Translational Kinetic Energy — Question 9

Question 9

The same mean energy does not specify a speed distribution

Two isotropic populations consist of particles of identical mass $m$. Population A has every speed equal to $v_0$. In B, half the particles are at rest and half have speed $\sqrt2v_0$. Both have zero mean velocity. No Maxwell-distribution assumption is made.

1. Compare their mean kinetic energies, rms speeds and mean speeds.

2. Calculate the kinetic-energy variance in each population. Can assigning the same energy-equivalent temperature prove either distribution is thermal equilibrium?

Question 9 - Solution

The same mean energy does not specify a speed distribution

(a) Method and calculation

$$\langle v^2\rangle_A=v_0^2,\quad\langle v^2\rangle_B=\tfrac12(0)+\tfrac12(2v_0^2)=v_0^2.$$$$\langle K\rangle_A=\langle K\rangle_B=K_0=\tfrac12mv_0^2,\quad v_{{\mathrm{rms}},A}=v_{{\mathrm{rms}},B}=v_0.$$$$\langle v\rangle_A=v_0,\quad\langle v\rangle_B=v_0/\sqrt2.$$

(b) Interpretation and further calculation

$$\operatorname{Var}_A(K)=0,\quad\operatorname{Var}_B(K)=\tfrac12(0-K_0)^2+\tfrac12(2K_0-K_0)^2=K_0^2.$$$$T_{\mathrm{energy}}=2K_0/(3k_B)\quad\text{is the same for both}.$$Equal mean energy fixes only one moment. These artificial discrete distributions need not be equilibrium distributions.

Physical check

Pressure in the isotropic kinetic model also agrees at equal number density, despite the different higher moments.