# Homework Help: Statistical mechanics: Particle on a spring

1. Mar 3, 2012

### SoggyBottoms

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 ($-\infty < x < \infty$). 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 $e^{-\frac{x^2}{2\sigma^2}}$.

a) Calculate the spring constant C as a function of m, $\sigma$, $k_b$ 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 $e^{-\frac{x^2}{2\sigma^2}}$. The energy of the spring is $\frac{1}{2}C x^2$, so the chance of finding the particle in position x is also proportional to $e^{-\frac{1}{2}C x^2 \beta}$. This gives $\frac{1}{2}C x^2 \beta = \frac{x^2}{2\sigma^2}$ or $C = \frac{k_B T}{\sigma^2}$

Is this correct?

b) Do they mean finding $\langle x^2 \rangle$?

Last edited: Mar 3, 2012
2. Mar 4, 2012

### Jolb

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.

3. Mar 4, 2012

### SoggyBottoms

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