Saturday, 2 March 2019

homework and exercises - Treating gravitational lensing as index of refraction



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?



Answer



ANSWER TO :


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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...