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What is the physical interpretation of

$$ \langle p_\theta^2\rangle = \dfrac{8\pi^2 I^2}{\beta^2} \tag{1}$$

and

$$\langle p_\varphi^2 \rangle = \dfrac{16\pi^2 I^2}{3\beta^2} \tag{2}$$

where the Hamiltonian (for $1$ particle) that describe the rotation of a diatomic molecule is

$$ H_1 (\theta, \varphi, p_\theta, p_\varphi)=\dfrac{1}{2I} \left(p_\theta^2 +\dfrac{p_\varphi^2}{\sin^2 \theta} \right) \tag3 $$

where $I$ is the inertia momentum of the molecule ($I=mr_0^2$). Are the units correct? What do the numerical values mean?

Note that $\beta =1/k_BT$ as usual in statistical mechanics. For the definition of the expected value in the canonical ensemble, I've used $$ \langle p_\theta^2 \rangle =\dfrac{1}{h^2}\dfrac{1}{Z(\beta,\alpha,1)}\int_0^{2\pi} \text{d}\varphi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\varphi \cdot p_\theta^2 \text{ e}^{-\beta H_1},$$ where $$ Z(\beta,\alpha,1)= \int_0^{2\pi} \text{d}\varphi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\varphi \text{ e}^{-\beta H_1}.$$ I don't know if this is right or not.

For example: [1] Statistical Mechanics, Ryogo Kubo, p. 200, nº 3.

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  • $\begingroup$ The units in your expressions look wrong to me. From the expression for the Hamiltonian $p^2$ has units of $IH$ for both $\theta$ and $\phi$, but your expectation values have units of $(Ik_BT)^2$ and $k_BT$ has units of energy $\endgroup$ – By Symmetry Nov 16 '16 at 0:03
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If an particle has moment of inertia $I$ and angular momentum $L=I\omega$ about some axis, its rotational kinetic energy is $K=\frac12 I\omega^2 = \frac{L^2}{2I}$. So your source seems to be using $p_\theta$ to mean something like "angular momentum associated with motion in the $\hat\theta$ direction," which isn't too surprising. If you use the usual "physics" spherical polar coordinates where the coordinate $\theta$ is the angle with the $z$-axis and the coordinate $\phi$ is the angle from the $x$-$z$ plane, then a given angular momentum $p_\phi$ about the $z$-axis is associated with more energy when the moment of inertia is reduced by having the rotor near the $z$-axis.

We expect an angular momentum to have units of joule-seconds, which is consistent with your definition of the Hamiltonian: $$ [p^2] = [IH] = [\rm {kg\,m^2}][J] = [J\cdot s]^2 $$

However, your expectation values for the angular momenta have units equivalent to $[IH]^2$, which is $[p^4]$. That's because you haven't normalized your expectation value by the "zeroth moment" of your probability distribution. Instead you should have something more like $$ \left<p^2\right> = \frac{ \int d\theta\ d\phi\ dp_\theta\ dp_\phi\ p^2 \exp{-\beta H} }{ \int d\theta\ d\phi\ dp_\theta\ dp_\phi\ \exp{-\beta H} } $$ Remember when you do these sorts of integrals that the differentials (here, $dp$) also carry their own units, which must be accounted for in the result.

My naïve expectation is that you should find $$ \frac{\left<p^2\right>}{2I} = \frac12 kT $$ for both angular momenta since, in thermal equilibrium, the average energy in each quadratic degree of freedom is $\frac12 kT$. I don't know whether you're doing the right integral yet or not, in that case.

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  • $\begingroup$ I think the answer is perfect. Thanks a lot, @rob! $\endgroup$ – Clare Francis Nov 16 '16 at 0:49
  • $\begingroup$ Feel free to click the "upvote" button if you think the answer is useful, and/or the green "accept answer" icon if you think your problem is solved to your satisfaction. $\endgroup$ – rob Nov 16 '16 at 0:55
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By definition we have $$ \langle a^2 \rangle = \dfrac{\int_0^{2\pi} \text{d}\phi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\phi \cdot a^2 \text{ e}^{-\beta H}}{\int_0^{2\pi} \text{d}\phi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\phi \text{ e}^{-\beta H}} \tag{S.1}$$ but, for $a=p_\theta$, this expression can be simplified $$\langle p_\theta^2 \rangle = \dfrac{\int_{-\infty}^\infty p_\theta^2 \text{ e}^{-\beta p_\theta^2/(2I)} \text{ d}p_\theta}{\int_{-\infty}^\infty \text{e}^{-\beta p_\theta^2/(2I)}\text{ d}p_\theta}. \tag{S.2}$$ For the Hamiltonian $(3)$ we obtain $$ Z(\beta,\alpha,1)=8\pi^2 \dfrac{I}{\beta} \tag{S.3}$$ and using this expression $$\int_\mathbb{R} \text{d}x \text{ e}^{-ax^2} = 2a\int_\mathbb{R} \text{d}x \text{ } x^2 \text{ e}^{-ax^2}=\sqrt{\dfrac{\pi}{a}}\tag{S.4}$$ into $(S.2)$ we have $$ \boxed{\langle p_\theta^2 \rangle =\dfrac{I}{\beta}= k_B TI}. \tag{S.5}$$ and, for $a=p_\phi$, the expression $(S.1)$ turns into $$\langle p_\phi^2 \rangle = \dfrac{\int_{0}^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\phi \text{ } p_\phi^2 \text{ e}^{-\beta p_\phi^2/(2I\sin^2 \theta)}}{\int_{0}^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\phi \text{ }\text{e}^{-\beta p_\phi^2/(2I\sin^2 \theta)}}. \tag{S.6}$$ We obtain $$ \boxed{\langle p_\phi^2\rangle = \dfrac{2}{3}\dfrac{I}{\beta}=\dfrac{2}{3} k_B TI}. \tag{S.7}$$ Equations $(S.5)-(S.7)$ are $\textbf{the dispersion of each (conjugate) angular momentum}$ because $$ \langle p_\theta \rangle = \langle p_\phi \rangle =0.$$ Therefore, $$ \sigma_{p_\theta}^2 = \langle p_\theta^2 \rangle= k_B TI > \sigma_{p_\phi}^2 = \langle p_\phi^2 \rangle = \dfrac{2}{3}k_B TI = \dfrac{2}{3}\langle p_\theta^2 \rangle \geq 0.$$ The physical interpretation is that the $\textbf{dispersion}$ of $\theta$-$\textbf{angular momentum}$ $p_\theta$ is bigger than the $\textbf{dispersion}$ of $\phi$-$\textbf{angular momentum} $ $p_\phi$ in a factor of $3/2$: $$ \boxed{\sigma_{p_\theta}^2 = \dfrac{3}{2}\sigma_{p_\phi}^2}. \tag{S.8}$$ According to @rob the energy of one atom is $$ E_1 = \langle H_1 \rangle = \dfrac{1}{2I} \left( k_B TI + k_B TI\right)=k_B T \tag{S.9}$$ and for the molecule $$ E_{1+1}=\langle 2H_1 \rangle = 2k_B T.\tag{S.10}$$

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