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General Relativity - Deflection of light

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the deflection of light given this metric, along null geodesics.

    2dj2phi.png


    2. Relevant equations


    3. The attempt at a solution

    Conserved quantities are:

    [tex]e \equiv -\zeta \cdot u = \left( 1 - \frac{2GM}{c^2r} \right) c \frac{dt}{d\lambda} [/tex]
    [tex]l \equiv \eta \cdot u = r^2 \left( 1 - \frac{2GM}{c^2r} \right) \frac{d\phi}{d\lambda} [/tex]


    For a null, "time-like" vector, ##u \cdot u = 0##:
    [tex] - \left( 1 - \frac{2GM}{c^2r} \right)^{-1} e^2 + \left( 1 - \frac{2GM}{c^2r} \right) \left( \frac{dr}{d\lambda} \right)^2 + \frac{l^2}{r^2} \left(1 - \frac{2GM}{c^2r} \right)^{-1} = 0 [/tex]

    Letting ##b = \frac{l}{e}##:

    [tex]\frac{1}{b^2} = \frac{1}{l^2} \left( 1 -\frac{2GM}{c^2r} \right)^2 \left( \frac{dr}{d\lambda} \right)^2 + \frac{1}{r^2} [/tex]

    Does this mean that the effective potential ##W_{eff} = \frac{1}{r^2}##?

    [tex]\frac{1}{b^2} = \frac{1}{l^2} \left( 1 -\frac{2GM}{c^2r} \right)^2 \left( \frac{dr}{d\lambda} \right)^2 + W_{eff} [/tex]

    To find deflection, we find ##\frac{d\phi}{dr}##:
    [tex]\frac{d\phi}{dr} = \frac{\frac{d\phi}{d\lambda}}{ \frac{dr}{d\lambda}} = \frac{1}{r^2 \sqrt{ \frac{1}{b^2} - W_{eff} }} [/tex]

    [tex]\Delta \phi = 2 \int_{r_1}^{\infty} \frac{1}{r^2 \sqrt{ \frac{1}{b^2} - W_{eff} }} dr [/tex]

    The turning point is when ##W_eff(r_1) = \frac{1}{b^2}##, which means ##b=r_1##.

    [tex] = 2 \int_1^{0} \frac{w^2}{b^2} \left( \frac{1}{b^2} - \frac{w^2}{b^2} \right)^{-\frac{1}{2}} \cdot \frac{-b}{w^2} dw [/tex]

    [tex] = 2 \int_0^1 \left(1 - w^2\right)^{-\frac{1}{2}} dw [/tex]

    [tex] = 2 \left( \frac{\pi}{2} \right) [/tex]

    [tex]\Delta \phi = \pi [/tex]

    So this means no deflection, which is strange.
     
    Last edited: Feb 2, 2015
  2. jcsd
  3. Feb 2, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I'm not knowledgeable in this field, but I will just make a comment. The metric you are working with is an example of a "conformally flat" metric. In general, such a metric can be written in the form ##g_{\mu \nu} = f \eta_{\mu \nu}## where ##\eta_{\mu \nu}## is the Minkowski (flat) metric and ##f## is a positive function defined on the space-time manifold. It can be shown that there is no bending of light for a conformally flat metric. This is not too surprising if you note that any null direction for the flat metric ##\eta_{\mu \nu}## is automatically a null direction for the metric ##g_{\mu \nu}## and vice versa.
     
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