这是一份liverpool利物浦大学PHYS393/MATH327的成功案例

$$
Z_{1}^{\mathrm{tr}}=\frac{V}{h^{3}}\left(2 \pi m k_{\mathrm{B}} T\right)^{\frac{3}{2}}
$$
The probability, $P_{s}$, of a particle being in a particular state $s$ (with energy $\varepsilon_{s}^{\mathrm{tr}}$ ) of translational motion is given by the Boltzmann distribution
$$
P_{s}=\frac{\exp \left(-\frac{\varepsilon_{s}^{\mathrm{tr}}}{k_{\mathrm{B}} T}\right)}{Z_{1}^{\mathrm{tr}}}
$$
When there are $N$ particles, the mean occupation number, $\bar{n}{s}$, of energy levels (average number of particles in state $s$ ) can be expressed as $$ \bar{n}{s}=N P_{s}
$$
Now we use the definition of the semi-classical system. In particular the fact, that the number of particles is much smaller than the number of available energy levels hence
$$
\bar{n}_{s} \ll<1 \text { for all } s
$$

PHYS393/MATH327 COURSE NOTES ：

most probable speed (max. of $P(v))$
$$
v_{\mathrm{m} . \mathrm{p} .}=\sqrt{\frac{2 k_{\mathrm{B}} T}{m}}
$$
mean speed:
$$
\langle v\rangle=\int_{0}^{\infty} v P(v) \mathrm{d} v=\sqrt{\frac{8 k_{\mathrm{B}} T}{\pi m}}
$$
second moment:
$$
\left\langle v^{2}\right\rangle=\int_{0}^{\infty} v^{2} P(v) \mathrm{d} v=3 \frac{k_{\mathrm{B}} T}{m}
$$
note: this is equivalent to $\bar{E}=\frac{1}{2} m\left\langle v^{2}\right\rangle=\frac{3}{2} k_{\mathrm{B}} T$
and the rms (root-mean-square) speed: $v_{r m s}=\sqrt{\left\langle v^{2}\right\rangle}=\sqrt{\frac{3 k_{\mathrm{B}} T}{m}}$