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Line element under coordinate transformation to get polar form

  1. Jun 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello Guys, I am reading Hobson's General Relativity and I have come across an exercise problem, part of which frustrates me:
    3.20 (P. 91)
    In the 2-space with line element
    [itex]ds^2=\frac{dr^{2}+r^{2}d\theta^{2}}{r^{2}-a^{2}}-\frac{r^{2}dr^{2}}{{(r^{2}-a^{2})}^{2}}
    [/itex]

    and given [itex]r{\frac{d\theta}{dr}}=tan\phi [/itex]
    show that the space is mapped to a Euclidean plane in which (r, phi) are taken as polar coordinates.

    2. Relevant equations



    3. The attempt at a solution

    So I attempted to express [itex]d\theta [/itex] as a l.c. of dr and dphi, but I don't know how to handle the [itex]\frac{d\theta}{dr}[/itex] the given relation [itex]r{\frac{d\theta}{dr}}=tan\phi [/itex]

    to express the [itex]d\theta[/itex] in given line element in terms of dphi and dr

    Thanks for any help =]
     
  2. jcsd
  3. Sep 1, 2013 #2
    I can't finished the problem too.

    We can write the matrix representation of the metric:
    [itex] g_{\mu\nu} = -\dfrac{a^2}{(r^2-a^2)^2} dr^2 + \dfrac{r^2}{r^2-a^2} d\theta^2 [/itex]

    From which we can write the matrix representation and the inverse and use

    [itex] \Gamma^a_{bc} = \dfrac{1}{2}g^{af}.... [/itex]

    to obtain the following connections:

    [itex] \Gamma^r_{rr}=-\dfrac{2r}{r^2-a^2} \; \Gamma^r_{\theta\theta}=-r [/itex]
    [itex] \Gamma^\theta_{r\theta}=\Gamma^\theta_{\theta r}=-\dfrac{a^2}{r(r^2-a^2)}[/itex]

    When I go to geodesic eq: [itex] \dfrac{d^2 x^a}{d\lambda^2} + \Gamma^a_{\alpha\beta} \dfrac{dx^\alpha}{d\lambda} \dfrac{dx^\beta}{d\lambda}=0 [/itex]

    and substitute I can't find a way to obtain the equation wanted in the problem:
    [itex] a^2 \left(\dfrac{dr}{d\theta}\right)^2 + a^2r^2 = Kr^4 [/itex]
    where K is a constant such if K = 1 then geodesic is null....

    Thanks in advance.
     
  4. Sep 2, 2013 #3
    Sorry, forgot to post the second part of the answer.

    We can see that the equation [itex] a^2 \left(\dfrac{d^2r}{d\theta^2}\right)^2 + \ldots [/itex] is a line in polar coordinates by substituting directly with [itex] \dfrac{dr}{d\theta}=cotg\phi [/itex].

    We obtain something like [itex] cotg\phi + 1 = K\left(\dfrac{r}{a}\right)^2 [/itex]

    If we write the equation for line in Cartesian [itex] y = ax + b[/itex] with [itex] a,b [/itex] parameters and make the coordinate change to polar coordinates we get [itex] \dfrac{b}{r} = 1 - acotg\phi [/itex]

    We can see the two equations have the same form to error of free parameters [itex]a,b[/itex]
     
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