Interacting Einstein solids: rectangular peak in multiplicity graph

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 2.22.

In a system composed of two interacting Einstein solids, the multiplicity function, which gives the number of microstates as a function of the number of energy quanta {q_{A}} in solid {A}, is very sharply peaked about the point where the quanta are evenly distributed between the two solids. The width of the peak is approximately {q/\sqrt{N}} for a system containing a total of {q} energy quanta and {N} oscillators.

We can get another, somewhat rougher, estimate of this width by first calculating the total number of microstates {\Omega_{total}} accessible to the system (that is, the total number of microstates summed over all possible macrostates), and then finding the number of microstates {\Omega_{mp}} in the most probable macrostate (even distribution of energy quanta). If we then assume that the peak in the graph is rectangular rather than Gaussian, then the width {w} of the peak is found from the area of the rectangular peak, according to

\displaystyle  w=\frac{\Omega_{total}}{\Omega_{mp}} \ \ \ \ \ (1)

To apply this, let’s consider a simple system where the two solids each have {N} oscillators and the total number of quanta is {q=2N}. For large {N} and {q}, we can use the approximation derived earlier for the number of microstates in a solid with {q} quanta and {n} oscillators (I’m using a lowercase {n} here to distinguish it from the {N} in the problem):

\displaystyle  \Omega\approx\sqrt{\frac{n}{2\pi q\left(q+n\right)}}\left(\frac{q+n}{q}\right)^{q}\left(\frac{q+n}{n}\right)^{n} \ \ \ \ \ (2)

To find {\Omega_{total}}, we can combine the two solids into one composite solid since we’re interested in all microstates, no matter how the quanta are divided up between the two solids. In this case {q=2N} and {n=2N}, so we get

\displaystyle   \Omega_{total} \displaystyle  \approx \displaystyle  \sqrt{\frac{2N}{2\pi\left(2N\right)\left(4N\right)}}\left(\frac{4N}{2N}\right)^{2N}\left(\frac{4N}{2N}\right)^{2N}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \frac{2^{4N}}{\sqrt{8\pi N}} \ \ \ \ \ (4)

To find {\Omega_{mp}}, we need to separate the solid into its constituent parts {A} and {B}, and assign quanta so that {q_{A}=q_{B}=\frac{q}{2}=N}. The total number of microstates for this particular macrostate is then

\displaystyle   \Omega_{mp} \displaystyle  = \displaystyle  \left[\sqrt{\frac{N}{2\pi\left(N\right)\left(N\right)}}\left(\frac{2N}{N}\right)^{N}\left(\frac{2N}{N}\right)^{N}\right]^{2}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{2^{4N}}{4\pi N} \ \ \ \ \ (6)

The term in square brackets is {\Omega} for one of the solids, since it contains {n=N} oscillators and {q_{A}=N} energy quanta. Since the other solid is identical, we just square the result to get {\Omega_{mp}}.

The width of the peak is then

\displaystyle  w=\frac{4\pi N}{\sqrt{8\pi N}}=\sqrt{2\pi N} \ \ \ \ \ (7)

[The Gaussian width is {q/\sqrt{N}=2\sqrt{N}} and since {\sqrt{2\pi N}\approx2.5\sqrt{N}} the rectangular approximation isn’t actually all that bad.]

The total number of macrostates in the two-solid system is {q+1=2N+1\approx2N}, so the fraction of macrostates that have reasonably large probabilities is

\displaystyle  \frac{\sqrt{2\pi N}}{2N}=\sqrt{\frac{\pi}{2N}} \ \ \ \ \ (8)

For a macroscopic solid with {N=10^{23}}, this fraction comes out to around {4\times10^{-12}}. This shows just how unlikely it is that a macroscopic system will ever be found with an energy distribution significantly different from the most probable case.

3 thoughts on “Interacting Einstein solids: rectangular peak in multiplicity graph

  1. Pingback: Entropy | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *