Yet another explanation of negative temperatures

Our toy model consists of 2 dice. The macrostates are charaterized by the energy of the system given by the sum [ n ] of the dots.

The number [ Ω = Ω(n) ] of compatible microstates for a given macrostate is the number of ways you can roll this sum.

We have

  Ω(2) = Ω(12) = 1
  Ω(3) = Ω(11) = 2
  Ω(4) = Ω(10) = 3
  Ω(5) = Ω(9)  = 4
  Ω(6) = Ω(8)  = 5
  Ω(7)         = 6

or

  Ω(2≤n≤7)  = n - 1
  Ω(7≤n≤12) = 13 - n

For convenience, we also set

  Ω(n<2) = Ω(n>12) = 0

In statistical mechanics, temperature is not defined as an average energy, but rather (in suitable units) as the reciprocal of thermodynamic [ β ] and thus

  T = Ω/Ω'

For our discrete system, we replace the derivative with the mean of the forward and backward difference quotient of step size 1, ie

  Ω'(n) = 1/2 { (Ω(n+1) - Ω(n))/1 + (Ω(n) - Ω(n-1))/1 }
        = 1/2 { Ω(n+1) - Ω(n-1) }

A short calculation shows

  Ω'(2≤n<7)  = 1
  Ω'(n=7)    = 0
  Ω'(7<n≤12) = -1

and thus

  T(2≤n<7)  = Ω(n) = n - 1 > 0
  T(n=7)    = ∞
  T(7<n≤12) = -Ω(n) = n - 13 < 0

Even though our model has no notion of kinetic energy, there's still a nice relationship between total energy and temperature.

This is particularly instructive if we offset our energy scale:

For [ E = n - 7 ] we have

  E(T>0) = T - 6 ∈ [-5,0[
  E(T=∞) = 0
  E(T<0) = T + 6 ∈ ]0,5]

As we see here, negative temperatures correspond to higher energies, and it only takes a finite amount of energy to go from positive to infinite and from infinite to negative temperature states.