Mean elongation of spring with temperature

1. Dec 11, 2008

jorgen

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

$$F = -\alpha*x +m*g$$

which I can integrate for an expression of the energy

$$E = \frac{-\alpha*x^2}{2}+m*g*x$$

I can factorize this into

$$E = (x-\frac{m*g}{\alpha})^2$$

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

$$Z = \int_{-\infty}^{\infty}exp(-E*\beta)$$

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

Best
J

2. Dec 11, 2008

Redbelly98

Staff Emeritus
I don't think so. If you expand that expression, you get something different than
$$E = \frac{-\alpha*x^2}{2}+m*g*x$$

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

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

3. Dec 11, 2008

jorgen

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

4. Dec 11, 2008

Redbelly98

Staff Emeritus
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?

5. Dec 12, 2008

jorgen

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

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

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

6. Dec 12, 2008

Redbelly98

Staff Emeritus
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.