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Transformation of the metric tensor from polar to cartesian coords

  1. Dec 29, 2013 #1
    I'm working on a problem that requires me to take the cartesian metric in 2D [1 0;0 1] and convert (using the transformation equations b/w polar and cartesian coords) it to the polar metric. I have done this without issue using the partial derivatives of the transformation equations and have come up with the metric in polar coordinates [1 0;0 r^2].

    Just for grins, I decided to use the partial derivatives and convert back to cartesian using the polar metric, expecting to come up with the exact same thing I started with, namely [1 0;0 1]. Unfortunately, that is not what happened. Shouldn't this work? Can anyone help me in where my thought process is wrong here?

    Note, this is not a HW question; I am a degreed engineer teaching myself relativity from a workbook.
     
  2. jcsd
  3. Dec 29, 2013 #2

    PeterDonis

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    It should work, yes, but you will need to post more details before we can give any feedback on where you might have gone wrong.
     
  4. Dec 29, 2013 #3
    For example, in the attached word file, I've given the equation to convert the metric from primed (r, theta) to unprimed (x, y) coordinates. I have also listed the partial derivatives I've used.

    In this file, you can see gxx will not (unless my math is failing me) give 1, which would be the gxx component of the cartesian metric in 2D.
     

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  5. Dec 29, 2013 #4

    PeterDonis

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    Yes, it does. You have:

    $$
    g_{xx} = \left( \frac{\partial r}{\partial x} \right)^2 g_{rr} + \left( \frac{\partial \theta}{\partial x} \right)^2 g_{\theta \theta}
    $$

    We have (rewriting your equations slightly to make the math easier to see) ##\partial r / \partial x = x / r## and ##\partial \theta / \partial x = - y / r^2##, and the metric coefficients are ##g_{rr} = 1## and ##g_{\theta \theta} = r^2## (the latter may be where you went wrong in your math). Plugging everything in gives

    $$
    g_{xx} = \frac{x^2}{r^2} + \frac{y^2}{r^4} r^2 = \frac{x^2 + y^2}{r^2} = 1
    $$
     
  6. Dec 29, 2013 #5
    Ahhh, I failed to see that I could use r and (x^2+y^2)^.5 interchangeably. Thank you very much for this clarification!

    As a side note, how are you able to type the equations directly into the post?

    Edit: nevermind, I've just discovered Latex.
     
  7. Dec 29, 2013 #6

    PeterDonis

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