Mean elongation of spring with temperature

[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

 

[Redbelly98]

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

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

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

 

[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

 

[Redbelly98]

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?

 

[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

 

[Redbelly98]

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?

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​

... - 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

Actually, the integral in the numerator is not zero.

and any hints how to proceed to find the variance?

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