Gravitational deflection of light

Remarks

For a null geodesic with $u = 1/r$, the orbit equation is $u'' + u = 3GM u^2/c^2$. The zeroth-order solution is the straight line $u_0 = \sin\varphi / b$. Inserting it into the right side and solving to first order gives $u = \sin\varphi/b + (GM/(c^2 b^2))(1 + \cos^2\varphi)$, and setting $u = 0$ at the two ends gives the deflection $4GM/(c^2 b)$. A Newtonian computation that treats light as particles moving at speed $c$ gives only half this value, $2GM/(c^2b)$. For a ray grazing the Sun the formula gives about $1.75$ arcseconds; the eclipse expedition of 1919 led by Eddington confirmed a value of this size. In the parameterized post-Newtonian formalism the deflection is $\frac{1 + \gamma}{2} \cdot \frac{4GM}{c^2 b}$.

Going deeper

For a student

The bending of light is the basis of gravitational lensing. A galaxy or a cluster of galaxies between us and a distant source can produce several images of the source, arcs, or a full ring when the alignment is perfect. Half of the general-relativistic deflection comes from the time part of the metric, as in the Newtonian picture, and the other half from the curvature of space.