# General solution of grad r^n

1. Feb 12, 2005

### Reshma

I worked out this problem from Griffith's book. The problem is to find the general expression for $$\nabla(r^n)$$. This is how I worked it out:

If $$\vec r = \hat x x+\hat y y+\hat z z$$

r is the separation vector whose magnitude is given by $$\sqrt{x^2+y^2+z^2}$$

Hence $$r^n = (x^2+y^2+z^2)^\frac{n}{2}$$

I applied the $$\nabla$$ operator to it and this is solution I got:

$$\nabla(r^n) = n(r^2)^\frac{2n-2}{2}\vec r$$

Is this the right way to find the solution or is there another generalised solution?

2. Feb 12, 2005

### kanato

It seems to me that you shouldn't have r^2 in the parenthesis, that should be the unit vector at the end $$\hat{r}$$.. also you can reduce the fraction in your exponent. It is much easier to do something like that in spherical coordinates, see http://mathworld.wolfram.com/SphericalCoordinates.html

3. Feb 12, 2005

### dextercioby

$$\nabla (r^{n})=n r^{n-1} \nabla r$$

Compute the gradient of "r" and see whether the whole result matches yours...

Daniel.

4. Feb 12, 2005

### dextercioby

Reshma,just an advice,at the end of your calculation try to see whether your formula fits special cases.For example,n=1.
Check whether your formula "delivers the goods" for this simple case...

Daniel.

5. Feb 12, 2005

### cepheid

Staff Emeritus

$$\nabla(r^n)$$

Don't hijack other people's threads with random stuff! Post it in General Discussion. :grumpy:

Reshma,

Since the function you have been given is a function of r, it may be easier to compute the gradient in spherical coordinates. See the inside front cover of Griffiths for the formula. That way, you can at least check your answer using another method.

6. Feb 12, 2005

### dextercioby

I think it's much easier in CARTESIAN COORDINATES...

Daniel.

7. Feb 12, 2005

### cepheid

Staff Emeritus
First of all, I just re-read the thread, and realised that kanato already made the claim that it's easier in spherical coords. Second of all, I agree because the derivatives wrt phi and theta are zero, and you're only calculating one derivative!

8. Feb 12, 2005

### dextercioby

Okay.Have it your way.Though i think the expression for "nabla" in spherical coordinates is much harder to remember (in an exam,for instance) than the one in cartesian coordinates...

Daniel.

9. Feb 12, 2005

### Gokul43201

Staff Emeritus
If Jackson is good for anything it is this.

10. Feb 12, 2005

### dextercioby

And the inside covers of Cohen-Tannoudji.And the first 100 pages from Griffiths.

Daniel.

11. Feb 12, 2005

### kanato

It can be, but it's really useful to remember the gradiant and divergence for special situations where you have no angular dependence, because then they are really simple, and they take a problem that would be a pain in cartesian coordinates and make it an easy problem in one variable.

12. Feb 12, 2005

### dextercioby

I see no point in memorizing only the "r" derivative-thing from any of the traditional diff.operators...That's why i don't know them and have to use Cohen-Tannoudji every time...Which i don't mind at all...

Daniel.

13. Feb 12, 2005

### Gokul43201

Staff Emeritus
Come on. There's no memorizing involved in remembering the radial components of the gradient in spherical/cylindrical co-ords. It's the other components that need memorizing.

And this problem is clearly an example where knowing just the radial component (instead of having to do it in cartesian coords) make things a lot easier on yourself.

14. Feb 13, 2005

### Reshma

Daniel,

I am EXTREMELY sorry for my typographical error. I reviewed my result yesterday and it is actually

$$\nabla(r^n) = n(r^2)^\frac{n-2}{2}\vec r$$

I applied the result for n=2 and I got the result as

$$\nabla(r^2) = 2\vec r$$

which tallies with the general solution.
Well if you are referring to the spherical coordinates, it would rather be a "specific solution''. I've mentioned $$\vec r$$ is just a separation vector which can belong to any coordinate system.