# Most probable macrostates of two Einstien solids

1. Jan 27, 2013

### Yuriick

1. The problem statement, all variables and given/known data

Consider two Einstein solid A and B with $N_{A} = 200$ and $N_{B} = 100$. Suppose now that the oscillations of solids A and B are different. The energies of oscillators in solid A are measured in units of $ε_{A}$, so that the energy of solid A is $U_{A} = q_{A} ε_{A}$. Likewise, for solid B, the energy is $U_{B} = q_{B} \epsilon_{B}$. Assume that $ε_{A}$ = 2$ε_{B}$, and that the total energy is 100 $ε_{A}$. 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: $\ln n! = n\ln n - n$

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

3. The attempt at a solution

$U_{Total} = U_{A} + U_{B} = q_{A} ε_{A} + q_{B} ε_{B} = 100 ε_{A}$
or
$q_{A} ε_{A} + 2 q_{B} ε_{A} = 100 ε_{A}$
$q_{A} + 2 q_{B} = 100$

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

The total microstate for partition of energy $q_{A}, q_{B}$ is
$Ω_{A}(q_{A})Ω_{B}(q_{B})$

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

I have no idea if what I'm doing is right. Any insight would be really appreciated, thanks.

2. Jan 29, 2013

### Yuriick

Figured it out. For anyone who finds this thread:
Once you have $Ω_{A}$ and $Ω_{B}$
Use

$S = kln(Ω_{A}) +kln(Ω_{B})$

Then we know that the most probable macrostate is when,

$\frac{δS}{δq_{A}} = 0$

So just solve for $q_{A}$,then use it to solve for $q_{B}$.

I got
$q_{A} = \frac{200}{3}$
$q_{B} = \frac{50}{3}$