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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?

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    $\begingroup$ The notes you refer to in your previous post don't really explain the machinery you're trying to interpret particularly well. It would be worth reading some actual lecture notes dedicated to statistical mechanics first, at least the basics. $\endgroup$
    – Charlie
    Mar 21, 2020 at 1:29

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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. $\endgroup$ Jul 19, 2018 at 23:33
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    $\begingroup$ It sounds like you're confusing the canonical and microcanonical ensemble. In the microcanonical ensemble the energy of the entire system is fixed. In the canonical ensemble the system can exchange energy with the surrounding "heat bath". The equation you've given reduces to $\frac{1}{\Omega(E)}$ for the microcanonical ensemble. $\endgroup$
    – Charlie
    Mar 21, 2020 at 1:26
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The connection between thermodynamic beta and inverse temperature can be established as follows (notation from Pathria, 3rd Ed.). Take two systems in thermal contact, each with $NVE$ macrostates (microcanonical ensembles). $$E^{(0)}=E_1+E_2\tag{1}$$ We can see that the total number of microstates, $\Omega^{(0)}$, is really just a function of either $E_1$ or $E_2$, $$\Omega^{(0)}=\Omega_1(E_1)\Omega_2(E_2)=\Omega_1(E_1)\Omega_2(E^{(0)}-E_1)\equiv\Omega^{(0)}(E_1) \tag{2}$$ An axiom necessary in establishing equilibrium is that the total number of microstates of the combined system must be at a maximum, or in other words, $$\frac{\partial \Omega^{(0)}}{\partial E_1}=0 \tag{3}$$ Therefore, $$\frac{\partial \Omega^{(0)}}{\partial E_1}=\Omega_1\frac{\partial \Omega_2}{\partial E_1}+\Omega_2 \frac{\partial \Omega_1}{E_1}=0$$ $$\frac{\partial \Omega_2}{\partial E_1} = \frac{\partial \Omega_2}{\partial E_2}\frac{\partial E_2}{\partial E_1}=-\frac{\partial \Omega_2}{\partial E_2}$$ $$\frac{1}{\Omega_1}\frac{\partial \Omega_1}{\partial E_1}=\frac{1}{\Omega_2}\frac{\partial \Omega_2}{\partial E_2}$$ $$\frac{\partial \ln \Omega_1}{\partial E_1}=\frac{\partial \ln \Omega_2}{\partial E_2} \equiv \beta_1=\beta_2 \tag{4}$$ Generalizing, $$\beta = \frac{\partial \ln \Omega}{\partial E} \tag{5}$$ Where (4) is established by definition. We also know that, $$\left ( \frac{\partial S}{\partial E} \right)_{N,V} = \frac{1}{T} \tag{6}$$ If we divide (6) by (5) and take non-differential changes, $$\frac{\Delta S}{\Delta \ln \Omega}=\frac{1}{\beta T} = constant = k_B \tag{7}$$ This is probably the most fundamental way to establish the relationship between $\beta$ and $T$, but there are other links between them.

Edit: also worth noting that it can be shown that Gibbs' entropy generalizes to (7) when using $P_r = 1 / \Omega$. So (7) really is the entropy formula (with no additive constant), but just for the microcanonical case.

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This seems like confusing notation. In a system of $N$ particles you don't normally talk about the energy of a single particle. A system of energy $E$ has some number of possible microstates, $\Omega(E)$. The probability that the system, at any given moment, is in one particular microstate $r$ is (assuming the principle of a priori probability) $$P_r=\frac{1}{\Omega(E)}. \tag{1}$$ This doesn't tell you anything about the energy of the individual particles. If you're talking about a specific microstate in which one particle has most of the energy that microstate still has a probability given by eq.(1) of existing at any given moment.

Secondly the entropy is usually defined as $$S=k_b\ln{\Omega(E)}, \tag{2}$$ at least where I've seen it, which is essentially just a measure of the number of microstates (and hence contains information about how likely you are to "guess" the exact microstate of a system at energy $E$ at any given moment).

Your argument from here loses me, it seems like you might have misunderstood the fundamentals. It's probably worth you revisiting them.

I am by no means an expert.

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