statistical mechanics - Understanding "natural variables" of the thermodynamic potentials using the example of the ideal gas

I'm struggling with the concept of "natural variables" in thermodynamics. Textbooks say that the internal energy is "naturally" expressed as

$$U = U(S,V,N)$$

For an ideal gas, I could take the Sackur–Tetrode equation - which gives me $S(U,V,N)$ - and solve for $U$ to get

$$U = \frac{3Nh^2}{4 \pi m} \left( \frac{N}{V} \left( \exp \left( \frac{S}{Nk} \right) - \frac{5}{2} \right) \right) ^{2/3} = U(S,V,N)$$

However, I have never seen this expression before. Usually, people invoke the equipartition theorem to get

$$U = \frac{3}{2}N k_B T = U(T,N)$$ Or they use the ideal gas law to get $$U = \frac{3}{2}N k_B T = \frac{3}{2}pV = U(p,V)$$

So sticking with the example of the ideal gas, this motivates the following questions:

• What is "natural" about $U(S,V,N)$ compared to $U(T,N)$ and $U(p,V)$?


• Can I derive the expressions for $U(T,N)$ and $U(p,V)$ from $U(S,V,N)$?


• Can I derive $U(S,V,N)$ from $U(T,N)$ and $U(p,V)$?

Note that this question is not about the Legendre transformation between different thermodynamic potentials but about expressing the same thermodynamic potential $U$ in terms of different variables.

Answer

I'll answer your questions one by one.

1. About what makes $U(S,V,N)$ natural:

Keep in mind that thermodynamics started out as an experimental science, with people looking to use it for practical purposes (such as Carnot, etc.). Now as energy has always had center stage in physics, since it is conserved in many systems, so it made for a good starting point. One observation people made about energy is that it $\textbf{scales with the system}$. What that means is that by increasing the size of the system (all its parameters), the energy also increased in proportion. This means that when looking for an equation for the energy, we need to look for parameters which also scale with the system, or mathematically speaking, we need an equation that is $\textbf{first order homogeneous}$, i.e.

$$U(\lambda X_1, \lambda X_2 ...) = \lambda U(X_1, X_2...)$$ Where $\lambda \in \mathbb{R}$ and $X_1,X_2...$ are the parameters. The parameters for which the above is true are known as $\textbf{extensive parameters}$. Intuitively, these are the volume, $V$ and the number of particles, $N$. Less intuitively, the entropy, $S$ is also extensive. There are other extensive parameters which are used to calculate $U$, such as magnetic moment. But for most simple systems, $S,V,N$ are enough.

2. About deriving $U(T,p,N)$

Given $U(S,V,N)$, you can find $T$ and $-p$ as taking the partial derivative of $U$ with respect to $S$ and $V$ respectively. Note that $T$ and $-p$ are $\textbf{intensive parameters}$, wherein they are NOT homogeneous first order. The equation would look something like:

$$T(\lambda X_1, \lambda X_2 ...) = T(X_1, X_2...)$$ Now since we have partial derivatives, you could write down functions $U$ as a function of $T$ and $p$, but in doing so, you would be losing information. Suppose you gave your new function $U(T,p,N)$ to a friend and he wanted to find $U(S,V,N)$, he would have to conduct 2 integrals, which would leave him wanting 2 constants of integration to nail down the function. So you see you can 'derive' $U(S,V,N)$ from $U(T,p,N)$ only upto a constant. This is why the Legendre transformations are helpful. They are a way to change coordinates without losing information.

$\textbf{When I say he has to conduct 2 integrals, I mean the following:}$

Suppose you expressed $U$ as a function of $T,p,N$, and using the following relations:

$$T=\left(\frac{\partial U}{\partial S}\right)_{V,N} , -p=\left(\frac{\partial U}{\partial V}\right)_{S,N} ,$$ Then, you'll have the following type of equation $$U=f\left(\frac{\partial U}{\partial S},\frac{\partial U}{\partial V},N\right)$$ Which is a partial differential equation ($f$ is some function of the variables). In the best case scenario that this is separable, 2 integrals would have to be conducted and hence 2 integration constants would be needed to obtain the exact function $U(S,V,N)$.

$\textbf{About Ideal Gases:}$ There are many ways to work out fundamental relations for ideal gases. The one you referenced in your question actually comes from Statistical Mechanics after considering phase space volumes. Thermodynamics alone is incomplete and requires many manipulations to get the results we want. In most cases, we only have expressions for $T$ and/or $p$, known as $\textbf{equations of state}$. What is really important to understand is the difference between the $\textbf{energy representation}$ and $\textbf{entropy representation}$. When the fundamental relation takes the form $U(S,V,N)$, we are working in the energy representation. If it is of the form $S(U,V,N)$, we have the entropy representation. Now the equation of state (relating to T) for the entropy representation is

$$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)_{V,N}$$ In the above it is clear that the right hand side must be a function of $U,V,N$. So the equipartition equation you wrote down is not a fundamental equation, but an equation of state in the entropy representation!