In Einstein's theory of gravity, an electromagnetic wave passing near a massive object is bent from its rectilinear path. We may regard this bending equivalently as due to a medium of refractive index $$\mu_g = 1 - \frac{2\phi}{c^2}$$
where $\phi$ is the gravitational potential due to an object of mass $M$.

Consider radiation for a Quasar Q (treated as point like object) reaching observer P as shown in figure. Here $b$ is the distance of closes approach to gravitational object $M$. Show that the optical path length for path QP is given by $$d = d_1 + d_2 + \frac{2MG}{c^2}\left(\ln\frac{4 d_1 d_2}{b^2}\right)$$ enter image description here

  • What does "we may regard this bending equivalently as due to a medium of refractive index" mean? If we can treat it as a medium, them what do we take the angle of incidence, refraction, the medium interface etc to be?
  • By optical path length of path QP, should we find $\int n(s)ds$ over the line segment QP or should we do it over the actual path taken by light to move from Q to P? And how do we show the last part?
  • $\begingroup$ By the way you've worded this question ("Show that ..") it appears as though it's a homework exercise. If so, please add the tag "homework and exercises". Thanks $\endgroup$
    – docscience
    Oct 20, 2016 at 17:58
  • $\begingroup$ Yeah, it was an olympiad question, I did add that tag $\endgroup$ Oct 21, 2016 at 4:21

1 Answer 1



DOUBT 1: They Are basically modelling the curving light ray to be the same scenario as when a light beam passes through a medium having continuously varying R.I for each infinitesimal layer(like the atmosphere suppose). The function of R.I here is given as a function of the potential, which is a function of the distance from the object M.

DOUBT 2: You have to find the optical path length $\Delta$, which by definition is:

$\Delta = \int_{C} n(s).ds$ where C denotes the curve signifying its actual path traversed.

  • $\begingroup$ a)But then what do we take as the plane of medium? Is it the position vector w.r.t M ? And B) what is the strategy to find C? $\endgroup$ Oct 20, 2016 at 16:45
  • $\begingroup$ if I am not very mistaken, notice that the R.I as defined in the problem changes only if the radial distance of the photon changes from M. Clearly, if we consider circles of radius R, everytime the light ray passes through one such circle, the R.i changes. So i think, the surfaces at which refraction will occur are circular, centered at M, and having instantaneous radius R. $\endgroup$
    – Lelouch
    Oct 20, 2016 at 16:49
  • $\begingroup$ If that is the case, shouldn't the light bend away from M as RI<1? $\endgroup$ Oct 20, 2016 at 16:51
  • $\begingroup$ $\phi$ is negative. $\endgroup$
    – Lelouch
    Oct 20, 2016 at 16:52
  • $\begingroup$ Oh yeah, I didn't see that. Thanks for the help, I'll try to solve it now $\endgroup$ Oct 20, 2016 at 16:54

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