Gravitational lensing

Builds on: Sections 10.2, 5.4 and 4.2
Reading time: 45 min
Exercises: 3
Olympiad problems: 2

During the total eclipse of 29 May 1919 two British expeditions, one to the island of Príncipe and one to Sobral in Brazil, photographed the stars of the Hyades round the darkened Sun. Compared with photographs of the same stars taken at night, the stars were pushed outwards by amounts that, scaled to a star at the Sun's very edge, came to about 2.0″ at Sobral and 1.6″ at Príncipe. Einstein's general relativity had predicted 1.75″, twice what a calculation with Newton's gravity gives. In 1979 astronomers found two quasars in Ursa Major only 6.1″ apart with identical spectra. They are two images of one quasar, whose light has been bent round a galaxy that lies in front of it. In this section we see how a mass acts as a lens.

10.6.1The bending of light

Before you readMake a guess first

Starlight that just grazes the edge of the Sun is bent by its gravity. About how large is the bending?

In Section 10.2 we saw rays of light swing round a black hole. Far from the mass the same pull bends a ray only slightly, and the bending is largest for rays that pass closest.

ResultThe bending of light by a mass(10.14)

\hat{α} = \frac{4 G M}{c^{2} b} = \frac{2 R_{s}}{b}

$\hat{α}$	the angle in radians by which the ray is bent
$M$	the mass of the lens
$b$	the impact parameter, the distance by which the ray passes the centre of the mass
$R_{s}$	the Schwarzschild radius of the mass

This holds when

b

is much larger than

R_{s}

, so that the bending is small. A calculation with Newton's gravity, treating light as fast particles, gives half this angle, and the eclipse of 1919 was the first test between the two.

For a ray grazing the Sun, $b = R_{⊙}$ , and the bending is $4 G M_{⊙} / (c^{2} R_{⊙}) = 8.49 \times 10^{- 6}$ rad, which is 1.75″. Newton's half would be 0.88″. A star seen 10 solar radii from the Sun is moved outwards by only 0.175″.

10.6.2The Einstein ring and the two images

Picture a source S far behind a lens L, with an observer O on the near side. Let the lens be at the distance $D_{L}$ , the source at $D_{S}$ , and the source a further distance $D_{LS}$ behind the lens. In a space that is not expanding $D_{LS} = D_{S} - D_{L}$ . Over cosmological distances this no longer holds, as the twin quasar below shows. Without the lens the source would be seen at the small angle $β$ from the lens. Both angles are measured along the line through the lens and the source, and are positive on the side of the source. With it, light that leaves the source on one side of the lens is bent towards the observer, and the source is seen at a different angle $θ$ . All the angles are tiny, so the geometry of the bent ray gives the lens equation.

ResultThe lens equation of a point mass(10.15)

\begin{aligned} β & = θ - \frac{θ_{E}^{2}}{θ} \\ θ_{E} & = \sqrt{\frac{4 G M}{c^{2}} \frac{D_{LS}}{D_{L} D_{S}}} \end{aligned}

$β$	the angle between the lens and the true position of the source, positive on the source's side
$θ$	the angle between the lens and an image of the source, negative for an image on the other side
$θ_{E}$	the Einstein radius
$D_{L}, D_{S}$	the distances of the lens and of the source
$D_{LS}$	the distance from the lens to the source

The lens is a single point mass, and all the angles are small. The equation has two solutions,

θ_{\pm} = \frac{1}{2} (β \pm \sqrt{β^{2} + 4 θ_{E}^{2}})

, so a point lens always makes two images, one outside the Einstein radius on the same side as the source and one inside it on the other side.

Show the derivation

A ray that passes the lens at the distance $b = D_{L} θ$ is bent by $\hat{α} = 4 G M / (c^{2} D_{L} θ)$ . Measured at the distance of the source, the ray appears to come from a point $D_{S} θ$ from the line to the lens, while the source is at $D_{S} β$ , and the bending accounts for the difference $D_{LS} \hat{α}$ . So $D_{S} θ = D_{S} β + D_{LS} \hat{α}$ , which gives $β = θ - (D_{LS} / D_{S}) \hat{α}$ . Putting in $\hat{α}$ gives the result, with $θ_{E}^{2} = 4 G M D_{LS} / (c^{2} D_{L} D_{S})$ .

When the source lies exactly behind the lens, $β = 0$ , every direction round the lens is alike and the image becomes a ring of radius $θ_{E}$ , called an Einstein ring. Light is neither made nor destroyed by a lens, but the images are stretched along the ring and squeezed across it. Since the brightness of each small patch of a source is unchanged, the stretched images together are brighter than the source would be alone.

ResultThe magnification of a point lens(10.16)

A = \frac{u^{2} + 2}{u \sqrt{u^{2} + 4}}, u = \frac{β}{θ_{E}}

$A$	the factor by which the two images together are brighter than the source alone
$u$	the angle between the lens and the source, in units of the Einstein radius

u = 1

the source is 1.34 times brighter, or 0.32 magnitudes. When

u

is small,

A

is close to

1 / u

, so at

u = 0.1

it is 10.0 times brighter, 2.5 magnitudes. The source is assumed to be much smaller than the Einstein radius.

Quick checkA source half an Einstein radius away

A source lies at $u = 0.5$ from a point lens. Where are its two images, and how many times brighter are they together?

10.6.3Microlensing

When the lens is a star in our own Galaxy, the Einstein radius is far too small to see. A faint star of 0.3 solar masses 4 kpc away, in front of a star 8 kpc away in the bulge of the Galaxy, has an Einstein radius of 0.55 milliarcseconds, so its two images are far too close to be seen apart with an ordinary telescope. What we see is the brightening. Stars in the Galaxy move, so the lens drifts across the background star over weeks, and the background star brightens and fades again in a smooth, symmetric curve with the same shape in every colour. This is microlensing. The time the source takes to move by one Einstein radius relative to the lens is the Einstein time $t_{E}$ , which sets the length of the event.

Figure 10.6A star passing behind a lens

Closest approach

u_{0}

Time

Watch

Source from lens
Images
Brighter by

The figure shows, on the left, the sky round a lens L, with the dotted circle of its Einstein radius. A source S, drawn as a small open circle where it really is, moves along the dashed line and passes the lens at the smallest separation

u_{0}

, in Einstein radii. The lens makes two images of it, A outside the ring on the source's side and B inside it on the other side, each the shape of the source mapped through the lens equation. On the right, the heavy curve shows how many magnitudes brighter the source looks as time passes, with time in Einstein times. Running the event moves the source and the point on the curve together. The figure can be set to , , or .

Example 10.11A microlensing event towards the bulge

A star of 0.3 solar masses 4 kpc away passes in front of a star in the bulge 8 kpc away, moving across it at 5 milliarcseconds a year. Find the Einstein radius and the Einstein time.

Solution

Here $D_{LS} = 4$ kpc, so by (10.15),

θ_{E}^{2} = \frac{4 G \times 0.3 M_{⊙}}{c^{2}} \frac{4 kpc}{4 kpc \times 8 kpc},

which gives $θ_{E} = 0.55$ milliarcseconds. At 4 kpc this radius is 2.2 au. Moving at 5 milliarcseconds a year, the lens crosses one Einstein radius in $0.55 / 5$ of a year, which is 40 days.

AnswerThe Einstein radius is 0.55 milliarcseconds and the Einstein time about 40 days. A source that passes nearly behind the lens stays more than 0.32 magnitudes brighter for about 40 days on each side of the peak.

Surveys watch hundreds of millions of stars in the bulge to catch the few thousand that brighten in this way each year. A planet round the lens star adds a short extra spike to the curve, and the lens of one event of 2011 was shown in 2022 to be a black hole of several solar masses, wandering alone and seen only by its gravity.

Example 10.12The mass of the lens of the twin quasarfinish it

The two images of the twin quasar are 6.1″ apart. For any round lens, $c^{2} θ_{E}^{2} D_{L} D_{S} / (4 G D_{LS})$ is the mass inside the Einstein radius, and for a galaxy whose mass grows in proportion to the radius the two images lie at $β \pm θ_{E}$ , so half their separation is the Einstein radius. Take $θ_{E} = {3.05}^{″}$ , and take $D_{L} = 1.04$ Gpc, $D_{S} = 1.74$ Gpc and $D_{LS} = 1.15$ Gpc, the distances that the model of Chapter 13 gives for the redshifts of the galaxy and the quasar. Estimate the mass of the lensing galaxy.

The start

Solving (10.15) for the mass gives $M = c^{2} θ_{E}^{2} D_{L} D_{S} / (4 G D_{LS})$ , with $θ_{E} = {3.05}^{″} = 1.48 \times 10^{- 5}$ rad.

Your stepNow work out the mass in solar masses.

× 10¹² solar masses

Common errorsSection 10.6

The bending $4 G M / (c^{2} b)$ is twice what Newton's gravity gives for light. Using $2 G M / (c^{2} b)$ gives the wrong answer.
A point lens always makes two images. When the source is far from the lens, the inner image is very faint and close to the lens, but it is still there.
In the lens equation $D_{LS}$ is the distance from the lens to the source, not from the observer.
It is wrong to think that a lens adds light of its own. Its stretched images cover more of the sky than the source would, and that is why the total brightness grows.

ExercisesSection 10.63 questions

10.6.1
A star is seen 10 solar radii from the centre of the Sun during an eclipse. By how many arcseconds is it moved outwards?
arcseconds
10.6.2
A microlensing event reaches its peak with the source 0.1 Einstein radii from the lens. By what factor does the source brighten?
10.6.3
The light curve of a star brightens by 2 magnitudes and fades again over a few weeks, in exactly the same way in blue and in red light. What is the most likely cause?

Olympiad problemsPast IOAA problems, restated2 problems

2007The Einstein ring10 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 4. We have restated it in our own words and kept the original data, adding the wavelength of 500 nm that the official solution uses for part (c). The official solution prints the Einstein radius of part (b) as 0.956 × 10⁻⁹ rad (1.97 × 10⁻⁴″), and the data give 0.978 × 10⁻⁹ rad (2.02 × 10⁻⁴″). The solution was last checked on 23 October 2026.

A spherical mass $M$ bends a ray that passes it at the distance $ξ$ by the small angle $4 G M / (ξ c^{2})$ . A source S, a lens L and an observer O are nearly in line, with the lens at the distance $D_{L}$ and the source at $D_{S}$ . The source is seen at the small angle $θ$ from the lens, where without the lens it would be at $β$ .

(a) When the source is exactly behind the lens, $β = 0$ , show that the image is a ring of angular radius $θ_{E} = \sqrt{(4 G M / c^{2}) (D_{S} - D_{L}) / (D_{L} D_{S})}$ . (2 points)

(b) A star is about 50 kpc away, and a lens of one solar mass is 10 kpc in front of it. Find the angular radius of the Einstein ring. (1 point)

(c) What is the resolution of the Hubble Space Telescope, with its mirror of 2.4 m, at 500 nm? Could it resolve this ring? (2 points)

(d) For a source not exactly behind the lens there are two images. Find their positions $θ_{1}$ and $θ_{2}$ in terms of $β$ and $θ_{E}$ . (2 points)

(e) With $η = β / θ_{E}$ , find $θ_{1} / β$ and $θ_{2} / β$ in terms of $η$ . (2 points)

(f) Find the ratio $Δ θ / Δ β$ for each image, when the source moves by a small $Δ β$ . (1 point)

I solved it without the solution

Answer(a) As shown. (b) $9.78 \times 10^{- 10}$ rad, or $2.02 \times 10^{- 4}$ arcseconds. (c) $2.54 \times 10^{- 7}$ rad, so the ring cannot be resolved. (d) $θ_{1, 2} = β / 2 \pm \sqrt{β^{2} / 4 + θ_{E}^{2}}$ . (e) $\frac{1}{2} (1 \pm \sqrt{η^{2} + 4} / η)$ . (f) $\frac{1}{2} (1 \pm η / \sqrt{η^{2} + 4})$ .

2019Bending radio waves by Jupiter and the Moon10 points

Adapted from IOAA 2019 (Keszthely), theory problem 2. We have restated it in our own words and kept the original data, adding the masses and radii of Jupiter and the Moon, which the original left to the data sheet. The solution was last checked on 23 October 2026.

In the eclipse of 1919 starlight passing the edge of the Sun was found to be bent by 1.75″. A ray passing a body at the distance $d$ is bent by an angle proportional to $4 G M / (d c^{2})$ . Radio interferometers spread over Earth can now measure positions to 0.1 milliarcseconds. Could they detect the bending of radio waves from a quasar by (a) Jupiter and (b) the Moon? Estimate the bending in each case. Take Jupiter's mass and radius as $1.898 \times 10^{27}$ kg and $6.991 \times 10^{7}$ m, and the Moon's as $7.342 \times 10^{22}$ kg and $1.737 \times 10^{6}$ m.

I solved it without the solution

Answer(a) Yes, Jupiter bends the radio waves by about 17 milliarcseconds. (b) No, the Moon bends them by only 0.026 milliarcseconds.