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I'm going through some old engineering lecture notes. I've already spotted some errors in the notes. In an important part of a derivation, the lecturer did the following:

$Tds = du + pd(1/\rho)$

can be rewritten as

$T\nabla s = \nabla u + p\nabla(1/\rho)$

The equation is the second law of thermodynamics where s, u, p, and $\rho$ are the entropy, internal energy, pressure, and density: all scalars, and all state variables (that do not depend on the path taken to get from point a to point b).

I've seen similar transformations done for the material derivative:

$TDs/Dt = Du/Dt + pD(1/\rho)/Dt$

where

$Dg/Dt = \partial g/\partial t + \vec{u}\bullet\nabla g$

My question is:

When are these forms of taking a derivative interchangeable or not (for a scalar in 3D Cartesian space)? Or do you know of a handy online reference that explains it?

Take a more general case (still scalar variables):

If I can write,

$dg = (\partial g/\partial a)da + (\partial g/\partial b)db$

I get the sense that I can't just swap "d" for $\nabla$... about notation: let's stick with x,y,z as spatial coordinates, t is time, everything else is an arbitrary scalar, and $\vec u$ is a vector.

Eric
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1 Answers1

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Recall that $$df=\nabla f\cdot d\vec r$$

Then, note that we can write for any displacement vector $d\vec r$

$$\begin{align} ds&=\nabla s\cdot d\vec r\\\\ du&=\nabla u\cdot d\vec r\\\\ d(1/p)&=\nabla (1/p)\cdot d\vec r \tag 1 \end{align}$$

Using the expressions in $(1)$, we can write

$$\begin{align} Tds=du+pd(1/p)\implies &T(\nabla s\cdot \vec dr)=(\nabla u\cdot d\vec r) +p(\nabla (1/p)\cdot d\vec r)\\\\ \implies &(T\nabla s)\cdot \vec dr=(\nabla u+p\nabla (1/p))\cdot d\vec r\\\\ \implies &(T\nabla s-\nabla u-p\nabla (1/p))\cdot d\vec r=0 \end{align}$$

As this holds for all displacement vectors, then we conclude

$$\bbox[5px,border:2px solid #C0A000]{T\nabla s=\nabla u+p\nabla (1/p))}$$

Mark Viola
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