Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $N\sim 10^{23}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.