Converting a unit vector from cartesian to cylindrical

In summary, the conversation discusses converting a unit vector from cartesian coordinates to cylindrical geometry. The conversion involves using sine and cosine definitions, a transformation matrix, and a system of equations. The resulting cylindrical coordinates for the given unit vector are (1, pi/2, 0). The same theory applies for a direction vector as for a position vector.
  • #1
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Hi, I was wondering if anyone could help with a vector question that I have.

If I have a unit vector defined in cartesian co-ordinates as p= (0,1,0) how would I go about converting this vector to a cylindrical geometry.

I understand that I will probably need to use p_r=sqrt(px^2+py^2) and p_theta=arctan(py/px) but also want to convert the vector to a cylindrical geometry and i don't think it can be as simple as this

thanks
 
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  • #2
You need to use the sine and cosine definitions. I'll draw a picture.
 
  • #3
thank you
 
  • #4
Look at the definitions, and see if you can figure out what to do. If not, post back and we'll take it from there.
 

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  • #5
By the way, welcome to physics forums vector problems.
 
  • #6
thank you very much for the welcome :)

just to clarify in case this changes things. The vector i am working with is a direction vector rather than a position vector. i.e. it describes the orientation of a particle with preferred orientation (0,1,0).

Will the same theory apply, thanks for the help. This is really outside my wheel house
 
  • #7
Yea, it doesn't matter, as far as math is concerned, a vector is a vector. And you're welcome for the welcome. The definition I provided in the image holds true for any position on that circle.
 
  • #8
from the picture it would seem that p_polar=(r,theta,z)=(1,pi/2/0) but this just seems too simple.

I would have thought i would have needed to use some sort of transformation matrix to transform from cartesian to polar e.g. p_r=cos(theta)p_x+sin(theta)p_y, p_theta=-sin(theta)p_x+cos(theta)p_y but is this instead only for a flow with multiple components.

Thanks again
 
  • #9
You can use a transformation matrix, but it's also equivalent to a system of equations relating the coordinates. Since y is independant of x as well as for z in any order, the matrix is unnecessary. Also, what you have is correct. (1,pi/2,0) is the correct set of coordinates for (0,1,0).
 
  • #10
so if i had a more general point in cartesian co-ordinates; p_cart=(px,py,pz) to express this in a cylindrical geometry do I just do p_polar=(sqrt(px^2+py^2),arctan(py/px),pz) or will I need a transformation matrix in this case

thanks
 
  • #11
nope, that's good.
 
  • #12
ok that's great, thank you very much for your help
 
  • #13
No problemo, Glad to help.
 

Related to Converting a unit vector from cartesian to cylindrical

What is a unit vector?

A unit vector is a vector with a magnitude of 1, used to indicate direction in a coordinate system.

What is the difference between cartesian and cylindrical coordinate systems?

Cartesian coordinates use x, y, and z coordinates to represent a point in 3D space, while cylindrical coordinates use a distance from the origin, an angle, and a height to represent a point.

How do you convert a unit vector from cartesian to cylindrical coordinates?

To convert a unit vector from cartesian to cylindrical coordinates, you can use the following formula:
r = sqrt(x^2 + y^2)
θ = atan2(y, x)
z = z
This will give you the corresponding cylindrical coordinates for the given cartesian unit vector.

Can you convert a unit vector with a magnitude other than 1?

Yes, you can convert a unit vector with any magnitude to cylindrical coordinates. However, the resulting vector will no longer be a unit vector.

Why is converting a unit vector from cartesian to cylindrical coordinates useful?

Converting a unit vector from cartesian to cylindrical coordinates can be useful in many applications, such as in physics and engineering, where cylindrical coordinates are often used to describe rotational motion. It can also make certain calculations and equations easier to work with in certain situations.

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