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Most probable macrostates of two Einstien solids

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider two Einstein solid A and B with [itex]N_{A} = 200[/itex] and [itex]N_{B} = 100[/itex]. Suppose now that the oscillations of solids A and B are different. The energies of oscillators in solid A are measured in units of [itex]ε_{A}[/itex], so that the energy of solid A is [itex]U_{A} = q_{A} ε_{A}[/itex]. Likewise, for solid B, the energy is [itex]U_{B} = q_{B} \epsilon_{B}[/itex]. Assume that [itex]ε_{A}[/itex] = 2[itex]ε_{B}[/itex], and that the total energy is 100 [itex]ε_{A}[/itex]. Compute most probable macrostate

    (Hint: to do this, you should write multiplicity/entropy of each solid and will have to and if/where one needs to take into account the fact that the units of energy of the two solids are different).


    2. Relevant equations

    (1) Sterlings approximation: [itex]\ln n! = n\ln n - n [/itex]

    (2) Multiplicity: [itex]Ω = \frac{(q+N-1)!}{(q)!(N-1)!} [/itex]
    There another forumla that says
    (3) [itex]Ω = (\frac{eq}{N})^{N} [/itex]

    3. The attempt at a solution


    [itex]U_{Total} = U_{A} + U_{B} = q_{A} ε_{A} + q_{B} ε_{B} = 100 ε_{A}[/itex]
    or
    [itex]q_{A} ε_{A} + 2 q_{B} ε_{A} = 100 ε_{A}[/itex]
    [itex]q_{A} + 2 q_{B} = 100 [/itex]

    This means that for each [itex]Ω_{A}(q_{A}) [/itex] microstates of energy [itex]q_{A}[/itex], there are [itex]Ω_{B}(50 -\frac{q_{A}}{2})[/itex] microstates accesible to B.

    The total microstate for partition of energy [itex] q_{A}, q_{B} [/itex] is
    [itex]Ω_{A}(q_{A})Ω_{B}(q_{B})[/itex]

    So the we can use (3),
    [itex]Ω_{A}(q_{A})Ω_{B}(q_{B}) = (\frac{e q_{A}}{N_{A}})^{N_{A}}
    (\frac{e(50 - \frac{q_{A}}{2})}{\frac{N_{A}}{2}})^{\frac{N_{A}}{2}}

    [/itex]


    I have no idea if what I'm doing is right. Any insight would be really appreciated, thanks.
     
  2. jcsd
  3. Jan 29, 2013 #2
    Figured it out. For anyone who finds this thread:
    Once you have [itex]Ω_{A}[/itex] and [itex]Ω_{B}[/itex]
    Use

    [itex]S = kln(Ω_{A}) +kln(Ω_{B})[/itex]

    Then we know that the most probable macrostate is when,

    [itex] \frac{δS}{δq_{A}} = 0[/itex]

    So just solve for [itex]q_{A}[/itex],then use it to solve for [itex]q_{B}[/itex].

    I got
    [itex]q_{A} = \frac{200}{3}[/itex]
    [itex]q_{B} = \frac{50}{3}[/itex]
     
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