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Fig. 17.5 Experimental phase diagram of the liquid He 4-He 3
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Fig. 17.6 The graph on the left is redrawn from the data of
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Fig. 17.5 to facilitate comparison with the prediction of mean-field theory, as shown on the right.
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The Gaussian model is obtained by keeping only terms up to order m 2 (x) in the Landau free energy. The model makes sense only if ro > 0, or t > 0; but it has the virtue of being exactly soluble, because the partition function is a Gaussian functional integral:
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e- G/ kT
% f(Dm)
e-E[m, H]
(17.78) 2(x) - m(x)h(x)] E[m, H] = f(dx)[ -tm(x)'\72m(x) + trom
where h(x) = H(x)/kT. We can read off the answer from (17.14), with K = - '\7 2 + ro: Q=%(detK)-1/2 exp Hh,K- 1 h) (17.79) The eigenvectors of K are exp(ik x), with corresponding eigenvalue k 2 + roo The determinant of K is just the product of the eigenvalues: detK= n(k 2 +ro ) (17.80)
To calculate (h, K-1h), we expand h(x) in a Fourier series as in (17.8), obtaining
(h, K-1h) == f(dx)h(X)(
+ ror1h(x)
h(-k)h(k) k2 + ro
Substituting these results into (17.79), we obtain, up to an additive constant, log Q = - 1
L log ( k 2 + ro) + -2 L 2 k V k
+ ro
(17 .82)
Let us first set h = O. In the limit V ~ 00 the Gibbs free energy is, up to an additive constant, given by G kT 2 V = - 2 f(dk)log(k + aot) (17.83) The singular part of the specific heat is obtained by taking the second derivative with respect to t, replacing T by a constant Tc : C k d- 1 d-l dk _ t(d-4)/21A'(I) ds _s _-=_ V 0 (k 2 + a ot)2 0 (S2 + 1)2
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Note that this integral has two possible types of divergences. It may diverge at the upper limit when A ~ 00. This is called an "ultraviolet divergence." The integral may also diverge at the lower limit when t ~ O. This would be an "infrared divergence." The critical exponents are determined by the nature of the
infrared divergence. As we can see from (17.84), a part of the infrared divergence is already isolated in the factor t(d-4)/2. Whether this is the entire contribution depends on how the integral behaves at the upper limit, the ultraviolet domain. By simple power counting, we see that the integral converges for d < 4, and diverges for d ~ 4. These cases lead to different infrared behavior, and have to be discussed separately:
( a) d < 4. The integral converges when A
~ 00 :
C _ V
(b) d
= 4. The integral diverges logarithmically when A ~ 00:
- - const. + log t - log t
( c) d > 4. The integral diverges like a power when A ~ 00: - - const. V These are all consistent with the assignment
Dt = 2
where the last value is obtained from (16.59). The value of a agrees with the mean-field value a = for d ~ 4, but disagrees violently for d < 4. In Fig. 17.7 we contrast the behavior of the specific heat in mean-field theory with that in the Gaussian model for d < 4. The disagreement indicates that in less than four dimensions we cannot ignore the spatial fluctuations of the order parameter. The free energy (17.83) is ultraviolet-divergent. But the divergence does not affect the nature of the infrared singularities, thanks to the triviality of the model. However, their very presence might make one feel uneasy about the consistency of the model. In the, next chapter we shall learn how to rid the free energy of ultraviolet divergences (see Section 18.6). We now calculate the correlation function. From the definition (16.18), we obtain its Fourier transform as
f(k) = (m(k)m(O) - (m(0)\2'1T)d S(k)
Now m(O) is the sum of all Fourier components. When h = 0, we have
(m(k) = 0, (m(k)ii1(p) = SK(k + p)
because the Landau free energy depends only on the modulus of m(k). Thus
model (d<4)
Fig. 17.7 Comparison of the heat capacity in mean-field theory and in the Gaussian model.
when h = 0 we have
f(k) =
~(m(-k)m(k) v
3 log Q
This can be calculated by temporarily restoring the field h:
r(k) = 3i1(k) 3i1( -k)
Using this with (17.82), we obtain _ r( k) = const.
----=-2-k TO
r(x) -
d X 2-
This is the Ornstein-Zernike form we obtained earlier in mean-field theory. Therefore v = !, "1/ = 0, as in mean-field theory. Using "1/ = 0 in (16.59), we find d+2 D =-(17.90) h 2 All other critical exponents can values of Dt and Dh . The results 4-d a = -2-' f3 =
v=!, "1/=0
now be found from (16.59), using the known are summarized below: d-2 d+2 1)= - y = 1, -4-' d - 2' (17.91)