How many different ways can you distribute three indistinguishable particles in a 3 x 3 ensemble of distinguishable boxes?
Hint: Allow up to 3 particles per box and require at least 1 particle per occupied box to organize yourself. Here are some subscenarios:
1) All particles in the same box
2) Each particle in its own box
3) No two boxes in the same row contain the same number of particles.
4) No two boxes in the same column contain the same number of particles.
5) No two boxes contain the same number of particles.
As an easier example to work out for practice, show that for 3 indistinguishable particles in 4 distinguishable boxes, you would find 20 microstates.
Hint: Allow up to 3 particles per box and require at least 1 particle per occupied box to organize yourself. Here are some subscenarios:
As an easier example to work out for practice, show that for 3 indistinguishable particles in 4 distinguishable boxes, you would find
1 Answer
I get
DISCLAIMER: LONG ANSWER! (obviously)
METHOD ONE: DRAWING OUT MACROSTATES
I'm going to do this a bit out of order, because having one particle per box is the most confusing one.
Here is the representative macrostate:
With
9 boxes, you getbb9 ways to have three-particle boxes.
Here is the representative macrostate:
With two particles in the same box, you have
6 configurations of the remaining particle in its own box within the same row. With3 rows, that givesbb18 configurations.
Here is the representative macrostate:
Rotate the box
90^@ , then repeat(3) forbb18 more, except we would have technically done it column-wise because the boxes are distinguishable.
Here is the representative macrostate:
You should get
4 configurations with one two-particle box in the upper left, times three columns for the two-particle box within the same row equals12 configurations. Multiply by the three rows to getbb36 configurations.
Play around with this. You should get:
a) All particles in the same row:3 configurationsHere is the representative macrostate:
b) Two particles in the same row:2 configurations if two particles are in the first two columns,2 configurations if the particles are in columns1 and3 ,2 configurations if the particles are in the last two columns, times3 rows, for18 total.Here is the representative macrostate:
c) All particles in different rows:2 diagonal configurations,4 configurations with two particles on the off-diagonal, for a total of6 .Here is the representative macrostate:
Apparently, I get
3 + 18 + 6 = bb27 here.
Play around with this. You should get:
a) All particles in the same column:3 configurationsHere is the representative macrostate:
b) Two particles in the same column:2 configurations if two particles are in the first two rows,2 configurations if the particles are in rows1 and3 ,2 configurations if the particles are in the last two rows, multiplied by the3 columns, for18 total.Here is the representative macrostate:
c) All particles in different columns: We don't count these, because they are redundant.Apparently, I get
3 + 18 = bb21 here.
However, that only accounts for
METHOD TWO: DERIVING AN EQUATION
We begin with the assumption of
They would each be separated by
"x x" cdots | "x x x" cdots | "x x" cdots
The walls are distinguishable, and thus could be swapped to obtain distinctly different configurations. With some amount of distinguishable things, there are a factorial number of ways to arrange them.
So, there are
To avoid double counting, we then divide by
This gives
bb(t = ((g + N - 1)!)/(N!(g-1)!)" microstates")
when the
To check this equation, we put
color(blue)(t) = ((4 + 3 - 1)!)/(3! (4 - 1)!) = (6!)/(6 cdot 6) = (5! cdot cancel6)/(cancel6 cdot 6)
= (120)/6 = color(blue)(20) microstatescolor(blue)(sqrt"") ,
just like the example shows.
In your case, with
color(blue)(t) = ((9 + 3 - 1)!)/(3! (9 - 1)!) = (11!)/(6 cdot 8!) = (cancel(8!) cdot 9 cdot 10 cdot 11)/(6 cdot cancel(8!))
= color(blue)(165) microstates
So we were short by