3
$\begingroup$

I'm familiar with the approximation of a sum over energy states by an integral over phase space: $$ \sum_n f(\epsilon_n) \approx \frac{1}{h^3}\int f\left(H(\textbf{r},\textbf{p})\right) \, d^3\textbf{r} \, d^3\textbf{p}, $$ where $H(\textbf{r},\textbf{p})$ is the Hamiltonian of the system. This approximation is useful for calculating partition functions, densities of states, etc. When the energy states just depend on momentum, this approximation makes perfect sense to me. For example, if I consider the partition function of an ideal gas (considering only one particle for now), I can make the conversion

$$Z_1 = \sum_n e^{-\beta \epsilon_n} \approx \frac{1}{\left(\Delta p\right)^3} \int e^{-\beta \left|\textbf{p}\right|^2/2m} \, d^3\textbf{p}.$$

Since the momentum in each direction is quantized due to the boundary conditions ($p_n = hn/2L$), I see that $\Delta p = h/2L$. (If you use periodic boundary conditions instead, you get $p_n = hn/L$ and $\Delta p = h/L$, but then the bounds on the integral are different.) Thus we find

$$Z_1 \approx \frac{8V}{h^3} \int_0^{\infty}\int_0^{\infty}\int_0^{\infty} e^{-\beta \left|\textbf{p}\right|^2/2m} \, dp_x \, dp_y\, dp_z,$$

or equivalently (and more naturally with periodic boundary conditions),

$$Z_1 \approx \frac{V}{h^3} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-\beta \left|\textbf{p}\right|^2/2m} \, dp_x \, dp_y\, dp_z. $$

This is equivalent to the first phase space integral I wrote above if the Hamiltonian doesn't depend on position, since the spatial integral just yields a factor of the volume.

All of this seems reasonable. However, when applying it to a system whose energy states depend on position as well, e.g., an ideal gas in a gravitational field, I can't see how to apply this logic to the spatial integrals. If know the sum over $n$ in the first expression is really an integral when dealing with a continuous system, but don't we still need to divide the integral by some factor of $\Delta x \Delta y \Delta z$ to get the right units? For the case of an ideal gas in a gravitational field I would expect something like $$ Z_1 \approx \frac{V}{h^3} \frac{1}{\left(\Delta x \Delta y \Delta z\right)} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta \left(\frac{\left|\textbf{p}\right|^2}{2m} + mgz\right)} \, d^3\textbf{r}\, d^3\textbf{p},$$ where the $V/h^3$ comes from the $\Delta p$ as before, but I don't see what the $\Delta x_i$ would be.

To yield the first expression, the factors of $\Delta p$ and $\Delta x$ must combine to yield a $h$. How does this happen? Everything I've seen just pulls the phase space integral out of thin air, using the uncertainty principle to justify some sort of "minimal phase space volume," to put in a factor of $h$ to make the units work out. That seems unsatisfactory to me, particularly since we can derive it so completely and rigorously in the case where there is no spatial dependence.

I hope this question is clear, but let me know if it's not, and I can edit it.

$\endgroup$
  • $\begingroup$ Upon further reflection, I realize that in the example I gave (the ideal gas in a gravitational field), the motion in the $z$-direction is no longer that of a free particle, so $\Delta p_z \neq h/L_z$. This only complicates things further. $\endgroup$ – Klein Four Jan 18 '17 at 13:55
1
$\begingroup$

lets define $\Gamma(E)$ as the number of states till energy E, and define $\frac{d\Gamma(E)}{dE}=g\left(E\right)$ .

By definition in a 3 dimensional world

$\Gamma\left(E\right)=\frac{1}{h^{3}}\iint d^{3}pd^{3}q$

why the factor $h^{3}$? because for every dimensin you have $\Delta x\Delta p\propto h $

here you need the number of states for a given energy E :

so you take the phase space volume of the system with the energy E and basically divide it by the volume that 1 state takes $\left(h^{3}\right)$.

now note that $Z=\sum_{i}g_{i}e^{-\beta E_{i}} $where $g_{i} $is the degeneracy of each E_{i}

when taking the energy spectrum to a continuous spectrum you get that

$ Z=\int g\left(E\right)e^{-\beta E}dE$

However as we definded $g\left(E\right)$ above we can see that $g\left(E\right)dE=d\Gamma\left(E\right)$

and by the definition of $\Gamma\left(E\right)$ we get that $d\Gamma\left(E\right)=\frac{1}{h^{3}}d^{3}pd^{3}q $

plug into Z and get

$Z=\frac{1}{h^{3}}\iint e^{-\beta E(q,p)}d^{3}pd^{3}q $

In case of the ideal gas that is in the gravitational field this is pretty straight forward:

$Z_{1}=\frac{1}{h^{3}}\iint e^{-\beta\frac{p^{2}}{2m}-\beta mgz}d^{3}pd^{3}q$

$Z_{1}=\frac{1}{h^{3}}\left[4\pi\int_{0}^{\infty}e^{-\beta\frac{p^{2}}{2m}}p^{2}dp\right]*\left[A\int_{0}^{L}e^{-\beta mgz}dz\right]$

where A is the area of the container the gas is contained in and L is the height

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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