物理

{栖岸计划}热力统计学2

补一段笔记，上述推导中，有$\bar v$这个数的取值没打出来，故在这里推导

$\bar{v} = v \int_0^\infty f(v) \cdot 4\pi v^2 dv \sim v \int_0^\infty dp$

$\bar{v} = \int_0^\infty f(v) \cdot 4\pi v^3 dv$

$\bar{v} = 4\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \int_0^\infty e^{-\frac{m}{2kT} v^2} v^3 dv$

$G_4\left(\frac{m}{2kT}\right) = \frac{1}{2a^2}$

$\bar{v} = 4\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \cdot \frac{1}{2 \left( \frac{m}{2kT} \right)^2}$

$\bar{v} = 2\pi \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \left( \frac{m}{2kT} \right)^{-2}$

$\bar{v} = 2π(π)^{-\frac{3}{2}} (\frac{m}{2kT})^{\frac{3}{2}}(\frac{m}{2kT})^{-2} = 2\pi ( \pi )^{-\frac{3}{2}} (\frac{m}{2kT})^{-\frac{1}{2}}$

$=2π (π)^{-\frac{3}{2}} (\frac{2kT}{m})^{-\frac{1}{2}}$

$= 2 π^{-\frac{1}{2}} \left( \frac{2kT}{m} \right)^{-\frac{1}{2}}$

$= \left( \frac{8kT}{m\pi} \right)^{\frac{1}{2}}$