Canonical ensemble of a simplified DNA representation

zexxa
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Question
Form the canoncial partition using the following conditions:

  • 2 N-particles long strands can join each other at the i-th particle to form a double helix chain.
  • Otherwise, the i-th particle of each strand can also be left unattached, leaving the chain "open"
  • An "open" link gives the strand ##\epsilon## amount of energy where ##\epsilon > 0##
  • A "closed" link gives the strand no energy
  • For ##m < N##, the strand must be "open" for ##i \leq m## and "closed" for ##m < i \leq N##
  • Note that ##m \neq N##
  • Each particle are independent of each other and they weakly interact

Attempt

I treated the energy states at ##i /leq m## and ## i > m## as a simple one energy state i.e.
## E = \sum_i ^N n_i \epsilon , {n_i} = \begin{cases} 1, & \text{if $i \leq m$}.\\
0, & \text{otherwise} \end{cases}##
Therefore,
##Z(\beta , V , N) = \{ exp[ - \beta \epsilon ] \} ^m \{ exp[ 0 ]\} ^{N-m} = exp[ - \beta \epsilon ] ^m##

Does this make sense?
 
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zexxa said:
Does this make sense?
No. What is the base equation for the partition function?
 
The base equation?
## Z ( \beta, V, N) = \sum_\alpha exp[ -\beta E ] = \sum_\alpha exp [ -\beta \sum_i^N n_i \epsilon ]
= \sum_\alpha exp [ -\beta m \epsilon]
= \{exp[ -\beta \epsilon ]\}^m
= exp [-\beta \epsilon] ^m##

The base equation is the first 2 equations, and the rest are the manipulations I did to get to my answer from before.
 
zexxa said:
## Z ( \beta, V, N) = \sum_\alpha exp[ -\beta E ] = \sum_\alpha exp [ -\beta \sum_i^N n_i \epsilon ]
= \sum_\alpha exp [ -\beta m \epsilon]
= \{exp[ -\beta \epsilon ]\}^m
= exp [-\beta \epsilon] ^m##
I don't understand how you got from the 4th to the 5th terms in there. Also, shouldn't there be a link between ##\alpha## and ##m##?
 
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