How to find gradient of $ f = \frac{\mathrm{ln}(kr)}{r} $

Question

I have to find gradient of: $$ f = \frac{\mathrm{ln}(kr)}{r} $$

It is well-known that:

$$\mathrm{grad} = \frac{\partial}{\partial{x}}\vec{i} + \frac{\partial}{\partial{y}}\vec{j} + \frac{\partial}{\partial{z}}\vec{k}$$

but neither $r$ nor $k$ is vector here, (it is also well known that division of vectors is normally undefined, so $r$ cannot be the vector by defenition), how should I handle this then?

Answer 1

grad($f$)$=(\hat i\frac{\delta}{\delta x}+\hat j\frac{\delta}{\delta y}+\hat k\frac{\delta}{\delta z})f$

Now, $\frac{\delta}{\delta x}(\frac{\mathrm{ln}(kr)}{r})=\frac{1}{r^2}\frac{\delta r}{\delta x}+\mathrm{ln}(kr) (\frac{-1}{r^2})\frac{\delta r}{\delta x}$

As $r^2=x^2+y^2+z^2$, On differentiating partially w.r.t. $x$ you will get $\frac{\delta r}{\delta x}=\frac{x}{r}$. Plug it in the above to get the value of partial derivative $\frac{\delta}{\delta x}(\frac{\mathrm {In}(kr)}{r}) $. Similarly you can obtain $\frac{\delta}{\delta y}(\frac{\mathrm{In}(kr)}{r})$ and $\frac{\delta}{\delta z}(\frac{\mathrm{In}(kr)}{r})$.

Answer 2

I suspect your $f$ is given in spherical coordinates, with $r$ specifying the distance from the origin.

Then you have $$\nabla f=\frac{\partial f}{\partial r}\hat{r},$$ as $f$ has no angular components.

Also beware that you cannot take the gradient of a vector; it is only defined for scalar fields.

Try solving the problem now. If it is still causing you difficulties, let me know.

Answer 3

Take the gradient in polar coordinates $$\nabla=\frac{\partial}{\partial{r}}\hat{r}+\frac{1}{r}\frac{\partial}{\partial\phi}+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}$$ So, $$\nabla{f(r)}=(\frac{1}{r^{3}}-\frac{\ln(kr)}{r^{2}})\hat{r}$$