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Derivation of mazwell boltzman statistics.

  1. Sep 22, 2013 #1
    Let number of energy levels be e(subscribe 0), e(subscribe 1)...........e(subscribe r)
    N be the total number of particles and n(subscribe i) be the number of particle in with energy level where i =0,1,2,.......r
    Let Ω be the number of ways attaining a given microstate
    Ω=N!/[ n(subscribe 0)!*n(subscribe 1)!*..........n(subscribe r)! ]
    For the most probable state Ω should be maximum i.e.,
    dΩ/d(n subscribe i) =0

    lnΩ=ln[N!]-ln [summation of n(subscribe i)! ] where i=0,1,.......r.
    Applying stirlings approximation
    lnΩ={ N*ln[N]-N }- [summation of n(subscribe i)*ln [n(subscribe i)]-n(subscribe i)
    Taking derivative with respective n(subscribe i)
    Let n(subscribe i)=n(i)

    (1/Ω)*dΩ/dn(i)=-[summation of ln(n(i)) ]

    At maximum dΩ/dn(i) =0 => dΩ =- summation of lnn(i) dn(i) =0
    => summation of lnn(i) dn(i)=0 let this be first equation
    Total energy is constant i.e.,
    Summation of n(i)*e(i) = constant, => summation of e(i)dn(i)=0 ( 2nd equation) because the summation of change in the total energy level of each state is to zero

    Total number of particle is constant
    Summation of dn(i)=0 be the third equation
    By lagranges method of undetermined multiplication
    Multiply 2nd equation by b and 3rd equations by a and add the three equations
    We get => summation of [ ln(n (i)) +a + b*e(i) ] d( ni ) =0

    And the next step is [ln(n(i)) +a + b*e(i) ] =0....., how is it?

    We can't divide both sides dn(i) since the equation is in summation.
     
    Last edited: Sep 22, 2013
  2. jcsd
  3. Sep 22, 2013 #2
    Please use the underscore key "_" instead of the word "subscribe". It's standard notation and much easier to read. "e_1" instead of "e(subscribe 1)" and "e^1" instead of "e(superscribe 1)"
     
  4. Sep 22, 2013 #3

    stevendaryl

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    Even better, write [ i t e x ] e_1 [ / i t e x ] or [ i t e x ] e^1 [ / i t e x ] (without the spaces) to make [itex]e_1[/itex] and [itex]e^1[/itex]
     
  5. Sep 22, 2013 #4

    dlgoff

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    Even the https://www.physicsforums.com/Nexus/editor/superscript.png [Broken] and https://www.physicsforums.com/Nexus/editor/subscript.png [Broken] from the "Go Advanced" method is better. e.g. e1 or e1
     
    Last edited by a moderator: May 6, 2017
  6. Sep 22, 2013 #5

    stevendaryl

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    Here's what I think you're saying, but it's hard to read:

    Let [itex]\Omega = \dfrac{N!}{n_1! n_2! \ldots n_r!}[/itex]

    Using Sterling's approximation, we can write this as:

    [itex]ln(\Omega) = ln(N!) - \sum_i n_i ln(n_i) - n_i[/itex]

    Now, we want to maximize this subject to the constraints:
    • [itex]\sum_i n_i = N[/itex], the total number of particles.
    • [itex]\sum_i n_i E_i = E[/itex], the total energy.

    So we use Lagrange multipliers, and maximize
    [itex]ln(\Omega) - a (\sum_i n_i) - b (\sum_i n_i E_i)[/itex]
    [itex]= ln(N!) - \sum_i (n_i ln(n_i) - n_i - a n_i - b n_i E_i)[/itex]

    Now, to maximize, we vary each of the [itex]n_i[/itex] and set the variation to zero. The variation is:

    [itex]\sum_i \delta n_i (ln(n_i) - a - b E_i)[/itex]

    For this to be zero regardless of the values of [itex]\delta n_i[/itex], each term in the summation must be zero. So for every [itex]i[/itex], we must have:

    [itex]ln(n_i) - a - b E_i = 0[/itex]
     
  7. Sep 27, 2013 #6
    Even if ln(ni)-a-be¡ doesn't equal to zero, then there is chance that summation will be zero doesn't it?why don't we consider that case?
     
    Last edited: Sep 27, 2013
  8. Sep 27, 2013 #7

    Bill_K

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    Because, as stevendaryl said, the summation must be zero for arbitrary δni, and this can only be true if each term is zero separately.

     
  9. Sep 28, 2013 #8
    Say for example we have ab+cd +ef =0
    We have three cases
    1st is (a,c,e)=0
    2nd one is (b,c,e)=0
    3rd is say for example let ab=3 , cd=5 and ef=-8

    What we have done is regardless if value of del(n¡), the second part will zero, yeah i get that but what about other possible cases and why did we took this particular line and preceded?
     
    Last edited: Sep 28, 2013
  10. Sep 28, 2013 #9

    stevendaryl

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    The equation we want to solve is:

    [itex] \sum_i \delta n_i (ln(n_i) - a - b E_i) = 0[/itex]

    We want this to be true for EVERY possible value of [itex]\delta n_i[/itex]. So let's pick a particular value of [itex]\delta n_i[/itex]:

    [itex]\delta n_{17} = 1[/itex]
    [itex]\delta n_0 = \delta n_1 = \ldots = \delta n_{16} = \delta n_{18} = \ldots = 0[/itex]

    So for that value of [itex]\delta n_i[/itex], the sum simplifies to:

    [itex]ln(n_{17}) - a - b E_{17} = 0[/itex]

    The only way to make the sum equal to zero for every possible value of [itex]\delta n_i[/itex] is if every term is zero.
     
    Last edited: Sep 28, 2013
  11. Sep 28, 2013 #10
    You made a mistake the total number of particle is constant therefore if dn_17th is 1 there must be another state dn_i=(-1) that is when you remove on particle from a state you have to add it up to a another system so that
    Summation of dn_i=0
     
  12. Sep 28, 2013 #11

    stevendaryl

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    You're forgetting what the whole point of lagrange multipliers is: Introducing lagrange multipliers allows you to treat the parameters as independent, for the purposes of minimization.
     
  13. Sep 28, 2013 #12

    stevendaryl

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    There's one way to see how this works. To start off with, you have [itex]r[/itex] parameters, [itex]n_1, n_2, \ldots n_r[/itex] and you have two constraints: the total number of particles, and the total energy. So there are [itex]r-2[/itex] independent parameters.

    Now, we introduce two new parameters, the two Lagrange multipliers. We don't introduce any new constraints. So now the situation is that we have [itex]r+2[/itex] parameters, and we have [itex]2[/itex] constraints. So there are [itex]r[/itex] independent parameters. We can let those [itex]r[/itex] independent parameters be [itex]n_1, n_2, \ldots n_r[/itex]. Then the two lagrange multipliers are determined by those.

    So introducing lagrange multipliers allows us to treat the [itex]n_i[/itex] as independent.
     
  14. Sep 28, 2013 #13

    stevendaryl

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    We can work out the very simplest case: just two energy levels [itex]e_1[/itex] and [itex]e_2[/itex]. In this case, we don't need lagrange multipliers, because the constraints uniquely determine [itex]n_1[/itex] and [itex]n_2[/itex]:

    [itex]n_1 + n_2 = N[/itex]
    [itex]n_1 e_1 + n_2 e_2 = E[/itex]

    So

    [itex]n_1 = \dfrac{N e_2 - E}{e_2 - e_1}[/itex]
    [itex]n_2 = \dfrac{E - N e_1}{e_2 - e_1}[/itex]

    The Lagrange multiplier approach gives us:

    [itex]ln(n_1) - a - b e_1 = 0[/itex]
    [itex]ln(n_2) - a - b e_2 = 0[/itex]

    So

    [itex]n_1 = e^{a+b e_1}[/itex]
    [itex]n_2 = e^{a + b e_2}[/itex]

    We still have the same constraints:

    [itex]n_1 + n_2 = N[/itex]
    [itex]n_1 e_1 + n_2 e_2 = E[/itex]

    but these become constraints on [itex]a[/itex] and [itex]b[/itex]:

    [itex]e^{a + b e_1} + e^{a + b e_2} = N[/itex]
    [itex]e_1 e^{a + b e_1} + e_2 e^{a + b e_2} = E[/itex]
     
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