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Statistical mechanics -

  1. May 3, 2006 #1
    I've seen an approximation in a statistical mechanics book, given without any proof:
    [tex]\frac{<N!>}{<(N-n)!>} = <N^n> (1 + \mathcal{O}( \frac{1}{<N!>} ))[/tex]

    I've been trying to work out a proof, but I'm simply stuck :yuck: any ideas how it can be proved?
     
    Last edited: May 3, 2006
  2. jcsd
  3. May 3, 2006 #2

    0rthodontist

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    What do the pointy brackets mean? If you ignore them, the equation couldn't possibly be true. Maybe if it were 1 - O(1/N!) instead of 1 + O(1/N!).
     
  4. May 4, 2006 #3
    Pffff. Silence is hella better if you don't have anything useful.
    The equation is correct.
     
  5. May 4, 2006 #4

    Hurkyl

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    This leads one to wonder why you didn't keep silent. :tongue:
     
  6. May 4, 2006 #5
    If you take it this way, same goes for you as well :biggrin:
    But I did have a point in saying that: I do have a valid case, and case is still unsolved.
     
  7. May 4, 2006 #6

    Hurkyl

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    Yes, but I'm not the one telling people to keep quiet! And besides, 0rthodontist's response is useful:


    (1) It points out that you should define your terms. (what are the pointy brackets? :tongue: I have my guesses, but those require additional contextual information you have not supplied)

    (2) Are you sure your problem is correct? Teachers and books make mistakes.

    (3) Maybe 0rthodontist is right about what happens if you remove the pointy brackets. That might give significant information towards what you'd have to do to solve your problem.


    All of that aside, your snide response surely has turned people away from aiding on this problem. At the very least, it has put me off quite a bit -- normally I would have jumped right onto the problem the instant I saw it and asked if you had considered Stirling's formula.
     
  8. May 4, 2006 #7

    nrqed

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    Please keep in mind that even if a reply is not what you wanted or what you deem useful, this is still someone who took the time to reply, someone who had no obligation to do so and who gets nothing financially in return for trying to help.
     
  9. May 4, 2006 #8

    Physics Monkey

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    Hi gulsen,

    Before I can say anything I need to get some details from you. My assumption is that you are working in the grand canonical ensemble. Is this right?
     
  10. May 10, 2006 #9
    Well, I didn't say what to do. I said what would be better.
    Plus I guess what those brackets are obvious to anyone who has practice in QM either statistical mechanics. Otherwise, I'd end up defining what + means as well.
    This wasn't a homework problem and that pissed me off; the thread was moved to homework forum :/

    I'm not sure whether if it's correct or not, and that'd be revealed within this thread. Maybe it won't go that way, though. However, it was just what the book says, exactly.

    As for Stirling's formula, yes I did.

    Yes.
    And really huge thanks for trying to work-out a solution, unlike the other guys!
     
  11. May 10, 2006 #10

    Hurkyl

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    (1) People who don't practice QM or statistical mechanics might be able to help you. I've seen asymptotic analysis studied far more often in computer science than in any other subject... and guess what? That's what 0rthodontist studies.


    (2) I have enough familiarity to know that the pointy brackets probably mean to take the expected value of a random number. I would even guess from your notation that N is the random variable, and n is just some constant.

    But without knowing the distribution of the random variable, I can't do anything. In fact, [itex]\mathcal{O}(1 / \langle N! \rangle)[/itex] doesn't even make sense unless the distribution involves some parameter that doesn't even appear in your post.
     
  12. May 11, 2006 #11
    Actually, I was expecting help from physics department, and this was originally posted under physics forum, with the title "statistical mechanics -". You see, I'd try maths department as my last chance because they're so pedantic, You'll have to define whether if N is real or complex, explain your notation first, tell about the problem... These are all "obvious" to a physicist who has studied the topic.
    I'd try to ask a mathematician unless I get a solution from a fella studying physics. That'd save a bunch of time.

    Well, could we cut this off and let the thread live with it's title.
    Please.
     
  13. Jun 25, 2008 #12
    have you tried to use Stirling's approximation.
    it says that [tex]N![/tex] roughly is equal to [tex]N^{N}\cdot e^{-N}[/tex].

    i've seen through a lot of confusion in statistical mechanics books with that guy, also this is the first order version which it looks like you need. The second order mulitplies that whole thing by [tex]\sqrt{2\pi N}}[/tex]
     
  14. Jun 25, 2008 #13
    have you tried to use Stirling's approximation.
    it says that [tex]N![/tex] roughly is equal to [tex]N^{N}\cdot e^{N}[/tex].
    i've seen through a lot of confusion with that guy, also this is the first order version which it looks like you need. The second order mulitplies that whole thing by [tex]\sqrt{2\pi N}}[/tex]
     
  15. Jun 26, 2008 #14
    What is [tex]N[/tex] defined as? If it is just any number, it's wrong, as N^N is clearly always greater than N! (for positive integers, at least...).
     
  16. Jun 26, 2008 #15

    Mute

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    This thread is over two years old - I doubt the original poster is going to be returning any time soon.

    At any rate, I'm not sure what the average brackets are supposed to mean (I would speculate that perhaps the author was working in the grand canonical ensemble, and was working with average particle numbers, and the factorial symbol should be outside the angled brackets), but even leaving them out, the expression

    [tex]\frac{N!}{(N-n)!} \simeq N^n[/tex]

    is correct, as can be shown using Stirling's approximation. The next order term is negative, so I would guess that the book the original expression was taken from just didn't care about the sign, which might explain 0rthodontist's original statement that it was wrong. Of course, using stirling's approx I found the correction would be a multiplicative factor [itex]\exp(-n^2/N)[/itex], which of course Taylor expands to [itex]1 - n^2/N[/itex], which isn't clearly [itex]\mathcal{O}(1/N!)[/itex] - I would guess that either there is an error in that part of the original formula or terms I neglected from Stirling's approx combine to bump this correction to this order.
     
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