Converting Sums to Phase Space Integrals in Statistical Physics - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-17T21:12:11Z https://physics.stackexchange.com/feeds/question/305855 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/305855 3 Converting Sums to Phase Space Integrals in Statistical Physics Klein Four https://physics.stackexchange.com/users/46051 2017-01-18T08:26:05Z 2019-01-27T15:26:44Z <p>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</p> <p>$$Z_1 = \sum_n e^{-\beta \epsilon_n} \approx \frac{1}{\left(\Delta p\right)^3} \int e^{-\beta \left|\textbf{p}\right|^2/2m} \, d^3\textbf{p}.$$</p> <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 </p> <p>$$Z_1 \approx \frac{8V}{h^3} \int_0^{\infty}\int_0^{\infty}\int_0^{\infty} e^{-\beta \left|\textbf{p}\right|^2/2m} \, dp_x \, dp_y\, dp_z,$$</p> <p>or equivalently (and more naturally with periodic boundary conditions),</p> <p>$$Z_1 \approx \frac{V}{h^3} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-\beta \left|\textbf{p}\right|^2/2m} \, dp_x \, dp_y\, dp_z.$$</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>I hope this question is clear, but let me know if it's not, and I can edit it.</p> https://physics.stackexchange.com/questions/305855/-/457067#457067 1 Answer by Pavel Penshin for Converting Sums to Phase Space Integrals in Statistical Physics Pavel Penshin https://physics.stackexchange.com/users/112232 2019-01-27T14:45:26Z 2019-01-27T14:45:26Z <p>lets define <span class="math-container">$\Gamma(E)$</span> as the number of states till energy E, and define <span class="math-container">$\frac{d\Gamma(E)}{dE}=g\left(E\right)$</span> .</p> <p>By definition in a 3 dimensional world </p> <p><span class="math-container">$\Gamma\left(E\right)=\frac{1}{h^{3}}\iint d^{3}pd^{3}q$</span></p> <p>why the factor <span class="math-container">$h^{3}$</span>? because for every dimensin you have <span class="math-container">$\Delta x\Delta p\propto h$</span></p> <p>here you need the number of states for a given energy E :</p> <p>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 <span class="math-container">$\left(h^{3}\right)$</span>. </p> <p>now note that <span class="math-container">$Z=\sum_{i}g_{i}e^{-\beta E_{i}}$</span>where <span class="math-container">$g_{i}$</span>is the degeneracy of each E_{i} </p> <p>when taking the energy spectrum to a continuous spectrum you get that </p> <p><span class="math-container">$Z=\int g\left(E\right)e^{-\beta E}dE$</span></p> <p>However as we definded <span class="math-container">$g\left(E\right)$</span> above we can see that <span class="math-container">$g\left(E\right)dE=d\Gamma\left(E\right)$</span></p> <p>and by the definition of <span class="math-container">$\Gamma\left(E\right)$</span> we get that <span class="math-container">$d\Gamma\left(E\right)=\frac{1}{h^{3}}d^{3}pd^{3}q$</span></p> <p>plug into Z and get </p> <p><span class="math-container">$Z=\frac{1}{h^{3}}\iint e^{-\beta E(q,p)}d^{3}pd^{3}q$</span></p> <p>In case of the ideal gas that is in the gravitational field this is pretty straight forward:</p> <p><span class="math-container">$Z_{1}=\frac{1}{h^{3}}\iint e^{-\beta\frac{p^{2}}{2m}-\beta mgz}d^{3}pd^{3}q$</span></p> <p><span class="math-container">$Z_{1}=\frac{1}{h^{3}}\left[4\pi\int_{0}^{\infty}e^{-\beta\frac{p^{2}}{2m}}p^{2}dp\right]*\left[A\int_{0}^{L}e^{-\beta mgz}dz\right]$</span></p> <p>where A is the area of the container the gas is contained in and L is the height</p>