- #1

- 466

- 5

the table on this image shows a system of two einstein solids isolated from the environment. with three oscillators and a total of 6 units of energies (hf). can someone explain to me how they got they're Ω

_{A}and Ω

_{B}values?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter iScience
- Start date

In summary, the conversation discusses a system of two Einstein solids sharing six units of energy and the calculation of the number of possible microstates for each solid. The method involves breaking up the cases and using combinatorial identities such as the binomial coefficient bracket. The total number of microstates for the system is found to be 90.

- #1

- 466

- 5

the table on this image shows a system of two einstein solids isolated from the environment. with three oscillators and a total of 6 units of energies (hf). can someone explain to me how they got they're Ω

Physics news on Phys.org

- #2

- 419

- 29

Since there are two solids sharing six units of energy, they first break up the cases (solid A has 0 energy units, solid B has 6 energy units)=(0,6), (1,5), (2,4) , (3,3), (4, 2), (5,1), (6,0). Now your question is: How do I calculate the number of possible microstates there are for each solid in each of those cases? By microstates, I mean number of distinct ways to distribute the energy units among the solids' energy levels.

So let's say we're dealing with the case (2,4). How do we calculate how many possible microstates there are for each solid? Solid A has three oscillators and 2 units of energy. How can we distribute 2 units of energy among the three oscillators of solid A? The possibilities would look something like (1,1,0),(0,1,1),(1,0,1), (2,0,0),(0,2,0),(0,0,2). This means there are six possible microstates for solid A, so Ω_{A}=6. Since there are six total units of energy, the remaining four units of energy are in solid B, which is also made of three oscillators. So we can calculate the number of possible microstates for B by the same counting method: the possible microstates of B are (4,0,0),(0,4,0),(0,0,4),(3,1,0),(3,0,1),(0,3,1),(1,3,0),(1,0,3),(0,1,3),(2,1,1),(1,2,1),(1,1,2),(2,2,0),(0,2,2),(2,0,2), which implies Ω_{B}=15. Thus the total number of microstates for the composite system of both A and B is Ω_{A}Ω_{B}=6*15=90 microstates (remembering we have assumed that A has 2 and B has 4).

As the photo you attached suggests, you can use quicker combinatorial identities to derive numbers like this, e.g. the binomial coefficient bracket. As an example of a real crafty combinatorics identity, if we want to calculate how many ways there are to distribute the 6 units of energy among the 6 oscillators (with no assumption about how many each subsystem has), we use the multiset binomial coefficient [see http://en.wikipedia.org/wiki/Binomial_coefficient#Multiset_.28rising.29_binomial_coefficient ] according to :

[tex]\Omega = \left ( \begin{pmatrix} 6 \\ 6\end{pmatrix}\right)= \begin{pmatrix}6+6-1 \\ 6\end{pmatrix} = \frac{(6+6-1)!}{((6+6-1)-6)!6!}=\frac{11!}{5!6!} = 462[/tex] as your book states.

If we wanted to do the example I did earlier--where we assume A has 2 and B has 4--we could write:

[tex]\Omega_A\Omega_B = \left ( \begin{pmatrix} 3 \\ 2\end{pmatrix}\right)\left ( \begin{pmatrix} 3 \\ 4\end{pmatrix}\right)= \begin{pmatrix}3+2-1 \\ 2\end{pmatrix} \begin{pmatrix}3+4-1 \\ 4\end{pmatrix}=\frac{4!}{(4-2)!2!}\frac{6!}{(6-4)!4!}=6 \cdot 15 = 90[/tex]

So let's say we're dealing with the case (2,4). How do we calculate how many possible microstates there are for each solid? Solid A has three oscillators and 2 units of energy. How can we distribute 2 units of energy among the three oscillators of solid A? The possibilities would look something like (1,1,0),(0,1,1),(1,0,1), (2,0,0),(0,2,0),(0,0,2). This means there are six possible microstates for solid A, so Ω

As the photo you attached suggests, you can use quicker combinatorial identities to derive numbers like this, e.g. the binomial coefficient bracket. As an example of a real crafty combinatorics identity, if we want to calculate how many ways there are to distribute the 6 units of energy among the 6 oscillators (with no assumption about how many each subsystem has), we use the multiset binomial coefficient [see http://en.wikipedia.org/wiki/Binomial_coefficient#Multiset_.28rising.29_binomial_coefficient ] according to :

[tex]\Omega = \left ( \begin{pmatrix} 6 \\ 6\end{pmatrix}\right)= \begin{pmatrix}6+6-1 \\ 6\end{pmatrix} = \frac{(6+6-1)!}{((6+6-1)-6)!6!}=\frac{11!}{5!6!} = 462[/tex] as your book states.

If we wanted to do the example I did earlier--where we assume A has 2 and B has 4--we could write:

[tex]\Omega_A\Omega_B = \left ( \begin{pmatrix} 3 \\ 2\end{pmatrix}\right)\left ( \begin{pmatrix} 3 \\ 4\end{pmatrix}\right)= \begin{pmatrix}3+2-1 \\ 2\end{pmatrix} \begin{pmatrix}3+4-1 \\ 4\end{pmatrix}=\frac{4!}{(4-2)!2!}\frac{6!}{(6-4)!4!}=6 \cdot 15 = 90[/tex]

Last edited:

Share:

- Replies
- 5

- Views
- 639

- Replies
- 1

- Views
- 797

- Replies
- 11

- Views
- 1K

- Replies
- 3

- Views
- 2K

- Replies
- 2

- Views
- 626

- Replies
- 66

- Views
- 3K

- Replies
- 2

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 5

- Views
- 1K

- Replies
- 39

- Views
- 3K