What is the General Solution for the Gradient of r^n?

[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?

 

[kanato]

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?

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

 

[dextercioby]

To see whether your solution is correct or not,follow this
\nabla (r^{n})=n r^{n-1} \nabla r

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

Daniel.

 

[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.

 

[cepheid]

Biology, this is a thread about computing:

\nabla(r^n)

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

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.

 

[dextercioby]

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

Daniel.

 

[cepheid]

First of all, I just re-read the thread, and realized 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!

 

[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.

 

[Gokul43201]

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.

If Jackson is good for anything it is this.

 

[dextercioby]

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

Daniel.

 

[kanato]

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.

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.

 

[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.

 

[Gokul43201]

I see no point in memorizing only the "r" derivative-thing from any of the traditional diff.operators.

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.

 

[Reshma]

To see whether your solution is correct or not,follow this
\nabla (r^{n})=n r^{n-1} \nabla r

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

Daniel.

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.
Thanks for your help.

Regards,
Reshma

 

[Reshma]

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

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.