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Homework Help: Prove 2-dimensional Riemann manifold is conformally flat

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Establish the theorem that any 2-dimensional Riemann manifold is conformally flat in the case of a metric of signature 0.
    Hint: Use null curves as coordinate curves, that is, change to new coordinate curves
    [itex]\lambda[/itex] = [itex]\lambda[/itex](x0, x1), [itex]\nu[/itex] = [itex]\nu[/itex](x0, x1)
    gab[itex]\lambda[/itex],a[itex]\lambda[/itex],b = gab[itex]\nu[/itex],a[itex]\nu[/itex],b = 0
    and show that the line element reduces to the form
    ds2 = e2[itex]\mu[/itex]d[itex]\lambda[/itex]d[itex]\nu[/itex]
    and finally introduce new coordinates [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] + [itex]\nu[/itex]) and [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] - [itex]\nu[/itex])

    2. Relevant equations
    Conformally flat metric: gab = [itex]\Omega[/itex]2[itex]\eta[/itex]ab ([itex]\eta[/itex]ab is a flat metric)
    ds2 = gabdxadxb

    3. The attempt at a solution
    It says in the hint that the metric has a signature of 0, so it must be flat in a specific set of coordinates, but there are 3 different coordinate systems the hint tells me to use in this problem:
    (x0, x1), ([itex]\lambda[/itex], [itex]\nu[/itex]), and [[itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] + [itex]\nu[/itex]), [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] - [itex]\nu[/itex])]
    so I don't know in which coordinate system I should make the assumption that the metric is flat.

    I know that the line element equation in the ([itex]\lambda[/itex], [itex]\nu[/itex]) coordinate system is:
    ds2 = g[itex]\lambda\lambda[/itex]d[itex]\lambda[/itex]2 + g[itex]\lambda\nu[/itex]d[itex]\lambda[/itex]d[itex]\nu[/itex] + g[itex]\nu\lambda[/itex]d[itex]\nu[/itex]d[itex]\lambda[/itex] + g[itex]\nu\nu[/itex]d[itex]\nu[/itex]2
    and if I assume the metric is flat in this coordinate system, then that equation can be reduced to:
    ds2 = g[itex]\lambda\lambda[/itex](d[itex]\lambda[/itex]2 - d[itex]\nu[/itex]2) + 2g[itex]\lambda\nu[/itex]d[itex]\lambda[/itex]d[itex]\nu[/itex]
    where I used the symmetry of the metric. If I am to reduce this equation to
    ds2 = e2[itex]\mu[/itex]d[itex]\lambda[/itex]d[itex]\nu[/itex]
    then I need to show that
    1. g[itex]\lambda\lambda[/itex]d[itex]\lambda[/itex]2 + g[itex]\nu\nu[/itex]d[itex]\nu[/itex]2 = 0
    2. 2g[itex]\lambda\nu[/itex] = e2[itex]\mu[/itex].

    I don't know how to do either. I have substituted d[itex]\lambda[/itex] with ([itex]\partial\lambda[/itex]/[itex]\partial[/itex]x0)dx0 + ([itex]\partial\lambda[/itex]/[itex]\partial[/itex]x1)dx1 and tried all sorts of algebraic manipulations, but I have not been able to cancel out those 2 terms or figure out how 2g[itex]\lambda\nu[/itex] = e2[itex]\mu[/itex]. Could I get some help, please?
  2. jcsd
  3. Oct 14, 2011 #2
    I've managed to work through the steps in the hint, but it didn't lead me to the gab = [itex]\Omega[/itex]2[itex]\eta[/itex]ab equation that I should be getting. The only way I see to eliminate the d[itex]\lambda[/itex]2 and d[itex]\nu[/itex]2 terms in the line element equation is to assume that d[itex]\lambda[/itex]2 = d[itex]\nu[/itex]2. Since the ([itex]\lambda[/itex], [itex]\nu[/itex]) coordinate system is arbitrary, I should be allowed to make this assumption.
    Then, I set [itex]\mu[/itex] = [itex]\frac{1}{2}[/itex][itex]\int[/itex]([itex]\partial[/itex][itex]\lambda[/itex]/[itex]\partial[/itex]x0)2[(dx0 + dx1)/(dx0 - dx1)], and this allows me to define g[itex]\lambda\nu[/itex] + g[itex]\nu\lambda[/itex] = e2[itex]\mu[/itex]. I came up with the equation for [itex]\mu[/itex] from the identity [itex]\frac{d}{dx}[/itex]ln x = [itex]\frac{1}{x}[/itex].

    I did the coordinate transformation to [[itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] + [itex]\nu[/itex]), [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] - [itex]\nu[/itex])] and came up with this:
    g[itex]\lambda\lambda[/itex] = gyz/2
    g[itex]\nu\nu[/itex] = -gyz/2
    g[itex]\lambda\nu[/itex] = gyy/2
    where y = [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] + [itex]\nu[/itex]) and z = [itex]\frac{1}{2}[/itex]([itex]\lambda[/itex] - [itex]\nu[/itex])

    There are 2 problems with this. One, g[itex]\lambda\lambda[/itex] should be a function of gyy, g[itex]\lambda\nu[/itex] should be a function of gyz, etc. if I am to show that gab = [itex]\Omega[/itex]2[itex]\eta[/itex]ab.
    Two, even if the above were true, the metric in the ([itex]\lambda[/itex], [itex]\nu[/itex]) coordinate system is not flat, because if it was, then g[itex]\lambda\nu[/itex] would be 0, not e2[itex]\mu[/itex]/2, and ds2 would be 0. So all I would have proven is that [itex]\overline{g}[/itex]ab = [itex]\Omega[/itex]2gab, not gab = [itex]\Omega[/itex]2[itex]\eta[/itex]ab. Is there someone who can help?
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