3. 2D & 3D coherent magnetization rotation model

We shall add to the previous presentation a vector generalization of the Stoner-Wohlfarth model, developed by Thiaville [8].

To start, one presents the 2D generalization. The same principles will apply to the 3D generalization. The total free energy density will be written in the general form:

(1)
where is the anisotropy energy density, which depends in the 2D case on the angle between the easy axis and the polarization vector. This expression could be written as:
(2)
if one makes the notations:
(3)
(4)
(5)
(6)
(7)
where is a constant (in the uniaxial case the anisotropy constant), and is a unit vector on polarization vector direction.
The equilibrium and stability conditions are obtained deriving (2); one obtains:

(e) (8)
and
(s) (9)

where
(10)
and is a unit vector, perpendicular to .

If one consider the inverse problem (for a given find the locus of the applied field vector tip), (e) and (s) represent the equations of two straight lines:
(e) is a line perpendicular to the vector (parallel to the vector)
(s) is a line perpendicular to the vector (parallel to the vector).
So, the (e) and (s) lines are reciprocally perpendicular, (e) being parallel to the equilibrium orientation of the polarization vector. The system defining the critical curve is given by:
(11)
and has the solution:
. (12)
Example: For the uniaxial case:



and
. (13)
which represents the equation of an astroid.
If one makes the derivative of the in order to find the direction of the tangent to the critical curve, one finds:
(14)
that is, the tangent is parallel to the equilibrium orientation of the polarization vector, as the (e) line. We can introduce an orientation of the critical curve: the positive sense of the critical curve may be given by.

Note
One observes that all the properties we derived for the uniaxial case in the previous sections applies to the general 2D case.

Constant energy curves

The locus of the applied field vector tip for one given value of the energy, w, in the equilibrium state is the solution of the system:
(15)
that is,
. (16)

The tangent to the constant energy curve is given by:
(17)
that is, it is perpendicular to the equilibrium orientation, (see Figure 16).
Defining by the arc-length on the constant energy curve we have:
(18)

Fig. 16

where is the curvature radius of the constant energy curve.
In order to find the position of the curvature centre, noted with, we can use the relation (see Figure 16):
. (19)
Introducing in (19) the expressions (16) and (18) one obtains:
(20)
that is, the curvature centre is on the critical curve. From the mathematical point of view one may say that the critical curve is the common evolute of all constant energy curves.

Geometric construction for energy calculations

Thiaville [9] also proved that it is possible to use a simple geometric construction to estimate the free energy density when a field is applied to the ferromagnetic particle. The normalized total free energy density may be expressed as:
(21)
and if one uses (11), one obtains:
(22)
Taking into account (14), the arc-length on the critical curve is given by:
(23)
which is equivalent to:
. (24)
In conformity with (24) the difference between the values of in two equilibrium states on the critical curve characterized by q₀ and q is twice the arc-length measured on the critical curve. This property could be used in (22) for expressing the sum between the first and the second term. The final term in (22) is minus twice the length of the segment on the equilibrium line (tangent to the critical curve) linking the point on the critical curve with the tip of the applied field vector (see Figure). In fact, in this way one can calculate the difference of energy for two states characterized by two values of the applied field. For example, when the field is applied as presented in the Figure, the equilibrium (stable) orientation of the polarization vector is the orientation of the vector . Using the consequences of the previous geometrical discussion we can say that the difference of energy between the “C” state and the “A” state (see Figure, the “A” state is characterized by an applied field vector whose tip is on the point “A”) is twice the sum of the lengths of the arc AB and the segment BC, that is:
(25)

Fig. 17

Note:
The points characterized by (A, see Figure 18) and (B) have the same energy; the points A and B could be connected by two quarters of the astroid one positive (AC) and one negative (CB).

Fig. 18

Fig.19

When the tip of the applied field vector is inside the astroid there are three tangents to the critical curve passing through it: two for stable equilibrium states and one for an unstable equilibrium state (see Figure 19).

3D coherent magnetization rotation model

The method applied in the previous section could be used for developing a similar set of geometrical rules for the 3D coherent magnetization model [10].
The most general form of the free energy density (normalized to 2K this time in order to obtain a simpler form) is:
(25)
where:
(26)
is the unit vector of the polarization vector.
To obtain the stable equilibrium conditions we need the following expressions:
(27)
(28)
(29)
(30)
(31)
In the calculus we used:
(32)
(33)
(34)
(35)
(36)
(37)
where is a set of three unit vectors which are an orthogonal basis.
From the equilibrium conditions:
(38)
(39)
one can find a general expression for the vector. From (38) and (39), which represent the equations of two planes, we can calculate the components on and directions:
(40)
(41)
The intersection of these planes is the straight line oriented on the direction. The general solution of the system (38)-(39) is:
(42)
If (,) are the spherical angles defining the orientation of the polarization vector at the equilibrium the free energy density could be developed in the Taylor series near this equilibrium orientation as follows:
(43)
where
, (44)
and
. (45)
Taking into account the equilibrium conditions (38) and (39) the stability condition,
(46)
may be written as:
(47)
or, in a simpler form;
(48)
where
(49)
In order to have (48) for any value of x, the determinant should be negative and the coefficient of should be positive:
(50)
Introducing the equilibrium condition given by (42) in the first equation in (50) we are founding two solutions:
(51)
From the other equation in (50) we can say that the stable equilibrium part of the line defined by (42) is given by the condition:
. (52)
Introducing in (42) one finds the equations of two surfaces, .