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I have been attempting to calculate the entropy increase of a monatomic ideal gas upon addition of a small amount of heat, and have been unable to resolve an apparent contradiction between the requirement that $dS = dQ/T$ and Boltzmann's law, $dS \propto d\log \Omega$. I describe my dillemma below.

A monatomic ideal gas of constant total energy, $H$, and with $n/3$ particles each of unit mass has Hamiltonian given by:

$$ H = \sum_{i=1}^n \frac{p_i^2}{2}. $$

This may be re-written as

$$ (\sqrt{2H})^2 \equiv R^2 = \sum_{i=1}^n p_i^2 $$

Written this way, we see that $R=\sqrt{2H}$ is the radius of an $n$-sphere which the $p_i$ coordinates of the phase space of the system are restricted to. We refer to the accessible volume of the system within the phase space as $\Omega$ (this accessible volume also includes the position coordinates, however we may ignore them, as they will remain unchanged throughout this analysis, and these will only contribute a constant factor $\propto V^n$ in the integral over the accessible phase space).

According to Boltzmann, the entropy of a system is related to the accessible phase volume, $\Omega$, by

$$ S = k_B \log \Omega $$

And, in particular, for a small change in the entropy, $dS \propto d\log \Omega$. Now suppose our monatomic ideal is at unit temperature, and suppose we add a small amount of heat, $dQ$, to the gas while keeping its volume constant. We also suppose the energy supplied is so small the temperature may be treated as constant. This will increase the total energy, $H$, by $dH = dQ$, and for a gas at unit temperature this will increase the entropy by $dS = dQ = dH$ as well. Hence we expect the accessible volume of phase space to increase in proportion with $d\log \Omega$. The accessible region of the momentum phase space will now be an $n$-sphere of slightly greater radius:

$$ (\sqrt{2(H + dH)})^2 = \sum_{i=1}^n p_i^2 $$

Because the radius of the accessible $n$-sphere has increased, the accessible volume of the phase space, $\Omega$, will also have increased. In particular, the surface area of a unit sphere is proportional to $R^{n-1}$, and so the increase in the volume of the accessible phase space upon the addition of heat $dQ = dH$ will be

$$ d\Omega \propto (2H+2dH)^{(n-1)/2} - (2H)^{(n-1)/2} \approx ndH = ndS $$

However, Boltzmann's law leads us to expect that the accessible volume will increase as

$$ dS = d\log \Omega. $$

Effectively, this naive calculation shows a proportionality between the phase volume and the heat supplied to the system (albeit with a very large constant):

$$ d\Omega \propto dS $$

whereas Boltzmann's law suggests an exponential dependence:

$$ d\Omega \propto d(\exp S ) $$

I'm rather confused about this. I had expected that Boltzmann's law would work in this simplest-possible-case. Why isn't the phase volume growing exponentially as energy is added, as Boltzmann's law predicts?

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