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  How does gravitational lensing account for Einstein's Cross?

+ 12 like - 0 dislike

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.

enter image description here (Image source)

enter image description here (Image source)

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

asked Aug 28, 2011 in Theoretical Physics by Dale (60 points) [ revision history ]
edited Aug 25, 2019 by Dilaton
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

5 Answers

+ 12 like - 0 dislike

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.

The odd number paradox

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.


(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:

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

Where $R_s = {2GM\over c^2}$$ is the Schwartschild radius, b is the impact parameter, the distance of closest approach, and everything is determined by dimensional analysis except the prefactor, which is as I gave it. The General Relativity deflection is twice this, because the space-space metric components contribute an equal amount, as is easiest to see in Schwartschild coodinates in the large radius region, and this is a famous prediction of GR.

When the deflections are small, and they always small fractions of a degree in the actual images, the total deflection is additive over the point masses that make up the lensing mass. Further, the path of the light ray from the distant light source is only near the lensing source for a very small fraction of the total transit, and this lensing region is much smaller than the distance to us, or the distance between the light source and the lensing mass. These two observations mean that you can squash all the material in the lensing mass into a single x-y plane, and get the same deflection, up to corrections which go as the ratio of the radius of a galaxy to the distance from us/source to the galaxy, both of which are safely infinitesimal. The radius of a galaxy and dark-matter cloud is a million light years, while the light source and us are a billion light years distant.

You convert $\Delta\theta$ to x-y plane coordinates I am using by multiplying by a unit distance. This gives the amount and direction of deflection from a given point mass. The total deflection of the light ray at distance B is given by the sum over all point masses in the galaxy and its associated dark-matter of this vector contribution, which is four time the mass (twice the Schwartschild radius) divided by the distance, pointing directly toward the mass. This sum is $\Delta v$.

What is important to note is that this sum is equal to the solution to a completely different problem, namely the 2-d gravitational field of (four times) the squashed planar mass. In 2d, gravity goes like 1/r. The planar gravitational field of the planar mass distribution gives $\Delta v$, and it is most important to note that this means that $\Delta v$ is the gradient of the 2d gravitational potential:

$$ \Delta V= - \nabla \phi$$


$$\phi(x) = \int \rho(u) \ln(|x-u|) d^2u$$

where the two dimensional density $\rho(u)$ is the integral of the three dimensional density in the z direction (times $4G\over c^2$). This is important, because you can easily determine $\phi$ from the mass distribution by well known methods for solving Laplace's equation in 2d, and there are many exact solutions.

The impact parameter B is equal to $vR_1$, the original direction the light ray goes times the distance from the light-source to the lensing object, and the position this light ray reaches when it gets to us is:

$$ v'(v) = v(R_1 + R_2) + \Delta v(vR_1) R_2$$

Choosing a new normalization for v so that vR_1 is the new v, and choosing a normalization for v' so that v'(v) is v at large distances:

$$ v'(v) = v - {R_1\over R_1+R_2} \nabla \phi(v) $$

This is important, because it means that the whole thing is a gradient, the gradient of:

$$ v'(v) = - \nabla(\phi'(v))$$ $$\phi'(v) = {R_1\over R_1+R_2} \phi(v) - {v^2\over 2} $$

The resulting potential also has a 2d interpretation--- it is the gravitational potential of the planar squashed mass distribution in a Newton-Hooke background, where objects are pushed outward by a force proportional to their distance.

The 2-d gravity potential is easy to calculate, often in closed form, and to find the lensing profile, you just look for the maxima, minima, and saddles of the 2-d potential plus a quadratically falling potential.

This solves the problem for all practical astrophysical situations. I found it remarkable that the deflection field is integrable, but perhaps there is a simpler way of understanding this.

Point mass

The 2d potential of a point mass is

$$ \phi(v) = \ln(|v|)$$

and for an object directly behind it, you get

$$ \phi'(v) = A \ln(|v|) - |v|^2$$

This gives a central singularity (or if you spread out the mass in the center, a dim image right on top of the mass) plus a perfect ring where $r=\sqrt{A}$. This is the ring image.

Moving the light source off center just shifts the relative position of the two potential centers. The new potential is:

$$ \phi'(v) = {A\over 2} \ln(x^2 + y^2) - {(x-a)^2 + y^2\over 2}$$

Setting the x and y derivatives of the potential to zero, you find two critical points (not counting the singular behavior at x=y=0). The two points both have a singular Jacobian, so they give very large magnifications, and smears or arcs.

The two images occur at $$y=0$$,

$$x= {a\over 2} \pm \sqrt{A^2 - {a\over 2}^2}$$

So the smear to the side where the object is at is moved further, at large values of a, the second image is right on top of the lesing mass, and at small values of a, the two images are moved in the direction of the displacement by half the amount of displacement.

Quadrupole mass distribution

Consider two masses of size {1\over 2} at position $\pm a$. This gives a potential which is a superposition of the two masses:

$$ \phi(x,y) = {1\over 4} \ln((x-a)^2 + y^2 ) + {1\over 4} \ln((x+a)^2 + y^2) = {1\over 2} \ln(r^2) + a^2 { x^2 - y^2\over 2r^2}$$

The part in addition to the ordinary $M\ln(r)$ potential of a point source is a quadrupole. Lensing in a quadrupole has a simple algebraic solution. Differentiating, and subtracting the linear part gives

$$ {Ax\over r^2} ( 1 + {a^2\over r^4}(6y^2 - 2x^2) - {r^2\over A} ) =0$$ $$ {Ay\over r^2} ( 1 + {a^2\over r^4}(2y^2 - 6x^2) - {r^2\over A} ) =0$$

The x=0,y=0 point is at the singular position. The real critical points are at the other simultaneous solutions:

$$ x=0, y= \pm\sqrt{A}\sqrt{{1\over 2} \pm \sqrt{{1\over 4} + {2a^2\over A}}}$$ $$ y=0, x= \pm\sqrt{A}\sqrt{{1\over 2} \pm \sqrt{{1\over 4} - {2a^2\over A}}}$$

Of these eight points, two are imaginary (taking the minus sign inside the square root for y), and two are outside the domain of validity of the solution (taking the minus sign inside the square root for x--- the point is $\sqrt{2}a$), which is right by the point masses making the quadrupole). This leaves four points. But they are all local maxima, none of these are saddles. The saddles are found by solving the nontrivial equations in parentheses for x and y.

Taking the difference of the two equations reveals that $x=\pm y$, which gives the four saddle solutions:

$$ \pm x=\pm y = \sqrt{A}$$

There are eight images for a near-center source lensed by a quadrupole mass. For small values of a, The two images along the line of the two masses are pulled nearer by a fractional change which is $a^2\over A$, the two images perpendicular to the line of the two masses are pulled apart by a fractional change of $a^2\over A$, while the four images on the diagonals are at the location of the point-source disk.

For me, this was surprising, but it is obvious in hindsight. The quadrupole field and the Newton-Hooke fields both point along the lines y=x on the diagonal, and their goes from inpointing near the origin to outpointing far away, so it must have a zero. The zeros are topological and stable to small deformations, so if you believe that the field of the galaxy is spherical plus quadrupole, the Einstein cross light source has to be far enough off-center to change the topology of the critical points.

Quadrupole mass distribution/off-center source

To analyze moving off center qualitatively, it helps to understand how saddles and sources respond to movement. If you move the light source, you move the Newton-Hooke center. The result is that the points that were previously sources and saddles now have a nonzero vector value.

When the position of a source slowly gets a nonzero vector value, that means that the source is moving in the direction opposite this value. If a saddle gets a nonzero value, the saddle is moving in the direction of this value reflected in the attracting axis of the saddle.

This means that if you start with a very asymmetric quadrupole, and you slide the source along the long-axis of the source-saddle-source-saddle-source-saddle-source-saddle ellipse towards one of the sources at the end of the long axis, one of the short axis sources and the short axis saddles approach each other. They annihilate when they touch, and they touch at a finite displacement, since the result must smoothly approach the spherically symmetric solution.

Right after the sources and saddles annihilate, you get a cross, but it doesn't look too much like Einstein's cross--- the surviving two saddles and two sources are more asymmetric, and the narrow arm is much narrower than the wide arm.

Line source

For the lensing from a line source, you write the 2-d potential for a line oriented along the y-axis (it's the same as a plane source in 3d, a point source in 1d, or a d-hyperplane source in d+1 dimensions--- a constant field pointing toward the object on either side):

$$ \phi(x) = B|x|$$

And subtract off the Newton-Hooke source part, with a center at $x=a$.

$$\phi'(x) = B|x| - {1\over 2} ((x-a)^2 + y^2)$$

The critical points are on the y-axis by symmetry, and they are very simple to find:

$$ y=0, x= B+a$$ $$ y=0, x= -B+a$$

These are the two images from a long dark-matter filament, or any other linear extended source. Cosmic strings give the same sort of lensing, but the string-model of cosmic strings give ultra-relativistic sources which produce a conical deficit angle, and are technically not covered by the formalism here. But the result is the same--- doubled images.

If you spread out the line source so that it is a uniform density between two lines parallel to the y-axis (this would come from squashing a square beam of uniform mass density into a plane), the lensing outside the two lines is unaffected, by the 2-d Gauss's Law. The interior is no longer singular, and you get a third imag

answered Sep 13, 2011 by Ron Maimon (7,730 points) [ no revision ]
That is the most detailed answer I've ever seen on stackexchange :-)

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user BarsMonster
Thanks! I finally used something from Differential Equations!!!

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Dale
Tiny pic has removed your image =( booo!

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Dale
@JoeHobbit: Unfortunately, I don't have a copy--- it's described in words--- draw two saddles and two sources and link them to the origin and infinity in the unique way with flow-lines.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Ron Maimon
For future readers: the two saddles were on opposite sides of the origin and corresponding to two of the 4 light points making up Einstein's cross. The other two light points were at the sources.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Dale
+ 1 like - 0 dislike

The most important factor in creating these sorts of distributions is the non spherically symmetric aspects of the galaxy, creating a very warped lens. Not only is the visible part of a galaxy usually disc-like in nature, the majority of the mass is located in dark-matter halos located around the galaxy. Observational evidence suggests that these halos are "flat," in that they are oblong and not spherical, creating a somewhat counter intuitive lens shape.

The limiting case of this is the idea of a "cosmic string," a theoretical 1-d topological defect in space-time which is essentially a long dense string. There was quite a bit of news in astronomy/astrophysics circles about the so called Twin Quasar which was thought to be evidence of gravitational lensing by such an object (essentially, an extremely mirror symmetric lens), although since then it has been showed otherwise.

Look at the simulation here: http://www-ra.phys.utas.edu.au/~jlovell/simlens/

Here is another article with some examples : http://www.aeos.ulg.ac.be/lens_en.php

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Benjamin Horowitz
answered Aug 28, 2011 by Benjamin Horowitz (195 points) [ no revision ]
Given that galaxies are plates (and clusters I thought could be plate-like too), this would seem to explain a dual image very well. While your references show an example of a 4-image, it's not clear what shape produced that.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Alan Rominger
This is not the explanation--- every spherical symmetric source produces arcs and smears, never points. The failure here is spherical symmetry--- the mass distribution is extended along the long direction.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Ron Maimon
Ah, my wording was imprecise, "multiple images" is a confusing term. The main point of the answer was the second sentence. I have edited my answer to be more clear

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Benjamin Horowitz
Your wording was not imprecise, it was completely precise, it was just wrong. Your answer is correct now, after your edits. But you still didn't explain the saddle point images to the right and to the left.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Ron Maimon
Two crests are definitely considered "multiple images". You really don't need to be quite so hostile to mis-interpretations of words.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Benjamin Horowitz
I am not hostile, just honest. I thought the same before I worked it out. Your original answer, minus links: "Rings only form in perfect alignment. If the alignment is not precise, multiple images can form. Also you are assuming it is a point mass, when actually it is an extended galaxy, creating edge effects." This is not true-- arcs form in any spherically symmetric distribution, extended or point. the assumption that fails is spherical symmetry, not perfect alignment. The "edge effects" you claim are not edge effects, they are bulk effects--- every far source is lensed to a cross.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Ron Maimon
+ 1 like - 0 dislike

We don't know enough about this galaxy acting as a "lens" to tell for sure. It could be any number of things.

It could be a ring, formed by the galactic core acting as a lens, but cut in 4 chunks by the spiral arms. It could be caused by non-uniform distribution of mass in the galaxy. There might be other causes as well.

Some simulations:

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Florin Andrei
answered Sep 13, 2011 by Florin Andrei (10 points) [ no revision ]
It cannot be a ring cut in chunks--- this requires a massive conspiracy to make the four pointlike quasar-copies. Rings require perfect alignment, and work with sources with width, where a part of the source is aligned. Quasars are point sources.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user Ron Maimon
+ 1 like - 0 dislike

This thread came up some years ago, I saw it referenced and would like to posit an idea. My belief, backed up with math here, is that this cannot be lens. Lensing always results in rings, sometimes faint, sometimes not. There would be some evidence of a ring in the Hubble image, which is quite deep and quite fully resolved. There is none. So what could this be?

Arp's idea of ejection of quasars from AGN is very interesting. He only considered pairwise ejection - obviously in opposite directions. Presumably some physics magic in the dense matter of an AGN results in a bipolar instability, and the lobes disconnect and run off at some considerable velocity.

If this were admitted as possible, then it would also seem possible to develop a quadrupolar instability with the lobes at the vertices of a tetrahedron. Thus the object here could be an example of tetrahedral ejection. By considering, say, an interactive Java model of methane, one can rotate this model to exactly match the configuration of objects in the Cross. This is either a refection of reality, or a fantastic coincidence.

For what it's worth, I did publish an extension of GR that allows for new physics in dense matter conditions;


Downloadable copy here



This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user D R Lunsford
answered Mar 8, 2014 by D R Lunsford (10 points) [ no revision ]
+ 0 like - 3 dislike

Indeed dear friend, this form is the result of a phenomenon called gravitational lensing. Fortunately, between the earth and a quasar located at 8,000 million light years away is a galaxy 400 million light years. The gravity of the galaxy acts as a huge but imperfect lens heading in different paths from the quasar light, which is like a point, so there are four images that are around the galaxy. In this case the gravitational lensing effect produces a symmetrical cross because the lensing galaxy is almost exactly in our line of sight of the quasar. This cross is named after Albert Einstein, whose theory of relativity predicted this phenomenon.

This post imported from StackExchange Physics at 2019-08-25 16:21 (UTC), posted by SE-user jormansandoval
answered Aug 28, 2011 by jormansandoval (-10 points) [ no revision ]
It seems like we have ample answers and references offering that "it is a cross because of an imperfect lens," but the question was asked already possessing this knowledge.

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

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