# Divergence in spherical coordinates

1. May 13, 2010

### billiards

I am stuck on this problem.

Use these equations:

$$\textbf{v}(\textbf{r}) = f(r)\textbf{r}$$

$$\frac{\partial r}{\partial x} = \frac{x}{r}$$

And the chain rule for differentiation, show that:

$$(\nabla\cdot\textbf{v}) = 2f(r) + r\frac{df}{dr}$$
(cylindrical coordinates)

Any help greatly appreciated, I will post my progres so far in a following post.

Cheers

2. May 13, 2010

### billiards

So far I have tried to crack this by subbing in

$$\textbf{v}(\textbf{r}) = f(r)\textbf{r}$$

in the bottom equation I get:

$$(\nabla\cdot f(r)\textbf{r}) = 2f(r) + r\frac{df}{dr}$$

But I think that if we write out the left hand side of the equation and separate out the differentials we get this:

$$(\nabla\cdot f(r)\textbf{r}) = r_{x}\frac{\partial f(r)}{\partial x} + f(r)\frac{\partial r_{x}}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y} + f(r)\frac{\partial r_{y}}{\partial y}$$

Which I think reduces further to:

$$(\nabla\cdot f(r)\textbf{r}) = 2f(r) + r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y} = 2f(r) + r\frac{df}{dr}$$

Which IF that is right (and it's a big IF) basically means that I'm left to prove that:

$$r\frac{df}{dr} = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}$$

That's where I'm stuck. I have a feeling that the total derivative is needed next and I am thinking along those lines, I still haven't used the chain rule yet so I think that will come in to play in the total derivative context. It could well be that I've strayed off and am heading nowhere with this, which is why I need a bit of guidance.

Thanks

3. May 13, 2010

### LCKurtz

I assume you know, or can prove, that for scalar f and vector V in rectangular coordinates,

$$\nabla\cdot f\vec V = \nabla f \cdot \vec V + f(\nabla \cdot\vec V)$$

Also, I assume the standard notation

$$\vec r = \langle x, y \rangle,\ r = |\vec r|$$

Then by the first identity

$$\nabla\cdot f(r)\vec r = \nabla f(r) \cdot \vec r + f(r)(\nabla \cdot \vec r)$$

The only place you need the chain rule is to calculate the first term on the right.

 Corrected typos.

Last edited: May 13, 2010
4. May 14, 2010

### billiards

If I understand you correctly then I have more or less followed your reasoning thus far.

I am stuck on the bit you mention in your last line, I am struggling to calculate the first term on the right.

The way I see it is that

$$\nabla f(r) \cdot \vec r = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}$$

Is that correct?

That's my last line in my earlier post. How do I invoke the chain rule here to get this into the form I am after?

Thanks

5. May 14, 2010

### LCKurtz

When you differentiate f(r) with respect to, for example, x, you get f'(r)rx. And do the dot product last.

6. May 16, 2010

### billiards

Okay, I think I've got it now.

Use the chain rule:

$$\frac{\partial f(r)}{\partial x} = \frac{df}{dr}\times \frac{\partial r}{\partial x}$$

And the result:

$$\frac{\partial r}{\partial x} = \frac{x}{r}$$

To get:

$$r\frac{df}{dr} = r_{x}\frac{df}{dr}\frac{x}{r} + r_{y}\frac{df}{dr}\frac{y}{r}$$

Now rx=x. Use this result as well as the fact that in cylindrical coordinates r2 = x2+y2,and the term on the right reduces to the term on the left. Is this OK?

Cheers

Last edited: May 16, 2010
7. May 16, 2010

### LCKurtz

Yes, I think you have it figured out now.

8. May 17, 2010

### billiards

Ahhh good. For some reason that one had me scratching my head for a while, but it seems quite simple now.

Cheers