# Derivation of the Debye Specific Heat Capacity

1. Sep 26, 2011

### neophysicist

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

Hello everyone!

I'm using the text:
"Elements of Solid State Physics - JP Srivastava (2006)"

I have followed the argument leading up to the derivation of the Debye formula for specific heat capacity, so we now have;

$$C_V = \frac{9N}{\omega_D^3} \frac{\partial}{\partial T} \int_0^{\omega_D}\frac{\hbar \omega^3}{exp(\frac{\hbar \omega}{k_BT})-1}d\omega$$

The next equation presented is the final form which I am having difficulty deriving.

$$C_V = 9Nk_B(\frac{T}{\theta_D})^3 \int_0^{\frac{\theta_D}{T}}\frac{x^4e^x}{(e^x-1)^2}dx$$

2. Relevant equations

We are supposed to use the substitutions;

$$\theta_D = \frac{\hbar\omega_D}{k_B}$$

and

$$x = \frac{\hbar\omega}{k_BT}$$

3. The attempt at a solution

$$d\omega = \frac{k_BTx}{\hbar}dx$$
$$\therefore \ C_V = \frac{9Nk_B}{\theta_D^3} \int_0^{\frac{\theta_D}{T}} \frac{\partial}{\partial T}(\frac{T^4x^3}{e^x-1})dx$$

Now is where I run into difficulty. Applying the quotient rule I get;

$$C_V = 9Nk_B(\frac{T}{\theta_D})^3 \int_0^{\frac{\theta_D}{T}} \frac{x^3(e^x-1)+x^4e^x}{(e^x-1)^2})dx$$

So I can see that I am tantalizingly close but clearly I must be making a dumb mistake somewhere.
I would be grateful if anybody could help me out as this is really bugging me and it's chewed up enough of my revision time already!

2. Sep 26, 2011

### kreil

3. Sep 27, 2011

### neophysicist

Hi Kreil,

Thanks for the response.

Yes, I did try the Leibniz rule earlier but it seemed to get even messier, unless I am applying it incorrectly (in which case please point it out). Here is what I got;

$$\frac {\partial}{\partial T} \int_0^{\frac {\theta_D}{T}} f(x,T)dx = \int_0^{\frac {\theta_D}{T}}\frac {\partial f}{\partial T} dx + f(\frac {\theta_D}{T},T).(-\frac {\theta_D}{T^2}) - f(0,T).0$$

$$=\int_0^{\frac {\theta_D}{T}}\frac {\partial f}{\partial T} dx - \frac {\theta_D^4}{T(exp(\frac {\theta_D}{T})-1)}$$

$$\int_0^{\frac{\theta_D}{T}} \frac{x^3(e^x-1)+x^4e^x}{(e^x-1)^2}) dx - \frac {\theta_D^4}{T(exp(\frac {\theta_D}{T})-1)}$$

where $$f(x,T) = \frac {T^4x^3}{e^x-1}$$

From this point, I don't see how it is possible to bring the 2nd term into the integrand so that it hopefully cancels out the extraneous terms in my first attempt.

Any ideas? anyone?

Last edited: Sep 27, 2011
4. Oct 1, 2011

### G01

Hi Divya,

Welcome to PF. Your efforts to help on the forum are very much appreciated. However, it is against forum rules to post full solutions. Be keep this in mind for the future.

5. Oct 2, 2011

### Jagdish Pd.

Apply Leibniz rule first to the partial derivative of integral ( ∫ -- dω ) and then make substitutions --- changing the variable ω to x . Notice that partial temperature derivative of Debye frequency is zero. This gives the result.

6. Oct 4, 2011

@Divya