Find gradient in spherical and cartesian coordinates

[naele]

Homework Statement

Find the gradient of [itex]3r^2[/itex] in spherical coordinates, then do it in Cartesian coordinates

Homework Equations

[tex]
\nabla f=\hat r \frac{\partial f}{\partial r} + \hat \theta \frac{1}{r} \frac{\partial f}{\partial \theta}+ \hat \phi \frac{1}{r\sin \theta}\frac{\partial f}{\partial \phi}
[/tex]
[tex]z=r \cos \theta[/tex]

The Attempt at a Solution

Since there's no [itex]\theta, \phi[/itex] then the gradient is simply [itex]6r \hat r[/itex]. Transforming to cartesian coordinates gives [tex]\frac{z}{6}\hat z[/tex] because cos 0 = 1. Any of the other coordinate transforms involve [itex] \sin \theta [/itex] or [itex] \sin \phi[/itex] so z is the only non-zero coordinate.

 

[Dick]

Are you given that theta=0? If so that's ok, except how did the 6 move from the numerator to the denominator?

 

[naele]

It wasn't necessarily given, I just assumed it was since the [itex]3r^2[/itex] has no theta or phi term to just consider them as zero when I took the gradient. That second part with z/6 I suspect is completely wrong because I don't understand how to transform from spherical to cartesian.

 

[Dick]

It wasn't necessarily given, I just assumed it was since the [itex]3r^2[/itex] has no theta or phi term to just consider them as zero when I took the gradient. That second part with z/6 I suspect is completely wrong because I don't understand how to transform from spherical to cartesian.

Then you can't put theta=0. [itex]6 r \hat r[/itex] is fine. What's r in cartesian coordinates? What's [itex]\hat r[/itex] in cartesian coordinates? You must know at least one of those. Look them up if you don't. Alternatively, express 3r^2 in cartesian coordinates and do it directly.

 

[naele]

r is just the magnitude so in Cartesian that's [tex]\sqrt{x^2 + y^2 +z^2}[/tex]. I think that [tex]\hat r = \sin \theta \cos \phi \hat x + \sin \theta \sin \phi \hat y + \cos \theta \hat z[/tex]

 

[Dick]

That's ok. Then 3r^2=3*(x^2+y^2+z^2). Can you use that to find the gradient directly in cartesian coordinates? But why not write
[tex]
\hat r = \sin \theta \cos \phi \hat x + \sin \theta \sin \phi \hat y + \cos \theta \hat z
[/tex]
as
[tex]
\hat r = (x \hat x + y \hat y + z \hat z)/r
[/tex]
That's the same thing isn't it?

 

[naele]

Yes, I see it now. Doing it both ways gives me the same result
[tex]6x \hat x + 6y \hat y + 6z \hat z[/tex]