Aspin-^12 solid 33For nuclear spinsitwill bea‘dilution refrigerator’, adevice capableofreaching
3 –10 mK whose working substance isa^3 He–^4 He mixture which phase separates
at temperaturesbelow 0.7 K. The magnet nowadaysis usually a superconducting
solenoidoperatedinthemainheliumbathforbothexperiments.
The CMN coolant is often in the form of a powdered slurry around a bunch of fine
copper wiresfor thermalcontact. The active particlesfor nuclear cooling are the^63 Cu
and^65 Cu nucleiin copper metal(bothofthese nucleihavingspin^32 ).Thecopper can
be in the form of wires, plates or powder.
Thefinalelementinthe experimentistheheat switchandthermallink. We require
an arrangement which willgive excellent thermal contact in the precoolingstage
(A→B) but which can give near perfect isolation in the adiabatic cooling stage
(B→C). Mechanicalsystemsgive too muchfrictionalheating,anduse oflow
pressure heliumgas as a heat exchange medium has serious drawbacks (e.g.itis
difficult to remove for the second stage). The modern solution is to use a thermal
linkof high-conductivitymetal(silver or copper)brokenbya superconductingheat
switch. This consists of apure metal (aluminium or tin) which is a superconductor.
When a (fairly small) magnetic field is applied to the switch, the superconductivity
isdestroyedandthe metalisagoodconductor ofheat, as wantedinthe precooling
leg. However, when this field is removed and the switch becomes superconducting,
itbecomes a very poor conductor ofheat–the superconducting electrons movein
anorderedway,sothat althoughtheycarrychargewithout resistance theyhave no
entropy to carry! A superconducting heat switch can have a conductance ratio of 10^5
orhigher at 10 mK.
Suppose we wishto cool liquid^3 Heinto theμK temperature region, a worthwhile
tasksinceitbecomesa‘superfluid’below1–2 mK.Clearlywecanusethearrangement
of Fig. 3. 6 to cool some Cu spins. The major technical problem remaining is to achieve
adequate thermalcontactbetween theCuspins andthe sampleof liquid^3 He.And
this turns out to be the most severe problem of all. The trick is to cut down extraneous
heatinputs (to thepWlevel)withantivibration measures andwithcarefulelectrical
screening, and also to maximize the area of contact between the copper metal and the
liquid with ultrafine sintered silver powder. In this way it is possible, at the time of
writing,toachievehelium temperatures ofbelow 9 0 μK.
What limits the final temperature? This is a question worth asking. The problem
canbe statedsimply.IfItake (3.7) seriously,thenisit not possibletomakethefinal
temperatureTTT 2 as small as I wish, simply by reducingB 2 to zero? In fact can the spins
notbe cooledto theabsolute zero?
The answeris ‘no, theycannot’, one ofmanywaysofstatingthethirdlaw of
thermodynamics. And the reason lies in the meaning of (3.7). That equation arose
from thefact that the occupation numbers mustbe unchangedinanadiabatic process,
i.e.that 2 μB/kkkBTmust remainatafixedvalue. The numerator 2μBisthe energy level
spacingε, and it is this quantity which cannot reach zero. The two spin states can never
have the same energy (‘the groundstate ofa system can neverbedegenerate’ might
be another wayofstatingthethirdlaw). Insteadthere willalwaysbe some residual