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Homework Help: Statistical mechanics: Particle on a spring

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data

    A classical particle with mass m is in thermal equilibrium with a fluid at temperature T. The particle is stuck to a harmonic ('Hookean') spring and can only move on a horizontal line ([itex]-\infty < x < \infty[/itex]). The position of the particle is x = 0 if the spring is in its equilibrium position, but thermal movement can cause it to stretch or compress. The probability distribution of finding the particle in position x is proportional to [itex]e^{-\frac{x^2}{2\sigma^2}}[/itex].

    a) Calculate the spring constant C as a function of m, [itex]\sigma[/itex], [itex]k_b[/itex] and/or T.

    b) Calculate the average quadratic displacement of the particle.

    3. The attempt at a solution

    The probability of finding the particle in position x is proportional to [itex]e^{-\frac{x^2}{2\sigma^2}}[/itex]. The energy of the spring is [itex]\frac{1}{2}C x^2[/itex], so the chance of finding the particle in position x is also proportional to [itex]e^{-\frac{1}{2}C x^2 \beta}[/itex]. This gives [itex]\frac{1}{2}C x^2 \beta = \frac{x^2}{2\sigma^2}[/itex] or [itex]C = \frac{k_B T}{\sigma^2}[/itex]

    Is this correct?

    b) Do they mean finding [itex]\langle x^2 \rangle[/itex]?
    Last edited: Mar 3, 2012
  2. jcsd
  3. Mar 4, 2012 #2
    This is a weird question, but I think based on what the question says, you're giving the right answer. I actually think the question must be screwed up.

    It's weird because:
    I've never heard the term "average quadratic displacement" but <x2> sounds right.

    The question is thermodynamically nonsense because:
    Here σ is supposedly a constant, and the spring constant depends on temperature. (Problems I've seen have a constant spring constant, and the probability distribution has thermal dependence.) Since σ is nonzero, even at zero temperature, there are fluctuations in the particle's position. If the fluctuations are thermal, this is a contradiction, since at zero temperature there is no thermal motion. This also implies that at zero temperature, the spring constant is zero, which is very odd.
  4. Mar 4, 2012 #3
    Thanks. The problem might lie in my translation of the question, but I made sure to do it as faithfully as possible and looking at the original question it is pretty much a word for word translation. I will have to look at it some more and come back to it.

    (If anyone speaks Dutch it's questions 4a and 4b http://www.a-eskwadraat.nl/tentamens/NS-201b/NS-201b.2009-03-16.tent.pdf [Broken].)
    Last edited by a moderator: May 5, 2017
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