Problem: Consider placing a fictitious open surface in an equilibrium ideal gas at temperature \(T\); although the net particle current density through such a surface would be \(\textbf J=\textbf 0\), if one only counts the particles that go through the surface from one side to the other, then show that the resulting unidirectional particle current density \(J\) is non-zero, and given by:
\[J=\frac{1}{4}n\langle v\rangle\]
where \(n=p/k_BT\) is the number density and \(\langle v\rangle=\sqrt{8k_BT/\pi m}\) the average speed.
Solution:
Problem: By an analogous calculation, show that the unidirectional kinetic energy current density \(S\) for an ideal gas (which one might also think of as a heat flux \(S=q\)) is given by:
\[S=\frac{1}{2}nk_BT\langle v\rangle\]
And hence, show that the average kinetic energy of particles hitting a wall is enhanced by a Bayes’ factor of \(4/3\) compared to the bulk kinetic energy \(\frac{3}{2}k_BT\) per particle.
Solution:
