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Switching coordinate system of a field

  1. Apr 13, 2017 #1
    1. The problem statement, all variables and given/known data
    Say I have some sort of a vector field in the cylindrical coordinate system [tex] \vec{F}(r, \Theta, z) = f(\vec{A}(r,\Theta,z),\vec{B}(r,\Theta,z)) [/tex]

    How do I switch to the Cartesian coordinates? More precisely, how do I transform [tex] A_r = g(A_x,A_y,A_z), A_\Theta = h(A_x,A_y,A_z)[/tex] and so on?
    2. Relevant equations

    https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

    3. The attempt at a solution

    I understand that [tex]{\left(A_z\right)}_{cyl.} = {\left(A_z\right)}_{Cart.} [/tex]
    and similarly for B and that
    [tex] A_x = A_r\cdot\cos(\Theta), A_y = A_r\cdot\sin(\Theta [/tex].

    However, what do I do with [tex] A_\Theta[/tex]?
     
  2. jcsd
  3. Apr 13, 2017 #2

    kuruman

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    ##\vec{A}=A_r~\hat{r}+A_{\theta}~\hat{\theta}##
    ##\hat{r}=\cos \theta~\hat{x}+\sin \theta~\hat{y}## and ##\hat{\theta}=-\sin \theta~\hat{x}+\cos \theta~\hat{y}##
    Therefore,
    ##\vec{A}=A_r~(\cos \theta~\hat{x}+\sin \theta~\hat{y})+A_{\theta}~(-\sin \theta~\hat{x}+\cos \theta~\hat{y})##
    From which
    ##A_x = A_r~\cos \theta - A_{\theta}~\sin \theta~;~~ A_y = A_r~\sin \theta+A_{\theta}\cos \theta##
    Is this what you are looking for?
     
  4. Apr 13, 2017 #3
    I don't see [tex]A_z[/tex] in your equations. Why is that? Also, one can express [tex]\Theta[/tex] as [tex] = \arctan(\frac{y}{x})[/tex] correct? But what's the meaning of it when x = 0? If I am solving a physics problem then surely the set of points for which x = 0 aren't necessarily impossible.
     
  5. Apr 13, 2017 #4

    kuruman

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    You don't see Az because it is the same in Cartesian and cylindrical coordinates. I omitted it to save time, but you can assume it's there.
    What makes you think these points might be impossible? When x = 0, the vector points either along the +y axis in which case θ = π/2 or along the -y axis in which case θ = 3π/2.
     
  6. Apr 13, 2017 #5
    Thank you for your answers. One more question. Assuming [tex]A_x = A_r\cdot\cos(\Theta), A_\Theta = 0[/tex] If I wish to find how Ax changes with time is the following approach correct?
    [tex] \frac{dA_x}{dt} = \frac{dA_r}{dt}\cdot\cos(\Theta) - A_r\cdot\sin(\Theta)\cdot\frac{d\Theta}{dt}[/tex]
     
  7. Apr 13, 2017 #6

    kuruman

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    That is correct. It says that when ##A_x## changes with time, it can happen either because ##A_r## changes or because the angle changes or both.
     
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