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Curves under conformal trans.

  1. Mar 2, 2014 #1
    Hi. This problem is about General Relativity. I am not actually taking a course, but studying myself. I tried to solve Hobson 2.7:

    1. The problem statement, all variables and given/known data
    A conformal transformation is not a change of coordinates but an actual change in
    the geometry of a manifold such that the metric tensor transforms as [itex]\bar{g}_{ab}(x)=\Omega^2(x) g_{ab}(x) [/itex], where [itex]\Omega (x) [/itex] is some non-vanishing scalar function of position. In a pseudo-Riemannian manifold, show that if [itex]x^a(\lambda) [/itex] is a null curve with respect to [itex]g_{ab} [/itex] (i.e. [itex]ds^2=0 [/itex] along the curve), then it is also a null curve with respect to [itex]\bar{g}_{ab} [/itex]. Is this true for timelike curves?

    2. Relevant equations

    for null curves:
    ds^2=g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2=0
    \therefore g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}=0

    3. The attempt at a solution
    after the conformal transformation, we have
    ds^2=\Omega^2 g_{\mu \nu}dx^\mu dx^\nu
    =g_{\mu \nu}(\Omega dx^\mu) (\Omega dx^\nu)
    and along the path [itex] x^\mu(\lambda) [/itex]:

    ds^2=g_{\mu \nu} \left[\frac{d}{d\lambda}(\Omega dx^\mu) \frac{d}{d\lambda}(\Omega dx^\nu) \right] (d\lambda)^2

    \\=g_{\mu \nu}\left(\frac{d\Omega}{d\lambda}dx^\mu +\Omega \frac{dx^\mu}{d\lambda} \right)
    \left(\frac{d\Omega}{d\lambda}dx^\nu +\Omega \frac{dx^\nu}{d\lambda} \right) (d\lambda)^2

    \\=g_{\mu \nu} \left\{ \left(\frac{d\Omega}{d\lambda}\right)^2 dx^\mu dx^\nu
    +\Omega \frac{d\Omega}{d\lambda} \left(dx^\mu \frac{dx^\nu}{d\lambda} + \frac{dx^\mu}{d\lambda} dx^\nu \right) +\Omega^2 \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} \right\} (d\lambda)^2

    \\= g_{\mu \nu} \left\{ \left(\frac{d\Omega}{d\lambda}\right)^2 \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^4
    + 2\Omega \frac{d\Omega}{d\lambda} \left(\frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} \right) (d\lambda)^3
    + \Omega^2 \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2 \right\}

    \\= \left( g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} \right)
    \left( \left(\frac{d\Omega}{d\lambda}\right)^2(d\lambda)^2+ 2\Omega \frac{d\Omega}{d\lambda}d\lambda +\Omega^2 \right)(d\lambda)^2 = 0

    now in the last line, the first term vanishes because of the null curve for the original [itex]g_{\mu \nu} [/itex], so the whole [itex]ds^2 [/itex] after the transformation vanishes, which means it is a null curve.

    Is this solution correct?
    Last edited: Mar 2, 2014
  2. jcsd
  3. Mar 2, 2014 #2


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    Homework Helper
    Gold Member

    Hello, spookyfish.

    It seems to me to be more straightforward. Before the transformation you have a curve for which any infinitesimal displacement ##dx^\mu## along the curve satisfies
    ds^2= g_{\mu \nu}dx^\mu dx^\nu
    = 0
    After the transformation you have a new metric ##\overline{g}_{\mu \nu} = \Omega^2 g_{\mu \nu}##. But the spacetime coordinates of points in the manifold have not changed their values.

    So, what is the value of ##d\overline{s}^2= \overline{g}_{\mu \nu}dx^\mu dx^\nu## for the same ##dx^\mu##?
  4. Mar 2, 2014 #3
    The value would be
    d\bar{s}^2 =\Omega^2 g_{\mu \nu} dx^\mu dx^\nu
    but we need to check [itex] d\bar{s}^2 [/itex] of a curve. Can I say that
    d\bar{s}^2 =\Omega^2 g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2=0
    because [itex] g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2=ds^2=0 [/itex]?
    Also, using this would not imply that timelike curves and spacelike curves remain the same, because [itex] \Omega^2 [/itex] doesn't change the sign of [itex]ds^2 [/itex]?
  5. Mar 2, 2014 #4


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    Homework Helper
    Gold Member

    Yes, that all sounds correct.

    I don't see the need to explicitly introduce the curve parameter ##\lambda##, because ##d\bar{s}^2 =\Omega^2 g_{\mu \nu} dx^\mu dx^\nu## and ##d\bar{s}^2 =\Omega^2 g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2## have the same value. But, you can if you want.
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