# Converting differential to gradient

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)?

• Relevant: differential vs. derivative. In a nutshell: $df=\nabla f\cdot \mathbf{dx}$. Commented Sep 26, 2022 at 9:12

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 $$\nabla H=v\nabla p$$.
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.