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Gravitational index of refraction

  1. Oct 15, 2008 #1
    Suppose an experimenter firstly is entirely ignorant of general relativity and non-euclidean space. He finds (say) a point-source gravitational field. He fires light beams about and notices the deflection. He concludes that there's a not-equals-one index of refraction n about the source point. Let's say he is able to recognize and make a formula for n with no problems. What is the function n( r ) ? SURELY this has been formulated! Does anyone have this formula?
  2. jcsd
  3. Oct 15, 2008 #2
    Price et al. Problem book in GR.
  4. Oct 15, 2008 #3


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  5. Oct 15, 2008 #4
    Thanks atyy

    The formula given is

    n(r) = ( 1 + [tex]m/2r[/tex])[tex]^{3}[/tex]/( 1 - [tex]m/2r[/tex])​

  6. Oct 15, 2008 #5


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    He's got the interesting remark at the end that it's not entirely self consistent, because the circumference of a circle with radius [itex]r[/itex] is not [itex]2\pi r[/itex]. So I guess it's a pretty good approximation when [itex]m/r[/itex] is small.
  7. Oct 15, 2008 #6
    I noticed that too. Also "is not" is not as informative as greater than or less than. While dozing off last night, it occured to me that the equation is not dimensionally consistent ( mass divided by length ) so it might be in those "geometric units" which always bugs me. I would think the normal form of the equation would have to contain the GRAVITATIONAL CONSTANT?!
  8. Oct 15, 2008 #7

    Jonathan Scott

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    Yes, that's the isotropic metric form of the expression in geometric units, where c and G are set to 1.

    In ordinary units and in the "weak field" approximation, that would be better expressed as follows:

    n(r) = 1 + \frac{2Gm}{rc^2}

    That is to say, the effective speed of light is decreased by a fraction [itex]2Gm/rc^2[/itex] relative to an isotropic coordinate system.
  9. Aug 27, 2011 #8
    For a general metric [itex]c^2d\tau^2=\alpha c^2dt^2-\beta dx^2[/itex], the index of refraction in the x direction is [itex]n_x=\sqrt{\frac{\alpha}{\beta}}[/itex].

    An observer basing their measurements on a euclidean space would use a radius scaled so that the circumference is [itex]2\pi r[/itex]. Thus we must use the Schwarzschild solution. This is not isotropic, so the index of refraction is not the same in each direction.

    Radially: [itex]n_r(r)=(1-\frac{r_s}{r})^{-1} \approx 1+\frac{2GM}{c^2r}[/itex]

    Normal: [itex]n_\theta(r)=(1-\frac{r_s}{r})^\frac{-1}{2} \approx 1+\frac{GM}{c^2r}[/itex]

    The formulas given above are for an isotropic coordinate system, sacrificing the Euclidean radius to circumference ratio.
  10. Aug 27, 2011 #9


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    You'll need some sort of hodgepodge of slowing clocks, shrinking (or would it be expanding?) rulers, as well as the "index of refreaction" (and I'm not sure how it would tie in to the other effects, you'd need to start sepcifying a lot of coordinate dependent stuff) to fully duplicate the effects of a curved space-time geometry.

    This is because the spatial slice (with a Schwarzschild clock synchronization) is curved. Hence the need to mess with rulers.

    Einstein's heated ruler thought experiment shows that it may be possible in principle to imagine a geometry with changing rulers and a (not-really-physical) underlying flat background.

    Gravitational time dilation shows the need to mess with time, and you'd need to specify how the index of refraction was to be measured (obviously if you use local clocks and rulers you'll always get an index of one, and that wouldn't give the desired deflection).
  11. Aug 27, 2011 #10
    If we are assuming an observer unaware of general relativity, the logical choice would be to assume the observer is effectively using a ruler and clock equivalent to an observer at infinity, i.e. the Swartzchild metric.

    Now in GR light travels along null lines, so the metric is 0. Solving for [itex]\frac{cdt}{dx}[/itex] in the radial and transverse directions gives the index of refractions above.
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