Statistical Physics, Second Revised and Enlarged Edition

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Superfluid states in Fermi–Dirac systems 173

Temperature (mK)

Pressure

(b

ar
)

Normal liquid

0

0

10

20

30

40

123

Superfluiduperflui
B-phaseB-phase

Solid

A

Fig. 15.2Thephasediagram of^3 Heatmillikelvin temperatures, andin zero appliedmagneticfield.The
A-andB-phases aredifferent superfluidphases.


3 .Inthe A-phase, these vectors correspondtoaspin(S)direction andan orbital
(L)direction, sharedbyallpairsin a particular region ofthefluid. Hence the
A-phase has many similarities with liquid crystals, which also order with a highly
anisotropic vector orderingparameter. (Themagnitudeoftheorder parameter
gives the strength of the ordering; the direction of the vectorgives the direction
of alignment of the liquid crystal molecules.)


  1. In the B-phase, theS=1,L= 1 pairs collect togetherin a muchmore uniform
    way, and the anisotropyis practicallynon-existent (although it can have secondary
    manifestations). Many experimental properties of the B-phase are described sim-
    ply in terms ofasingle scalarparameter. Figure 15.3gives an example of this
    type of behaviour, and it also serves as a dramatic illustration of the existence of
    superfluidity.As noted above in section 14.2.2,^3 Heisahighlyviscousliquidabove
    TTTC,so that thedampingofavibratingwireis verylarge. When theliquid is cooled
    belowTTTC, however, the damping falls off rapidly, varying as exp(−/kkkBT),the
    usualBoltzmannfactor, atlow temperatures as the normalfluid isfrozen out. At
    thelowest temperatures thevibratorbehaves asif in a vacuum withadamping
    coefficient 100 000 times less than atTTTC.

  2. The fact that the pairs haveS= 1 means that they are magnetic. Thebehaviour
    ofthe superfluid isthus profoundly influencedbyappliedmagneticfields. This
    is so even in the B-phase, which becomes unstable if a large enough field (of
    order 0.4T)is applied,the superfluidreverting to the more magneticA-phase. The
    wholephasediagram changes, whichiswhytheB= 0 qualification was needed

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