# How does the fundamental assumption of statistical physics make sense?

Consider two systems A and B in thermal contact. System A has $N_A=3$ simple harmonic oscillators and the system B has $N_B=3$ simple harmonic oscillators as well. Each system has a number of energy units $q$ (macrostates of the statistical system) such that $q=q_A+q_B=6$.

By applying the formula for the multiplicity of the systems, $$\Omega(N,q)=\frac{(q+(N-1))!}{q! (N-1)!}$$ and the fact that the total number of multiplicities is given by $\Omega_{total}=\Omega_A\Omega_B$ we obtain the following combinations of energy units {$q_A,q_B$}:

{0,6} = 28,

{1,5} = 63,

{2,4} = 90,

{3,3} = 100,

{4,2} = 90,

{5,1} = 63,

{6,0} = 28.

As far as my understanding goes, this then implies that the most likely outcome for the interacting systems is each having 3 energy units. However, the fundamental assumption of statistical physics states the following:

In an isolated system in thermal equilibrium, all accessible microstates are equally probable.

Surely this would then imply that {0,6} is no less likely to occur than {3,3}, since all accessible microstates are equally probable, thereby making q=0 and q=6 just as likely to occur as q=3 and q=3.

Can someone please explain to me where my misunderstanding lies?

• I don't quite understand your setup, but I think your problem is that {3,3} is a macrostate, not a microstate. The microstates corresponding to {3,3} in you system are each of the 100 states that you noted down by the multiplicity. So if those are equally likely your macrostate will be weighted by the multiplicity. Does that help? – Wolpertinger Feb 9 '16 at 15:24
• @Numrok - I have borrowed your distinction of "micro" vs "macro" states to clarify my answer. Thanks for the inspiration. – Floris Feb 9 '16 at 15:26
• @Floris: good answer, now I also understand what the initial setup was :) +1 – Wolpertinger Feb 9 '16 at 15:28

You have to distinguish between "different states" and "number of states" - or, in the words of @Numrok, between "macrostates" and "microstates".

The fundamental theorem refers to "accessible micro states". If I have three white balls and two buckets to put them in, I could put two balls in one and one in the other (that is a macro state); there are in fact many ways (microstates) in which I could achieve that distribution. If I number the balls 1-2-3, the six ways are

bucket
#1 #2
1  23
2  13
3  12
12   3
13   2
23   1


On the other hand, "three balls in one bucket" would only have two states:

bucket
#1  #2
-    123
123    -


So "two in one and one in the other" is more likely - there are more ways in which that distribution can be achieved. It doesn't mean that the other arrangement will never happen; just that it's less likely to be observed if you randomly look at the system.

• The next thing to do is to consider the how the strenght of the preference for equilibrium macrostates and the width of the distribution changes as the size and energy of the system grows towards macroscopic size. You find that the preference for the equilibrium macrostates grows ever more extreme and the width shrinks toward a case that is distinguishable only with the most sensitive of experiments. It's a worthwhile exercise. – dmckee --- ex-moderator kitten Feb 9 '16 at 15:39
• @dmckee - I agree that it's a good exercise, but it seems to go rather beyond the scope of the original question. Was your comment a suggestion for me to edit the answer, or intended as an encouragement to OP ? – Floris Feb 9 '16 at 15:41
• Sorry. I wasn't suggesting that the work should go here (because you're right the size of the project grows quickly), but instead offering a comment for any reader who wanted to know more. – dmckee --- ex-moderator kitten Feb 9 '16 at 15:42

I know it's late but maybe someone will find it useful someday. I agree, this can be a bit confusing. But try to think it that way. You have 10 coins, and imagine that you toss them in a way that they make a line. Then a microstate HHHHHHHHHH is as probable as microstate HTTTHHTHTT - think of an order, what is the chance that you get exactly HTTTHHTHTT sequence?? But when you count heads and tails, you 'reduce' it to a macrostate. Now 10 heads is not as probable as 4 heads and 6 tails, because if you just count them there are many microstates corresponding to a 4 heads and 6 tails macrostate, but only one microstate correstponding to 10 heads macrostate. Hope this helps.