JohnDenker
commented
9 years ago The book systematically mostly avoids multi-dimensional derivatives. Instead it uses the integral formulation. I can see good pedagogical reasons for that choice (even though when I'm doing calculations it's not my favorite way of doing things). In any case, I'm not going to second-guess the decision.
Previously I said "not every vector field is the gradient of some potential" ... which expresses the idea using the derivative formulation. To be a good team player, I should rephrase my suggestion using the integral formulation. It goes like this:
Suppose we are integrating along some path. Actually, suppose there are two paths, Γa and Γb. They share the same starting-point and ending-point. That is:
Γa(1) = Γb(1) = x1, and
Γa(2) = Γb(2) = x2The interesting thing is, sometimes the value of the integral depends only on the endpoints x1 and x2. We say the value is a function of x1 and x2. However, sometimes the value of the integral depends on every detail of the path. We say the value is a _functional_ of Γ.
What's even more remarkable is that in the expression
ΔE == E(2) - E(1) = W + QE itself is a function of the thermodynamic state, and therefore ΔE is a function of two states, independent of how the path goes from 1 to 2 ... whereas neither W nor Q is a function of state(†). W and Q are not state functions and cannot be written in terms of state functions, not even approximately, because they depend on every detail of the path Γ(†).
(†) means: except in trivial casesFor about a dozen reasons (pedagogical and otherwise) I recommend making this dependence explicit, every time W and Q appear. That is, make a habit of writing them as functionals, explicitly: W[Γ] and Q[Γ]. As a point of notation: Observe that a function such as E(1) is written using parentheses, whereas a functional such as W[Γ] is written square brackets. For example:
ΔE(1,2) == E(2) - E(1) = W[Γ] + Q[Γ]This is important. Every freshman who looks at thermodynamics tries to invent W-like and Q-like functions that are functions of state. Sometimes they think they have succeeded, but of course they never really succeed.
ericmazur
Chapter 20, page 547: There is a string of problems here, centered around the concept of «thermal energy». This is in some ways related to issue #35.
In a previous chapter, «thermal energy» was defined to be "energy that was transferred thermally". In other words,Δ(Eth) := (ΔE)th [by definition of Δ(Eth)] [1]That is risky business, because the «thermal energy» that we see in parentheses on the LHS exists only in rather special situations, such as the infamous Slinktato™. The bugs come home to roost on page 547. If we plug equation 1 into equation 20.2, we find that (ΔE)th = W + Q. However, Q is equal to (ΔE)th [by definition of Q], so equation 20.2 is evidently a wrong equation. It says Q = W + Q. It is grossly violated whenever there is nonzero W.So, there is tremendous potential for confusion here. For example, consider a gas that expands against a thermally-insulating piston, doing work against another gas. In accordance with equation 20.2, this is a transfer of thermal energy, but it is not a thermal transfer of energy.
The same bug appears in a slightly different guise in equations 20.3 and 20.4, which allege that E_th is a function of state. It might be for a Slinktato™ ... but it absolutely cannot be for a gas (ideal or otherwise).
This is a classic misconception. Every freshman who has toyed with thermodynamics has tried to develop a theory in which «heat» and «work» are thermodynamic potentials, i.e. functions of state. All-too-often, they think they have succeeded ... but of course they never really succeed. The physics says that neither TdS nor PdV is the gradient of any potential, except in trivial cases.
In the PER literature a holy war was waged against the word «heat», arguing the «heat» is not a noun. As the saying goes, this was the wrong war, at the wrong place, at the wrong time, and with the wrong enemy. The problem is not with the parts of speech; the problem is with the concept.
In particular, using «thermal energy» as a euphemism for «heat» is not helpful, not even a little bit, because the idea is still wrong. The question is not whether TdS is a noun; the question is whether it is a scalar. Hint: It's not.
Constructive suggestion: Oddly enough, the equations on page 547 are rather easy to fix. The LHS of equations 20.3 and 20.4 is properly E, the plain old energy. You can differentiate it to obtain dE = T dS - P dV (subject to mild restrictions), which more-or-less corresponds to equation 20.2. (Repairing the wording that surrounds the equations will be a somewhat bigger job.)
Bedrock suggestion: It would help to get students to understand that not every vector field is the gradient of some potential. This will require a major effort. This is central to understanding what you can (and cannot!) do with Kirchhoff's so-called «laws», and it is also central to understanding what you can (and cannot!) do with «heat» and «work». Otherwise you're just playing whack-a-mole against an endless succession of misconceptions.
Some diagrams and other tricks that may help with this can be found at https://www.av8n.com/physics/non-grady.htm