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    Average length of polymer as a function of tension

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    The partition function is

    Z = Integral of exp(-beta U) sin(theta_1)...sin(theta_N) dtheta_1 dphi_1...dtheta_N dphi_N (1)

    The integrals over theta are over the interval from zero to pi. The phi integrals are from zero to 2 pi. Note that this covers all directions the segments of the polymer can point in. The potential energy function is:

    U = -Sum of F L cos(theta_j) from j = 1 to N (2)

    I'm using capital L for the segment length instead of lower case "l" because of readability (lowercase "l" looks a bit like the number 1, and the letter i ).

    Let's examine the term Exp(-beta U). Clearly if you substitute the summaton (2) in the exponential, you get a summation in the exponential which can be written as a product of the exponential of the individual terms in the summation:

    Exp(-beta U) = Product of exp(beta F L cos(theta_j) (3)

    If we now substitute (3) in (1) what you get, as far as the theta integrals are concerned, is something of the form:

    Integral of f(x_1)f(x_2)f(x_3)f(x_4)...f(x_M)dx_1 dx_2 dx_3...dx_N

    You can always rewrite this as:

    Integral of f(x_1)dx_1 Integral of f(x_2)dx_2 Integral of f(x_3)dx_3 ...Integral of f(x_N)dx_N

    All the integrals are thus the same; it doesn't matter how ...

    Solution Summary

    The average length of polymer as a function of tension is determined.

    $2.19

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