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What this notation mean? (I know that is a partial derivative, but I don't understand the meaning of the evaluation bar at the right)

$$\frac{\partial g}{\partial T}\Big|_{SA,p}$$

Is this relation true?

$$\frac{\partial g}{\partial T}\Big|_{SA,p}=\left(\frac{\partial g}{\partial T}\right) _{SA,p}$$

Also used here

$$d\rho=\left(\frac{\partial \rho}{\partial T}\right)_{S,p}+dT\left(\frac{\partial \rho}{\partial S}\right)_{T,p}+dS\left(\frac{\partial \rho}{\partial p}\right)_{T,S}dp$$

And another thing, the $$d\rho$$ must be intended as the differential of the function $$\rho$$?

So if I have understood well, the equality can be rewritten as:

\begin{equation*} \frac{1}{\rho}\left(\frac{\partial \rho}{\partial T}dT+\frac{\partial \rho}{\partial S}dS+\frac{\partial \rho}{\partial p}dp\right)=-\rho\left(\frac{\partial \rho^{-1}}{\partial T}dT+\frac{\partial \rho^{-1}}{\partial S}dS+\frac{\partial \rho^{-1}}{\partial p}dp\right) \end{equation*}

\begin{equation*} \frac{1}{\rho}\sum_{i=1}^3\frac{\partial \rho}{\partial x_i}dx_i=-\rho\sum_{i=1}^3\frac{\partial \rho^{-1}}{\partial x_i}dx_i \end{equation*}


Proof:

$$ \frac{d\rho}{\rho}=-\frac{d\alpha}{\alpha}=-\rho d\rho^{-1}=-\rho\sum_{i=1}^3\frac{\partial \rho^{-1}}{\partial x_i}dx_i $$ Where $$ \alpha=\frac{1}{\rho}\\ \rho=f(S,T,p)\\ $$

$\Box$

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    $\begingroup$ Usually, the bar means that the partial derivative is taken at a particular, and constant, value of, say, $SA,p$ as in the first example. $\endgroup$
    – Jas Ter
    Commented May 5, 2013 at 20:00
  • $\begingroup$ Simen K and Ted Shifrin has already answered your first question. About $\mathrm{d}\rho$, what else can you imagine it to be? And I have absolutely no idea what you meant by the two displayed equations under "So if I have understood well..." Where did you get those two displayed inequalities and why do you think they are true? $\endgroup$ Commented May 6, 2013 at 9:03
  • $\begingroup$ @WillieWong Of course in fact I've left a positive comment under the Simen's answer. For the equations that you've mentioned, please try to click on "Also used here" and follow the explanation "It is convenient to express the equation of state in differential form as follows" $\endgroup$
    – Aurelius
    Commented May 6, 2013 at 9:12

1 Answer 1

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The notation $\left(\dfrac{\partial V}{\partial p}\right)_T$, for example, is used by chemists and others. The variable(s) outside the parentheses are to be kept fixed. This is useful notation when there are variables that are implicitly related and each can be in turn thought of as a function of the others.

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  • $\begingroup$ Good answer but partial. Please, can you look at my updated question and try to answer at it? $\endgroup$
    – Aurelius
    Commented May 6, 2013 at 8:54
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    $\begingroup$ Your first equation fir $d\rho$ has $dT$ and $dS$ in the wrong places. The evaluation bar is ordinarily used as @Simen K. indicated. These are two different notations. The rest is correct. $\endgroup$ Commented May 6, 2013 at 12:57

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