There was a question in Statistical Mechanics 3rd ed by RK Pathria and PD Beale, section 3.8, asking to show that the harmonic oscillator obeys the equipartition theorem. It was well proven. But at the end of the proof there was an equation out of the blue with no introduction called "the density of states". I put the equation below. $$ g(E)= \frac{1}{(\hbar\omega)^N} \frac{1}{2\pi i} \int_{\beta'-i\infty}^{\beta'+i\infty} \frac{e^{\beta E}}{\beta^N} \, d\beta \qquad (\beta'>0). $$ Would you please explain what are those? What does it mean by the variables?

  • 1
    $\begingroup$ Hi, can you provide the according references, please. $\endgroup$ – mikuszefski Apr 9 at 6:59
  • $\begingroup$ books.google.com.bd/… $\endgroup$ – F.sharmin Apr 9 at 7:22
  • $\begingroup$ here is the link sir $\endgroup$ – F.sharmin Apr 9 at 7:25
  • $\begingroup$ I have fixed up your equation, added a proper reference to the book, and removed the image. If you are not happy with this you can revert to the previous version (using the "edit" link). Please refer to MathJax to learn how to do this yourself in future. $\endgroup$ – user197851 Apr 9 at 9:12

That equation doesn't come "out of the blue". It is explained in Pathria and Beale section 3.4, eqn (7), as the inverse Laplace transform of the partition function $Q(\beta)$. So, we write $$ Q(\beta) = \int_0^\infty dE \, e^{-\beta E} g(E) $$ where $g(E)$ is the density of states (number of states per unit energy). This can be regarded as the Laplace transform of the function $g(E)$. Therefore, formally, the equation can be inverted, to write $g(E)$ as an integral involving $Q(\beta)$: $$ g(E)=\frac{1}{2\pi i} \int_{\beta'-i\infty}^{\beta'+i\infty} d\beta \,e^{\beta E} Q(\beta). $$ That is your equation, for the particular case of harmonic oscillators, where $Q(\beta)$ is known.

A proper understanding of the integral, though, requires some background in complex analysis. The variable $\beta$ is allowed to take complex values, with real and imaginary parts. It is an integral in the complex plane, along a line slightly to the right of the imaginary axis.

  • $\begingroup$ Can not manage to thank you properly sir! Thank you! $\endgroup$ – F.sharmin Apr 9 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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