Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Schwarzschild metric in terms of refractive index

  1. Oct 26, 2013 #1
    This is a spin off from another thread:

    First there are a couple of mathpages http://mathpages.com/rr/s8-04/8-04.htm and http://mathpages.com/rr/s8-03/8-03.htm that discuss the refractive index model and highlights the differences.

    The first obvious objection is that the 'medium' must have a refractive index that is independent of frequency, because there is no rainbow effect. In nature all known refractive materials are frequency dependent so the metric can't be interpreted in terms of any normal material. Then again, a vacuum can't be treated as a normal material as it does not cause any friction on objects passing through it.

    The second objection is that the refractive index cannot be isotropic. In the radial direction, the refractive index has to be ##N_r = 1/(1=2m/r)## and in the horizontal direction the refractive index has to be ##N_x = 1/\sqrt{1-2m/r}##. This objection is not as strong as the previous one as many materials have non isotropic refractive index. See http://en.wikipedia.org/wiki/Double_refraction. Some artificial composite materials called meta-materials have refractive properties never seen in nature, such as negative refractive indexes.

    Mathpages uses an isotropic refractive index of ##N = 1/(1=2m/r)## to approximate the Schwarzschild metric and finds some differences, mainly that the deflection is not quite right and that the photon orbit is at r=4m rather than the usual 3m.

    Mathpages does not explore the non isotropic model in any detail and I was curious what that would look like, starting with this simplified model with parallel thin layers:


    The incident ray passes from layer r to layer r' with the refractive index and velocity of the ray broken down into orthogonal components. Here ##\gamma(r)## and ##\gamma(r')## mean ##1/\sqrt{1-2m/r}## and ##1/\sqrt{1-2m/r'}## respectively. Everything is measured in the coordinates of the observer at infinity. Layer r' is lower in the gravitational field than layer r and has a higher refractive index. By simple trigonometry it is easy to determine that:

    [tex]\frac{\tan(\theta)}{\tan(\theta')} = \frac{V_x/V_r}{V_x'/V_r'} = \frac{\gamma(r)}{\gamma(r')} = \frac{N_x}{N_x'}[/tex]

    This is an interesting, yet still surprisingly simple version of Snell's law.

    Mathpages mentions that the optical model does not take account of the additional curvature due to the curvature of spacetime. If we take the above simplified model with flat parallel layers and adapt it to the form of concentric spherical shells, it seems almost certain that their will be additional curvature of the incident ray due to the curvature of the refractive index shells, over and above the the parallel model, but whether that is enough to duplicate what is actually observed remains to be analysed.

    P.S. The point of this exercise is to try and produce a mathematical model that may simplify some calculations or provide insight.

    Attached Files:

    Last edited: Oct 26, 2013
  2. jcsd
  3. Oct 27, 2013 #2
    Good reference as it is usually the case from mathpages.
    They conclude in the first reference that even though using isotropic coordinates of the Schwarzschild metric (with the radial coordinate defined as provided in the other thread) gives formally the same equations of motion for light paths that Fermat's optical model as shown in the Padmanabhan book and as confirmed in the form of the Shapiro delay by the Cassini probe, that resemblance is purely formal for many theoretical reasons(one of them independence from frequency as you point out, another that Fermat's optical model is Euclidean while GR's is obviously non-euclidean) (see the last 3 paragraphs of the first reference).
    Last edited: Oct 27, 2013
  4. Oct 27, 2013 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    I haven't gotten into the mathematics yet, but is the basic idea that you want to find a coordinate system where the spatial slices are flat, so that all the curvature is thrown into the time component? That isn't always possible, is it?
  5. Oct 27, 2013 #4
    The Euclidean/non Euclidean aspect is a valid concern and I would like to investigate this further as below. (P.S. I don't think it is necessary or desirable to quote the entire OP in the first reply.)

    I was considering things purely from the point of view of the effects on light such as deflection of light grazing the sun or the Shapiro delay. Mathpages does seem to concede that by using isotropic coordinates or a non-isotropic refractive index, that those effects can be exactly duplicated. I would like as a mathematical exercise try to do that with a non isotropic refractive index as mathpages did not actually attempt that.

    The trouble is that when it comes to local measurements using natural rulers and clocks, things do not work so well. The isotropic version of the Schwarzschild metric uses 'unnatural' rulers where a vertical ruler will not match a horizontal ruler when rotated to the horizontal. I agree with stevendaryl that we cannot exactly duplicate the Schwarzschild metric locally with a model that depends only on the time component. Basically, the optical model can explain light paths in coordinate terms, but has to invoke gravitational time dilation and gravitational length contraction to explain local measurements and then becomes redundant. Nevertheless an optical model might be a useful mathematical tool for motion in a gravitational field. To handle timelike motion, presumably the equations will require a velocity dependent component.
  6. Oct 27, 2013 #5


    Staff: Mentor

    How do you figure that?
  7. Oct 27, 2013 #6
    Looks like I was getting coordinate and local measurements mixed up. I withdraw that bit.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook