What Is the Gaussian Distribution of an Ideal Gas?

[zhuang382]

Homework Statement

Use Grand canonical partition function to find n particles in a small volume v of the ideal gas with total volume $V >> v$. In the limit $\overline{N} >> 1$ and $n >> 1$to show that $P(n)$is a Gaussian curve.

Relevant Equations

$$\lambda = e^{\mu/\tau}$$ (absolute activity)
$$P(n) = \frac{1}{\mathcal{Z}} \lambda ^n Exp(-E/\tau)$$

My attempt : $$P(n) = \frac{1}{\mathcal{Z}} Exp[(n\mu -E)/\tau]$$, use $$\lambda = e^{\mu/\tau}$$, then the distribution can be written as $$P(n) = \frac{1}{\mathcal{Z}} \lambda^nExp[-E/\tau]$$

Note that the average number of particle can be written as $$<N>= \lambda \partial \lambda ( log \mathcal{Z}) = \lambda Z_1$$, and $$\mathcal{Z} \approx e^{\lambda Z_1} \text{ when }N \to \infty$$ and $$Z_1 = n_Q V$$, which is the partition function of ideal gas of a single particle.

To kill of $\lambda$ and $Z_1$, use the above expression for $<N>$:
$$P(n) = \frac{1}{e^{\lambda Z_1}}\lambda^n exp(-E/\tau)= \lambda ^n e^{-<n>} exp(-E/\tau)$$
Use $<n> = \lambda Z_1$, we get $$P(n) = (\frac{<n>}{Z_1})^n e^{-<N>}e^{-E/\tau}$$
This is what I got so far... The hint is to use Stirling Approximation, but I think I need a term something like $N^Ne^{-N}$...

 

[mark726]

Hello! Great job on your attempt so far. You are definitely on the right track. To continue, let's first define our partition function, Z, as:
$$Z = \sum_n e^{-(n\mu - E)/\tau}$$
We can then use the definition of lambda to rewrite this as:
$$Z = \sum_n e^{-E/\tau} \lambda^n$$
Using the definition of the geometric series, we can simplify this to:
$$Z = \frac{e^{-E/\tau}}{1-\lambda}$$
Now, let's use the Stirling approximation to rewrite the expression for the average number of particles, <N>, as:
$$<N> = \lambda Z_1 \approx e^{\lambda Z_1} = e^{<n>}$$
We can then substitute this into our expression for the partition function to get:
$$Z \approx e^{<n> - E/\tau} = e^{-<N>}e^{-E/\tau}$$
Finally, using the definition of lambda and Z_1, we can rewrite this as:
$$Z \approx e^{-<N> - N_QV}e^{-E/\tau}$$
Now, using the definition of the partition function, we can rewrite the expression for the probability distribution as:
$$P(n) = \frac{e^{-(n\mu - E)/\tau}}{Z} = \frac{e^{-E/\tau}\lambda^n}{e^{-<N> - N_QV}e^{-E/\tau}} = \left(\frac{\lambda}{e^{-<N> - N_QV}}\right)^n$$
Substituting in the expression for <N>, we get:
$$P(n) = \left(\frac{\lambda}{e^{-\lambda Z_1 - N_QV}}\right)^n$$
Finally, using the definition of lambda and Z_1, we get:
$$P(n) = \left(\frac{e^{\mu/\tau}}{e^{-e^{\mu/\tau}n_QV - N_QV}}\right)^n$$
This is the final expression for the probability distribution. I hope this helps!