Transform a vector from Cartesian to Cylindrical coordinates

[VinnyCee]

Homework Statement

Transform the vector below from Cartesian to Cylindrical coordinates:

Q\,=\,\frac{\sqrt{x^2\,+\,y^2}}{\sqrt{x^2\,+\,y^2\,+\,z^2}}\,\hat{x}\,-\,\frac{y\,z}{x^2\,+\,y^2\,+\,z^2}\,\hat{z}

Homework Equations

Use these equations:

A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A}\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi

A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi

A_z\,=\,A_z

x\,=\,\rho\,cos\,\phi

y\,=\,\rho\,sin\,\phi

The Attempt at a Solution

Q_{\rho}\,=\,\frac{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2}}{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2\,+\,z^2}}\,cos\,\phi

Q_{\rho}\,=\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,cos\,\phi

Q_{\phi}\,=\,-\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,sin\,\phi

Q_z\,=\,-\,\frac{z\,\rho\,sin\,\phi}{\rho^2\,+\,z^2}

Does the above look correct?

 

[tiny-tim]

Hi VinnyCee!

A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A}\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi

A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi

No … these are the equations for an ordinary rotation of the x and y axes through an angle phi.

Q_{\rho}\,=\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,cos\,\phi

Q_{\phi}\,=\,-\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,sin\,\phi

hmm … you want polar coordinates of a vector along the x-axis … so what will its phi coordinate be?

 

[VinnyCee]

Really? The equations for transforming from Cartesian to Cylindrical were given in the notes for the class in a matrix form and then solved out to a formula, the one below.

Here is a more detailed version for the \rho coordinate transformation only:

A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A} \,=\,\left(\hat{\rho}\,\cdot\,\hat{x}\right)\,A_x\,+\,\left(\hat{\rho}\,\cdot\,\hat{y}\right)\,A_y\,+\,\left(\hat{\rho}\,\cdot\,\hat{z}\right)\,A_z\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi

Is that incorrect? If so, what is the correct way to transform from Cartesian to Cylindrical?I think the \phi coordinate would be zero in this case since there is no y-component, right?

 

[Redbelly98]

A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A}\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi

x\,=\,\rho\,cos\,\phi

y\,=\,\rho\,sin\,\phi

The Attempt at a Solution

Q_{\rho}\,=\,\frac{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2}}{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2\,+\,z^2}}\,cos\,\phi

Here you've included the Ax cos(phi) term, but forgot the Ay sin(phi) term

 

[VinnyCee]

A_y is zero, so it need not be included, right?

 

[Redbelly98]

Oops, I was confusing Ay and Az. My bad.

Your results from post #1 look okay to me. My understanding is the dot product of the vector Q with the coordinate unit-vector is the component of Q in that coordinate direction, which is exactly what you did.

On the other hand I'm hesitant to disagree with tiny-tim, since he knows this stuff pretty well also.

 

[tiny-tim]

A_y is zero

That's right! for x > 0 and y = 0, phi must be zero.

, so it need not be included, right?

No!

A three-dimensional vector needs three coordinates …

if one coordinate is 0, you have to say so!

Here is a more detailed version for the \rho coordinate transformation only:

A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A} \,=\,\left(\hat{\rho}\,\cdot\,\hat{x}\right)\,A_x\,+\,\left(\hat{\rho}\,\cdot\,\hat{y}\right)\,A_y\,+\,\left(\hat{\rho}\,\cdot\,\hat{z}\right)\,A_z\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi

Is that incorrect?

No, that is correct … the other one is wrong.

 

[Redbelly98]

Isn't that the same expression as in post #1 for A_ρ ?

 

[tiny-tim]

Isn't that the same expression as in post #1 for A_ρ ?

ah … I meant the phi-coordinate A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi is wrong.

 

[Redbelly98]

Um, if we use the convention that φ's unit vector points counterclockwise, and φ is taken counterclockwise from the positive x-axis, then I agree with VinnyCee's formula.

 

[VinnyCee]

So, does that mean that the original post contains the correct answer? Or is the Q_{\phi} variable still going to be zero even though the formulas given produce the expression I wrote in the first post?

The expression written from a matrix in class is:

A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A} \,=\,\left(\hat{\phi}\,\cdot\,\hat{x}\right)\,A_x\,+\,\left(\hat{\phi}\,\cdot\,\hat{y}\right)\,A_y\,+\,\left(\hat{\phi}\,\cdot\,\hat{z}\right)\,A_z\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi

Since...

\hat{x}\,\cdot\,\hat{\phi}\,=\,-\,sin\,\phi

\hat{y}\,\cdot\,\hat{\phi}\,=\,cos\,\phi

\hat{z}\,\cdot\,\hat{\phi}\,=\,0

Right?

 

[tiny-tim]

sorry!

Um, if we use the convention that φ's unit vector points counterclockwise, and φ is taken counterclockwise from the positive x-axis, then I agree with VinnyCee's formula.

So, does that mean that the original post contains the correct answer?

Hi VinnyCee and Redbelly98!

Yes, I completely mis-read the original question.

Your original answer was totally right.

Please ignore my previous posts.

Sorry.