Why do we use different differential notation for heat and work?

Just recently started studying Thermodynamics, and I am confused by something we were told, I understand we use the inexact differential notation because work and heat are not state functions, but we are told that the '$$df$$' notation is only for functions and that the infinitesimal heat and work are 'not changes is anything' surely they can be expressed as functions of something? and they are still changes as they do change? What is the thermodynamic reason for describing them as not being changes in anything?

I believe the good way to present thermodynamics is through the formalism of differential geometry.

When the thermodynamic process is reversible it can be described as a curve on a manifold of equilibrium states (because each intermediate step is equilibrated). Then $$\delta W = -p dV$$ and $$\delta Q = T dS$$ are differential forms - covectors tangential to the manifold of equilibrium states. However, (in general) they are not exterior derivatives $$d f$$ of any state function $$f$$. There is no function of state $$W$$ such that $$dW = \delta W$$.

• Are there any introductory texts to thermodynamics that use differential geometry, aimed at undergraduate students? Mar 28, 2022 at 10:53
• @HansWurst see Bamberg and Sternberg's text for a brief introduction to thermodynamics (its the last chapter). The necessary background is developed in chapter 5 (differential calculus in several variables and introduction to 1-forms), 7 (integration of 1-forms along curves, arclength of curves) and 15 (k-forms, exterior derivatives, Stokes theorem). Having said this, the text is filled with many other interesting topics all explained at the 2nd year undergraduate level using introductory differential geometry, whenever applicable. Mar 28, 2022 at 11:07
• @HansWurst see also Florian Scheck. Statistical Theory of Heat. The corresponding chapter is only a few pages (I think 25 pages) long, tho. Mar 28, 2022 at 11:45
• Your answer is misleading. In general, neither work nor heat is a differential form. In order to be that, its coefficients should be functions of the state variables. In a general non-quasi-static change, this is false. Mar 28, 2022 at 16:59
• You're right, I corrected my answer Mar 29, 2022 at 7:54

Notation
Sometimes heat and work are marked by special signs to underscore that they are not real differentials, such as differentials with a stroke, as shown here or something like $$\text dU = \delta Q + \delta W.$$ However, there is no single established notation here, and most of the time one simply does not bother to use any special symbols - the risk of misunderstanding is very low (after one has understood the basics of stat. mech.)

Are work and heat functions?
Work and heat are, of course functions, but they depend not only on the variables of the system, and therefore they are not functions of the state variables alone. E.g., if we work in $$p,V$$ variables, then there are many paths connecting states $$p_1,V_1$$ and $$p_2,V_2$$ - each such path corresponds to a different combination of work and heat, although the internal energy at the end of the path is always the same. This means that heat and work are not differentials in strictly mathematical sense, whereas the internal energy is (see here for the difference between a derivative and a differential).

• What are work and heat functions of then? Mar 28, 2022 at 10:00
• @BioPhysicist e.g., they could be functions of time - if we are given $p(t)$ and $V(t)$. But in this case we have specified a path. Perhaps, I was incorrect saying that they are functions of more variables than just the state variables. Mar 28, 2022 at 10:03
• Could you give me an example of why they aren't strictly differentials? Mar 28, 2022 at 12:29
• @user1007028 pay attention to what I say: they are not differentials of state variables They can be functions, e.g., of time, and be differentials of their variables - but not of the state variables. E.g., take two paths in PV plane corresponding to the same change of $U$ (i.e., its differential) but different work, and you have proven it. The problem here is not with work and heat, but with what differential means in math. Mar 28, 2022 at 12:35
• Okay, I'm starting to see, thanks for that, to be honest the reason I'm confused is my module is quite basic and they did not explain the paths. Mar 28, 2022 at 12:42

"What is the thermodynamic reason for describing them as not being changes in anything?"

Well, what would they be changes in? There isn't some quantity of heat belonging to something of which $$\delta Q$$ is a change; it's only heat while the energy is flowing. A similar remark applies to work, which is also energy in transit.

• Are the heat and work not changes in the energy of the system? I'm not very far in thermodynamics (and our lectures aren't great in all honesty) but surely an infinitesimal amount of work done and heat released are infinitesimal changes in its internal energy? Mar 28, 2022 at 12:24
• Could you even think of the total work as a quantity that can change infinitesimally?, I'll have another look. Mar 28, 2022 at 12:33
• "Are the heat and work not changes in the energy of the system?" They do contribute to changes in the internal energy (according to the First Law of Thermodynamics), but you can't equate heat or work individually to changes in internal energy because it is their sum, that so equates. Mar 28, 2022 at 17:07

$$df,\,\Delta f$$ and $$\delta f$$ are associated with the idea of a change from an initial value to a final value.

So $$\Delta f$$ or $$\delta f$$ are equal to $$f_{\rm final}-f_{\rm initial}$$ and $$df$$ when the change is infinitesimal.

As you have pointed out with work and heat there are no initial and final states but it is useful to use some sort of notation for amount of work done and heat supplied.

Some people use $$\delta f$$ explaining that it represents a "small" amount whereas others use the lower case delta with a bar through it (I cannot find the Latex symbol for this although there is one for $$h$$, $$\hbar$$) to make the distinction more obvious.
So "deltabar Q" might equal $$m\,C\,\delta T$$ and "delta bar W" might equal $$F\,\Delta x$$ where $$\delta T = T_{\rm final}-T_{\rm initial}$$ and $$\Delta x = x_{\rm final}-x_{\rm initial}$$.