# Why kinetic energy is omitted from the Boltzmann factor?

I am studying an example of application of Boltzmann's distribution Concepts in Thermal Physics (specifically the example 4.4 p.43). It is stated that given an isothermal atmosphere, the probability of finding a particle at height $$z$$ is given by:

$$P(z) \propto \exp\left({\frac{-mgz}{kT}}\right)$$

What I can't understand is why the kinetic energy is omitted if the Boltzmann distribution is generally given as:

$$P(E) = \frac{1}{Z} \exp\left({\frac{-E}{kT}}\right)$$

where $$E = K + V$$. I tried to justify the first equation by trying to derive the marginal distribution $$P(z)$$. Here is what I did:

$$P(E) = \frac{1}{Z} \exp\left({\frac{-E}{kT}}\right)= \frac{1}{Z} \exp\left({\frac{-K}{kT}}\right) \cdot \exp\left({\frac{-V}{kT}}\right)$$

In order to simplify the calculations I assumed discrete energy levels for both kinetic and potential energy. Therefore, the marginal distribution $$P(z)$$ or $$P(V)$$ is given by:

$$\sum_i \frac{1}{Z}\exp\left({\frac{-V}{kT}}\right) \exp\left({\frac{-K_i}{kT}}\right)$$

Because any possible combination of $$V_j$$ and $$K_i$$ is possible the partition function can be written as:

$$Z = Z_K \cdot Z_V$$

therefore the above sum reduces to:

$$P(z) =\frac{1}{Z_V} \exp\left({\frac{-mgz}{kT}}\right)$$

Is my reasoning correct? What if the energy levels weren't discrete?

• Kinetic energy term is not omitted in Boltzmann distribution. In kinetic theory of gases : $${\frac {1}{2}}m{\overline {v^{2}}}={\frac {3}{2}}k_{\mathrm {B} }T.$$, so you have it there already. Commented Apr 19, 2022 at 16:31
• Does the kinetic energy depend on where the particle is? Commented Apr 19, 2022 at 16:31
• @DanielSank I have assumed that kinetic energy depends just on velocity and potential on position. Can kinetic energy depends on position? Commented Apr 19, 2022 at 19:07
• @AgniusVasiliauskas That is why I asked the question. In the example, the Boltzmann factor contains just the potential energy (gravitational potential energy). Commented Apr 19, 2022 at 19:09
• @AntoniosSarikas In this case the kinetic energy does not depend on position (kinetic energy depending on position is rather unusual), which is the point. As the kinetic energy doesn't depend on position, the probability of finding particles at a given height shouldn't involve the kinetic energy. Commented Apr 19, 2022 at 22:36

Splitting kinetic and potential energies: \begin{align} p &= \frac 1q e^{-\left(\frac{E_k}{kT}+\frac{E_p}{kT}\right)} \\ &= \frac 1q e^{-\frac{E_k}{kT}-\frac{E_p}{kT}} \\ &= \frac 1q \frac{~~e^{-\frac{E_k}{kT}}}{e^{\frac{E_p}{kT}}} \end{align}
Acknowledging that $$E_k = 3/2~kT$$, let's further to reduce equation to $$p = \frac{1}{e^{3/2}q} e^{-\frac{E_p}{kT}} = \frac1Q e^{-\frac{mgz}{kT}} \, .$$
So, basically kinetic energy part is factored out into normalization factor $$Q$$. Yet, there's another kinetic energy term inside exponent denominator, so that decreasing molecular concentration due to greater potential energy can still be compensated with molecule higher average speeds, i.e. bigger gas temperature if any.
• I can't understand why you substitute $E_k = 3/2 kT$ (the mean kinetic energy). Commented Apr 21, 2022 at 12:02
• Anyways probability of finding particle at given $z$ is already a ratio between potential and kinetic energies. I.E. $$\exp\left({\frac{-mgz}{kT}}\right)$$, can be re-stated as $$\exp\left({-3/2\frac{E_p}{E_k}}\right)$$ by multiplying exponent numerator and denominator by 3/2. Commented Apr 21, 2022 at 12:37