So uniform ensemble is the ensemble of all possible states with energy less than E, and since entropy derived from both are equivalent with each other in the thermodynamic limit, many functions derived from them are the same, however, I am wondering why that is the case since apparently the uniform ensemble contains more states and thus more information than the microcanonical ensemble.

# [Physics] Why uniform ensemble and microcanonical ensemble are the same in thermodynamic limit

thermodynamics

#### Related Solutions

The microcanonical ensemble is the (maximum entropy) probability distribution *for a given specified total energy*. What you've calculated is actually the maximum entropy distribution with *no constraints* on the energy, which is the same as the canonical distribution at infinite temperature ($\beta = 0$).

To correctly calculate the microcanonical entropy, first observe that in this system, the total energy $E=\epsilon n_1$, or $n_1 = E/\epsilon$, as long as $E$ is an integer multiple of $\epsilon$.

Since the total energy completely determines $n_1$, we can calculate the number of states in the microcanonical ensemble just as you did, but without the summation over $n_1$:

$$\Omega(E) = \frac{N!}{(N-E/\epsilon)!(E/\epsilon)!}$$

This is only defined when $E$ is an integer multiple of $\epsilon$, but in an appropriate limit of large $N$ and small $\epsilon$ you should be able to use Stirling's approximation to write $S(E) = \log \Omega(E)$ as if it were a continuous function of $E$.

Once this is done, the temperature enters through the usual thermodynamic definition, $$ \frac{1}{T} = \frac{\partial S}{\partial E}, $$ which ought to be fairly straightforward to calculate.

I hope this helps, and sorry for not working through all the details myself.

Consider first the microcanonical ensemble. The total energy of the $N$ oscillators is denoted $$E=\hbar\omega\left(n+{N\over 2}\right) \ \Leftrightarrow\ n={E\over\hbar\omega}-{N\over 2}$$ i.e. $n$ is the sum of the $N$ quantum numbers $n_i$ of each oscillators. $n$ can be interpreted as the number of energy quanta in the system. The number of stationary states with energy $E$ is, as you wrote, $$\Omega(E)={(N+n-1)!\over n!(N-1)!}$$ Therefore the microcanonical energy is $$S(E)=k_B\ln\Omega(E)=k_B\big[\ln(N+n-1)!-\ln n!-\ln (N-1)!\big]$$ Using Stirling approximation $\ln M!\sim M\ln M-M$, we get after cancellation of the three linear terms $$S(E)\simeq k_B\big[(N+n-1)\ln(N+n-1)-n\ln n-(N-1)\ln(N-1)\big)$$ The microcanonical temperature is defined as $${1\over T}={\partial S\over\partial E}={\partial S\over\partial n}{dn\over dE}={k_B\over\hbar\omega}\big[\ln (N+n-1)-\ln n\big]$$ Invert this relation to write $n$ as a function of $\beta=1/k_BT$ $$\beta={1\over\hbar\omega}\ln{N+n-1\over n}\ \Leftrightarrow\ n={N-1\over e^{\beta\hbar\omega}-1}$$ i.e. the Bose-Einstein distribution ! Noting that $$N-1+n=(N-1){e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}$$ the microcanonical entropy now reads $$\eqalign{ S(E)&\simeq k_B\big[(N+n-1)\ln(N+n-1)-n\ln n-(N-1)\ln(N-1)\big)\cr &=k_B\left[(N-1){e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1} \ln{(N-1)e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{N-1\over e^{\beta\hbar\omega}-1}\ln{N-1\over e^{\beta\hbar\omega}-1}-(N-1)\ln(N-1)\right]\cr &=(N-1)k_B\Big[\Big(\underbrace{{e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{1\over e^{\beta\hbar\omega}-1}-1}_{=0}\Big)\ln(N-1)\Big.\cr &\quad\quad\quad\quad\quad\quad\Big. +{e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}\ln{e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{1\over e^{\beta\hbar\omega}-1}\ln{1\over e^{\beta\hbar\omega}-1}\Big]\cr &={(N-1)k_B\over e^{\beta\hbar\omega}-1}\Big[\beta\hbar\omega e^{\beta\hbar\omega}+\Big(e^{\beta\hbar\omega}-1\Big)\ln{1\over e^{\beta\hbar\omega}-1}\Big]\cr }$$ The logarithm can be written as $$\ln{1\over e^{\beta\hbar\omega}-1} =\ln{e^{-\beta\hbar\omega/2}\over e^{\beta\hbar\omega/2}-e^{\beta\hbar\omega/2}}=-{\beta\hbar\omega\over 2}-\ln 2\sinh{\beta\hbar\omega\over 2}$$ The microcanonical entropy is now $$\eqalign{ S&=(N-1)k_B\Big[\beta\hbar\omega {e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{\beta\hbar\omega\over 2}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{e^{\beta\hbar\omega}+1\over e^{\beta\hbar\omega}-1}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{e^{\beta\hbar\omega/2}+e^{-\beta\hbar\omega/2}\over e^{\beta\hbar\omega/2}-e^{-\beta\hbar\omega/2}}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{\cosh\beta\hbar\omega/2\over\sinh\beta\hbar/2}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{\rm cotanh}\ \!{\beta\hbar\omega\over 2}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr }$$ which is precisely the expression of the entropy in the canonical ensemble with the assumption $N-1\simeq N$.

## Best Answer

When you calculate the partition function in the uniform ensemble,

$\bar\Omega(E) = \frac{1}{h^3}\int d^3x_1d^3x_2\dots d^3x_N\int d^3p_1d^3p_2\dots d^3p_N\ \Theta(E-H(\vec x_1,\dots,\vec x_N,\vec p_1,\dots))$

where $\Theta$ is the Heaviside function and $H(x,p)$ is the classical Hamilton function, i.e. the energy. The corresponding microcanonical partition function would then be given by

$\Omega(E,\Delta E) = \bar\Omega(E+\Delta E)-\bar\Omega(E)$

where $[E,E+\Delta E]$ is a small energy interval which is small enough that all energies within it are macroscopically indistinguishable from $E$ but large enough that it contains a large number of microstates.

Geometrically, you can take $\bar\Omega$ to be the volume of the region in $6N$-dimensional phase space enclosed by the hypersurface $E=H(x,p)$. E.g. for a harmonic oscillator, the hypersurface would be a $6N$-dimensional hyperellipsoid or, given proper rescaling, a hypersphere. For an ideal gas, $H(p)$ but independent on position, so one could integrate the position out to get the volume and be left with an integral over a $3N$-dimensional hypersphere in momentum space.

While the uniform partition function is the volume of the region enclosed by that hypersurface, the microcanonical partition function is the volume of a shell around that hypersurface, since its the region enclosed by $E+\Delta E=H(x,p)$ and $E=H(x,p)$ and $\Delta E\ll E$.

Now here's the catch: Generically, the further you increase the dimension of the space, the more volume will be located in close proximity of the enclosing surface. Take the circle of radius $R$ as an example: if you decompose it into concentric rings of thickness $\Delta r\ll R$, each ring of radii $r$ and $r+\Delta r$ will cover an area of $2\pi r\cdot \Delta r$. Thus, the closer $r$ is to the outer rim $R$, the bigger its contribution to the total area $\pi R^2$. More formally this is expressed by the functional determinant $dx\,dy=2\pi r\,dr$. Analogously, for a 3-dimensional sphere, we have $d^3x=4\pi r^2\,dr$ and in $d$ dimensions, $d^dx\propto r^{d-1}\,dr$ holds.So now that's why the phase space volume of the region enclosed by a $6N$ (or $3N$) dimensional hypersurface is highly concentrated in the proximity of that hypersurface. It is so highly concentrated that, in fact, we can neglect the inner volume we would have to subtract to get the microcanonical partition function from the uniform partition function, thus $\Omega(E,\Delta E)\approx\bar\Omega(E)$ in the thermodynamic limit where $N\sim 10^{23}$.