Contrast in multicomponent systems

Physica

102A(1980) 120-130 @ North-Holland

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J. des CLOIZEAUX Service de Physique

SYSTEMS

and G. JANNINK

Thkorique et Service de Physique du Solide et de Rksonance Centre d’Etudes Nucl&aires de Saclay*

Magnitique,

Received 12 October 1979

The forward scattering of radiation by multicomponent systems is expressed in terms of the chemical potentials and of the collision lengths associated with the components. Contrast lengths can be defined in various unequivalent ways. The formula giving the scattering cross section in terms of such contrasts is derived geometrically. This approach allows a discussion concerning the best definition of the contrast lengths.

1. Introduction

In a scattering experiment by multicomponent systems, the forward scattered intensity is mostly produced by the fluctuations of compositions in the system. The scattering experiment is therefore used as a tool for the structure analysis of such systems. An important concept derived in the calculation of the cross section, is the contrast factor of one component versus the other ones. Actually, an observation of the composition fluctuations is possible only if such contrast factors are large enough. The contrast factors do not only depend on the collision lengths of the scatterers or their polarizability, but also on extensive properties of the system. Here, we derive and discuss these properties. The analysis of composition fluctuations in binary systems is straightforward. It is known to be difficult’) in multicomponent systems. The difficulty comes, in part, from the fact that it is possible to express the cross sections in several equivalent forms; to each of these forms, however, correspond different contrasts factors. We explore several forms and define the optimum decomposition. Since around 1970, neutron scattering is used for the study of multicomponent systems*). This type of scattering is very convenient because combinations of certain collision lengths yield larger contrast factors than in light or X ray scattering. As a consequence, several expressions for the cross section of multicomponent systems have been published2*3) recently. In their * CEN Saclay, Boite Postale No. 2, 91190 Gif-s/-Yvette, France. 120

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articles, the authors deal with dilute solutions of macromolecules in a single solvent. Interaction between the macromolecules is neglected. The macromolecules are tight molecular structures made of several component molecules. Well known examples are the polystyrene-polysoprene block copolymers4) in benzene, and the ferritin5) globular structures in water. The composition of the macromolecule may fluctuate as a result of dispersion, in molecular weight, of the component molecules. The scattered intensity in the limit of zero angle is very sensitive to these fluctuations in composition, and an important polydispersion effect has been established4.‘). Here we consider a more general situation where the solute components are not necessarily bound to each other, but interact as a result of chemical affinity. Thus, in particular, we can account for the interaction between macromolecules. Another situation of interest may also be studied. This consists of the macromolecules dispersed in a mixture of two solvents, one of which is preferentially absorbed on certain components of the macromolecule. Calculation of the cross section for such systems has been given earlier, when scattering experiments were made using only light of X ray scattering. Actually, one of our results is identical to a result first obtained by H.C. Brinkmann and J.J. Hermans6) in 1948, and rederived in 1950 by J.G. Kirkwood and R.J. Goldberg’) and simultaneously by W. StockmayeP). Let us also mention the work of A. Vrij and J. Th. G. Overbeek’), who rediscussed the problem in 1962. All these derivations lead to the same essential result, but they are not as general as they could be and do not emphasize the properties which make the final result so remarkable. For this reason we give here a new derivation of this result. As a tool, we use simple vector analysis, which has the advantage of giving a geometric interpretation of the problem and which allows a convenient discussion.

2. The problem The scattering system is made of m components. Molecules of a given component L~((Y= 1,. . . , m) are identical. To each molecule of a component cr corresponds a collision length a,. This length is the sum of the collision lengths aia of all the atoms i belonging to a molecule a, =

2

U&e

i E molecule of component

(Y

(1)

The system is enclosed in a sample holder at a given pressure p. The incident beam intercepts a volume fraction V of the sample.

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AND G. JANNINK

We consider the scattering cross section at a given scattering angle 6

(2) where summations are made over all atoms i belonging to a given component (Y in volume V, and over all components: ria is the position of atom i in a molecule of component o, q is the wave vector transfer (4 = (27r/h) sin O/2), A is the wavelength of the incident radiation. The bracket ( )” indicates that the average value has to be taken for all possible configurations in the volume V. In the limit of zero-scattering angle, we can write this cross section as follows (3) where summations are made over all molecules I belonging to a given component B in V, and where the a, are given by eq. (1). Introducing the average number N,, of molecules (Y in V

and the corresponding as follows’“*9)

chemical potential EL,,we can write the cross section B

(4) where N, is considered as a function of V and the set {p} of pa, (Y= 1,. . . , m. The system is supposed to be compressible; as a consequence, the partial derivatives

(5) exist. They are related to the partial derivatives the Jacobi relation

appearing in formula (4) by

(6) The derivatives

i&L aN,

(5) also satisfy the following relation (see the Appendix)

> f!$> V,(N)

=

p,(N}+?$'

(7)

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where u, is the partial volume >V

va=arv,>P,,N)’ and x is the compressibility 1 av -vap >p,{N)

x=

(8)

The volume V is a homogeneous

xCXN,v,.

V=

The derivatives the N,. Therefore ?

function of degree 1, of the N,. Therefore (9)

a~JLJN,),,~Nj are homogeneous

functions of degree zero, of

Na%$p,,N) =O

(10)

which is known as the Gibbs-Duhem relation. Since the thermodynamic properties of the system, obtained in other kinds of experiments, are measured at constant pressure or at constant numbers of molecules, it is desirable to express the cross section as a function of the quantities on the right side of eq. (7). The problem is therefore the determination of the explicit form of 2 as a function of x and the &AJc?N,&,,,,. We now formulate the problem using notations appropriate to vector space analysis. The space is of dimension p. We define the 3 vectors 1 a),

of component

a,,

1 n),

of component

N,,

1 u),

of component

v,/V Vx,

and the 2 operators

a=l,...,m

(11)

G and P, by the matrix elements

Using (4), (6) and (7), we find for the cross section I B a=( a 1G+P

II) a9

where I is the unit matrix. The Gibbs-Duhem relation is written as

(12)

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G 1n) =

AND G. JANNINK

0.

By construction,

(13)

P is the projector

P=lv)(v(.

(14)

By definition, the vectors ing to (9) and (11)

) n) and I o ) are such that (v I n) # 0, since accord-

(v I n) = (V/x)?

(13

The problem is now to calculate the cross section 2 in terms of quadratic forms which contain separately G and P.

3. The solution It is convenient Ib)=la)-A

to introduce

a vector I b),

(16)

(4,

where A is an arbitrary number, to be determined The cross section (12) can be written as

later.

(17)

This formula can be easily simplified. From (13) and (14), we get immediately (G+P)In)=lvHv(n), and therefore

Id=

&

I v>(v I n>,

(18)

from which we deduce

+FIv)=l-

(19)

Using (18) and (17), we write (16) as

(20) Another simplification is introduced i.e.

by requiring that 1 b) is orthogonal to 1 n),

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(6 I n) = 0,

(21)

*=(n (n +

(22)

then a> u)’

PBz = (M)‘+(6

I&&)*

(23)

Let us now transform the second term in the right side of (23). For this purpose, we introduce an arbitrary 1 u}. The relations (G+P)Iu)=GIu)+lv)(uIu), (G+J’)jn)=(u)(+) can be conbined to give

(24)

(25) and since (G + P) has an inverse ((v I n) # 0)

Multiplying by (b I on the left, we obtain

(26) Now, we would like to identify G 1 u) and 1 b), in order expression for the second term of (23). Therefore, we can try to solve the equation G 1 u) = 1 b).

to find a new

(27)

The vector 1 b) belongs to the vector subspace E. orthogonal to I n) (see (21)) and since I n) is an eigenvector of G, G I u) belongs to Eo. On the other hand, the operator GEo, which is the restriction of G in the space E. and acts in E,, has an inverse in Eo. (Notice that G has no inverse.) If I u) belongs to &, eq. (27) can be written G&u)=lW

(28)

and has a unique solution /~)=Gz/b).

(2%

Thus 1 u) exists and we have (b 1 U) = (b 1 Gz 1 b).

(30)

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Let us now use eq. (27) in eq. (26). By comparing with (30), we get

(b ( u>=(b I&

1b) = (b 1GE; 16).

(31)

Bringing this result in (23), and using (15), we find the result (32) which gives the cross section as a sum of a term proportional to the compressibility and of a term related to the chemical potentials at constant pressure. The vector 1 b) is (33) Its components,

are the contrast lengths compressibility x.

and we note that they

are independent

of the

4. Generalization For using the result (32), we have to calculate the quadratic form (b I GE: 1 b), starting from a given matrix G and from given vectors I a), I n), 1 v). We show in this section that there are several ways of achieving such a calculation and we give different expressions for the same result. For this purpose, we generalize the method used in section 3 and consider the subspace E orthogonal to a vector I t) such that (n ( t) f 0. In particular, the vectors ( u’) = I u)-

1 np

0 +

u, n>

(35)

( b’) = ( b)-

1 t)(’ +0

b, t>

(36)

and

belong to E. Eq. (35) gives Gju’)=Glu)=jb) and by combining this equality and (36), we obtain

(37)

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r)#.

b’) = G 1u’) - 1

t

Let GE be the operator defined in subspace subspace. From eq. (28), we deduce

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(38) E and equivalent

GE 1 u’) = 1 b’).

to G in that

(39)

This equation can now be solved in space E, since GE is regular 1 u’) = GE’ 1 b’)

(40)

and we have (b’ 1 u’) = (b’ 1 GE’ 1b’).

(41)

On the other hand, eqs. (35) and (36) show that (b’ 1 u’) = (b 1 u).

(42)

Finally by comparing eqs. (31), (41), and (42), we find that (bI&b)=(b’IC,‘Ib’).

(43)

The cross section can be written now

PBS= (a 1u)‘+ + (b’ 1G,’ 1b’),

(4)

where (45) A more condensed

form of (45) is

1b’)= 1b)- 1t$$,

W)

where (47) Form (44) leads now directly to calculable results

5. Discussion

The vector 1 t) can be chosen arbitrarily, with the restriction (n I t) # 0. Thus ( t) may point in the direction of the axis a! corresponding to the component (Y.For instance, we choose a = 1

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(48)

where the elements of GE, are

This is the result obtained by Brinkmann and Hermans4) and similar results can be obtained by giving other values to (Y. We may also try for 1 t) another choice. Consider for instance (50)

I t> = I v>*

In that case

The quantities N, do not appear anymore in the definition of 1 6). Thus, in this case, the contrast lengths are independent of the composition of the system. This form can be useful for discussing experiments in which the composition varies. We may also note that the total constant length ( (b’ 1b’) I”* has a maximum but no minimum. Thus, the contrast is maximum when 1b’) = 1b) and this case corresponds to the solution of section 3.

Acknowledgement

One of us (G.J.) thanks R. Bidaux for helpful discussions.

Appendix

We want to prove that

(Al) where 8V v0 =anr, >P. (Nk x=

---

1 av v @

p.(N)’

642)

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Let us consider the partial derivatives k aNI >p.(N)’ Let us calculate these quantities by replacing the variables variables V,(N). We obtain with the help of (A2)

p,(N)

by the

(A3) On the other hand, we have

CL,

aF

aF -> aN,

=

’

V,(N)’

=

-a

v,(N)’

(A4)

where F is the Helmoltz free energy. Using these definitions we obtain

MEL,, av

>

=

-- ap

(A9

> V, {Nl’

ax

V,(N)

From (A3) and (AS), we deduce &!E. aNp >

V,(N)=

The pressure

p=f$. {

> p.(N)+

$

“%

is an homogeneous

> V,(N)’

function of the variables

1

(A7)

Putting

where c, = NJV, we deduce from (A7)

I =-62 I av aEJ, I ap

aV

V. IN

_@

=+fh,

aNa V,{N)

Vfh

p,(N)

(A6)

=$.fi’

Nafh,

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By combining these equations,

AND G. JANNINK

and by using (A2) we obtain

Finally, by combining (A6) and (AS), we obtain the announced

result (Al).

References 1) A. Miinster, Statistical Thermodynamics, Chapter 17, Section 4 (Academic Press, New York, London, 1974). 2) J.P. Cotton and H. Benoit, J. Physique 36 (1975) 905. 3) H. Stuhrmann and R. Kirste, Z. Phys. Chem. 56 (1967). 4) L.M. Ionescu, Thesis Strasbourg 1976. 5) H. Stuhrmann, J. Appl. Cryst. 7 (1974) 173. 6) H.C. Brinkmann and J.J. Hermans, J. Chem. Phys. 17 (1949) 574. 7) J.G. Kirkwood and R.J. Goldberg, J. Chem. Phys. 18 (1950) 54. 8) W.H. Stockmayer, J. Chem. Phys. 18 (1950) 58. 9) A. Vrij and J. Th. G. Overbeek, J. of Colloid Science 17 (1962) 570. 10) F. Zernicke, Dissertation Amsterdam, 1918.