Extra Material for Problem Set 2

I do not expect you to understand the material presented here. I wanted to post this so that those of you who are confused about the sign convention in Problem Set 2 Problem 1 can be satisfied and happy.

The sneaky thing in Problem 1 is that we are changing ensembles. We’re working in an isothermal-isobaric (well, really, iso-tension…?) ensemble. The parition function we derived in class is useful for $N,V,T$ ensembles! But we need something for an $N,P,T$ ensemble.

You see, we assumed that no external work was being done on the system. However, we have another constraint in our Lagrangian now, which is that the tension must be constant: $\langle f\rangle = F$. You could work through another nasty Lagrange multiplier problem (try it if you feel ambitious!) and you could find the correct partition function that way.

But there’s a neat-o trick for switching ensembles super-fast. It’s called a Legendre Transformation.

Legendre Transformations

How does a Legendre transform work? The key idea is to use the product rule. If $(J,x)$ is a conjugate pair of variables, then $d(Jx) = xdJ + Jdx$ relates the variation $dJ$ in quantity $J$ to the variation $dx$ in quantity $x$.

Let’s go through the details. Consider a function of two independent variables, call it $f(x,y)$. It’s differential is

$$df = \frac{\partial f}{\partial x}\bigg)_y dx + \frac{\partial f}{\partial y}\bigg)_x dy$$

Let’s define $u\equiv\frac{\partial f}{\partial x}\bigg)_y$ and $w\equiv\frac{\partial f}{\partial y}\bigg)_x$. I’m only doing this to keep the ugly math symbols to a minimum. You can write everything out with partials if you wish!

Our equation can now be rewritten as

$$df = udx + wdy$$

Much cleaner.

Now here’s the clever part: use the product rule to compute the differential

$$d(wy) = ydw + wdy$$

and subtract this equation from $df$ to get

$$df - d(wy) = udx + wdy - ydw - wdy$$ $$= udx - ydw$$

Call this cool new function the Legendre transformation $g$:

$dg = udx - ydw$.

To summarize, we have done a Legendre transformation from an original function $f(x,y)$ to a new function $g(x,w)$ by switching from variable $y$ to its conjugate variable $w$. Of course, one could instead switch $x$ to $u$ to obtain $h(u,y)$ or one could switch both independent variables to get $k(u,w)$. We see therefore that for two variables, there are four possible variants on the function. To make the connection back to thermodynamics, we might call these various functions the potentials.

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Example: Gibbs Free Energy as a Legendre transformation of Helmholtz Free Energy

Recall the differential of Helmholtz Free Energy:

$$dF = Jdx - S dT + \mu dN$$

This differential is extremized when $(N, x, T)$ are constant. What if we want to find a new differential that is extremized when $(N, J, T)$ are constant? We need to do a Legendre transformation of $J, x$.

So take $d(Jx)$ and subtract from the Helmholtz:

$dF - d(Jx) = Jdx - SdT + \mu dN - Jdx - xdJ$ $= -SdT + \mu dN - x dJ$

Hey, that’s the differential of the Gibbs Free Energy! I.e., $dG = dF - d(Jx)$.

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In your homework, you need to work in the Gibbs ensemble - you need a new partition function. Let’s start with the “$F=ma$” of statistical mechanics:

$$F = -k_B T \ln Z(N,x,T)\Rightarrow Z(N,x,T) = e^{-\beta F}$$

Now the Legendre transformation! Multiply both sides by $e^{\beta(Jx)}$:

$$Z(N,x,T)e^{\beta Jx} = e^{-\beta(F + JX)} = e^{-\beta G}$$

So the new partition function for Homework Problem 1 is:

$$Z(N,J,T) = Z(N,x,T)e^{\beta Jx}$$