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## Pathria chapter 4 diatomic molecule problem

Posted by peeterjoot on March 18, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

## Question: Diatomic molecule ([1] pr 4.7)

Consider a classical system of non-interacting, diatomic molecules enclosed in a box of volume $V$ at temperature $T$. The Hamiltonian of a single molecule is given by

\begin{aligned}H(\mathbf{r}_1, \mathbf{r}_2, \mathbf{p}_1, \mathbf{p}_2) = \frac{1}{{2m}} \left( \mathbf{p}_1^2 + \mathbf{p}_2^2 \right)+\frac{1}{{2}} K \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2.\end{aligned} \hspace{\stretch{1}}(1.0.1)

Study the thermodynamics of this system, including the dependence of the quantity $\left\langle{{r_{12}^2}}\right\rangle$ on $T$.

Partition function
First consider the partition function for a single diatomic pair

\begin{aligned}Z_1 &= \frac{1}{{h^6}} \int d^6 \mathbf{p} d^6 \mathbf{r} e^{-\beta \frac{ \mathbf{p}_1^2 + \mathbf{p}_2^2 }{2m}} e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}} \\ &= \frac{1}{{h^6}} \left( \frac{2 \pi m}{\beta} \right)^{6/2}\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Now we can make a change of variables to simplify the exponential. Let’s write

\begin{aligned}\mathbf{u} = \mathbf{r}_1 - \mathbf{r}_2\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\mathbf{v} = \mathbf{r}_2,\end{aligned} \hspace{\stretch{1}}(1.0.3b)

or

\begin{aligned}\mathbf{r}_2 = \mathbf{v}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\mathbf{r}_1=\mathbf{u} + \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

Our volume element is

\begin{aligned}d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 = d^3 \mathbf{u} d^3 \mathbf{v} \frac{\partial(\mathbf{r}_1, \mathbf{r}_2)}{\partial(\mathbf{u}, \mathbf{v})}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

It wasn’t obvious to me that this change of variables preserves the volume element, but a quick Jacobian calculation shows this to be the case

\begin{aligned}\frac{\partial(\mathbf{r}_1, \mathbf{r}_2)}{\partial(\mathbf{u}, \mathbf{v})} &= \begin{vmatrix}\partial r_{11}/\partial u_1 & \partial r_{11}/\partial u_2 &\partial r_{11}/\partial u_3 &\partial r_{11}/\partial v_1 &\partial r_{11}/\partial v_2 &\partial r_{11}/\partial v_3 \\ \partial r_{12}/\partial u_1 & \partial r_{12}/\partial u_2 &\partial r_{12}/\partial u_3 &\partial r_{12}/\partial v_1 &\partial r_{12}/\partial v_2 &\partial r_{12}/\partial v_3 \\ \partial r_{13}/\partial u_1 & \partial r_{13}/\partial u_2 &\partial r_{13}/\partial u_3 &\partial r_{13}/\partial v_1 &\partial r_{13}/\partial v_2 &\partial r_{13}/\partial v_3 \\ \partial r_{21}/\partial u_1 & \partial r_{21}/\partial u_2 &\partial r_{21}/\partial u_3 &\partial r_{21}/\partial v_1 &\partial r_{21}/\partial v_2 &\partial r_{21}/\partial v_3 \\ \partial r_{22}/\partial u_1 & \partial r_{22}/\partial u_2 &\partial r_{22}/\partial u_3 &\partial r_{22}/\partial v_1 &\partial r_{22}/\partial v_2 &\partial r_{22}/\partial v_3 \\ \partial r_{23}/\partial u_1 & \partial r_{23}/\partial u_2 &\partial r_{23}/\partial u_3 &\partial r_{23}/\partial v_1 &\partial r_{23}/\partial v_2 &\partial r_{23}/\partial v_3 \end{vmatrix} \\ &= \begin{vmatrix}\partial r_{11}/\partial u_1 & \partial r_{11}/\partial u_2 &\partial r_{11}/\partial u_3 &\partial r_{11}/\partial v_1 &\partial r_{11}/\partial v_2 &\partial r_{11}/\partial v_3 \\ \partial r_{12}/\partial u_1 & \partial r_{12}/\partial u_2 &\partial r_{12}/\partial u_3 &\partial r_{12}/\partial v_1 &\partial r_{12}/\partial v_2 &\partial r_{12}/\partial v_3 \\ \partial r_{13}/\partial u_1 & \partial r_{13}/\partial u_2 &\partial r_{13}/\partial u_3 &\partial r_{13}/\partial v_1 &\partial r_{13}/\partial v_2 &\partial r_{13}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{21}/\partial v_1 &\partial r_{21}/\partial v_2 &\partial r_{21}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{22}/\partial v_1 &\partial r_{22}/\partial v_2 &\partial r_{22}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{23}/\partial v_1 &\partial r_{23}/\partial v_2 &\partial r_{23}/\partial v_3 \end{vmatrix} \\ &= 1.\end{aligned} \hspace{\stretch{1}}(1.0.6)

Our remaining integral can now be evaluated

\begin{aligned}\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}} &= \int d^3 \mathbf{u} d^3 \mathbf{v} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \int d^3 \mathbf{u} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \int d^3 \mathbf{u} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \left( \frac{ 2 \pi }{ K \beta } \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Our partition function is now completely evaluated

\begin{aligned}Z_1 = V\frac{1}{{h^6}} \left( \frac{2 \pi m}{\beta} \right)^{3}\left( \frac{ 2 \pi }{ K \beta } \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

As a function of $V$ and $T$ as in the text, we write

\begin{aligned}Z_1 = V f(T)\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}f(T) = \left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.9b)

Gibbs sum

Our Gibbs sum, summing over the number of molecules (not atoms), is

\begin{aligned}Z_{\mathrm{G}} &= \sum_{N_r = 0}^\infty \frac{z^{N_r}}{N_r!} Z_1^{N_r} \\ &= e^{ z V f(T) },\end{aligned} \hspace{\stretch{1}}(1.0.10)

or

\begin{aligned}q &= \ln Z_{\mathrm{G}} \\ &= z V f(T) \\ &= P V \beta.\end{aligned} \hspace{\stretch{1}}(1.0.11)

The fact that we can sum this as an exponential series so nicely looks like it’s one of the main advantages to this grand partition function (Gibbs sum). We can avoid any of the large $N!$ approximations that we have to use when the number of particles is explicitly fixed.

Pressure

The pressure follows

\begin{aligned}P &= z f(T) k_{\mathrm{B}} T \\ &= e^{\mu/k_{\mathrm{B}} T}\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{11/2}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

Average energy

\begin{aligned}\left\langle{{H}}\right\rangle &= -\frac{\partial {q}}{\partial {\beta}} \\ &= - z V \frac{9}{2} \frac{f(T)}{T} \frac{\partial {T}}{\partial {\beta}} \\ &= z V \frac{9}{2} \frac{f(T)}{T^3} \frac{1}{{k_{\mathrm{B}}}},\end{aligned} \hspace{\stretch{1}}(1.0.13)

or

\begin{aligned}\left\langle{{H}}\right\rangle = e^{\mu/k_{\mathrm{B}} T} V \frac{9}{2} k_{\mathrm{B}}^2 \left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Average occupancy

\begin{aligned}\left\langle{{N}}\right\rangle &= z \frac{\partial {}}{\partial {z}} \ln Z_{\mathrm{G}} \\ &= z \frac{\partial {}}{\partial {z}} \left( z V f(T) \right) \\ &= z V f(T)\end{aligned} \hspace{\stretch{1}}(1.0.15)

but this is just $q$, or

\begin{aligned}\left\langle{{N}}\right\rangle &= e^{\mu/k_{\mathrm{B}} T} V\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Free energy

\begin{aligned}F &= - k_{\mathrm{B}} T \ln \frac{ Z_{\mathrm{G}} }{z^N} \\ &= - k_{\mathrm{B}} T \left( q - N \ln z \right) \\ &= N k_{\mathrm{B}} T \beta \mu - k_{\mathrm{B}} T q \\ &= z V f(T) \mu - k_{\mathrm{B}} T z V f(T) \\ &= z V f(T) \left( \mu - k_{\mathrm{B}} T \right)\end{aligned} \hspace{\stretch{1}}(1.0.17)

\begin{aligned}F = e^{\mu/k_{\mathrm{B}} T} V \left( \mu - k_{\mathrm{B}} T \right)\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

Entropy

\begin{aligned}S &= \frac{U - F}{T} \\ &= \frac{V}{T} e^{\mu/k_{\mathrm{B}} T} \left( k_{\mathrm{B}} T \right)^{3/2}\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( \frac{9}{2} k_{\mathrm{B}}^2 - \left( \mu - k_{\mathrm{B}} T \right) \left( k_{\mathrm{B}} T \right)^3 \right).\end{aligned} \hspace{\stretch{1}}(1.0.19)

Expectation of atomic separation

The momentum portions of the average will just cancel out, leaving just

\begin{aligned}\left\langle{r_{12}^2}\right\rangle &= \frac{\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 e^{-\beta K \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 /2 }}{\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 /2 }} \\ &= \frac{ \int d^3 \mathbf{u} \mathbf{u}^2 e^{-\beta K \mathbf{u}^2 /2 }}{\int d^3 \mathbf{u} e^{-\beta K \mathbf{u}^2 /2 }} \\ &= \frac{\int da db dc \left( a^2 + b^2 + c^2 \right) e^{-\beta K \left( a^2 + b^2 + c^2 \right) /2}}{\int e^{-\beta K \left( a^2 + b^2 + c^2 \right)/2}} \\ &= 3 \frac{\int da a^2 e^{-\beta K a^2/2}\int db dc e^{-\beta K \left( b^2 + c^2 \right) /2}}{\int e^{-\beta K \left( a^2 + b^2 + c^2 \right)/2 }} \\ &= 3 \frac{\int da a^2 e^{-\beta K a^2/2}}{\int e^{-\beta K a^2/2}}\end{aligned} \hspace{\stretch{1}}(1.0.20)

Expanding the numerator by parts we have

\begin{aligned}\int da a^2 e^{-\beta K a^2/2} \\ &= \int a d\frac{ e^{-\beta K a^2/2}}{- 2 \beta K/2} \\ &= \frac{1}{\beta K}\int e^{-\beta K a^2/2}.\end{aligned} \hspace{\stretch{1}}(1.0.21)

This gives us

\begin{aligned}\boxed{\left\langle r_{12}^2 \right\rangle = \frac{3}{\beta K} = \frac{3 k_{\mathrm{B}} T}{K}.}\end{aligned} \hspace{\stretch{1}}(1.0.22)

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.