1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transforming divergence from cartesian to cylindrical coordinates

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Compute the divergence in cylindrical coordinates by transforming the expression for divergence in cartestian coordinates.

    2. Relevant equations

    F = F_x i + F_y j + F_z k
    div F = ∂F_x/∂x + ∂F_y/∂y + ∂F_z/∂z .......... (divergence in Cartesian coordinates)

    I need to transform this into

    divF = (1/rho)(∂(rho*F_rho)/∂rho) + (1/rho)(∂F_theta/∂theta) + ∂F_z/∂z ...... (divergence in cylindrical coordinates)

    3. The attempt at a solution

    Using the chain rule,
    ∂F_x/∂x = (∂F_x/∂rho)(∂rho/∂x) + (∂F_x/∂theta)(∂theta/∂x) + (∂F_x/∂z)(∂z/∂x)
    ∂F_y/∂y = (∂F_y/∂rho)(∂rho/∂y) + (∂F_y/∂theta)(∂theta/∂y) + (∂F_y/∂z)(∂z/∂y)
    ∂F_z/∂z = (∂F_z/∂rho)(∂rho/∂z) + (∂F_z/∂theta)(∂theta/∂z) + (∂F_z/∂z)(∂z/∂z)

    ∂rho/∂x = x/∂ = costheta
    ∂theta/∂x = -y/rho^2 = -sintheta/rho
    ∂z/∂x = 0
    ∂rho/∂y = y/∂ = sintheta
    etc. (these are the transformational equations)

    Then I try inputing this into the cartesian definition for divergence and obtain
    divF = [(∂F_x/∂rho)costheta + (∂F_x/∂theta)(-sintheta/rho)] + [(∂F_y/∂rho)sintheta + (∂F_y/∂theta)(costheta/rho)] + ∂F_z/∂z

    But how does that simplify to the expression in cylindrical coordinates?
  2. jcsd
  3. Aug 30, 2009 #2
    any 1 here plzzzzzzzzzzzzzz. solve this!
    i also need divergence in spherical!
  4. Aug 30, 2009 #3


    User Avatar
    Science Advisor

    You haven't said anything about what [itex]F_\rho[/itex] and [itex]F_\theta[/itex] are in terms of [itex]F_x[/itex] and [itex]F_y[/itex].
  5. Aug 30, 2009 #4
    lets assume that fx=f_rho*cos(theta)-f_theta*sin(theta)
  6. Aug 30, 2009 #5


    User Avatar
    Science Advisor
    Homework Helper

    oooh, this is all virtually unreadable :rolleyes:

    can everybody please use the usual symbols? … :smile:
  7. Aug 30, 2009 #6
  8. Aug 31, 2009 #7
    Nay 1 here pleaseeeeeeeeeee solve this
    i need it badly!
  9. Aug 31, 2009 #8


    User Avatar
    Homework Helper
    Gold Member

    Okay, then
    [tex]\frac{\partial f_x}{\partial\rho}=\frac{\partial}{\partial\rho}(f_\rho\cos\theta-f_\theta\sin\theta)=\frac{\partial f_\rho}{\partial\rho}\cos\theta-\frac{\partial f_\theta}{\partial\rho}\sin\theta[/tex]

    Calculate [tex]\frac{\partial f_y}{\partial\rho}[/tex], [tex]\frac{\partial f_x}{\partial\theta}[/tex] and [tex]\frac{\partial f_y}{\partial\theta}[/tex] the same way and substitute them all into this expression.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Transforming divergence cartesian Date
Fourier Transformation of ODE Thursday at 7:22 AM
Convergence of a series Monday at 1:39 PM
Fourier transform of exponential function Mar 15, 2018
Diffusion equation in polar coordinates Mar 14, 2018
Proving a statement about the rank of transformations Feb 22, 2018