Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

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

Consider an ideal gas contained in a volume V at temperature T. If all particles are identical the Grand canonical partition function can be calculated using $$Z_g(V,T,z) := \sum_{N=0}^\infty z^N Z_c(N,V,T)$$ where $z$ is the fugacity, and $$Z_c(N,V,T):= \frac{1}{N!h^{3N}} \int_\Gamma e^{- \beta H(X)} dX$$ is the canonical partition function.

For identical particles I can compute the Grand canonical partition function. Now I am asked about two different kind of particles with masses $m_2= 2m_1$ and energies of $$h_1(p) = \frac{p^2}{2m_1}$$ and $$h_2(p) = \frac{p^2}{2m_2} + \Delta$$ where $\Delta > 0$ is a constant. The fugacitys are given as $z_{1,2}=e^{\beta \mu_{1,2}}$, where $\mu_{1,2}$ are the corresponding chemical potentials. The task is to find the partition function.

I am sorry to admit, but I have absolutly no clue where to start. Since no potential is present the Hamiltionian should be $$H(X) = \sum_{i=1}^{N_1} \frac{p_i^2}{2m_1} + \sum_{i=1}^{N_2} \frac{p_j^2}{2m_2} + N_2 \Delta$$ where $N_1+N_2=N$ are the numbers of particles of the differen kind. First I don't understand where that $\Delta$ comes from? And I have no idea how to continue. Simply inserting $H$ in $Z_c$ does not give any usefull results.

share|cite|improve this question

Here's my shot at it and my whole thought process so we are checking each other.

If we put in the Hamiltonian into the exponent and talk it through, we get the following: $$ \int e^{-\beta( \sum_i p^{2}_i/2m_1 +\sum_j p^{2}_j/2m_2)} $$ but what is the measure of integration? well they are non interacting, so we have to look at each individual particle, of each type, and then sum over all the possible momenta they could have $$ \int e^{-\beta( \sum_i p^{2}_i/2m_1 +\sum_j p^{2}_j/2m_2)}\Pi_i d^3p_i \Pi_j d^3p_j $$ where I have ignored anything to do with space recklessly. now my guess for the density of states part is $$ \frac{d^3p d^3 x}{\hbar^3}=\frac{4\pi V p^2 dp}{\hbar^3} $$ (there is no spatial dependence, so I just integrated over the whole volume) so really I had $$ \int e^{-\beta H}\frac{d^{3n}p d^{3n} x}{\hbar^{3n}} $$ and I need to put that in $$ \frac{(4\pi)^{N_1 +N_2}V^{N_1+N_2}}{\hbar^{3(N_1+N_2)}} \int_{0}^{\infty} e^{-\beta( \sum_i p^{2}_i/2m_1 +\sum_j p^{2}_j/2m_2)}(\Pi_i p^{2}_i dp_i)( \Pi_j p^{2}_j dp_j) $$ This is a ton of do-able individual integrals, all multiplied, because the sum in the exponents are multiplied down below. I have been really sloppy about numerical factors (like $e^{-\beta N_2 \Delta}$) and constants, sorry. I hope this helps/is correct.

share|cite|improve this answer
Thanks for your help. I think your thoughts are correct. However I doubt, this is the finial result I am suppost to get. There should be a way to get rid of the integrals I suppose. – Haatschii Jun 19 '12 at 8:32
Why do you think that? The integrals are variants on a Gaussian integral, you can have Alpha do it :) – kηives Jun 19 '12 at 17:06

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.