Divergence of a radial field ##F=\hat{r}/r^{2+\varepsilon}##

[PhysicsKush]

Homework Statement

Let's say we have a radial field $F=\hat{r}/r^{2+\varepsilon}$. What is the divergence of this field ? You may ignore the origin.

Relevant Equations

$$\vec{\nabla} \cdot \vec{F} = \left(\left( \frac{1}{r^2 }\right)\frac{\partial(F_{r} r^2)}{\partial r}
+ \left( \frac{1}{r \sin \theta}\right)\frac{\partial(f_{\theta}\sin \theta)}{\partial \theta} + \left( \frac{1}{r\sin \theta}\right)\frac{\partial f_{\phi}}{\partial \phi}\right) dr d\theta d\phi$$

Following (1),
\begin{align*}
\text{div} F = \vec{\nabla} \cdot \vec{F} &= \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 F_{r}\right) \\ &= \frac{1}{r^2} \frac{\partial }{\partial r} \left( r^2 \frac{1}{r^{2+\varepsilon}}\right) \\ &= \frac{1}{r^2} \frac{\partial}{\partial r} (r^{-\varepsilon}) \\ &= \left( \frac{-\varepsilon}{r^2}\right) \frac{1}{r^{\varepsilon +1}} \\ &= -\frac{\varepsilon}{r^{\varepsilon +3}} \qquad , \text{for} \ \ r \in \mathbb{R} \setminus \{ 0\}
\end{align*}

The solution I'm providing seems to be to simplistic. Perhaps I have missed a step somewhere? Thank you in advance.

 

[PeroK]

Why do you think it should be more complicated than that?

 

[PhysicsKush]

Why do you think it should be more complicated than that?

Thank you for your reply. Quite frankly this is my final answer but the reasoning seems quite simple. I just wanted to make sure that I don't miss anything out :)!

 

[PeroK]

Thank you for your reply. Quite frankly this is my final answer but the reasoning seems quite simple. I just wanted to make sure that I don't miss anything out :)!

There's nothing missing that I can see.

 

[vela]

The solution I'm providing seems to be to simplistic.

You mean "too simple." Simplistic means "oversimplified." The phrase "too simplistic" means "too oversimplified" (as opposed to being oversimplified just enough). It just doesn't make sense.

https://brians.wsu.edu/2016/05/31/simplistic/

 

[PhysicsKush]

You mean "too simple." Simplistic means "oversimplified." The phrase "too simplistic" means "too oversimplified" (as opposed to being oversimplified just enough). It just doesn't make sense.

https://brians.wsu.edu/2016/05/31/simplistic/

Ah English is my fourth language therefore I still make common mistakes, thanks for pointing out this grammatical mistake !

 

[Delta2]

Like the others i don't see anything wrong in your work. Maybe it is kind of simple because it is done in spherical coordinates. Try to do it in cartesian coordinates i think it would be one page long or even more!.

 

[pasmith]

Like the others i don't see anything wrong in your work. Maybe it is kind of simple because it is done in spherical coordinates. Try to do it in cartesian coordinates i think it would be one page long or even more!.

$$\sum_i\frac{\partial}{\partial x_i} \left(\frac{x_i}{r^{3+\epsilon}}\right) = \sum_i\left(\frac{1}{r^{3 + \epsilon}}\frac{\partial x_i}{\partial x_i} - (3 +\epsilon) \frac{x_i}{r^{4 + \epsilon}}\frac{\partial r}{\partial x_i} \right) = \frac{3}{r^{3 + \epsilon}} - (3 +\epsilon) \frac{\sum_i x_i x_i}{r^{5 + \epsilon}} = \frac{3 - (3 + \epsilon)}{r^{3 + \epsilon}} = - \frac{\epsilon}{r^{3 + \epsilon}}$$ in view of the obvious results $$\sum_i x_ix_i = r^2,\qquad\sum_i\frac{\partial x_i}{\partial x_i} = 3,\qquad \frac{\partial r}{\partial x_i} = \frac {x_i} r$$

 

[Delta2]

$$\sum_i\frac{\partial}{\partial x_i} \left(\frac{x_i}{r^{3+\epsilon}}\right) = \sum_i\left(\frac{1}{r^{3 + \epsilon}}\frac{\partial x_i}{\partial x_i} - (3 +\epsilon) \frac{x_i}{r^{4 + \epsilon}}\frac{\partial r}{\partial x_i} \right) = \frac{3}{r^{3 + \epsilon}} - (3 +\epsilon) \frac{\sum_i x_i x_i}{r^{5 + \epsilon}} = \frac{3 - (3 + \epsilon)}{r^{3 + \epsilon}} = - \frac{\epsilon}{r^{3 + \epsilon}}$$ in view of the obvious results $$\sum_i x_ix_i = r^2,\qquad\sum_i\frac{\partial x_i}{\partial x_i} = 3,\qquad \frac{\partial r}{\partial x_i} = \frac {x_i} r$$

Well you have compactified some steps (for example i would like to see the representation of the radial field in cartesian coordinates and the derivatives of $r=\sqrt{\sum x_i^2}$ with respect to each cartesian coordinate) but ok i am more than surprised that it can be done in less than half a page.

 

[pasmith]

Everything follows from $$r^2 = \sum_j x_j^2$$ and thus by the chain rule $$2r \frac{\partial r}{\partial x_i} = \sum_j 2x_j \frac{\partial x_j}{\partial x_i} = 2x_i.$$
As the radial basis vector is $x_i/r$, functions of $r$ alone are not too difficult to deal with in cartesian coordinates; it's when functions depend on $\theta$ and $\phi$ that things get messy.

Naturally, if you write $(x,y,z)$ instead of $(x_1,x_2,x_3)$ then you will use more paper.

 

[vanhees71]

To save even more paper you can also use the Einstein summation convention and drop the summation symbols.