# Relativity

## Gravitational Lensing

### By Mark Lawrence

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### Introduction

In 1916, when Einstein first released his Theory of General Relativity, one of his first predictions was that light would bend in a gravitation field. He suggest that if a star were photographed right at the edge of the sun during a solar eclipse, it would appear to move. Today, this effect is called Gravitational Lensing. A team of astronomers dutifully set out to take this picture, and the effect was found and recorded.

Figure 1: Light from a distant star is bent by the Sun's gravity, making the star appear to move

Since Einstein's success, gravitational lensing has become a favored effect in astronomy. Objects have been photographed as showing two images, one above and one below a galaxy due, we're told, to gravitational lensing. In some pictures, entire galaxies appear to be stretched and curved. Recently the Hubble telescope has been used to photograph lensing events in the Large Magellanic Clouds. What does this mean? Does this make any sense at all? Can we predict these effects?

### Arithmetic Precision

Astronomy is not a science where numbers are known to high accuracy. We can count things very well, but that's about it. Distances to remote objects must be deduced from logical arguments which include uncertainties. Masses of remote objects are even more uncertain. So, in this paper we'll only be computing to within about 10%. We can use modern calculators to perform arithmetic to 15 decimal places, but when the variables put into these calculations are uncertain by large factors, the resulting decimal places are meaningless. In this paper 32 = π2 = 3 * π = 10. We'll be working with extremely small angles, so we'll also presume sin θ = θ.

We're going to need some numeric values. In accord with our limitations on precision, we will use simplified values of the constants to keep the arithmetic clear. Here they are:

 RearthOrbit = 1 AU = 500 light seconds = 1.5 * 1011 meters Msun = 2 * 1030 kg Rsun = 7 * 108 m Mmoon = ¾ * 1023 kg Rmoon = 1.7 * 106 m Rmoonorbit = 3.8 * 108 m = 1¼ light second Mgalaxy = 200 billion solar masses Rgalaxy = 50,000 light years G = 2/3 * 10-10 N m2 / kg2 = 2/3 * 10-10 m3 / kg s2 c = speed of light = 3 * 108 meters per second G / c2 = ¾ *10-27 m / kg 1 year = π * 107 seconds 1 light year = π * 107 seconds * 3 * 108 meters per second = 1016 meters 1 radian = 3600 * 180 / π seconds = 206000 seconds of arc

Table 1: Constants

G, the gravitational constant, can be thought of as the conversion factor from mass to black hole radius. So, if we multiply G / c2 by 2 Msun, we get 3,000 meters, which is the radius of a black hole with the mass of the sun.

### Calculating the shift of a star image

Einstein found the basic equation for light deflecting around the Sun:

 4 G M deflection = R c2

This formula is only good for small angles of deflection. To duplicate Einstein's deflection calculation, we use

 4 G M 4 * ¾ * 10-27 * 2 * 1030 6 * 103 deflection = = = R c2 7 * 108 7 * 108

= .00000857 radians = .0004° = 1.77 seconds of arc. The photographs found almost this precise number. This great success in prediction has been an inspiration to many. This being the case, let's look a little closer at this result.

Consider how a camera works. Suppose you're to photograph a tree:

Figure 2: Photographing a tree.

If the tree is 100 yards away from the camera and 100 feet tall, then the image of the tree at 50 yards away from the camera is about 50 feet tall, and the image of the tree at 10 yards away from the camera is about 10 feet tall. Now, let's think more closely about the star being photographed. We'll presume this is a normal average star about the same size as our sun, so it's about 109 meters across. We'll presume the star is not too far away from us, say about 15 light years away from us. The earth is about 500 light seconds away from the sun. 15 light years is about 500 million light seconds.

Figure 3: Observing a distant star (not to scale).

We'll use "L" to represent the distance from the Earth to the Sun, and "S" for the distance from the Sun to the light source. "b" will be the distance from the light ray to the center of the Sun at closest approach, also known as the "impact parameter."

The image of our distant star at the sun is about a millionth of the original size, so at the distance of the sun the image is about 1,000 meters across. We can see immediately that gravitational lensing de-focuses the image: the portion of the image which is closer to the sun gets deflected a little bit more, and the portion which is farther away from the sun gets deflected a little less. Let's calculate how much.

 d deflection -4 G M -4 * ¾ * 2 * 10-27 * 1030 6 * 103 = = = = 1.2 * 10-14 radians / meter d b b2 c2 50 * 1016 50 * 1016

The image we're trying to photograph is 1000 meters tall at the sun, so the difference in deflection from the top to the bottom is 1.2 * 10-11 radians. This deflection, multiplied by the distance to the earth is 500 * 3 * 108 * 1.2 * 10-11 = 1800 * 10-3 = 1.8 meters. This means that if the astronomers were using a 1 meter telescope, almost half the light from the star was deflected away from their mirror. Fortunately astronomers are used to photographing very dim objects, and losing half the light is not a big deal to them.

### Einstein Rings

One way to notice Gravitational Lensing is a shift in the apparent position of a star, as above. Another way to notice it is if the lens and the star are perfectly lined up. In this case, the star can appear as a ring, and can also appear considerable brighter than without the lensing.

We're talking here about something that seems pretty strange, so it's worth while to pause and ask if this actually happens. Incredibly, yes. The picture below was taken by Rémi Cabanac using the European Southern Observatory Very Large Telescope, and was released to the public in June of 2005.

The lens in this case is a galaxy about 8 billion light years from us, while the source galaxy is about 12 billion light years away. This picture is looking back in time 80% of the way to the Big Bang. To our eyes this ring extends about 1/3 of the way around a circle - the galaxies aren't perfectly lined up. However, the more sensitive instruments attached to the telescope see the ring as extending 3/4 of the way around a circle.

You can see just at a glance that the Einstein Ring is quite a bit brighter than the galaxy would be without the lens. This is because a lot of light that would normally be sent all over space is now being focused directly at us. Without the lens the distant galaxy would be just a tiny dot; with the lens it appears as a large bright ring. We can calculate how much brighter a gravity lens makes a galaxy or star.

Let's consider gravitational lensing of a star in the Large Magellanic Cloud (LMC). Our galaxy, the Milky Way, has a few very small galaxies that are in orbit around us. The Large and Small Magellanic Clouds are two such small galaxies. They are only visible from the southern hemisphere, so Magellan was the first European to record them. Again, we'll use a typical star, about 109 meters across. The LMC is about 200,000 light years away, so that's about 2*1021 meters away. Let's presume the lens is inside our galaxy - a nearby black hole, for example, about 30,000 light years away from us.

Figure 4: A black hole causes an Einstein Ring image.

The image of our star at 30,000 light years is 30/(30+170) = 15% full size, about 108 meters. If the lensing object were a black hole of mass 5 solar masses, the radius of the black hole would be about 15,000 meters, about 10 miles. Near the surface of the black hole, the deflection of light is so great that no light can come directly from the star to us - any light getting anywhere near the black hole will be captured or scattered off in other directions. How far from the black hole does an image appear?

We know the formula for the deflection of light: φ = 4GM/bc2. This time, however, Earth, the lens, and the source are all in a straight line. We need to figure out how the deflection angle divides on each side of the lens.

Figure 5: Dividing up the deflection angle

φE + φS = φ
L φE = b = S φS
φE +  L/S φE = φ
φE = S φ / (L+S)

Now we're ready to find b, the distance of closest approach of the light ray to the black hole. The black hole has five solar masses, so it weighs 1031kg.

b = L φE = L S φ / (L+S)

 4 G M 4 G M (L+S) φ = = c2 b c2 L S φ

 √ 4 G M (L+S) √ 4 * ¾ * 10-27 * 1031 * (30,000 +170,000) * 1016 φ = = c2 L S 30,000 * 1016 * 170,000 * 1016

φ = 1.1 * 10-8  = .003 arc seconds

b = L S φ / (L+S) = 30,000 * 1016 * 170,000 * 1016 * 1.1 * 10-8 / (30,000 +170,000) * 1016
= 3 * 1012 meters

The image of the star is shifted to the sides of the black hole by 3 trillion meters, about 20AU, which roughly corresponds to the orbit of Uranus.

### Brightness of an Einstein Ring

We're now in a position to see if this ring image of the star is brighter than the star itself without the intervening black hole. We need to know how thick the radiating ring is. This is a little tricky. Normally, there is a simple cone of light from the star to our telescope mirror. But due to the defocussing of the gravitational lens, the cone no longer has a simple shape. We're going to have to find that shape in order to find the light intensity coming from the star.

Figure 5: The image of the star at the black hole, I1, is smaller than if there were no black hole in the way.

Without the lens, the area of the star at the lens distance is easy, it's just 4π(R L / (L + S))2 , where R is the radius of our star. The area of the Einstein Ring is also easy, it's just 2π b L θ, where θ is the angular width of the star image at the Einstein Ring. However, the lens spreads the light rays, so now we have to solve for two angles, θ and Δφ. We get Δφ by differentiating the equation for φ:

 d φ -4 G M = d b b2 c2

 √ 4 G M (L+S) φ = c2 L S
b = L S φ / (L+S)

 L S √ 4 G M (L+S) √ 4 G M L S b = = L + S c2 L S c2(L + S)

 d φ -4 G M -4 G M c2 (L + S) L + S = = = d b b2 c2 c2 4 G M L S L S

 d φ L + S ( L + S ) θ Δφ = Δb = L θ = d b L S S
Now we need the width of the ring so as to calculate its area:

2R = (L + S) θ + S Δφ = 2 (L + S) θ

θ = R / (L + S)

Ar = 2π b L θ = 2π b L R / (L + S)

A = 4πR2 L2 / (L + S)2

 Ar 2π b L R (L + S)2 b ( L + S ) 3 * 1012 * (30,000 + 170,000) * 1016 = = = = 10,000 A (L + S) 4πR2 L2 2 R L 2 *109 * 30,000 * 1016

Our gravitational lens, a black hole in our galaxy, is making a star in the Large Magellanic Cloud 10,000 times brighter. We see now how astronomers look for this: watch the LMC, and see if every now and then it flashes a lot brighter for a couple days. This flash indicates a black hole drifted precisely between us and one of the LMC stars. In fact, these flashes have been observed.

### Gravitational Lensing by a Diffuse Lens

Suppose the lens is not a black hole, but something diffuse, like a very large cloud of dust, or an entire galaxy. Although it's hard to see light coming through a galaxy, it's sometimes possible to see radio waves or x-rays. Recently a very large object was found due to gravitational lensing, but the object itself is completely invisible. It's currently thought that this very large object is a cloud of dark matter - something very mysterious, not at all understood by physics, that only interacts with us through gravity.

If the object is diffuse, then if the light goes through the object, the mass M we use is not the mass of the object, but rather fraction of the mass that's inside the orbit of the light. If we presume a uniform density D, then the mass of an object out to a radius b is 4πb3D. We just have to remember when we're done with our calculations to check that b is less than the radius of the lens; if it's larger than we use the total mass M.

 4 G 4/3 πb3D 16 G πb2 D 16 G π (L S φ)2 D deflection = φ = = = b c2 3c2 3 (L + S)2 c2

 3 (L + S)2 c2 φ = 16 G π (L S)2 D

 3 (L + S) c2 b = 16 π G L S D

Let's try these formula out on the moon. We already know the answer - during an eclipse, there's no ring of visible sun.

volume of moon = 2.2 * 1019 m3
mass of moon = ¾ * 1023 kg
average density of moon = 1/3*104 kg / m3
L+S = 501 ls = 1.5*1011 m
LS = 500 ls2 = 5 * 1019 m2

 3 * 1.5*1011 b = = ¾ *1014 m 16 * 3 * ¾ *10-27 * 5 * 1019 * 1/3*104

The moon is only 106 meters in radius, so our answer of 1014 m makes no sense, just as we expected.

### Galaxies and Orbits

NGC 1300: A Barred Spiral Galaxy

There are about 200 billion galaxies visible to our telescopes. About three quarters of all galaxies are spiral galaxies. About a quarter of all spiral galaxies are barred spiral galaxies; a barred spiral is shown above. Barred spirals have large fraction of their stars arranged in a straight bar, then spiral arms coming off the ends of the bar. Our own galaxy, the Milky Way, is currently thought to be a barred spiral.

The dynamics of spiral galaxies have been a mystery for almost a century. If we think of a galaxy as a large number of stars orbiting around a central mass of stars, then the picture above makes no sense. Consider our solar system:

Sun, Mercury, Earth, Mars Lined Up; and 44 Days Later

If the planets were to ever line up, as shown above, this would not last. Mercury makes a complete orbit around the Sun in 88 days; the Earth tales a year, 365 days; and Mars takes 686 days to complete an orbit. In 44 days, Mercury has moved to the other side of the sun, the Earth has moved about 45 degrees in its orbit, and Mars has moved about 23 degrees in its orbit. Pluto, which takes 248 years to complete a single orbit, basically doesn't move at all in 44 days. We see that the planets all lining up is a rare and short lived thing - so special it has a name, syzygy. However, in the barred spiral shown above, the stars in the bar are all lined up and apparently all staying that way. If we saw one such galaxy, it would perhaps be luck. However, we see about 35 billion such galaxies, which pretty much precludes luck.

There are three mysteries here: how do stars in such different orbits wind up going around the center of the galaxy in the same time; how did the stars get lined up in the first place; and why are there nearly always two arms which are on exact opposite sides of the galaxy. In galaxy time, lining up the stars must happen very quickly: a star typically makes a full orbit of its galaxy in about 100 million years, and the universe is only about 14 billion years old, so the stars have only made about 100 or so orbits since their galaxy formed. The fact that there are nearly always two arms, nearly always exactly opposite each other is a big clue: this clearly demonstrates that the two arms are somehow coupled to each other. Our job is to try to deduce the source of this coupling.

### Gravitational Lensing and Curved Space-Time

We can understand how these galaxies formed by remembering how light bends around a black hole, forming an Einstein ring. Up until now we've been quite sloppy with our language, and we'll have improve. In fact, just as you know from your every day experience, light travels in straight lines. We very commonly use lasers today to trace out a perfectly straight line. So, how can we explain the path of the light as shown below? Einstein and General Relativity assure us that this is the straightest of all possible lines: it's space-time that's curved, not the light ray.

Imagine you followed the light ray with a very fast space ship. As you got close to the black hole, the space ship would curve around the black hole just as the light ray does. The space ship would do this naturally: there would be no steering by the navigator, and the crew would not feel they were going around a corner, as you feel when you go around a corner in a car. In fact, if the space ship had no windows, it would be nearly impossible for the crew to tell when they were near the black hole. Just as the space shuttle astronauts in orbit around the Earth do not feel their weight or the fact of their curved orbit, the crew of our space ship would not feel their weight or the fact of their curved path around the black hole. Near the black hole, the paths taken by your laser beams are bent, your meter sticks are bent, the path taken by your space ship is bent. If you try to make a different definition of a straight line, I think you will quickly get confused.

The path looks curved to us because we have the intuition that space-time is flat, just like this piece of paper; that triangles have angles that sum to 180 degrees; that right triangles with sides a, b, c have them in the proportion a2 + b2 = c2; that the circumference of a circle is π times the diameter. In fact, in our universe near a large mass, none of these statements are true. Triangles have angles that sum to more than 180 degrees. The Pythagorean Theorem must be replaced with something rather more complicated. Circles have circumferences that are less than π times their diameter. At the surface of the Earth, these effects are typically on the order of one part in a billion, so we didn't notice them until very recently. At the surface of the Sun, the effects are more like one part in a million, and can be observed with our best telescopes. Near the surface of a black hole these effects become very strong, so strong that even light can orbit a black hole - the straight line of the light ray follows space-time which is so strongly curved that the path of the light ray forms a complete circle.

### Gravitational Lensing and Spiral Arm Formation

We have seen that a gravitational lens can make a star appear brighter by a factor of 10,000, perhaps more in special circumstances. Not only is this true, but gravity follows the same path as the light rays. When the star looks 10,000 times brighter, it also has 10,000 times the gravitational pull. In the case of our individual stars in the Large Magellanic Cloud, the gravitational pull from an individual star 200,000 light years away is so tiny that even 10,000 times the pull is completely immeasurable.

It's different in a galaxy. We're just learning about galactic structure, but currently it's thought that typical spiral galaxies have black holes in their centers that weigh 50 million to 500 million times as much as our sun. In addition, the galactic core, the spherical group of stars in the center of the galaxy, contains about ¼ of the total mass of the galaxy. The lensing properties of such a huge mass are quite strong. Below is a Hubble Space Telescope picture of dozens of remote galaxies having their images curved and distorted by the more nearby galactic cluster Abell 2218.

We must not only explain how it is that the stars in a spiral arm all orbit the galactic core with the same orbital period; we must also explain how they got lined up in the first place. The answer to this is now more clear: initially, a galaxy is most likely a relatively unstructured disk of stars orbiting the core, much like the rings of Saturn. However, stars naturally clump up in the galactic disk. These clumps are called globular clusters, and are more or less mini-galaxies trapped within our own galaxy. A typical globular cluster has 10,000 to 1 million stars. Globular clusters are much lower in heavy element abundance than our Sun, so globular clusters are believed to have formed from the more primordial matter present in the young galaxy just after its formation. Globular clusters formed at about the same time as the galaxy.

It's currently believed that most globular clusters have a black hole at their core which contains roughly 5% of the cluster's mass. As such a globular cluster orbits the galactic core, stars on the opposite side of the core will feel a sudden increase in gravity when they are precisely lined up opposite the globular cluster with the galactic core exactly in line between. The gravity of the cluster is magnified just as the light from our individual star was. This is like the mass of the galaxy suddenly increasing when everything is lined up. It's now easy to see how such clusters on opposite sides of the galactic core will lock each other into synchronized orbits, and any stars that come between them will also be locked into the same orbit.

The Globular Cluster M15 in Pegasus; about 1,000,000 stars grouped together in our Galaxy

### Calculating the Gravity Coupling

Let's consider a globular cluster containing a million stars that's about half way out an arm, 25,000 light years from the core, and calculate the gravity enhancement at 25,000 light years on the other side of the core. The core consists of a massive black hole, perhaps 10 million to 500 million solar masses, and a whole bunch of stars, perhaps 50 billion, with an average separation of about a light year. First we'll see if the stars make any difference using our formula for lensing by diffuse mass.

 3 (L + S) c2 b = 16 π G L S D

The stars have a typical mass of 1030 kg, so the density is 1030 / 1 light year3 = 1030 / 1048 = 10-18. We presume the lens (the galactic core) is 25,000 light years from us, and the source is 25,000 light years from the lens. Now, the impact parameter, b, is

 3 (L + S) c2 3 *5 * 1020 5 b = = = ≅ 1024 m = 108 light years 16 π G L S D 16 π * ¾ *10-27 * 6 * 1040 * 10-18 24 π * 10-7 * 10-18

The galactic core is only about 10,000 light years across, so this answer makes no sense. This means we can ignore the stars in the galactic core and only consider the black hole in the core. Again, we'll calculate for the cluster being 25,000 light years out from the core, and we'll calculate the gravity enhancement 25,000 light years on the other side of the core. We'll use 100 million solar masses for the core black hole.

 √ 4 G M L S √ 4 * ¾ *10-27 * 108 * 2 * 1030 * 625 * 1038 b = = = 1016 m = 1 light year c2(L + S) 50 * 1019

Now we'll calculate the gravity enhancement on the cluster's black hole. We'll presume the cluster has a mass of about 1 million solar masses, and the cluster's black hole has a mass of 50,000 solar masses (5%). Such a black hole has a radius of  150 million meters.

 Ar b ( L + S ) 1016 * 50,000 * 1016 = = ≅ 108 A 2 R L 2 * 150 * 106 * 25,000 * 1016

When we are aligned with both the core black hole and the cluster black hole, we see the apparent mass of the galaxy increase rather dramatically. The cluster black hole is twice as far from us as the core black hole, so the gravitational force is ¼ as much. The mass of the galactic core is about 1011 solar masses; the focused mass of the cluster black hole looks like 1012 solar masses. When the cluster black hole, the core black hole, and the Earth are all lined up we actually feel more gravity from the cluster black hole than we do from the entire galactic core.