# Divergence in spherical polar coordinates

1. Aug 18, 2011

### Idoubt

I took the divergence of the function 1/r2$\widehat{r}$ in spherical coordinate system and immediately got the answer as zero, but when I do it in cartesian coordiantes I get the answer as 5/r3.

for $\widehat{r}$ I used (xi+yj+zk)/(x2+y2+z2)1/2

what am i missing?

2. Aug 18, 2011

### IttyBittyBit

What you're missing is the handling of the singularity at r=0.

3. Aug 18, 2011

### I like Serena

Hi Idoubt!

If I take the cartesian derivative with respect to x I get:
$${\partial \over \partial x}{x \over (x^2+y^2+z^2)^{3/2}} = {-2 x^2+y^2+z^2 \over (x^2+y^2+z^2)^{5/2}}$$
Of course the derivatives wrt to y and z have to be added yet...

How did you get 5/r3?

4. Aug 19, 2011

### Idoubt

sorry I made a silly error in my calculation

$\frac{\partial}{\partial x}$ x(x2+y2+z2)-$\frac{3}{2}$ =

(x2+y2+z2)-$\frac{3}{2}$ - 3x2(x2+y2+z2)-$\frac{5}{2}$

When i add y & z parts it cancels and becomes zero, so the div of $\frac{1}{r^2}$$\widehat{r}$ is zero.

The reason that confused me was Electric fields can have a non zero divergence. Is that because of the singularity at r=0?

5. Aug 19, 2011

### I like Serena

Not every electric field is generated by exactly 1 symmetrical charge.
And in practice there is no singularity, since the charge would have to be a point charge for that.

Last edited: Aug 19, 2011
6. Aug 19, 2011

### Idoubt

Isn't the field due to just one charged assumed to be a field created by a point charge? ( I mean it makes no difference if we make that assumption right? )

But, how can an electric field have a non zero divergence then?

7. Aug 19, 2011

### I like Serena

Yes. It would only make a difference when you get very close to where the point charge is supposed to be.

Take for instance 2 point charges and you'll have non-zero divergence.

8. Aug 19, 2011

### Idoubt

But when we take two charges we no longer have the $\frac{1}{r^2}$ function ( unless we approximate at large distances)

It is my understanding that even the electric field of a point charge has non zero divergence is it not so?

9. Aug 19, 2011

### I like Serena

Right on both counts. :)

My point about point charge is unrelated to divergence.
Only that a real electric field has no singularity.

10. Aug 19, 2011

### Idoubt

my question is , how can a field which is scalar multiple of the function $\frac{1}{r^2}$$\hat{r}$ have a divergence when i just proved that the divergence of this function is zero?

11. Aug 19, 2011

### vanhees71

The divergence is not 0, it's $4 \pi \delta^{(3)}(\vec{x})$. It's not too tricky to prove with help of Green's first integral theorem applied to the whole space with a little sphere around the origin taken out and then taking its radius to 0.

12. Aug 19, 2011

### Idoubt

This by definition is a singularity isn't it?

13. Aug 19, 2011

### Galron

In physics it's often modelled thusly AAMOI:

I know not relevant but...

14. Aug 19, 2011

### Idoubt

greek to me :)

15. Aug 19, 2011

### Galron

meh a and b are just constants and in this case F(X) is obviously the Function of X.

The Greek symbol that looks like an O with a hat means Tensor.

E(X) I would assume means the Energy of the system in question in terms of the vector.

16. Aug 19, 2011

### Idoubt

greek with a few constants and a function of x now =)

17. Aug 19, 2011

### Galron

If you've done matrices you'll know a tensor is expressed as a matrix of a matrix.

A vector in this case is simply a matrix with a scalar.

http://en.wikipedia.org/wiki/Vector_space

So to steal more wiki imagery a tensor is:

Last edited by a moderator: Apr 26, 2017
18. Aug 19, 2011

### Idoubt

hmm ok ok so thats a little advanced for me, but my question was about the divergence of the $\frac{1}{r^2}$$\hat{r}$ function. In simpler terms can you tell me the reason for non-zero divergence in the electric field of a point charge?

19. Aug 19, 2011

### vanhees71

No, it's a distribution, the socalled Dirac $\delta$ distribution. It's a very useful concepts. It's even so useful that the mathematicians built a whole new subfield of mathematics to define this concept rigorously, functional analysis!

20. Aug 19, 2011

### I like Serena

I suspect you're referring to Maxwell's law that says:
$$\textrm{div }\vec E = \frac {\rho} {\epsilon_0}$$

The divergence of a spherically symmetric charge is zero everywhere except at the place where the charge actually is.
A point charge is a special case, since the charge density would be infinite (actually a Dirac delta-function).
In practice the charge would take in a certain volume.
The divergence of the electric field is not zero at the place where the charge density is not zero.