Deflection of Light

With no mass $M$, the equation reduces to $u'' +u=0$ which has the solution $u={1\over r_0}\cos\varphi$ which is the equation of a straight line written in polar coordinates. Lets solve the full equation to first order in the small parameter ${r_s\over r_0}$. The full solution will be the solution for with $M=0$ plus a small perturbation due to the mass.

$$u={1\over r_0}\cos\varphi+D D$$

$$u'' +u-{3r_s\over 2}u^2 = 0$$ the differential equation

$$D'' +D-{3r_s\over 2}\left({1\over r_0}\cos\varphi+D\right)^2= 0 u$$

$$D'' +D-{3r_s\over 2}\left({1\over r_0^2}\cos^2\varphi\right)= 0 {r_s\over r_0}$$

$$D=-{r_s\over 2r_0^2}\cos^2\varphi+{r_s\over r_0^2}$$ guess solution, plug in hw

$$u(\varphi)={\cos\varphi\over r_0}-{r_s\over 2r_0^2}\cos^2\varphi+{r_s\over r_0^2}$$

In the last two lines we have guessed the solution to the differential equation. It can be tested just by plugging it in.

We can rather simply compute the deflection of light from this. At large $r$, $u\rightarrow 0$ and for small ${r_s\over r_0}$ we expect $\varphi$ to be near to ${\pi\over 2}$ there.

$${\cos\varphi\over r_0}-{r_s\over 2r_0^2}\cos^2\varphi+{r_s\over r_0^2}=0$$

$${\cos\left({\pi\over 2}+\delta\right)\over r_0}+{r_s\over r_0^2}=0$$

$${\delta\over r_0}={r_s\over r_0^2}$$

$$\delta={r_s\over r_0}$$

This is half the deflection of light. The trajectory is symmetric and starts at $-{\pi\over 2}+\delta$ so the total angular deflection of light is $2\delta$.

$$\Delta=2{r_s\over r_0}$$

This is 1.75 arc seconds for light nearly grazing the outside of the sun as measured during solar eclipse.