What Determines the Maximum Number of Microstates at Equilibrium?

[I_laff]

$\Omega(E_1)$ is the number of microstates accessible to a system when it has an energy $E_1$ and $\Omega(E_2)$ is the number of microstates accessible to the system when it has an energy $E_2$. I understand that each microstate has equal probability of being occupied, but could someone explain at equilibrium why $\Omega(E_1)\Omega(E_2)$ is the maximum number of microstates?

 

[Lord Jestocost]

Such a product term means the total number of states

$\Omega(E_1,E_2)=\Omega(E_1)\Omega(E_2)$

of two non-interacting systems with energies $E_1$ and $E_2$, respectively.

 

[I_laff]

Such a product term means the total number of states

$\Omega(E_1,E_2)=\Omega(E_1)\Omega(E_2)$

of two non-interacting systems with energies $E_1$ and $E_2$, respectively.

Apologies if the answer is obvious, but why does $\Omega(E_1)\Omega(E_2)$ give the total number of states?

 

[Nugatory]

The first system may be in any of $\Omega(E_1)$ states; for any of these the second may be in any of $\Omega(E_2)$ states. It’s really no different than a system of consisting of two standard six-sided dice: the first may be in any of six states, the second may be in any of six states, so there are six times six equals thirty-six possible states for the two-dice system.

 

[I_laff]

The first system may be in any of $\Omega(E_1)$ states; for any of these the second may be in any of $\Omega(E_2)$ states. It’s really no different than a system of consisting of two standard six-sided dice: the first may be in any of six states, the second may be in any of six states, so there are six times six equals thirty-six possible states for the two-dice system.

Ah of course. Thanks, that cleared things up.