# Deriving Gibbs ensemble. Why is $\frac{\partial S}{\partial x} = -\frac{f}{T}$?

In Gibbs ensemble, we assume that along with energy exchanges (as heat), there is also an extensive parameter $$x$$ that changes in a field $$f$$ (so that $$dE = dQ + fdx$$) such that $$E_{\text{sys}}+E_{\text{env}}=E_0$$ and $$x_{\text{sys}}+x_{\text{env}}=x_0$$ where $$E_0$$ and $$x_0$$ are both constants.

Then we observe that the probability $$P(\mu)$$ of finding the system in a microstate $$\mu$$ is proportional to $$\mathcal N_{\text{env}}(E_0-E_\mu, x_0-x_\mu)$$.

Then using that the number of particles in the system is much smaller than that of those in the environment (to ignore the higher derivatives), and using that $$\partial S/\partial E = 1/T$$, we get that $$P(\mu)$$ is proportional to $$\exp\left(-\frac{E_\mu}{k_BT} -\frac{1}{k_B}\frac{\partial S}{\partial x}x_\mu\right).$$

Now, all that remains to show is that $$\frac{\partial S}{\partial x}= -\frac{f}{T}$$, and we'll get the celebrated $$e^{-\beta(E_\mu-fx)}$$. But I've no clue how to prove this.

I tried considering two systems labeled 1 and 2, isolated from the rest of the world, exchanging energy and $$x$$ such that \begin{align} E_1 + E_2 &= \text{const}, \quad\text{and}\\ x_1+x_2 &=\text{const}. \end{align}

Then we need to extremize $$S_1(E_1, x_1) + S_2(E_2, x_2)-\lambda(E_1+E_2)-\mu(x_1+x_2),$$ which yields \begin{alignat}{2} \frac{\partial S_1}{\partial E_1} &= \frac{\partial S_2}{\partial E_2} &= \lambda &\stackrel{\text{def}}{=}\frac{1}{T},\quad\text{and}\\ \frac{\partial S_1}{\partial x_1} &= \frac{\partial S_2}{\partial x_2} &= \mu. \end{alignat}

But this was a dead end as I could not show $$\mu=-f/T$$ here.

• I have not studied statistical mechanics till. But in classical thermodynamics, we can find $\frac{\partial S}{\partial x}$ as follows. If $S(E,x)$, then by first law of thermodynamics, $TdS=dE-fdx\implies dS=\frac{dE}{T}-\frac{fdx}{T}$. Hence, $\frac{\partial S}{\partial x}=-f/T$. I have not studied ensembles etc. But in classical thermodynamics this identity is a consequence of first law when fundamental relation of entropy is given.
– Iti
Apr 22 at 10:14
• @Iti I would have been content with that if I knew why $dQ = TdS$ holds, which I don't, at least from statistical mechanics. :(
– Atom
Apr 22 at 10:29
• @Atom It is much simpler to work with the thermodynamic identity than trying to prove with statistical mechanics. In thermodynamics $dQ = T dS$ directly comes from the definition of entropy. Apr 22 at 13:37
• @Atom Not really. Think of a Carnot cycle. Heat is exchanged along the two isotherms. At the same time, some work is done. Apr 22 at 20:06
• @Atom In your proof, what you are trying to use is not statistical mechanics, but thermodynamics, in the last paragraph. Indeed you can proof that the parameter $T$ that comes up from this $\frac{\partial E}{\partial S} = T$ has the usual property of temperature. I don't think you can prof the equation you would like to within the framework of Stat. Mech. The same for $fdx$. Hope I explain my self correctly. Apr 23 at 1:28

I think @Iti, @GiorgioP, and @Mark_Bell are right in their comments. The connection, $$\beta\rightarrow\frac{1}{T}$$ $$S_{\textrm{ensemble}}\rightarrow S_{\textrm{Carnot}}$$ is not obvious. It's genius on the part of Bolzmann and Gibbs. There's a reason why people write things in stone. And Botzmann's intuition is written on his tombstone:
You need thermodynamics to make this connection. To be more specific, that kind of "circular calculus" that characterises it. For your system, $$dE=TdS+fdx$$ Then, it must be, $$dE=\left.\frac{\partial E}{\partial S}\right|_{x}dS+\left.\frac{\partial E}{\partial x}\right|_{S}dx$$ Then, $$\left.\frac{\partial E}{\partial x}\right|_{S}=f$$ But also, $$dS=\frac{1}{T}dE-\frac{f}{T}dx$$ And then, $$\left.\frac{\partial S}{\partial E}\right|_{x}=\frac{1}{T}$$ $$\left.\frac{\partial S}{\partial x}\right|_{E}=-\frac{f}{T}$$ If you're not convinced, given that your system is entirely analogous to a $$P$$, $$V$$, $$T$$ system, go to your most reliable book on statistical mechanics and see how they derive the concept of pressure from a "purely statistical" point of view. You will find the same thermodynamical argument with just a change of a minus sign. $$f$$ will be $$-P$$, and $$x$$ will be $$V$$.