretical course for the selected value of \(K\). Agreement of the curve with the points gives good confirmation of the correctness of the theory. The following value of the boundary energy corresponds to this value of \(K\):

\[ \gamma = 2.7\ \mathrm{erg}/\mathrm{cm}^{2}. \]

Fig. 4. Relation between \(H_s' - H_0\) and the thickness of the nucleus \(d\) (see Fig. 11 in Siksus’s article). 1 — Siksus’s experimental points, 2 — theoretical course for \(\gamma = 2.7\ \mathrm{erg}/\mathrm{cm}\)

Let us now compare this quantity with the theoretical value according to Bloch. The material with which Siksus worked \((15\%\,\mathrm{Ni} — 85\%\,\mathrm{Fe})\) has a body-centered lattice. Therefore we have

\[ \gamma = 2\sqrt{2}\sqrt{\frac{I C}{a}}. \]

\[ C=\frac{3}{2}\lambda\sigma. \]

The tension for these experiments was equal to \(\sigma = 65\ \mathrm{kg}/\mathrm{mm}^{2} = 6.5\cdot 10^{9}\ \mathrm{dyn}/\mathrm{cm}^{2}\). The exchange integral, according to Heisenberg, is determined by the Curie temperature; for a body-centered lattice we have, in particular,

\[ 2I = k\theta. \]

Taking further \(\theta = 1000^\circ\ \mathrm{abs.};\ \lambda = 2.5\cdot 10^{-5};\ a = 2.7\ \text{Å},\) we find

\[ \gamma = 2.1\ \mathrm{erg}/\mathrm{cm}^{2}. \]

Taking into account the excessively rough estimate of the theoretical value, it may be considered that good agreement is obtained.

To all this one more remark must be added. We have everywhere assumed that the nucleus lies inside the material, and not on the surface of the wire. In several cases Siksus was able to prove this assumption directly. He took a wire in which there was a frozen-in nucleus and etched it from the surface. In this process no change in the thickness of the nucleus was observed\(^6\).

Therefore, the nucleus lay inside the material, and not on the surface of the wire.

This brings us to the question of how, in general, a nucleus capable of growth can arise inside an auxiliary coil. According to all observations known up to now, very small nuclei never lead to growth; on the contrary, they always have a tendency to decrease, since for very small nuclei growth always requires the expenditure of energy to create a surface, which is proportional to the increase in the volume of the nucleus. This is completely analogous to the state of affairs that obtains in the case of condensation of supersaturated vapor, where, owing to surface tension, the smallest droplets always have a tendency to evaporate, and not to grow by condensation. In the case of gases, condensation always occurs on ions, dust particles, or at other places where this difficulty of nucleation is absent. If only it were possible to remove completely all these factors facilitating the growth of nuclei, the supersaturation could be increased so greatly that appreciable spontaneous nucleation would begin, caused by density fluctuations. The latter, owing to the atomistic structure of matter, always occur; however, they are too small, and only at a very high degree of supersaturation do these fluctuations lead to the formation of nuclei of such a large size that they can grow. In this case the second law of thermodynamics is violated each time, since, strictly speaking, such a process is impossible according to this law. By analogy with this, one may suppose that the process of remagnetization arises at places where the homogeneity of the crystal is disturbed, for example where the boundary energy is very small, or else where there remain in the material residual regions from previous magnetizations, which are themselves the nuclei of remagnetization. However, we still know very little about this question. One can only say with certainty that the spontaneous formation of nuclei, in which the difficulty of nucleation at a given field is overcome, has not yet been discussed.

In these experiments one more unexpected fact was observed. The auxiliary coil with whose help Sixtus obtained the nucleus was approximately 1 cm long. In spite of this, the frozen nucleus that is obtained upon brief switching-on of the starting field \(H_s\) in this short coil has a length of up to 13 cm. Its ends are thus outside the auxiliary field. The ends of a positively magnetized nucleus are therefore able, while the auxiliary field is switched on, to grow inside a negatively magnetized material; but they lose this ability when this auxiliary field is switched off, although the field itself exerts no noticeable action at the ends of the nucleus. This can again be easily understood with the aid of Fig. 3. Such a nucleus, whose ends have emerged from the auxiliary field, is for the most part in the main field, which retards its growth; the latter ceases when the auxiliary field is switched off. This can

to present it in this way. The short additional coil, for its part, stops inhibiting the growth of the thickness. Therefore a noticeable displacement of the lateral boundaries of the nucleus begins. The phase point in the \(l-d\) plane crosses the boundary surface \(OD\) and enters region II, where limited growth of the length in the main field is possible. Thus the growth of the length occurs indirectly, because, as a consequence of the growth of the thickness under the action of the field of the additional coil, the ratio of the dimensions of the nucleus, \(\frac{l}{d}\), decreases, which makes possible growth of the length in the main field. When the additional field is switched off, the growth of the thickness is again inhibited. And the length then continues to grow until the phase point reaches the curve \(OD\) corresponding to the main field, where it will be inhibited, and the growth of the length will stop.

Fig. 5. Position of the three largest nuclei on the \((l-d)\) plane

Whether these arguments are correct can be decided by checking whether the actual phase points of these elongated nuclei lie on the curve \(OD\). The result of the comparison is shown in Fig. 5. The field in which Sixtus carried out experiments with the largest nuclei was \(H-H_0 = 0.28\) oersted. In Fig. 5 the boundary curve \(OD\) for this field is shown, \(\gamma = 2.7\ \mathrm{erg/cm^2}\). The phase points are plotted for the three largest nuclei. Two of them lie exactly on the curve \(OD\). The third evidently corresponds to too short a nucleus, for which this theory is not applicable, since a large part of it lies inside the additional coil.

Thus in this case as well we obtain good confirmation of the theory developed here. For further substantiation of the theory we do not yet have experimental material. Sixtus’s measurements have so far been the only ones in this field. But since, when they were set up, these theoretical considerations were not present, it is natural that they cannot fully serve for the purposes of verifying this theory. In connection with this, we are undertaking a series of experimental investigations.

References

1. K. J. Sixtus and L. Tonks, Phys. Rev., 37, 930 (1931); 42, 419, 1932; 43, 70, 1933; 43, 931, 1933; preceding article, see p. 000.
2. F. Bloch, Z. Physik, 74, 295, 1932.
3. K. J. Sixtus, Phys. Rev., 48, 425, 1935.
4. W. Döring, Z. Physik, 108, 137, 1938.
5. M. Kersten, Probleme der technischen Magnetisierungskurve, J. Springer, 1938, p. 42.
6. Personal communication from Dr. Sixtus concerning unpublished work.

APPENDIX I.

CALCULATION OF THE TIME OF PENETRATION OF A MAGNETIC FIELD INTO A WIRE DURING MAGNETIZATION1

Formula (2) for the time of penetration of a magnetic field into a wire being remagnetized, cited by Sixtus in § 3, can be obtained from the following simple considerations1.

Choose the \(x\)-axis along the axis of the wire. Let \(R = R(x)\) give us the equation of the unknown, for the time being, section by a plane passing through the axis of the wire, bounding the surface between the antiparallel domains. Our first task is to determine this function \(R(x)\).

The change of flux in the transverse section of the wire during the time \(dt\) is equal to \(2\pi R\,dR \cdot 4\pi \Delta I\), where \(4\pi \Delta I\) is the change of induction between the two domains. The circulation of the induced electric field around the wire at a distance \(r > R\) from the axis is equal to

\[ 2\pi rE=\left(8\pi^2R\,\frac{\Delta I}{c}\right)\frac{dR}{dt}. \tag{I,1} \]

(of course, for \(r < R\), \(E=0\)). Introducing the longitudinal propagation velocity \(v=\dfrac{dx}{dt}\) and denoting \(\dfrac{dR}{dx}\) by \(R'\), we find that

\[ E=\frac{4\pi vRR'\Delta I}{rc}. \tag{I,2} \]

Since the current density \(j=\dfrac{E}{\rho}\) (\(\rho\) is the specific resistance), the total circular current per unit length of the wire is equal to

\[ \int_R^a \frac{E}{\rho}\,dr = \frac{4\pi vRR'\Delta I}{\rho c}\ln\frac{a}{R}. \tag{I,3} \]

where \(a\) is the radius of the wire.

This alternating current, in turn, creates a magnetic field \(H_e\), which at the distance \(R\) is equal to

\[ H_e=-\frac{16\pi^2v\Delta I}{\rho c^2}\,RR'\ln\frac{a}{R}. \tag{I,4} \]

The negative sign indicates that \(H_e\) is antiparallel to the external field \(H\), and is thus a braking field.

In addition to this field there is also the field \(H_p\) from the free “poles” on the boundary. Therefore the resultant field \(H_m\) is equal to

\[ H_m = H + H_e + H_p . \tag{I,5} \]

Motion of the boundary will occur if

\[ H_m > H_0, \]

where \(H_0\) is the critical field; thus

\[ H_e + H_p = -(H - H_0) = -\Delta H . \]

Neglecting \(H_p\) (see Sixtus, § 3) and eliminating \(H_e\) from (I,4), we find

\[ RR' \ln \frac{a}{R} = \frac{\Delta H \cdot \rho c^2}{16\pi^2 v \cdot \Delta I}. \tag{I,6} \]

Integrating (I,6), we obtain the equation of the cross section \(R = R(x)\)

\[ R^2\left[\ln \frac{a^2}{R^2} + 1\right] = \frac{\Delta H \cdot \rho c^2 x}{4\pi^2 v \cdot \Delta I}, \tag{I,7} \]

where the constant of integration is chosen so that the curve \(R(x)\) intersects the axis of the wire at the point \(x=0\). The length of the boundary “funnel” \(\lambda\) can be obtained from (I,7), if one sets \(R=a\), since between \(\vartheta t\), \(\lambda\), and \(v\) there is the obvious relation

\[ v\vartheta t = \lambda . \tag{I,8} \]

Substituting \(\lambda\) from (I,7), we immediately arrive at the formula of Sixtus § 3 (2).

ADDENDUM II

THE LAW OF CHANGE OF THE DIRECTION OF MAGNETIZATION IN THE TRANSITION ZONE BETWEEN REGIONS OF SPONTANEOUS MAGNETIZATION\(^2\)

The distribution of the directions of the magnetic moment in the boundary layer between two antiparallel magnetized regions can be found as follows. Let us first enumerate all the energies possessed by a ferromagnet. These will be: 1) the positive exchange energy, whose appearance is caused by inhomogeneities in the distribution of the directions of the magnetic moments. The density of this energy is equal to\(^2\)

\[ F_1 = \frac{I}{4a}\left[(\Delta \alpha_x)^2 + (\Delta \alpha_y)^2 + \Delta \alpha_z^2\right], \tag{II,1} \]

where \(\alpha_x\), \(\alpha_y\), and \(\alpha_z\) are the direction cosines of the spontaneous magnetization with respect to the coordinate axes, \(I\) is the exchange integral, and \(a\) is the lattice constant.

2) The energy of magnetic anisotropy, for example for iron, has the following form[^8]:

\[ F_2 = K_1(\alpha_x^2\alpha_y^2+\alpha_y^2\alpha_z^2+\alpha_z^2\alpha_x^2)+K_2\alpha_x^2\alpha_y^2\alpha_z^2, \tag{II, 2} \]

where \(K_1, K_2\) are the so-called constants of magnetic anisotropy.

3) The magnetoelastic energy of external stresses

\[ F_3=-\frac{3}{2}\sigma\lambda_{100}(\alpha_x\beta_x+\alpha_y\beta_y+\alpha_z\beta_z)^2+ \]

\[ +3\sigma(\lambda_{100}-\lambda_{111})\cdot(\alpha_x\alpha_y\beta_x\beta_y+\alpha_y\alpha_z\beta_y\beta_z+\alpha_z\alpha_x\beta_z\beta_x), \tag{II, 3} \]

where \(\lambda_{100}\) and \(\lambda_{111}\) are the magnetostriction constants along the crystallographic directions (100) and (111), and \(\beta_x,\ \beta_y\) and \(\beta_z\) are the direction cosines of the stress with respect to the coordinate axes.

In Sixtus’ experiments the tensile stresses \(\sigma\) are so large that we may neglect the energy \(F_2\) in comparison with \(F_3\) \(\left(\dfrac{3}{2}\sigma\lambda_{100}\gg K_1\right)\). Thus the entire anisotropy is practically determined by \(F_3\); precisely for this reason a uniform tensile stress \(\sigma\) throughout the whole volume of the wire turns the latter, in the magnetic sense, into a “uniaxial” single crystal (i.e. creates one axis of easiest magnetization, parallel to the direction \(\vec{\sigma}\)). We take the axis of the wire as the \(z\)-axis; then \(\beta_x=\beta_y=0,\ \beta_z=1\), and, consequently,

\[ F_3=-\frac{3}{2}\sigma\lambda_{100}\alpha_z^2. \tag{II, 4} \]

We find the distribution of magnetic moments between two regions with antiparallel magnetization while neglecting the effects caused by the surface of the specimen. They are essential only for determining the sizes of the spontaneous regions, but practically do not affect the sign of the change in the direction of magnetization in the boundary zone between them. In addition, we assume beforehand that no magnetic “charges” arise in the layer, i.e. that everywhere we have \(\operatorname{div}\mathbf I=0\) and there is no jump in the normal component of the magnetization \(I_n\) for any surface drawn in the boundary layer. This means that if we choose the \(x\)-axis perpendicular to the axis of the wire and normal to the boundary between spontaneous regions, then the distribution of the magnetic moments will depend only on the coordinate \(x\). The magnetization vector, having a direction along the positive \(z\)-axis in one region, rotates in the layer through \(180^\circ\) and points in the other region along the negative \(z\)-axis. We find the law of distribution of the moments from the condition that the total energy of the crystal (in the approximations indicated above) has a minimum, i.e.

\[ \int\left\{\frac{I}{4a}\left[\left(\frac{\partial\alpha_x}{\partial x}\right)^2+\left(\frac{\partial\alpha_y}{\partial x}\right)^2+\left(\frac{\partial\alpha_z}{\partial x}\right)^2\right]-\frac{3}{2}\sigma\lambda_{100}\alpha_z^2\right\}\,dV=\min. \tag{II, 5} \]

Let us denote the angle between the magnetization vector \(\mathbf I\) and the \(z\)-axis by \(\vartheta\); then, since the vector \(\mathbf I\) lies in the \(yz\)-plane, we have

\[ \alpha_x=0,\quad \alpha_y=\sin\vartheta,\quad \alpha_z=\cos\vartheta, \]

where \(\vartheta\) is a function only of \(x\), and, consequently, (II, 5) takes the form

\[ \int\left[\frac{I}{4a}\left(\frac{\partial\vartheta}{\partial x}\right)^2-\frac{3}{2}\sigma\lambda_{100}\cos^2\vartheta\right]dx=\min, \]

The Euler equations read

\[ \frac{1}{2}\cdot\frac{l}{a}\cdot\frac{\partial^2 \vartheta}{\partial x^2} -3\delta\lambda_{100}\sin\vartheta\cos\vartheta=0, \]

whence, integrating, we find

\[ \left(\frac{\partial\vartheta}{\partial x}\right)^2 -\frac{6\delta\lambda_{100}a}{l}\sin^2\vartheta =\mathrm{const}. \tag{II, 6} \]

The dimensions of the spontaneous regions are large in comparison with the width of the boundary layer. Therefore, as boundary conditions one may take

\[ \vartheta=0 \text{ for } x=-\infty;\qquad \vartheta=\pi \text{ for } x=+\infty \]

and

\[ \vartheta'=0 \text{ for } x=\pm\infty \text{ or for } \vartheta=0,\pi. \tag{II, 7} \]

Then the constant in (II, 6) is equal to zero, and we obtain

\[ \left(\frac{\partial\vartheta}{\partial x}\right)^2 =\frac{6\delta\lambda_{100}a}{l}\sin^2\vartheta. \tag{II, 8} \]

Integrating this equation, we find the solution satisfying conditions (II, 7) in the following form:

\[ \cos\vartheta=-\operatorname{tgh}\sqrt{\frac{3\delta\lambda_{100}a}{2l}}\cdot 2x. \tag{II, 9} \]

Noting further that, in Döring’s notation[^3] (see the preceding article),

\[ \frac{3}{2}\delta\lambda_{100}=C \]

we see that (II, 9) coincides with his formula (4).

ADDENDUM III

NEW EXPERIMENTS ON THE DETERMINATION OF THE BOUNDARY ENERGY \(\gamma\) AT LARGE BARKHAUSEN JUMPS

After the publication of the collection on the technical magnetization curve, in which the above-cited articles by Sixtus and Döring were included, a paper by Döring and Haake[^4] appeared, giving new experimental estimates of the energy of the boundary layer and, consequently, further confirmation of the theory.

The experiments were carried out with an elastically stretched wire, with a cross-sectional diameter of \(0.3\) mm, made of an Fe—Ni alloy (40% Fe, 60% Ni). Already at loads of \(5\ \mathrm{kg/mm^2}\), almost completely rectangular hysteresis loops were obtained.

A large number of frozen-in nuclei were obtained, and the dependence of the pole strength on their cross section was measured. According to Döring’s formulas (15) and (16), we have

\[ H_s' = H_0+\frac{5\pi}{8}\cdot\frac{\gamma}{I_s}\cdot\frac{1}{d}. \]

In Fig. 1 the results of measurements are given for two values of the load \(\sigma\). The scatter of the points is determined mainly by the inhomogeneities of the values of \(H_0\) for different portions of the wire. The slope of the straight lines pro-

is proportional to the boundary energy \(\gamma\). Already from Fig. 1 it is evident that \(\gamma\) increases with \(\sigma\), as is required by Bloch’s theory. Kondorskii [5] in Döring’s article, noting that

\[ c \simeq \frac{3}{2}\lambda_s . \]

Fig. 1. Dependence of the starting field \(H_S\) on the reciprocal thickness of the nucleus, \(\frac{1}{d}\).

Figure 2 shows the dependence of \(\gamma\) on \(\sigma\) for five values of the stresses. The scatter of the points is again determined by fluctuations. The curve gives the theoretical variation of \(\gamma\) with \(\sigma\) according to formula (3). The exchange integral was calculated according to Heisenberg from the Curie point. The agreement is sufficiently good for such approximate estimates. The authors point out that they are continuing work on the further development of the theory.

Fig. 2. Dependence of the boundary energy \(\gamma\) on the stress \(\sigma\).
1 — theoretical curve according to Bloch, 2 — experimental points.

REFERENCES

1. L. Landau and E. Lifschitz, Sow. Phys., 8, 157, 1935.
2. K. J. Sixtus and L. Tonks, Phys. Rev., 42, 419, 1932.
3. R. Gans, Ann. Phys., 24, 680, 1935.
4. W. Döring and H. Haake, Physik., 7, 39, 865, 1938.
5. Kondorskii, Zhurnal eksperimental’noi i teoreticheskoi fiziki, 7, 1117, 1937; Sow. Phys., 11, 597, 1937.

1. See § 3 of Sixtus’s article, pp. 67–69. ↩↩↩