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In Fermi's "Thermodynamics" there's a proof of the formula: $$W=\int _{V_1} ^{V_2} p\,\text dV,$$that is, the work done by the pressure of a gas that expands from a volume $V_1$ to a volume $V_2$ on the surface that contains it is equal to the integral above. The proof goes like this:

Consider a surface element $\text d \sigma $ and let $\text d n$ be its displacement in the direction normal to it. The infinitesimal work done on this element during the expansion is given by $$F_\perp \text d n=p\,\text d\sigma \, \text d n.$$ Since the pressure is assumed to be constant everywhere, this gives:$$\text d W=p\int \text d \sigma \, \text d n.$$ On the other hand, the variation $\text d V$ is given by the surface integral:$$\text d V=\int \text d \sigma \, \text d n$$ and so the formula.

I don't think it is unrespectful to Fermi to call this a fake proof, at least by the mathematician's point of view. I was wondering how could one rigorously justify all the passages, starting from the usual definition of work:$$W=\int _{\mathbf r _1} ^{\mathbf r _2} \mathbf F\cdot \text d\mathbf r .$$ In particular, how could I make sense of the (very puzzling) formula $\text d V=\int \text d \sigma \, \text d n$?

This is one of a billion cases, in elementary physics, where is used an infinitesimal reasoning to get a finite result (where there's no “differential forms” or other sophisticated technology implied) and I think it would be interesting to hear a mathematician point of view.

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  • $\begingroup$ There are really two things I don't understand here: 1) In the formula $F_\perp dn = p d\sigma dn$, we would seem to have a $1$-form $F_\perp dn$ equal to a $2$-form $p d \sigma dn$. Is $F_\perp$ itself "infinitesimal," then, and thus a $1$-form? 2) In the expression $\int d\sigma dn$, which $1$-form is presumably being "integrated out"—one can, if I recall correctly, make rigorous sense of this—to result in a $1$-form? $\endgroup$ Oct 6 '13 at 19:52
  • $\begingroup$ Yes, the $F_\perp=d \sigma$ is an infinitesimal. The integral $\int d\sigma dn$ is with respect to the $\sigma$ variable. $\endgroup$
    – pppqqq
    Oct 6 '13 at 19:57
  • $\begingroup$ @BranimirĆaćić is right - the thing translates straightforwardly into differential forms. But why start from $W=\int _{\mathbf r _1} ^{\mathbf r _2} \mathbf F\cdot \text d\mathbf r$? The rigorous physical point is that work is reversible. A classic way to make this sharp is to define it as convertible to lifting or lowering a weight. You can then do $W = \int {\rm d}{\bf r \cdot F}$, on that weight. But an electrical definition would have given $W = \int {\rm d}Q\ V$. $\endgroup$ Oct 7 '13 at 5:04
  • $\begingroup$ I'd start from there because, as far as I know, that's the mechanical definition of work. Do you mean that it would be easier, for the problem considered, to use a operative definition of work as the lifting of a weight? $\endgroup$
    – pppqqq
    Oct 7 '13 at 13:48
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    $\begingroup$ I recommend the definition using a weight (or some-such) because it is more rigorous - not in terms of calculus, but in terms of physics. While your formula is the mechanical definition of work, it is not the thermodynamic definition of work. $\endgroup$ Oct 8 '13 at 7:17
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I know this is not really what the OP wanted, but to me $W = \int {\rm d}{\bf r \cdot F}$ is a bad starting point. It's a bit like a schoolboy saying Pythagoras' theorem is $a^2 + b^2 = c^2$ and forgetting about triangles. I am sure you can sharpen it up by defining just what manifold to apply the formula too, but the classic definition of thermodynamic work is more physical:

Work is done by a system on the surroundings if the sole effect on the surroundings could have been raising a weight.

I think my university textbook was much more pedantic. In particular it would have taken care of the weight falling as well. But I will leave that as an exercise.

So here is my proof of $\Delta W = P\Delta V$, for any constant-pressure process (turning this into $dW = PdV$ really is about maths). Go to the theoretical physics lab (place of thought experiments) and:

  • Place the system of interest inside a bath filled with an incompressible fluid of negligible density.
  • Seal the that bath with a piston of weight $mg$, have a vacuum above the piston.
  • Allow any thermodynamic process to occur in your system, and observe the change in height $\Delta y$ of the piston/weight after it has come to rest.

proof: If $\Delta y >0$, then by the definition above the system has done work $W = mg\Delta y$. By Pascal's law, the pressure of the fluid on the piston is the same as $P$ for the system under test. Since the fluid is incompressible, the volume change in the system $\Delta V$ is also the volume change in the whole bath. But if the piston's area is $A$, then $\Delta V = A\Delta Y$. But the hydrodynamic force on the piston has to balance gravity, thus $PA = mg$. Therefore $P\Delta V = mg\Delta Y = W$.

I like to think that proof is even more aggravating to mathematician's than Fermi's "fake" proof. But what interests me is that it probably buries $W = \int {\rm d}{\bf r \cdot F}$ underneath some physical assumption. My guess is that it is in Pascal's law. But then you need Pascal's law to make sense of the formula in the first place.

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  • $\begingroup$ Thank you, I really like this line of reasoning. I'm not going to accept the answer because my point was how could I make rigorous Fermi's proof by the analytic point of view, but this is surely adding something interesting to the post. +1 $\endgroup$
    – pppqqq
    Oct 8 '13 at 12:29
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Indeed, $\vec{F}$ $\large\left(~\mbox{due to the pressure}~\right)$ is perpendicular to the surface. That defines the pressure $P$ according to $\vec{F} \equiv P\,{\rm d}\vec{S}$. $P$ is always a scalar. In addition, the displacement ${\rm d}\vec{r}$ is parallel to ${\rm d}\vec{S}$. Then, $$ \vec{F}\cdot{\rm d}\vec{r} = P\,{\rm d}\vec{S}\cdot{\rm d}\vec{r} = P\ \overbrace{\quad% \left\vert{\rm d}\vec{S}\right\vert \cdot \left\vert{\rm d}\vec{r}\right\vert\cos\left(0\right)\quad} ^{\displaystyle{{\rm d}V}} = P\,{\rm d}V $$ The result is misleading since $P\,{\rm d}V$ is evaluated at the surface and $P\,{\rm d}V$ is not an exact differential.

Whenever $\vec{F}$ isn't parallel to ${\rm d}\vec{S}$ is due to other physical contributions as the viscosity which is added to the contribution from the pressure. In any case, $P$ remains as a scalar. By the way, $\eta\nabla^{2}\vec{v}$ is the force per unity of volume due to the viscosity. $\eta > 0$: Viscosity coefficient. $\vec{v}$: Fluid velocity.

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  • $\begingroup$ Thank you, I understand the idea behind this. I just would like to know if it's possible to formulate with rigour the above theorem without the aid of differential forms, or nonstandard analysis and stuff like that. $\endgroup$
    – pppqqq
    Oct 7 '13 at 18:58

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