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I Killing's vectors

  1. Jun 29, 2017 #1
    I have tried to find the three Killing vectors for the metric $$ds^2 = dr^2 + r^2d \theta^2$$ that is, the Euclidean metric of ##\mathbb{R}^2## written in polar coordinates. I found these to be

    $$\bigg(\text{first}\bigg) \ \ \xi_r = \text{Cos} \theta \\
    \xi_\theta = -\text{rSin} \theta \\

    \bigg(\text{second}\bigg) \ \ \xi_r = \text{Sin} \theta \\
    {\xi_\theta = \text{rCos} \theta} \\

    \bigg(\text{third}\bigg) \ \ \xi_r = 0 \\
    \xi_\theta = \text{r²}$$ As I have found solutions only for 3d on web, I would like to know whether these are correct or not.
     
  2. jcsd
  3. Jun 29, 2017 #2

    Orodruin

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    Why don't you check whether or not they satisfy the Killing equations?
     
  4. Jun 29, 2017 #3
    I did
    And they do satisfy the Killing equation.
     
  5. Jun 29, 2017 #4

    Orodruin

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    And thus they are Killing vector fields ...
     
  6. Jun 29, 2017 #5
    :biggrin:
     
  7. Jun 29, 2017 #6
    What bothers me is that in 2d we should have only two independent vectors. So I should be able to get one of those three above by a linear combination of the other two, but when I do that, I get non constant coefficients multiplying them.
     
  8. Jun 29, 2017 #7

    Orodruin

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    These are vector fields, not vectors.
     
  9. Jun 29, 2017 #8
    So they are'nt vectors? Can you say a bit more on this please
     
  10. Jun 29, 2017 #9

    Orodruin

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    There is no such thing as a "Killing vector". The Killing equation is a differential equation and as such describes vector fields, ie, assignments of one vector to each point in the space.
     
  11. Jun 29, 2017 #10
    For instance, what could be such one vector?
     
  12. Jun 29, 2017 #11

    Orodruin

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    You wrote down several vector fields (in coordinate basis) in the firs post.
     
  13. Jun 29, 2017 #12
    But you say they aren't vectors. I asked for an example of assigment of a vector by a vector field
     
  14. Jun 29, 2017 #13

    Orodruin

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    Your field is an assignment of a (dual) vector to every point in space!

    For example, for ##\theta = 0## (and arbitrary r) your first field takes the value ##\xi = dr##, where ##dr## is the coordinate basis dual vector.
     
  15. Jun 29, 2017 #14
    I got it. Thanks.
     
  16. Jun 29, 2017 #15
    Is it correct to say that they are three independent vector fields? In the sense that one cannot be expressed as a multiple of another one.

    Also, if we evaluate any of them at a particular point ##(r, \theta)## do they form three linearly dependent vectors?
     
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