Multiplicity/Probability: Einstein model of Solids

[WWCY]

Homework Statement

I'm having some trouble with this problem as it seems my concept of the problem is completely wrong. Could someone look through my solutions and point out what I understood wrongly?

Thanks in advance.

Homework Equations

The Attempt at a Solution

a) 21 macrostates - this was correct

b) If we take the entire system to be one big oscillator, this would be $\binom{q + N - 1}{q} = \binom{20 + 20 -1}{20} = 6.89 \times 10^{10}$ which was correct too

The following parts were wrong:

c) I thought I'd try to model the problem as a coin toss. For each quanta of energy, it could either be in oscillator A or oscillator B, which corresponds to heads or tails for a single coin.

By this logic, the multiplicity of having 20 quanta in either oscillator is 1. The total multiplicity is $2^N = 2^{20}$. The probability is then $1/{20^{20}}$

c) Again, by the same logic, the multiplicity associated with having 10 quanta in each is $\binom{20}{10}$. Probability is then $\frac{\binom{20}{10}}{20^{20}}$

I have successfully used this method to solve a problem regarding ideal gases, but am not sure why it can't be used here. The problem regarding the ideal gas is shown below.

 

[DrClaude]

For (c), how can you get a multiplicity that is greater than the number of micro states you calculated in (b)? You need the fundamental principle that all microstates are equally probable.

Once you solve (c) correctly, the same approach can be used for (d).

 

[WWCY]

Thanks for the response.

For (c), how can you get a multiplicity that is greater than the number of micro states you calculated in (b)? You need the fundamental principle that all microstates are equally probable.

Once you solve (c) correctly, the same approach can be used for (d).

I've managed to work it out, but am struggling to rationalise the applicability of different approaches to solve the problem.

In terms of the logic behind the methods used, why can I use the "heads/tails" analogy in the ideal gas case but not in the Einstein Solid case?

 

[haruspex]

why can I use the "heads/tails" analogy in the ideal gas case but not in the Einstein Solid case?

For the ideal gas, the energy is tied to molecules, which have individuality.
Consider two molecules which can be in any of four boxes, independently, two boxes to the left and two to the right. They can both be on the left in four ways: both in the far box, both in the near box, A in the far and B in the near, or A in the near and B in the far.
Packets of energy are different. Changing the above to be two packets of energy in four machines, there are only three microstates for both packets being in the machines on the left: both in the far, both in the near, and one in each.

 

[WWCY]

Thanks for the response!

For the ideal gas, the energy is tied to molecules, which have individuality.
Consider two molecules which can be in any of four boxes, independently, two boxes to the left and two to the right. They can both be on the left in four ways: both in the far box, both in the near box, A in the far and B in the near, or A in the near and B in the far.
Packets of energy are different. Changing the above to be two packets of energy in four machines, there are only three microstates for both packets being in the machines on the left: both in the far, both in the near, and one in each.

However, if we talk about the gas particles being of the same species, won't there be 3 microstates as well? Since we'd be double-counting the cases with one particle in each box.

 

[haruspex]

Thanks for the response!

However, if we talk about the gas particles being of the same species, won't there be 3 microstates as well? Since we'd be double-counting the cases with one particle in each box.

That is certainly not true of macroscopic objects, and I do not think it is true of molecules at everyday temperatures, even if they are the same species. Maybe it kicks in at low temperatures, where the molecules have fewer possible states, but I have never studied quantum theory, so I wouldn't know.

 

[WWCY]

That is certainly not true of macroscopic objects, and I do not think it is true of molecules at everyday temperatures, even if they are the same species. Maybe it kicks in at low temperatures, where the molecules have fewer possible states, but I have never studied quantum theory, so I wouldn't know.

Allow me to rephrase what you've said to see if I understood it right.

Molecules are like coins in the sense that they can be "labelled" as molecules 1,2, and 3. This means that I can definitively say that the microstate where molecule 2 is in Box A and 3 is in Box B is different from the microstate where 3 is in A and 2 is in B. With quanta and oscillators I can't give these packets any label, a packet in oscillator A is merely a packet in A; it makes no difference which packet it is. Solving the problem with the coin analogy would have therefore caused overcounting.

Is this somewhat correct?

 

[haruspex]

Allow me to rephrase what you've said to see if I understood it right.

Molecules are like coins in the sense that they can be "labelled" as molecules 1,2, and 3. This means that I can definitively say that the microstate where molecule 2 is in Box A and 3 is in Box B is different from the microstate where 3 is in A and 2 is in B. With quanta and oscillators I can't give these packets any label, a packet in oscillator A is merely a packet in A; it makes no difference which packet it is. Solving the problem with the coin analogy would have therefore caused overcounting.

Is this somewhat correct?

That is my understanding.

 

[WWCY]

That is my understanding.

Thanks for your help!