I'm trying to understand what are the required conditions for an ideal gas in one spatial dimension to act as a heat reservoir. I would appreciate answers that don't use the terms heat, temperature, heat capacity, entropy unless they are well defined for the system described in the question. (I'm trying to find reasoning from first principles)

I'll start with a helpful calculation: An ideal gas with $N$ particles in one dimension has energy which is proportional to $\sum_{i=1}^Np_i^2$, where $p_i$ is the momentum of the $i$'th particle. Therefore, the number of microstates of the ideal gas with a total energy $E$ is proportional to $E^{N-1}$. Let's denote the constant of proportionality $\alpha^{N-1}$, therefore the number of microstates with energy $E$ is $\Omega(E)=(\alpha E)^{N-1}$. These three equations follow: $$\log \Omega(E)=(N-1)\log(\alpha E)$$ $$\frac{\partial}{\partial E}\log \Omega(E)=\frac{N-1}{E}$$ $$\frac{\partial^2}{\partial E^2}\log \Omega(E)=-\frac{N-1}{E^2}$$

Now, let's suppose that some thermodynamic system $S$ is coupled to this one-dimensional ideal gas, which I denote by $S'$, and that the combined system is isolated with total energy $E_\text{tot}$. The coupling lets energy flow between the systems, but the interface between them does not allow the "flow" of volume, particles etc..

Using the principle of equal a priori probabilities, we assume that all microstates of the combined system have the same probability. It follows that the probability $p(s)$ of a microstate $s$ of $S$, with energy $E_s$, is proportional to $\Omega({E_\text{tot}-E_s)}$. Taylor expanding around $E_s=0$ we get: $$p(s)\propto e^{\log\Omega({E_\text{tot}-E_s)}}=e^{\log\Omega({E_\text{tot})}-\frac{E_s}{E_\text{tot}}(N-1)+\mathcal{O}\left(\left(\frac{E_s}{E_\text{tot}}\right)^2\right)}$$ where the leading coefficient of the $\mathcal{O}\left(\left(\frac{E_s}{E_\text{tot}}\right)^2\right)$ term is $N-1$.

Denoting $\beta=\frac{N-1}{E_\text{tot}}$, we get: $$p(s)\propto e^{-\beta E_s+\mathcal{O}\left(\left(\frac{E_s}{E_\text{tot}}\right)^2\right)}$$

(The proportionality means that the factor of proportion does not depend on $s$).

You are probably aware that I'm following the derivation of the canonical ensemble. At this point of the derivation, usually the $\mathcal{O}\left(\left(\frac{E_s}{E_\text{tot}}\right)^2\right)$ is omited, owing to the fact that $S'$ is a heat reservoir. This is the point I'd like to better understand, and here are my questions:

  1. When is it justified to omit the term? I understand that it is required that $\left(\frac{E_s}{E_\text{tot}}\right)^2$ is small compared to its coefficient $N-1$. But when can I expect this condition to hold (especially noting that usually $N-1$ is taken very large)? I mean, suppose that the only variable that I directly control in the experiment is $E_\text{tot}$. How can I know which values of $E_\text{tot}$ I should provide in order for the energy of the system to be distributed between $S$ and $S'$ such that $\left(\frac{E_s}{E_\text{tot}}\right)^2$ is indeed small enough?
  2. When can the ideal gas behave as a heat reservoir for small $N$, such as $N=1,2$?
  3. Does the mass of the particles in the ideal gas matter?
  4. What are we assuming about the interaction between the systems $S$ and $S'$?
  5. Suppose that the system $S$ is also an ideal gas (in three dimensions), and that its interaction with $S'$ is such that $S'$ can actually only influence only one specific particle in $S$. In other words, the energy flow from $S'$ to $S$ allows for the energy to go only to one specific particle in $S$. Could $S'$ still be considered a heat reservoir in this case? If no, what condition is not satisfied?
  6. For a general system $S'$ (not necessarily and ideal gas), what are the conditions on $S'$, $S$ and the interactions between them that allow the treatment of $S'$ as a heat reservoir? I'd appreciate if the answer to this question (question 6) is given in two versions: one without referring to heat, temperature, heat capacity, entropy and the other which does includes these terms (or some of them)

Edit 1:

An additional question:

  1. Does the question whether a system behaves as a heat reservoir depend also on the specific calculation of interest? For example, if we are interested in the extensive property $A$ which is defined as the sum over all particles $i$ of $E_i^{10^{10^{10}}}$, where $E_i$ is the energy of the $i$'th particle. Calculating its expected value involves integrating the integrand $A\cdot p(s)\text{d}s$. Here, clearly the integrand might be dominated by microstates $s$ with high energy, which seems to suggest that we can't use the standard canonical ensemble.
  • $\begingroup$ Your analysis has important, but hidden points related to dynamics of such systems. In order to exchange energy ( heat), systems S and S' has to exchange kinetic energy between particles. It is important to note that if such exchange is not efficient, the whole process may take time, and I would not exclude particular causes when thermodynamic equilibrium is even not possible here. For example for 1 dimensional systems, exchange occur only in one point! $\endgroup$
    – kakaz
    Dec 17, 2017 at 9:39

1 Answer 1


This is a partial answer.

In statistical physics, we are ultimately interested in calculating statistical magnitudes such as the mean energy of the system $S$. This is given by: $$\langle E\rangle=\int E_s\cdot p(s)\text{d}s$$ where the integral is over all microstates $s$ of $S$. The question of whether $S'$ can be considered as a heat reservoir is equivalent to the question of whether we can use the canonical ensemble relation $p(s)\propto e^{-\beta E_s}$ (without the higher order terms). Now, when integrating over $s$, surely we will encounter states with $E_s$ such that $\frac{E_s}{E_\text{tot}}$is too large for us to be able to omit the higher order terms in $p(s)$. However, it is actually not so straight forward, since the integral can be over many different values of $s$ with the same energy $E_s$, and this multiplicity can be large enough to amplify even the higher orders of $E_s$ in $p(s)$. This multiplicity is $\Omega_S(E_s)$, and the integral can be written as: $$\langle E\rangle=\int E\cdot \Omega_S(E)p(s_E)\text{d}E$$ where $s_E$ is a representative microstate of $S$ with energy $E$. This integral is helpful since it samples each value of $p(s)$ exactly once. It shows us that the contribution to the integral of a specific energy $E_s$ is $E_s \Omega_S(E_s)p(s)$. From here we conclude that we can neglect higher order terms $e^{\mathcal{O}\left(\left(\frac{E_s}{E\text{tot}}\right)^2\right)}$ if $E_s \Omega_S(E_s)e^{\mathcal{O}\left(\left(\frac{E_s}{E\text{tot}}\right)^2\right)}$ is very small compared to other energies. $\Omega_S(E_s)p(s)$ is exactly $P(E)$ and is given by: $$P(E)\propto \Omega_S(E)\Omega_{S'}(E_\text{tot}-E)$$ It happens that usually $P(E)$ is much larger for some energy that the other energies, and thus this energy dominates the integral. Usually this energy is small enough that the higher order terms of $p(s)$ can be neglected.

The conclusion is that in order that we can use the canonical ensemble rule, we need to count the number of states $\Omega_S(E)$ and $\Omega_{S'}(E_\text{tot}-E)$, and make sure that $E\cdot\Omega_S(E)\Omega_{S'}(E_\text{tot}-E)$ is highly dominated by a small enough energy $E$. Obviously, this shows that there is dependence on the number of degrees of freedom in both $S$ and $S'$.

  • $\begingroup$ This analysis looks completely like computation of exponential integral by stepest phase method ( I am not sure if this is correct English terminology. It is typical method used for following from Faynman path integral to classical physics) $\endgroup$
    – kakaz
    Dec 17, 2017 at 9:42

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