Question

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by n(r) = 1 + 2 GM/rc² where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.

Answer 1

Let us consider two spherical surfaces of radius r and r + dr. Let the light be incident at an angle θ at the surface at r and leave r + dr at an angle θ + dθ. Then from snell's law,

`n(r) sin θ = n(r + dr) sin(θ + dθ) = (n(r) + ((dn)/(dr)) dr) (sin θ * cos dθ +  cos θ * sin dθ)`

⇒ `n(r) sin θ = (n(r) + ((dn)/(dr)) dr) (sin θ + cos θ * dθ)`

For small angle, sin dθ ≈ dθ and cos dθ ≈ 1

Ignoring the product of differentials

⇒ `n(r) sin θ = n(r) * sin θ + ((dn)/(dr)) dr * sin θ + n(r) * cos θ * dθ`

or We have, `- (dn)/(dr) tan θ = n(r) (dθ)/(dr)`

`(2  GM)/(r^2c^2) tan θ = (1 + (2  GM)/(rc^2)) (dθ)/(dr) ≈ (dθ)/(dr)`

`int_0^(θ_0) dθ = (2  GM)/c^2 int_(-oo)^(oo)  (tan θ  dr)/r^2`

Now, `r^2 = x^2 + R^2` and tan θ = `R/x`

`2rdr = 2xdx`

Now substitution for integrals, we have

`int_0^(θ_0)  dθ = (2  GM)/c^2 int_(-oo)^(oo) R/x (xdx)/(x^2 + R^2)^(3/2)`

Put `x = R tan phi`

`dx = R sec^2 phi d phi`

∴ θ₀ = `(2 GMR)/c^2 int_((- pi)/2)^(pi/2)  (R sec^2 phi d phi)/(R^3 sec^3 phi)`

θ₀ = `(2 GM)/(Rc^2) int_((- pi)/2)^(pi/2) cos phi d phi = (4 GM)/(Rc^2)`

⇒ θ₀ = `(4 GM)/(Rc^2)`. This is the required proof.