# Most probable macrostates of two Einstein solids

• Yuriick
In summary, the problem involves two Einstein solids, A and B, with different numbers of oscillators and energies measured in different units. The total energy is known to be 100 times the energy of solid A, and the task is to compute the most probable macrostate for the system. Using the formula for multiplicity, Ω, and applying the principle of maximum entropy, the values of q_{A} and q_{B} can be determined to be q_{A} = \frac{200}{3} and q_{B} = \frac{50}{3}, respectively.
Yuriick

## Homework Statement

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).

## Homework 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) $Ω = (\frac{eq}{N})^{N}$

## 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.

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

## What is a macrostate?

A macrostate is a set of conditions that define the overall state of a system, such as the total energy or number of particles.

## What are Einstein solids?

Einstein solids are idealized models used to study the behavior of solids at low temperatures. They consist of a finite number of particles (atoms or molecules) that are confined to a lattice structure and can only vibrate with discrete energy levels.

## How are the most probable macrostates of two Einstein solids determined?

The most probable macrostates of two Einstein solids are determined by considering the total number of microstates (possible arrangements of particles) that correspond to a particular macrostate. The macrostate with the highest number of microstates is considered the most probable.

## What is the significance of the most probable macrostate in two Einstein solids?

The most probable macrostate in two Einstein solids represents the state that the system is most likely to be in at a given time. It can provide valuable insights into the behavior of solids at low temperatures and help to predict their properties.

## Can the most probable macrostate change over time?

Yes, the most probable macrostate can change over time as the system undergoes thermal fluctuations. However, the overall trend is for the system to eventually reach equilibrium and settle into the most probable macrostate.

Replies
1
Views
1K
Replies
2
Views
1K
Replies
7
Views
2K
Replies
10
Views
5K
Replies
1
Views
3K
Replies
3
Views
5K
• Classical Physics
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K