13.07.2015 Views

Statistical models of elasticity in main chain and smectic liquid ...

Statistical models of elasticity in main chain and smectic liquid ...

Statistical models of elasticity in main chain and smectic liquid ...

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

2.2. MODEL OF HAIRPIN CHAINS 17The number <strong>of</strong> hairp<strong>in</strong>s on the cha<strong>in</strong> is governed by the temperature <strong>of</strong> thesystem <strong>and</strong> the energy <strong>of</strong> a hairp<strong>in</strong> defect. The partition function can becalculated by summ<strong>in</strong>g all the configurations multiplied by their associatedBoltzmann factors. This is most naturally done by splitt<strong>in</strong>g up the sums foreven <strong>and</strong> odd numbers <strong>of</strong> defects.Z hp = Z even +Z odd (2.26)Z even = ∑ 2N n−1 (n(n−2)!! 2 1− ( )n−2)z 2 2Ne −nβu h(2.27)even nZ odd = ∑odd n2N n−1 ((n−1)!! 2 1− ( zN) 2)n−12e −nβu h. (2.28)Fortunately these series can be summed up to <strong>in</strong>f<strong>in</strong>ity result<strong>in</strong>g <strong>in</strong> modifiedBessel functions. The power series expansion <strong>of</strong> modified Bessel function isgiven by∞∑I ν (z) = e −π 2 νi (z/2) ν+2kJ ν (iz) =(2.29)(k!)(ν +k)!k=0In summ<strong>in</strong>g the odd series, the follow<strong>in</strong>g result is useful(n−1)!! = 2.4....(n−1) (2.30)( )= 2 n−1 n−12 ! (2.31)2Substitut<strong>in</strong>g n = 2p+1 reduces the odd part <strong>of</strong> the partition function to(∞∑ 1− ( ) )z 2 pN2pNZ odd = 2(p!) 2 2 2p e −(2p+1)βu h(2.32)p=0= 2e −βu hI 0(e −βu hNSimilarly for the even series, the result(n−2)!! = 2 n−22√1− ( ))z 2N. (2.33)( n−22)! (2.34)enables the series to be summed. Substitut<strong>in</strong>g n = 2p gives( ( ∞∑ N 2p−1 1− z) 2) p−1NZ even = 2e −2pβu h. (2.35)2(p−1)!p! 4p=1Substitut<strong>in</strong>g p ′ = p−1 enables this expression to be rewritten as( ( ∞∑ N 2p′ +1 1− z) 2) p ′NZ even = 22(p ′ +1)!p ′ e −2(p′ +1)βu h(2.36)! 4p ′ =0

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!