I had a question on one of the details of the derivation of the second law of thermodynamics starting from the phase space volume. I'll type out what I understand so far:

Letting the Hamiltonian depend on some external parameter $a$ so that $H=H(a)$. The phase space volume can be written as

$$\bar{\Omega}(E,a)=\int d\Gamma \Theta(E-H(a)),$$

where $\Theta$ is the heaviside function. The total differential is

$$d\bar{\Omega}(E,a)=\int d\Gamma \delta(E-H(a))(dE-\frac{\partial H}{\partial a}da)=\Omega(E,a)(dE-\langle\frac{\partial H}{\partial a}\rangle da).$$

In the equation above $$\Omega=\int d\Gamma \delta(E-H(a)).$$

Using the logarithmic derivative

$$d \log\bar{\Omega}=\frac{\Omega}{\bar{\Omega}}(dE-\langle\frac{\partial H}{\partial a}\rangle da).$$

Using the definition of entropy $S=k\log\bar{\Omega}$

$$k\ dS=\frac{\Omega}{\bar{\Omega}}(dE-\langle\frac{\partial H}{\partial a}\rangle da).$$

At this point the book jumps to

$$ dS=\frac{1}{T}(dE-\langle\frac{\partial H}{\partial a}\rangle da)$$

without much explanation. My question is how the $1/T$ follows from the previous line?


The expression $$ k_B \frac{\Omega}{\bar{\Omega}} $$ equals $$ k_B\frac{1}{\bar{\Omega}}\frac{d\bar{\Omega}}{dE} $$ which equals $$ \frac{dS}{dE}. $$ In thermodynamics, where $S$ is the Clausius entropy, this is equal to $1/T$ where $T$ is the Kelvin temperature. In statistical physics, this expression can be taken as a definition of $1/T$ of a system from the microcanonical ensemble $E,a$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.