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\centerline{\textbf{The Diatomic Ideal Gas}}
\bigskip

  Yesterday we worked out the internal energy per molecule of a
  diatomic gas associated with translational kinetic energy,
  rotational kinetic energy, and vibrational energy (which has both a
  kinetic and potential component).  For each case (except
  translation), you considered both the low- and high- temperature
  limits.

  In each case you had sums that looked like
  $$
    \sum_{i} \; \text{something} \; e^{-\beta E_i}
  $$
  and $\beta (E_1 - E_0)$ was either large or small.

  For the high-temperature limits, you needed to convert summations
  into integrals, which was a reasonable approximation because the
  change of the thing being summed (summand?) was small as you changed
  the quantum numbers by one, so treating it as a continuum was okay.

  For the low-temperature limit, you had an easier scenario, as all
  the Boltzmann factors were all very small compared with the ground
  state.  So you could just truncate the sum after a few terms.

  \paragraph{Translation at high T}
  $$
    \frac{U}{N} = \sum_{n_xn_yn_z} \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2} \frac{1}{Z} e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2}} 
  $$
  $$
    \approx \int_0^\infty \int_0^\infty \int_0^\infty
    \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2} \frac{1}{Z} e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2}} dn_x dn_y dn_z 
  $$
  $$
    = \frac{\hbar^2 \pi^2}{2mL^2} \frac{1}{Z}
    \int_0^\infty \int_0^\infty \int_0^\infty
    \left(n_x^2 + n_y^2 +
    n_z^2\right)e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
        n_z^2\right)}{2mL^2}} dn_x dn_y dn_z 
  $$
  $$
    u_x = \sqrt\beta\frac{\hbar\pi}{\sqrt{2m}L} n_x 
  $$
  $$
    \frac{U}{N} = \frac{\hbar^2 \pi^2}{2mL^2} \frac{1}{Z}
    \left(\frac{\sqrt{2m} L \sqrt{k_BT}}{\hbar\pi}\right)^3
    \int_0^\infty \int_0^\infty \int_0^\infty
    \left(u_x^2 + u_y^2 +
    u_z^2\right)e^{-\left(u_x^2 + u_y^2 + u_z^2\right)} du_x du_y du_z
  $$
  $$
    = \frac{\sqrt{2m} L (k_BT)^{\frac32}}{\hbar\pi} \frac{1}{Z}
    \int_0^\infty \int_0^\infty \int_0^\infty
    \left(u_x^2 + u_y^2 +
    u_z^2\right)e^{-\left(u_x^2 + u_y^2 + u_z^2\right)} du_x du_y du_z
  $$
  At this point we've extracted the physics from the integral.  It's
  clearly not zero, and it also isn't infinite, so it's just some
  number that we can work out later.  But we still need $Z$...
  $$
    Z = \sum_{n_xn_yn_z} e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 +
      n_z^2\right)}{2mL^2}} 
  $$
  $$
    \approx \int_0^\infty \int_0^\infty \int_0^\infty
     e^{-\beta \frac{\hbar^2 \pi^2 \left(n_x^2 + n_y^2 + n_z^2\right)}{2mL^2}} dn_x dn_y dn_z \\
    u_x = \sqrt\frac{\beta}{2m}\frac{\hbar\pi}{L} n_x 
  $$
  $$
    Z = \left(\frac{\sqrt{2mk_BT} L}{\hbar\pi}\right)
    \int_0^\infty \int_0^\infty \int_0^\infty
    e^{-\left(u_x^2 + u_y^2 + u_z^2\right)} du_x du_y du_z 
  $$
  $$
    = \left(\frac{\sqrt{2mk_BT} L}{\hbar\pi}\right)
    \left(\int_0^\infty e^{-u^2} du\right)^3
  $$
  Once again, we've extracted the physics from the integral, leaving a
  dry, dimensionless husk.  In this case, I cleaned that husk up a
  bit, so it'll be a bit more compact.  Putting these together (with a
  minimum of simplification, we get:
  $$
    \frac{U}{N} = \frac{
      \frac{\sqrt{2m} L (k_BT)^{\frac32}}{\hbar\pi}
      3 \left(\int_0^\infty u^2e^{-u^2}du\right)
      \left(\int_0^\infty e^{-u^2}du \right)^2
    }{
      \left(\frac{\sqrt{2mk_BT} L}{\hbar\pi}\right)
      \left(\int_0^\infty e^{-u^2} du\right)^3
    } 
  $$
  $$
    = k_B T 3 \left(\int_0^\infty u^2e^{-u^2}du\right)
    \left(\int_0^\infty e^{-u^2}du \right)
  $$
  $$
    = \frac32 k_BT
  $$

  \paragraph{Rotation at low T}
  I'll skip over this, as it isn't very exciting (and it's taking a
  long time to \LaTeX\ this).  I'll just mention that it drops
  exponentially to zero at low temperature.  This is a universal
  property of systems with a ``gap,'' which is to say, with a finite
  energy difference between the ground state and the first excited
  state.
  \paragraph{Rotation at high T}
  $$
    Z = \sum_{l=0}^\infty \sum_{m=-l}^l e^{-\beta \frac{\hbar^2 l(l+1)}{2I}} 
  $$
  $$
    = \sum_{l=0}^\infty (2l+1) e^{-\beta \frac{\hbar^2 l(l+1)}{2I}}
  $$
  $$
    \approx \int_{0}^\infty (2l+1) e^{-\beta \frac{\hbar^2
        l(l+1)}{2I}} dl
  $$
  As it turns out, we can relatively easily do this integral.
  However, the ``$+1$'' terms are insignificant, since $l$ is
  integrated up to infinity, and the large $l$ contribution
  dominates.  So we can:
  $$
    Z \approx \int_{0}^\infty 2l e^{-\beta \frac{\hbar^2 l^2}{2I}} dl
  $$
  $$
    u = \frac{\hbar l}{\sqrt{2I k_BT}} 
  $$
  $$
    Z = \frac{8I k_BT}{\hbar^2}\int_{0}^\infty u e^{-u^2} u 
  $$
  The integral is easy to do, but there's no urgent need to do so:  we
  have already taken the physics out of the integral.
  $$
    \frac{U}{N} \approx \frac{1}{Z} \int_{0}^\infty 2l \frac{\hbar^2l^2}{2I} e^{-\beta \frac{\hbar^2 l^2}{2I}} dl
  $$
  $$
    = \frac{1}{Z} 2\frac{\hbar^2}{2I} \left(\frac{4I k_BT}{\hbar^2}\right)^2
    \int_{0}^\infty u^3 e^{-u^2} du 
  $$
  $$
    = \frac{
      2\frac{\hbar^2}{2I} \left(\frac{4I k_BT}{\hbar^2}\right)^2
      \int_{0}^\infty u^3 e^{-u^2} du
    }{
      \frac{8I k_BT}{\hbar^2}\int_{0}^\infty u e^{-u^2} u
    } 
  $$
  $$
    = 4 k_BT \frac{
      \int_{0}^\infty u^3 e^{-u^2} du
    }{
      \int_{0}^\infty u e^{-u^2} u
    } 
  $$
  $$
    = kT
  $$
  Here we can see much of the essential physics by recognizing that
  the energy is proportional to $k_BT$ without performing the
  integrals.  I probably have mistakes somewhere in the above, but the
  final answer is correct.

  \paragraph{Harmonic oscillator at high T}
  For the harmonic oscillator, I'll demonstrate a different approach,
  since I think showing the same approach for the third time in a row
  is a bit boring.
  $$
    Z = \sum_{n=0}^\infty e^{-\beta\left(n+\frac12\right)\hbar\omega_0} 
  $$
  $$
    = e^{-\beta\frac12 \hbar\omega_0} \sum_{n=0}^\infty e^{-\beta n\hbar\omega_0} 
  $$
  $$
    = e^{-\beta \frac12 \hbar\omega_0} \sum_{n=0}^\infty \left(e^{-\beta \hbar\omega_0}\right)^n
  $$
  This is just a harmonic series, so we can solve it using the
  standard trick, where I'll call the series $s$:
  $$
    s \equiv \sum_{n=0}^\infty \left(e^{-\beta \hbar\omega_0}\right)^n 
  $$
  $$
    e^{-\beta \hbar\omega_0}s = e^{-\beta \hbar\omega_0}\sum_{n=0}^\infty
    \left(e^{-\beta \hbar\omega_0}\right)^n 
  $$
  $$
    = \sum_{n=1}^\infty \left(e^{-\beta \hbar\omega_0}\right)^n 
  $$
  $$
    = \left(\sum_{n=0}^\infty
    \left(e^{-\beta \hbar\omega_0}\right)^n\right) - 1 
  $$
  $$
    e^{-\beta \hbar\omega_0}s = s - 1 
  $$
  $$
    1 = \left(1 - e^{-\beta \hbar\omega_0}\right) s 
  $$
  $$
    s = \frac{1}{1 - e^{-\beta \hbar\omega_0}} 
  $$
  $$
    Z = \frac{e^{-\beta \frac12 \hbar\omega_0}}{1 - e^{-\beta \hbar\omega_0}}
  $$
  So that's nice.  Of course, we still want to find the energy.  To do
  this, we can employ yet another trick---although it's not so hard to
  do in the high-temperature limit the same way we solved the previous
  problem.  We can recognize that
  $$
    U = \sum_i P_i E_i 
  $$
  $$
    = \sum_i E_i \frac{e^{-\beta E_i}}{Z} 
  $$
  $$
    = \frac{1}{Z} \sum_i E_i e^{-\beta E_i} 
  $$
  $$
    Z = \sum_i e^{-\beta E_i} 
  $$
  $$
    \left(\frac{\partial Z}{\partial \beta}\right)_{E_{i}}{\beta}{E_i} = \sum_i -E_i e^{-\beta E_i} 
  $$
  $$
    U = -\frac{\left(\frac{\partial Z}{\partial \beta}\right)_{E_{i}}{\beta}{E_i}}{Z}
  $$
  So once we have the partition function, we could just take a
  derivative to find the internal energy.  So for the simple harmonic
  oscillator, we have:
  $$
    \left(\frac{\partial Z}{\partial \beta}\right)_{E_{i}}{\beta}{E_i} =
    -\frac12 \hbar\omega_0 Z -
    \frac{e^{-\beta \frac12 \hbar\omega_0}}{\left(1 - e^{-\beta
        \hbar\omega_0}\right)^2}
    e^{-\beta
        \hbar\omega_0} \hbar\omega_0 
  $$
  $$
    =
    -\frac12 \hbar\omega_0 Z -
    \frac{1}{1 - e^{-\beta\hbar\omega_0}}
    e^{-\beta
        \hbar\omega_0} \hbar\omega_0 Z 
  $$
  $$
    =
    -\frac12 \hbar\omega_0 Z -
    \frac{\hbar\omega_0}{e^{\beta\hbar\omega_0} - 1} Z 
  $$
  $$
    \frac{U}{N} = \frac12 \hbar\omega_0 + \frac{\hbar\omega_0}{e^{\beta\hbar\omega_0} - 1}
  $$
  This gives us an exact solution for the internal energy of a simple
  harmonic oscillator, but we still haven't found the high-temperature
  limit.  To find that, we have to take $\beta\hbar\omega_0 \ll 1$.
  In this case, we can just use a simple Taylor's expansion approach:
  $$
    \frac{U}{N} = \frac12 \hbar\omega_0 +
    \frac{\hbar\omega_0}{e^{\beta\hbar\omega_0} - 1}
  $$
  $$
    \approx \frac12 \hbar\omega_0 +
    \frac{\hbar\omega_0}{1 + \beta\hbar\omega_0 - 1}
  $$
  $$
    = \frac12 \hbar\omega_0 +
    \frac{\hbar\omega_0}{\beta\hbar\omega_0} 
  $$
  $$
    = \frac12 \hbar\omega_0 + k_BT 
  $$
  $$
    \approx k_BT
  $$
  This tells us that like the rotational energy, the vibrational
  energy approaches $k_BT$ per molecule at high temperatures.

  There is a general rule which occurs in the classical limit (which
  is the high-temperature limit), that any quadratic term in the
  energy ends up providing $\frac{1}{2}k_BT$ to the internal energy.  This
  is called the \emph{equipartition theorem}.  Since there are three
  translational degrees of freedom ($v^2$ in each direction), the
  kinetic energy gives us $\frac32 k_BT$.  There are two ways to
  rotate a diatomic molecule, which gives us an additional $k_BT$.
  And finally, the vibration has both kinetic and potential energy,
  which each provide half of the final $k_BT$.
\vfill
\leftline{\textit{by David Roundy}}
\leftline{\copyright DATE David Roundy}
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