- #1

rohanprabhu

- 414

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**[SOLVED] Surprising Gradient not 'Surprising Enough'**

## Homework Statement

Q] Sketch the vector function and

[tex]

v = \frac{\hat{r}}{r^2}

[/tex]

and compute it's divergence.

**The answer may surprise you... can you explain it?**

['r' is the position vector in the Euclidean space]

## Homework Equations

[tex]

\nabla \cdot v = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}

[/tex]

## The Attempt at a Solution

[tex]

\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}

[/tex]

and,

[tex]

\mathbf{\hat{r}} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{\sqrt{x^2 + y^2 + z^2}}

[/tex]

Hence,

[tex]

\frac{\mathbf{\hat{r}}}{r^2} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}

[/tex]

[tex]

\nabla \cdot v = \frac{(x^2 + y^2 + z^2)^{\frac{1}{2}}((x^2 + y^2 + z^2)^3 - 3x^2)}{(x^2 + y^2 + z^2)^3} + \frac{(x^2 + y^2 + z^2)^{\frac{1}{2}}((x^2 + y^2 + z^2)^3 - 3y^2)}{(x^2 + y^2 + z^2)^3} + \frac{(x^2 + y^2 + z^2)^{\frac{1}{2}}((x^2 + y^2 + z^2)^3 - 3z^2)}{(x^2 + y^2 + z^2)^3}

[/tex]

on using Simplify in Mathematica:

[tex]

\nabla \cdot v = \frac{3(-1 + x^4 + y^4 + 2y^2 z^2 + z^4 + 2x^2 y^2 + 2x^2 z^2))}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}

[/tex]

But there was nothing 'surprising' about this result. What am I doing it wrong?

Also, this problem is from D. Griffith's 'Introduction to Electrodynamics'. And I am at the 2nd chapter which deals with Calculus used in Electrodynamics. The thing is that I am heavily confused between a Gradient, Divergence and Curl. Hence, to overcome that, I think the only way is practicing a lot of questions. Could anyone please show me any source where I could get lots of questions on Gradient, Divergence and Curl to practice [it'd be nice if it were a free resource].

thanks