Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 37B / a

The radial first integral for an equatorial null geodesic in Schwarzschild spacetime isSince $L=r^2\dot{\varphi }$, regarding $r$ as a function of $\varphi$ givesSet $y=1/r$. Then $dr/d\varphi =-y^{\prime }/y^2$ and therefore $\dot{r}=-L{y}^{\prime }$. Division by $L^2$ gives the equatorial null orbit in Schwarzschild spacetimewhere $b=L/E$ is the impact parameter. Differentiating with respect to $\varphi$ givesBy continuity this yields

In the flat-spacetime limit $M=0$, the orbit equation is the harmonic oscillator equationIts general solution is $y=A\cos \varphi +B\sin \varphi$. The conditions $r\to \infty$, hence $y\to 0$, at $\varphi =0$ and $\varphi =\pi$ set $A=0$. Choosing $r=b$ at closest approach $\varphi =\pi /2$ fixes $B=1/b$, soThis is a straight line at perpendicular distance $b$ from the origin in polar coordinates.

Writewhere $\Delta y$ is first order in $M$. Substitution into $y^{\prime \prime }+y=3M{y}^2$ and omission of terms of order $M\Delta y$ and $(\Delta y{)}^2$ givesThis is the linearized orbit equation used in the perturbative derivation of Schwarzschild light deflection.

The homogeneous solution isFor the constant and $\cos 2\varphi$ forcing terms, a particular solution isIndeed, applying $d^2/d{\varphi }^2+1$ multiplies the $\cos 2\varphi$ coefficient by $-3$. ThusThe $B\sin \varphi$ term only changes the definition of $b$ at this order, so one may set $B=0$.

Demanding that the incoming ray comes from infinity at $\varphi =0$ means $y(0)=0$. The solution from part (iv) then givesAt the outgoing end write $\varphi =\pi +\delta \varphi$, where $\delta \varphi$ is first order in $M$. ThenwhereasThe condition $y(\pi +\delta \varphi )=0$ therefore givesRestoring units gives $4GM/(b{c}^2)$, the leading Schwarzschild light deflection.