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Statistical Mechanics - Physics at Oregon State University

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9.7. RENORMALIZATION GROUP THEORY. 221<br />

which gives us ˜ J ′ . Finally, we divide the first and third equ<strong>at</strong>ion to get<br />

e 2˜ h cosh(2 ˜ J + ˜ h)<br />

cosh(−2 ˜ J + ˜ h) = e2˜ h ′<br />

which gives us ˜ h ′ . The solutions are<br />

˜J ′ ( ˜ J, ˜ h) = 1<br />

4 log<br />

<br />

cosh(2 ˜ J + ˜ h) cosh(−2 ˜ J + ˜ h)<br />

cosh 2 ( ˜ <br />

h)<br />

˜h ′ ( ˜ J, ˜ h) = ˜ h + 1<br />

2 log<br />

<br />

cosh(2 ˜ J + ˜ h)<br />

cosh(−2 ˜ J + ˜ <br />

h)<br />

g( ˜ J, ˜ h) = 1<br />

16 log<br />

<br />

16 cosh(2 ˜ J + ˜ h) cosh(−2 ˜ J + ˜ h) cosh 2 ( ˜ <br />

h)<br />

(9.159)<br />

(9.160)<br />

(9.161)<br />

(9.162)<br />

We are now able to rel<strong>at</strong>e the original partition function to a new partition<br />

function describing half the particles on twice the length scale. Because of our<br />

redefining the variables, the partition function depends on the variables N, ˜ J,<br />

and ˜ h. This is equivalent to the original set N, T, h. We have redefined the<br />

temper<strong>at</strong>ure scale by using βJ and redefined the field by using βh. Hence the<br />

partition function is a function Z(N, ˜ J, ˜ h). We have expressed this in the form of<br />

a partition function with new coupling constants and fields <strong>at</strong> half the number<br />

of particles, We needed to add an energy shift. This transl<strong>at</strong>es to<br />

Z(N, ˜ J, ˜ h) = e Ng( ˜ J, ˜ h) Z( 1<br />

2 N, ˜ J ′ , ˜ h ′ ) (9.163)<br />

This equ<strong>at</strong>ion rel<strong>at</strong>es the properties of the original system to the properties of<br />

a system with double the length scale, and renormalized interaction strengths.<br />

Because the partition functions are directly rel<strong>at</strong>ed, thermodynamic properties<br />

are similar! Our goal is now to repe<strong>at</strong> this process over and over again. We<br />

have<br />

Z( 1<br />

2 N, ˜ J ′ , ˜ h ′ ) = e 1<br />

2 Ng( ˜ J ′ , ˜ h ′ ) Z( 1<br />

4 N, ˜ J”, ˜ h”) (9.164)<br />

Note th<strong>at</strong> the functional forms g(x, y), ˜ J”(x, y), and ˜ h”(x, y) are the same as<br />

before!<br />

When we apply this procedure repe<strong>at</strong>edly we therefore find<br />

Z(N, ˜ J, ˜ h) = e Ng( ˜ J, ˜ h) e 1<br />

2 Ng( ˜ J ′ , ˜ h ′ ) · · · Z(0, ˜ J ∞ , ˜ h ∞ ) (9.165)<br />

The free energy follows from log(Z) = −βG. Therefore we have<br />

−βG = N<br />

<br />

g( ˜ J, ˜ h) + 1<br />

2 g( ˜ J ′ , ˜ h ′ ) +<br />

2 1<br />

g(<br />

2<br />

˜ J”, ˜ <br />

h”) + · · ·<br />

(9.166)

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