# Connection between Entropy, Energy and Zustandsum

1. Sep 30, 2014

### Nikitin

Hi. Say you have a canonical ensemble, and its zustandssumme is $Z = \sum_j e^{- \beta E_j}$.

Then $$d \: ln(Z) = \frac{-d\beta}{Z} \sum_j e^{- \beta E_j} E_j - \frac{-\beta}{Z} \sum_j e^{- \beta E_j} dE_j$$

Further, my book says the second term is given by the work $dW$ done on the system. Why exactly is that so? What about added heat? As for the first term, it's $-<E> d \beta$.

So, going on, we get $$d(ln(Z) + \beta <E>) = \beta (dW + d<E>) = \beta dQ$$, where $dQ$ is added heat.

OK, now my book makes the jump and declares $$S= k_B (ln(Z) + \beta E)$$ and $$\beta = 1/k_B T$$

Uhm, but why is this so? Why can't just as well $S= C \cdot k_B (ln(Z) + \beta E)$ and $\beta = 1/ C \cdot k_B T$, where C is any number?

2. Sep 30, 2014

### Nikitin

On second reading I see that my post was very poorly formulated, as was short on time. I apologize. At any rate, what I am curious about is:

1) Why does $dW = \frac{- \beta}{Z} \sum_j e^{- \beta E_j} dE_j = dW$ (in the OP I had a sign typo in front of this term, btw. Sorry.)

2) How come C = 1 in $S= C \cdot k_B (ln(Z) + \beta E)$ and $\beta = 1/ C \cdot k_B T$?

3. Sep 30, 2014

### ShayanJ

I can only answer this part.
The $E_j$s are actually energy of different energy levels. They depend on the volume of the system and other things which are factors that affect the interactions the particles are involved in. From these other things, only positions of system's particles are properties of the system but in statistical mechanics, we're not considering the system in that much detail and so that's irrelevant. So the only relevant characteristic of the system which the energy of the energy levels depend on, is the system's volume. But exchange of energy by changing the volume of the system, is called work.

Last edited: Sep 30, 2014
4. Sep 30, 2014

### Nikitin

But why do you ignore the potential energy between the particles and their temperature? Surely if you added heat or did work to a system of particles their energy would increase either way. And if you applied an electric field onto a system of charged particles, their potential energy would increase in relation to each other..

5. Oct 1, 2014

### Nikitin

Guys, I found answers here: Statistical Mechanics for idiots. http://www.physics.dcu.ie/~jpm/PS302/Lect2.pdf

Just in case I confused somebody else along with me, the guy in the link above will explain everything in a clear fashion (for once!!).

6. Oct 1, 2014

### Jano L.

I wondered about this too. Here are my thoughts.

2) Comparing the equation of state of a gas $pV=Nk_B T$ (where $T$ is empiric temperature in Kelvins) to the equation of state that can be calculated for canonical ensemble of a point particle model of ideal gas, it turns out that there is agreement when we postulate that $\beta$ gives $\frac{1}{k_B T}$. Interpretation of $\beta$ as giving $1/(k_B T)$ is then assumed for all other physical models.

1) To connect this probabilistic theory with discrete states $j$ to thermodynamics, some additional assumptions are necessary. One possible assumption is that expected average work done on the system in quasi-static process can be expressed as

$$dW = \sum_j p_jdE_j$$

where $$p_j$$

is probability that system is in the state $j$.

Since $U = \sum_j p_j E_j$, then obviously expected average accepted heat is

$$dQ = \sum_j E_jdp_j.$$

This assumption does not seem trivial and may not be completely general, but it seems to work well in this case, enabling us to identify $k_B \sum_k -p_j \ln p_j$ with thermodynamic entropy $S$.

Some motivation for the above assumption can be found in the calculations based on Schroedinger's equation. Let work be done on the system; the Hamiltonian operator then depends on some parameter, say volume of the box $V$. How do the Born probabilities $|c_j|^2$ and the Hamiltonian eigenvalues $E_j$ defined by

$$\psi = \sum_j c_j\phi_j$$
$$\hat{H}\phi_j = E_j \phi_j$$
change when $V$ changes slowly in time? It turns out the Born probabilities $|c_j|^2$ remain constant, while the eigenvalues $E_j$ change with the parameter $V$.

Alternatively, we may assume that thermodynamic entropy is given by the formula $S=k_B\sum -p_j\ln p_j$ with the Boltzmann probabilities, derive the relation $TdS = \sum_j E_j dp_j$ and thus also $dQ = \sum_j E_jdp_j$ for quasi-static process.

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