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Jacobian Matrix for Polar Coordinates

  1. Jun 6, 2012 #1
    Hi,

    I need some help understanding the solution to a problem.

    Equations:

    x = r.cos(θ)
    y = r.sin(θ)

    r = x2 + y2
    theta = arctan(y/x)


    Question:
    Determine the Jacobian Matrix for (x,y)T and for (r, θ)T

    SOLUTION:

    I understand and can compute by myself the Jacobian for (x,y)T, but the solution to for J(r, θ) i dont understand.

    J(r,θ) = ( (@r/@x, @r/@y) , (@θ/@x, @θ,@y) )T

    Ok so that makes sense...

    Then they gave me this...

    (@r/@x, @r/@y) =( x / sqrt( x2 + y2 ) , y / sqrt(x2 + y2) )

    Why isn't it just 2x and 2y respectively? why does it resemble something similar to a magnitude?

    thanks for your help.
     
  2. jcsd
  3. Jun 6, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, a "Jacobian" is not at all what you say. The Jacobian of a transformation is a matrix but what you wrote has only two components. If [itex]x= f(r,\theta)[/itex], [itex]y= g(r\theta)[/itex] then the Jacobian is
    [tex]\frac{\partial x, y}{\partial r,\theta}= \begin{bmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{bmatrix}[/tex]

    In this case, that would be
    [tex]\begin{bmatrix}cos(\theta) & sin(\theta) \\ r sin(\theta) & r cos(\theta)\end{bmatrix}[/tex]

    The Jacobian the other way would be
    [tex]\begin{bmatrix}\frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y}\end{bmatrix}[/tex]
    Here, we have [itex]r= (x^2+ y^2)^{1/2}[/itex] and [itex]\theta= tan^{-1}(y/x)[/itex] so that the Jacobian is
    [tex]\begin{bmatrix}cos(\theta) & sin(\theta) \\ -\frac{1}{r} sin(\theta) & \frac{1}{r} cos(\theta)\end{bmatrix}[/tex]
    which is, of course, the inverse matrix to the previous matrix.
     
  4. Jun 6, 2012 #3

    Ok, so my understanding of Jacobian Matrix is how you have described. Ive got no idea how to post matrices etc on here so my descriptions above may have been a bit off, but I'll try suss it out for this post...

    Ok so, my understanding is fine all the way up until you talk about the Jacobian "the other way around"

    Why isnt it the case that....

    [tex]\frac{\partial r, \theta}{\partial x, y}= \begin{bmatrix}\frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y}\end{bmatrix}[/tex]

    ?



    Just getting back to my original question, and I'm sorry if I have missed a larger point you were trying to make, but I'm still trying to work out how


    [tex]\frac{\partial r}{\partial x}= [/tex] x / √(x2 + y2)


    Since the original equation is...

    r = x2 + y2

    I would have expected dr/dx = 2x


    Thanks very much for your help
    My answers for the lower half of the matrix match the solutions given, just not the top half.
     
  5. Jun 6, 2012 #4
    [itex]r = \sqrt{x^2 + y^2}[/itex], not [itex]x^2 + y^2[/itex].
     
  6. Jun 7, 2012 #5
    Yes, that makes much more sense. But I've been given r = x^2 + y^2... must have been a mistake in the question itself.


    Thanks for your help guys.
     
  7. Nov 10, 2014 #6
    the first row of inverse Jacobian matrix should be <cosθ , -sinθ/r>
    and the second row is <sinθ, cosθ/r>.
     
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