# How does gravitational lensing account for Einstein's Cross?

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Einstein's Cross has been attributed to gravitational lensing. However, most examples of gravitational lensing are crescents known as Einstein's rings. I can easily understand the rings and crescents, but I struggle to comprehend the explanation that gravitational lensing accounts for Einstein's cross. I found this explanation, but it was not satisfactory.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Dale

edited Aug 25, 2019
Relevant thought: are the four dots mirror images of the original light source or are they distorted chunks of a crescent where the rest of the crescent has been blocked? Perhaps there are other possibilities I have not considered.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Dale
I would add... I'm also interested to know if the Einstein cross was a specific prediction that people had obtained from GR or if the term was coined after the image was discovered. That could help explain a lot.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Alan Rominger
More on Einstein's cross: hyperphysics.phy-astr.gsu.edu/hbase/Astro/eincros.html

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Dale

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The middle galaxy in Einstein's cross has an elliptical mass distribution which is wider in the direction of the short leg of the cross (originally, this said long leg), with a center of mass where you see the galaxy. The object is slightly to the right of the center of the ellipse, in the direction of the long leg of the cross (the original answer had the direction reversed). This type of lensing is acheivable in such a configuration, when the lensing object is relatively close to us, so that the rays pass the central region, where the quadrupole moment asymmetry of the gravitational field is apparent.

### Lensing map

Given a light source, call the line between us and the source the z axis, and parametrize outgoing light rays by the x-y coordinates of their intesection with an x-y plane one unit distance away from the source in our direction. This is a good parametrization for the tiny angles one is dealing with. The light rays are parametrized by a two dimensional vector v.

These light rays then go through a lensing region, and come out going in another direction. Call their intersection point with the x-y plane going through our position v'. The lensing problem is completely determined once you know v' as a function of v.

We can only see those rays that come to us, that is, those rays where v'(v) is zero. The number and type of images are entirely determined by the number and type of zeros of this vector field. This is why it is difficult to reconstruct the mass distribution from strong-lensing--- many different vector fields can have the same zeros.

The only thing we can observe is the number of zeros, and the Jacobian of the vector field v' at the zero. The Jacobian tells you the linear map between the source and the observed image, the shear, magnification, inversion.

The lensing map is always asymptotically linear, v'(v)= v for large v, because far away rays are not lensed, and the scale of v is adjusted to make this constant 1.

### Generic strong lensing

In a generic strong lensing problem, the vector field v has only simple zeros. The Jacobian is a diagonalizable matrix with nonzero eigenvalues. This means that each image is perfectly well defined, not arced or smeared. The image is arced only in the infinitely improbable case that you have a singular Jacobian.

But we see gravitational arcs all the time! The reason for this is that for the special case of a spherically symmetric source, the Jacobian is always singular. The source, the center of symmetry, and us make a plane, and this plane includes the z-axis, and necessarily includes the direction of the image. The Jacobian at a zero of v' always has a zero eigenvalue in the direction perpendicular to this plane.

This means that the spherically-symmetric far-field of any compact source will produce arcs, or smears. When the lensing object is very far, the rays that get to us are far away from the source, and we see far-field arcs and smears. When the lensing galaxy is close, the lensing field has no special symmetry, and we see points with no smearing.

So despite the intuition from point sources and everyday lenses, Einstein's cross is the generic case for lensing, the arcs and smears are special cases. You can see this by holding a pen-light next to a funhouse-mirror. Generically, at any distance, you will see the pen light reflected at multiple images, but only near special points do you get smearing or arcing.

### Topological considerations

There is a simple topological theorem about this vector field v'. If you make a large circle in the v plane, and go around it counterclockwise, the value v'(v) along this circle makes a counterclockwise loop once around. This is the winding number of the loop.

You can easily prove the following properties of the winding number:

• Every loop has a winding number
• if you divide a loop in two, the winding number of the two parts add up to the winding number of the loop.
• the winding number of a small circle is always 0, unless the vector field is zero at inside the circle.

Together they tell you what type of zeros can occur in a vector field based on its behavior at infinity.

The winding number of the vector field in a small circle around a zero is called its index. The index is always +1 or -1 generically, because any other index happens only when these types of index zeros collide, so it is infinitely improbable. I will call the +1 zeros "sources" although they can be sources sinks or rotation/spiral points. The -1 zeros are called "saddles". The images at saddles are reflected. The images at sources are not.

These observations prove the zero theorem: the number of sources plus the number of saddles is equal to the winding number of a very large circle. This means that there are always an odd number of images in a generic vector field, and always one more source than saddle.

A quick search reveals that this theorem is known as the "odd number theorem" in the strong lensing community.

This theorem is very strange, because it is exactly the opposite of what you always see! The generic images, like Einstein's cross, almost always have an even number of images. The only time you see an odd number of images is when you see exactly one image. What's the deal?

The reason can be understood by going to one dimension less, and considering the one-dimensional vector field x'(x). In two dimensions, the light-ray map is defined by a zeros of a real valued function. These zeros also obey the odd-number theorem--- the asymptotic value of x'(x) is negative for negative x and positive for positive x, so there are an odd number of zero crossings.

But if you place a point-source between you and the object, you generically see exactly two images! The ray above that is deflected down, and they ray below that is deflected up. You don't ever see an odd number. How does the theorem fail?

The reason is that the point source has tremendously large deflections when you get close, so that the vector field is discontinuous there. Light rays that pass very close above the point are deflected very far down, and light rays that pass very close below are deflected far up. the discontinuity has a +1 index, and it fixes the theorem. If you smooth out the point source into a concentrated mass distribution, the vector field becomes continuous again, but one of the images is forced to be right behind the continuous mass distribution, with extremely small magnification.

So the Einstein cross has five images: there are four visible images, and one invisible images right behind the foreground galaxy. This requires no fine tuning--- the fifth image occurs where the mass distribution is most concentrated, which is also where the galaxy is. Even if the galaxy were somehow transparent, the fifth image would be extremely dim, because it is where the gradient of the v field is biggest, and the smaller this gradien, the bigger the magnification.

### Einstein's cross

After analyzing the general case, it is straightforward to work out qualitatively what is happening in Einstein's cross. There is a central mass, as in all astrophysical lenses, so there is an invisible central singularity/image with index +1. the remaining images must have 2 sources and 2 saddles. The most likely configuration is that the two sources are the left and right points on the long leg of the cross, and the two saddles are the top and bottom points (in my original answer, I had the orientation backwards. To justify the orientation choice, see the quantitative analysis below)

You can fill in the qualitative structure of the v'(v) vector field by drawing its flow lines. The image below is the result. It is only a qualitative picture, but you get to see which way the light is deflecting (I changed the image to reflect the correct physics):

alt text The flow lines start at the two sources, and get deflected around the two saddles, with some lines going off to infinity and some lines going into the central singularity/sink. There is a special box going around source-saddle-source-saddle which cuts the plane in two, and inside the box, all the source flows end on the central singularity/image and outside all the source flows end at infinity.

The flow shows that the apparent fourfold symmetry is not there at all. The two sources are completely different from the two saddles. The direction of light deflection is downward towards the long axis of the cross, and inward toward the center. This is the expected deflection from a source which is elliptical oriented along the long-direction of the galaxy.

### Model

(The stuff in this section was wrong. The correct stuff is below)

### General Astrophysical Lensing

The general problem is easy to solve, and gives more insight into what you can extract from strong lensing observations. The first thing to note is that the deflection of a particle moving at the speed of light past a point mass in Newton's theory, when the deflection is small is given by the integral of the force over a straight line, divided by the nearly constant speed c, and this straightforward integral gives a deflection which is:

$$\Delta\theta = -{R_s\over b}$$