Statistical Mechanics

Using Classical Mechanics, the average energy for an oscillator is :
< E > = κT
< E > = < (1/2)mv² > + < (1/2)κx²> = (1/2)κT + (1/2)κT = κT

< E > = κT

Using Statistical Mechanics, we can write:
< E > = Σ n_i E_i = NΣ E_i exp[- βE_i]/Z = (N/Z)Σ E_i exp[- βE_i]= N ΣE_i P(E_i)

"i" is the ith level P(Ei) = exp[-βEi]/ Z Z = iΣ0 → ∞ exp[-βEi] = 1/β

Where: P(E_i) is the probability for n_i particles in the level "i" to have the energy E_i. The sum is related to all the states "i".

The normalization of this probability can be written as:
Σ P(E_i) = 1. or : ∫ dE P(E)= 1 [from: 0 → ∞]
∫ dE P(E) = (1/Z)∫ dE exp[- βE] = (1/Z)[- 1/β]exp[- βE](From: 0 → ∞)= β/Z = 1
Thus:
β = 1/Z    
We can write:
P(E_i) = β exp[- βE]    

We have:
< E > = Σ E_i P(E_i) = Σ E_i exp[- βE_i]/Z = (1/Z)∫ E exp[- βE]dE

∫ x exp[-β x] dx = (1/β²) ∫ y exp[- y ]dy
Where y = β x
By part;
∫ y exp[- y ]dy; = - y exp [- y] - exp [- y] = - (y + 1)exp[- y]
(form : 0 → n, n is the number of states). We take n large, then:

∫ y exp[- y ]dy = [- (y + 1)exp[- y]] [: 0 → n] = 1

Hence:
∫ x exp[-β x] dx = 1/β², and < E > = 1/Z . β²
We have then: < E > = 1/β
And: < E > = 1/β = κT. Hence:
β = 1/κT

The relationship for the probability becomes:
P(E_i) = (1/κT) exp[- E/κT]