Converting differential to gradient

Question

Landau & Lifschitz's fluid mechanics book proposes the following statement for an isentropic proccess:

$$dH=vdp \Rightarrow \nabla H=v\nabla p$$

What's the rigorous way to get this result (converting differentials to gradients)?

Answer 1

The Cartan differential basically encodes the gradient by: \begin{align*} \mathrm d H &=\partial_1H\mathrm dx_1 +\partial_2H\mathrm dx_2 +\partial_3H\mathrm dx_3 \\ =v\mathrm d p &=v(\partial_1p\mathrm dx_1 +\partial_2p\mathrm dx_2 +\partial_3p\mathrm dx_3). \end{align*} By applying this to $x_1$, $x_2$ and $x_3$ as well as using $\mathrm dx_i(x_j)=\delta_{ij}$, we get $\partial_iH=v\partial_ip$ and therefore $\nabla H=v\nabla p$.

Answer 2

The relationship $dH = v dp$ is valid along any reversible path. Suppose that path is a function of space, e.g., $x=x(t)$. Along each step $dx$ of the path the above equation is satisfied, then dividing by $dx$ we obtain $$\frac{dH}{dx} = v \frac{dp}{dx}$$ The isentropic process is reversible, so the above holds.