# Bending of Light in General Relativity using Perturbation

It is standard textbook calculation (e.g. Schutz's First Course in General Relativity page 294) that we can find a total angular change in light deflection due to gravity to be $$$$\Delta\phi=\frac{4GM}{bc^{2}}$$$$ where $$b$$ is impact parameter.

However, I am trying to do this via perturbation method, since for typical stars like our Sun, $$\frac{GM}{rc^{2}}\ll 1$$. The idea is that from Schwarzschild metric, we can show that

$$$$\frac{d\phi}{dr}=\frac{1}{r^{2}}\left(\frac{1}{b^{2}}-\frac{1}{r^{2}}\left(1-\frac{2GM}{rc^2}\right)\right)^{-1/2}$$$$

From here, we consider $$$$\frac{d\phi}{dr}=\frac{d\phi}{dr}\Bigr|_{M=0}+\delta(r)$$$$ with $$|\delta|\ll 1$$. Then it should be true that $$\Delta\phi=2\int_\infty^b\delta(r)dr$$, with the extra factor of 2 accounting for symmetry of deflection. So the perturbation should be done making use of $$\frac{GM}{rc^{2}}\ll 1$$ on $$\delta(r)$$.

I tried all sorts of approximations but it does not seem to work. In one case the integral diverges while in other cases I am off by at least a factor of 4. Textbooks like Schutz or Hobson never do it this way too. Anyone could help?

• I'm a bit surprised this attracted a close vote. Wanting to do an easy calculation the hard way purely out of curiousity seems perfectly reasonable to me as there may be unexpected insights to be gained. I do this myself :-) Commented Mar 31, 2016 at 7:12

There's an important subtlety: in the Schwarzschild metric, the impact parameter $b$ is not equal to the radius of closest approach. Let's start with the geodesic equation $$\frac{1}{2}\dot{r}^2 + \frac{L^2}{2r^2}\left(1-\frac{2GM}{c^2r}\right) = \frac{1}{2}E^2.$$ At the radius of closest approach $R_0$ we have $$\frac{1}{2}E^2 = \frac{L^2}{2R_0^2}\left(1-\frac{2GM}{c^2R_0}\right)$$ or $$b^2 = R_0^2\left(1-\frac{2GM}{c^2R_0}\right)^{\!-1} \tag{1},$$ where $b = L/E$. In other words, $b$ depends on $M$, and we need to take this into account. If we substitute (1) into $$\frac{d\phi}{dr}=\frac{1}{r^{2}}\left[\frac{1}{b^{2}}-\frac{1}{r^{2}}\left(1-\frac{2GM}{c^2r}\right)\right]^{-1/2},$$ we get \begin{align} \frac{d\phi}{dr}&=\frac{1}{r^{2}}\left[\frac{1}{R_0^{2}}-\frac{1}{r^{2}} -\frac{2GM}{c^2}\left(\frac{1}{R_0^3}-\frac{1}{r^3}\right)\right]^{-1/2}\\ &= \frac{1}{r^{2}}\left[\frac{1}{R_0^{2}}-\frac{1}{r^{2}}\right]^{-1/2} \left[1-\frac{2GM}{c^2}\left(\frac{R_0^{-3}-r^{-3}}{R_0^{-2}-r^{-2}}\right)\right]^{-1/2}. \end{align} Now we can use the first-order approach $$\frac{d\phi}{dr}\approx\left.\frac{d\phi}{dr}\right|_{M=0}+\delta(r),$$ with $$\left.\frac{d\phi}{dr}\right|_{M=0} = r^{-2}\left(R_0^{-2}-r^{-2}\right)^{-1/2},\\ \delta(r) = \frac{GM}{c^2}r^{-2}\left(R_0^{-3}-r^{-3}\right)\left(R_0^{-2}-r^{-2}\right)^{-3/2}.$$ Therefore $$\Delta\phi|_{M=0} = 2\int_{R_0}^{\infty}\frac{\text d r}{r^2\left(R_0^{-2}-r^{-2}\right)^{1/2}} = 2\int_0^{1}\frac{\text d u}{\left(1-u^{2}\right)^{1/2}} = 2\sin^{-1}(1) = \pi,$$ where we used $u=R_0r^{-1}$, and \begin{align} \Delta\phi|_\delta &= \frac{2GM}{c^2}\int_{R_0}^{\infty}\frac{R_0^{-3}-r^{-3}}{r^2\left(R_0^{-2}-r^{-2}\right)^{3/2}}\text d r\\ &= \frac{2GM}{c^2R_0}\int_0^1\frac{1-u^{3}}{\left(1-u^{2}\right)^{3/2}}\text d u\\ &= \frac{2GM}{c^2R_0}\int_0^{\pi/2}\frac{1-\sin^{3} x}{\cos^2 x}\text d x\\ &= \frac{2GM}{c^2R_0}\left[\int_0^{\pi/2}\frac{1-\sin x}{\cos^2 x}\text d x + \int_0^{\pi/2}\sin x\,\text d x\right] \\ &= \frac{2GM}{c^2R_0}\left[\int_0^1 \frac{2\text d t}{(1+t)^2} + 1\right] \\ &= \frac{4GM}{c^2R_0}, \end{align} where we used $t = \tan(x/2)$ and the tangent half-angle formulae. Finally, we can use (1) again to replace $R_0$ with $b$, so that $$\Delta\phi|_\delta = \frac{4GM}{c^2b},$$ to first order in $M$.
Let us rearrange the derivative as $$\frac{d\phi}{dr} = - \frac{\left(-\frac{b}{r^2}\right)}{\sqrt{1- \left( \frac{b}{r}\right)^2 + a \left( \frac{b}{r}\right)^3}}$$ with $a = \frac{2GM}{bc^2}$, such that the desired integral becomes $$- \int_\infty^b{dr\; \frac{\left(-\frac{b}{r^2}\right)}{\sqrt{1- \left( \frac{b}{r}\right)^2 + a \left( \frac{b}{r}\right)^3}}} = - \int_0^1{d\xi \frac{1}{\sqrt{1- \xi^2 + a \xi^3}}}$$
Your problem is that although the integrand has a nice expansion in $a$, $$\frac{1}{\sqrt{1- \xi^2 + a \xi^3}} = \frac{1}{\sqrt{1- \xi^2}} - \frac{\xi^3}{2\left(1 - \xi^2 \right)^{3/2}}\;a + \dots$$ the integral does not, since the coefficient of $a$ becomes infinite on integration: $$\int_0^1{d\xi \frac{\xi^3}{2\left(1 - \xi^2 \right)^{3/2}}} = \frac{1}{4}\int_0^1{d\eta \frac{\eta}{(1 - \eta)^{3/2}}} \rightarrow \infty\;\;\;$$ The right way to handle this is to extract $\frac{d\phi}{d(b/r)}$ from $\frac{d\phi}{dr} = \frac{d\phi}{d(b/r)} \frac{d(b/r)}{dr} \equiv \frac{d\phi}{d(b/r)} \left(-\frac{b}{r^2}\right)$ and then switch to $\xi = b/r$ and $$\left( \frac{d\xi}{d\phi} \right)^2 = 1 - \xi^2 + a \xi^3$$ which is indeed amenable to a perturbation approach. See for instance these notes.