Partition Function Derivation: Where Did I Go Wrong?

In summary, the conversation discusses the use of the Boltzmann factor and its relationship to the 1st law of thermodynamics. The speaker seems to have made a mistake in their calculations and is seeking clarification. They also mention the importance of the total number of accessible microstates and its relationship to the partition function. The conversation also touches on the magnetic dipole example and the possible error in assuming that the degeneracy and the number of microstates are the same. The speaker asks for further explanation and clarification on the different steps and the ensemble being used.
  • #1
thelaxiankey
5
1
Self-repost from physics.SE; I underestimated how dead it was.

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.
 
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  • #2
Could you explain a bit more what you are doing at the different steps? Also, in what ensemble are you working?
 
  • #3
Michael Sandler said:
But we know by the 1st law that
dSR=dU/T Why is this so? No work done?
(All the other terms are 0 or negligible). So, for isothermic situations, (Is that an isothermal process?)
U=TS These are functions of state, so this equation is describing a state, and not a process.
 

Related to Partition Function Derivation: Where Did I Go Wrong?

1. What is a partition function in statistical mechanics?

The partition function is a mathematical concept used in statistical mechanics to describe the distribution of particles in a system. It is a sum of all possible states of a system, each weighted by the probability of that state occurring. In other words, it represents the total number of ways the system can be arranged.

2. How is the partition function derived?

The partition function is derived using the principles of statistical mechanics and the laws of thermodynamics. It involves calculating the energy levels and degeneracies of the system and using them to determine the probabilities of each state. The final expression for the partition function is dependent on the specific system being studied.

3. What is the significance of the partition function in statistical mechanics?

The partition function plays a crucial role in statistical mechanics as it allows for the calculation of important thermodynamic quantities such as entropy, free energy, and heat capacity. It also provides a way to connect microscopic properties of a system to macroscopic observables.

4. How is the partition function related to the Boltzmann distribution?

The partition function is directly related to the Boltzmann distribution, which describes the probability of a system being in a certain state. The partition function is used to normalize the Boltzmann distribution, ensuring that the sum of all probabilities is equal to one.

5. Can the partition function be applied to any type of system?

The partition function can be applied to any system that obeys the laws of thermodynamics and has discrete energy levels. It is commonly used in statistical mechanics to study gases, liquids, and solids, but can also be applied to more complex systems such as biological molecules and even the entire universe.

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