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If I am calculating the partition function for $H=cp$, ultrarelativistic gas in three dimensions. And by breaking down $d \Gamma$ into $dq$ and $dq$ and further using spherical coordinates I will get $d^3p =p^2 dp d\Omega $.

What bothers me now is that the integral for dp is taken $\int_{0}^{\infty}dp p^2 \exp(\beta cp) $ but $\int dq $ is indefinite and represents the volume.

So why is the dp intergral taken from 0 to inf and not some indefinite integral like dq?

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  • $\begingroup$ integrating from $0$ to $\infty$ is a definite integral? I'm a bit confused. The reason for the bounds just comes from the usual spherical coordinate transformation of the region $\endgroup$ Feb 5, 2019 at 18:37
  • $\begingroup$ IntertialObserver: sorry typo, I mean indefinite. dp is definite, dq is indefinite which I dont understand. $\endgroup$ Feb 5, 2019 at 18:38
  • $\begingroup$ The integral over space is definite though. You integrate over the entire volume. $\endgroup$ Feb 5, 2019 at 18:42
  • $\begingroup$ Then the partition function would be infinity since the space would be infinity, I dont really understand this now... $\endgroup$ Feb 5, 2019 at 18:45

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The $dq$ integral isn't indefinite, it's definite - it integrates over the whole space in which the gas is confined. Usually this is taken to be some finite box.

If the 'box' is infinite along one or more dimensions, then there is no meaningful sense in which it can be compressed / expanded along said dimensions, and so we can treat its volume as an 'infinite constant' which doesn't contribute to any observable.

E.g. in the canonical ensemble all observables are given by derivatives of $\log(Z)$ in which the infinite volume drops off. If the volume was finite, then one could think of varying with respect to it (e.g. compressing) to get observables like pressure.

Think of various observables you can compute and you'll see the volume never matters.

There is one exception I can think of: Let's say the $x$ direction is infinite, then what is the probability of finding a a particle in a particular finite interval? Here the answer will be 0, and it comes from dividing by the infinite volume:

$$ P = \frac{\intop _{\rm interval} dx}{\intop _{-\infty}^{\infty} dx} = 0$$

Hence the infinite volume reflected the fact that finding a particular particle in a finite range is probability 0.

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  • $\begingroup$ Okay so you are saying it should be definite. But when I chose a region of space then I write $\int d\Gamma =$ constant (Landau), but then I dont understand where the integration range from 0 to inf comes to the dp term $\endgroup$ Feb 5, 2019 at 20:49
  • $\begingroup$ The range for $|p|$ is $0$ to $\infty$ simply because $\boldsymbol{p}$ is unbounded. $\endgroup$ Feb 5, 2019 at 22:41

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