I'm trying to show that for an ideal gas in a uniform gravitational field $\vec{g} = (0, 0, -g)$, confined to a box of height $L$ and base area $A$, the mean potential energy per particle is $$\langle v (\vec{r_i}) \rangle \approx \frac{mgL}{2},$$ provided that $k_B T \gg mgL$.

Thus far I've found that the distribution for the height of a particle is

$$P_z(z_i) = \frac{\beta mg}{1 - \exp(-\beta mgL)} \exp(-\beta mgz_i),$$

and so, the mean potential energy is given by

$$\langle v (\vec{r_i}) \rangle = \langle mgz_i \rangle = \frac{1}{\beta} - \frac{mgL}{\exp(\beta mgL) - 1}.$$

Now, that the mean potential energy should be $mgL/2$ when $k_BT \gg mgL$ seems very intuitive to me, however, I'm having trouble demonstrating this using the expression I've found. Any help would be greatly appreciated!


1 Answer 1


If the thermal energy of a particle is much larger than the change in potential energy when it moves from bottom to top (that's what $k_BT>>mgL$ means) then it doesn't care about gravity. So the density of the gas is uniform. So the center of mass is halfway up.

  • $\begingroup$ Yes, that is clear to me, and is what I meant by this result appearing very intuitive. What I'd like to do is to demonstrate that the expression for the potential energy $\langle v( \vec{r_i} ) rangle\$ reduces to this. $\endgroup$
    – Gauss57
    Commented Feb 7, 2021 at 7:56

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