- #1

thelaxiankey

- 5

- 1

So this follows Schroeder's Intro to Thermal Physics equations 6.1-6.7, but the question isn't book specific. Please let me be clear: I know for a fact I'm wrong. However, it feels like performing seemingly allowed manipulations, I arrive at an incorrect conclusion... what gives?

We say that:

$$ \text{Boltzmann factor} = e^{-E \beta} $$

Where $$ \beta = 1/k_b T$$

But we know by the 1st law that $$ dS_R = dU / T$$ (All the other terms are 0 or negligible). So, for isothermic situations, $$ U = TS$$ Additionally, $$ S(E) = k_b \ln \Omega(E) $$ (all of these are used by Schroeder so I think they're right), and then:

$$ E = -U = -T k_b \ln(\Omega(E))$$

And if we solve for omega, we get:

$$\Omega(E) = e^{-E / (k_b T)} = e^{-E \beta} = \text{Boltzmann factor} $$

But this is nowhere mentioned in the book, and seems important and/or horribly wrong! Moreover, this means that $$ Z = \text{ Total # of accessible microstates at all energies} $$ Which is also nowhere mentioned, and feels very important! Where did I go wrong? This feels very important and yet doesn't seem to be mentioned... anywhere.

This all falls apart even harder with the magnetic dipole example, because we start to get stuff like:

$$ P(E) = \frac{ e^{-E \beta} \Omega(E) }{Z} = \frac{\Omega^2(E)}{\text{Total # of Microstates}} $$

I suspect that in this case the error is due to the fact $\Omega(E) \neq \text{Degeneracy}(E)$, but I'm not sure about that, because it seems to indicate that the $\Omega$ represents something that isn't the number of microstates, but some other number entirely.