# Deflection of light by a mass

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 13; Problem 13.1.

We’ve seen how to derive the equations of motion for a photon in a gravitational field, so we can now apply these equations to study the deflection of light as it passes a massive object. The equations of motion are

 $\displaystyle \frac{d\phi}{dt}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}\left(1-\frac{2GM}{r}\right)b\ \ \ \ \ (1)$ $\displaystyle \left[\frac{1}{b}\left(1-\frac{2GM}{r}\right)^{-1}\frac{dr}{dt}\right]^{2}+\frac{1}{r^{2}}\left(1-\frac{2GM}{r}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{b^{2}} \ \ \ \ \ (2)$

These equations describe a photon’s motion for ${r>2GM}$, but when we’re discussing the deflection of light as it passes a star such as the sun, typically ${r\gg2GM}$. For example, for the Sun, ${2GM=2.954\mbox{ km}}$ and the radius of the sun is ${6.955\times10^{5}\mbox{ km}}$ so a photon grazing the surface of the Sun as it passed has a value of ${r/2GM=4.25\times10^{-6}}$. In this limiting case, we can make a few approximations to get an idea of how much light is deflected as it passes a massive object.

Things are a bit easier if we work with the variable ${u\equiv1/r}$, so we can use the chain rule to write

 $\displaystyle \frac{dr}{dt}$ $\displaystyle =$ $\displaystyle \frac{dr}{du}\frac{du}{d\phi}\frac{d\phi}{dt}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{u^{2}}\frac{du}{d\phi}u^{2}\left(1-2GMu\right)b\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(1-2GMu\right)b\frac{du}{d\phi} \ \ \ \ \ (5)$

Plugging this into the radial equation of motion above, we get

 $\displaystyle \left(\frac{du}{d\phi}\right)^{2}+u^{2}\left(1-2GMu\right)$ $\displaystyle =$ $\displaystyle \frac{1}{b^{2}}\ \ \ \ \ (6)$ $\displaystyle \left(\frac{du}{d\phi}\right)^{2}+u^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{b^{2}}+2GMu^{3} \ \ \ \ \ (7)$

We can take the derivative of this with respect to ${\phi}$ and get

 $\displaystyle 2\frac{du}{d\phi}\frac{d^{2}u}{d\phi^{2}}+2u\frac{du}{d\phi}$ $\displaystyle =$ $\displaystyle 6GMu^{2}\frac{du}{d\phi}\ \ \ \ \ (8)$ $\displaystyle \frac{d^{2}u}{d\phi^{2}}+u$ $\displaystyle =$ $\displaystyle 3GMu^{2} \ \ \ \ \ (9)$

If ${M=0}$ (that is, space is flat), this equation reduces to

$\displaystyle \frac{d^{2}u}{d\phi^{2}}+u=0 \ \ \ \ \ (10)$

which has the general solution

$\displaystyle u=A\sin\phi+B\cos\phi \ \ \ \ \ (11)$

If we orient the coordinate system so that ${\phi=\frac{\pi}{2}}$ is the angle of closest approach to the origin, then ${u=1/r}$ is a maximum at that angle so

 $\displaystyle A\cos\frac{\pi}{2}-B\sin\frac{\pi}{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (13)$

and we can set ${A\equiv1/r_{c}}$ where ${r_{c}}$ is the distance of closest approach to the origin:

$\displaystyle u=\frac{1}{r}=\frac{1}{r_{c}}\sin\phi \ \ \ \ \ (14)$

This is, in fact, the equation of a straight line in polar coordinates as can be verified by drawing the right-angled triangle with sides ${r}$ (as the hypotenuse) and ${r_{c}=r\sin\phi}$ (the third side is just ${r\cos\phi}$). From the equation above, with ${M=0}$, we have

$\displaystyle \left(\frac{du}{d\phi}\right)^{2}+u^{2}=\frac{1}{r_{c}^{2}}\left(\cos^{2}\phi+\sin^{2}\phi\right)=\frac{1}{r_{c}^{2}}=\frac{1}{b^{2}} \ \ \ \ \ (15)$

Thus ${r_{c}=b}$, which is the impact parameter, and gives the distance of closest approach.

Now let’s put the mass back in and apply the assumption ${r_{c}\gg2GM}$. In this case, we expect that ${u}$ will be fairly close to the flat space solution, so we can try a solution of the form

$\displaystyle u=\frac{1}{r_{c}}\sin\phi+\frac{w\left(\phi\right)}{r_{c}} \ \ \ \ \ (16)$

where ${w}$ is some function that should be small for all values of ${\phi}$. Plugging this into the equation of motion above:

 $\displaystyle \frac{d^{2}u}{d\phi^{2}}+u$ $\displaystyle =$ $\displaystyle 3GMu^{2}\ \ \ \ \ (17)$ $\displaystyle -\frac{1}{r_{c}}\sin\phi+\frac{1}{r_{c}}\frac{d^{2}w}{d\phi^{2}}+\frac{1}{r_{c}}\sin\phi+\frac{w\left(\phi\right)}{r_{c}}$ $\displaystyle =$ $\displaystyle \frac{3GM}{r_{c}^{2}}\left(\sin^{2}\phi+2w\sin\phi+w^{2}\right) \ \ \ \ \ (18)$

Because of our assumption ${r_{c}\gg2GM}$, ${r_{c}\gg3GM}$ as well, so if we’re saving only up to first-order terms in the ‘small’ quantities, only the first term on the RHS will contribute. We then get

 $\displaystyle \frac{d^{2}w}{d\phi^{2}}+w$ $\displaystyle \approx$ $\displaystyle \frac{3GM}{r_{c}}\sin^{2}\phi\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3GM}{2r_{c}}\left(1-\cos2\phi\right) \ \ \ \ \ (20)$

This differential equation can be solved by assuming a solution of form ${w=A+B\cos2\phi}$:

$\displaystyle -4B\cos2\phi+A+B\cos2\phi=\frac{3GM}{2r_{c}}\left(1-\cos2\phi\right) \ \ \ \ \ (21)$

from which we get

 $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{3GM}{2r_{c}}\ \ \ \ \ (22)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{GM}{2r_{c}} \ \ \ \ \ (23)$

So

$\displaystyle u=\frac{1}{r_{c}}\left(\sin\phi+\frac{3GM}{2r_{c}}\left(1+\frac{1}{3}\cos2\phi\right)\right) \ \ \ \ \ (24)$

We saw above that in flat space, ${r_{c}=b}$, the impact parameter. How about when there is a mass present? We can find out by substituting this equation into the equation of motion (above)

$\displaystyle \left(\frac{du}{d\phi}\right)^{2}+u^{2}=\frac{1}{b^{2}}+2GMu^{3} \ \ \ \ \ (25)$

and saving only terms linear in ${GM/r_{c}}$. We have

 $\displaystyle \frac{du}{d\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{r_{c}}\left(\cos\phi-\frac{GM}{r_{c}}\sin2\phi\right)\ \ \ \ \ (26)$ $\displaystyle \left(\cos\phi-\frac{GM}{r_{c}}\sin2\phi\right)^{2}+\left(\sin\phi+\frac{3GM}{2r_{c}}\left(1+\frac{1}{3}\cos2\phi\right)\right)^{2}$ $\displaystyle =$ $\displaystyle \frac{r_{c}^{2}}{b^{2}}+\frac{2GM}{r_{c}}\left(\sin\phi+\frac{3GM}{2r_{c}}\left(1+\frac{1}{3}\cos2\phi\right)\right)^{3} \ \ \ \ \ (27)$

We can now save only terms up to those involving ${GM/r_{c}}$:

$\displaystyle \cos^{2}\phi-2\frac{GM}{r_{c}}\cos\phi\sin2\phi+\sin^{2}\phi+\frac{3GM}{r_{c}}\sin\phi\left(1+\frac{1}{3}\cos2\phi\right)\approx\frac{r_{c}^{2}}{b^{2}}+\frac{2GM}{r_{c}}\sin^{3}\phi \ \ \ \ \ (28)$

Collecting terms, we get

$\displaystyle \frac{r_{c}^{2}}{b^{2}}\approx1+\frac{GM}{r_{c}}\left[-2\cos\phi\sin2\phi+3\sin\phi+\sin\phi\cos2\phi-2\sin^{3}\phi\right] \ \ \ \ \ (29)$

Using the shorthand notation ${s\equiv\sin\phi}$ and ${c\equiv\cos\phi}$, and the identities ${\sin2\phi=2sc}$ and ${\cos2\phi=1-2s^{2}}$, we get

 $\displaystyle -2\cos\phi\sin2\phi+3\sin\phi+\sin\phi\cos2\phi-2\sin^{3}\phi$ $\displaystyle =$ $\displaystyle -4sc^{2}+3s+s-2s^{3}-2s^{3}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -4s\left(1-s^{2}\right)+3s+s-2s^{3}-2s^{3}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -4s+4s+4s^{3}-4s^{3}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (33)$

Thus to this level of approximation, ${r_{c}=b}$.

Finally, in our coordinate system the closest approach to the mass occurs at ${\phi=\frac{\pi}{2}}$ so from 24 this distance is

 $\displaystyle \frac{1}{r_{min}}$ $\displaystyle =$ $\displaystyle \frac{1}{r_{c}}\left(1+\frac{3GM}{2r_{c}}\left(1-\frac{1}{3}\right)\right)\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r_{c}}\left(1+\frac{GM}{r_{c}}\right) \ \ \ \ \ (35)$

Thus

 $\displaystyle r_{min}$ $\displaystyle =$ $\displaystyle r_{c}\left(1+\frac{GM}{r_{c}}\right)^{-1}\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle r_{c}\left(1-\frac{GM}{r_{c}}\right) \ \ \ \ \ (37)$

where the last line uses a Taylor expansion to first order. The closest approach is therefore less than the impact parameter ${b=r_{c}}$.