# Multiplication of phase spaces for entropy

I have a problem understanding the phase space volume of entropy in the following example:

To make it easy let's say we have a system with one dimension, two particles & and a rough discretisation of the impulse values. Let the total value of the absolute impulses be $$2$$. so in regard to the impulses we have a certain amount of microstates that can appear:

$$+2,0$$/

$$-2,0/$$

$$0,+2/$$

$$0,-2/$$

$$+1,+1/$$

$$+1,-1/$$

$$-1,+1/$$

$$-1,+1/$$

This gives a total of 8 accessible microstates. Now, let's say we add a second system with the same properties. If we combine those two systems each system of course can take on one of its $$8$$ microstates, which leads to a total $$8\cdot 8=64$$ possible options of microstates. This is the same as $$\Omega(\text{total}) = \Omega(1) \cdot \Omega(2)$$

But it is now also possible that i.e. a single particle takes on an impulse of $$4$$, while the other three particles do not move, or a particle has an impulse of $$3$$ and one of the other particles an impulse of $$1$$, etc.

So the total number of accessible microstates is higher than just the multiplication of the two separate systems. In thermodynamics the total amount of accessible microstates of two combined systems systems (with equal temperature) is only the multiplication and doesn't include the the additionally gained microstates. No entropy should be produced, but in our thought experiment additional microstates are produced? Where am I making a mistake?

Kind regards.

• I haven't thought about this enough so this is a comment not an answer, but the usual multiplicication normally follows from some sort of assumption that things are weakly interacting, which might be the way out here. – jacob1729 Jun 10 at 22:33

$$\Omega(E)$$ is the number of microstates of the system with energy $$E$$. If the system can be divided into two non-interacting parts $$A$$ and $$B$$ with respective energies $$E_A$$ and $$E_B$$, then the total number of microstates accessible to the composite system is $$\Omega_A(E_A) \cdot \Omega_B(E_B)$$.
If the two parts can exchange energy, than we can say only that the total energy is $$E=E_A+E_B$$, which means that the total number of microstates of the composite system is $$\Omega_{tot}(E) = \sum_{E_A} \Omega_A(E_A) \cdot \Omega_B(E-E_A)$$ where $$E_A$$ ranges over all values accessible to subsystem $$A$$.
I'm not sure why you are formulating this problem in terms of impulses. This doesn't make physical sense to me. If instead you say that the system consists of two particles each of which can have energy $$E=0,\epsilon,2\epsilon,3\epsilon,\ldots$$ then we would have $$\Omega(0)=1, \Omega(\epsilon)=2, \Omega(2\epsilon) = 3, \Omega(3\epsilon)=4$$, and $$\Omega(4\epsilon)=5$$.
If you have two systems in contact with total energy $$4\epsilon$$, then the total number of microstates accessible to the composite system is
$$\Omega_{tot}(4\epsilon)=\Omega(0)\Omega(4\epsilon) + \Omega(\epsilon)\Omega(3\epsilon)+ \Omega(2\epsilon)\Omega(2\epsilon)+\Omega(3\epsilon)\Omega(\epsilon)+\Omega(4\epsilon)\Omega(0)$$ $$= 5+8+9+8+5 = 35$$ The total number of ways to split $$n$$ parcels of energy among $$p$$ individuals is $${n+p-1}\choose{p-1}$$. For 4 particles and 4 total units of energy, this becomes $$\frac{7!}{3!4!} = 35$$, which matches our previous analysis.