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Following on from my previous question:

Exponential form of Boltzmann Distribution

I am now trying to understand the relationship between the thermodynamic beta and the inverse temperature.

Ignoring the Boltzmann constant for now, consider a system with $N$ particles and $M$ discrete energy levels $E_0, E_1, E_2, \ldots, E_M$. We shall use the notation $f(E_i)$ to denote the probability that a given particle is in energy level $E_i$.

Suppose that the total energy of the system is $E$. Then:

$$ f(E) = \frac{1}{\Omega(E)} $$

where $\Omega(E)$ is the total number of microstates with energy $E$. Why is this true? Because if one particle has all the energy, then all the other particles must be in the energy level $E_0$, and there is only one microstate in which this configuration occurs.

Next:

$$ \log(f(E)) = \log(1/\Omega(E)) = - \log(\Omega(E)) = -S$$

But we know that $f(E)$ is proportional to $e^{-\beta E}$ for some $\beta$, so:

$$ 1/T = \frac{d}{dE} S = - \frac{d}{dE} (-\beta E) = \beta $$

This argument is not quite correct, because I have ignored the constant of proportionality. But can it be made correct?

Also, somewhat unrelated, it would appear from the form of the Boltzmann distribution that for a system with fixed energy $E$ the probability $f(E_i)$ is always non-zero even when $E_i > E$ (though it may be very small). Isn't this a problem?

Thanks.

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Since you are dealing with macrostates of constant energy, I assume you are dealing with microcanonical ensembles. The system is totally isolated and there is no exchange of heat, work or particles with the surroundings.

Suppose that the total energy of the system is E. Then:

f(E)=1/Ω(E)

where Ω(E) is the total number of microstates with energy E. Why is this true?

It is just the definition of the microcanonical ensemble wherein equal probability is assigned to all microstates of a given macrostate. Nothing surprising.

You then go on to cite a situation where all the energy belongs to one particle and all other particles have zero energy. This is not a macrostate, but a microstate where you have complete information about your system, there is no need for any statistical physics here.

Also from the reference in the linked question, the Boltzmann function is the probability distribution for individual energies in the macrostate, a higher probability for the lower energy particles and with an exponential decay. You should not be relating this with the probability of a microstate in the ensemble.

It would appear from the form of the Boltzmann distribution that for a system with fixed energy E the probability f(Ei) is always non-zero even when Ei>E (though it may be very small). Isn't this a problem?

Also the microcanonical ensemble assigns equal probability to all microstates that have the given energy E of the macrostate and zero probability to all other configurations. Hence there is no way any particle in any of the microstates will have an energy higher than E.

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  • $\begingroup$ I am using $f(E)$ to denote the probability that a single particle has energy $E$ given that the total energy of the system is $E_0$. I agree from the definition in terms of microstates we must have $f(E) = 0$ if $E > E_0$. However, from the form of the Boltzmann distribution $f(E) = \frac{e^{-\beta E}}{Z}$ there is nothing to stop you "plugging in" a value of $E$ which is greater than $E_0$ and getting a non-zero answer. $\endgroup$ – Andrew Davidson Jul 19 '18 at 23:33

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