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Probability of the polymer chain

  1. Aug 3, 2017 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    I want to check if I think it right!

    3. The attempt at a solution
    N=1: ← or → (2 configurations/ each length is l)
    N=2: ← or → or ←← or ←→ or →← or →→
    (6 configurations/ folded polymer's length is l/2 andunfolded polymer's length is l)
    Thus, the configuration written in the function of x:
    2^{N} \hspace{1cm}\text{ for } x=l \\
    (l/x)! \hspace{1cm}\textrm{ otherwise}

  2. jcsd
  3. Aug 3, 2017 #2
    If N is the number of links of length l in the chain and x is the total length in a given configuration, how many links are in one direction and how many are in the other direction? How many ways are there are selecting the combinations of links that do this? (This is like a coin flipping problem).
  4. Aug 3, 2017 #3
    oh!So the length l doesn't mean the total length?
  5. Aug 3, 2017 #4
    NO, it's the length of each link.
  6. Aug 3, 2017 #5
    Thanks! Reading mistake!
  7. Aug 3, 2017 #6
    Actually, I have another question. How come the second question asks about the total number of configurations? In my opinion, there are infinite configurations depend on unlimited x. Does the question ask about the average configuration number?
  8. Aug 3, 2017 #7
    For a specified x, there are not an infinite number of configurations (at least not in 1D).
  9. Aug 4, 2017 #8
    But, I think the first question asks the total number of specific x. Or, can the polymer be folded? This really make the answers of two questions different.
  10. Aug 4, 2017 #9
    It seems to me the implication is that it is folded along a straight line.
  11. Aug 6, 2017 #10

    Is it what the question means?
  12. Aug 6, 2017 #11
    I don't think so. You need to figure out for a chain of identical joined links that can be folded along the x axis. So, for x = L, and N= 2, there are zero configurations. Leave off the arrows. For x = 2L and N = 2, there is only one configuration. For x = L and N = 1, there is only one configuration.
  13. Aug 6, 2017 #12
    But the question mentions that each segment can be orientated in positive or negative directions. Don't we consider the direction (arrow) in different cases?
  14. Aug 6, 2017 #13
    Also, why can't N=2 chain be folded? Thanks!!
  15. Aug 6, 2017 #14
    No. It's just if there is a fold or not at each jumction. The focus should be in the junctions, not the links.
  16. Aug 6, 2017 #15
    Yes If N=2, you can have x=1 with a fold.
  17. Aug 6, 2017 #16
    So you mean if N=2 there are 3 configurations?

    Attached Files:

  18. Aug 6, 2017 #17
    If N is 2 and x is 2, then there is 1 configuration. If N is 2 and x is 1, I would have to decide whether there are 2 configurations or 1. My inclination would be to count it as 1 configuration.
  19. Aug 12, 2017 #18
    I think this is more viable!

    Attached Files:

  20. Aug 12, 2017 #19
    For the entropy in the system,
    we get
    S\approx -k\int{P(x)lnP(x)}dx=...=(-\frac{k}{2})(1-ln(\frac{2}{\pi N}))\approx -\frac{k}{2} \\
    A=-\int S dT = \frac{1}{2}kT \\
    *A=free enegy; U=internal energy
    So weird that U=0. Am I doing this wrong?? Thank you!!
  21. Aug 12, 2017 #20


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    Suppose r of the N are oriented one way and the remaining N-r the other way. What would the total length be?

    By the way, the expression on the right in b) cannot correct. The argument to exp() must be dimensionless, but x is a length. I would rewrite it by replacing x by x/l everywhere.
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