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Recently I had gone through a statistical physics course and I learned about phase space. A point in phase space representing the state of the system of $``N"$ particles. $``N"$ can be any number. And if we have the phase space distribution function, say $f(p,q)$, $p$ is the canonical momentum and $q$ is the generalized coordinate, then $$f(p,q)d\tau$$ gives the probability of finding the system in that particular state in the $dp$$dq$ range. $$\int f(p,q)d\tau=1$$ But now after reading Landau's Physical Kinetics which is Vol. $10$. It is stating otherwise.Landau Physical Kinetics Vol. 10 Why is there two other definition of distribution functions in phase space? The doubt is in the product of distribution function and the volume element, sometimes it says the mean number of molecules and sometimes it says the probability. How is this possible with phase space distributions?

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  • $\begingroup$ I am having trouble understanding your question. What two other definitions are you referring to exactly? $\endgroup$ – By Symmetry Jun 20 '18 at 12:26
  • $\begingroup$ @BySymmetry I have updated the question. My doubt was in understand the product in the distribution function and the phase space volume element, am I suppose to understand it as probability or number of molecules? $\endgroup$ – Jyotishraj Thoudam Jun 20 '18 at 12:33
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These are the same definitions. I think there is a misunderstanding in the use of the language when you say "product of distribution function and the volume element"

The first equation you have written defines the phase space to present it in 2D, based upon the coordinates and the momentum, for the N particles. The second equation is the statement that you have a normalized probability distribution over that phase space so that the densities are actual probabilities that do not need re-normalization; this phase space exhausted by the integral over $d\tau$. In the passage you cite the 'p' and 'q' are defined to represent the same quantities. Since $f(p,q)d\tau$ ($f d\tau$ in the passage) is a continuous distribution, we need to refer to expected value (mean value as stated in that passage since I assume uniformity in the phase space) for a region. This product you see is the need to get a value from the continuous probability distribution into a probability for the space. Over the trajectory of the phase space, some molecules will be in the state f(t,p,q) and some in f(t',p',q') and the probabilities will not be the same.

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  • $\begingroup$ Phase space can have any dimension. The first equation does not necessarily mean $2D$ because $f(p.q)d\tau$ general expression of every phase space distribution multiplied by $d\tau$ which is $d\tau = dpdq$. My doubt is still there, if $f$ is the probability distribution then $fd\tau$ should be a probability, but in the attached image of Landau's book it is referring to the "Mean Number of Molecules" which is far from representing a probability. $\endgroup$ – Jyotishraj Thoudam Jun 28 '18 at 13:44
  • $\begingroup$ @jyotishrajthoudam, this for 2D can be an example, the microstate definition can be in higher dimensions. There can be more than $(p,q)$ in the phase space, depends upon the macrostate definition. The 'density' in the phase space is what is being referred to. If that phase space is normalized (sums to one) a region in the phase space is the expected (mean for uniform) region identified by the integral. $\endgroup$ – Vass Jun 28 '18 at 14:11
  • $\begingroup$ @jyotishrajthoudam, reading over your question again a couple of times, I now just understood exactly what you are referring to... if the integral covers the whole phase space, the mean number of molecules would be N, rather than a subset. The probability is used to calculate and expected number of molecules as in the manner used to find the number of 4s in a certain number of dice tosses. $\endgroup$ – Vass Jun 28 '18 at 14:23

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