Statistical mechanics: Particle on a spring

AI Thread Summary
A classical particle in thermal equilibrium with a fluid is attached to a harmonic spring, allowing it to move along a horizontal line. The probability distribution for the particle's position is given by e^{-\frac{x^2}{2\sigma^2}}, leading to the conclusion that the spring constant C can be expressed as C = \frac{k_B T}{\sigma^2}. There is confusion regarding the term "average quadratic displacement," which is interpreted as the expectation value ⟨x²⟩. Concerns are raised about the thermodynamic implications of the problem, particularly regarding the constant nature of σ and the behavior of the spring constant at zero temperature. The translation of the original question is confirmed to be accurate, but further clarification may be needed.
SoggyBottoms
Messages
53
Reaction score
0

Homework Statement



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.

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:
Physics news on Phys.org
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.
 
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 .)
 
Last edited by a moderator:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top