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}}.

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