# Converting from differential form to substantial derivative form (Gibb's Equation)

Problem: My problem is understanding how to go from the traditional form of Gibb's equation, to the substantial derivative form:

$$Tds=de+pdv \rightarrow T\frac{Ds}{Dt} = \frac{De}{Dt} + p\frac{Dv}{Dt}$$

I've looked through at least 8 or 9 different textbooks, and all of them skip this step. I do not understand how you can simply replace the differential forms with substantial derivatives.

My attempt:

Beginning with $$s=s(e,v)$$ and taking the time derivative $$\frac{\partial s}{\partial t} = \frac{\partial s}{\partial e}\frac{\partial e}{\partial t} + \frac{\partial s}{\partial v}\frac{\partial v}{\partial t}$$

From Gibb's differential form above, we know:

$$\frac{\partial s}{\partial e} = \frac{1}{T} \ \ and \ \ \frac{\partial s}{\partial v} = \frac{P}{T}$$

Substituting, we have: $$\frac{\partial s}{\partial t} = \frac{1}{T} \frac{\partial e}{\partial t} + \frac{p}{T}\frac{\partial v}{\partial t}$$

This is where I become stuck. I know the substantial derivative is the form:

$$\frac{D}{Dt} = \frac{\partial}{\partial t} + \vec{u} \cdot \nabla ()$$

If I substitute in for the $$\frac{\partial}{\partial t}$$ terms for $$s, e, \ \ and \ \ v$$, I end up with extra $$\vec{u} \cdot \nabla$$ terms. $$\frac{Ds}{Dt} = \frac{1}{T}\frac{De}{Dt} + \frac{p}{T}\frac{Dv}{Dt} - \frac{1}{T}\vec{u} \cdot \nabla e - \frac{p}{T}\vec{u} \cdot \nabla v + \vec{u} \cdot \nabla s$$

Is there some complicated process no one ever talks about, where you can show these extra terms cancel, or am I missing something fundamental? I have seen textbooks literally say that "one simply replaces the differentials in the Gibbs equation with the substantial derivative." If someone could please offer some insight as to why someone can "simply replace" the differential terms with the substantial derivative terms, I would greatly appreciate it.

After expanding the time derivative, expand entropy in terms of its spatial derivatives as well. This will give you an expression for, $$\nabla s = \frac{1}{T} \nabla e + \frac{p}{T} \nabla v.$$ Take the substantial derivative of s, and then substitute in for the partial time derivative (using the expressions I obtained in the original post) and gradient of s (using the expression I obtained in this post). This should give you the expected solution from the original post. Now you can derive it without "simply replacing" terms and maintain your sanity (or maybe that's just me).