6. Laminar model for flat slab in tilted field

The simplest of experientially observed equilibrium domain structures of the IS is onedimensional laminar lattice in slab-like specimens placed in a tilted field. Therefore such a specimen/field configuration is the most convenient for modeling. A laminar model for tilted field (LMTF) was developed in [25, 26]. Schematics of the specimen in the LMFT is shown in Figure 13.

Setting of the model is:

Meissner state [37]. In Figure 11 we reproduce typical magneto-optical images obtained for a 2.5-μm-thick film. The most unexpected result revealed with this specimen is that in perpendicular field the critical field Hcr ≈ 0:4Hc at T ! 0. A typical magnetization curve obtained with another (3.86-μm-thick) film is shown in Figure 12. Hcr for this specimen at 2.5 K is 0.65Hc and

102 Superfluids and Superconductors

Figure 11. Magneto-optical images taken with 2.5-μm-thick in film at 2.5 K. [H∥, H<sup>⊥</sup> in Oe]: (a) [0, 1], (b) [60, 8], (c) [100, 6],

Figure 12. Magnetization curve of 3.86-μm-thick indium film measured in perpendicular field at 2.5 K. Green (orange) circles represent the data measured at increasing (decreasing) field. Shadowed area represents the specimen condensation energy (1/2 in the reduced coordinates of this graph). Hc was determined from magnetization curve in parallel field, and

the specimen volume was determined from the slope of that curve. After Kozhevnikov et al. [25].

(d) [110, 3], and (e) [115, 1.3]. Superconducting regions are black. After Kozhevnikov et al. [25].


We start from construction of a thermodynamic potential F T <sup>~</sup>ð Þ ; <sup>V</sup>; Hi , which is the Legendre transform of the Helmholtz free energy F Tð Þ ; V; B to the variables ð Þ T; V; Hi . It is often referred to as the Gibbs free energy<sup>4</sup> :

$$\tilde{F} = F - \frac{\mathbf{B} \cdot \mathbf{H}\_i}{4\pi} V = F - \frac{B\_{\parallel} H\_{i\parallel}}{4\pi} V - \frac{B\_{\perp} H\_{i\perp}}{4\pi} V = F - \frac{B\_{\parallel} H\_{i\parallel}}{4\pi} V = F - \frac{H\_{\parallel}^2}{4\pi} V,\tag{13}$$

where F Tð Þ ; V; B is Helmoltz free energy. The term ð Þ B � Hi=4π V reflects work done by the magnet power supply to keep the set field H when the flux in the system changes [2]. In our case the flux of the perpendicular component is fixed, and therefore the term ð Þ B⊥Hi⊥=4π V

<sup>4</sup> It should be remembered that canonical Gibbs free energy is function of pressure, but not volume, as in this case.

Figure 13. Cross-sectional view of the specimen/field configuration in the laminar model for tilted field. H<sup>∥</sup> and H<sup>⊥</sup> are parallel and perpendicular components of the applied field, respectively. Domains are rectangular parallelepipeds extended along H<sup>∥</sup> (y-axis). The healing length Lh is the characteristic distance over which the disturbed field relaxes to the uniformly distributed state. In N domains the parallel component of the induction B<sup>∥</sup> ¼ H∥, while the perpendicular component B<sup>⊥</sup> ¼ H⊥=rn, where r<sup>n</sup> ¼ Dn=D ¼ Vn=V is volume fraction of the N phase and Vn is the total volume of the N phase. After Kozhevnikov et al. [26].

drops out. On that reason in pure perpendicular field, <sup>F</sup><sup>~</sup> <sup>¼</sup> <sup>F</sup> [2, 3, 8]. On the other hand, Hi<sup>∥</sup> ¼ H∥, due to the specimen geometry (see setting (iv) above).

To transform F T <sup>~</sup>ð Þ ; <sup>V</sup>; Hi to the total free energy F T <sup>~</sup>ð Þ ; <sup>V</sup>; <sup>H</sup> <sup>M</sup>, we need to add terms associated with energy of interaction of the applied field H with the specimen. In pure parallel case, this term is <sup>þ</sup> <sup>H</sup><sup>2</sup> <sup>=</sup>8<sup>π</sup> � �<sup>V</sup> [2]. In pure perpendicular case, it is � <sup>H</sup><sup>2</sup> =8π � �V (see appendix in [26] and/ or [2]). Therefore in our case the total free energy of the specimen is

$$
\tilde{F}\_M = \tilde{F} + \left[\frac{H\_{\parallel}^2}{8\pi} - \frac{H\_{\perp}^2}{8\pi}\right] V. \tag{14}
$$

<sup>~</sup><sup>f</sup> <sup>M</sup> <sup>¼</sup> <sup>f</sup> <sup>n</sup><sup>0</sup> � <sup>1</sup> � <sup>r</sup><sup>n</sup> ð Þ <sup>H</sup><sup>2</sup>

c 8π þ H2 ⊥ 8πr<sup>n</sup> � r<sup>n</sup> H2 ∥ 8π þ 2 H2 c 8π δ <sup>D</sup> <sup>þ</sup> <sup>2</sup>

<sup>D</sup><sup>2</sup> <sup>¼</sup> <sup>d</sup><sup>δ</sup> r2

> c 8π

After plugging this optimal D into Eq. (15), the latter takes form:

r2 <sup>n</sup> <sup>¼</sup> <sup>h</sup><sup>2</sup>

hcr<sup>⊥</sup> ¼

<sup>b</sup><sup>2</sup> <sup>¼</sup> <sup>b</sup><sup>2</sup>

<sup>⊥</sup> ¼ 1 � 4h<sup>⊥</sup>

<sup>~</sup><sup>f</sup> <sup>M</sup> <sup>¼</sup> <sup>f</sup> <sup>n</sup><sup>0</sup> � <sup>H</sup><sup>2</sup>

q

And magnitude of the reduced induction b ¼ B=Hc in the N domains is

<sup>⊥</sup> <sup>þ</sup> <sup>b</sup><sup>2</sup>

<sup>∥</sup> <sup>¼</sup> <sup>h</sup><sup>2</sup> ⊥=r<sup>2</sup> <sup>n</sup> <sup>þ</sup> <sup>h</sup><sup>2</sup>

ffiffiffiffiffiffiffi <sup>δ</sup>=<sup>d</sup> <sup>p</sup> � <sup>h</sup><sup>2</sup>

c <sup>8</sup><sup>π</sup> ð Þ <sup>b</sup><sup>⊥</sup> � <sup>h</sup><sup>⊥</sup>

<sup>M</sup> � �∇<sup>H</sup> <sup>F</sup>~<sup>M</sup>

where y and z are unit vectors along the y and z axes, respectively.

<sup>~</sup><sup>f</sup> <sup>M</sup> <sup>¼</sup> <sup>f</sup> <sup>n</sup><sup>0</sup> � <sup>H</sup><sup>2</sup>

two contributions in the specimen free energy.

At the IS/N transition rn=1, hence

using b<sup>⊥</sup> from Eq. (20): b<sup>2</sup>

Eq. (3):

Then, minimizing <sup>~</sup><sup>f</sup> <sup>M</sup> with respect to <sup>D</sup>, one finds equilibrium period of the structure

H2 c B2 ⊥

<sup>1</sup> � <sup>r</sup><sup>n</sup> ð Þ <sup>1</sup> � <sup>h</sup><sup>2</sup>

where h<sup>⊥</sup> and h<sup>∥</sup> are reduced components of the applied field H⊥=Hc and H∥=Hc, respectively. Important to note that with the optimal D the terms related to the S/N interfaces and to the field inhomogeneity near the surface are equal. This means that "responsibility" for deviation of the properties of real specimens from those in the PL model is equally shared between these

Minimizing <sup>~</sup><sup>f</sup> <sup>M</sup> with respect to <sup>r</sup>n, one finds equilibrium volume fraction of the N component:

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>4</sup>ð Þþ <sup>δ</sup>=<sup>d</sup> <sup>1</sup> � <sup>h</sup><sup>2</sup>

Before calculating the magnetic moment, we transform Eq. (17) substituting r<sup>n</sup> from Eq. (18) and

Now one can calculate the specimen magnetic moment from the definitive relationship

� � ¼ � <sup>∂</sup>F~<sup>M</sup>

ffiffiffiffiffiffiffi <sup>δ</sup>=<sup>d</sup> <sup>p</sup> � <sup>h</sup><sup>2</sup>

∥

<sup>∥</sup> ¼ 1 � 4h<sup>⊥</sup>

<sup>2</sup> <sup>¼</sup> <sup>f</sup> <sup>n</sup><sup>0</sup> � <sup>B</sup><sup>2</sup>

∂H<sup>∥</sup> y þ ∂F~<sup>M</sup> ∂H<sup>⊥</sup> z

!

⊥ 8π

∥

� <sup>2</sup> ffiffiffiffiffiffiffi

<sup>⊥</sup>= 1 � 4h<sup>⊥</sup>

<sup>¼</sup> <sup>d</sup><sup>δ</sup> <sup>1</sup> � <sup>r</sup><sup>n</sup> ð Þ<sup>2</sup>

> <sup>∥</sup> � <sup>h</sup><sup>2</sup> ⊥ rn � 4h<sup>⊥</sup>

" # r

<sup>n</sup> <sup>1</sup> � <sup>r</sup><sup>n</sup> ð Þ<sup>2</sup>

H2 ⊥ 8π D

> H2 c H2 ⊥

<sup>d</sup> <sup>1</sup> � <sup>r</sup><sup>n</sup> ð Þ<sup>2</sup> <sup>þ</sup>

ffiffiffi δ d

� �: (18)

ffiffiffiffiffiffiffi

<sup>∥</sup>. Then Eq. (17) becomes very compact:

<sup>1</sup> � <sup>r</sup><sup>n</sup> ð Þ<sup>2</sup>

δ=d p : (19)

δ=d p : (20)

: (21)

, (22)

H2 ∥ <sup>8</sup><sup>π</sup> � <sup>H</sup><sup>2</sup> ⊥ <sup>8</sup><sup>π</sup> : (15) 105

Intermediate State in Type-I Superconductors http://dx.doi.org/10.5772/intechopen.75742

: (16)

, (17)

Now, summing:


Intermediate State in Type-I Superconductors http://dx.doi.org/10.5772/intechopen.75742 105

$$\tilde{f}\_M = f\_{n0} - (1 - \rho\_n) \frac{H\_c^2}{8\pi} + \frac{H\_\perp^2}{8\pi\rho\_n} - \rho\_n \frac{H\_\parallel^2}{8\pi} + 2\frac{H\_c^2}{8\pi} \frac{\delta}{D} + 2\frac{H\_\perp^2}{8\pi} \frac{D}{d} (1 - \rho\_n)^2 + \frac{H\_\parallel^2}{8\pi} - \frac{H\_\perp^2}{8\pi}.\tag{15}$$

Then, minimizing <sup>~</sup><sup>f</sup> <sup>M</sup> with respect to <sup>D</sup>, one finds equilibrium period of the structure

$$D^2 = \frac{d\delta}{\rho\_n^2 (1 - \rho\_n)^2} \frac{H\_c^2}{B\_\perp^2} = \frac{d\delta}{(1 - \rho\_n)^2} \frac{H\_c^2}{H\_\perp^2}. \tag{16}$$

After plugging this optimal D into Eq. (15), the latter takes form:

$$\tilde{f}\_M = f\_{n0} - \frac{H\_c^2}{8\pi} (1 - \rho\_n) \left[ 1 - h\_{\parallel}^2 - \frac{h\_{\perp}^2}{\rho\_n} - 4h\_{\perp} \sqrt{\frac{\delta}{d}} \right],\tag{17}$$

where h<sup>⊥</sup> and h<sup>∥</sup> are reduced components of the applied field H⊥=Hc and H∥=Hc, respectively. Important to note that with the optimal D the terms related to the S/N interfaces and to the field inhomogeneity near the surface are equal. This means that "responsibility" for deviation of the properties of real specimens from those in the PL model is equally shared between these two contributions in the specimen free energy.

Minimizing <sup>~</sup><sup>f</sup> <sup>M</sup> with respect to <sup>r</sup>n, one finds equilibrium volume fraction of the N component:

$$
\rho\_n^2 = h\_\perp^2 / \left(1 - 4h\_\perp \sqrt{\delta/d} - h\_\parallel^2\right). \tag{18}
$$

At the IS/N transition rn=1, hence

drops out. On that reason in pure perpendicular field, <sup>F</sup><sup>~</sup> <sup>¼</sup> <sup>F</sup> [2, 3, 8]. On the other hand,

Figure 13. Cross-sectional view of the specimen/field configuration in the laminar model for tilted field. H<sup>∥</sup> and H<sup>⊥</sup> are parallel and perpendicular components of the applied field, respectively. Domains are rectangular parallelepipeds extended along H<sup>∥</sup> (y-axis). The healing length Lh is the characteristic distance over which the disturbed field relaxes to the uniformly distributed state. In N domains the parallel component of the induction B<sup>∥</sup> ¼ H∥, while the perpendicular component B<sup>⊥</sup> ¼ H⊥=rn, where r<sup>n</sup> ¼ Dn=D ¼ Vn=V is volume fraction of the N phase and Vn is the total volume of the N

To transform F T <sup>~</sup>ð Þ ; <sup>V</sup>; Hi to the total free energy F T <sup>~</sup>ð Þ ; <sup>V</sup>; <sup>H</sup> <sup>M</sup>, we need to add terms associated with energy of interaction of the applied field H with the specimen. In pure parallel case, this

> H2 ∥ <sup>8</sup><sup>π</sup> � <sup>H</sup><sup>2</sup> ⊥ 8π

" #

<sup>⊥</sup> <sup>þ</sup> <sup>B</sup><sup>2</sup> ∥ � �

<sup>c</sup> <sup>1</sup> � <sup>r</sup><sup>n</sup> ð Þ=8<sup>π</sup> � ��, where <sup>f</sup> <sup>n</sup><sup>0</sup> <sup>¼</sup> Fn0=<sup>V</sup> is free energy density

<sup>⊥</sup> � <sup>H</sup><sup>2</sup> ⊥

=8π.

=8π � �V (see appendix in [26] and/

V: (14)

� �=8πd, and plugging all

Hi<sup>∥</sup> ¼ H∥, due to the specimen geometry (see setting (iv) above).

<sup>=</sup>8<sup>π</sup> � �<sup>V</sup> [2]. In pure perpendicular case, it is � <sup>H</sup><sup>2</sup>

<sup>F</sup>~<sup>M</sup> <sup>¼</sup> <sup>F</sup><sup>~</sup> <sup>þ</sup>

<sup>c</sup>δ=8πD.

or [2]). Therefore in our case the total free energy of the specimen is

term is <sup>þ</sup> <sup>H</sup><sup>2</sup>

phase. After Kozhevnikov et al. [26].

104 Superfluids and Superconductors

Now, summing:

a. Free energy at zero field V f <sup>n</sup><sup>0</sup> � <sup>H</sup><sup>2</sup>

c. Energy of the S/N interfaces 2VH<sup>2</sup>

b. Energy of the field B in the N domains Vr<sup>n</sup> B<sup>2</sup>

in Eq. (14), one obtains for <sup>~</sup><sup>f</sup> <sup>M</sup> <sup>¼</sup> <sup>F</sup>~M=V:

d. Excess energy of the field over the healing length 2VLh rnB<sup>2</sup>

of the N state in zero field.

$$h\_{\sigma\perp} = \sqrt{4(\delta/d) + 1 - h\_{\parallel}^2} - 2\sqrt{\delta/d}.\tag{19}$$

And magnitude of the reduced induction b ¼ B=Hc in the N domains is

$$b^2 = b\_{\perp}^2 + b\_{\parallel}^2 = h\_{\perp}^2/\rho\_n^2 + h\_{\parallel}^2 = 1 - 4h\_{\perp}\sqrt{\delta/d}.\tag{20}$$

Before calculating the magnetic moment, we transform Eq. (17) substituting r<sup>n</sup> from Eq. (18) and using b<sup>⊥</sup> from Eq. (20): b<sup>2</sup> <sup>⊥</sup> ¼ 1 � 4h<sup>⊥</sup> ffiffiffiffiffiffiffi <sup>δ</sup>=<sup>d</sup> <sup>p</sup> � <sup>h</sup><sup>2</sup> <sup>∥</sup>. Then Eq. (17) becomes very compact:

$$\tilde{f}\_M = f\_{n0} - \frac{H\_c^2}{8\pi} (h\_\perp - h\_\perp)^2 = f\_{n0} - \frac{B\_\perp^2}{8\pi} (1 - \rho\_n)^2. \tag{21}$$

Now one can calculate the specimen magnetic moment from the definitive relationship Eq. (3):

$$\mathbf{M} \equiv -\nabla\_H(\tilde{F}\_M) = -\left(\frac{\partial \tilde{F}\_M}{\partial H\_{\parallel}} y + \frac{\partial \tilde{F}\_M}{\partial H\_{\perp}} z\right)\_{\prime} \tag{22}$$

where y and z are unit vectors along the y and z axes, respectively.

The first term in Eq. (22) is

$$\frac{\partial \tilde{F}\_M}{\partial H\_{\parallel}} = \frac{V}{H\_c} \cdot \frac{\partial \tilde{f}\_M}{\partial h\_{\parallel}} = \frac{V}{H\_c} \cdot \frac{H\_c^2}{8\pi} \cdot 2(h\_\perp - h\_\perp) \frac{\partial b\_\perp}{\partial h\_{\parallel}}.\tag{23}$$

Specifically, in the thick slabs r<sup>n</sup> ¼ H=Hc (Eq. (18)), and therefore (a) B ¼ h=r<sup>n</sup> ð ÞHc ¼ Hc and Hi ¼ B=μ ¼ Hc, and (b) 4πM=V ¼ �Hc þ H, meaning that 4πMð Þ0 =V ¼ �Hc and Hcr ¼ Hc, exactly as it takes place in the PL model (Figure 4b). The third condition of Eq. (11) B ¼ H follows from the law of the flux conservation always valid for infinite slabs. Thus the LMTF model explains why the PL model works the best for thick specimens: because in such case combined contributions due to near-surface field inhomogeneity and due to the S/N interfaces (both are characterized by the ratio δ=d) are negligible compared to the bulk terms in Eqs. (15)

Now, when we are convinced in correctness of the formulas for magnetization (see more in [25, 26]), we can rewrite Eq. (17) in its canonical form coinciding with the mandatory form

> ð<sup>H</sup> 0

More than three decades starting from the 1930s, the problem of the IS was in the main focus of experimental and theoretical researches on superconductivity. This resulted in significant progress reached in understanding properties of the IS as well as properties of superconducting state as a whole. Excellent reviews of these researches are available in [1, 12]. However some puzzles in the IS properties remained open until their possible explanations emerged in studies of recent

In particular, we discussed a recently developed phenomenological model of the IS composed for infinite slabs in arbitrary tilted magnetic field. Naturally, this model is not and cannot be free of disadvantages. One of them can be associated with the use of an oversimplified Tinkham approximation for the field distribution and domain shape near the surface through which the flux enters and leaves the specimen. We believe that modern experimental capabilities associated, e.g., with muon spectroscopy and noninvasive scanning magnetic microscopy, can help to resolve this important and very interesting issue, which we discussed in the Section IV. The new model discussed in Section VI is restricted by the slab-like specimens. Its extension to all ellipsoidal shapes covered in the model of Peierls and London is another possible avenue

Finally, it is important to remind that the IS is one of two inhomogeneous superconducting states. The second state is the mixed state in type-II superconductors, taking place in vast majority of superconducting materials, including those used in practical applications. Therefore understanding of properties of the IS can help to understand properties of the mixed state. As an example, the field distribution and shape of the normal domains (vortices in type-II materials) near the specimen surface should be similar in both these inhomogeneous states.

<sup>M</sup> � <sup>d</sup><sup>H</sup> <sup>¼</sup> Fn<sup>0</sup> � <sup>H</sup><sup>2</sup>

c 8π V � ð<sup>H</sup> 0

Intermediate State in Type-I Superconductors http://dx.doi.org/10.5772/intechopen.75742 107

M � dH (29)

and (17).

for the total free energy Eq. (4):

7. Concluding remarks

of research on the IS.

<sup>F</sup>~<sup>M</sup> <sup>¼</sup> <sup>F</sup>~Mð Þ� <sup>H</sup> <sup>¼</sup> <sup>0</sup>

ð<sup>H</sup> 0

where the components of M are given by Eqs. (24 and 25).

years. In this chapter we mostly focused at results of these studies.

M � dH ¼ Fs<sup>0</sup> �

Since ∂b⊥=∂h<sup>∥</sup> ¼ h∥=b<sup>⊥</sup> (see Eq. (20)), the final form of the parallel component of the specimen magnetic moment is

$$-M\_{\parallel} = \frac{\partial \bar{F}\_M}{\partial H\_{\parallel}} = \frac{V}{H\_c} \frac{H\_c^2}{8\pi} 2\left(b\_\perp - h\_\perp\right) \frac{h\_{\parallel}}{b\_\perp} = \frac{V H\_c}{4\pi} \left(1 - \frac{h\_\perp}{b\_\perp}\right) h\_{\parallel} = \frac{V}{4\pi} \left(1 - \rho\_n\right) H\_{\parallel}.\tag{24}$$

And the perpendicular component of the moment is

$$-M\_{\perp} = \frac{\partial \bar{F}\_M}{\partial H\_{\perp}} = \frac{V}{H\_{\varepsilon}} \frac{\partial \bar{f}\_M}{\partial h\_{\perp}} = -\frac{V}{H\_{\varepsilon}} \frac{2H\_{\varepsilon}^2}{8\pi} \left(b\_{\perp} - h\_{\perp}\right) \left(\frac{\partial b\_{\perp}}{\partial h\_{\perp}} - 1\right) = \frac{V}{4\pi} \left(1 - \rho\_{\pi}\right) \left(1 - \frac{\partial B\_{\perp}}{\partial H\_{\perp}}\right) B\_{\perp}.\tag{25}$$

All obtained formulas are analyzed in detail in [25, 26], where it is shown that the model correctly describes experimental data. In particular, the coherence length calculated from measured D using Eq. (16) agrees well with that obtained from the magnetic field profile measured in [36]. Here we confine our discussion by limiting cases.

In parallel field (H<sup>⊥</sup> ¼ 0) the model (Eq. (18)) yields r<sup>n</sup> = 0, meaning that the specimen is in the Meissner state where the N phase is absent. Then <sup>~</sup><sup>f</sup> <sup>M</sup> (Eq. (17)) converts to Eq. (5):

$$\tilde{f}\_M = f\_n - \frac{H\_c^2}{8\pi} \left(1 - h^2\right) = f\_n - \frac{H\_c^2}{8\pi} + \frac{H^2}{8\pi},\tag{26}$$

and M ¼ M<sup>∥</sup> (Eq. (24)) converts to Eq. (1):

$$M\_{\parallel} = M = -\frac{V}{4\pi}H.\tag{27}$$

In perpendicular field (H<sup>∥</sup> ¼ 0) one can see that hcrð Þ ¼ hcr<sup>⊥</sup> decreases with decreasing thickness d (Eq. (19)) in accord with the experimental data [25, 26, 33], and the induction Bð Þ ¼ b � Hc in N domains equals to Hc at H ¼ HI ¼ 0 and decreases with increasing H (Eq. (20)), as it was found experimentally in [34]. For magnetization 4πM=V at H ! 0, when r<sup>n</sup> ¼ 0, the model (Eq. (25)) yields

$$\frac{4\pi M(0)}{V} = \frac{V}{4\pi} \left(1 - \frac{\partial B\_{\perp}}{\partial H\_{\perp}}\right) H\_c. \tag{28}$$

Since B decreases with increasing H, ð Þ ∂B⊥=∂H<sup>⊥</sup> < 0, and therefore the expression in parentheses is greater than unity. This makes 4πMð Þ0 =V greater than Hc, thus explaining appearance of the excess magnetization at HI as it is seen, e.g., in Figures 9 and 12.

The infinite slab in perpendicular field represents ellipsoid with η = 1. If the slab is thick (i.e., d ≫ δ), the LMTF model converts to the PL model for specimens with unity demagnetization. Specifically, in the thick slabs r<sup>n</sup> ¼ H=Hc (Eq. (18)), and therefore (a) B ¼ h=r<sup>n</sup> ð ÞHc ¼ Hc and Hi ¼ B=μ ¼ Hc, and (b) 4πM=V ¼ �Hc þ H, meaning that 4πMð Þ0 =V ¼ �Hc and Hcr ¼ Hc, exactly as it takes place in the PL model (Figure 4b). The third condition of Eq. (11) B ¼ H follows from the law of the flux conservation always valid for infinite slabs. Thus the LMTF model explains why the PL model works the best for thick specimens: because in such case combined contributions due to near-surface field inhomogeneity and due to the S/N interfaces (both are characterized by the ratio δ=d) are negligible compared to the bulk terms in Eqs. (15) and (17).

Now, when we are convinced in correctness of the formulas for magnetization (see more in [25, 26]), we can rewrite Eq. (17) in its canonical form coinciding with the mandatory form for the total free energy Eq. (4):

$$\tilde{F}\_M = \tilde{F}\_M(H=0) - \int\_0^H \mathbf{M} \cdot d\mathbf{H} = F\_{s0} - \int\_0^H \mathbf{M} \cdot d\mathbf{H} = F\_{n0} - \frac{H\_c^2}{8\pi}V - \int\_0^H \mathbf{M} \cdot d\mathbf{H} \tag{29}$$

where the components of M are given by Eqs. (24 and 25).
