-1
$\begingroup$

As far as I understand, when calculating the partition function we are summing over all of phase space. So, for ideal gas for example we have

$$H =\sum\limits_i\frac{p^2_i}{2m}$$

and so

$$Z=\int\limits_{-\infty}^\infty e^{\sum\limits_i \beta p^2_i/2m}d^{3N}qd^{3N}p = \int\limits_{-\infty}^\infty d^{3N}q \cdot \int\limits_{-\infty}^\infty e^{\sum\limits_i \beta p^2_i/2m}d^{3N}p$$

at this point the textbook makes the claim that $$ \int\limits_{-\infty}^\infty d^{3N}q = V^N$$

I would understand this claim if the integrand had some indicator function that zeroes all states outside our "box", but I don't see any, so how does that make sense? If this is some "approximation" we do to compensate for the approximated hamiltonian (no interactions between particles), is there a general rule for when I can just ignore a region in phase space?

$\endgroup$
7
  • 2
    $\begingroup$ Which textbook? If the particles are in a box with volume $V$, then indeed you integrate over an indicator function. $\endgroup$ Commented Aug 3 at 16:05
  • $\begingroup$ a general rule... probably something close to "physicists are notoriously sloppy/imprecise when it comes to maths" $\endgroup$
    – Kyle Kanos
    Commented Aug 3 at 18:00
  • $\begingroup$ @TobiasFünke My course uses many books, and I can't find the one the derivation is taken from, but it is very simillar to this: farside.ph.utexas.edu/teaching/sm1/lectures/node65.html Could you maybe point out where would I insert the indicator function? $\endgroup$ Commented Aug 3 at 18:12
  • $\begingroup$ Are you sure of your formula? For an ideal gaz, you should rather get $V^N$. Your domain is just a cartesian product so the integrals factorize. $\endgroup$
    – LPZ
    Commented Aug 3 at 18:43
  • $\begingroup$ @LPZ Yes that is correct, I made a typo, thank you. However my question remains. $\endgroup$ Commented Aug 3 at 19:09

1 Answer 1

2
$\begingroup$

On the page linked in a comment, there is a clear indication that one is working with $N$ particles in a volume $V$.

Statistical Mechanics of $N$-particle systems is always done by confining the $N$ particles in a finite volume V because we can only work with a finite number of particles (to avoid divergencies), and a finite number of particles in an infinite volume would imply a vanishing particle density.

However, such a confinement can be obtained in different ways. The most common method used in modern textbooks is confining the positions inside a finite volume $V$ with a one-body potential equal to zero inside the volume and equal to $ +\infty$ outside. In such a case, the integration of the positions could be formally extended to the whole space, But all points outside the volume will give a vanishing contribution and then the integration could be limited to the volume $V^N$. An alternative possibility, exploited in the past, would be to add a smooth confining potential, like a harmonic potential, such that the average number density in the central region is close to the number density of the state one is interested in. In such a case, integration on the position variables is performed on the whole space. In both cases, thermodynamics is recovered only at the thermodynamic limit. I.e., in the limit where volume and number of particles both diverge by keeping their ratio constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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