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Partition function of a gas of $N$ identical classical particles is given by

$$ Z~=~\frac {1}{N! h^{3N}} \int \exp[-\beta H(p_1.......p_n, x_1....x_n)]d^3p_1...d^3p_n,d^3x_1...d^3x_n $$

in this above equation we use $N!$ as the total number of sub-systems of a system of identical particles. and $ h^{3N} $ to make the partition function dimensionless. I am not clear how $ h^{3N} $ is used to make it dimensionless.

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3 Answers 3

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The easiest way to think about it is that $\exp(\dots)$ is just a number and doesn't affect the dimension. However, you still have $3N$ factors of the momenta and the position lying around that will give you dimensions of [Length x Momentum]${}^{3N}$. Planck's constant has the units of Length x Momentum, so the $3N$ factors of $h$ cancel the $3N$ factors coming from the integral.

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Notice that the $e^{-\beta H}$ is dimensionless, while each factor of $dp$ contributes one factor with the dimensions of momentum while each $dx$ contributes one factor with the dimensions of length. Therefore each factor $dp dx$ contributes a factor with dimensions of angular momentum. Since there are $3N$ of these factors (N particles and 3 dimensions) in the integration measure, the integral has a total dimension of angular momentum to the power of $3N$. On the other hand, $h$ has dimensions of angular momentum, so dividing by $h^{3N}$ makes the full expression dimensionless.

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and why did we divide the integration by N!. –  Bibi Fatema Feb 6 '13 at 20:16
    
Read, for example, this paper: link.springer.com/article/10.1023%2FA%3A1015161825292?LI=true . –  Yvan Velenik Feb 7 '13 at 13:05
    
@fatema Sorry for the late reply. The factor is basically because for identical particles, any permutation of the particles leaves the state of the system the same, and since there are $N!$ permutations of $N$ particles, we divide by $N!$ to make sure we don't overcount states in the partition function. –  joshphysics Feb 7 '13 at 16:53
    
Thank u sir @ joshphysics –  Bibi Fatema Feb 7 '13 at 17:24
    
i am failed to access into this side @ Yvan Velenik. if anyone has this paper will u post it for me? –  Bibi Fatema Feb 7 '13 at 17:42

The phase space of coordinates and momenta components of the N particles has certain size in that space, there is a number of microstates inside this size determined by the cell size in the quantized phase space (due to uncertainity h for each degree of freedom dx dp and for 3N dergrees of freedom of the N particles) it will be h^3N. The number of possible arrangements of these N distinguishable, particles is N! repeated in the numerator), but they are indistigwishable we must correct it by dividing the quantity b N!.

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