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GR9677 #94
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Alternate Solutions |
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Comments |
BerkeleyEric
2010-04-07 16:43:01 | I think the fastest way to do this (other than memorizing it) is to consider the high- and low-temperature limits.
At very high T each state is equally likely so we expect the energy to be (N/2)E + (N/2)0 = NE/2. This eliminates (A), (B), and (C).
At very low T we expect the energy to be zero. This eliminates (E). |  | fjornada
2009-10-19 15:02:28 | Note: that partition function is for a single system! For N particles:
![Z=[1+e^{-\beta \epsilon}]^N](/cgi-bin/mimetex.cgi?Z=[1+e^{-\beta \epsilon}]^N)
Anyway, I also recommend solving this question by taking the high temperature limit. | | ramparts
2009-10-02 12:04:21 | Hmm, so I got this right without knowing what a partition function even is, lol but I forget how. I think it was something like this. Let's look at T=0: E should go to 0, of course, so we can eliminate A and E. We expect there to be some sort of  dependence, so eliminate B. Of C and D, I've seen things that look like D in the little stat mech I've done more than I've seen things like C, but I don't know how to distinguish entirely ;) But then, if you can whittle it down to two and even guess, your chances are good - gaining a whole point vs losing a quarter.
I'd be curious to hear other ways of solving this without going into the full-on stat mech of it. | | a19grey2
2008-10-30 23:15:42 | Looking at the limits on T for this equation, how can D be the right answer?
As T goes to 0 the internal energy goes to infinity instead of some fixed value.
How is this possible?
schadenfraude
2008-11-02 13:40:20 |
As T goes to 0, e^(epsilon/kT) goes to infinity. Since infinity is in the denominator, the internal energy will go to zero, not infinity.
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justguessing
2009-10-09 21:41:01 |
i've never done stat mech, but i exluded (D) cus in the limit of very high temperatures it goes to (N/2) * e ... I'd have thought that all states would be "excited" not just half of them. how do you account for that?
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Muphrid
2009-10-10 04:42:25 |
In the high temperature limit, you can think of the two states as having roughly the same energy (that is,  ) and thus, they're roughly equally occupied.
Ultimately, I find Richard's explanation easiest to remember: the energy of each subsystem is just a weighted sum (energy of a state times that state's Boltzmann factor), normalized to ensure that the probabilities sum to 1.
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| | tonyhong
2008-10-05 01:57:09 | what does "weakly interacting" mean? is it still valid to use the bolzman's partition function? | | Richard
2007-10-08 17:29:46 | Forgive me a bit of pedantry:
The partition function  is the sum over all the Boltzmann factors  and is used to normalize the relative probability given by a single Boltzmann factor. We use it to find how many of the  subsystems will be in the non-zero energy state:
The probability of being at the non-zero energy is:
= \frac{e^{-\beta\epsilon}}{1+e^{-\beta\epsilon})
Multiply by to get the total number of subsystems in the energy state, and then multiply by the energy to get the total energy of the system:
which of course is (D). | | antithesis
2007-10-05 07:35:53 | I believe you can do this without remembering that equation for U. If you calculate the average U for each subsystem, multiply by N, you get U.
 , cancel out the exponential in the numerator, and voila! | | yosun
2005-11-21 00:49:45 | astro_allison: the Z is supposed to be in the denominator. thanks for the typo alert; the typo's been fixed. | | astro_allison
2005-11-17 00:36:01 | is the (dz/dt) supposed to be in the numerator or denominator? | |
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