Question

evaluate ∇ × r(vector) / r(magnitude)

Answer 1

Answer on Question #42009 - Math - Vector Calculus

Vector $\vec{r}$ has following components in Descartes coordinate system: $\vec{r} = (x; y; z)$, $r = \sqrt{x^2 + y^2 + z^2}$.

Let us evaluate this cross product using the Leibniz rule for nabla operator:

$$[\nabla, \frac{\vec{r}}{r}] = \frac{1}{r} [\nabla, \vec{r}] + [\nabla (\frac{1}{r}), \vec{r}]$$

Let us use the property of nabla operator $\nabla f(r) = \frac{\partial f(r)}{\partial r} \cdot \frac{\vec{r}}{r}$. It is easy to prove it. Let us take x-component of nabla operator: $\frac{\partial}{\partial x} f\left(\sqrt{\left(x^2 + y^2 + z^2\right)}\right) = \frac{\partial f(r)}{\partial \sqrt{\left(x^2 + y^2 + z^2\right)}} \cdot \frac{2x}{2\sqrt{\left(x^2 + y^2 + z^2\right)}} = \frac{\partial f(r)}{\partial r} \cdot \frac{x}{r}$. The same result takes place for y and z derivative (with y and z in numerator respectively).

Thus, $\nabla \left(\frac{1}{r}\right) = \frac{-1}{r^2} \cdot \frac{\vec{r}}{r} = \frac{-\vec{r}}{r^3}$, so $[\nabla \left(\frac{1}{r}\right), \vec{r}] = \frac{-1}{r^3} [\vec{r}, \vec{r}] = 0$ (according to the properties of cross product, $[\vec{a}, \vec{a}] = 0$).

The term $[\nabla, \vec{r}] \equiv \text{rot} \, r$ is equal to zero. By definition, $[\nabla, \vec{r}]_k = \varepsilon_{ijk} \nabla_i (\vec{r})_j$. Since Levi-Civita tensor is not equal to zero only if its indexes are different, and $\nabla_i \vec{r}_j = \delta_{ij}$, all components of $[\nabla, \vec{r}]$ are equal to zero, thus $[\nabla, \vec{r}] = 0$.

Finally, $[\nabla, \frac{\vec{r}}{r}] = \frac{1}{r} [\nabla, \vec{r}] + [\nabla (\frac{1}{r}), \vec{r}] = 0$.

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