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Unit vectors in polar co ordinates

  1. Aug 26, 2013 #1
    I have two questions
    1) How the radial and traversal unit vectors are vector funcitons of scalar variable θ (angle between the position vector and polar axis.
    2) To find velocity and accleration in polar co ordinates why it is need to write the traversal and radial unit vectors by transforming its basis to cartesian co ordinate basis i,j? I have little bit idea that it is for our convenience that we are transforming basis to simplify the problem to get the velocity, but i don't know tell how it is convenient
    Last edited: Aug 26, 2013
  2. jcsd
  3. Aug 26, 2013 #2


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    Staff: Mentor

    Answering #2 first: We're trying to calculate the magnitude of the velocity and acceleration vectors, and it's really easy to calculate the magnitude of a vector if you know its components in the Cartesian basis: ##S=\sqrt{x^2+y^2}##. If you have the components in a different basis such as polar coordinates, you can still calculate the magnitude but you have to use a thing called the "metric tensor" - you can't just use the simple Pythagorean theorem - and that's often just as much or more work than converting into Cartesian coordinates. (Actually, there's nothing special about Cartesian coordinates, it just so happens that the metric tensor in that coordinate system is so trivial that you don't notice it).

    And for #1: the unit vector in the ##r## direction at a point is a unit vector in the direction that ##\theta## is not changing, and the unit vector in the ##\theta## direction is a unit vector pointing in the direction that ##r## is not changing. These vectors will point in different directions at different points in the plane, but their components in the ##(r,\theta)## basis will not; the vectors are ##(1,0)## and ##(0,1)## everywhere when using polar coordinates.
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