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I am having problems in understanding the logic of this distribution:

$P(\Psi_{j})=\displaystyle\frac{e^{-E_{j}/kT}}{\displaystyle\sum_{j'}e^{-E_{j'}/kT}}$

The book I am studying use the case of a sample in contact with a reservoir at thermal equilibrium to derive this distribution. I understand the derivation, but I don't understand the logic of the distribution itself. The aspect I'm having problem with is the fact that the lowest the energy of the sample, the highest the probability. What I don't understand is why this happens even thought there is an average energy given by the temperature which I thought should be more probable then any energy lower than this for a given particle. This doubt implies that I am looking at $P(\Psi)$ as the probability for a given particle, wich does not seem to be the case, but if I think of it as the probability for the hole sample, it makes even less sense for me since the energy should be totaly given by the temperature, so it wouldn't make sense to make a distribution of it if the temperature is considered constant.

Thanks in advance

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    $\begingroup$ Question very similar to this one. You can check the answers there and update your question accordingly. $\endgroup$ – gatsu Nov 4 '13 at 0:11
  • $\begingroup$ @gatsu sorry, I hadn't seen that question, but the answers there seem to confuse me more than help. I'm looking for answers that show in what points exactly my thinking is wrong, while keeping more conceptual then mathematical. $\endgroup$ – Ivan Lerner Nov 4 '13 at 15:49
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Maybe your intuition about energy and temperature need to be revisited. Your system can exchange energy with the reservoir at a given temperature. The system+reservoir will iterate through all microstates with equal probability (total energy being fixed), but you can show by using entropy arguments, that the probability of the system being in a state with energy $E$ is given by your first equation. The average energy of your system is then

$\langle E \rangle = \sum_i E_i P(E_i) $.

That is not the same as the most probable energy, which is $E_0$.

When deriving the Boltzmann factor from the reservoir argument, there are corrections to the factor when the reservoir is finite. You can write, for the system in uniquely labelled states $i$ and $j$,

$\frac{P(i)}{P(j)} = \frac{\Omega_R(i)}{\Omega_R(j)}$

where $\Omega_R(i)$ is the number of microstates of the reservoir when the system is in state $i$. Writing this in terms of entropy gives a more fundamental form

$\frac{P(i)}{P(j)} = e^{-\frac{S_R(i) - S_R(j)}{k_B} }$

If you label the states of the system by energy, and allow it to be continuous, then you can Taylor expand $S_R(E_i)$ around $E=0$, since by assumption the energy of the reservoir is vastly bigger than that of the system. Changing variables to the energy of the reservoir $U_{res}$, the linear term is $-\frac{\partial S_R}{\partial U_{res}} E_i = -\frac{E_i}{T}$, from which the Bolzmann factor follows.

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  • $\begingroup$ Thanks, this answer helped me a lot, for some reason I hadn't realized that statement about the average not being the most probable, but once you pointed it out everything else slowly came into place in my head. $\endgroup$ – Ivan Lerner Nov 4 '13 at 16:56
  • $\begingroup$ Actually the most probable energy tends to be about the same as the average energy, at least in large systems. Definitely however, the most probable microstate is not at all related to the most probable energy. This is because at the most probable energy there are a huge number of microstates, who together have a very large probability even though each individual has a tiny probability. $\endgroup$ – Nanite Nov 11 '13 at 23:02
  • $\begingroup$ @Nanite, I don't believe this to be the case for something like an ideal gas. The most probable energy is 0 and the average energy is given by the equipartition theorem, both of which follow from the Boltzmann distribution. $\endgroup$ – lionelbrits Nov 12 '13 at 1:48
  • $\begingroup$ Hmm, I think it depends on what ensemble you are in. When we say a gas at temperature T, we don't really mean it is in perfect thermal equilibrium with an infinite heat bath at temperature T, but rather it is in the same macrostate as such a gas, but in an ensemble where the total energy is fixed (microcanonical). $\endgroup$ – lionelbrits Nov 12 '13 at 1:58

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