In Einstein's theory of General Relativity gravity is understood as a manifestation of the curvature of spacetime. One consequence of this is that light rays should bend near any mass. In particular, Einstein predicted that the light from a star should bend as it goes past a massive object like the sun. That prediction was first tested by Sir Arthur Eddington in 1919 during an total eclipse of the sun. A star that would otherwise not be visible, because it was behind the sun, was observed close to the sun's outer edge (its limb) confirming the bending of light by gravity.

In 1937 Zwicky (Phys. Rev. 51, 1937, 290) noted that the bending of light by massive objects should lead to gravitational lensing: the focusing of light from a distant object by a mass that lies in the line-of-sight between the distant object and the observer. At the time the prediction was nothing more than an interesting curiosity because the technical means to test it simply did not exist, and would not exist for another sixty years! Today, gravitational lensing not only affords a beautiful test of Einstein's theory, but also provides a tool for investigating the halo of dark matter thought to surround the Milky Way.

The basic idea, as pointed out by the astrophysicist Bohdan Paczynski (Princeton) in 1986, is that if the galactic halo contains dark objects---called massive compact halo objects (machos), with masses between that of Jupiter (about 1/1000 of a solar mass) and very dim stars (e.g., brown dwarfs) with masses too small (less than 1/10 of a solar mass) to trigger thermonuclear ignition, then once in a while these massive halo objects should cross the line-of-sight between the earth and a more distant luminous star. If the halo object gets close enough to the line-of-sight we should see a noticeable lensing effect, characterized by a temporary brightening of the light from the distant star. This effect, which I describe in detail below, is called microlensing

Searching for halo objects

Four experimental teams, EROS, MACHO, OGLE and DUO are now searching for halo objects using the stars in the Large Magellanic Cloud (LMC), a satellite galaxy of the Milky Way, as the back-drop against which the halo objects move. The basic idea is illustrated in the figure below.

Figure 1.

 Light from a star (the yellow dot in the figure) in the LMC is deflected, indeed focused, by a passing halo object (the brown dot). These objects act as gravitational lenses. In principle, by measuring the distribution and rate of microlensing events we could learn something about the nature and distribution of these halo objects. In practice, this is very difficult because the accidental line-ups occur infrequently. The EROS collaboration, for example, has measured a microlensing rate of only 7.3 x 10-7 events/star/year (astro-ph/9511073). So, if you looked at about 10 million stars continuously for one year you might expect to detect only one or two microlensing events! Clearly, this research is very challenging. The EROS team is able to monitor about 4 million stars, simultaneously. In a period of three years they found two events that can be interpreted as microlensing events. The MACHO team has found several possible microlensing candidates.

Many stars are known to brighten and dim; some in a regular and repeating pattern, while others do so irregularly. These are the variable stars. The brightness versus time of a stellar object is called its light curve. With so many variable stellar objects in the sky how can we possibly distinguish the light curve of microlensing events from that of, say, variable stars?

The answer is that general relativity predicts a precise form for the light curve of a microlensed star. This light curve may even be unique. Also, microlensing events are singular events: a star appears to brighten then dim, and this sequence is not repeated. Another interesting signature of microlensing is that the effect is achromatic: that is, it does not depend on the color of the light (provided the source and lens are compact). Therefore, if you look at a star through two filters, usually blue and red, the blue and red light curves should be identical. We know of no stellar phenomena, other than microlensing, which would produce such behaviour. Here is the light curve of one of the possible microlensing events found by the MACHO team.

Figure 2.

Notice the perfectly symmetrical nature of the light curves and the fact that the red and blue light curves are indeed the same. This is very convincing evidence that this is, in fact, the light curve of a microlensing event.

The Einstein Ring

We shall now work through the mathematics of microlensing. What we want to calculate is the angle between the star's image and the halo object. Happily, the mathematics is not so bad; honest! It requires just a bit of basic geometry and one formula from general relativity. Oh---and a spot of calculus! Consider the diagram below.

Figure 3.

Let's suppose that the earth (the blue dot on the left), a halo object (the brown dot in the middle) and a light source, say a star (the white dot on the right), are almost lined up. In the absence of gravitational lensing the light from the star would travel from the star directly along the line-of-sight to the earth. However, the presence of the halo object causes light from the star, which would otherwise have missed the earth, to bend towards the earth. The light now appears to come from the direction of the yellow dot, which is an image of the star. As we shall see shortly there is another image on the other side of the halo object. (As drawn, the second image would lie below the halo object.)

The distance between the earth and the light source is denoted by L---this would be 55 kpc for a star in the LMC. The distance between the earth and the halo object is R, maybe 10 kpc. The distance between the image (the yellow dot) and the halo object (the brown dot) is d. The angle between the halo object and the star is denoted by b, while the angle between the halo object and the star's image is denoted by q. Obviously, the angle between the star and its image is q-b. The light from the star bends by an angle a, which according to general relativity is given by

(1) a = 4GM/c2d.

This is the only formula we shall need from general relativiy! Notice, the formula can be written in terms of the Schwarzchild radius, Rs = 2GM/c2:

(2) a = 2RS/d.

For simplicity, even though this is not a good approximation for the Large Magellanic Cloud, we shall assume that the star is very much further away from earth than the halo object; that is, we shall assume that L >> R. With this assumption we may write the approximate equation a = q-b. When we combine it with Eq. (2) we get

(3) q-b = 2RS/d.

Now we need to figure out how the distance d is related to q. For small angles this is easy to work out; the desired relationship is

(4) d = Rq,

which, when combined with Eq. (3) gives

(5) q-b = 2RS/Rq.

A slight re-arrangement of Eq. (5) leads to the quadratic equation

(6) q2 - bq - 2RS/R = 0,

whose solutions q+ and q- are

(7a) q+ = b/2 + (b2/4 + 2RS/R )1/2,

(7b) q- = b/2 - (b2/4 + 2RS/R )1/2.

So here are the formulas we wanted. Notice that the q+ solution is slightly larger than the angle b, the angular separation between the star and the halo object. This solution corresponds to an image that is displaced away from the line-of-sight (at the location of the yellow dot in the figure above). The q- solution corresponds to a second image displaced away from the halo object. The angle q- is actually negative! All this means is that the second image is diametrically apposite the first image, but on the other side of the halo object. In the figure above the second image would lie below the gravitational lens. The diagram below shows what this might look like on the sky. (For convenience, we have oriented the lens and the star so that a line drawn through their centers is horizontal.)

Figure 4.

Imagine a halo object (the lens) moving across the sky, close to a star (the light source). At some instance the lens will be an angle b away from the star. An image of the star would form at an angle q+ to the left of the lens; a second image would form at an angle |q-| to the right of the lens. The vertical bars | | simply means that we just consider the size of the angle and forget about its sign. Note, since the halo object is moving (upwards in the diagram) the size of the angle b will change, as will the relative orientation of the two images. The left image will move along a downward arc, while the right image will move along an upward arc. These microlensing events can last for days; it would be quite spectacular to make a movie of them. Unfortunately, for halo objects, the angles involved are, at present, too small to be resolved.

What happens when the distant star and the halo object (that is, the lens) are perfectly aligned? Perfect alignment means that the angle b between the star and the halo is zero. In that case, we find from Eq. (7)

(8a) q+(b=0) = + (2RS/R )1/2,

(8b) q-(b=0) = - (2RS/R )1/2.

The images, not surprisingly, are displaced symmetrically about the lens. In fact, because of the rotational symmetry about the line-of-sight (you can rotate everything about the line-of-sight without changing the relative orientation of the halo object and the star) there would, in fact, be a ring of images about the halo object. The halo object would have a halo! This ring is called an Einstein Ring. The radius of the ring RE, at the position of the halo object, is given by

(9) RE = Rq+(b=0) = (2RSR)1/2.

The radius RE is called the Einstein Radius.


Let's work out the Einstein radius for a halo object that is 10 kpc away, and which has a Schwarzchild radius of RS = 1/10 km. This Schwarzchild radius corresponds to a halo object whose mass is about 1/30 that of the sun. (The sun's Schwarzchild radius is about 3 km.) One parsec is equal to approximately 31 trillion kilometers, so 10 kiloparsecs is equal to 3.1 x 1017 km. Therefore, the Einstein radius is

RE = (2x10-1x3.1x1017)1/2 = 249 million km.

This is about 1.7 AU; therefore, the ring would be slightly larger than the orbit of Mars. How large would the diameter of the ring appear in the sky in arcseconds? (One arcsecond = 1/3600 of a degree.) This answer is

Angular diameter = (2RE/R)*(360/2p)*3600 arcseconds

= (2 * 2.49 x 108/3.1 x 1017)*(360/2p)*3600,

= 3.2 x 10-4 arcseconds.

That is, about 1/10 of a milliarcsecond! At present, such a tiny angle is utterly beyond our ability to resolve.


In the above discussion we considered a gravitational lens at a fixed angular separation b from a star. In reality, of course, the angular separation would always be changing because of the relative motion of the lens and the star. As we noted above, this causes the total brightness of the two images to change in a very special way. In this section we shall derive the form of the microlensing light curve. The amount of light we receive from a star is determined by the solid angle subtended by the star. The solid angle is just the apparent angular area of the star in the sky. The lensing effect increases the solid angle over which we receive light, thereby increasing the amount of light we receive. So if we can calculate the solid angle of the star in the absence of lensing and the solid angle with lensing the light magnification would be simply

magnification = total solid angle with lensing/solid angle without lensing.

To see how this works consider the figure below

Figure 5.

The brown dot represents the gravitational lens. Far behind the lens is a star. Consider a strip of width Du on the star's surface. Let u be the distance between the lens and the points A and B on the strip. Light from these points travels out in all directions; however, the only rays that we need consider are the ones that travel to points A+, B+, A- and B-. In the absence of lensing the light from the points A and B would come straight towards us. But because of lensing the light from point A appears to come from the points A+ and A-; the light from point B appears to come from the points B+ and B-. The lensing effect therefore creates two image strips of width Dy+ and Dy-, to the left and right of the lens, as illustrated. Let the radial distance from the lens to the points A+ and B+ be y+ and let y- be the radial distance from the lens to the points A- and B-.

The amount of light E from each strip is just the flux f (the energy per unit area per second) times the area of the strip:

(10a) E = f *Area = f u f Du

(10a) E+ = f *Area = f y+f Dy+

(10a) E- = f *Area = f y-f Dy-,

where f is the angle between the points A and B, relative to the lens. (We have used the fact that the length of an arc = its radius times its angular size.) Therefore, the magnification m is just

(11) m = (E+ + E-)/E = (y+/u)(Dy+ /Du) + (y-/u)Dy- /Du).

Notice that the flux f and angle f cancel out. So all we need do now is to relate u, y+ and y- to b, q+ and q-, respectively, and to calculate the ratios Dy+ /Du and Dy- /Du. The first set of relations are easy:

(12) u = Rb, y+ = Rq+ , y- = Rq-.

The second set (the ratios) needs a bit of calculus! Here are the results

(13a) Dy+ /Du = 1/2[1 + b/(b2 + 8RS/R )1/2]

= 1/2[1 + u/(u2 + 4RE2)1/2]

(13b) Dy- /Du = 1/2[b/(b2 + 8RS/R )1/2 - 1]

= 1/2[u/(u2 + 4RE2)1/2- 1].

In Eq. (13) we have substituted u for b. From Eqs. (12) and (7) we can write

(14a) y+ = Rq+ = Rb/2 + R(b2/4 + 2RS/R )1/2

= 1/2[u + (u2 + 4RE2)1/2]

(14b) y- = Rq- = R(b2/4 + 2RS/R )1/2 - Rb/2

= 1/2[(u2 + 4RE2)1/2 - u].

We now substitute Eqs. (13) and (14) into Eq. (11), and after a bit of algebra we get the formula for the magnification m as a function of the distance u of the lens from the line-of-sight to the star:

(15) m = (u2 + 2RE2)/[u(u2 + 4RE2)1/2].

We can simplify this formula further if we measure u in units of the Einstein radius RE. To do that we just divide u by the Einstein radius in Eq. (15). This gives, finally, the simple formula

(16) m = (u2 + 2)/[u(u2 + 4)1/2].

However, we are not yet quite done! As the lens moves across the sky, or if we keep the lens fixed and imagine the star moving, the distance u will change; it will be a function u(t) of the time t. What is that function? We shall assume that the lens (or the star) moves in a straight line across the sky at a constant speed V. The distance traveled in time t is just D = Vt. At some time, let's say t = 0, the star and lens will be as close in the sky as they can get. Let's denote this distance of closest approach by u0 = u(t=0). Note that this distance is at right-angles to the direction of motion of the lens (or the star, if we consider the lens to be fixed). See Fig. 4. At any other time t, the distance u(t) will be the hypotenuse of a right-angled triangle; the other side is u0 and the third is Vt/RE. We divide Vt by RE because we are measuring lengths in units of the Einstein radius. From Pythagorus' theorem we have for any other time t

(17) u(t) = (u02 + (V/RE)2 t2)1/2.

The time Dt =RE/V to cross a distance equal to one Einstein radius is called the lensing time. This can be anywhere from hours to weeks. Usually, we write u(t) in terms of the lensing time:

(18) u(t) = (u02 + (t/Dt)2)1/2.

The formulas given by Eqs. (16) and (18) provide a precise description of the microlensing light curve, examples of which are shown in Fig. 2.

Last updated April 6, 1998, Harrison B. Prosper