So, just reproducing that. And now I want to recall for you,
something from last week. It was in video 5.6, and that was that
dU, was equal to this expression. And we then expanded this, noting that
energy depended on volume, and we got an expression of probability times
derivative of energy with respect to volume, dv.
Plus here's this term again, same term here, as here.
We associated this first term with work, and this second term with heat.
Right? And in particular then, this was Delq
reversible. So at this stage, if I make this
substitution for this sum, as Delq reversible.
I get that ds is equal to Boltzmann's constant times beta times Delq
reversible. Mind you the definition of ds is, it's
Delq reversible divided by t. And as a result, Kb times Boltzmann's
constant must be this 1 over t. Ergo, beta has to be one over kT.
Quod erat demonstrandum. So we finally accomplished a proof.
And this was really how Boltzmann related, Boltzmann did this statistical
analysis. Noted the similarity between his
equations, and the classical thermodynamic equations.
And was able then to establish the value of his constant, and it's relationship to
classical thermodynamics. So a, a tour de force piece of work on,
on Boltzmann's part. And thus we've established a connection
between statistical and classical thermodynamics.
So this was a short proof, that's all there is to this video.