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Properties of the condensed phases of helium and hydrogen
Physica
IV,
no 4
April
1937
PROPERTIES OF THE CONDENSED PHASES OF HELIUM AND HYDROGEN by A. BI JL (55th
Communication
of the ,,v n n d e r W a ill s-Fund”)
Summary The influence of the zero point motion on the density of a crystal is discussed and the result applicated to hydrogen and deuterium. The zero point motion affects also the specific heat, which is at the lowest temperatures determined by the velocity of sound. The latter could be calculated by considering the influence of an elastic deformation on the wavefunction of the crystal. The result is that the square of the velocity of sound can be obtained as a sum of the contributions of the potential % the kinetic energy; in the case of solid helium and hydrogen energy and good agreement with experimental results is obtained if the molecules are treated as hard spheres. As a consequence it is shown that at very small densities a negative ,‘thermal expansion may be expected.
5 1. Introdmtiort. Liquid helium differs in some of its properties from the condensed phases of heavier elements. The most remarkable differences are : a) the density of condensed helium is very small, so that the intermolecular distances are considerably larger than the gaskinetic diameter of the atom; b) the thermal expansion at the lowest temperatures is negative; c) solid helium excists only under pressure ; d) liquid helium shows a phase transition of the second order. As long as a reliable theory for liquids does not excist, it is hardly possible to expect an explanation of these anomalies. Therefore in this article the question will be posed: In how far do the properties of the crystals of light elements differ from those of ‘normal’ crystals? ‘It will appear that the differences mentioned ab,ove under (a) and (b) can also be expected for crystals.
-
329 -
330
A. BIJL
A general argument can be given to show why the common theory of crystals cannot be applied here. This theory only considers ‘harmonic’ binding forces between the particles and so it can be applied only in the case that the amplitude of the zero point motion is not larger than the region in which this assumption holds. It has beenpointedoutby Clusius and BartholomC1) thatfor hydrogen and helium the assumption of harmonic forces is no longer valid. If this assumption is abandoned, however, no rigourous theory of the crystal can be given as the energy can no longer be written as the sum of the energies of independent oscillators. By using special methods the influence of the zero point motion on the density, the specific heat and the thermal expansion could be discussed *). $ 2. The crystal structure and the density. The method commonly used to obtain the energy of a lattice is to calculate the sum of the mutual potential energies over the points of a point lattice. This calculation is carried out for different types of lattices and different lattice parameters; the lattice with the smallest energy will be stable. In an exact wave mechanical theory this method is not correct. For although a crystal is defined by the fact that some of its properties are periodic functions in space, it can no longer be assumed that all particles have fixed positions. It is only correct to say that the probability for finding a particle is a periodic function in space. So we have to deal with a probability distribution for the intermolecular distances. A rough idea of this distribution can be obtained by considering the wave function of one particle in the field of its neighbours which are considered to be fixed in their equilibrium positions. This simplification is in some way analogous to the one made by E i n s t e i n to attaok the problem of the specific-heat. In this one particle problem the particle is surrounded by impenetrable molecules whose radii are large enough, for all realistic values -of the lattice parameters to provide a’potential’barrier for the inclosed particle. So this particle cannot escape and its lowest wave function is determined unambiguously. *) The influence of the zero SiA’on and hy London; t h o 1 o m 6 (see references).
point motion on the density was already discussed by and on the specific heat by C 1 us i u s and B a r-
33 1 . This wave function is zero at the boundary of the potential ‘hole’ and will in general have a maximum in the center. This form is still more pronounced for the probability of finding the particle, which is the square of the wave function. Therefore the particle will nearly always be found in the neighbourhood of the center. In calculating the potential energy it is therefore permissable to assume in a first approximation that the particle is always exactly in the center of the hole. This simplification was used in the calculations of F. L’o nd o n “). It gives appreciable errors only’ in the case that the intermolecular distances are large, as than the maximum of the wave function is less sharp. Therefore in cases with smaller intermolecular distances the wave mechanical treatment gives the same result for the potential energy as the ordinary treatment. Differences arise, however, with the calculation of the kinetic zero point energy of the lattice. For most crystals this energy can be neglected, being small compared with the potential energy. For He, Hz and D,, however, it is very important as was first pointed out by F. S i m o n 3). This kinetic energy can also be studied in the one particle model. Its ‘mean value’ is determined in this model by the shape of the wave function. If the v an d e r W a a 1 s-forces and the mass of the particle are small enough, this wave function and therefore the kinetic energy will not alter much if the surrounding molecules are replaced by hard spheres and the v a n d e r W a a 1 s-forces are neglected. A further simplification is possible by replacing the ‘hole’ formed by the surrounding molecules by an impenetrable spherical boundary, whose diameter is equal to the shortest diameter of the ‘hole’. This is permissable because at the points outside the sphere the particle is in the close neighbourhood of two or more surfaces of the surrounding molecules, and the wave function is consequently very small. So we obtain as an estimate for the kinetic energy the same value as for the lowest energy state-of a particle enclosed in a sphere with radius a - 6: h2 T= (1) 8m(a - CJ)”’ PROPERTIES
OF
CONDENSED
PHASES
OF
HELIUM
(a = distance between nearest neighbours; meter, which is not defined rigourously.)
AND
HYDROGEN
o = molecular
dia-
332
A. BIJL
If the v a n d e r W a a 1 s-forces cannot be neglected this value will generally be larger because the wave function will be more concentrated in the center. This picture enables us to draw some conclusions as to the crystal structure and the density if we may assume that the arrangement of of the neighbours which gives the smallest total energy will be approximately identical with the most probable arrangement of the many particle system at the absolute zero. If n is the number of nearest neighbours, the potential energy per atom is equal to:
v = 5v(u),
(2)
where ~(a) is the mutal potential energy of two molecules at distance a. (Interactions between molecules which are not nearest neighbours will always be neglected.) The crystal symmetry allows only a limited number of nearest neighbours, which numbers are independent of the value of a. It is therefore obvious that for any given value of a for which v (a) < 0, n will have its highest value because the potential energy is then a minimum. So we may expect that n = 12, i.e. a cubic face centered or a hexagonal close packed structure will be the most stable one. a is now determined from the condition that the total energy T + V is a minimum as a function of a. The calculation of the density along these lines has been carried out for He by F. L o n d o n “). It can be applied also for H, and D, (see note ‘1) Some remarks about the accuracy obtained with this model are given in note 2. 5 3. Sfiecific heat and velocity of sotind. After D e b y e .and B o r n the heat motion in a crystal can be approximately described by a superposition of harmonic undamped waves. The specific heat is obtained as a sum over the contributions of all wave numbers. The contribution of one single oscillator of frequency v depends on the only the oscillators of low freratio hv/kT. At low temperatures quency are of importance for the specific heat. .As in the case considered the potential energy can no longer be described as a quadratic function of the coordinates of the particles, the analysis of the heat motion into harmonic waves is not possible, However as at low temperatures the high frequencies are degenerated,
PROPERTIES
OF CONDENSED
PHASES
OF HELIUM
ANP
HYDROGEN
333
we have to deal only with the long waves (the sound waves) fo; which we may assume that they are harmonic and undamped. We may therefore calculate the specific heat at the lowest temperatures by the ordinary treatment of D e b y e. In this method the specific heat can be conveniently expressed as a function of the velocity of sound, as this velocity determines the relation between wave number and frequency. The velocity of sound is connected with the specific heat by the formulae: (3) and
C”=q$q;)3
(4’
where : c = velocity of sound V = molecular volume. In a rigourous treatment longitudinal and transversal waves ought to be treated separately. As in this paper no exact numerical result can be obtained it seems justified to neglect the difference between those two velocities. 5 4. Sciwoedinger eyuatipz for a deformed crysia!. To calculate the velocity of sound it is sufficient to know this velocity for one large wave length. This velocity is ‘determined by the energy necessary for the deformation as a function of the amplitude and of the mass of the molecules. At low temperatures this deformation energy is usually confined, to a change in the potential energy of the lattice. The fundamental difference between the usual case and that considered here is, however, that here also the kinetic zero point motion changes with the distance between the particles, as was shown in $ 2. It would be possible to draw some conclusions from these considerations as to the influence of the kinetic zero point motion on the velocity of sound. As for this purpose the knowledge of dTa/da2 is required, which can only be deduced from (1) with a very low accuracy, it seems justified to analyse in more detail the change of the S c h r o e d i n g e r wave function by the deformation, We will therefore consider a crystal deformed by a density wave along the x-axis (the x-axis being placed in the same direction as one
334
A. BIJL
of the lines connecting nearest neighbours). The wave lenght of the wave will be called a/q, where q is a very small number. The amplitude and phase of the wave can be indicated by the complex amplitude A, : A, = $:
e(2xiqk) x,.
(x, is the deviation of the particle If the crystal is described by a A, can be treated as a quantum value - which is for a macroscopic value -- is given by:
k from its mean position) wave function Y, the amplitude mechanical quantity. Its mean operator always equal to its real
(dw = dq dy, dzl . . . . d+) The problem is solved if & is known as a function of time, because then its frequency and hence the velocity of sound is known. A,(t), however, can be obtained from (6), if Y(i) is known. Y(t) is determined by the equation of motion: h ay =H-Y, 2Tci-z where H is the
H a m i 1 t o n operator
for the crystal:
The only ambiguity left now is in the chaise of the initial wave function Y(t = 0). It must be so chosen that for & = 0 it describes a state of the crystal near the absolute zero. It is therefore logical to take this initial Y in such a way that for & + 0 it becomes identical with the lowest eigenfunction. This can be obtained by choosing Y(t = 0) in such a way that it is among all wave functions having the same value for 2, the one with the lowest energy. If Y(t = 0) is chosen in this way a differential equation can be obtained for it. For if Y satisfies the condition that: U=/dwY*hY
(9)
PROPERTIES
OF CONDENSED
is a minimum,
PHASES
OF
with the additional
HELIUM
AND
conditions
/dw*YY*=
HYDROGEN
335
that
1
and jdw.Y”AY=A,,
it must satisfy the equation: HY
+ ;
(QA, + a*Af)
Y = hY.
(10)
5 5. Discussion of the wave eqzcation. We will first find the meaning of the constants a and A; it proves that there is a relation between a and the frequency and between A and the deformation energy. Substitution of:
a = r - e(icp)
(‘1)
in (10) gives : H\r
+ 2r c xk co?, (hqk
+ ‘p) ’ ‘r = hy.
k
(14
. In this equation the second term on the left hand side is responsible for the deformation. If in a crystal an external force of magnitude - 29’ cos (2nqk + ‘p) derived from a potential 2??& cos (2x@ f cp), acts on every k’th particle, this system (i.e. crystal and external forces) would have the operator (12) for the total energy. The equilibrium condition for a particle in this system is that the sum of the external forces and the crystal forces is zero, hence: + 2r cos (27cqk + $3) + g
= 0. k
(13)
This relation can be obtained directly from (12) by differentiating with respect to +, multiplication with Y* and subsequent integration. As a consequence of the equation of motion (7) we have: ::
W
(14
mXk=-Jz-
By substitution of (I 4) into (13), multiplying summation over k we obtain:
,*=“j
2
rl’
with e (ixiqk)
and ,
(15)
336
A. BIJI.
As we may assume that 2, is a periodic v2A, = - A, and hence :
function
of the time, (16)
Thus the value of v2 in which we are interested the ratio between A, and ct.
is determined
by
The relation between A and the deformation energy can -be obtained by differentiating (12) with respect to r, multiplying with Y* and integrating. The result is: 6%
z=-zy
2N
hence :
(18) On the other hand we obtain
directly
from (10) :
E(r) = A - ; (a& + u*A,*),
(19)
or, with (17) :
(20)
E(Y) = h + So the deformation
energy is equal to:
E(r) = E(o) + N$,
A, 12
or: E(r) which formula crystal.
E(o) = -
ij Ij Xk “xk .
is the same as is obtained
(21)
in the usual theory of the
of the frequency and the speiific heat. As the phase
3 6. Calculation
of the density’wave is of no importance, is real. (12) than becomes: m
we will chose it so that u ’
+ 2x f & cos (2x44 \r = hy.
(22)
PROPERTIES
OF CONPENSED
PHASES
OF
HELIUM
AND
If we expand Yin powers of a, we obtainby and higher powers of a:
337
HYDROGEN
neglecting
the second
(H-EE,)g=-2(~cos(2rrqk)~x,9Y0.
(23)
A solution of this equation is already sufficient for the calculation of the frequency. For if we neglect higher powers of a, the amplitude can be written:
which gives, by (17) : 1 yjF=
--m/dw
Y$A,g
.
Equation (23) will be discussed first under the restriction that the potential energy can be represented by a quadratic function of the coordinates, and later in the general case. The potential
energy for a linear
chain can be written:
If this potential energy is introduced in the Hamilton operator H in (23), a solution for aY?/aa is:
ay -&
I
operator
1 a\r, E cos’2xqk . ax. = 2s - sin2 nq k k
(26)
For proving this it is convenient to write the left hand side of (23) in another form by going over from the xk to the expressions give twice the distance between the 2xk xk-1 -. xk+ 1, which particle k ad the center of gravity of its neighbours. So we have: 2x
COS
(2scqk) . xk =
k
Introducing
1 )= (2x, 2 sin2 Tcq k
xk+l -
&,)
cos 2nqk.
(26) in the left hand side of (23) we obtain:
(H - E,) g
= -
2 si!2 xq Y,, f g.
cos 2lcqk,
(27)
or, with (25) : (H--E,)
Physicn
g
IV
r
-
2 .12
Tcq
Y,y;cos25rqk.
(2xk-xk+l-xk--l).
(28)
22
338
A. BIJL
The frequency is now.calculated with the help of (26) and (24); partial integration gives : 1 m (29) P = 4s * sin2 xq ’ For q --f 0, this leads to the known
result:
The meaning of (26) is that the disturbed .wave function can be obtained from the undisturbed one by a translation. So in ,the harmonic case the probability distribution is not deformed but only displaced. ’ In the general case no general solution of (23) can be obtained. A useful approximation can be based on the following analysis of the deformation of a crystal by a long wave. When the particle 9 lies at the maximum of the deformation, the mean distance between the particles 9 - k and p + k. is exactly equal to 2ka. The deformed crystal can therefore be obtained from the undeformed one by successive deformations. First a force is applied on the particle fi, the mean positions of the particles fi - 1 and p + 1 being kept constant, untill2$ - Z#+, - %Pp--lis equal to the value in the deformed state. Than the group of the particles p, p + 1, and $J- 1 is displaced with unaltered mutual mean distances relative to the particles p f’ 2 and p - 2, and so on, until1 the final deformation is obtained. . If we neglect forces between particles that are not nearest neighbours, it may be assumed that the relative positions of the particles in a group are determined by the forces on the inner particles and are independent of the forces on the end particles (as the mean distance between the end particles is fixed). So it is not necessary to introduce additional forces on the inner particles to keep their relative positions constant during the displacement of a group. Moreover, the relation between the mean displacement of a group and the total force applied on it is independent of the size of the.group. We may therefore determine this relation for the case that the group contains one particle. This reduction of the deformation energy to a one particle problem is not restricted to a linear case but may’ also be applied to a simple
PROPERTIES
OF CONDENSED
PHASES
OF
HELIUM
AND
HYDROGEN
339 .
cubic lattice. It is clear that the index k’than refers to a. plane of lattice points perpendicular to the x-axis and not to a single particle. It cannot be applied to a’more complicated structure, for instance a cubic face centered lattice. In this case there is between two lattice planes k and 4 + 1 with distance a another lattice plane which we give the index k + 4. The nearest neighbours of the particles in the plane k lie as well in the plane k + 4 as in h + 1. So the assumption that only lattice planes which have the smallest possible distance interact is certainly not valid. And as a wave function which describes the correlation between the positions of different particles is lacking, the deformation energy cannot be calculated. We will therefore obtain the velocity of sound for a linear chain, or a simple cubic lattice. To this result we will apply some estimatkd corrections to obtain a value also for the face centered lattice. To the wave equation which determines the motion of a particle in the field of its neighbours for the linear case we must add a perturbing potential energy -ax. The perturburation equation then becomes :
(31) (compare (22) and (23)). In the case of a quadratic potential function the effect of the perturbation was a simple displacement of the wave function. In the case considered, however, the potential function is very steep at the boundary and the “displacement” of the wave function must be zero at the boundary. We will therefore write:
(32) where f must .be a function which is zero at the boundary and is constant in the center: These conditions are satisfied by the function: f = [l +$($y]-*>
03)
in which l,~ is a constant which-can still be chosen at will. For at the boundary 1 aY/ax I> 0, and at the center aY?/& is zero. Introducing (33) into the left hand side of (31), and writing (l/Y?) aY/& =‘x, we obtain.
340
A.
BIJL
The expression at the right hand side of (34) is an odd function of x, which remains finite at the boundary. If we develop in powers of x, lo must be so chosen that the term with x3 disappears. The form between brackets can then be written approximately as -fix, and (31) is a solution of (32) if kp = 1. If the v a n d e r W a a 1 s-forces are neglected, the particle moves between impenetrable walls with a distance 2 (u - cr) or 2J. As ‘37 is known in this case, the second term.in the bracket can be evaluated. ~~ must be equal to Q 1’12 and we obtain:
P=
7?h2 27.
Ml4 *
(35)
The frequency of the wave is given by ( 16), which equat.ion can be transformed to give : v2 = 2 sin2 7cq .
Fk rn(22h -
5&+, -z&J
.
where Fk is the force on the k’th particle and is in our case equal to a; 2gk - Z$+i - & is given by the integral:
As f N 1 at places where Y is large this integral the result becomes, using kp = 1:
is equal to ka. So
v2 = sin2 xq - ;.2P For a cubic face centered lattice some corrections to the value above obtained for v2 are necessary: a) As the crystal contains two lattice planes per unit lenght a, the mass m in (36) must be multiplied by 2. b) The compressibility of the crystal is smaller as compared with that of a simple cubic lattice. The effect of this influence cannot be calculated but it can roughly be estimated to increase v2 by a factor 2. c) The “wave lenght” of the particle is smaller than in the simple cubic case as the particle is inclosed in a sphere rather than in a cube. We will therefore multiply the second term of p in (34) by a factor 4/3, which is the ratio of the kinetic energies of a particle enclosed in a sphere and in the cube in which the’ sphere can be inscribed.
PROPERTIES
OF CONDENSED
PHASES
After applying these corrections the final result becomes:
OF HELIUM
AND
HYDROGEN
and using the relation
02=&-&&.
341 (3) for 0:
(37)
This formula has been applied to calculate the specific heat for solid helium, hydrogen and deuterium. The result is shown in table I. TABLE
I
The good agreement between experimental and calculated values is accidental, because O2 is only certain & 40%. In the ratio between the O-values for He and H, many incertainties are eliminated and it is satifying that this figure agrees with the experimental one. In the case of D2 a positive correction should be applied to account for the influence of the v a n d e r W a a 1 s-forces. The calculation of this correction has not been carried out because of the uncertainties in the potential function. $7. The thermal expalzsiolz. If the specific heat is known for different volumes, the thermal expansion is determined by thermodynamical relations. For this quantity is given by:
(38) As the entropy
S is given by
the quantity considered is determined by the change of c, with volume, i.e. of 0 with V. It is therefore possible to see w.ether the formula obtained for O2 gives some information about the cause of the abnormal sign of the thermal expansion of liquid He II, - although it must be stressed
342
A. BIJL
again that the formula only holds for crystals of a not too small density. If the v a n d e r W a a 1 s-forces are not neglected, we must start from formula (34). The first term at the right hand side is approximately equal to x . (LPV/&~),_,, and therefore O2 may be written in the form:
(39) where b is a numerical constant which is in the hard sphere case equal to x2/l 92. The second term exists because a displacement of the mean position of the molecule also causes a deformation of its probability distribution. This deformation will be larger the larger is the slope of the wave function at the boundary of the field. So if the v a n d e r W a a 1 s-forces have the effect to decrease this slope relatively to its value in the hard sphere case, they will decrease the value of b. If we want only to determine the sign of a@/&~, we may determine instead the sign of a@/%.z. The equation (39) gives:
(40) If the density is small, we may put : V = that a@/& sahll be positive becomes: 8.21 cc a8
4 >
1 db blz2
( G-bz
) y&c’
On the other hand O2 must be positive, 21u -
u/P, and the condition
therefore:
ml4 *
as a necessary condition: $-+$
or: 2(a-CJ)
> a--
+$-.
(41)
About db/dl we may expect, if anything, that it is positive. For the potential field in the one particle model has the form given in fig. 1. The wave function will therefore be rather flat at the center, and have a large slope at the boundary. An increase of 1 will mainly cause a further decrease of the wave function in the center, but alter only slightly the slope at the boundary. So ‘this slope will increase com-
PROPERTIES
OF CONDENSED
PHASES OF HELIUM
AND HYDROGEN
343
pared with the slope of the wave function in the hard sphere case, i.e. b increases. So we may expect a negative expansion if 2(a - C) N a. The density of liquid He II is somewhat higher then given by this condition. Numerical calculation, using S 1 a t e r’s formula for the potential energy shows that the two terms in (40) are of the same order of magnitude. But as the 0 values obtained with the formula (39) are not equal to the experimental ones, it is not yet certain that this possibility for a negative expansion is realised in liquid helium. V
Fig. 1. Potential energy of particle between fixed neighbours at smaller density.
The author is much indebted to discussions on this subject with Dr. A. M i c h e 1 s and Drs. J. de Boer and to their valuable suggestions which made it possible to overcome some difficulties. Note
I.
Starting from atoms, viz :
S 1 a t e r’s formula
for the potential
energy
of two
He
we obtain an approximation for the potential energy of H, and D, in the following way: The coefficient 0.149 is proportional to the product of the polarisability and the ionisation energy; so the corresponding figure for hydrogen is equal to 1.30. The first term is due to the overlapping of the wave functions of the two molecules; so the coefficient of I in the exponential may be expected to be inverse proportional to the radius of the molecule if the number of electrons is constant; for hydrogen it will therefore have the value -3.8, if we accept the diameter of the hydrogen molecule to be 2.80 A:
344
PROPERTIES
The condition the exponential,
OF CONDENSED that V = 0 for which becomes
p-=t
Y
PHASES OF HELIUM
AND HYDROGEN
= 2.80 A determines the coefficient equal to 113. So the result is:
~,3e-x8r-!g?
J
10-11
Accepting for the kinetic energy the formula which the total energy becomes a minimum can obtain: for H, : a = 3,93”A (experimental for D, : a = 3.63”A (experimental For the minimum energy is found: H t : 96 cal./mol (exp. 184 cal./mol) Dz : 225 cal./mol (exp. 274 cal.!mol)
of
evg.
(I), the value be calculated,
of a for and we
3.88’A) 3.63”A)
I)
Note II. The question may be posed whether the energy value obtained by the “one particle picture” is a good approximation. It must be first remarked that both the calculations of the kinetic energy and the potential energy have errors and there is no reason to expect that these errors are correlated So especially in the case that these energies are nearly equal and opposite a large relative error in the total energy may be expected. In a case where the many p&ticle problem can be solved exactly we may compare the approximated energy value with the correct one. So we may for instance consider a linear gas of impenetrable molecules. If the “diameter” of the molecules is o, and the volume is equal to: N(Z + a), the smallest value for the total energy is: $+-&(1.+2.+
whereas
in the one particle
. .. . +
N2).
model it is equal to:
This value is 25% lower than the correct value. It is highly probable that in a more dimensional problem the error will be less as the chance that the surrounding molecules deviate all at the same time seriously from their average configuration becomes smaller if the number of nearest neighbours increases. Moreover the potential energy will also oppose larger deformations, which has been left out enterily in the above example. So the error mentioned may be expected to give an upper limit. Received March 2, 1937. REFERENCES 1) 2) 3) 4) 5)
K. F. F. W. E.
Clusius und E. Bartholom6, Z.phys.Chem.B30,237, 1935. L o n d o n, Proc. roy. Sot. London, 183, 576, 1936. Simon, Nature, 133, 529, 1934. H. Keesom and A. P. Keesom, Phpsica, 3, 105, 1936. B art h o 1 o m 6 und A. E II c k e n, Z. Eleclrochem. 42, 547, 1936.
IV,
no 4
April
1937
PROPERTIES OF THE CONDENSED PHASES OF HELIUM AND HYDROGEN by A. BI JL (55th
Communication
of the ,,v n n d e r W a ill s-Fund”)
Summary The influence of the zero point motion on the density of a crystal is discussed and the result applicated to hydrogen and deuterium. The zero point motion affects also the specific heat, which is at the lowest temperatures determined by the velocity of sound. The latter could be calculated by considering the influence of an elastic deformation on the wavefunction of the crystal. The result is that the square of the velocity of sound can be obtained as a sum of the contributions of the potential % the kinetic energy; in the case of solid helium and hydrogen energy and good agreement with experimental results is obtained if the molecules are treated as hard spheres. As a consequence it is shown that at very small densities a negative ,‘thermal expansion may be expected.
5 1. Introdmtiort. Liquid helium differs in some of its properties from the condensed phases of heavier elements. The most remarkable differences are : a) the density of condensed helium is very small, so that the intermolecular distances are considerably larger than the gaskinetic diameter of the atom; b) the thermal expansion at the lowest temperatures is negative; c) solid helium excists only under pressure ; d) liquid helium shows a phase transition of the second order. As long as a reliable theory for liquids does not excist, it is hardly possible to expect an explanation of these anomalies. Therefore in this article the question will be posed: In how far do the properties of the crystals of light elements differ from those of ‘normal’ crystals? ‘It will appear that the differences mentioned ab,ove under (a) and (b) can also be expected for crystals.
-
329 -
330
A. BIJL
A general argument can be given to show why the common theory of crystals cannot be applied here. This theory only considers ‘harmonic’ binding forces between the particles and so it can be applied only in the case that the amplitude of the zero point motion is not larger than the region in which this assumption holds. It has beenpointedoutby Clusius and BartholomC1) thatfor hydrogen and helium the assumption of harmonic forces is no longer valid. If this assumption is abandoned, however, no rigourous theory of the crystal can be given as the energy can no longer be written as the sum of the energies of independent oscillators. By using special methods the influence of the zero point motion on the density, the specific heat and the thermal expansion could be discussed *). $ 2. The crystal structure and the density. The method commonly used to obtain the energy of a lattice is to calculate the sum of the mutual potential energies over the points of a point lattice. This calculation is carried out for different types of lattices and different lattice parameters; the lattice with the smallest energy will be stable. In an exact wave mechanical theory this method is not correct. For although a crystal is defined by the fact that some of its properties are periodic functions in space, it can no longer be assumed that all particles have fixed positions. It is only correct to say that the probability for finding a particle is a periodic function in space. So we have to deal with a probability distribution for the intermolecular distances. A rough idea of this distribution can be obtained by considering the wave function of one particle in the field of its neighbours which are considered to be fixed in their equilibrium positions. This simplification is in some way analogous to the one made by E i n s t e i n to attaok the problem of the specific-heat. In this one particle problem the particle is surrounded by impenetrable molecules whose radii are large enough, for all realistic values -of the lattice parameters to provide a’potential’barrier for the inclosed particle. So this particle cannot escape and its lowest wave function is determined unambiguously. *) The influence of the zero SiA’on and hy London; t h o 1 o m 6 (see references).
point motion on the density was already discussed by and on the specific heat by C 1 us i u s and B a r-
33 1 . This wave function is zero at the boundary of the potential ‘hole’ and will in general have a maximum in the center. This form is still more pronounced for the probability of finding the particle, which is the square of the wave function. Therefore the particle will nearly always be found in the neighbourhood of the center. In calculating the potential energy it is therefore permissable to assume in a first approximation that the particle is always exactly in the center of the hole. This simplification was used in the calculations of F. L’o nd o n “). It gives appreciable errors only’ in the case that the intermolecular distances are large, as than the maximum of the wave function is less sharp. Therefore in cases with smaller intermolecular distances the wave mechanical treatment gives the same result for the potential energy as the ordinary treatment. Differences arise, however, with the calculation of the kinetic zero point energy of the lattice. For most crystals this energy can be neglected, being small compared with the potential energy. For He, Hz and D,, however, it is very important as was first pointed out by F. S i m o n 3). This kinetic energy can also be studied in the one particle model. Its ‘mean value’ is determined in this model by the shape of the wave function. If the v an d e r W a a 1 s-forces and the mass of the particle are small enough, this wave function and therefore the kinetic energy will not alter much if the surrounding molecules are replaced by hard spheres and the v a n d e r W a a 1 s-forces are neglected. A further simplification is possible by replacing the ‘hole’ formed by the surrounding molecules by an impenetrable spherical boundary, whose diameter is equal to the shortest diameter of the ‘hole’. This is permissable because at the points outside the sphere the particle is in the close neighbourhood of two or more surfaces of the surrounding molecules, and the wave function is consequently very small. So we obtain as an estimate for the kinetic energy the same value as for the lowest energy state-of a particle enclosed in a sphere with radius a - 6: h2 T= (1) 8m(a - CJ)”’ PROPERTIES
OF
CONDENSED
PHASES
OF
HELIUM
(a = distance between nearest neighbours; meter, which is not defined rigourously.)
AND
HYDROGEN
o = molecular
dia-
332
A. BIJL
If the v a n d e r W a a 1 s-forces cannot be neglected this value will generally be larger because the wave function will be more concentrated in the center. This picture enables us to draw some conclusions as to the crystal structure and the density if we may assume that the arrangement of of the neighbours which gives the smallest total energy will be approximately identical with the most probable arrangement of the many particle system at the absolute zero. If n is the number of nearest neighbours, the potential energy per atom is equal to:
v = 5v(u),
(2)
where ~(a) is the mutal potential energy of two molecules at distance a. (Interactions between molecules which are not nearest neighbours will always be neglected.) The crystal symmetry allows only a limited number of nearest neighbours, which numbers are independent of the value of a. It is therefore obvious that for any given value of a for which v (a) < 0, n will have its highest value because the potential energy is then a minimum. So we may expect that n = 12, i.e. a cubic face centered or a hexagonal close packed structure will be the most stable one. a is now determined from the condition that the total energy T + V is a minimum as a function of a. The calculation of the density along these lines has been carried out for He by F. L o n d o n “). It can be applied also for H, and D, (see note ‘1) Some remarks about the accuracy obtained with this model are given in note 2. 5 3. Sfiecific heat and velocity of sotind. After D e b y e .and B o r n the heat motion in a crystal can be approximately described by a superposition of harmonic undamped waves. The specific heat is obtained as a sum over the contributions of all wave numbers. The contribution of one single oscillator of frequency v depends on the only the oscillators of low freratio hv/kT. At low temperatures quency are of importance for the specific heat. .As in the case considered the potential energy can no longer be described as a quadratic function of the coordinates of the particles, the analysis of the heat motion into harmonic waves is not possible, However as at low temperatures the high frequencies are degenerated,
PROPERTIES
OF CONDENSED
PHASES
OF HELIUM
ANP
HYDROGEN
333
we have to deal only with the long waves (the sound waves) fo; which we may assume that they are harmonic and undamped. We may therefore calculate the specific heat at the lowest temperatures by the ordinary treatment of D e b y e. In this method the specific heat can be conveniently expressed as a function of the velocity of sound, as this velocity determines the relation between wave number and frequency. The velocity of sound is connected with the specific heat by the formulae: (3) and
C”=q$q;)3
(4’
where : c = velocity of sound V = molecular volume. In a rigourous treatment longitudinal and transversal waves ought to be treated separately. As in this paper no exact numerical result can be obtained it seems justified to neglect the difference between those two velocities. 5 4. Sciwoedinger eyuatipz for a deformed crysia!. To calculate the velocity of sound it is sufficient to know this velocity for one large wave length. This velocity is ‘determined by the energy necessary for the deformation as a function of the amplitude and of the mass of the molecules. At low temperatures this deformation energy is usually confined, to a change in the potential energy of the lattice. The fundamental difference between the usual case and that considered here is, however, that here also the kinetic zero point motion changes with the distance between the particles, as was shown in $ 2. It would be possible to draw some conclusions from these considerations as to the influence of the kinetic zero point motion on the velocity of sound. As for this purpose the knowledge of dTa/da2 is required, which can only be deduced from (1) with a very low accuracy, it seems justified to analyse in more detail the change of the S c h r o e d i n g e r wave function by the deformation, We will therefore consider a crystal deformed by a density wave along the x-axis (the x-axis being placed in the same direction as one
334
A. BIJL
of the lines connecting nearest neighbours). The wave lenght of the wave will be called a/q, where q is a very small number. The amplitude and phase of the wave can be indicated by the complex amplitude A, : A, = $:
e(2xiqk) x,.
(x, is the deviation of the particle If the crystal is described by a A, can be treated as a quantum value - which is for a macroscopic value -- is given by:
k from its mean position) wave function Y, the amplitude mechanical quantity. Its mean operator always equal to its real
(dw = dq dy, dzl . . . . d+) The problem is solved if & is known as a function of time, because then its frequency and hence the velocity of sound is known. A,(t), however, can be obtained from (6), if Y(i) is known. Y(t) is determined by the equation of motion: h ay =H-Y, 2Tci-z where H is the
H a m i 1 t o n operator
for the crystal:
The only ambiguity left now is in the chaise of the initial wave function Y(t = 0). It must be so chosen that for & = 0 it describes a state of the crystal near the absolute zero. It is therefore logical to take this initial Y in such a way that for & + 0 it becomes identical with the lowest eigenfunction. This can be obtained by choosing Y(t = 0) in such a way that it is among all wave functions having the same value for 2, the one with the lowest energy. If Y(t = 0) is chosen in this way a differential equation can be obtained for it. For if Y satisfies the condition that: U=/dwY*hY
(9)
PROPERTIES
OF CONDENSED
is a minimum,
PHASES
OF
with the additional
HELIUM
AND
conditions
/dw*YY*=
HYDROGEN
335
that
1
and jdw.Y”AY=A,,
it must satisfy the equation: HY
+ ;
(QA, + a*Af)
Y = hY.
(10)
5 5. Discussion of the wave eqzcation. We will first find the meaning of the constants a and A; it proves that there is a relation between a and the frequency and between A and the deformation energy. Substitution of:
a = r - e(icp)
(‘1)
in (10) gives : H\r
+ 2r c xk co?, (hqk
+ ‘p) ’ ‘r = hy.
k
(14
. In this equation the second term on the left hand side is responsible for the deformation. If in a crystal an external force of magnitude - 29’ cos (2nqk + ‘p) derived from a potential 2??& cos (2x@ f cp), acts on every k’th particle, this system (i.e. crystal and external forces) would have the operator (12) for the total energy. The equilibrium condition for a particle in this system is that the sum of the external forces and the crystal forces is zero, hence: + 2r cos (27cqk + $3) + g
= 0. k
(13)
This relation can be obtained directly from (12) by differentiating with respect to +, multiplication with Y* and subsequent integration. As a consequence of the equation of motion (7) we have: ::
W
(14
mXk=-Jz-
By substitution of (I 4) into (13), multiplying summation over k we obtain:
,*=“j
2
rl’
with e (ixiqk)
and ,
(15)
336
A. BIJI.
As we may assume that 2, is a periodic v2A, = - A, and hence :
function
of the time, (16)
Thus the value of v2 in which we are interested the ratio between A, and ct.
is determined
by
The relation between A and the deformation energy can -be obtained by differentiating (12) with respect to r, multiplying with Y* and integrating. The result is: 6%
z=-zy
2N
hence :
(18) On the other hand we obtain
directly
from (10) :
E(r) = A - ; (a& + u*A,*),
(19)
or, with (17) :
(20)
E(Y) = h + So the deformation
energy is equal to:
E(r) = E(o) + N$,
A, 12
or: E(r) which formula crystal.
E(o) = -
ij Ij Xk “xk .
is the same as is obtained
(21)
in the usual theory of the
of the frequency and the speiific heat. As the phase
3 6. Calculation
of the density’wave is of no importance, is real. (12) than becomes: m
we will chose it so that u ’
+ 2x f & cos (2x44 \r = hy.
(22)
PROPERTIES
OF CONPENSED
PHASES
OF
HELIUM
AND
If we expand Yin powers of a, we obtainby and higher powers of a:
337
HYDROGEN
neglecting
the second
(H-EE,)g=-2(~cos(2rrqk)~x,9Y0.
(23)
A solution of this equation is already sufficient for the calculation of the frequency. For if we neglect higher powers of a, the amplitude can be written:
which gives, by (17) : 1 yjF=
--m/dw
Y$A,g
.
Equation (23) will be discussed first under the restriction that the potential energy can be represented by a quadratic function of the coordinates, and later in the general case. The potential
energy for a linear
chain can be written:
If this potential energy is introduced in the Hamilton operator H in (23), a solution for aY?/aa is:
ay -&
I
operator
1 a\r, E cos’2xqk . ax. = 2s - sin2 nq k k
(26)
For proving this it is convenient to write the left hand side of (23) in another form by going over from the xk to the expressions give twice the distance between the 2xk xk-1 -. xk+ 1, which particle k ad the center of gravity of its neighbours. So we have: 2x
COS
(2scqk) . xk =
k
Introducing
1 )= (2x, 2 sin2 Tcq k
xk+l -
&,)
cos 2nqk.
(26) in the left hand side of (23) we obtain:
(H - E,) g
= -
2 si!2 xq Y,, f g.
cos 2lcqk,
(27)
or, with (25) : (H--E,)
Physicn
g
IV
r
-
2 .12
Tcq
Y,y;cos25rqk.
(2xk-xk+l-xk--l).
(28)
22
338
A. BIJL
The frequency is now.calculated with the help of (26) and (24); partial integration gives : 1 m (29) P = 4s * sin2 xq ’ For q --f 0, this leads to the known
result:
The meaning of (26) is that the disturbed .wave function can be obtained from the undisturbed one by a translation. So in ,the harmonic case the probability distribution is not deformed but only displaced. ’ In the general case no general solution of (23) can be obtained. A useful approximation can be based on the following analysis of the deformation of a crystal by a long wave. When the particle 9 lies at the maximum of the deformation, the mean distance between the particles 9 - k and p + k. is exactly equal to 2ka. The deformed crystal can therefore be obtained from the undeformed one by successive deformations. First a force is applied on the particle fi, the mean positions of the particles fi - 1 and p + 1 being kept constant, untill2$ - Z#+, - %Pp--lis equal to the value in the deformed state. Than the group of the particles p, p + 1, and $J- 1 is displaced with unaltered mutual mean distances relative to the particles p f’ 2 and p - 2, and so on, until1 the final deformation is obtained. . If we neglect forces between particles that are not nearest neighbours, it may be assumed that the relative positions of the particles in a group are determined by the forces on the inner particles and are independent of the forces on the end particles (as the mean distance between the end particles is fixed). So it is not necessary to introduce additional forces on the inner particles to keep their relative positions constant during the displacement of a group. Moreover, the relation between the mean displacement of a group and the total force applied on it is independent of the size of the.group. We may therefore determine this relation for the case that the group contains one particle. This reduction of the deformation energy to a one particle problem is not restricted to a linear case but may’ also be applied to a simple
PROPERTIES
OF CONDENSED
PHASES
OF
HELIUM
AND
HYDROGEN
339 .
cubic lattice. It is clear that the index k’than refers to a. plane of lattice points perpendicular to the x-axis and not to a single particle. It cannot be applied to a’more complicated structure, for instance a cubic face centered lattice. In this case there is between two lattice planes k and 4 + 1 with distance a another lattice plane which we give the index k + 4. The nearest neighbours of the particles in the plane k lie as well in the plane k + 4 as in h + 1. So the assumption that only lattice planes which have the smallest possible distance interact is certainly not valid. And as a wave function which describes the correlation between the positions of different particles is lacking, the deformation energy cannot be calculated. We will therefore obtain the velocity of sound for a linear chain, or a simple cubic lattice. To this result we will apply some estimatkd corrections to obtain a value also for the face centered lattice. To the wave equation which determines the motion of a particle in the field of its neighbours for the linear case we must add a perturbing potential energy -ax. The perturburation equation then becomes :
(31) (compare (22) and (23)). In the case of a quadratic potential function the effect of the perturbation was a simple displacement of the wave function. In the case considered, however, the potential function is very steep at the boundary and the “displacement” of the wave function must be zero at the boundary. We will therefore write:
(32) where f must .be a function which is zero at the boundary and is constant in the center: These conditions are satisfied by the function: f = [l +$($y]-*>
03)
in which l,~ is a constant which-can still be chosen at will. For at the boundary 1 aY/ax I> 0, and at the center aY?/& is zero. Introducing (33) into the left hand side of (31), and writing (l/Y?) aY/& =‘x, we obtain.
340
A.
BIJL
The expression at the right hand side of (34) is an odd function of x, which remains finite at the boundary. If we develop in powers of x, lo must be so chosen that the term with x3 disappears. The form between brackets can then be written approximately as -fix, and (31) is a solution of (32) if kp = 1. If the v a n d e r W a a 1 s-forces are neglected, the particle moves between impenetrable walls with a distance 2 (u - cr) or 2J. As ‘37 is known in this case, the second term.in the bracket can be evaluated. ~~ must be equal to Q 1’12 and we obtain:
P=
7?h2 27.
Ml4 *
(35)
The frequency of the wave is given by ( 16), which equat.ion can be transformed to give : v2 = 2 sin2 7cq .
Fk rn(22h -
5&+, -z&J
.
where Fk is the force on the k’th particle and is in our case equal to a; 2gk - Z$+i - & is given by the integral:
As f N 1 at places where Y is large this integral the result becomes, using kp = 1:
is equal to ka. So
v2 = sin2 xq - ;.2P For a cubic face centered lattice some corrections to the value above obtained for v2 are necessary: a) As the crystal contains two lattice planes per unit lenght a, the mass m in (36) must be multiplied by 2. b) The compressibility of the crystal is smaller as compared with that of a simple cubic lattice. The effect of this influence cannot be calculated but it can roughly be estimated to increase v2 by a factor 2. c) The “wave lenght” of the particle is smaller than in the simple cubic case as the particle is inclosed in a sphere rather than in a cube. We will therefore multiply the second term of p in (34) by a factor 4/3, which is the ratio of the kinetic energies of a particle enclosed in a sphere and in the cube in which the’ sphere can be inscribed.
PROPERTIES
OF CONDENSED
PHASES
After applying these corrections the final result becomes:
OF HELIUM
AND
HYDROGEN
and using the relation
02=&-&&.
341 (3) for 0:
(37)
This formula has been applied to calculate the specific heat for solid helium, hydrogen and deuterium. The result is shown in table I. TABLE
I
The good agreement between experimental and calculated values is accidental, because O2 is only certain & 40%. In the ratio between the O-values for He and H, many incertainties are eliminated and it is satifying that this figure agrees with the experimental one. In the case of D2 a positive correction should be applied to account for the influence of the v a n d e r W a a 1 s-forces. The calculation of this correction has not been carried out because of the uncertainties in the potential function. $7. The thermal expalzsiolz. If the specific heat is known for different volumes, the thermal expansion is determined by thermodynamical relations. For this quantity is given by:
(38) As the entropy
S is given by
the quantity considered is determined by the change of c, with volume, i.e. of 0 with V. It is therefore possible to see w.ether the formula obtained for O2 gives some information about the cause of the abnormal sign of the thermal expansion of liquid He II, - although it must be stressed
342
A. BIJL
again that the formula only holds for crystals of a not too small density. If the v a n d e r W a a 1 s-forces are not neglected, we must start from formula (34). The first term at the right hand side is approximately equal to x . (LPV/&~),_,, and therefore O2 may be written in the form:
(39) where b is a numerical constant which is in the hard sphere case equal to x2/l 92. The second term exists because a displacement of the mean position of the molecule also causes a deformation of its probability distribution. This deformation will be larger the larger is the slope of the wave function at the boundary of the field. So if the v a n d e r W a a 1 s-forces have the effect to decrease this slope relatively to its value in the hard sphere case, they will decrease the value of b. If we want only to determine the sign of a@/&~, we may determine instead the sign of a@/%.z. The equation (39) gives:
(40) If the density is small, we may put : V = that a@/& sahll be positive becomes: 8.21 cc a8
4 >
1 db blz2
( G-bz
) y&c’
On the other hand O2 must be positive, 21u -
u/P, and the condition
therefore:
ml4 *
as a necessary condition: $-+$
or: 2(a-CJ)
> a--
+$-.
(41)
About db/dl we may expect, if anything, that it is positive. For the potential field in the one particle model has the form given in fig. 1. The wave function will therefore be rather flat at the center, and have a large slope at the boundary. An increase of 1 will mainly cause a further decrease of the wave function in the center, but alter only slightly the slope at the boundary. So ‘this slope will increase com-
PROPERTIES
OF CONDENSED
PHASES OF HELIUM
AND HYDROGEN
343
pared with the slope of the wave function in the hard sphere case, i.e. b increases. So we may expect a negative expansion if 2(a - C) N a. The density of liquid He II is somewhat higher then given by this condition. Numerical calculation, using S 1 a t e r’s formula for the potential energy shows that the two terms in (40) are of the same order of magnitude. But as the 0 values obtained with the formula (39) are not equal to the experimental ones, it is not yet certain that this possibility for a negative expansion is realised in liquid helium. V
Fig. 1. Potential energy of particle between fixed neighbours at smaller density.
The author is much indebted to discussions on this subject with Dr. A. M i c h e 1 s and Drs. J. de Boer and to their valuable suggestions which made it possible to overcome some difficulties. Note
I.
Starting from atoms, viz :
S 1 a t e r’s formula
for the potential
energy
of two
He
we obtain an approximation for the potential energy of H, and D, in the following way: The coefficient 0.149 is proportional to the product of the polarisability and the ionisation energy; so the corresponding figure for hydrogen is equal to 1.30. The first term is due to the overlapping of the wave functions of the two molecules; so the coefficient of I in the exponential may be expected to be inverse proportional to the radius of the molecule if the number of electrons is constant; for hydrogen it will therefore have the value -3.8, if we accept the diameter of the hydrogen molecule to be 2.80 A:
344
PROPERTIES
The condition the exponential,
OF CONDENSED that V = 0 for which becomes
p-=t
Y
PHASES OF HELIUM
AND HYDROGEN
= 2.80 A determines the coefficient equal to 113. So the result is:
~,3e-x8r-!g?
J
10-11
Accepting for the kinetic energy the formula which the total energy becomes a minimum can obtain: for H, : a = 3,93”A (experimental for D, : a = 3.63”A (experimental For the minimum energy is found: H t : 96 cal./mol (exp. 184 cal./mol) Dz : 225 cal./mol (exp. 274 cal.!mol)
of
evg.
(I), the value be calculated,
of a for and we
3.88’A) 3.63”A)
I)
Note II. The question may be posed whether the energy value obtained by the “one particle picture” is a good approximation. It must be first remarked that both the calculations of the kinetic energy and the potential energy have errors and there is no reason to expect that these errors are correlated So especially in the case that these energies are nearly equal and opposite a large relative error in the total energy may be expected. In a case where the many p&ticle problem can be solved exactly we may compare the approximated energy value with the correct one. So we may for instance consider a linear gas of impenetrable molecules. If the “diameter” of the molecules is o, and the volume is equal to: N(Z + a), the smallest value for the total energy is: $+-&(1.+2.+
whereas
in the one particle
. .. . +
N2).
model it is equal to:
This value is 25% lower than the correct value. It is highly probable that in a more dimensional problem the error will be less as the chance that the surrounding molecules deviate all at the same time seriously from their average configuration becomes smaller if the number of nearest neighbours increases. Moreover the potential energy will also oppose larger deformations, which has been left out enterily in the above example. So the error mentioned may be expected to give an upper limit. Received March 2, 1937. REFERENCES 1) 2) 3) 4) 5)
K. F. F. W. E.
Clusius und E. Bartholom6, Z.phys.Chem.B30,237, 1935. L o n d o n, Proc. roy. Sot. London, 183, 576, 1936. Simon, Nature, 133, 529, 1934. H. Keesom and A. P. Keesom, Phpsica, 3, 105, 1936. B art h o 1 o m 6 und A. E II c k e n, Z. Eleclrochem. 42, 547, 1936.