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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?

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  • $\begingroup$ 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. $\endgroup$ Apr 19 at 16:31
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    $\begingroup$ Does the kinetic energy depend on where the particle is? $\endgroup$
    – DanielSank
    Apr 19 at 16:31
  • $\begingroup$ @DanielSank I have assumed that kinetic energy depends just on velocity and potential on position. Can kinetic energy depends on position? $\endgroup$ Apr 19 at 19:07
  • $\begingroup$ @AgniusVasiliauskas That is why I asked the question. In the example, the Boltzmann factor contains just the potential energy (gravitational potential energy). $\endgroup$ Apr 19 at 19:09
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    $\begingroup$ @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. $\endgroup$
    – DanielSank
    Apr 19 at 22:36

1 Answer 1

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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.

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  • $\begingroup$ I can't understand why you substitute $E_k = 3/2 kT$ (the mean kinetic energy). $\endgroup$ Apr 21 at 12:02
  • $\begingroup$ What other kinetic energy expression you would suggest if not the one offered by kinetic theory of gases ? $\endgroup$ Apr 21 at 12:23
  • $\begingroup$ 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. $\endgroup$ Apr 21 at 12:37
  • $\begingroup$ The Boltzmann factor doesn't contain average energy that is what troubles me. $\endgroup$ Apr 21 at 18:54
  • $\begingroup$ So according to you temperature of the system is not related with average kinetic energy of molecule at all? $\endgroup$ Apr 21 at 21:39

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