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.