:2)(V-~)=~T in Java

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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" )
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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
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~ = ~{~TIog(V _ ~) 2
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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
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Fig. 17.2 Landau free energy that leads to the Van der Waals equation of state.
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volumes VI and V2, satisfying the conditions
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(a1/J/aV)V~Vl = (a1/J/aV)V~V2 = 0
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1/J(V1 ) = 1/J(V2) The first of these requires P(VI ) = P(V2), which leads to
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V/ -
With the help of (17.50), the second condition can be rewritten
1/J(V1 ) -1/J(V2)
V2) - jV1PdV = 0
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
Approaching the critical point from T> 1 along V = 1, we find BP/ Hence the susceptibility is
av =
- 6t.
1 ( -av)
which gives y = 1.
Near the critical point, put
T = 1
+ t,
P = 1 + p,
Then the equation of state can be rewritten as
v 3 + p(1 +
V)3 -
H p + 8t)(1
+ V)2 =
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
we obtain (17.58)
Substituting these into (17.52) and keeping only terms to order x
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.
Within the mean-field approximation, consider
tf(m, H) = !am 2 + ~bm4 + icm6 - mH/kT
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
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
+ bm 3 + ems t/JI/(m) = a + 3bm 2 + 5em 4
t/J'(m) = am
Setting Bt/J/Bm = 0, we obtain five roots, as befits a quintic polynomial:
m~ = -2 (-b Vb 2 e
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