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Answers to Physics 176 One-Minute Questionnaires Lecture date: February 24 and March 15, 2011 Why does the term ( ∂S R /∂E ) N,v ( - E S ) become e - E s / ( Kat R ) in cal- culating the probability for the system to be in a state with en- ergy E s ? Please take a look at my class lecture notes dated March 15, pages 10-12, where I hopefully answer this question to your satisfaction. If not, I will be glad to meet with you and go through the details. In general, how does one calculate the partition function Z You will soon see several examples for which the partition function can be evaluated in a concise mathematical form, say for the Einstein solid, the paramagnet, for an ideal gas consisting of molecules rather than atoms, and for a long polymer that assembles out of monomers. For most cases, however, one either tries to obtain an approximation to the exact partition function or one tries to evaluate the partition function numerically. The latter is fairly easy to do even for physical systems that are much too hard to study mathematically (analytically) in a useful way. How do we measure the mass of stars such as Betelgeuse? The preferred way to measure the mass of some remote star is to see if it is part of a binary star system (two stars orbiting each other, most stars in the sky are members of multiple star systems, our Sun is an exception). By observing several features of the stars such as the Doppler shifts of their light with respect to Earth (which gives the component of the speeds of motion toward or away from Earth) and especially whether one star passes in front of and behind the other star, one can deduce various details of the orbits of the stars and so the relative mass of the stars. Betelgeuse does not have an orbiting companion so instead one deduces its mass less accurately (in the 10-20% range) by using an important as- tronomical plot called the Hertzsprung-Russell diagram (see the Wikipedia article “Hertzsprung-Russell diagram”). This is a plot of a star’s luminosity (total light power emitted per second) versus the star’s surface temperature (which can be deduced from a knowledge of the distribution of light fre- quencies, since most stars act as ideal blackbody radiators of light). When one plots many stars on this diagram, rather remarkably the points do not lie spread out over an area but lie along a thin well-deﬁned curve known as 1

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