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Confused about tangent to a circle (polar/Cartesian convert)

  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data
    The problem is from D'Inverno's book on GR, problem 5.6. We're using the Jacobian/transformation matrix to convert the tangent to a circle centered at the origin of radius A from Cartesian to polar coordinates. I can do the problem and get the book answer, that's okay. What confuses me is when I try to do the conversion without using the matrix and just do it the old-fashioned calculus class way I get an extra factor of A.

    2. Relevant equations
    Polar coordinate system is primed, Cartesian is unprimed.

    [tex]x=a cos(\phi)\\
    y=a sin(\phi)\\


    I don't know why this last one won't render, I've tried everything I could think of.
    Mod note: It's fixed now. Instead of \dot\hat{r}, use \dot{\hat{r}}
    3. The attempt at a solution
    When I try to convert [tex]
    \vec{T}[/tex] to polar coordinates by substituting x and y in terms of [tex]r, \phi[/tex] I get:[tex]
    \vec{T}=-(a sin(\phi))\dot{\phi}\hat{x}+(a cos(\phi))\dot{\phi}\hat{y}=a\dot{\phi}\hat{\phi}=\vec{T'}\\[/tex]

    The problem is that using the transformation matrix I get [tex]\vec{T'}=\dot{\phi}\hat{\phi}[/tex], no factor of a, and this is what the book has as an answer too. Where am I going wrong? Thanks.
    Last edited by a moderator: Jan 22, 2017
  2. jcsd
  3. Jan 22, 2017 #2


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    I do not have d'Inverno, but probably your problem is that you have normalised your polar basis vectors.
    \dot\hat{x} is not a good TeX command. You must do \dot{\hat{x}} or \dot{\hat x}
  4. Jan 22, 2017 #3


    Staff: Mentor

    I edited the OP before noticing that you had addressed this problem.
  5. Jan 22, 2017 #4
    Yes but my Cartesian basis vectors (xhat and yhat) are also normalized. I get the feeling that the book answer is wrong, just by considering units, and that there should be a factor of a in the transformed tangent vector.
  6. Jan 22, 2017 #5


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    Not the Cartesian basis, the polar basis. The Cartesian basis is normalised by default.
  7. Jan 23, 2017 #6
    Aren't the polar vectors always normalized? Where would the normalization enter in to the derivation of the transformation matrix?
  8. Jan 23, 2017 #7


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    No. Why would they be?

    Edit: A normalised basis would mean a displacement rate of one when changing the corresponding coordinate. The displacement rate when changing the polar angle (and therefore the tangential basis vector) depends on the radius. Of course, you can then normalise the basis, but then you must take this into account and your transfer coefficients will not be given by the partial derivatives.
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