Coordinate Systems Homework: Prove \nabla.\vec{r}=3

[shyta]

Homework Statement

For the cartesian, cylindrical, spherical coordinate system,
prove that
\nabla.\vec{r} = 3 and \nablax\vec{r}=0

Homework Equations

For cylindrical coord system,

\vec{r} = s\vec{s} + z\vec{z}

\nabla = \vec{s} \delta/\deltas + \vec{\varphi}\frac{1}{\varphi}\delta/\delta\varphi + \vec{z} \delta/\deltaz

The Attempt at a Solution

Hi guys, I managed to do the cartesian coord part, but I'm having trouble with the cylindrical/spherical parts. Using these 2 equations I tried to do a dot product but I'm getting a 2 instead. What am I doing wrongly here?
Please help.

 

[I like Serena]

Hi shyta!

You appear to have a wrong formula there.
Compare for instance with this page:
http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

Furthermore, the gradient of r is not zero, perhaps you intended the curl?

 

[shyta]

Ops, I meant the cross of r. (edited that, thanks for pointing out) I'm pretty sure the gradient formula is correct as checked on the wiki link you gave. Do you mean my r vector is wrong? \vec{r} = s\vec{s} + z\vec{z}

Well I looked through my workings and I'm quite sure I got that right as well. Perhaps you could be more specific?

Thanks i like serena!

 

[I like Serena]

The formula for divergence in cylindrical coordinates is:
\nabla \cdot \vec A = <br /> {1 \over s}{\partial \left( s A_s \right) \over \partial s} <br /> + {1 \over s}{\partial A_\phi \over \partial \phi} <br /> + {\partial A_z \over \partial z}
As you can see this does not match what you have.

In your case this becomes:
\nabla \cdot \vec r = <br /> {1 \over s}{\partial \left( s \cdot s \right) \over \partial s} <br /> + {\partial z \over \partial z}

 

[shyta]

Hmmmmm okay I can see how that would give me the correct answer.. But still I am a little confused about this formula on wiki because in my textbook, the gradient in cylindrical coordinates formula is as the one I stated above. In particular,

<br /> {1 \over s}{\partial \left( s A_s \right) \over \partial s} <br />

vs just <br /> {\partial \left( A_s \right) \over \partial s} <br />

Thanks for the replies!

 

[I like Serena]

On the same wiki page you can see that you have the formula for gradient in cylindrical coordinates almost right.
Except for you putting a phi, where there should be an "s".

Note that the formulas for gradient, divergence and curl are not trivial in cylindrical or spherical coordinates.
It takes some work to derive these formulas.
Are they not in your textbook?

 

[shyta]

ohh! I got it! omg lol thanks alot!