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Homework Help: Transformation of expoential to hyperbolic

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data

    The book has it exp(-MgbH/KT) =(sinh(2S+1)x/2)/(sinh(x/2)) for M=2S+1, and x = gbH/(kt).

    2. Relevant equations



    3. The attempt at a solution

    I'd have it as cosh(Mx)-sinh(Mx). How did they get the above result? Help please. Thanks.
     
    Last edited: Jul 28, 2011
  2. jcsd
  3. Jul 27, 2011 #2

    Dick

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    If x=0 and y=1 then the book's formula gives exp(0)=0. 1=0 is nonsense. You don't need to ask where it comes from. Or is there a typo in there? Your formula is certainly correct.
     
    Last edited: Jul 27, 2011
  4. Jul 27, 2011 #3
    Sorry I have made a mistake in writing that down. The book actually have it as exp(-MgbH/KT). M = 2S+1. The book let gbH/KT be x. And then it went ahead to say exp(-MgbH/KT) = sinh[(2S+1)x/2]/sinh(x/2). Which is sinh[Mx/2]/sinh(x/2) . Sorry for the inconvenience.
     
  5. Jul 28, 2011 #4

    uart

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    Well for [itex]x=2[/itex] and [itex]M=1[/itex] that equation yields [itex]\exp(-2) = 1[/itex]?

    Here I'm assuming k=K.

    BTW. Does (sinh(2S+1)x/2) mean (x/2) sinh(2S+1). Or do you mean sinh((2S+1)x/2) ?
     
  6. Jul 28, 2011 #5
    K is a boltzmann constant. M is magnetisation. b = beta.
    I meant the last one. The equation should read: exp(-MgbgH/KT) = sinh[(2S+1)x/2]/sinh(x/2) with M and x as defined above.
    Cheers.
     
    Last edited: Jul 29, 2011
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