# Integrating Differentials in Thermodynamics

“My bad!”, he said.

I’ve just realised I made a mistake when teaching my statistical physics course last term. Fortunately it was a minor and careless maths mistake, rather than any lack of physics clarity. But it’s time to set the record straight!

Often in thermodynamics you will derive equations in differential form. For example, you might be given some equations of state and asked to derive the entropy of a system using the first law

$\displaystyle dE = TdS - pdV$

My error pertained to exactly such a situation. My students had derived the equation

$\displaystyle dS = (V/E)^{1/2}dE+(E/V)^{1/2}dV$

and were asked to integrate this up to find $S$. Naively you simply integrate each separately and add the answers. But of course this is wrong! Or more precisely this is only correct if you get the limits of integration exactly right.

Let’s return to my cryptic comment about limits of integration later, and for now I’ll recap the correct way to go about the problem. There are four steps.

1. Rewrite it as a system of partial DEs

This is easy – we just have

$\displaystyle \partial S/\partial E = (V/E)^{1/2} \textrm{ and } \partial S / \partial V = (E/V)^{1/2}$

2. Integrate w.r.t. E adding an integration function $g(V)$

Again we do what it says on the tin, namely

$\displaystyle S(E,V) = 2 (EV)^{1/2} + g(V)$

3. Substitute in the $\partial S/\partial V$ equation to derive an ODE for $g$

We get $dG/dV = 0$ in this case, easy as.

4. Solve this ODE and write down the full answer

Immediately we know that $g$ is just a constant function, so we can write

$\displaystyle S(E,V) = 2 (EV)^{1/2} + \textrm{const}$

Contrast this with the wrong answer from naively integrating up and adding each term. This would have produced $4(EV)^{1/2}$, a factor of $2$ out!

So what of my mysterious comment about limits above. Well, because $dS$ is an exact differential, we know that we can integrate it over any path and will get the same answer. This path independence is an important reason that the entropy is a genuine physical quantity, whereas there’s no absolute notion of heat. In particular we can find $S$ by integrating along the $x'$ axis to $x' = x$ then in the $y'$ direction from $(x',y')=(x,0)$ to $(x',y')=(x,y)$.

Mathematically this looks like

$\displaystyle S(E,V) = \int_{(0,0)}^{(E,V)} dS = \int_{(0,0)}^{(E,0)}(V'/E')^{1/2}dE' + \int_{(E,0)}^{(E,V)}(E'/V')^{1/2}dV'$

The first integral now gives $0$ since $V=0$ along the $E$ axis. The second term gives the correct answer $S(E,V) = 2(EV)^{1/2}$ as required.

In case you want a third way to solve this problem correctly, check out this webpage which proves another means of integrating differentials correctly!

So there you have it – your health warning for today. Take care when integrating your differentials!

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