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Statistical Physics number of particles with spin 1/2 in a B-field

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data

    therm52.jpg

    2. Relevant equations



    3. The attempt at a solution

    To be honest I'm clueless. I've missed a large amount of the course and just struggling to find any sources that explain this. I don't want the answer to the question, I want to figure that out for myself. What I really need is someone to point me in the right direction by either explaining the question in part or providing a good source of information such as a website or book.

    I don't even know how to visualize the question. Would it be as a sheet of N particles orientated along the magnetic field direction? The particles can be orientated either positive to the direction of the magnetic field or opposed to it. When it is said that there are n up does this mean that the number of rows could be thought of as n? Or am I completely wrong?

    Any and all information would be greatly appreciated. Thanks
     
  2. jcsd
  3. Apr 18, 2013 #2

    mfb

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    Something like that. You have N independent particles, each of them can be oriented along the magnetic field ("up", ms=+1/2) or in the opposite direction ("down", ms=-1/2). Spins are additive, so if you have n "up" and N-n "down", ...

    (b) is combinatorics (which particles are "up" and which are "down"?). Once you have that, you can use it to calculate (c) and (d).
     
  4. Apr 18, 2013 #3
    Thank you so much for the reply. That's made it much clearer.

    So part a is simply:

    m_s = n(1/2) + (N-n)(-1/2)
    m_s = n/2 - (1/2)(N-n)
    m_s = (1/2)(2n-N)

    I'm having a bit of trouble getting part b as a function of n. Because isn't the number of states simply (number of up + number of down)^N?

    This gives N^N as the answer. I'm obviously missing something.
     
  5. Apr 18, 2013 #4

    mfb

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    Let's consider a simple example: N=3, n=1. You have the options (up, down, down), (down, up, down) and (down, down, up), for a total of 3 options.
     
  6. Apr 21, 2013 #5
    I'm sorry but the concept still isn't clear to me. Whichever way I try to define the number of possible states I have to use the term N. What am I missing?

    Thanks again for the help :)
     
  7. Apr 21, 2013 #6

    mfb

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    Yes, N will show up in the result. Where is the problem?
     
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