# 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.

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.

1. 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!

• I like your answer! I understand what you write about $U(T,p,N)$ but I can't immediately connect it to the ideal gas. There, I have the function $U(T,N)$. Also could you elaborate on the "2 integrals"part? – Marc Jan 24 '17 at 23:24
• I added to the answer to elaborate :) – Sreekar Voleti Jan 25 '17 at 0:41
• Thank you, this answers my question :-) [little latex-tip in case you don't know: if you write \left( instead of ( the size of your bracket is matched to your expression] – Marc Jan 25 '17 at 13:53
• Glad your question is answered. Thanks for the latex tip as well, I always found that small bracket size annoying! – Sreekar Voleti Jan 27 '17 at 0:39
• I don't think point (1) is correct at all. While the natural variables for $U$ happen to all be extensive, this is not true for the other thermodynamic potentials. For example, the natural variables for $G$ are $p$ and $T$, which are both intensive. I think the real answer is given by By Symmetry below. – knzhou Jun 18 '18 at 10:59

The first law of thermodynamics states that we can write the change in internal energy as the heat applied plus the work done on the system $$\mathrm{d}U = \delta Q + \delta W\,.$$

In order to make some use of this expression we need to find some expression for $\delta Q$ and $\delta W$. In general this is difficult, not least because work and heat are path dependent quantities. In the special case of a reversible path, however we make some progress. In particular the reversible work $\mathrm{d} W_\mathrm{rev} = -p \mathrm{d}V + \mu \mathrm{d}N$ and from the definition of entropy we have $\mathrm{d} Q_\mathrm{rev} = T\mathrm{d}S$. Pluggin gthis into the fist law gives us $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V+\mu \mathrm{d}N\,.$$ Moreover because all the variables in this expression are functions of state, this expression must be path independent, and so holds on all paths and not just reversible ones.

Notice that this differential is exceptionally simple. It only contains the thermodynamic variables or their differentials, not complicated functions of them. It also allows us to write expressions like $p = -\left(\frac{\partial U}{\partial V}\right)_S$, giving us an easy way to calculate the pressure or temperature.

This is what is meant when we say that $U$ is naturally a function of $S$, $V$ and $N$. If $U=U(S,V,N)$ it has an exceptionally simple differential that allows us to relate the different thermodynamic variables. We could write $U$ as a function of say $T$ and $P$ $$\mathrm{d}U = \left(\frac{\mathrm{d}U}{\mathrm{d} T}\right)_p\mathrm{d} T + \left(\frac{\mathrm{d}U}{\mathrm{d} P}\right)_T\mathrm{p}$$ but $\left(\frac{\mathrm{d}U}{\mathrm{d} T}\right)_p$ would generally be some complicated function of the different thermodynamic variables.

Notice as well that this derivation used only generally properties of any thermodynamic system, it did not depend on the exact details of the system being considered. If we tied to define "natural" variables based on the explicit form of $U$ as a function of the variables, rather than using the differentials we would not find much general behaviour.

Similarly when we look at quantities such as the enthalpy or the Gibbs free energy we can derive similar simple differentials, allowing us to relate the other thermodynamic variables through the derivatives of the potentials.

• Thanks. What I understood from your answer is that for my function $U(T,N)$, i is difficult to calculate a partial derivative where I keep $p$ fixed. This makes sense to me. – Marc Jan 24 '17 at 23:20

Probably not the most satisfying answer, but with my experience I guess that this are physics books. It is "natural" because that is what you have control over in a physics thought experiment. You put a fixed amount of particles in a fixed volume where you have a defined amount of accessible states. Physicists, hence, like to work with internal energy. In contrast, in chemistry you'll find the free energy, free enthalpy, etc, because the way they "cook" provides other "natural" variables. In an open test tube you control temperature and pressure and the numbers of particles is most likely not constant. So I would call "natural" as most convenient to control. Problem is that most of the time you cant control $S$,$N$,$V$ in real life. For $N$ vs $\mu$ and $P$ vs $V$ this is somewhat easy to see, and I understand that due to the difficulties with entropy it is harder to understand, why this is the case for $S$ vs $T$. But from a pure physics point of view $S$ is probably more convenient and, therefore "natural".

Having read Callen's Thermodynamics I take $S(U,V,N)$ to be the fundamental equation and $U,V,N,$ to be the natural quantities. I think this is appropriate because $U,V,N,$ may be assigned to any system, whether in equilibrium or not; this implies that they may be unambiguously defined under arbitrary conditions. Also they are additive over sub-systems comprising the main system. That one can invert the roles of $S$ and $U$ is a consequence of the postulate that relation between them is one-to-one.

• Both are considered fundamental equations in Callen, only the representation differs, i.e. in the Energy Representation, the fundamental equation is U(S,V,N) and in the Entropy representation, S(U,V,N) is. – Sreekar Voleti Jan 24 '17 at 5:38
• Yes but he begins with $S(U,V,N)$, which beginning seems more natural to me. In fact any of its Legendre transformations may be considered fundamental because they are all equivalent. – Deep Jan 24 '17 at 7:31