**Microlensing**

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**.

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.

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/c

^{2}d.

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/c^{2}:

(2) a = 2R

_{S}/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 = 2R

_{S}/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 = 2R

_{S}/Rq.

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

(6) q

^{2 }- bq - 2R_{S}/R = 0,

whose solutions q_{+ }and q_{- }are

(7a) q

_{+}= b/2 + (b^{2}/4 + 2R_{S}/R )^{1/2},(7b) q

_{-}= b/2 - (b^{2}/4 + 2R_{S}/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) = + (2R_{S}/R )^{1/2},(8b) q

_{-}(b=0) = - (2R_{S}/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 R_{E}, *at the position of the halo
object*, is given by

(9) R

_{E}= Rq_{+}(b=0) = (2R_{S}R)^{1/2}.

The radius R_{E} is called the **Einstein
Radius**.

**Example**:

Let's work out the Einstein radius for a halo
object that is 10 kpc away, and which has a Schwarzchild radius
of R_{S }= 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 10^{17} km. Therefore, the Einstein radius is

R

_{E}= (2x10^{-1}x3.1x10^{17})^{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 = (2R

_{E}/R)*(360/2p)*3600 arcseconds= (2 * 2.49 x 10

^{8}/3.1 x 10^{17})*(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/(b^{2 }+ 8R_{S}/R )^{1/2}]= 1/2[1 + u/(u

^{2}+ 4R_{E}^{2})^{1/2}](13b) Dy

_{-}/Du = 1/2[b/(b^{2 }+ 8R_{S}/R )^{1/2}- 1]= 1/2[u/(u

^{2}+ 4R_{E}^{2})^{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(b^{2}/4 + 2R_{S}/R )^{1/2}= 1/2[u + (u

^{2}+ 4R_{E}^{2})^{1/2}](14b) y

_{-}= Rq_{- }= R(b^{2}/4 + 2R_{S}/R )^{1/2}- Rb/2= 1/2[(u

^{2}+ 4R_{E}^{2})^{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 = (u

^{2}+ 2R_{E}^{2})/[u(u^{2}+ 4R_{E}^{2})^{1/2}].

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

(16) m = (u

^{2}+ 2)/[u(u^{2}+ 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/R_{E}. We divide Vt by R_{E
}because we are
measuring lengths in units of the Einstein radius. From
Pythagorus' theorem we have for any other time t

(17) u(t) = (u0

^{2}+ (V/R_{E})^{2}t^{2})^{1/2}.

The time Dt =R_{E}/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) = (u0

^{2}+ (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*