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 refrative index of the medium given by
$\text{n}(\text{r})=1+2\frac{\text{GM}}{\text{rc}^2}$
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.
$\text{n}(\text{r})=1+2\frac{\text{GM}}{\text{rc}^2}$
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.
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Let us consider two spherical surfaces of radius r and r + dr. Let the light be incident at an angle $\theta$ at the surface at r and leave r + dr at an angle $\theta+\text{d}\theta$. Then from Snell's law,
$\text{n}(\text{r})\sin\theta=\text{n}(\text{r}+\text{dr})\sin(\theta+\text{d}\theta)$
$=\bigg(\text{n}(\text{r})+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}\bigg)(\sin\theta.\cos\text{d}\theta+\cos\theta.\sin\text{d}\theta)$
$\Rightarrow\ \text{n}(\text{r})\sin\theta=\bigg(\text{n}(\text{r})+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}\bigg)(\sin\theta+\cos\theta\text{d}\theta)$
For small angle, $\sin\text{d}\theta\approx\text{ and }\cos\text{d}\theta\approx1$
Ignoring the product of differentials
$\Rightarrow\ \text{n}(\text{r})\sin\theta=\text{n}(\text{r}).\sin\theta+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}.\sin\theta+\text{n}(\text{r}).\cos\theta.\text{d}\theta$
or we have, $-\frac{\text{dn}}{\text{dr}}\tan\theta=\text{n}(\text{r})\frac{\text{d}\theta}{\text{dr}}$
$\frac{2\text{GM}}{\text{r}^2\text{c}^2}\tan\theta=\Big(1+\frac{2\text{GM}}{\text{rc}^2}\Big)\frac{\text{d}\theta}{\text{dr}}\approx\frac{\text{d}\theta}{\text{dr}}$
$\int_0^\theta\text{d}\theta=\frac{2\text{GM}}{\text{c}^2}\int_{-\infty}^{\infty}\frac{\tan\theta\text{dr}}{\text{r}^2}$
Now, $\text{r}^2=\text{x}^2+\text{R}^2\text{ and }\tan\theta=\frac{\text{R}}{\text{x}}$
$2\text{rdr}=2\text{xdx}$
Now subsituting for integtals, we have
$\int_0^\theta\text{d}\theta=\frac{2\text{GM}}{\text{c}^2}\int_{-\infty}^{\infty}\frac{\text{R}}{\text{x}}\frac{\text{xdx}}{(\text{x}^2+\text{R}^2)^\frac{3}{2}}$
Put $\text{x}=\text{R}\tan\phi$
$\text{dx}=\text{R}\sec^2\phi\text{d}\phi$
$\therefore\ \theta_0=\frac{2\text{GMR}}{\text{c}^2}\int_{\frac{-\text{x}}{2}}^{\frac{\text{x}}{2}}\frac{\text{R}\sec^2\phi\text{d}\phi}{\text{R}^2\sec^3\phi}$
$\theta_0=\frac{2\text{GM}}{\text{Rc}^2}\int_{\frac{-\text{x}}{2}}^{\frac{\text{x}}{2}}\cos\phi\text{d}\phi=\frac{4\text{GM}}{\text{Rc}^2}$
$\Rightarrow\ \theta_0=\frac{4\text{GM}}{\text{Rc}^2}$. This is the required proof.
$\text{n}(\text{r})\sin\theta=\text{n}(\text{r}+\text{dr})\sin(\theta+\text{d}\theta)$
$=\bigg(\text{n}(\text{r})+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}\bigg)(\sin\theta.\cos\text{d}\theta+\cos\theta.\sin\text{d}\theta)$
$\Rightarrow\ \text{n}(\text{r})\sin\theta=\bigg(\text{n}(\text{r})+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}\bigg)(\sin\theta+\cos\theta\text{d}\theta)$
For small angle, $\sin\text{d}\theta\approx\text{ and }\cos\text{d}\theta\approx1$
Ignoring the product of differentials
$\Rightarrow\ \text{n}(\text{r})\sin\theta=\text{n}(\text{r}).\sin\theta+\Big(\frac{\text{dn}}{\text{dr}}\Big)\text{dr}.\sin\theta+\text{n}(\text{r}).\cos\theta.\text{d}\theta$
or we have, $-\frac{\text{dn}}{\text{dr}}\tan\theta=\text{n}(\text{r})\frac{\text{d}\theta}{\text{dr}}$
$\frac{2\text{GM}}{\text{r}^2\text{c}^2}\tan\theta=\Big(1+\frac{2\text{GM}}{\text{rc}^2}\Big)\frac{\text{d}\theta}{\text{dr}}\approx\frac{\text{d}\theta}{\text{dr}}$
$\int_0^\theta\text{d}\theta=\frac{2\text{GM}}{\text{c}^2}\int_{-\infty}^{\infty}\frac{\tan\theta\text{dr}}{\text{r}^2}$
Now, $\text{r}^2=\text{x}^2+\text{R}^2\text{ and }\tan\theta=\frac{\text{R}}{\text{x}}$
$2\text{rdr}=2\text{xdx}$
Now subsituting for integtals, we have
$\int_0^\theta\text{d}\theta=\frac{2\text{GM}}{\text{c}^2}\int_{-\infty}^{\infty}\frac{\text{R}}{\text{x}}\frac{\text{xdx}}{(\text{x}^2+\text{R}^2)^\frac{3}{2}}$
Put $\text{x}=\text{R}\tan\phi$
$\text{dx}=\text{R}\sec^2\phi\text{d}\phi$
$\therefore\ \theta_0=\frac{2\text{GMR}}{\text{c}^2}\int_{\frac{-\text{x}}{2}}^{\frac{\text{x}}{2}}\frac{\text{R}\sec^2\phi\text{d}\phi}{\text{R}^2\sec^3\phi}$
$\theta_0=\frac{2\text{GM}}{\text{Rc}^2}\int_{\frac{-\text{x}}{2}}^{\frac{\text{x}}{2}}\cos\phi\text{d}\phi=\frac{4\text{GM}}{\text{Rc}^2}$
$\Rightarrow\ \theta_0=\frac{4\text{GM}}{\text{Rc}^2}$. This is the required proof.
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