Solved Assignment 6 for Thermal Physics | PHYS 224, Assignments of Thermal Physics

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224 problem set 6 solutions

224 problem set 6 solutions Problem 3.4. A “miserly” system (A) can certainly be in thermal equilibrium with another system (B)--they just need to be at the sarne temperature. Usually, however, the equilibrium will not be stable. If system B is also “miserly,” then any small flow of energy from B to A will cause the temperature of B to increase while the temperature of A decreases. We then get a run-away effect, as more and more energy spontaneously flows from B to A. And if the initial fluctuation results in energy flowing from A to B, the run-away effect goes in the opposite direction. (This instability is sometimes offered as a proof that miserly systems cannot even exist. Imagine splitting a single miserly system into two parts, A and B; then since A and B cannot be in stable equilibrium with each other, the system as a whole is unstable. However, this “proof” has a loophole: If there are long-range forces between the particles in the original system, then we cannot mentally divide it into two miserly subsystems that interact only thermally. Gravitationally bound systems like stars can be miserly precisely because of the long-range gravitational forces between the particles.) But what if system B is not ‘miserly’? If it is a large “reservoir” whose temperature doesn’t change significantly when it absorbs or emits energy, then again any small transfer of energy from B to A will result in A becoming colder than B so we get a run-away effect. The only way for the equilibrium to be stable is if system 2 is “normal? 911 sufficiently smaul (more nrecierly ‘ise « safP ciently lew beat capacity! that a spontaneous transfer of energy from 8 to A causes B to cool off more more than A does. Then A will become a bit hotter than B and the energy will spontaneously flow back. Problem 3.8. For an Einstein solid in the low-temperature limit, U = Nee ‘/*? and therefore , Cy = Ne eer = Ne(Sa)e" _ nk( 5) eer ee aT kre)” NET Even though the prefactor ¢/#T blows up as T — 0, the whole expression goes to zero because of the exponentially small exponential factor. ‘To plot the heat capacity T gave the instruction Plot [(1/t)"2*Exp[(-1/t] ,{t,0,.2},PlotRange->All] to Mathematica, and it produced the graph below, showing the exponentially suppressed heat capacity as T-0 \ 0.15] val 0.4 0.075} 0.05} 9.025} Cy iNk