Read Experimental techniques: Methods for cooling below 300 mK text version

Journal of Low Temperature Physics, Vol. 87, Nos. 3/4, 1992

Experimental Techniques: Methods for Cooling Below 300 mK

G. Frossati

Kamerlingh Onnes Laboratory,Leiden University, Nieuwsteeg 18, 2311 SB Leiden, The Netherlands

There are at present three methods for cooling samples to temperatures below 300 mK: dilution, Pomeranchuk, and nuclear refrigeration. We give the basic principles of these methods with more details concerning dilutions refrigerators. This should allow the construction of a simple all plastic refrigerator for temperatures lower than 15 mK, or an even simpler Pomeranchuk cell. The source of heat leaks and other important points for reaching temperatures in the microkelvin range with nuclear refrigerators are given in the lecture by F. Pobell.

1. INTRODUCTION In the course of these two lectures we will discuss the techniques for achieving temperatures below 300 mK, the so-called ultralow temperature region. For a comprehensive review of such techniques, including thermometry, thermal contact, etc., one should refer to the textbooks, Refs. 1-4. There are at present three main methods of cooling below 200 mK: (1) dilution refrigeration, (2) Pomeranchuk refrigeration, and (3) nuclear refrigeration. Dilution and Pomeranchuk refrigeration use the peculiar properties of 4He and 3He in the milliKelvin range.With the help of some phase diagrams we will point out some relevant features of such systems. Helium liquids are called quantum liquids, because of the large zero point energy of their light atoms combined with very weak electronic interaction (noble gases). Figure 1 shows the Lennard-Jones potential of helium (equal for both isotopes), defined as

595

0022-2291/92/0500-0595506.50/0 9 1992 Plenum Publishing Corporation

596

G. Frossati

2(3 ..-.15 "f_ 0 1 > 5 0

_ S P -G.zs~ -

7,

LR~o2..86~RE=3.65~

'1";3 I 4 , ,:

':MZ_j

Fig. 1. The Lennard-Jones potential of helium.

crH is the hard core distance, and the other parameters are clear from the figure. The attractive term (van der Waals attraction) tends to bind the atoms together causing a phase transformation from gas to liquid or to solid according to the intensity of the attractive force. The repulsive part reflects the impenetrability of matter. The kinetic energy of the system is the sum of ordinary kinetic energy, which for monoatomic gases is

E k =

3NkT

with that, of quantum nature, due to zero point motion h 2 ( N ~ 2/3

~~

\ v/

where N/V is the atomic density At low temperatures Ek goes to zero but Eko, which depends only on the atomic distance for a given mass, might become very large and counterbalance the attractive energy. This is the reason why helium is liquid even at T = 0. Figure 2 shows the phase diagrams of 3He (a, b) and 4He (c).

Methods for Cooling Below 300 mK

3H~ 4He

597

Q

20

9 Pn n 10

~':C'P' ~"

~

solid

13-

20 ---, 10

h.c.p, solid

b.c.c.

2 1 0

2 8 9 M Pe 0.32K

fa

~

2

normal

Uquid 2

T(K)

~u,d\

0

i 1

t 2 r (K)

t 3

1

1

3

3H~z

b

~_ 3.2 ~.3 . 0 G

2.8

liquid

, ,, ,~1

,~

\

,~1

,

[

,1,

,,~

I

I0

tO0 T(mK)

I000

Fig. 2. The phase diagrams of 3He (a, b) and 4He (c).

4He solidifies at P > 2.5 MPa for T < 1 K and becomes superfluid below 2.17 K at P = 0, the superfluid temperature decreasing somewhat for higher pressures. In the superfluid state the entropy of 4He goes quickly to zero since all the atoms (bosons) will tend to occupy the same ground state. (Because 4He has an even number of nuclei the spin is zero.) 3He, which has two protons, two electrons and one neutron has spin I = - 89 and is thus a fermion. The nucleus has a magnetic moment so that 3He is a magnetic liquid. Like all fermions at T < TF (degenerate systems) the entropy goes to zero proportionally to the temperature. The solid phase which is a collection of localized spins has a constant entropy given by S = R l n ( 2 I + l ) . This entropy is then larger than the liquid entropy at T < 300 m K and the melting curve displays a minimum. This behavior is used in Pomeranchuk cooling as we will see later. Liquid 3He also becomes superfluid but at a temperature nearly 100 times lower than 4He. Figure 3 gives an expanded view of the 3He phase diagram where the superfluid transition is shown below 2.5 mK. Since liquid 3He is composed of fermions it becomes superttuid by a mechanism similar to the Cooperpairing of electrons in a metal. Instead of pairing in the singlet state like the electrons, that is, with opposite spins so that the total spin S and angular momentum l are zero, 3He atoms pair in a spin triplet state with a total

598

G. Frossati

h.c.p.- solid

n

CI

f. m.(?)

10

b.c.c.-solid

~t

n

B_

perfluid nor, ma I liquid

1 10 -~

10 ~

10 ~

10 2

10 3

10 4

T(mK)

Fig. 3. View of the 3He phase diagram where the superfluid transition is shown below 2.5 mK.

angular m o m e n t u m l = 1. The wave functions describing superfluid 3He has thus many components (18), depending on the spin and angular m o m e n t u m quantum numbers ms and mr, which can take the values +1 and zero. There are thus several possible phases unlike electrons or 4He which have only one (see Leggett's lectures). 2. R E F R I G E R A T I O N BY D I L U T I O N OF 3He IN 4He

2.1. Principles

3He can be diluted in 4He at any concentration above about 0.85 K. Below that, phase separation occurs. 5 Figure 4 shows the phase diagram of 3He-4He mixtures in equilibrium with their vapour. In the phase separation region, the equilibrium concentrations at a particular temperature can be read off at the intersections of the horizontal isotherm and the curve as indicated for T = 0.5 K. The concentration X3 = N3/(N3+ N4) on the diluted 3He side of the diagram, goes to a

Methods for Cooling Below 300 mK

j5 ~

599

nopmal homogeneous

_ 2.0-

~

mixture

~d

1.0

superfluid ~ homogeneous ~

0.5

0

,/,

i I,,

L I,,

,I

~,

II,,

~"

0

0.2 0.4 0.6 0.8 3He conccntPation X

.0

Fig. 4. The phase diagram of 3He-4He mixtures in equilibrium with their vapour (from Ref. 2).

constant according to the expression 6 X3 = 0.0648( 1 + 8.4 T 2 + 9.4 T 3) while the 4He concentration on the right side of the diagram decreases exponentially as 7"s

X 4 = 0.85 T 2/3 e x p ( - 0 . 5 6 / T )

One sees that at very low temperatures 3He is an extremely pure system, all aHe being excluded from it. At T = 10 m K there should be only about one 4He atom every 1028 atoms of 3He.

600

G. Frosmtti

The dilution refrigerator principle was put forward by London 9 in 1951 and by London et al., ~~ in the present form, in 1962. The dilution process can be viewed as an upside-down evaporation (Fig. 5). At the temperatures we are interested in, 4He can be considered as a mechanical vacuum, provided the velocity of 3He atoms is such that no mutual friction between 4He and 3He atoms occurs, so that no 4He excitations are created. In this case 4He below about 100 m K has essentially no excitations. (see Ref. 11 for a review on the properties of helium, and Ref. 12 for the properties of 3He-4He mixtures). Figure 6 shows schematically how a dilution refrigerator works. 3He gas is removed from the still (D) by a pumping system at room temperature. In continuous operation 3He is reinjected, precooled at 4 K in the main bath and condensed at the 1 K plate (A). Heat is removed at the still (B) and at the heat exchanger (C). Phase separation occurs in the mixing chamber (E), the coldest part, where the sample is placed. The circuit is filled with a mixture of 3He and 4He which is precooled to 4 K, and condensed at = 1 K by giving heat to a pumped 4He bath, called the 1 K plate (or pot). The quantities of 3He and 4He are calculated so that in equilibrium at low temperatures (after phase separation) 3He fills completely one side of the refrigerator, called the concentrated side, the phase separation occurring in the coldest place called the mixing chamber. 4ne plus some fraction of the 3He must then fill the rest of the refrigerator up to the evaporator (also called "still" as a short for distillation pot). In the still, liquid-gas phase separation occurs. The sizes of the evaporator and mixing chamber are chosen so that there is reasonable margin in the choice of the initial 3He and 4He volumes. One can use the following figures for the gas-liquid ratios below 1 K. 1 cm 3 3He liquid = 662 cm 33He gas NTP 1 cm 3 4He liquid = 866 cm 3 4He gas NTP Typically 3He/4He ratios are about 0.2-0.3 (depending on the size of the mixing chamber). At the beginning then, one has the dilution refrigerator (D.R.) full of a homogeneous liquid mixture at 1.2-1.5 K. The still is filled to about half its volume. Condensation is achieved by allowing the mixture, arriving from a room temperature reservoir and precooled to 4 K to come in contact with the 1 K pot by means of heat exchange (H.E.). After the 1 K H.E. there is an impedance for the flow so that the pressure of the mixture inside the 1 K H.E. is always larger than the condensation pressure. If this condition is not fulfilled there will be only partial condensation and the rest of the gas will condense somewhere inside the D.R. sometimes causing instabilities, and an unwanted heat load in the lower temperature region.

Methods for Cooling Below 300 mK

601

"::i:i:i:i:!i"

i!iiiiiiiiiiiiiill il

O "q'u~

iiiiiiiiiiiii iii

"1"

602

G. Frossati

PUMP .'..("~=kl ,

I. He 300K

"1

iMPEDANCE (B)

~iTii".'i~I

OTK ~AT R(C," ~T~LL~E

CHAMBER

M N (E) XG II P AE O HS SPRT N EA AI

Fig. 6. Schematic view of a dilution refrigerator.

After condensation, the gas phase is removed from the still by means of a convement pump. The temperature drops and phase separation occurs in the still. 3He floats on top of 4He and is quickly eliminated and reinjected into the D.R. After some time, the concentrated side of the D.R. is filled with 3He-rich mixture at about 0.4-0.5 K with x 3 - 9 0 % , while 4He rich mixture occupies the rest of the D.R. (diluted side). Just as for any diluted solution, 3He diluted in 4He will behave nearly as gas ("quasi-gas") with osmotic pressure ~r. If the impedance of the channel connecting the mixing chamber (M.C.) to the still is small, then the osmotic pressure along the channel is nearly constant. Like in a ideal gas subject to a temperature gradient, if the still temperature is higher than that of the mixing chamber, the volume occupied by the gas will be larger which means that the 3He concentration will be smaller than the 6.48% limiting solubility (at P = 0 and in contact with 3He). Table I gives some values of P3, P4, P3 + P4 and x3 in % with the pressures in mbar (1 mbar = 100 Pa). One sees from Table I that the total gas pressure is quite low, below 0.4 K so that it is convenient to heat up the still (by means of a current through a resistance) to 0.6-0.7 K where P3 is an order of magnitude larger than at 0.4 K while the helium pumped is 98% 3He. In so doing a large amount of 3He can be distilled out and since x3o must remain constant

Methods for Cooling Below 300 mK TABLE 1

T (K) 0.4 0.5 0.55 0.60 0.65 0.70 0.75 0.8 0.85 0.90 1.0 1.2 P3 (mbar) 0.00286 0.0105 0.0173 0.0255 0.0361 0.0481 0.0624 0.077 0.095 O.103 0.150 0.203 Pa (mbar) 7 · 10 -6 1.1 x 10-5 5.6 x 10-5 2.25 · 1 0 - 4 6.92 x 10-4 0.0018 0.0043 0.0086 0.015 0.030 0.0752 0.203 P3 + P4 (mbar) 0.00286 0.0105 0.0173 0.0258 0.0368 0.0499 0.0667 0.0861 0.110 O.143 0.226 0.419 x~v (%) 99.99 99.8 99.6 99.1 98.1 96.4 93.6 90.7 86.3 78.9 66.6 50.0

603

x3L (%) 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.8 0.7 0.6 0.5

in the m i x i n g c h a m b e r , the s a m e a m o u n t o f 3He must c o m e f r o m the c o n c e n t r a t e d 3He phase. T h e e n t h a l p y o f d il u t e d 3He b e i n g larger t h a n that o f pure 3He, c o o l i n g will occur.

2.2. Enthalpy Balance

We can d e r i v e the c o o l i n g p o w e r at the m i x i n g c h a m b e r by m e a n s o f a e n t h a l p y b a l a n c e , with the h e l p o f Fig. 7 w h i c h represents the m i x i n g c h a m b e r o f Fig. 6.

mc i

Superleak Tmc

q

Fig. 7. Visual aid for the enthalpy balance at the mixing chamber (following Ref. 2). For explanations see text.

604

G. Frossati

Suppose the system to be at the temperature TMc. Piston 1 introduces in a reversible way 3He atoms at a rate ri3. Pistons 2 and 3 are displaced in such a way that the phase separation level remains constant. /~3 atoms are removed in the form of a diluted solution while zero enthalpy 4He atoms are introduced by piston 3 (through a superleak). The temperature is kept constant by the heater power (). Since the process is reversible, and since $4 = 0

O = ~iaTuc[So(T~c)-Sc(TMc)]

Since at thermodynamic equilibrium the chemical potential/~ of each constituent in one phase must be equal to that in the other phase, we have that (per mole of particles)

I..L3C = [J,,~3D ~,,L4 C ~- [.,~4D

or

Ho-Hc = T(So-Sc)=

that is

T~c(So-Sc)

(1)

(~ = ti3[Ho( TMc ) - Hc( TMc ) ]

I f we introduce 3He at a temperature Tc > Tmc this corresponds to a heat input

Qc = li3[n c ( T c ) - n c ( Tuc)]

(2)

so that to keep Tuc constant one should furnish an external power ()ext such that

Oox,:O-Oc

or, from (1) and (2)

Q0x, = ,i3[Ho(rM~)- H~ (T~)]

Let us now evaluate Ho and Hc. Experimentally the heat capacities of 3He below 50 m K are found to be linear in T and equal to C3 = 25 T J / m o l e K C is defined as C = ~-~

For a 6.48% mixture, C3 = 107T J / m o l e K (extrapolated from 5%) (see Refs. 13, 14). From dH = T d S + VdP, at constant pressure we have, since TdS = C d T

Hc(T) = He(0) +

C c d r = H ( 0 ) + 12.5 T2 J / m o l e K 2

with Hc(O) the enthalpy of 3He at T = 0 which will be taken equal to zero.

H c ( T ) = 12.5 T 2 J / m o l e K 2

Methods for Cooling Below 300 mK TABLE II

605

Pure Viscosity Thermal conductivity Specific heat Molar volume

3He

6% 3He dil. 4He 5 x 10-s T -2 2,4 x 10-4 T-1 107 T 430 · 10 -6 (Pa.S) W K -I m -~ J K -a mol -I m3 mole-~

1.8 x 10-7 T-2 3.3 x 10-4 T-1 25 T 36.38 · 10 -6

Since

Ho

with

= H C +

T ( S o - Sc )

Sc =

~o

rC3 --~ d T = 25 T J / m o l e K

($3(0) = 0)

SD =

Iore~ d T = 107T J / m o l e T

K

H n = 12.5 T 2 + 107 T 2 - 25 T 2 = 94.5 J / m o l e K

T h e n u m e r i c a l e x p r e s s i o n o f the c o o l i n g p o w e r is t h e n 0 = 94.5 T ~ c - 12.5 T ~ J / m o l e vi O n e sees f r o m this e x p r e s s i o n t h a t for a given h e a t l e a k Q the m i n i m u m t e m p e r a t u r e is o b t a i n e d w h e n Tc = T ~ c w h e r e

( TMC )min = 0-11~'~)

/

O\ ~/2

I f 0 = 0 . 1 / ~ W a n d vi = 10 -4 m o l e s / s e c t h e n Train = 2 x 10 -3 K. Also, if 0 = 0 TMc = 0.36 Tc. T a b l e I I s h o w s s o m e useful p r o p e r t i e s o f p u r e 3He a n d 3He d i l u t e d in 4He at 6% c o n c e n t r a t i o n , at P = 0 a n d T < 50 m K .

2.3. H e a t Exchangers

2.3.1. Kapitza Resistance

In o r d e r to o b t a i n very low t e m p e r a t u r e s one has to cool the i n c o m i n g 3He, w h i c h c a n b e d o n e b y using the e n t h a l p y o f the d i l u t e d c o l d p h a s e in its w a y u p to the still, u s u a l l y b y m e a n s o f h e a t e x c h a n g e r s . H e a t f r o m the w a r m i n c o m i n g 3He m u s t first be t r a n s f e r r e d to the h e a t e x c h a n g e r wall a n d t h e n to the o u t g o i n g c o l d liquid. This is d o n e in first a p p r o x i m a t i o n b y the p h o n o n s o f the l i q u i d c o l l i d i n g with the wall, c r e a t i n g

606

G. Frossati

phonons in the solid which are than coupled to the phonons and quasiparticles of the liquid on the other side of the wall. The process is schematically depicted in Fig. 8. The incident wave is reflected and refracted, with outging transverse and longitudinal phonons.

LIQUID SOLID

~,.~2t

Fig. 8. Phonon reflectionand refraction at a liquid-solid interfaxe.

Methods for Cooling Below 300 mK

607

Applying Shell's law to the longitudinal component, for instance, we have sin 0o sin 02, and for 02 a x = 90 o m

we

Co C2,

have 0~ a~ = arc sin ...... C2~

Co

where Co is the velocity of the incoming phonons in medium (1) and (?2, is the longitudinal velocity in medium (2). If medium (1) is 4He and medium (2) is copper then Co = 238 ms -~ and C21 ~ 5000 ms -~ which means that Oo< 30 (For 3He Co = 183 sec-~). For transverse phonons the result is very similar. This means that the incident energy flux is almost totally reflected. This phenomenon gives rise to an interface resistance which makes heat exchange very difficult at ultralow temperatures. The power transferred between two different media is similar to Stefan's radiation law Q=~ ( T~ - ~ c ) watt 4Kk where A is the area in m 2 and R r is the Kapitza resistivity in K 4 m2/W which depends on the nature of the two media in contact. For small T we have for the thermal resistance

U ~.........7~ ~

9

A

AT Q

RK

-

AT 3

-

so that the thermal resistance is proportional to T -3. Other temperature dependence are also possible. Figure 9 shows the Kapitza resistivity of several interfaces. For a general review on Kapitza resistance in metal powders see Harrison. ~5 The values for Kapton and Teflon are obtained from experiments with dilution refrigerators using the continuous H.E. equation. 16A7The Kapitza resistance of the Tokuriki powder was measured by the author, using a annealed silver plate 1 mm thick with two sintered silver sponges 1 mm thick, one on each side of the foil. The sponges were prepared by first sintering 0.2 mm thick XRP5 powder (Lyon Allemand) at 250~ and then the Tokuriki powder on top of it, at 160~ both sinterings being done in hydrogen, for 30 min, in a way similar to that described in Ref. 8.

60g 5

G. Frossati

i

I

i

I

I

~

I

.~

J

,xNx

T3r 10-1

3

/7

3He D - CuNi [6-1OI x 10- 3

3HeD -

10 - 2 5 3 2

BRASS 16-81 x IO - 3

y

- KAPTON FOILS(7#urnthick) 4 x I0-~

3HeD - TEFLON (O.lmm t h i c k } 1.5 x 10- 3

--

10-3

I

I

I

I

I

I

2 3

5

10

20

50 9 Trek

Fig. 9. The Kapitza resistivityof several interfaces.

2.3.2. Continuous Counterflow Heat Exchanger

A continuous counterflow H.E. is an exchanger where the body can support a thermal gradient, as opposed to a "step" or localized H.E. It can be described by two coupled differential equations, one for the diluted (D) and one for the concentrated (C) liquid with terms containing viscous heating, thermal conduction in the axial direction and heat exchanged (in the radial direction). Let us look at each term separately. 2.3.2.1. Thermal Conductivity and Viscosity. Both for 3Hec and 3HeD thermal conductivity and viscosity increase with decreasing temperature and in the degenerate region (T<< TF) they are proportional respectively to T -~ and T -2. The thermal conductivity of pure 3He and mixtures is shown in Fig. 1019'2~ and viscosity curves are shown in Fig. 11. 21 2.3.2.2. Intrinsic Heating. Consider a tube with axis x, along which flows 3He at a rate ti moles sec -~ (Fig. 12a). The pressure drop in the element Ax will be

A p = Az~II/

Methods for Cooling Below 300 mK

' '''"! 300 ' I ' ''l'"t

/

609

I '

'

'''"

4 X . l, / / /

..-..

",

"----"~,.' 1.3'/o 3Ht

Ioo

N,,

".

I t

"'

/i

9l J //

/

5 % 3 . e in*He

,o"

",,",,

\

"._\

/,'_-"

/I

i/

.-'"

.-"-----_

~Xe at 2.7 1

3, ,,,,,I 3 10 ~

4

P

a

~

,

I ~ ,,,,,l 30 100

l

300

, , ,,,, 1000

rtmK)

Fig. 10. Thermal conductivity of pure 3He and mixtures of 3He in 4He.

where AZ is the geometrical impedance of the element Ax and 12 is the volume flow, (equal to riV) where V is the molar volume of the fluid (Fig. 12b, 12c). For a cylindrical tube of diameter d we have A 128 Z = - ~ - ~ Ax while for a tube of rectangular section of smaller side b

AZ

= 1 2 Ax ab 3

The viscous heating will be given by A0vis c VA]O ~T~2AZ

= =

=

~TV2~2AZ

where ti is the molar flow (moles/sec). 2.3.2.3. Heating due to Thermal Conduction. The quantity of heat entering the volume element by thermal conduction will be

where s is the section of the tube.

610

G. Fromati

100000

m m

30000

10000

3000

a.

1000

i--,r/~~i '

95%

300

100

30-10

,

I

3

Z ll,,,l

,

I

,,,,,,I

,

I

,,,,,,,

10 30 1O0 temperature/inK

300

1000

Fig. 11. The viscosity of 3He and 3He-4He mixtures (from Ref. 21).

If no heat is exchanged (case of a connecting tube between two heat exchangers)

s~\ dx/+nv2~--

,--,/k

az

tiC--

dr

ax

(3)

2.3.2.4. Equation of a Continuous Heat Exchanger. The complete equation of a continuous counterflow heat exchanger is obtained by adding to (3) the term due to the heat transferred by Kapitza resistance which can be

Methods

for

Cooling

Below 300 mK

611

a | ............

I

I

I I I

"T. . . .

I I

I

I l t I I

I

fi ---X

r

b

~P~

d

D

C

U

Fig. 12. R e l e v a n t d i m e n s i o n s for c a l c u l a t i n g a heat e x c h a n g e r . For e x p l a n a t i o n see text.

written as

AQ = ac( T4c _ T4 ) Atrc

and

AQ = ao( T 4 - Tao) Acro

where Tw is the temperature of the wall. If

O'C ~ OrD ~ or

then

A---~ =

am( T 4

-

T4D)

where

OLcOI D O~m ~- Ct c + OtD 1 4Rkm

612

G. Frossati

is the average Kapitza conductivity and R k ra = R t r T 3 is the average Kapitza resistivity. I f

Ol c ~--- Ol D -~- Of

then

Rkm = 2R k.

Diluted side

SD -~xx( k D( T ) - d ~ ) + rld (/ 2 -d-~-x+ POtm( T4 - T'~D = li c D d--~-'~x~ )

Concentrated side

s c d ( kc( T) dTc~ + ~ldV2c dZc + pc~,.( T4c_ T40) = _ liCc dTc dx/ dx dx

where P = Acr/Ax is the effective perimeter of the heat exchanger.

2.3.2.5. Perfect Continuous Counter~low Heat Exchanger. There is no analytical solution for this set of equations. On the other hand it is always possible to design heat exchangers where the viscous and axial conduction terms are negligible as compared to the heat exchanged. In this case, the heat exchanger is called "perfect." In a perfect heat exchanger all the heat given by the concentrated phase is taken by the diluted phase, that is dHc = dHo

or

CcdTc = CodTo

or

(4)

since Cc = 25 T J / m o l = aT and Co = 107 T J / m o l = bT we have

aTcdTc = b TodTo

and with A t r = PAx we can write

aT 2 = bT 2

dHc= tiaTc dTc= ot,,,P( T4c - 7"4o) dx dx

or

amP 1 writing

T 4 = tiaTc dx

(5)

amP[1 - ( a / b ) 2] aFi Eq. (5) becomes

-A

Adx=

dT

Methods for Cooling Below 300 mK

613

integrating from x = 0 which is the point where the 3He enters the mixing chamber to x = L where L is the total length of the exchanger we have

2 A L = T ~ 2 ( 0 ) - T~2(L)

(6)

Since T c ( L ) is the temperature of the still, typically 0.6-0.7 K then Tc2(L) is negligible as c o m p a r e d to Tc2(0). Therefore we have

2AL = Tc 2

Since P L = or, the total exchange area, and neglecting ( a / b ) 2 --- 5 X 1 0 - 2 we obtain the following expression for Tc in absence of heat leaks (remembering that olm 1/(4Rkm))

=

ri T 2 = 50Rk,, -or

From

Q=94.5T~c-12.5T~

n

we have T 0 = 7.56 T ~ c - 0.08 --o ri

or

(7)

T 2 c = 6.61Rkm--+0.01 --

or

ri

(8)

The same deduction can be used in case of a Kapitza resistance proportional to T -2. In this case we find ri Tc = 75Rk,,-or

or, if

fl

TMc = 0.363 Tc (no heat leak)

TMc = 27 Rk,,, -or

which in the presence of a heat leak gives for TMc (form 4)

9 ti 2

27.3R

+0.010-

(9)

614

G. Frossati

2.3.3.3. Heat Leaks

The expressions (8), (9) allow the calculation of the exchange area necessary to achieve a certain minimum temperature TMc at a given 3He flow rate ri provided Rkm and t) are known, supposing also that some geometry for the H.E. has been devised that will make t)visc and ()co,d small compared to the heat exchanged ()exch. Let us first examine the heat leak to the mixing chamber t). If we measure Tc and TMc, we can determine () from --Q= 94.5 T~c - 12.5 T 2 J/mole Fi when the refrigerator has reached its equilibrium at a given flow rate ri. () can be divided in two terms, an external one and another which depends on the flow rate:

0 = 0res"['- 0(/i)

OrEs is a background heat leak coming from exchange gas, vibration, heat conduction along the supports, heat coming eventually from plastics, etc. Usually this heat leak decreases somewhat with time, because of internal relaxation effects, or because residual gas gets trapped on the cold surfaces. To decrease the ()r~s it is useful to mount the cryostat on a heavy concrete block (about 103 kg) resting on rubber pads. All rotary pumps and some large booster pumps could also been mounted on independent concrete blocks, as an additional precaution. The use of a shield at 30 to 50 m K also decreases the effect of radiation and exchange gas. One can use either 4He or 3He as exchange gas, typically 2-3 Torr at liquid nitrogen temperature which is then pumped overnight at 4K. In a well conceived system, residual heat leaks are usually smaller than 100 nW after a few days that the D.R. has been at low temperatures. In all cases, the problem of the residual heat leak can be overcome by using large flow rates. The flow dependent heat leak 0 ( h ) is actually much larger than ()r~sIn many cases ()(ri) is seen to increase roughly linearly with ri, that is (~/h = constant. This term can be attributed to a combination of viscosity and conduction effects and becomes important usually below 10 mK. Typically, when 0rr is dominating, the refrigerator reaches some minimum temperature upon decreasing the flow rate, and further decrease leads to an increase in TMC. When ()(h) is dominating, TMc will eventually stabilize for some range of ti values, and only when ri is low enough that 0r~ dominates, will TMc start to raise again. Q(h) can in principle be made as small as wanted.

Methods for Cooling Below 300 mK

615

In fact,

0 ( 1 ~ ) = Ovisr "1- 0 c o n d

For a tube of square section with side D and length L we have, for T and T + A T with A << T

Q(fi)=12rIV2ti2T-2LD4+KATD2L -1

or

Q(ti)=AD-4+BD 2

which has a minimum for D = Dop = (2A)/B) 1/6. Numerically, for each phase we have 0.217LEri2 ~ 1/6

OopD=(( TDhot.J[ T o ~ n ( ' ~ D h o t / 8.5 x 10 -3 L2ti

TOcold)/

,~ 1/2

cm

Dope = ( Tc.o, + TCoo,.) In( Tc.o./Tcoo,.)] 2 where we have taken

T Tho~ + T~o,d 2 and

cm

A T = Thor - T~omd

The heat leak O(fi) for D

= Dop

will be

0 ( . ) o = (Ov,s.+ 0oo.d)o

2.6 x 10 -7 Lri 2 2.4 X 10-6D 2 In ~ L Toco,d (W)

- D4(Toho,+ Tooo,J ~-

O(n)c = (0vise+ (Lo~

l a x 10-8Ln 2 3.3 x 10-6D 2 in T ~ ~ (W)

=D4(Tcho+ Tcoo,d)2 J-

L

o

with L and D in cm, T in Kelvin and ri in mol/sec. It is always possible to decrease O/Oi) by increasing D and L but the volume increases quickly9 If D' = D v ~ and L' = 2L, then Qcond- Qcond

9t Qvisc Qvisc --

2

but

V'=4V!

616

G. Frossati

The temperature distribution on both sides of a heat exchanger can be estimated using the appropriate formulas for Tc and TMc, assuming an initial temperature Tc Tstill~ 0.6 K and calculating the temperature drop for a silver exchanger area per unit length. A description of the D.R. which holds the record in continuous low temperatures (1.9 inK) is given in Ref. 22.

=

2.4. Example: Calculation and Construction of a Simple Plastic Dilution Refrigerator

A simple D.R. can be built out of readily available polyvinyl chloride (PVC) plastic, a material widely used for water and electricity pipes. The idea is to take a piece of thin walled (about 2 mm) PVC pipe and use it as the external envelope of the D.R. unit. Inside this pipe we will insert a threaded PVC rod which will provide the channel for the diluted phase. Inside this channel we will insert the heat exchanger which will serve to cool the concentrated phase. The bottom of the pipe will be the mixing chamber and the top will be the still. The diameter of the D.R. can be chosen according to several criteria: The size of the sample to be cooled, the amount of heat exchanger area which ultimately determines the final temperature to be reached, the need or not to fit the D.R. inside the bore o f a magnet, the tubes available in stock, etc. The length is usually more flexible. Once we have decided on the length of the mixing chamber, usually determined by the length of the sample or whatever has to go inside it, besides the sample itself (like thermometers, measuring coils etc.), we will have to allow for some length for the still, ideally somewhat more than the volume of 3He located inside the mixing chamber. The rest of the length will serve to accommodate the heat exchanger (H.E.), and this will depend on the total chosen area, as well as the form, etc. We will use a small diameter Teflon capillary of the type currently employed for wire insulation. Most laboratories have such "spaghetti" in stock. The advantage with respect to the more usual metal tube is that it comes in big lengths of several tens of meters, while small metal tubes are usually made in quite shorter lengths. This means that for reasonable exchange areas one would need to join several pieces together, which is delicate and can lead to leaks, besides being quite more expensive than Teflon. Another advantage is the smaller Kapitza resistance as compared to metals. From the data on Kapitza resistivities, (Fig. 9) we see that RK for Teflon is four to six times smaller than for CuNi or brass, which means that in principle for the same area the heat exchanged will be four to six times larger for a Teflon heat exchanger. A drawback is that the thermal conductivity is small than for metallic allows which means that at high temperatures the thermal conduc-

Methods for Cooling Below 300 mK

617

--~

Cu-NI tube Stainless steel vacuum c h a m b e r

--Stycast Stycast f e e d - t h r o u - PVC lid

lolnt

Capacitance level gauge C u - N i still H.E. Still thermometer Still heater

External PVC tube

plug ~ded PVC tube

Teflon ca i

finger or sample

ioint ' feed-through

Fig. 13. Plastic dilution refrigerator (see text).

618

G. Frossati

tion across the plastic will be smaller than the Kapitza conduction, thus decreasing the efficiency of the H.E. Nevertheless the temperature quickly drops in the beginning of the H.E. and the Kapitza conduction will be the bottleneck throughout most of the H.E. We will choose an external tube of inner diameter 50 mm and 250 mm in length. On top of it we will glue a PVC lid machined out of a solid rod, provided with a hole of 5 mm diameter and a plastic to metal joint, typically a 12 mm O.D., 0.5 mm wall thickness CuNi (or brass) tube glued with an epoxy glue (Stycast 1266, for instance) to the lid. The bottom is a three-piece conical joint also made of PVC, with a 5-10 ~ angle. Such a joint is sealed with vacuum grease, possibly Apiezon L, M, or N or silicone grease, and provides an easily dismantled access to the inside of the mixing chamber. The metal tube should be soft-soldered to the flange of a metallic vacuum chamber which will surround the D.R. unit. Care must be taken not to heat the glued joint, by using wet cotton around the tube. Inside the PVC tube of the D.R. we will insert a threaded rod which has been machined hollow, except for a disk which will form the top of the mixing chamber. A separate disk is machined and carefully glued with cyanoacrylate glue (instant glue), or PVC glue at about 40 mm from the top, so as to define a hollow leak-tight volume which provides thermal insulation of the mixing chamber from the higher temperature still. The thread has a square section, 4 x 4 mm in side, and runs all the way to the mixing chamber, except for the 4 cm of the still. The still wall is provided with several holes of about 5 mm diameter so as to have good communication between the inside and the outside of the tube, which will be glued to the plastic lid. We chose a Teflon capillary 0.8 mm I.D. with a wall thickness as small as possible, usually 0.1 mm, and we wind it on a steel or brass rod 2.5 mm in diameter, using a lathe. The resulting spiral is heated to 200-250~ to keep it from unwinding after removal of the metal rod. It is then wound in the threaded channel starting at the level of the still bottom and allowed to enter the mixing chamber as in Fig. 13. The beginning of the tube is connected to a CuNi capillary (I.D. = 0.5, O.D. = 0.8, 1 m length), snugly fitting inside the Teflon tube. A small amount of Apiezon grease is first applied to the tube which upon contraction provides good sealing. The metal tube is coiled around the still and allows good thermalization of the incoming 3He at the still temperature. The total number of threads for the PVC screw will be 40 for a wall thickness of 1 mm, with a length of about 142 mm per winding. We need 875 mm of Teflon tube for each 142 mm of coiled spiral so that the total length will be 35 m, with an area in contact with the concentrated side of 0.084 m 2.

Methods for Cooling Below 300 mK

619

The mixing chamber temperature is given by the perfect heat exchanger equation T2~c = 6.61Rkmli/o r + 0.01 Q / n

w i t h Rkm =

(10)

2 x 1.5 x 10 -3 K m2/W. We will assume a heat leak of the form Q = 1 0 - 3 t i + 10 -7 W. This heat leak is in fact just an educated guess and we will estimate it for the geometry which we have chosen. The mixing chamber temperature is shown in Fig. 14 as a function of ri for such a heat leak and also for other values. We see that it should be possible to cool below 10 m K with such a simple D.R. The flow dependent heat leak can be calculated for instance for each of the 40 windings both for the diluted and for the concentrated phase, but the formulas we derived are only valid for T well below TF, typically TF/10, thus below about 40 mK. We first calculated the area needed to reach 40 m K for ti = 10 -4 mol/s. From (10) we have 0.042 = 6.6 x 0.003 x 10-4/tr + 0.01 · (10-7+

10 -3 X

10-4)/(10 -4)

which gives cr = 1.25 x 10-am 2. The area per winding is 0 . 0 8 4 / 4 0 = 2 . 1 x 1 0 - a m 2 so that a mixing chamber placed after the first thread would reach already less than 40 mK. The heat conducted due to the thermal conductivity of the Teflon wall from the concentrated to the diluted phase can be estimated as follows: we saw before that along a perfect heat exchanger (far from the mixing chamber), AHc = AHD or 25T 2 = 107T 2 which means that Tc -~2Tn thus at the end of the first thread Tc --80 mk. The heat transferred by conduction will be

(~=A/L

f

0.04

k(T) dT

with

k(T)=5·

d 0.08

for A = 2 x 10-3m 2 and the thickness L = 10 4m we have A/L=20 and 0 = 3.8 x 10 - 6 W. The heat transferred by Kapitza conduction will be given by 0=A(T,-'*

T4.)/4RK = 2 x 10-3(0.084-0.044)/4 x 0.0015 = 1.28 x 10 -5 W

We see that the heat transferred in this high temperature part of the heat exchanger is limited by the p o o r thermal conduction of the Teflon. This could be improved by having a thinner tube or a length of CuNi tube. Nevertheless, the T 4 dependence of the Kapitza resistance will start dominating at lower temperatures so that for easiness we just use the Teflon capillary. As an exercise calculate at which temperature Tc the heat conducted by the Kapitza mechanism becomes smaller than that due to the thermal conductivity of Teflon.

620

G. Frossati

We can now estimate (~(ri) for the diluted side at ti = 10 -4 mol/s. T e n l d D will be the mixing chamber temperature Tmc obtained from (10) which from Fig. 13 is 6.5 m K for the assumed heat leak of 2 · 1 0 - 7 W. We have then (~(ti)o = (~visco+ 0condo = 3.8 X 10-9-1- 8.6 X 10 -9 = 1.24 X 10 -8 W, which is much smaller than the total heat leak we have assumed. This is not the case for the concentrated side, where (~visr is much larger due to the small diameter. As a second exercise calculate the heat generated in the concentrated side first for the 39 windings, then only for the last 10 and then with the total heat leak recalculate the final temperature for ri = 10-4mol/s and 1 0 - 3 mol/s. The dilution refrigerator must be connected to the pumping line and provided with a condensing impedance between the still and the 1 K bath as well as heaters and thermometers for the still and the mixing chamber. A capacitance level gauge made of two concentric brass tubes with a few tenths of mm gap can be used to monitor the liquid level inside the still. In order to precool the D.R., some helium exchange gas (a few cm 3 nTP) is allowed inside the stainless steel vacuum chamber. The D.R. is first precooled to 77 K (with liquid N2) and then to 4.2 K with liquid 4He. Once at 4 K the exchange gas is pumped away trough a pipe entering the vacuum chamber. The 1 K pot or bath are pumped to about 1-1.5 K and once the 3He-4He mixture is condensed the D.R. is ready to operate. Figure 14 shows the calculated M.C. temperature as a function of circulation rate ri for different heat leaks. 3. P O M E R A N C H U K REFRIGERATION

3.1. Principle

If we impose the condition of thermal equilibrium between solid and liquid 3He, the equality of chemical potentials implies

/x s = / z ~

Therefore

Gi Q

/21 /'i s

where G is the Gibbs function, and (since 3He is magnetic) we have

V~A P - StA T - M~A H = VsA P - SsA T - M s A H

Methods for Cooling Below 300 mK 100

621

I-

10

1

I

i0 "S

10- 4

n(mol s -I)

10-3

Fig. 14. The calculated M.C. temperature as a function of circulation rate ti for different heat leaks.

or ( Vt - V s ) A P = ( S t - S~)A T + (M~ - M ~ ) A H w h i c h in z e r o field b e c o m e s AP Sl - Ss

AT

V,-V~

s i n c e f o r 3He at T < 0.3 K St ~- y T a n d Ss = R In 2 w e h a v e , i n t e g r a t i n g in T ..... 8 9 2 - R T In 2 P(T) = t'/o) +

E-V~

622

G. Frossati

with Vi - V, > 0 so that P ( T ) has a minimum when St = Ss. In a high magnetic field the term (Mt-Ms)AH produces a change P(T) given by

AP = (V~-

Vs)-'

(M~-Ms) dH

Usually, in fields attainable in the laboratory, Ms > M~ a n d A P is negative and rather small, of the order of 400 mB at 5 m K at 9 T. If on the other hand the liquid is strongly polarized, the solid magnetization is saturated and the change in pressure AP will be governed by Mr. The function Mr(H) is not known and present speculations predict a depression AP --- 7 bar, while a much larger depression is not excluded. Figure 15 shows the entropy diagram of liquid and solid 3He as well as the melting curve.

S

In2

0.6

S/R Sotid

0,/.

0.2 0 P CMPa]

"

9

(e)

I

I

I~

~,Tm

I

32

9

(b)

Liquid' ~ ~ . . . . . ~ _ . ~ ~ . ~

I ' II I

~j~

T CK3

0

0.1

02

O.3

0A

Fig. 15. The entropy diagram and the melting curve of liquid and solid 3He.

Methods for Cooling Below 300 mK

623

The method known as Pomeranchuk cooling after Pomeranchuk 23 consists in precooling liquid 3He below 315 m K (where Ss = St) and then compressing the liquid until it reaches the melting curve when solid begins to form. Because of the negative slope, an increase in pressure corresponds to a decrease in temperature. The cooling power of the process is given approximately by

Q--= T ( R In 2 - vT) = 5 . 6 T - 4 0 T 2

ti and is to be c o m p a r e d with that of a D.R., given by at most Q = 82T 2 ti (when T~c = Tc = T)

At temperatures lower than 50 m K the cooling power of a Pomeranchuk cell is larger than that of a D.R. and at 2 m K it is nearly two orders of

m

_/

R

I

/,"

/

I

0.I

//

/" y - _

!~6T/y./~//;OT _ Y

///,

I..i/I

OtL'-4/, , ~ I,,il , , , ,,,,,

0.01

l

1

10

T(mK)

Fig. 16. The entropy of solid 3He in various fields. The dashed lines are calculated using the 9.2 T value as a starting point and assuming a paramagnetic behavior. The horizontal arrows are different possible Pomeranchuk trajectories.

624

!

G. Frossati

W (] 10~

103

lo2

,

I

a

I

,

0.1

1 TImK} 10

100

Fig. 17. Ratio of mechanical work W= P dV to the energy adsorbed per mole of 3He solidified.

magnitude larger. Solid orders at 1 m K (in zero field) and its entropy drops by about an order of magnitude thus reducing the cooling power. Figure 16 shows the entropy of solid 3He in fields up to 9.2T, extracted from the measured melting curve [24] and up to 30T (calculated). Another difficulty comes from the large mechanical work done by changing the volume from Vt to V~ (5% reduction). The ratio of this mechanical work P d V (where P ----34 bar at T = 5 mK) to the energy absorbed per mole of 3He solidified is given in Figure 17. One sees that at 1 m K P dV is 10 times larger than the energy absorbed by the solidification of 1 mole 3He. If some irreversibility in the compression cell is present then the heating can be larger than the cooling. Cooling stops, in practice, at temperatures around 0.8-0.7 mK. The presence of solid might also be a nuisance if the cell is used to cool something other than 3He because of the poor thermal conductivity of the solid, besides the fact that the method is not continuous. A very small Pomeranchuk cell provided with a sintered heat exchanger is a very good thermometer down to 1 mK, which can be calibrated with TA, TB, Tc, and

Tmin 9

3.2. Example of Experimental Cells

Because of the minimum of the melting curve one cannot pressurize 3He at temperatures below 315 m K above 29.3 bar by simply applying pressure from outside a via a capillary. There will be a solid plug in the region where the capillary is at 315 mK. For this reason a special cell with flexible walls must be used in such a way that deforming the wall by means

Methods for Cooling Below 300 mK

625

of, for instance 4He pressure, the volume decreases and hence the pressure increases. The cell is shown schematically in Fig. 18 copied from Ref. 25 and an example of a Pomeranchuk cell using BeCu bellows is shown in Fig. 19. This cell, used by the Cornell group 26 has a metallic body and a plastic tail which allows nuclear magnetic resonance to be done on the 3He. Since heating produces solid and since solid tends to form where the pressure is higher, it is here possible to have solid aHe at the bottom. On the other hand (as illustrated in Fig. 18) solid can also grow in other places. In the case where one is interested in studying solid 3He properties, like magnetization, it is impotant to know how much solid is formed inside the N M R coil and in this case such a cell is not always convenient. As for D.R. one can make a Pomeranchuk cell totally of plastic, which can be much smaller for the same useful volume and can be fitted totally inside an N M R coil.

FORCE

Fig. 18. Schematic diagram of a Pomeranchuk cell (from Ref. 25).

626

G. Frossati

DISPLACEMENT CAPACITOR TOP P L A T E - -

T,, CAPACITANCE )GE

r'ER CHAMBER SPACER-INTERMEDIATE CHAMBER ('He)

,)

IER CHAMBER .,) BON ISTOR

EPOXY TAIL

COPPER NMR SAMPLE

, NMR COIL NMR COIL Zcrn

I I

Fig. 19. Example of Pomeranchuk cell using BeCu bellows (from Ref. 26).

The assembly of a cylindrical plastic cell is shown in Fig. 20. It consists of two epoxy end caps and a cylindrical membrane made by gluing a few windings of thin Kapton foil. The membrane is reinforced at the ends by extra layers of Kapton to compensate for extra radial stress, which is twice the stress at the center. Once glued to the end caps and provided with a filling capillary the unit is glued inside an epoxy cylinder which also has a filling capillary, and will contain the pressurizing 4He. One of the end caps is drilled and will provide access for thermometers, heat exchangers, pressure gauge, etc. Figure 21 shows a section of an actual plastic Pomeranchuk cell used in experiments on polarized 3He.26 The calculation of a cylindrical Pomeranchuk cell requires the knowledge of the Young's modulus and of the maximum stress of the membrane material, which for Kapton and for BeCu is shown in Table III. To solidify liquid 3He AV v~- v~ V~ 1.31 25.54 -5.13%

Methods for Cooling Below 300 mK

627

Fig. 20. The assembly of a cylindrical plastic Pomeranchuk cell.

one can show that for a cylindrical membrane of radius R

--V = E-h AP34

where h is the thickness of the membrane and

AV

2R

ae34 = ( e 3 , - P 4 , ) - ( P 3 ~ - P 4 , )

Taking h = 40/~m,

P4, = 0,

P4, = 2.5 MPa, P3, = 3.35 MPa and (r = AV= 8.3% V

P3,Rh-1

(compare with a Be-Cu membrane where A V~ V = 0.7%). Superfluid 3He was discovered using a Pomeranchuk cell. 27 Since P = P ( T ) a calibrated pressure gauge is an ideal thermometer, provided the melting curve is known. The change in slope, the backstep and the flattening of P ( t ) (where t is the time of compression) are due to respectively the superfluid A transition, the B transition, and the solid ordering C. The transitions A and C are fixed in temperature respectively at 2.49 m K (+0.2 m K ) and 1.10 m K (+0.1 mK). The B transition during compression can undergo supercooling while during decompression it is reproducible

628

G. Frossati

a

b

C

cl

e

f

mg

3He space

mh mi

1ram

Fig. 21. A section of a plastic Pomeranchuk cell used in experiments on polarized 3He. (a) Silver rod for thermal anchoring to the nuclear stage; (b) stycast 2850 FT vacuum seal; (c) Kapton membrane; (d) araldite plug and wall; (e) Ag-Pt heat exchanger; (f) glass thermometer; (g) sapphire pressure gauge; (h) up viscometer; (i) carbon thermomenter; (j) down viscometer.

Methods for Cooling Below 300 mK

TABLE II1 Kapton polymide film E = 6 x 109 N m-2 pa o-= 3.4x 108 N m-2 pa BeCu (hardened) E =2.7· 10tl N m-2 pa o-= 1.3 x 109 N m-2 pa

629

and has a value of 2.0mK (+0.2 mK). These features are represented in Fig. 22. Compressing cold liquid 3He in a high field produces strongly polarized solid, which upon quick melting (a few seconds) produces polarized liquid 3He with a time constant of several minutes (the Castaing-Nozi~res effect). The rather long relaxation time allows to study a system which would otherwise be impossible to produce by simply cooling in a magnetic field. This is bringing important information concerning the nature of the 3He

_

2m,~

//r

B

Fes

A

Fig. 22. The superfluid A transition, the B transition, and the solid nuclear ordering C.

630

G. Frossati

liquid and of the 3He-3He interaction. Using a Pomeranchuk cell diluted mixtures of rather high 3He concentration (5.5 and 8%) have been polarized in a way similar to that of pure 3He, with relaxation times in excess of one hour at 300 mK. 2s 4. NUCLEAR REFRIGERATION We will briefly recall the principles of nuclear demagnetization as can be found in Refs 1 and 2. Prof. Pobell will discuss the important points for building a successful nuclear demagnetization system. The entropy of a system of paramagnetic (noninteracting) moments, electronic or nuclear, can be reduced by applying a magnetic field such that ~tB ~- K T If the heat is evacuated to some cold source at some temperature T~ and then the system is isolated by a convenient heat switch and the field removed adiabatically, then the temperature of the system will decrease. Spontaneous order will occur at nanoKelvin or picoKelvin temperatures if we are using nuclear moments and at MilliKelvin if we are dealing with electronic moments which are some 1000 times larger. Because dilution refrigerators now cover all the milliKelvin range, electronic demagnetization has lost most of its interest. We will then discuss (very briefly) the adiabatic demagnetization of nuclei (nuclear demagnetization). For nuclei of spin I in a magnetic field B there are 2 1 + 1 nuclear energy levels spaced equidistantly em= --/-*ngn mB where /~n = 5.05 x 1 0 - 2 7 J T -~ is the nuclear Bohr magneton, g, = Land6 factor for nuclei = 2 , and m = m a g n e t i c quantum number ( - I , - I 1 + ,...,I). The entropy of such a system is given by (see Launasmaa ~) S=nRln(2I+l) where ANoI(l+l 3kB 2 2 (12) nAB 2 2TEp.o (11)

is the molar nuclear Curie constant, n is the number of moles, No is Avogadro's number, and/.to = 4,r x 10 -7 V s/Am. The heat capacity at constant field is given by Ca= T(SS~ \ST/a

-- nAB2

I.*oT2

(13)

Methods for Cooling Below 300 mK

631

in order to cool the system of spins from T = oo to T = T~ one must extract an amount of heat AQ given b y

AQ =

i

t,

nAB 2 CB dT = tzoTi

(14)

Since the entropy is only a function of B~ T, if the field is changed adiabatically (S = const) then the ratio B~ T must remain constant. If the demagnetization starts at T~ in a field Bi and we reduce the field to By, then Ty = B~YT~ (15)

If Bi is zero or very low then one must consider the internal field Bi, so that the final field By will be substituted by (By+ Bint) and

= ~ (By + Bint) '/2

(16)

After demagnetization the system will warm up under the effect of the external heat leak. One can calculate the time to reach a temperature T > Ty. From

d__T=d__Q 1 x O dt dt CB - CB

A,---~-~o

_nAB}f T, T -2 dT = nAB} ( T j ' 7" /~o

T-')

(17)

The conduction electrons of the metal used for the demagnetization, couple the lattice to the nuclei and the nuclear-lattice relaxation time is given by z, Te = K, where Te is the conduction electron temperature and K is the Korriga constant, which for copper is 1.1sK. For Pt K = 0.029sK. The nuclei will absorb energy from the electrons which are at a temperature We, at a rate

(~- nA(B~B~) (T~- T.) gtok T.

The ratio between Te and T. for a given Q will be

-- = 1 + - -

(18)

Te T.

p,oKQ

nAB}

(19)

There is an optimum final field By(opt)which for a given (~ gives the minimum electronictemperature.

632

G. Frossati

From (14) and (18) we have

(_~___A) i/2 ~ nf(opt) = (20)

Figure 23 shows the entropy curves of copper in different magnetic fields compared to that of 3He and 3He diluted in 4He. In some rare-earth compounds known as Van-Vleck paramagnets the field seen by the rare-earth nuclei is enhanced with respect to the applied field. The enhancement can be quite large, typically a factor 5 to 100. 29 In such compounds the nuclear entropy decreases faster than in normal metals and thus smaller values of Bff T~ are necessary to obtain a large entropy decrease. Such behavior is shown in Fig. 24 for PrNi5 in a field of 6 T. The limiting low temperatures are nevertheless higher ( T = 300 mK) 3~ than for copper. PrNi5 has been used as a first demagnetization stage to precool a second stage, usually of copper. An electronic temperature of the copper stage of 48/zK was obtained in this way. 31 Using a good dilution refrigerator this intermediate nuclear stage can be eliminated. The lowest measured temperature obtained with a single copper nuclear stage is of 15 ~ K while 12/xK were obtained with a double

[-

C~.O,OgT Cu,O,gT C~,gT '\

-

~

lo~

I

-

-~

10-II/

3He_4He [ / ~ P

, , ,,,,.I

. " : 3H. ,

10 -2 lO-

~0-21 ~

10 -5

16 4

. , ,,,,..Yl

103

.............

T (K) Fig. 23. The entropycurves of copper in different magnetic fields compared to that of 3He, 3He diluted in 4He and PrNis (in 6 T).

Methods for Cooling Below 300 mK

633

nuclear stage. 32 Using a technique where copper plates covered with sintered silver are demagnetized directly in contact with liquid 3He a record liquid temperature lower than 100 p,K has been obtained. 33 REFERENCES

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