Statistical Mechanics - Physics at Oregon State University

8.3. MEAN FIELD RESULTS. 167
An interesting case is related to the possibility of having spontaneous magnetization.
When there is no external field we need to solve
m = tanh (β ∗ m) (8.41)
If we compare the slope of the hyperbolic tangent at the origin with the number
one, the slope of the left hand side, we see that for β ∗ > 1 there are three
solutions for m. When β ∗ 1, there is only one solution. Obviously, m = 0
is always a solution. When β ∗ is only slightly larger than one, the solutions of
the previous equation can be found by expanding the hyperbolic tangent, since
in this case all solutions have values for m which are very small. Therefore we
find
m ≈ β ∗ m − 1
3 (β∗m) 3
This leads to m = 0, as expected, or
(8.42)
m 2 ≈ 3(β∗ − 1)
β ∗3 ≈ 3(β ∗ − 1) (8.43)
If there is a non-zero solution which is very small, we need β∗ to be slightly
larger than one. Since such a small non-zero solution only occurs very close to
Tc, we find immediately that β∗ c = 1. Hence for T < Tc, where Tc is defined by
β∗ c = 1 or kBTc = qJ, we have a second solution that vanishes at Tc like
m ∝ 3(β ∗ − 1) =
3( Tc
T
− 1) =
3
T (Tc − T ) (8.44)
which gives a critical exponent β (do not confuse this β with kBT ) of 1
2 , just
as we argued in thermodynamics for mean field theory!
In order to find the thermodynamically stable state we have to compare the
free energies. From the result for the partition function we obtain
G(T, h = 0, N) = −NkBT log(2 cosh(β ∗ m)) + 1
2 NqJm2 (8.45)
Since m is close to zero for T near Tc the right hand side can be expanded
like
G(T, 0, N) ≈ −NkBT log(2) − NkBT log(1 + 1
2 (β∗ m) 2 ) + 1
2 NqJm2 ≈
−NkBT log 2 − NkBT 1
2 (β∗ m) 2 + 1
2 NqJm2 =
−NkBT log 2 − NqJ
2
kBT β 2 qJ − 1 m 2
−NkBT log 2 − NqJ
2kBT [qJ − kBT ] m 2