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Homework Help: The # of Microstates of a given Macrostate

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data
    So this question has been bugging me because I can't begin to start it. The question is, prove that [tex]\Omega[/tex], the number of microstates of the combination of two physical states in thermal contact is a Gaussian of the energy of one of the states. [tex]\Omega[/tex] is given here as [tex]\Omega_1*\Omega_2[/tex] where those are the respective number of microstates in the first and second systems, respectively. They are functions of the # of particles, Energy of the state, and Volume, of course, and, knowing the total energy E, we can express [tex]\Omega_2(E_2) = \Omega_2(E-E_1)[/tex] thus we only have one variable which effects [tex]\Omega[/tex]

    I've got no real attempt at a solution, I'm hoping I can get a starting off point from you guys, because I can't seem to find where I can establish that [tex]\Omega[/tex] is a Gaussian.

    The actual question reads thusly:
    1(a) : Show that, for two large systems in thermal contact, the number [tex]\Omega[/tex] can be expressed as a Gaussian in the variable [tex]E_1[/tex]. Determine the root-mean-square deviation of [tex]E_1[/tex] from the mean value in terms of other quantities pertaining to the problem.
    1(b) Make an explicit evaluation of the root-mean-square deviation in the special case where the systems are classical gases.

    I'd really appreciate some help, I just feel that I can't get off the ground here.

    The book is Pathria's Statistical Mechanics
  2. jcsd
  3. Jan 11, 2010 #2
    So I've gotten a little ways on this, at least in some poor way, I think. Anybody out there with help?

    Basically I have two ideas. The first involves the identification that we can regard the energy of a single particle as a continuous random variable with some distribution resulting from the system. With this, we can take the thermodynamic limit and make use of the Central Limit theorem to say that the distribution of energies for a single particle will resemble a Gaussian about the maximum value. Using this and the fundamental assumption, we can say that the actual energy will share this value multiplied by N, and this multiplication brings out somewhere a factor of N in the deviation. But this is not explicit, and I'm afraid I can't make it so.

    The second involves the idea that a binomial distribution limits to a Gaussian. I need to show, however, that we can call two systems in contact a binomial distribution in terms of the energy... which I can't seem to do. From there we take the limit and its a Gaussian in the energy, we are all set. I just can't think of a way to show that the energy of two systems in contact will look binomial in low numbers. A binary spin system has this binomial characteristic inbuilt, but it is in terms of the number of spins pointing in a given direction, not in terms of the energy of one of the systems.

    Help guys, you are the biggest array of experts I know on this stuff and I'm sure there are some ideas out there. Its a tough problem, so any input would be great!
  4. Jan 17, 2010 #3


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    Homework Helper

    You know that [itex]S = k\ln \Omega[/itex], so you can invert and write [itex]\Omega = e^{S/k}[/itex]. You also know that [itex]S = S_1(E_1,V) + S_2(E_2,V)[/itex]. What do you know about entropy in equilibrium? What condition must be satisfied, and what if you're not quite at that condition?

    Hope this helps you get started.
  5. Jan 24, 2010 #4
    Thanks for the help: your idea is basically the point I used to solve the problem (the only one in the class, mind you :)

    The best way to do this (at least mine) is to use a combination of the aforementioned idea, Taylor expansions, and thermodynamic principles, along with a statement of the Central Limit Theorem to give it a little rigor. Clearly if we consider the expansion of [tex]\Omega[/tex] about the (supposed) maximum we will encounter a Gaussian, as the first term is a constant, the second is zero (max) and the third is some non-vanishing, negative quadratic (of course, because this is the max and we expect it to be the only one). The O(3) terms fall off as we go to the thermodynamic limit as [tex]\Delta[/tex] E is very small. Thus when we exponentiate we get a Gaussian. We can claim the central limit theorem helps as well, as in the thermodynamic limit we have some distribution, which should be guaranteed Gaussian if it is a random pdf.

    Of course, this is the same way we all show everything is a SHO, so its rigor is definitely debateable, but as long as we keep our disturbances small, pretty much everything looks Gaussian if its the log of some quantity.

    Thanks to Mute for the jumping off point!
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