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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

    HallsofIvy

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    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)
    fy=f_rho*sin(theta)+f_theta*cos(theta)
     
  6. Aug 30, 2009 #5

    tiny-tim

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    oooh, this is all virtually unreadable :rolleyes:

    can everybody please use the usual symbols? … :smile:
     
  7. Aug 30, 2009 #6
    ok
    fx=fρ*cosφ-fφ*sinφ
    fy=fρ*sinφ+fφ*cosφ
     
  8. Aug 31, 2009 #7
    Nay 1 here pleaseeeeeeeeeee solve this
    i need it badly!
     
  9. Aug 31, 2009 #8

    gabbagabbahey

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    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]

    Right?
    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.
     
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