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A certain passage in my Fundamentals of Thermodynamics book is driving me crazy. I considered posting this is Physics.SE, but I think the question is eminently mathematical. Here is the passage:

As we know, $c_p$ is defined as: $$c_p = \left(\frac{\partial h}{\partial T} \right)_p \tag1$$ We also have seen that: $$T\ ds = dh - v\ dp \tag2$$ Therefore: $$c_p = T \left({\partial s \over \partial T} \right)_p \tag3$$

I've spent hours trying to figure out how to get to $(3)$. I really want to do this as rigorously as a graduated mechanical engineer can.

So far I have two "approaches," one that is fine for most of my friends but is very informal and another that is more correct, but leads me nowhere.

In case it's needed, let me give some background on the thermodynamic variables:

  • for curiosity's sake, $c_p$ means specific heat in a constant pressure process;
  • $p, v, T, u, h, s$ are properties, of which any one can be seen as a function of any other two;
  • the notation $\left({\partial a \over \partial b}\right)_c$ means that, in that derivative, $a = a(b,c)$, so its purpose is to make it clear that the variable held constant is $c$.

Approach 1: simple but "wrong"

From $(2)$, we have: $$dh = T\ ds + v\ dp \tag4$$

We plug that directly into $\partial h$ of $(1)$ to obtain: $$c_p = \left({T\ \partial s + v\ \partial p \over \partial T} \right)_p \tag5$$

Since $p$ is held constant, $\partial p = 0$, so $(3)$ emerges.

My problems with this:

  1. Making $\partial h = dh$ and plugging it in $(1)$ looks awful to me. I've never seen a "partial differential" outside of a partial derivative (e.g. $\partial h$ by itself). We usually take that liberty with "total differentials," but I've never seen that done with partials.
  2. $\partial p$ can't even exist inside the derivative of $(5)$, since it would mean $p = p(T, p)$, which... does it make any sense?! Anyway, it can be argued that the partial derivative would indeed be zero, but it sucks.

Approach 2: correct but dead end

If we make $h = h(T, p)$, then comes: $$dh = \left({\partial h \over \partial T} \right)_p dT + \left({\partial h \over \partial p} \right)_T dp\tag6$$

Equating with $(4)$ gives: $$\left({\partial h \over \partial T} \right)_p = T\frac{ds}{dT} + \frac{dp}{dT}\left[v - \left({\partial h \over \partial p} \right)_T \right] \tag7$$

So, from $(1)$: $$c_p = T\frac{ds}{dT} + \frac{dp}{dT}\left[v - \left({\partial h \over \partial p} \right)_T \right] \tag8$$

Which doesn't look like $(3)$ at all. In order to make it better, we could "open up" $\frac{ds}{dT}$ with $s(T,p)$ and eventually find: $$c_p = T \left({\partial s \over \partial T} \right)_p + \frac{dp}{dT}\left[v - \left({\partial h \over \partial p} \right)_T + T \left({\partial s \over \partial p} \right)_T \right] \tag9$$

We'd need the second term to be zero, but $\frac{dp}{dT}$ certainly isn't and I don't see why the $[\cdots]$ would be either.

Did I go wrong somewhere? Is Approach 2 really optimal? I appreciate any help.

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Here's how I would think about it:

Since we want to do this mathematically, let's be pedantic and not use the physicists' habit of letting the same letter denote a physical quantity and the function which describes its dependence on other physical quantities. (A convenient practice, but also a source of much confusion...)

Thus, if we wiew $h$ as a function of $T$ and $p$, we write $$h=H_1(T,p),$$ where $h$ is the value of a physical quantity and $H_1$ is a function of two variables. But if we wiew $h$ as a function of $s$ and $p$, it's a different function, so we use another symbol and write $$h=H_2(s,p)$$ instead. Now, $s$ may also be viewed as a function of $T$ and $p$, say $s=S(T,p)$, and if we insert this into $H_2$ we get an expression for $h$ as a function of $T$ and $p$, and this must of course agree with what the function $H_1$ says, so we have the relation $$ H_1(T,p) = H_2(S(T,p),p) \tag{a} . $$ What physicists write as $$ c_p = \left(\frac{\partial h}{\partial T} \right)_p , $$ a pure mathematician would write as $$ c_p(T,p) = \frac{\partial H_1}{\partial T}(T,p) , $$ and using the chain rule in (a), we see that we can also compute this as $$ c_p(T,p) = \frac{\partial H_2}{\partial s}(S(T,p),p) \, \frac{\partial S}{\partial T}(T,p) . \tag{b} $$ Next, the differential relation $$ dh = T \, ds + v \, dp $$ concerns $h$ as a function of $s$ and $p$, so it says something about the function $H_2(s,p)$, namely that if we write $T=\tau(s,p)$ and $v=V(s,p)$ to denote the dependence of $T$ and $v$ on $s$ and $p$, then we have $$ \tau(s,p) = \frac{\partial H_2}{\partial s}(s,p) \qquad\text{and}\qquad V(s,p) = \frac{\partial H_2}{\partial p}(s,p) . $$ Comparing the first of these two equalities with what we had in (b), we find that $$ c_p(T,p) = \tau(S(T,p),p) \, \frac{\partial S}{\partial T}(T,p) . $$ Here, finally, $\tau(S(T,p),p)$ is simply $T$ again (if you think about the definitions for a while), while $\frac{\partial S}{\partial T}(T,p)$ is what physicists write as $\left({\partial s \over \partial T} \right)_p$. And there you have it, $c_p = T \, \left({\partial s \over \partial T} \right)_p$.

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    $\begingroup$ Perfect! This is exactly what I was looking for. The distinction between $H_1(T,p)$ and $H_2(s,p)$, along with correct application of the chain rule, is the key. Thank you. (PS: sorry for the delay, I didn't mean it) $\endgroup$ – André Chalella Sep 11 '14 at 16:00
  • $\begingroup$ You're welcome! Glad to be of help. :-) $\endgroup$ – Hans Lundmark Sep 11 '14 at 17:25
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Approach 1 is not so bad. Think of dh as δh, a small change in h. You have a simple system. If p is fixed then there is effectively only one variable. In other words, these are not really partial derivatives, because only one variable can vary.

The use of partial derivatives is really a kind of aide-memoire, because in practice you have some formula for s involving maybe p and T and you have to remember that p is not a variable.

For example, for an ideal gas you might have $s=k\ln T-R\ln p+k'$ for some constants $k,k'$. If you keep $p$ fixed, then $s$ is just a function of $T$.

I am not sure why you think $\frac{\text{d}\,p}{\text{d}\,t}$ is definitely non-zero. $p$ is being kept fixed, so when $T$ varies it does not change. So the rate of change must be 0. Of course, there may be some mechanism going on behind the scenes to make that happen, but that is often the case.

I agree that it would be better if people did not use things like $\partial p$. It would be better to use the traditional physicist's approach of writing it as $\delta x$ - the equivalent of the mathematician's $\epsilon-\delta$ arguments.

In Approach 2, p is fixed, so it does not depend on anything, even T, so your argument above is not right.

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  • $\begingroup$ Now I see what you mean: as $p$ is a constant (i.e not a variable), effectively $h = h(T)$, so $\left({\partial h \over \partial T}\right)_p$ can really be seen as ${dh \over dT}$, $dh$ being $T\ ds$ (afterwards we'd revert $T\ {dh \over dT}$ back to partial form for completeness). This explains why the book was quick here. I thought ${dp \over dT}$ was nonzero because I took the function pressure (of any two variables) for the variable/constant which it really was (Hans' answer helped me in that regard). $\endgroup$ – André Chalella Sep 11 '14 at 16:27
  • $\begingroup$ Finally, can you elaborate on the theoretical difference between $\partial p$ and $\delta p$? Seems interesting. Thank you, @almagest, and sorry for taking so long. I really didn't mean it. $\endgroup$ – André Chalella Sep 11 '14 at 16:28
  • $\begingroup$ @AndréNeves see math.stackexchange.com/questions/143222/what-does-dx-mean?rq=1 Some of those answers (like the first one, accepted by the questioner) are misleading. But it introduces the subject of differential geometry, which is large and hard. dx has a respectable meaning as a differential form; but as it commonly appears in integrals and differential equations it is unhelpful. But it would take me more space than I have here to go into that. $\endgroup$ – almagest Sep 11 '14 at 19:41
  • $\begingroup$ @AndréNeves Another more general point. Mathematicians need physicists to keep them from going off the rails. The interaction between maths and its applications is subtle. $\endgroup$ – almagest Sep 11 '14 at 19:43
  • $\begingroup$ This imbroglio made me aware of the dangers of weak terminology in calculus. I had already read many questions about the subject (including the one you mentioned) and noticed that Differential Geometry is a name that comes up frequently. I wonder if studying it, even if just the basics, is a good way to get a good grasp in these matters, and also if it's doable for an engineer (that likes math very much). $\endgroup$ – André Chalella Sep 12 '14 at 9:41
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Differential of $f(x, y)$ in $(x_0, y_0)$ is a function defined like this: $$df(\delta x, \delta y)_{(x_0, y_0)} = \frac{\partial f}{\partial x}_{(x_0, y_0)} \delta x + \frac{\partial f}{\partial y}_{(x_0, y_0)} \delta y$$ for any $\delta x, \delta y \in \mathbb{R}$

Consider equation (2) and treat all functions as functions of $T, p$ then: $$T(\frac{\partial s}{\partial T}_{(T_0, p_0)} \delta T + \frac{\partial s}{\partial p}_{(T_0, p_0)} \delta p) = \frac{\partial h}{\partial T}_{(T_0, p_0)} \delta T + \frac{\partial h}{\partial p}_{(T_0, p_0)} \delta p - v \ dp(\delta T, \delta p)_{(T_0, p_0)} \ \ (*)$$ $d p$ is abuse of notation, for clarity let's get rid of it: $$d p(\delta T, \delta p)_{(T_0, p_0)} = d f (\delta T, \delta p)_{(T_0, p_0)},$$ where $f(T, p) = p$ so $$d f (\delta T, \delta p)_{(T_0, p_0)} = 1\ \delta p \ \ (**)$$

Now we can substitute (**) into $(*)$ and diffirentiate (*) by $\delta T$ to get result derived in your book. Actually this would be more strictly written approach 1: $$T\frac{\partial{ds}}{\partial \delta T} = \frac{\partial ds}{\partial \delta T} - v \frac{\partial d p}{\partial \delta T}$$

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    $\begingroup$ This was very good. Just some minor questions: you say we can zero ${\partial p \over \partial T}$ "since $p$ is constant", but couldn't we do it from the beginning since $p$ is an independent variable in the equation? Or isn't it, them being only $\delta T$ and $\delta p$ (not $T$ and $p$)? Also, if $p$ was not constant, that partial would be zero anyway, wouldn't it? Last thing: is there a typo in the last equation? $\endgroup$ – André Chalella Sep 12 '14 at 10:17
  • $\begingroup$ Yes, seems that "p is constant" does not matter here :) About variables: during first differentiation(when we evaluate differential), independent variables are $T$ and $p$, during second it is $\delta T$, $\delta p$. Also there actually was a typo in last formula, thanks for noticing! I've edited the answer to include all above considerations. $\endgroup$ – Artem Malykh Sep 12 '14 at 12:56
  • $\begingroup$ I think one typo remains, lol. It should be $\partial dh$ instead of $\partial ds$ on the right-hand side, right? Also, now I understand why ${\partial \over \partial \delta T}$ is "more strictly written approach 1." Thank you very, very much! However I guess I'd still leave the "zero in brackets" part. It was useful for my understanding that, in practice, that is what the ${\partial \over \partial \delta T}$ comes down to. $\endgroup$ – André Chalella Sep 13 '14 at 8:33
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You can derive pretty much everything from equation (2) you have

$Tds = dh - vdp$

Taking derivative with respect to T gives

$T\frac{\partial s}{\partial T}= \frac{\partial h}{\partial T} - v \frac{\partial p}{\partial T}$

(Note that we dropped a ds term from the LHS since it holds no significance in such equation)

Now keeping p constant you lose the last term and you get the desired result

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  • $\begingroup$ What exactly is "taking the derivative" here? Is it doing ${\partial\ (T\ ds) \over \partial T} = {\partial\ (dh - v\ dp) \over \partial T}$? If so, aside from $ds$, ${-\partial v \over \partial T}$ was left off too. Also, I feel uncomfortable doing things like ${\partial\ (dh) \over \partial T} = {\partial h \over \partial T}$, could you elaborate on its correctness? $\endgroup$ – André Chalella Sep 12 '14 at 9:51
  • $\begingroup$ As you pointed out, it's not exactly taking the derivative but sort of taking those infinitesimal in respect with T, it's the fastest way to think about this and also correct in many aspects, this is why both T and v didn't get derivative (since they are not infinitesimal) (in this explanation I'm withdrawing what I wrote in the "Note") $\endgroup$ – Avrham Aton Sep 12 '14 at 10:08

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