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Time derivative of 3D Spherical Coordinate

  1. Jun 3, 2015 #1
    When we obtain the velocity vector for position vector (r, θ, φ)
    Why do we take the time derivative of the radial part in the 3D Spherical Coordinate system only?
    Don't we need to consider the polar angle and azimuthal angle part like (dr/dt, dθ/dt, dφ/dt)?
     
  2. jcsd
  3. Jun 3, 2015 #2
    You also need to consider the other two parameters, unless ##\theta## and ##\psi## are invariant in time. It depends on the nature of your system or problem at hand.
     
  4. Jun 3, 2015 #3
    Yes. [tex]v = \frac{dr}{dt} + r\frac{d\theta}{dt} + r sin \theta \frac{d\phi}{dt} [/tex]

    See http://en.wikipedia.org/wiki/Spherical_coordinate_system#Kinematics, second equation.
     
  5. Jun 4, 2015 #4
    Thank you for the answers
    But there rises another question, why does the position vector has radial component only?
    Shouldn't it be rrθϕ? (r,θ,ϕ are unit vectors)
     
  6. Jun 4, 2015 #5
    The position vector for the spherical coordinate system is simply ##\boldsymbol{r} = r \boldsymbol{\hat{r}}##. You cannot use ##\theta## and ##\phi## as they are in a position vector. The scalar components of a position vector should have their units as distances. The units of ##\theta## and ##\phi## are in radians or degrees.
     
  7. Jun 4, 2015 #6
    Thank you!!
     
  8. Jun 4, 2015 #7
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