Discussion
For a position vector \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) the norm of the vector can be defined as \(|\vec{r}| = \sqrt{x^{2}+y^{2}+z^{2}}\). Given a function \(\phi = \ln|\vec{r}|\), its gradient \(\nabla\phi\) is
Answer: C
The gradient is \(\dfrac{\vec{r}}{\vec{r}\cdot\vec{r}}\).
With \(r = |\vec{r}|\) and \(\phi = \ln r\), the chain rule gives \(\nabla\phi = \frac{1}{r}\nabla r\), and \(\nabla r = \vec{r}/r\), the unit vector along \(\vec{r}\):
\[ \begin{aligned} \nabla\phi &= \frac{1}{r}\cdot\frac{\vec{r}}{r} \\ &= \frac{\vec{r}}{r^{2}} = \frac{\vec{r}}{\vec{r}\cdot\vec{r}} \end{aligned} \]
For instance, \(\partial\phi/\partial x = \frac{1}{2}\cdot\frac{2x}{x^{2}+y^{2}+z^{2}} = x/r^{2}\).
\(\vec{r}/|\vec{r}|\) is the gradient of r itself, not of ln r. \(\vec{r}/|\vec{r}|^{3}\) is minus the gradient of 1/r, the inverse-square field. \(\vec{r}\) is the gradient of \(r^{2}/2\).
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