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I am very new to statistical mechanics, so this question might seem easy for you. I am reading from Blundell and Blundell's Concepts in Thermal Physics, and my question is from chapter $4$ (Temperature and Boltzmann factor)

We introduce temperature via the $\frac{1}{k_BT} = \frac{d\Omega}{dE}$. In the derivations $\Omega$ is always a function of $E$, which sounds sensible, but the it has less trivial consequence $P(E) \propto e^{\frac{E}{k_BT}}$ , probability that in a canonical ensemble, system (i.e not the reservoir) has energy $E$. This still sounds reasonable, especially given some more examples.

What I don't get is one example author gives , where assuming isothermal atmosphere, find the distribution of density of molecules at height $z$: $\propto e^{-\frac{mgz}{k_BT}}$. While reading the derivation and definition of temperature, author uses systems that are in thermal contact, and looks for energy of individual systems that maximizes microstates. But I thought $E$ for the definition temperature of Boltzmann distribution is a sort of "thermal energy" in its nature (e.g internal energy of ideal gas). But how can we use $E=mgz$ , because it looks like being a bit above has nothing to do with "hotness" or "coldness" (and this is obviously reference frame dependent), or in general what is allowed $E$ for relating temperature to energy.

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When we consider the canonical ensemble (so constant temperature $T$) we have that given that the system consists of a set of energies ${E_i}$, the probability of finding the system in energy $E_j$ is proportional to $P \propto e^{E_j/k_b T}$. This can be any kind of energy $E$, as long as the system under consideration is in thermal equilibrium with an environment (also called a heat bath).

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  • $\begingroup$ Then it feels like temperature doesn't have to be about hotness : why being in higher altitude change your $T$ suddenly? $\endgroup$
    – Mahammad Yusifov
    Commented Mar 27 at 14:40
  • $\begingroup$ Being higher doesn't changes the temperature, but the probability $P$ of finding a molecule at high $z$ is low for low $T$. For a very high $T$ we thus expect a lot of $z$ "levels" to be occupied. The reason why high T means higher altitudes is that the molecules then have more kinetic energy (and are likely to go up further). $\endgroup$
    – OonyXx
    Commented Mar 27 at 15:19
  • $\begingroup$ Yeah that makes sense, I actually knew that. Somehow i found it strange that $E = mgz$ determines $\Omega(E)$, and that determines $T$. Thus $z$, or $mgz$ ,determines $T$. In a way your $z$ determining $T$ sounds kinda wierd , but the other way that if you have high $T$ you have high $z$ on avg, makes sense. In any case, I wanna think that $T$ is an abstract quantity that appreciating it as. a natural thing is somewhat biased then , because of $T(z)$(ish) dependence. $\endgroup$
    – Mahammad Yusifov
    Commented Mar 27 at 18:23
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    $\begingroup$ But a question, why all gases don't just fall down to ground? Certainly temperature doesn't exert a force. Possibly we will say we want high entropy, but that doesn't mean you can make things float for no reason. $\endgroup$
    – Mahammad Yusifov
    Commented Mar 27 at 18:26
  • $\begingroup$ I think the following post provides a clear answer to your question (if not let me know): physics.stackexchange.com/a/2039/319341 $\endgroup$
    – OonyXx
    Commented Mar 27 at 22:32

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