Материал: Маслов ИНТРОДУЦТИОН ТО ПХЫСИЦС ОФ СЕЦОНД-ОРДЕР МАГНЕТИЦ 2015

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T < θ the	solution x = 0 is	maximum.	Actually,	in	this case
∂2 F (x,T )	= N [−θ +T ]< 0 . For	x ≠ 0 let us	consider	two	cases for
∂x2

which we know the approximate analytical expressions for order parameter: 1) T < θ and T → θ− ; 2) T θ.

1. We earlier obtained (see § 14) that at T < θ and T → θ− the order

parameter was x ≈

−

. Therefore

3 1

∂2 F (x,T )

N −θ+

= N −θ+

∂x2

−

− 2

−3 1

(16.13)

=N −3T + 2θ+T ≈ 2N (θ−T )> 0.3Tθ − 2

So, this is the minimum.

2. We earlier obtained (see § 14) that at T θ the order parameter

was

≈1− 2e−2

. Therefore

∂

F (x,T )

= N −θ+

∂x2

−2

(16.14)

− 1− 4e

T 2 θ

=N −θ+ 4 e T > 0.

So, this is the minimum as well.

It can be shown that at any given temperature 0 ≤T < θ solution of the Curie-Weiss equation with x ≠ 0 corresponds to local minimum of

F = F (x,T ), and solution with x = 0 corresponds to local maximum.

Finally, we obtain that at T > θ the paramagnetic state is realized, and at T < θ the ferromagnetic state occurs.

17. Free energy of a ferromagnet near the critical temperature

We found the Helmholtz free energy																																	F = F (x,T, H ) at arbitrary T
and H :
F = F (T, H )+ N						−		x2			θ− xh +T 1+ x ln 1+ x																														+T 1− x ln 1			− x .
F = F (T, H )+ N						−					θ− xh +T 1+ x ln 1+ x																														+T 1− x ln 1			− x .
0						2																				2												2				2
						2																				2												2				2		2
																																												(17.1)
The order parameter							x = 0 at T > θ and																												x		1 at T < θ and T → θ− .
The order parameter							x = 0 at T > θ and																												x		1 at T < θ and T → θ− .
Therefore we can approximately expand																																				F = F (x,T, H )							into Taylor
series at T ≈ θ up to O(x4 )
		1+ x							+ x										1+ x																					+ x
		1+ x					1		+ x								=		1+ x						ln			(1+ x)−											1	+ x		ln 2 ≈
			2		ln					2							=			2					ln			(1+ x)−												2		ln 2 ≈
			2							2										2																				2
	≈		1 (1		+ x) x −										x2				+		x3				−				x4				−					1 (1+ x)ln 2 ≈
	≈		1 (1		+ x) x −														+						−								−					1 (1+ x)ln 2 ≈
			2													2						3								4								2						(17.2)
≈ 1				x	2			x		3				x			4										x			3					x	4				1	(1+ x)ln 2 =			(17.2)
					2					3							4	+ x2 −												3						4
	x −					+						−																				+							−
	x −					+						−																				+							−	2
2			2			3							4													2								3						2
					1							x	2					x		3				x		4								1		(1+ x)ln 2.
			=		1	x +						x				−		x				+		x								−		1
			=		2	x +						2				−		6				+										−		2
Analogously,					2							2						6						12										2
Analogously,
1− x			1− x									1				−x +							x2							x3							x4				1	(1− x)ln 2.		(17.3)
		ln							≈																	+								+						−
2		ln			2				≈			2										2				+					6			+		12				−	2
2					2							2										2									6					12					2

Combine


			1+ x					1+ x							1− x			1− x
				2		ln				2		+			2		ln		2	≈

			1				x2				x3			x4			1	(1+ x)ln 2 +						(17.4)
		≈		x +					−			+				−
			2				2				6			12			2
	1	−x +	x2			x3				x4			1		(1− x)ln 2 =					x2		x4
+	2				+			+				−									+		−ln 2.
			2			6				12			2							2		12

For the Helmholtz free energy we obtain

F = F	(T, H )+ N	−	x2	θ− xh +T	x2	+T	x4	,

0		2		2		12

where F0 (T, H )= F0 (T, H )−T ln 2 . And finally,

F (x,T, H )

F0 (T, H )

(T −θ)x

− xh.

In order not to increase accuracy, let us assume that T = θ

F (x,T, H )

F0 (T, H )

(T −θ)x

− xh.

(17.5)

(17.6) in the term

(17.7)

The Eq. (17.7) is correct at T > θ (x = 0) and at T → θ− . Let us analyze graphically the dependence F (x) at various T and H .

1. At T > θ and H = 0 , F (x) has the minimum at x = 0 (see Fig. 17.1), i.e., the atoms do not have the mean magnetic moments.

2. At T > θ and H > 0 , F (x)	has the minimum at x > 0 due to the
term −xh (see Fig. 17.2).	H < 0 , F (x) has the minimum at
3. Analogously, at T > θ and	H < 0 , F (x) has the minimum at
x < 0 (see Fig. 17.3).
4. At T < θ and H = 0 , F (x)	has two minima (see Fig. 17.4) divid-

ed by barrier. Thus, the ground state is doubly degenerated: x > 0 or x < 0 .

Fig. 17.1. Free energy dependence on x at T > θ and H = 0

Fig. 17.3. Free energy dependence on x at T > θ and H < 0

Fig. 17.2. Free energy dependence on x at T > θ and H > 0

Fig. 17.4. Free energy dependence on x at T <θ and H = 0

5. At T < θ and H ≠ 0 one of the minima becomes deeper (see Fig. 17.5), i.e., the magnetic field removes the degeneracy. Now, if the magnetic field tends to zero H →0 , the macroscopic system remains in the state corresponding to this deeper minimum, since the probability of tunneling through the barrier is very low.

Fig. 17.5. Free energy dependence on x at T <θ and H > 0

18. Spontaneous symmetry breaking

at the paramagnetic-ferromagnetic transition

Earlier we showed (see § 13) that at T < θ and H = 0 the order parameter x ≠ 0 . This corresponds to two physically unequivalent states

with magnetization M = NVµB x > 0 and M < 0 for the magnetic mo-

ments alignment along and against the selected axis z, respectively. In other words, at T < θ and H = 0 the system state is doubly degenerated, i.e., two different states have the same values of energy. The reason

is the parity of Helmholtz free energy function			F (x,T )= F (−x,T ) at
H = 0. This is apparent from the expression for F (x,T )					near the Curie
temperature
F(x,T ) = F0 (T )+ N T −θ x2	+		θ	x4 .		(18.1)
F(x,T ) = F0 (T )+ N T −θ x2	+	12		x4 .		(18.1)
2		12
Note that at H ≠ 0 the expression for F (x,T ) takes the form
F(x,T ) = F0 (T )+ N T −θ x2 +	θ		x4 − xh		,	(18.2)
F(x,T ) = F0 (T )+ N T −θ x2 +	12		x4 − xh		,	(18.2)
2	12
45

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