# Partition function of classical oscillator with small anharmonic factor

1. Aug 30, 2010

### castlemaster

1. The problem statement, all variables and given/known data

Having a unidemsional array of N oscillators with same frequency w and with an anharmonic factor $$ax^4$$ where 0 < a << 1

Calculate, up to the first order of a, the partition function.

2. Relevant equations

For one oscillator

$$Z=\frac{1}{h}\int{e^{-\beta(p^2/2m+1/2mw^2x^2+ax^4)}dpdx}$$

3. The attempt at a solution

Tried to

$$Z=\frac{K'}{h}\int{e^{\frac{-\beta*mw^2x^2}{2}(1+\frac{2a}{mw^2}x^2))}dpdx}$$

and I guess I can approximate $$(1+bx^2)$$ to something ... but I know more or less the solution and I can't figure out how to reach it.

2. Aug 30, 2010

### qbert

$$Z=\frac{1}{h}\int{e^{-\beta(p^2/2m+1/2mw^2x^2+ax^4)}dpdx}$$
$$=\frac{1}{h}\int{e^{-\beta(p^2/2m+1/2mw^2x^2)}e^{-\beta ax^4}dpdx}$$
$$=\frac{1}{h}\int{e^{-\beta(p^2/2m+1/2mw^2x^2)} (1 -\beta ax^4 + \cdots) }dpdx}$$

you should probably know how to calculate the last.
but here are some useful integrals

$$\int e^{-ax^2} = \sqrt{\pi/a}$$
$$\int x^2 e^{-ax^2} = \frac{1}{2} \sqrt{\frac{\pi}{a^3}}$$
$$\int x^4 e^{-ax^2} = \frac{3}{4} \sqrt{\frac{\pi}{a^5}}$$

3. Aug 30, 2010

### Dickfore

good luck.

4. Aug 30, 2010

### castlemaster

That's what I thought but part b of the problem say:

show that $$C_{v}=Nk(1-\frac{6*\alfa*k}{m^2w^4}T)$$

but
$$E=-\frac{d lnZ}{d\beta}$$
and
$$C_{v}=\frac{d E}{dT}$$

if I do all the integrals I get something like

$$ln Z=Nln(K_{1}\beta^{n})$$

and for the properties of the ln

$$ln Z=Nln(K_{1}) +Nnln(\beta)$$

and making the derivate over B will never give the Cv mentioned.

What I'm doing wrong?

5. Sep 2, 2010

### castlemaster

if I do

$$u=\frac{\beta mw^2x^2}{2}$$

then I get something like

$$\frac{1}{\beta mw^2}\int{e^{-u}e^{-\frac{4\alpha u^2}{\beta m^2w^4}}}$$

I think it approches what I need to end with, at least the variables are similar,

someone has a clue? maybe using the fact that the derivate of exp(ax^n) is nax^(n-1)exp(ax^n)

I'll appreciate any clue

thanks

6. Sep 2, 2010

### qbert

i already told you how to do this.

step 1. do the math.
you end up with $z = c_1 \beta^{-1} - a c_2 \beta^{-2}$
for constants c1, and c2.

step 2. still do the math.
find $E = - \frac{d \log z}{d\beta}$, this will still have an a in
the denominator, drop it (using small a approximation)

step 3. do even more math.
find $C_v = \frac{dE}{dT}$

step 4. ?

step 5. profit.

7. Sep 4, 2010

### castlemaster

You are right qbert,
I was just doing a stupid mistake everytime.

ln(a*b) = lna+lnb RIGHT
ln(a+b) = lna+lnb STUPID
ln(1+ax) aprox.= ax for a small

thanks for the patience