GR9677 #94



Alternate Solutions 
QuantumCat 20140918 14:42:35  An alternative expression for that isn't nearly as cumbersome is where is the standard and is the usual partition function.  

Comments 
QuantumCat 20140918 14:42:35  An alternative expression for that isn't nearly as cumbersome is where is the standard and is the usual partition function.   sina2 20131002 02:01:11  Because all particles are same, so, first, we find partition function for one particle.
We know:
  Nezumi 20130622 08:48:14  In the solution, one sees the expression:
.
Where does the go after that? It seems to just disappear. Is this a typo?
Nezumi 20130622 09:27:02 
Ah... you get a factor of when you differentiate wrt ! Silly me :)

  wavicle 20111015 13:08:50  Maybe I missed something. I agree that as T > there should be equal number of particles in both states. But I disagree that, in general (or specific), we expect that as T > 0, the internal energy to go to zero. What about a BoseEinstein condensate? If I'm not mistaken, below the Bose temperature, all the matter is in the ground state. Doesn't this imply that said matter has the ground state energy? which is different from zero   BerkeleyEric 20100407 16:43:01  I think the fastest way to do this (other than memorizing it) is to consider the high and lowtemperature 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).
timtamm 20110827 11:39:52 
i love this way of doing it... i'm just wondering if I need to memorize the formulas anyway just in case a problem can't be solved in this method... though it seems like that is unusual it could be useful as well

  fjornada 20091019 15:02:28  Note: that partition function is for a single system! For N particles:
Anyway, I also recommend solving this question by taking the high temperature limit.   ramparts 20091002 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 fullon stat mech of it.
Tbot 20111111 13:54:32 
I think that taking the limit as T> INFINITY narrows our options down to (D) and (E). To narrow it further, let . The energy of the system should be zero (because no thermal energy will be able to excite any of subsystems). This eliminates (E), leaving (D) as the answer. Is the a correct approach?

  a19grey2 20081030 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 20081102 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.

justguessing 20091009 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?

Muphrid 20091010 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.

  tonyhong 20081005 01:57:09  what does "weakly interacting" mean? is it still valid to use the bolzman's partition function?
keradeek 20110923 18:48:35 
'weakly interacting' means that they do not affect the energy levels of one another, but they can exchange energy with each other.

  Richard 20071008 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 nonzero energy state:
The probability of being at the nonzero energy is:
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).
flyboy621 20101105 23:40:03 
For those who don't know, Great explanation!

  antithesis 20071005 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 20051121 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 20051117 00:36:01  is the (dz/dt) supposed to be in the numerator or denominator?  

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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 fullon stat mech of it.

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