Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Boltzmann Distribution Derivation Question

  1. Oct 17, 2016 #1
    Hello, I have a question about Boltzmann Distribution.

    I wonder why partial N of Nj is 1 and partial U of Nj=Ej. because N is constant, partial N of Nj has to be 0 and Partial Nj of U has to be 0 as well.

    They are constants so, to make sense of the equation, alpha and beta have to be 0 but this is still absurd.

    Can anyone tell me why this is true?
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    With partial N of Nj you seem to mean ##\partial N\over \partial N_j##, right ?

    Well, if ##\ \ A = x+y+z\ \ ## then what is ##\ \ {\partial A\over \partial x}\ \ ## according to you ?
     
  4. Oct 17, 2016 #3
    if A is const, isnt it 0 where x is 0? or y,z is depedent on x where partial_x (y+z)=-1.
    N is not a function. it is a constant, according to the second and third screen shot. on the second screen shot, it says we are multiplying constants by zero.


    Well I thought, it would make sense if N is a function because we are dealing with some unknown N number of particles. then, partial_Nj (N)=1. but since the slide says Partial_Nj(N) is 0, so.. I got lost.
     
    Last edited: Oct 17, 2016
  5. Oct 18, 2016 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    N is ##\sum N_j## and as such has a partial derivative with respect to ##N_j##. Look up the definition of partial derivative. That doesn't change because a constraint (##N## is constant) is imposed.
    That would be the sheet with "13/28" and it's nonsense. (I understand your question a lot better now :smile:)

    If you look up the method of Lagrange multipliers perhaps it'll become a bit clearer, in the sense that at an extremum a necessary condition is that the gradients are linearly dependent. Hence these Lagrange multipliers ##\alpha## and ##\beta##.
    It certainly isn't true that all ##\ \ {\partial \phi\over \partial N_j} \ \ ## and ##\ \ {\partial \psi\over \partial N_j} \ \ ## are zero.
     
  6. Oct 18, 2016 #5
    Thanks a lot. I spent quite a lot of time on figuring out what went wrong in the note and what I did wrong.
    Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Boltzmann Distribution Derivation Question
Loading...