3.G.1. Continuous Phase Transitions

Since φ is a real scalar, if η is a complex number, vector, or tensor of odd rank, we must have α j =0  for all odd j because any term containing odd powers of η can never be a real scalar.  In which case, eq(3.63a) simplifies to

        ϕ( T,Y,η )= ϕ 0 ( T,Y )+ α 2 ( T,Y ) η 2 + α 4 ( T,Y ) η 4                   (3.64)

where, to ensure (formal) global stability, we must have

        α 4 ( T,Y )>0                             (3.66)

[ α 4 <0  implies global minima of φ at η±  ]

 

For fixed Y and T, each equilibrium state (phase) is a minimum of φ so that

        ( ϕ η ) TY =0                      ( 2 ϕ η 2 ) TY >0                             (3.65)

i.e.,

        2 α 2 η+4 α 4 η 3 =0              2 α 2 +12 α 4 η 2 >0                      (3.65a)

 

There are 2 solutions to eq(3.65a), namely,

        η=0                         with α 2 >0              (3.65b)

and

        η 2 = α 2 2 α 4                with  α 2 <0            (3.65c)

Obviously, (3.65b) and (3.65c) correspond to the disordered ( T> T C  ) and ordered ( T< T C  ) states, respectively.  Thus, the transition, or critical, point satisfies

α 2 ( T,Y )=0                     (3.65d)

For a given Y, let T C ( Y )  be the solution of eq(3.65d), i.e.,

α 2 [ T C ( Y ),Y ]=0             (3.65e)

Thus, T C ( Y )  describes a coexistence curve in the Y-T plane.  The conditions (3.65b,c,e) are guaranteed if we write

        α 2 ( T,Y )= α 0 ( T,Y )[ T T C ( Y ) ]                    (3.67)

where α 0 >0  is expected to be slowly varying in the neighborhood of the transition point.  Hence, eq(3.65c) becomes,

        η=± α 0 2 α 4 ( T C T )                                         (3.69)

Putting everything into (3.64) gives

        ϕ( T,Y,η )={ ϕ 0 ( T,Y ) ϕ 0 ( T,Y ) α 0 2 4 α 4 ( T C T ) 2            for       T> T C T< T C               (3.70)

Heat Capacity

Since φ is the Gibbs free energy, the molar heat capacity is given by,

        c Y =T ( 2 ϕ T 2 ) Y                               (3.71)

so that

        c Y ={ c Y0 c Y0 + α 0 2 2 α 4                 for       T> T C T< T C               (3.71a)

where

        c Y0 =T ( 2 ϕ 0 T 2 ) Y

Thus, there is a discontinuity at T C  of magnitude

        Δ c Y = c Y ( T C ) c Y ( T C + )= α 0 2 2 α 4                         (3.72)

which gives the c Y  vs T plot a shape of a λ (see Fig.3.21) and hence the reason for calling the critical point a λ-point.

Superfluid

The normal- superfluid transition of H e 4  is continuous.

The order parameter is the macroscopic superfluid (complex) wave function Ψ  so that

        ϕ( T,P,Ψ ) ϕ 0 ( T,P )+ α 2 | Ψ | 2 + α 4 | Ψ | 4                         (3.73)

with α 2 =( T T C ) α 0 .  Similar to (3.69), we have

        | Ψ | 2 = α 0 2 α 4 ( T C T )         Þ    Ψ= e iθ α 0 2 α 4 ( T C T )

where the phase θ is a real number that can be set to zero when current flow is absent.

External Force

In the presence of an external force f that couples to the order parameter η, the relevant free energy is [see (3.64)]

        ψ( T,Y,f )= ϕ 0 ( T,Y )+ α 2 η 2 + α 4 η 4 fη               (3.74)

which implies the Legendre transform (for equilibrium states),

        ϕ( T,Y,η )=ψ( T,Y,f )f ψ f

                        =ψ( T,Y,f )+fη

where ψ f =η .  Typical plots of ψ  can be found in Fig.3.22.

 

For fixed T,Y, and f, the equilibrium phases are minima of ψ so that

        ( ψ η ) TYf =2 α 2 η+4 α 4 η 3 f=0                     (3.75)

        ( 2 ψ η 2 ) TYf =2 α 2 +12 α 4 η 2 >0

See Fig.3.22 for a few typical solutions.  Note that (3.75) is simply the assertion that for equilibrium states, f is indeed the force conjugate to η, i.e., f= ( ϕ η ) TY .

More interesting is the susceptibility

        χ( f )= ( η f ) TY = ( 2 ψ f 2 ) TY

with the implicit assumption that all quantities take on their equilibrium values.  Taking the partial derivative of eq(3.75) gives

        2 α 2 ( η f ) TY +12 α 4 η 2 ( η f ) TY 1=0

Þ            χ( f )= 1 2 α 2 +12 α 4 η 2                              (3.76)

For f=0 , eq(3.65b,c) gives

        η={ 0 ± α 2 2 α 4          for   T> T C T< T C                     (3.76a)

so that with α 2 =( T T C ) α 0 , we have

        χ( 0 )={ 1 2 α 2 = 1 2( T T C ) α 0 1 4 α 2 = 1 4( T C T ) α 0               for   T> T C T< T C              (3.77)

which exhibits a divergence at T= T C .

Paramagnetic-Ferromagnetic Transition

In a para- to ferro- magnetic transition, critical values are called Curie values.

The order parameter is the magnetization vector M.

The symmetry that is broken is the rotational symmetry.  Thus,

        ϕ( T,M )= ϕ 0 ( T )+ α 2 MM+ α 4 ( MM ) 2

        ψ( T,H )= ϕ 0 ( T )+ α 2 MM+ α 4 ( MM ) 2 HM                 (3.78)

For H=0 , the analog of (3.76a) is

        M={ 0 ± M ˆ α 0 2 α 4 ( T C T )                 for   T> T C T< T C

where M ˆ  is a unit vector.

 

The heat capacity exhibits the λ shape as shown in Fig.3.23.