Degrees of Freedom and Molar Specific Heats — Question 7

Question 7

Equipartition gives unequal angular speeds for unequal moments of inertia

A classical rigid nonlinear molecule has rotational energy $E_{\mathrm{rot}}=\tfrac12I_1\omega_1^2+\tfrac12I_2\omega_2^2+\tfrac12I_3\omega_3^2$. Its principal moments satisfy $I_2=2I_1$ and $I_3=3I_1$. At the specified temperature, $\sqrt{k_BT/I_1}=2.00\times10^{12}$ rad/s. All three rotations are active.

1. Find each rms angular-velocity component and its mean energy contribution.

2. Compare the rms angular-momentum components $L_i=I_i\omega_i$. Does equal energy imply equal angular momentum or equal angular speed?

Question 7 - Solution

Equipartition gives unequal angular speeds for unequal moments of inertia

(a) Method and calculation

$$\tfrac12I_i\langle\omega_i^2\rangle=\tfrac12k_BT\ \Rightarrow\omega_{i,\mathrm{rms}}=\sqrt{k_BT/I_i}.$$$$(\omega_{1,\mathrm{rms}},\omega_{2,\mathrm{rms}},\omega_{3,\mathrm{rms}})=2.00\times10^{12}(1,1/\sqrt2,1/\sqrt3)\,\mathrm{rad/s}.$$

(b) Interpretation and further calculation

$$L_{i,\mathrm{rms}}=I_i\omega_{i,\mathrm{rms}}=\sqrt{I_ik_BT}.$$$$L_{1,\mathrm{rms}}:L_{2,\mathrm{rms}}:L_{3,\mathrm{rms}}=1:\sqrt2:\sqrt3,\quad\langle E_i\rangle=\tfrac12k_BT.$$A larger moment of inertia reduces angular-speed fluctuations while increasing angular-momentum fluctuations.

Physical check

Equipartition applies to the quadratic energies, not to the coordinates or momenta by themselves.