:2)(V-~)=~T

Quick Response Code integrated with javagenerate, create qr code none on java projects

(17.49)

Java bar code creation for javausing barcode generator for java control to generate, create bar code image in java applications.

where P, V, T respectively stand for pressure, volume, and temperature in units of the corresponding critical quantities. (Thus, for example, t = T - 1.) We shall take V as the order parameter and P as the conjugate field, and suppose that (17.49) is the result of minimizing a Landau free energy 1/1 (V, P, T), i.e., it is the equation a1/1/ aV = O. Thus, by integrating (17.49) at constant T and P, we obtain 3 8 1 1/1(V,P,T)=PV- V - 3"Tlog ( V- 3" )

Java bar code scanner for javaUsing Barcode scanner for Java Control to read, scan read, scan image in Java applications.

(17.50)

Control quick response code image with visual c#using .net tocreate denso qr bar code with asp.net web,windows application

up to an undetermined function of T. (The P dependence is correct because P couples linearly to the order parameter V) A qualitative sketch of this function is shown in Fig. 17.2. Minimizing it with respect to V yields

~_T_

Insert qr code iso/iec18004 with .netgenerate, create qr code jis x 0510 none on .net projects

3 V-

Control qr code iso/iec18004 data for visual basicto receive quick response code and qr code jis x 0510 data, size, image with visual basic barcode sdk

~ = ~{~TIog(V _ ~) 2

(17.51)

Control european article number 13 size on javato attach ean / ucc - 13 and gtin - 13 data, size, image with java barcode sdk

which is just the Van der Waals equation written in a different form. In the region of a first-order transition, 1/1 must have two equal minima, occurring at the

2d Matrix Barcode barcode library on javause java 2d barcode writer toattach 2d matrix barcode with java

THE LANDAU APPROACH

Control ucc - 12 size for java upc-a supplement 2 size on java

I/J(V,P)

Pfixed

Java postnet 3 of 5 printer on javause java postnet creator toincoporate usps postal numeric encoding technique barcode in java

---t"'----'-------------~V

Excel code-128 integrated for excelgenerate, create ansi/aim code 128 none with office excel projects

Fig. 17.2 Landau free energy that leads to the Van der Waals equation of state.

Linear 1d Barcode maker on .netusing barcode writer for .net for windows forms control to generate, create linear image in .net for windows forms applications.

volumes VI and V2, satisfying the conditions

RDLC Report Files code 128 code set c encoder in .netusing local reports rdlc todevelop code 128 code set a in asp.net web,windows application

(a1/J/aV)V~Vl = (a1/J/aV)V~V2 = 0

2d Matrix Barcode barcode library on .netusing local reports rdlc tocreate 2d barcode in asp.net web,windows application

1/J(V1 ) = 1/J(V2) The first of these requires P(VI ) = P(V2), which leads to

Control code 39 full ascii data with .netto render barcode code39 and barcode code39 data, size, image with .net barcode sdk

V2 - }

Bar Code barcode library in vbuse visual studio .net bar code generating toattach barcode on vb

1) -3 (1

Datamatrix 2d Barcode integration on vb.netgenerate, create gs1 datamatrix barcode none on visual basic.net projects

V/ -

(17.52)

With the help of (17.50), the second condition can be rewritten

1/J(V1 ) -1/J(V2)

P(V1

V2) - jV1PdV = 0

(17.53)

p(VI - V2) = jV1PdV

which is the Maxwell construction. We now calculate the critical exponents. Differentiating (17.49) with respect to V, we have ap 6 8 T

aV=VJ-3(V-t)2

Approaching the critical point from T> 1 along V = 1, we find BP/ Hence the susceptibility is

av =

- 6t.

1 ( -av)

(17.54)

which gives y = 1.

SPECIAL TOPICS IN STATISTICAL MECHANICS

Near the critical point, put

T = 1

+ t,

P = 1 + p,

V=I+v

(17.55)

Then the equation of state can be rewritten as

v 3 + p(1 +

V)3 -

H p + 8t)(1

+ V)2 =

(17.56)

When t = 0, this reduces to p = - 3v 3/2 for small v, which shows that the critical isotherm is an odd function about the critical point. For t < 0, the same approximation gives

which shows that the isotherm is a odd function about a point displaced from the critical point by 8t/3 along the p axis. Therefore, the Maxwell construction will yield coexisting volumes VI' V; that are symmetrically placed about the critical volume V = 1:

VI = 1

V2 = 1 - x

(17.57)

we obtain (17.58)

Substituting these into (17.52) and keeping only terms to order x

x=20

Hence f3 = !. The specific heat cannot be calculated because (17.50) is determined only up to a function of the temperature. Assuming it to be such as to give ideal gas behavior at large V, we obtain a = 0. In summary, the critical exponents have "classical" values. If we were interested only in the neighborhood of the critical point, we would write down a mean field theory identical to (17.37), with m identified with VI - v 2 , the specific-volume difference of the coexisting phases, and H with P - 1. We would arrive at the same critical exponents, although the model would be quite different. In the model we just discussed, VI - V 2 is a calculated quantity, and not a free variable. Mean-field theory, as we can see, has many guises, but they lead to the same critical exponents for theories of the same class.

1 7.6 THE TRICRITICAL POINT

Within the mean-field approximation, consider

tf(m, H) = !am 2 + ~bm4 + icm6 - mH/kT

(17.59)

For simplicity we take c to be a positive fixed constant; but a and b are variable parameters. We shall see that the system has a phase transition at a = 0, which is either of first or second order, depending on the sign of b. For definiteness, think of a and b as functions of temperature and pressure. Then the condition a = b = determines the critical temperature Tc and critical pressure Pc. Introducing as usual t = (T - TJ/~ and p = (P - PJ/PC' we shall take a and b to

THE LANDAU APPROACH

be linear in t and p

First put H

+ Bp + Dp 0. To find the minima of t/J, we need

a = At

b = Ct

(17.60)

+ bm 3 + ems t/JI/(m) = a + 3bm 2 + 5em 4

t/J'(m) = am

(17.61)

Setting Bt/J/Bm = 0, we obtain five roots, as befits a quintic polynomial:

(17.62)

where

m~ = -2 (-b Vb 2 e

4ae)

(17.63)

Noting that t/J(O) = a, we conclude that m = is a maximum if a < 0. This means that for a < 0, the only possible minima are m . Substituting (17.62) into (17.61), we obtain