Connection between the two definitions of entropy

[DaTario]

Hi All,

I would like to know how can one connect the two definitions of entropy
$\Delta S = \int_{T_i}^{T_f} \frac{dQ}{T}$ and $\Delta S = k_B \ln (\frac{W_f}{W_i})$,
particularly I am interested in how the logarithm emerges. Does it have to do with some linear dependence of the heat with T?
DaTario

 

[Baluncore]

https://en.wikipedia.org/wiki/Multiplicative_inverse#Calculus

 

[Lord Jestocost]

Summary:: Hi All, I would like to know how can we connect the two definitions of entropy
$\Delta S = \int_{T_i}^{T_f} \frac{dQ}{T}$ and $\Delta S = k_B \ln (\frac{W_f}{W_i})$

I have severe doubts that one can formally connect the Clausius entropy equation to the Boltzmann entropy equation, as the Boltzmann entropy equation actually holds only for systems in the so called microcanonical ensemble.

EDIT: An association between the Clausius entropy and the results form statistical phyiscs can be found when considering systems in the so called canonical ensemble.

 

[anuttarasammyak]

I think you had better read good texts to learn because we need some lines of mathematics.
Wikipedia https://en.wikipedia.org/wiki/Entropy_(statistical_thermodynamics)#cite_note-jaynes1965-1 shows it in brief in the column.

 

[Twigg]

As in above replies, the canonical ensemble provides the connection you need between the number of states $W$ and the thermodynamic entropy $\int \frac{dQ}{T}$. My personal favorite introduction to the subject is Schrödinger's book "Statistical Thermophysics". It's cheap (<$10) and short.

 

[DaTario]

https://en.wikipedia.org/wiki/Multiplicative_inverse#Calculus

It is quite interesting that thermodynamics definition has an integral with $T^{-1}$ inside and the Boltzmann entropy has a logarithm. I guess the connection is not likely to be so simple. If $dQ$ is always proportional to $dT$, then the integral yields a log function. But what is the countable physical parameter in $\int dQ/T$?

Perhaps, an explanation restricted to the case of a 1 mol sample of ideal gas would be a very good start.

 

[Twigg]

But what is the countable physical parameter in ∫dQ/T?

dQ is the change in the internal energy of the system due to heating. As the energy of the system is increased, the molecules can begin to occupy higher energy states. For example, as you raise the temperature a gas, you will find more molecules in high velocity states than before. Since more states become available (W increases, per Boltzmann definition), the entropy increases. To answer your question, the countable parameter is still W, the number of accessible microstates.

Perhaps, an explanation restricted to the case of a 1 mol sample of ideal gas would be a very good start.

The canonical ensemble says that the probability of finding the gas in a microstate with energy E is proportional to $e^{-E/kT}$. Since $p(E) \propto e^{-E/kT}$ and we know that probabilities have to sum to 1 ($\int p(E) = 1$), we know that $p(E) = \frac{1}{Z} e^{-E/kT}$ where $Z = \int e^{-E/kT}$. For a single gas particle, we know the kinetic energy is $\frac{1}{2} m v^2$, so $$Z = \int d^3 x \int d^3 v e^{-mv^2 / 2kT} = V\int d^3 v e^{-mv^2 / 2kT}$$ If you evaluate this integral, it gives $Z = \frac{V}{\lambda^3}$ where the constant $\lambda = \sqrt{\frac{m}{2\pi kT}}$ is the de Broglie wavelength of that particle. To get the partition function for N particles, take $Z_N = Z^N = \left( \frac{V}{\lambda^3} \right) ^N$. Using some trickery from the canonical ensemble, we have $$S = \frac{\partial}{\partial T} (kT \ln Z_N) = k \ln Z_N + kT \frac{\partial \ln Z_N}{\partial T}$$ Since internal energy is given by $U = -\frac{\partial }{\partial \beta} \ln Z_N = \frac{3}{2} NkT$ and since $\frac{\partial}{\partial \beta} = -kT^2 \frac{\partial}{\partial T}$, we have that $$S = k \ln Z_N + \frac{U}{T} = kN \left[\ln \left( \frac{V}{\lambda^3} \right) + \frac{3}{2} \right] \approx kN \ln \left( \frac{V}{\lambda^3} \right) $$

Notice that this last term is essentially the number of de Broglie wavelength-sized cubes you could stuff into a volume V, and this is essentially a number of microstates.

Note: all the ensembles have slight differences in the additive constant on the entropy. I forget which is the most accurate, but I would trust the Sackur-Tetrode equation (from the microcanonical ensemble).

 

[Lord Jestocost]

I think you had better read good texts to learn because we need some lines of mathematics.

Sorry! In case you need some lines of mathematics, please read section 1-4 "Canonical ensemble and thermodynamics" in Terrell L. Hill's book "An Introduction to Statistcal Thermodynamics".

 

[DaTario]

Thank you all. Special thanks to Twigg for these lines of mathematics. Reading all these comments, it came into my mind that the equivalence of these two definitions in a certain sense seems to point to a quantization procedure as Clausius definition is continuous in principle whereas Boltzmann is discrete.

 

[Twigg]

it came into my mind that the equivalence of these two definitions in a certain sense seems to point to a quantization procedure

That procedure is the microcanonical ensemble

 

[DaTario]

As in above replies, the canonical ensemble provides the connection you need between the number of states $W$ and the thermodynamic entropy $\int \frac{dQ}{T}$. My personal favorite introduction to the subject is Schrödinger's book "Statistical Thermophysics". It's cheap (<$10) and short.

Shame on me, when I was searching for it at Amazon's site and I saw its cover, I realized that I had it in my book shelf. Thank you for the indication anyway.

 

[Twigg]

I don't have it on me, but IIRC it doesn't cover the microcanonical ensemble. For that, you probably want a video lecture, because some of the geometrical arguments don't make sense without pictures (at least not to my pea brain).