Introductory Statistical Mechanics - counting number of microstates

In summary, the problem considers a system with 2 harmonic oscillators of frequencies w and 2w respectively. The total energy of the system is given by U=q * h_bar * w, where q is a positive odd integer. The number of microstates for each value of q can be determined by writing out a table of possible values for j1 and 2j2. The correct expression for the energy of the oscillators should be E_1 = h_bar * w * (n1 + 1/2) and E_2 = 2 * h_bar * w * (n2 + 1/2). The function for the number of microstates as a function of energy, g(q), is given by
  • #1
ausdreamer
23
0

Homework Statement



Consider a system composed of 2 harmonic oscillators with frequencies w and 2w respectively (w = omega). The total energy of the system is U=q * h_bar * w, where q is a positive negative integer, ie. q = {1, 3, 5, ...}.

Write down the number of microstates of the system for each value of q.

Homework Equations



-

The Attempt at a Solution



The energy of the first harmonic oscillator with frequency w is: E_1 = j1 * h_bar * w.
The energy of the second harmonic oscillator with frequency 2w is: E_2 = 2 * j2 * h_bar * w.

So now the total energy of the system is given by U = (j1 + 2j2) * h_bar * w = q * h_bar * w.

Say q = 1. So there's only1 microstate for this energy level because 1) my lecturer said that j1, j2 are integers and can represent the number of particles the harmonic oscillator, and by writing out a table for values of j1 and 2j2 we just get:

| j1 | 2j2|
----------
| 1 | 0 |
----------

Now say q = 3. By writing out the table of possible microstates we get the table:

| j1 | 2j2|
----------
| 3 | 0 |
| 1 | 2 |
----------

So for q=3, there are 2 possible microstates of the system. Repeating this a few more times, I get a table which looks like this:

(let g = number of microstates for energy q)

| q | g |
--------
| 1 | 1 |
| 3 | 2 |
| 5 | 3 |
| 6 | 4 |
| 7 | 5 |
. .
. .
--------

And so writing g(q) (number of microstates as a function of energy q) I get:

g(q) = CIELING(q/2)

However, this is apparently wrong according to my lecturer. Can someone see where I went wrong in my reasoning? Thanks
 
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  • #2
I don't see why q would be confined to odd values. Also, your expression for the energy of the oscillators is incorrect. They should be
\begin{align*}
E_1 &= \hbar\omega(n_1 + 1/2) \\
E_2 &= 2\hbar\omega(n_2 + 1/2)
\end{align*}
where ni=0, 1, ...
 
  • #3
Well the problem stated q as positive odd integers, and also my lecturer said we can ignore the 1/2h_bar * w term since all we're interested in is the difference in energy not E itself.
 
  • #4
Clearly if j1=0 and j2=1, you'd have q=2, which isn't odd. Either the problem is wrong or you're not accurately conveying the original problem statement.
 
  • #5
Dear student,

Your attempt at finding the number of microstates for each value of q is a good start. However, there are a few things that need to be clarified.

Firstly, the values of j1 and j2 represent the number of energy quanta in each oscillator, not the number of particles. So for q = 1, there is only one possible microstate because both oscillators can only have one energy quantum each (j1 = j2 = 1). Similarly, for q = 3, there are only two possible microstates because the first oscillator can have 3 energy quanta (j1 = 3) and the second oscillator can have 0 or 2 energy quanta (j2 = 0 or 1).

Secondly, the table you have created does not account for the fact that the oscillators are distinguishable. This means that the order in which the energy quanta are distributed between the oscillators matters. For example, for q = 6, the possible microstates are (j1 = 6, j2 = 0) and (j1 = 0, j2 = 3), but your table only shows (j1 = 3, j2 = 0) and (j1 = 1, j2 = 2).

To account for these factors, we can use the formula for the number of microstates in a system with distinguishable particles and distinguishable energy levels:

g(q) = (q + N - 1)! / (q! * (N - 1)!)

where N is the number of energy levels (in this case, 2) and q is the total energy of the system.

Using this formula, we can find the correct values for g(q):

| q | g |
--------
| 1 | 1 |
| 3 | 3 |
| 5 | 6 |
| 6 | 10 |
| 7 | 15 |
. .
. .
--------

I hope this helps clarify your understanding of the problem. Keep up the good work in your studies of statistical mechanics!

Best regards,

 

1. What is the concept of microstates in statistical mechanics?

The concept of microstates in statistical mechanics refers to the different ways in which the particles in a system can be arranged while still maintaining the same macroscopic properties. It is a fundamental concept used to understand the behavior of a large number of particles in a system, such as gas molecules in a container.

2. How is the number of microstates related to entropy?

In statistical mechanics, entropy is a measure of the randomness or disorder of a system. The number of microstates is directly related to entropy, as a larger number of microstates indicates a higher degree of disorder in the system, and therefore, a higher entropy.

3. What is the significance of counting the number of microstates?

Counting the number of microstates allows us to calculate the entropy and other thermodynamic properties of a system, which helps us understand the behavior of matter and energy at a microscopic level. It also allows us to make predictions about the behavior of a large number of particles in a system.

4. How does the number of microstates change with temperature?

The number of microstates increases with temperature, as the particles in a system gain more energy and have more freedom to move and arrange themselves in different ways. This leads to a higher entropy at higher temperatures.

5. Can the number of microstates ever decrease?

No, the number of microstates can never decrease in a closed system. This is due to the fundamental principle in statistical mechanics called the second law of thermodynamics, which states that the total entropy in a closed system will always increase or remain constant over time.

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