Find number of microstates of combined system

[MatinSAR]

Homework Statement

Imagine a system composed of four individual particles. These particles do not interact with each other, and the energy of each particle can be quantified as $0$, $e$, $2e$, and so on. This system is then divided into two subsystems: subsystem $A$, which includes particles $1$ and $2$, and subsystem $B$, which includes particles $3$ and $4$. Initially, subsystems $A$ and $B$ are isolated from each other. Subsystem $A$ has an energy of $5e$, while subsystem $B$ has an energy of $e$. Given these conditions, Answer following questions:

Relevant Equations

Statistical mechanics.

Question 1: Determine the number of possible microstates for the combined system.
My answer: 12 microstates.
Question 2: Suppose we have a system composed of two subsystems that have reached thermal equilibrium after being allowed to exchange heat and energy. Given that the total energy of the combined system before reaching equilibrium was 6e (5e from subsystem A and e from subsystem B), how many microstates would the system have after achieving equilibrium?
My answer: 43 microstates. I try to fix this.

Question 3: When subsystem A is in thermal equilibrium, what are the chances of it having energy levels of 0, e, 2e, …, 6e? Among these, which energy level has the highest probability?
Should I use Boltzmann distribution?

 

[MatinSAR]

Question 1: Determine the number of possible microstates for the combined system.

System A:
One particle with energy of $5e$ and one particle with energy of $0$ : 2 possibilities
One particle with energy of $4e$ and one particle with energy of $e$ : 2 possibilities
One particle with energy of $3e$ and one particle with energy of $2e$ : 2 possibilities
System B:
One particle with energy of $e$ and one particle with energy of $0$ : 2 possibilities

Number of microistates for the combined system = $6*2=12$

Question 2: Suppose we have a system composed of two subsystems that have reached thermal equilibrium after being allowed to exchange heat and energy. Given that the total energy of the combined system before reaching equilibrium was 6e (5e from subsystem A and e from subsystem B), how many microstates would the system have after achieving equilibrium?

Any ideas for this part? I’ve been diligently working on listing all possible microstates, and so far, I’ve identified more than 20.

Isn’t it true to draw a parallel between this scenario and the one where we’re trying to figure out the number of ways to distribute 5 apples among 3 boys? Now, we’re looking to distribute $6e$ among $4$ particles, resulting in $4^6$ microstates.

 

[haruspex]

Any ideas for this part?

What do you think two subsystems that have reached thermal equilibrium means for the total energy of each subsystem?

 

[MatinSAR]

What do you think two subsystems that have reached thermal equilibrium means for the total energy of each subsystem?

Their energy is constant and doesn't change by time.

 

[haruspex]

Their energy is constant and doesn't change by time.

I don't think that is what is meant.
It further explains "after being allowed to exchange heat and energy".
What have exchanged heat and energy, and what have been brought into equilibrium as a result?

 

[MatinSAR]

I don't think that is what is meant.
It further explains "after being allowed to exchange heat and energy".
What have exchanged heat and energy, and what have been brought into equilibrium as a result?

Please, Let me check my professor's lecture notes ...

 

[haruspex]

Please, Let me check my professor's lecture notes ...

I doubt that will help. It is a matter of interpretation of the question.
As I read it, the two subsystems have come into equilibrium with each other, so now have the same total energy.
This should increase the number of microstates somewhat, that representing an increase in entropy, but not hugely.

 

[MatinSAR]

As I read it, the two subsystems have come into equilibrium with each other, so now have the same total energy.

So both should have energy of $3e$ ?
If yes, Can I say that I should find how many ways are there to divide $3e$ between $2$ particles? This gives number of microstates foe one system.

 

[haruspex]

So both should have energy of $3e$ ?
If yes, Can I say that I should find how many ways are there to divide $3e$ between $2$ particles? This gives number of microstates foe one system.

Yes.

 

[MatinSAR]

Yes.

Thanks. I'll retry to solve using your help and I'll share the results.

 

[MatinSAR]

@haruspex For the second question, I believe we have $2^3=8$ microstates for each system. Therefore, the combined system has a total of 64 microstates. However, I have a question: In equilibrium, should systems A and B have equal energy? I thought that their temperatures should be equal …

Question 3: When subsystem A is in thermal equilibrium, what are the chances of it having energy levels of 0, e, 2e, …, 6e? Among these, which energy level has the highest probability?

According to you, Energy of system $A$ can be only $3e$. Am I right?

 

[haruspex]

@haruspex For the second question, I believe we have $2^3=8$ microstates for each system. Therefore, the combined system has a total of 64 microstates. However, I have a question: In equilibrium, should systems A and B have equal energy? I thought that their temperatures should be equal …


According to you, Energy of system $A$ can be only $3e$. Am I right?

I could well be wrong, but to me they read as independent questions.

In q2, the two subsystems are in thermal equilibrium with each other. That means the energy of each is in proportion to its number of particles. Since they have two particles each, each subsystem has energy 3e.

In q3, we are told that A in itself is in thermal equilibrium. According to https://www.physics.usu.edu/torre/3700_Spring_2015/Lectures/03.pdf, that means that all its microstates are equally likely. I don't think we are supposed to take A and B as also in equilibrium with each other, so now each has energy up to 6e.
Assuming that, in how many ways can A have energy 2e, for example?

 

[MatinSAR]

Assuming that, in how many ways can A have energy 2e, for example?

It has two particles so $2^2=4$. Can I continue like this for 3e, 4e and so on?

 

[Steve4Physics]

It has two particles so $2^2=4$. Can I continue like this for 3e, 4e and so on?

I'd like to sneak in as I think you have a misunderstanding.

Represent each microstate of subsystem-A as a pair of numbers - the energies of particles 1 and 2. E.g. (0, 2e) represents the microstate where particle-1 has zero energy and particle-2 has energy = 2e.

If subsystem-A has a total energy of 2e, the corresponding microstates are (0, 2e), (1e, 1e) and (2e, 0). There are only 3 (not 4) possible microstates.

 

[MatinSAR]

If subsystem-A has a total energy of 2e, the corresponding microstates are (0, 2e), (1e, 1e) and (2e, 0). There are only 3 (not 4) possible microstates.

I've counted (1e, 1e) two times, Thank you for pointing out. Should I write down all of the possible microstates to find the answer? It's too hard.

 

[Steve4Physics]

I've counted (1e, 1e) two times, Thank you for pointing out. Should I write down all of the possible microstates to find the answer? It's too hard.

Because each subsystem has only 2 particles, there is a very simple way to get the number of microstates of a subsystem for a given subsystem energy. There's no need to write-out the microstates once you see the pattern. Try this:

How many microstates does A have if A's energy is zero?
The only microstate is (0, 0). So there is one microstate.

How many microstates does A have if A's energy is 1e?
The microstates are (0, 1e) and (1e, 0). So there are two microstates.

How many microstates does A have if A's energy is 2e?
The microstates are (0, 2e), (1e, 1e) and (2e, 0 ). So there are three microstates.

Now you continue - do the above for A’s energy =3e. And, if necessary, for A’s energy = 4e.

Hopefully you will see the pattern (and why the pattern exists). Once you understand that, there is no need to write-out the microstates – you can immediately say how many there are for a given subsystem energy.

While I'm waffling, IMO question 3 makes no sense an/or is very badly worded. I suspect the intended question is something like this:

Question 3: Subsystems A an d B are able to freely exchange energy but the system total (6e) is constant. What are the chances of A having total energy of 0, e, 2e, …, 6e? Among these, which total energy of A has the highest probability?

I don't know if @haruspex would agree with this though,

 

[haruspex]

I'd like to sneak in as I think you have a misunderstanding.

Thanks for jumping in. I now see where I was misapplying a principle in post #12, now edited.

 

[MatinSAR]

Thank you for your time @Steve4Physics ...

Hopefully you will see the pattern (and why the pattern exists). Once you understand that, there is no need to write-out the microstates – you can immediately say how many there are for a given subsystem energy.

I will try to find the pattern tomorrow.

While I'm waffling, IMO question 3 makes no sense an/or is very badly worded. I suspect the intended question is something like this:

I will ask my professor about this next week. Are you suggesting that system A is not supposed to be in equilibrium?

I now see where I was misapplying a principle in post #12, now edited.

Thank you for your time @haruspex

 

[haruspex]

Are you suggesting that system A is not supposed to be in equilibrium?

In the light of @Steve4Physics' input, I would say the whole system is in equilibrium, which means that all microstates of the combined system are equally likely.

There is a trick for calculating how many ways the energy can be distributed. Suppose you have r identical items to put in n distinct boxes. Imagine n+r-1 'things' in a line. Select n-1 of them to represent box boundaries, the remaining r representing the items. That selection gives you a unique apportionment of the r items into the n boxes. It also works the other way around: any apportionment of the r items into the n boxes can be represented uniquely as a selection of n-1 things out of a line of n+r-1 things to be box boundaries.
Therefore the number of ways to distribute the r items is the same as selecting r things from n+r-1 things, $^{n+r-1}C_r$.

 

[Steve4Physics]

Are you suggesting that system A is not supposed to be in equilibrium?

It doesn’t mean anything to say that subsystem A is in equilibrium. With what is it in equilibrium? If it is in equilibrium with subsystem B, then why not say so explicitly?

However, you could make sense of question 3 if the question said:
a) Subsystems A and B are in equilibrium.
b) By ‘equilibrium’ we mean that the subsystems freely exchange energy and have reached a steady-state with the same 'temperature'.

But 'temperature' requires a large number of particles to have real physical meaning. Here we could interpret it as the time-average of the average-energy-per-particle.

EDIT: I prefer @haruspex's definition of equilibrium to mine: each possible microstate of the combined system is equally likely.

So I'm guessing that you are being asked how many microstates correspond to each of the 7 possible partitions of the system’s energy:
$E_A = 0 \text { and } E_B = 6e$
$E_A = 1e \text { and } E_B = 5e$
$E_A = 2e \text { and } E_B = 4e$
$E_A = 3e \text { and } E_B = 3e$
$E_A = 4e \text { and } E_B = 2e$
$E_A = 5e \text { and } E_B = 1e$
$E_A = 6e \text { and } E_B = 0$

Then, using your 7 microstate values*, find the probability of each of the 7 partitions and hence which one is most likely.

*Edit 2. The term 'microstate values' is unclear. By each 'microstate value', I mean the number of microstates in the partition. Thanks to @haruspex for pointing this out.

 

[haruspex]

using your 7 microstate values

Did you mean that? There are seven partitions of energy between A and B, as you wrote, but each partition groups many microstates of the whole system. Maybe by "microstate value" you mean the number of microstates in the partition.

 

[Steve4Physics]

Did you mean that? There are seven partitions of energy between A and B, as you wrote, but each partition groups many microstates of the whole system. Maybe by "microstate value" you mean the number of microstates in the partition.

Yes. I did indeed mean the number of microstates in the partition. I will clarify this with an edit.

 

[MatinSAR]

However, you could make sense of question 3 if the question said:
a) Subsystems A and B are in equilibrium.

I believe this is what the question is asking.

@haruspex and @Steve4Physics Thank you once again for your time. I think I can solve it now.