When, usually in text of physics or concerning thermodynamical aspects of chemistry, I find notations such as$$\mathrm{d}f=g\,\mathrm{d}t$$ I always interpretate it as $\frac{\mathrm{d}f(t)}{\mathrm{d}t}=g(t)$ in the sense that the derivative of $f$ at $t$ is $f'(t)=g(t)$ and always try to keep that in mind when I see steps of calculations involving such differentials because of my love for the spirit of mathematical proofs.

I have been told by friend of mine, who is a professor of mathematics, that, at a more or less elementary didactic level, the notations such as $\mathrm{d}A$ and $\delta A$ found in thermodynamics, are really intended to be "small quantities" with no pretence of mathematical rigour. Parenthetically, I notice that these differentials, as they are usually manipulated in elementary physics textbooks, cannot be the differential forms defined in differential geometry because the differential forms of differential geometry do not constitute an algebraic field, while these differentials usually are multiplied and divided by each other as if they were real numbers.

Nevertheless, I notice that integration is usually operated on such differentials and my love for the spirit of mathematical proof leads me to try to find a rigourous way to read such forms.

Can we then read a form such as, for example, $\delta A$, which is of the kind of forms that books integrate getting $\int \delta A=\int A_1\mathrm{d}x_1+\ldots+ A_n\mathrm{d}x_n$ , as $$A_1(x_1,\ldots,x_n)x_1'(s)\mathrm{d}s+A_2(x_1,\ldots,x_n)x_2'(s)\mathrm{d}s+\ldots +A_n(x_1,\ldots,x_n)x_n'(s)\mathrm{d}s$$where $s$ is the variable of a parametrisation of a path we are going to integrate on?

To give a practical example, since $p$, $V$ and $T$ are variables often used in thermodynamics, I would tend to read $\delta Q$ as $$Q_1(p,V,T)p'(s)\mathrm{d}s+Q_2(p,V,T)V'(s)\mathrm{d}s+Q_3(p,V,T)T'(s)\mathrm{d}s$$where the $Q_i$ are opportune functions of $(p,V,T)$ . Therefore, if $(p,V,T)=:\mathbf{r}:[a,b]\to\mathbb{R}^3$ parametrises the path of the transformation and we define $\mathbf{Q}:=(Q_1,Q_2,Q_3)$, I would read $\int\delta Q$ just as $\int_{\gamma}\mathbf{Q}\cdot\mathrm{d}\mathbf{r}$, i.e. as $$\int_a^bQ_1(p,V,T)p'(s)+Q_2(p,V,T)V'(s)+Q_3(p,V,T)T'(s)\,\mathrm{d}s$$where it is worth of note that one of the $Q_i$ usually is constantly $0$ because of the interdependence of the variables and where simplifications such as $\int_a^bQ_1(p,V,T)p'(s)\mathrm{d}s=\int_{p(a)}^{p(b)}Q_1(p,V,T)\mathrm{d}p$ are usually introduced when calculating .

I $\infty$-ly thank any answerer

• The hunt for perfect differentials has been an important part of thermodynamics from the beginning, hence the notation. This is due to the macroscopic formulation in terms of intensive and extensive (additive) quantities - the expressions need to be integrable for the extensive quantities like energy, entropy, etc. Greater care must be taken as the number of variables increases, and some are held constant. The chemist knows that holding temperature constant means the beaker is on a burner; constant volume means there is a lid on it; constant pressure means the lid is off ... – Peter Diehr Feb 6 '16 at 12:41
• While the mathematics professor is correct that physicists do not care about rigor (as long as the results are correct), differentials in calculations are differentials and not some "small quantities", otherwise one would have to use some form of interval arithmetic or carry Taylor series expansions and errors bounds around and nobody does that in theory (we do it as soon as we translate theory to experiment or observational data to theory). Can one formulate thermodynamics in the language of differential geometry? Certainly. Is it necessary? No. Will it lead to something new? Probably not. – CuriousOne Feb 6 '16 at 12:46
• I think this is an extremely interesting question. It is probably related to the question of just what constitutes a state---a poorly defined concept in classical thermodynamics. These two issues have.kept me from pursuing a real understanding of classical thermodynamics, so I will watch for good answers! And keep the faith in requiring understandable, yet rigorous justification. – user59591 Feb 6 '16 at 17:27
• "while these differentials usually are multiplied and divided by each other as if they were real numbers."...note that physicists always do that. They do it in GR when writing $\mathrm{d}s^2$ for the metric and then dividing that by $\mathrm{d}t^2$, they do it when solving differential equations by multiplying with $\mathrm{d}t$, they do it when deriving the form of a volume element, and I'm sure I can come up with more examples. And we always happily integrate the results. It's nothing special to thermodynamics. – ACuriousMind Feb 6 '16 at 18:41
• Yes, you can read the differential form as you suggested the parametrization in question being just the sequence of equilibrium states the process passes through. The parameter is completely arbitrary. Ever since Caratheodory's proof of the existence of an integrating factor (absolute temperature) following from the principle of adiabatic inaccessibility this is one way of describing the 2nd law of thermostatics. This is a good introduction web.ist.utl.pt/berberan/data/68.pdf – hyportnex Feb 6 '16 at 22:36