# Total multiplicity is product of individual multiplicities?

1. Oct 30, 2013

### mishima

Hi, I was looking for a proof or explanation of this. From Schroeder's Thermal Physics, pg 56, explaining interacting systems in equilibrium.

The example in the text is two 3-harmonic oscillators with a total of 6 units of energy. So one macrostate is where each has 3 units of energy. The individual multiplicity of each is 10, but the total multiplicity is 100 when considered together (10*10).

The explanation is "because the systems are independent of each other." I know there is something from introductory combinatorics which explains this (which is why he went no further), but I myself don't know.

Any help?

2. Oct 30, 2013

### jfizzix

For each microstate of the first oscillator, there are 10 microstates of the other oscillator, so there are 10 microstates of the oscillator-pair given that the first one is in a particular microstate.

Since there are 10 pair microstates for each microstate of the first oscillator, and there are 10 microstates of the first oscillator, there are 100 total microstates for the pair.

3. Oct 31, 2013

### dauto

Yes, if the system are independent the total multiplicity is the product of the individual multiplicity. Example: one die has 6 possible states [1; 2; 3; 4; 5; 6] and two dice have 6X6=36 states [(1,1); (1,2); (1,3); (1,4); (1,5); (1,6); (2,1); (2,2); (2,3); (2,4); (2,5); (2,6); (3,1); (3,2); (3,3); (3,4); (3,5); (3,6); (4,1); (4,2); (4,3); (4,4); (4,5); (4,6); (5,1); (5,2); (5,3); (5,4); (5,5); (5,6); (6,1); (6,2); (6,3); (6,4); (6,5); (6,6)].

4. Oct 31, 2013

### mishima

Hi, that makes sense. Is there a name for this rule?

5. Nov 1, 2013

### jfizzix

It's generally referred to as statistical independence. If two systems together have a specific probability distribution $P(X_{i},Y_{j})$for each being in given microstates $X_{i}$and $Y_{j}$ respectively, and these two systems are statistically independent of one another, then the probabilities will factor out.
$P(X_{i},Y_{j}) = P(X_{i})P(Y_{j})$

Where these probabilities are ratios of multiplicities:
$P(X_{i})=\frac{\Omega(X_{i})}{\Omega_{total}}$, $P(Y_{j})=\frac{\Omega(Y_{j})}{\Omega_{total}}$, $P(X_{i},Y_{j})=\frac{\Omega(X_{i},Y_{j})}{\Omega_{total}}$
If the probabilities factor, than so do the multiplicities.

If $P(X_{i},Y_{j}) = P(X_{i})P(Y_{j})$
then $\Omega(X_{i},Y_{j}) = \Omega(X_{i})\Omega(Y_{j})$

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