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

Question

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?

Answer 1

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 $dW = \delta W$.

Answer 2

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).

Answer 3

"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.

Answer 4

$ 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 $\Delta x = x_{\rm final}-x_{\rm initial}$.