# Why is the logarithm of the number of all possible states of a system differentiable?

Temperature of a system is defined as $$\left( \frac{\partial \ln(\Omega)}{ \partial E} \right)_{N, X_i} = \frac{1}{kT}$$ Where $\Omega$ is the number of all accessible states (ways) for the system. $\Omega$ can only take discrete values. What does this mean from a mathematical perspective? Many people say we have $10^{23}$ particles so $\Omega$ is almost continuous function of energy. Why is $10^{23}$ a nice number but $1000$ is not? When can one be sure they can differentiate $\ln(\Omega)$?

If you agree with me, do you know an alternative accurate definition for temperature?

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Dec 29 '15 at 12:25
• The definition of temperature that comes from Boltzmann equation does not depend on this kind of problems, but it is also thought in other context. – Hydro Guy Feb 3 '16 at 11:33
• You basically count states using factorials, and, for such large numbers, the Stirling's approximation works extremely nicely. That approximation is continuous. – FGSUZ Oct 29 '17 at 22:46

## 1 Answer

From a mathematical perspective this means that it is not differentiable. The problem is that you need the discreteness to be able to count states. If you replace the discreteness by something smooth you get something differentiable, but your definition of entropy no longer makes sense. This is just one of the points where mathematicians cringe, but it works perfectly fine in practice for the amount of particles dealt with in statistical physics.

Of course if you look at systems where few particles are involved this definition would not work. This is due to the fact that temperature is a macroscopic variable (as is for example pressure). You just don't know what the temperature or pressure of a single particle are.