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 in solid , is very sharply peaked about the point where the quanta are evenly distributed between the two solids. The width of the peak is approximately for a system containing a total of energy quanta and oscillators.

We can get another, somewhat rougher, estimate of this width by first calculating the total number of microstates accessible to the system (that is, the total number of microstates summed over all possible macrostates), and then finding the number of microstates 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 of the peak is found from the area of the rectangular peak, according to

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

To find , 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 and , so we get

To find , we need to separate the solid into its constituent parts and , and assign quanta so that . The total number of microstates for this particular macrostate is then

The term in square brackets is for one of the solids, since it contains oscillators and energy quanta. Since the other solid is identical, we just square the result to get .

The width of the peak is then

[The Gaussian width is and since the rectangular approximation isn’t actually all that bad.]

The total number of macrostates in the two-solid system is , so the fraction of macrostates that have reasonably large probabilities is

For a macroscopic solid with , this fraction comes out to around . 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.

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blippShould change “microstates” to “macrostates” in the second last paragraph there. Had me confused for a little while.

gwrowePost authorFixed now. Thanks.