Divergence of a vector function

[KEØM]

Homework Statement

Let's define the radial vector \vec{v}(r) = \hat{r}/r^{2} where \vec{r} = \vec{OP} (O being the origin of our coordinate system and P being our observation point at point (x, y, z)). Using spherical coordinates, demonstrate that \vec{\nabla<br /> } \cdot\vec{v}(r) = 0 everywhere except at r = 0. At r = 0, demonstrate that \vec{\nabla}\cdot\vec{v}(r) is going to infinity.

I can show the first part but I can't show how it goes to infinity when r = 0.

Homework Equations

\vec{\nabla}\cdot\vec{v} = \frac{1}{r^{2}}\frac{\partial(r^{2} v_{r})}{\partial r} + \frac{1}{r sin\theta}\frac{\partial}{\partial\theta}(sin\theta v_{\theta}) + \frac{1}{r sin\theta}\frac{\partial v_{\phi}}{\partial\phi}

The Attempt at a Solution

For the first part, because \vec{v} is only a function of r the two last terms in the divergence formula will equal zero. The first term becomes

\vec{\nabla}\cdot\vec{v}(r) = \frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\cdot\frac{1}{r^{2}}) = 0

zero as well, satisfying part a.

For part b, if we let r = 0 then \vec{v}(r) becomes infinitely large and I am not sure how to take the divergence of that.

Any help will be appreciated.

Thanks in advance,

KEØM

 

[vela]

Try applying Gauss's theorem where the volume is a sphere of radius R centered at the origin. Then let R go to zero and argue the result only makes sense if div v is infinite at the origin.

 

[KEØM]

Thanks for replying vela.

So I use Gauss's theorem and I get:

\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}(\vec{\nabla}\cdot\frac{\hat{r}}{r^{2}})(r^{2}sin\theta dr d\theta d\phi) = \int_{0}^{2\pi}\int_{0}^{\pi}(\frac{\hat{r}}{r^{2}})\cdot(r^{2}sin\theta d\theta d\phi \hat{r})

After evaluating the integral on the right hand side I get:

\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}(\vec{\nabla}\cdot\frac{\hat{r}}{r^{2}})(r^{2}sin\theta dr d\theta d\phi) = 4\pi

So now do I let R in the integration limit go to zero? I am not sure what to do with the integral after that happens.

Thanks again,

KEØM

 

[vela]

Now you want to argue that div v can't be bounded. Assume it is bounded and show that if that's true, the LHS will vanish as R goes to 0. But this contradicts the fact that the RHS is constant.

(This is a hand-waving argument since the divergence theorem only holds for fields that are continuously differentiable over the entire volume, which v obviously isn't.)

 

[KEØM]

So just as a recap of what you said. If I assume that div v is finite everywhere then I can show that as I let R go to zero my left hand side goes to zero (because any integral with both integration limits of the same value will be zero). But this assumption can't be right because the RHS is non-zero and constant.

Now from this how can I show that \vec{\nabla}\cdot\frac{\hat{r}}{r^{2}} = 4\pi \delta^{3}(\vec{r})?

Can I simply say that because the divergence is zero everywhere except at r = 0 (where it is infinite) and it's integral over any volume independent of r is 4\pi\cdot 1 (properties of the 3D dirac delta function) it must be equal 4\pi\delta^{3}(\vec{r})?

Thanks again vela.

KEØM

 

[vela]

Yup, you got it.