Converting vector in cartesian to cylindrical coordinates

• dingo_d
In summary: Yes, 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

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?

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 $$x_{\rho}$$

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Ok so you say I should use Jacobian, which is

$$J(r,\phi, z)=\begin{bmatrix} {dx\over dr} & {dx\over d\phi} &{dx\over dz} \\ {dy\over dr} & {dy\over d\phi} & {dy\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} {d(r\cos\phi)\over dr} & {d(r\cos\phi)\over d\phi} & {d(r\cos\phi)\over dz} \\ {d(r\sin\phi)\over dr} & {d(r\sin\phi)\over d\phi} & {d(r\sin\phi)\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} \cos\phi & -r\sin\phi & 0 \\ \sin\phi & r\cos\phi & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

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

Last edited:
dingo_d said:
Ok so you say I should use Jacobian, which is

$$J(r,\phi, z)=\begin{bmatrix} {dx\over dr} & {dx\over d\phi} &{dx\over dz} \\ {dy\over dr} & {dy\over d\phi} & {dy\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} {d(r\cos\phi)\over dr} & {d(r\cos\phi)\over d\phi} & {d(r\cos\phi)\over dz} \\ {d(r\sin\phi)\over dr} & {d(r\sin\phi)\over d\phi} & {d(r\sin\phi)\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} \cos\phi & -r\sin\phi & 0 \\ \sin\phi & r\cos\phi & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

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.

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 :\

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

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?

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1. What is the formula for converting a vector in cartesian coordinates to cylindrical coordinates?

The formula for converting a vector from cartesian to cylindrical coordinates is:
r = √(x² + y²)
θ = arctan(y/x)
z = z

2. How do I determine the direction of the vector in cylindrical coordinates?

The direction of the vector in cylindrical coordinates is determined by the angle θ, which is measured counterclockwise from the positive x-axis to the projection of the vector onto the xy-plane.

3. Can I convert a vector with negative values in cartesian coordinates to cylindrical coordinates?

Yes, you can convert a vector with negative values in cartesian coordinates to cylindrical coordinates. The formula for r will still be the same, but the angle θ may need to be adjusted depending on the quadrant in which the vector lies.

4. What is the purpose of converting vectors from cartesian to cylindrical coordinates?

Converting vectors from cartesian to cylindrical coordinates can be useful in certain applications, such as in physics and engineering, where cylindrical symmetry is present. It can simplify calculations and make it easier to visualize and understand the vector's properties.

5. Are there any limitations to converting vectors from cartesian to cylindrical coordinates?

One limitation of converting vectors from cartesian to cylindrical coordinates is that it only works for 3-dimensional vectors. It also may not be the most efficient method for certain vector operations, such as addition and subtraction. Additionally, it may be more challenging to work with cylindrical coordinates when dealing with curved surfaces or complex geometries.

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