Converting vector in cartesian to cylindrical coordinates

[dingo_d]

Homework Statement

This seems like a trivial question (because it is), and I'm just not sure if I'm doing it right.

I have vector in cartesian coordinate system:

\vec{a}=2y\vec{i}-z\vec{j}+3x\vec{k}

And I need to represent it in cylindrical and spherical coord. system

Homework Equations

a_\rho=a_x\cos\phi+a_y\sin\phi
a_\phi=-a_x\sin\phi+a_y\cos\phi
a_z=a_z

The Attempt at a Solution

What is cofusing me is this:
The formula for \phi is\phi=\arctan\frac{y}{x}. Are those x and y in fact a_x and a_y?

By some kind of reasoning it should be. But then \phi is \phi=\arctan\frac{-z}{2y} :\

Is this correct? And do I need to change the unit vectors too?

 

[planck42]

You can represent the \phi-component of a cylindrical/spherical vector in terms of \phi, like how you can represent the x-component of a Cartesian vector in terms of x. {\phi}=tan^{-1}\frac{y}{x} doesn't refer to the components of a vector [field]. Finally, unit vectors change according to the Jacobian matrix e.g. the transformation from the x unit vector to the \rho unit vector is the x\rho-component of the Jacobian, or x_{\rho}

 

[dingo_d]

Ok so you say I should use Jacobian, which is

J(r,\phi, z)=\begin{bmatrix} {dx\over dr} &amp; {dx\over d\phi} &amp;{dx\over dz} \\ {dy\over dr} &amp; {dy\over d\phi} &amp; {dy\over dz} \\ {dz\over dr} &amp; {dz\over d\phi} &amp; {dz\over dz}\end{bmatrix}<br /> =\begin{bmatrix} {d(r\cos\phi)\over dr} &amp; {d(r\cos\phi)\over d\phi} &amp; {d(r\cos\phi)\over dz} \\ {d(r\sin\phi)\over dr} &amp; {d(r\sin\phi)\over d\phi} &amp; {d(r\sin\phi)\over dz} \\ {dz\over dr} &amp; {dz\over d\phi} &amp; {dz\over dz}\end{bmatrix}<br /> =\begin{bmatrix} \cos\phi &amp; -r\sin\phi &amp; 0 \\ \sin\phi &amp; r\cos\phi &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{bmatrix}

But still where do I get the \phi? :\

 

[planck42]

Ok so you say I should use Jacobian, which is

J(r,\phi, z)=\begin{bmatrix} {dx\over dr} &amp; {dx\over d\phi} &amp;{dx\over dz} \\ {dy\over dr} &amp; {dy\over d\phi} &amp; {dy\over dz} \\ {dz\over dr} &amp; {dz\over d\phi} &amp; {dz\over dz}\end{bmatrix}<br /> =\begin{bmatrix} {d(r\cos\phi)\over dr} &amp; {d(r\cos\phi)\over d\phi} &amp; {d(r\cos\phi)\over dz} \\ {d(r\sin\phi)\over dr} &amp; {d(r\sin\phi)\over d\phi} &amp; {d(r\sin\phi)\over dz} \\ {dz\over dr} &amp; {dz\over d\phi} &amp; {dz\over dz}\end{bmatrix}<br /> =\begin{bmatrix} \cos\phi &amp; -r\sin\phi &amp; 0 \\ \sin\phi &amp; r\cos\phi &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{bmatrix}

But still where do I get the \phi? :\

Get the \phi for what? It's okay to leave it as it is when you're in cylindrical coordinates.

 

[dingo_d]

Oh, so I need to find a_\rho, and that's just

a_\rho=2y\cos\phi-z\sin\phi ?

Hmmm... I thought I had to calculate the phi :\

 

[planck42]

btw you should convert x's, y's, and z's to \rho's, \phi's, and z's/\theta's

 

[dingo_d]

So then the solution is:

a_\rho=2\rho\sin\phi\cos\phi-z\sin\phi
a_\phi=-2\rho\sin^2\phi-z\cos\phi
a_z=3\rho\cos\phi

And to obtain \vec{a} I just need to change unit vectors, and group them together to obtain the vector in cylindrical coordinate system?