Approximation of specific heat, Debye model

1. May 28, 2013

Lindsayyyy

Hi everyone

I have trouble with this task

1. The problem statement, all variables and given/known data
the specific heat cv is given by
$$c_v =\frac {N_A k_b \hbar^2}{{\Omega_D}^3 {k_b}^2 T^2} \int \limits_{0}^{\Omega_D} \! \frac {\Omega^4 exp\frac{\hbar \Omega}{k_b T}}{{(exp\frac{\hbar \Omega}{k_b T}-1})^2} \, d\Omega$$

I shall show that the approximation for high temperatures leads to:

$$c_v=3 N_A k_b [1-0.05 (\frac{\Theta_D}{T})^2]$$

2. Relevant equations

$$k_b \Theta_D= \hbar \Omega_D$$

3. The attempt at a solution

I tried to approximate my exponential function with the taylor seris, so I get for the exponential function something like:

$$exp(x) \approx 1+x$$ whereas x is equal to (h*omega)/(kbT)

I will only have x^2 in my denominator and x^2 in my numerator (because I neglect the h*omega/(kbT) in the numerator because T is very high).

But this only leads me to the Dulong Petit law. I have no idea where the minus aswell as the 0.05 should come from.

I also tried to approach the whole fraction via taylor but that didn't work out. Furthermore I tried to approximate via taylor until x^2, didn't work out either.

Can anyone help me out with this?

2. May 28, 2013

Staff: Mentor

You need to carry more terms in the expansions. You are trying to evaluate

$$\frac{x^4e^x}{(e^x-1)^2}$$

You need to keep the following terms:
$$e^x=1+x+\frac{x^2}{2}+...$$
$$(e^x-1)=x+\frac{x^2}{2}+\frac{x^3}{6}+...$$
$$(e^x-1)^2=x^2(1+x+\frac{7x^2}{12}+...)$$
$$\frac{x^4}{(e^x-1)^2}=x^2(1-x+\frac{5x^2}{12}+...)$$
Chet

3. May 29, 2013

Lindsayyyy

Ok, thank you very much.

I will try that later

4. May 29, 2013

Lindsayyyy

Well I tried it now, but I have a question.

We act on the asumption x^4/(...)^2 which means we already neglect the exponential function in the numerator. How can it be that I'm allowed to approximate my exponential function in the numerator then get to the expression x^4/(...)^2 and then I approximate the whole term. I don't see any logic behind that. Am I not supposed to approximate both exponential functions if I do it with one?

5. May 29, 2013

Staff: Mentor

I'm not sure I understand your question. The real point I was trying to emphasize is that you need to get the first three terms in the series exact.

I didn't intend to indicate that you only need one term for the exponential in numerator. I showed the three term series for the exponential in the numerator in my first equation. But I didn't show its product with x^4/(...)^2. I wanted to leave that for you to do. When you carry out this multiplication of the two series, you will find that one of the terms drops out. Once you do the integration, you will see where the 0.05 comes from.

Chet

6. May 29, 2013