Converting a unit vector from cartesian to cylindrical

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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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You need to use the sine and cosine definitions. I'll draw a picture.
 
thank you
 
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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By the way, welcome to physics forums vector problems.
 
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
 
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.
 
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
 
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.
 
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