## Transform a vector from Cartesian to Cylindrical coordinates

1. The problem statement, all variables and given/known data

Transform the vector below from Cartesian to Cylindrical coordinates:

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

2. Relevant equations

Use these equations:

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

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

$$A_z\,=\,A_z$$

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

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

3. 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?

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Hi VinnyCee!
 Quote by 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?

 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?

Mentor
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## Transform a vector from Cartesian to Cylindrical coordinates

 Quote by VinnyCee $$A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A}\,=\,A_x\,cos\,\phi\, +\,A_y\,sin\,\phi$$ $$x\,=\,\rho\,cos\,\phi$$ $$y\,=\,\rho\,sin\,\phi$$ 3. 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

 $A_y$ is zero, so it need not be included, right?
 Mentor Blog Entries: 10 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.

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 Quote by VinnyCee $A_y$ 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!
 Quote by VinnyCee 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.

 Mentor Blog Entries: 10 Isn't that the same expression as in post #1 for Aρ ?

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 Quote by Redbelly98 Isn't that the same expression as in post #1 for Aρ ?
ah … I meant the phi-coordinate $$A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi$$ is wrong.

 Mentor Blog Entries: 10 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? 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?

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 Quote by 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.
 Quote by VinnyCee So, does that mean that the original post contains the correct answer?
Hi VinnyCee and Redbelly98!

Yes, I completely mis-read the original question.