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Self-Consistent Field Theory and Its Applications by M. W. Matsen

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1.5 Mathematical Techniques <strong>and</strong> Approximations 23<br />

1.5.4 R<strong>and</strong>om-Phase Approximation<br />

Here, a L<strong>and</strong>au-Ginzburg free energy expression is derived for the entropic free energy, f e ,ofa<br />

polymer molecule in a weakly-varying field, w(r). As before in Section 1.5.2, this calculation<br />

will assume that the two ends of the chain are free, <strong>by</strong> imposing the initial condition in Eq.<br />

(1.87). We will also use the invarience discussed at the end of Section 1.2 to fix the spatial<br />

average of the field to ¯w =0. Given that w(r) only deviates slightly from zero, it is safe to<br />

assume that the ensemble-averaged segment concentration, φ α (r), never strays far from its<br />

spatial average,<br />

¯φ α ≡ 1 V<br />

∫<br />

dr φ α (r) =<br />

N<br />

ρ 0 V<br />

(1.90)<br />

The first step is to find an approximate solution for q(k ≠0,s). Since the integral in Eq.<br />

(1.85) involves w(k 1 ), which is small, the multiplicative factor of q(k − k 1 ,s) need not be<br />

particularly accurate. Approximating it <strong>by</strong> q(k − k 1 , 0) from Eq. (1.87) gives<br />

∂<br />

q(k,s)<br />

∂s<br />

≈<br />

∫<br />

1<br />

−xq(k,s) −<br />

(2π) 3 dk 1 w(k 1 )q(k − k 1 , 0) (1.91)<br />

= −xq(k,s) − w(k) (1.92)<br />

where x ≡ k 2 a 2 N/6. Given that q(k, 0) = 0 for k ≠0, the solution becomes<br />

where<br />

q(k,s) ≈−w(k)h(x, s) (1.93)<br />

h(x, s) ≡ 1 − exp(−sx)<br />

x<br />

(1.94)<br />

The second step is to solve for q(k =0,s), in which case Eq. (1.85) reduces to<br />

∂<br />

∂s q(k,s)=− 1 ∫<br />

(2π) 3 dk 1 w(k 1 )q(−k 1 ,s) (1.95)<br />

Since we have set ¯w =0, it follows that w(k 1 )=0for k 1 =0, which in turn implies that the<br />

approximation from Eq. (1.93) can be inserted inside the integral to give<br />

∂<br />

∂s q(k,s) ≈ 1 ∫<br />

(2π) 3 dk 1 w(k 1 )w(−k 1 )h(x 1 ,s) (1.96)<br />

The solution of this is<br />

where<br />

q(k,s) ≈ q(k, 0)<br />

g(x, s) ≡<br />

[<br />

∫<br />

1<br />

1+<br />

2(2π) 3 V<br />

]<br />

dk 1 w(−k 1 )w(k 1 )g(x 1 ,s)<br />

(1.97)<br />

2[exp(−sx)+sx − 1]<br />

x 2 (1.98)

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