What is $\nabla \times d\vec{r}$?

Question

Does $\nabla \times d\vec{r}$ have any meaning? How does one go about computing it?

Answer 1

For all $f$, we have $\nabla \times \nabla f=0$. We also have $\vec{r}=\nabla(r^2/2)$. $\nabla$ is independent of time and so commutes with time, so I'm thinking:

$\nabla \times d\vec{r}=dt \nabla \times \frac{d\vec{r}}{dt}=dt\frac{d}{dt}(\nabla \times \vec{r})=dt\frac{d}{dt}(\nabla \times \nabla(r^2/2))=0$

I get the same result in Einstein Notation.

$\nabla \times d\vec{r}=\epsilon_{ijk}\frac{\partial}{\partial x^j}dx^k$. Because order of differentiation doesn't matter, we have that $\frac{\partial}{\partial x^j}dx^k=\frac{\partial}{\partial x^k}dx^j$. Repeated indices implies a sum. The anti-symmetry of the Levi-Civita symbol means that those terms will be added and they will have opposite sign thus cancelling, again giving us zero.