I am trying to find the total energy of a mono-atomic 2D Debye solid. I started with the density of states: $$D(\omega)=\frac{A\omega}{\pi c^2} $$ where A is the area, $\omega$ the frequency and c the speed of sound.

Solving the following for the frequency of Debye: $$\int_0^{\omega_D} D(\omega)d\omega=2N$$ I obtained $\omega_D=2c\sqrt{\frac{N\pi}{A}}$ Now to obtain the total energy, I must integrate the following expression:

$$E= \int_0^{\omega_D}\frac{A\omega}{\pi c^2} \frac{\hbar\omega}{e^{\frac{\hbar\omega}{k_B T}} -1} d\omega$$

I use the change of variables $x=\frac{\hbar\omega}{k_B T}$ and $dx=\frac{\hbar d\omega}{k_B T}$

I finally arrive at $$E=\frac{A(k_BT)^2}{\pi c^2\hbar^2} \int_0^{\frac{\hbar \omega_D}{k_BT}} \frac{x^2}{e^x -1} dx$$

And I don't know hot to solve this. If the upper bound were infinity, it would be a know integral. How could I proceed?

  • 1
    $\begingroup$ You can't use the high-temperature approximation? $\endgroup$ Commented May 25, 2019 at 1:59
  • $\begingroup$ Yes, but somehow I am still having trouble solving it. Do you know if what I posted is correct so far? $\endgroup$ Commented May 25, 2019 at 2:01
  • $\begingroup$ Well, I'm uncertain about the derivation in 2D, but $e^x - 1 \approx x$ in the high temp. limit. The actual value of that integral is called the Debye function en.wikipedia.org/wiki/Debye_function $\endgroup$ Commented May 25, 2019 at 2:13
  • $\begingroup$ Thank you! I will try that approximation and see what I get $\endgroup$ Commented May 25, 2019 at 2:24

1 Answer 1


I think this might be a good idea to expand the $1/e^x-1$ as GP SERIES and then you can do individual $x^2e^{nx}$. I think it should work. Just like we do for packs formula for radiation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.