Statistical Physics, Second Revised and Enlarged Edition

122 Phase transitions

answeris

M=Nμtanh(μB/kkkBT)

=Nμtanh[μ(B 0 +λM)/kkkBT] (11.3)

The spontaneous magnetizationisobtainedbysettingthe appliedfieldB 0 = 0
in (11.3), givinga transcendental equation for M( 0 ). ReplacingM( 0 )bythe
dimensionless order parameterm,(11.1), we obtain

m=tanh(mTC/T) (11.4)

where

TTTC=Nμ^2 λ/kkkB (11.5)

Equation (11.4) canbe readilysolvedgraphically (or numerically) asinFig. 11.2.
Plottedas thex-variableonthegraphisx=mTTTC/T,so that (11.4)becomesxT/TTTC=
tanhx.The solution is simply the intersection of the curvem=tanhxwith the straight
linem=xT/TTTC.Since theslope oftanhxisunity close tox= 0 ,itis at once evident
that:

1. WhenT>TTTC,(11.4) has onlyone solution, namelym= 0. The spontaneous
   magnetization is zero, and the system is disordered.
   2 .T=TTTCis seen tobeadefinite transition temperature,inthatbelowTTTCother
   solutions arepossible.TTTCdepends linearlyon the mean field parameterλ.It is
   often called (for the ferromagnetic transition) the Curie temperature.
   3 .WhenT<TTTC,theequationhas threepossiblesolutions. We shallseelater that
   them=0 is now an unstable equilibrium. The solutions withm= 0 are the
   stable ones. Hence thesolid developsbelowTTTCaspontaneous magnetization with
   adefinite magnitude,but withanindefinitedirection.

T>TTTcT=TTTc

x

m

T<TTTc

+ 1

1

Fig. 11. 2 Findingthe spontaneous magnetization. The solution of (11.4) formisgiven bythe intersection
of the tanh curve with the straight line. Straight lines are drawn for these different temperatures.