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Mean elongation of spring with temperature

  1. Dec 11, 2008 #1
    Hi all,

    I have to determine the mean elongation of a spring with temperature. The spring constant is alpha and the gravitational acceleration is g. I have the following expression for the force

    [tex]F = -\alpha*x +m*g[/tex]

    which I can integrate for an expression of the energy

    [tex]E = \frac{-\alpha*x^2}{2}+m*g*x[/tex]

    I can factorize this into

    [tex]E = (x-\frac{m*g}{\alpha})^2[/tex]

    than I would like to put it into the Boltzman distribution getting the partition function and from that determine the mean elongation

    [tex]Z = \int_{-\infty}^{\infty}exp(-E*\beta)[/tex]

    but this gives me infinity - any help or advice appreciated thanks in advance,

    Best
    J
     
  2. jcsd
  3. Dec 11, 2008 #2

    Redbelly98

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    I don't think so. If you expand that expression, you get something different than
    [tex]E = \frac{-\alpha*x^2}{2}+m*g*x[/tex]
    which you had before.

    Also, the potential energy of the spring should be +½αx2

    Also, are you defining a positive displacement as upward or downward?

     
  4. Dec 11, 2008 #3
    hi,

    I am defining the displacement from -infinity to +infinity where the Boltzmann weights should be very small for none probable elongations.

    The potential energy is

    E = alpha*x^2/2 + m*g*x

    I need some hints on how to factorize the energy. Thanks in advance
     
  5. Dec 11, 2008 #4

    Redbelly98

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    I think I see where you need to go with this.

    Are you familiar with "completing the square" for quadratic expressions? That would be a good way to manipulate your energy expression, and get the integral in terms of something you can evaluate.

    BTW, is your integral expression missing a "dx" term?
     
  6. Dec 12, 2008 #5
    Hi again,

    I write the potential energy as (x-m*g/alpha)^2 and then substract the constant which is just a constant so it is not so important - I guess? - then I write it as a Boltzmann

    [tex]mean(x) = \frac{\int_{-\infty}^{\infty}x*exp(-\beta*(x-m*g/alpha)^2}{ \int_{-\infty}^{\infty}exp(-\beta*(x-m*g/alpha)^2)} [/tex]

    which due to the symmetry gives zeros - does this seem reasonable and any hints how to proceed to find the variance?

    Any hints appreciated thanks in advance

    Best
    J
     
  7. Dec 12, 2008 #6

    Redbelly98

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    Ah, you're right, the constant is not important and would cancel out later anyway.
    That being said, how would you feel about

    (alpha/2) (x + m g / alpha)2

    for the potential energy? If you're in doubt, try expanding that expression, also expand what you had, and compare them with

    alpha x2/2 + mgx​

    Actually, the integral in the numerator is not zero.

    I'd look at the definition of variance, and evaluate the integrals involved.
     
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