Derivation of the Debye Specific Heat Capacity

[neophysicist]

Homework Statement

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;

<br /> 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<br />

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

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

Homework Equations

We are supposed to use the substitutions;

<br /> \theta_D = \frac{\hbar\omega_D}{k_B}<br />

and

<br /> x = \frac{\hbar\omega}{k_BT}<br />

The Attempt at a Solution

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

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

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

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!

 

[kreil]

When taking a partial derivative of a definite integral, you need to use the Leibniz Integral Rule:

http://mathworld.wolfram.com/LeibnizIntegralRule.html

 

[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;

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

<br /> =\int_0^{\frac {\theta_D}{T}}\frac {\partial f}{\partial T} dx -<br /> \frac {\theta_D^4}{T(exp(\frac {\theta_D}{T})-1)}<br />

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

where <br /> f(x,T) = \frac {T^4x^3}{e^x-1}<br />

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?

 

[G01]

Hi,

My dad was perusing the forum and worked out the attached as a solution.

Hope it helps.

Divya

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.

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

 

[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.

 

[neophysicist]

@Divya

Please give my thanks to your father.

@Jagdish. Yes, I saw that from Divya's father's post. The critical part was applying the Leibniz Integral Rule FIRST. Then you can apply the substitution. This was what I was doing wrong. The results must be equivalent but only this method produces the result in a neat form. When done this way, the answer falls out nicely and with much less fuss as well.

It's amazing that varying the order of the steps (even though the procedure is technically correct in both instances) can make the difference between an analytically tractable and intractable solution.

Thanks once again.