Canonical ensemble of a simplified DNA representation

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Homework Help Overview

The discussion revolves around the canonical partition function in the context of a simplified DNA representation, specifically focusing on energy states of strands that can either be open or closed at certain particles. The problem involves understanding the implications of these states on the partition function and energy calculations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the formulation of the canonical partition function based on energy states defined by the attachment of particles in strands. There is an attempt to derive the partition function from energy considerations, leading to questions about the correctness of the approach and the underlying equations.

Discussion Status

The discussion is ongoing, with participants questioning the validity of the original poster's approach and seeking clarification on the foundational equations for the partition function. Some participants express confusion about specific transitions in the mathematical manipulations presented.

Contextual Notes

There appears to be a lack of consensus on the correct formulation of the partition function, and participants are actively engaging in clarifying the relationships between variables involved, particularly the connection between the index α and the parameter m.

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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