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> You could imagine, say, a 3D grid or lattice that you can shine a light through onto a piece of paper to get an irregular 2D pattern, which would be your quas

> You could imagine, say, a 3D grid or lattice that you can shine a light through onto a piece of paper to get an irregular 2D pattern, which would be your quasicrystal. This is where I lose it. I actually can't imagine such a thing. Every regular 3D crystal I imagine has a repeating pattern in its shadow. For every ray of light passing through one part of the 3D lattice, I can locate parallel rays that produce the same result in other parts of the lattice. What am I missing here? Just not imagining the right lattice types? Or are we assuming a point-source of light so that no 2 rays are parallel?

mensetmanusman 4y ago
If you angle the 3D lattice at an irrational angle relative to the 2D plane the points will not be perfectly periodic in the plane.
- boppo1 4y ago
  What is an irrational angle? Is this something I can actually do physically, or is it more of a theoretical math thing? For example, if I'm holding a toy that is a lattice showing the 3d structure of carbon between my dining room table and ceiling lamp, how do I rotate it such that it is irrational relative to my table?
  - McBeige 4y ago
    I'm guessing it's irrational as in rational vs irrational numbers. Rational means a fraction of whole numbers, so irrational numbers are those which cannot be represented as such a fraction. A 1/4 turn is rational, a 1/pi turn is irrational. I feel like the light has to be parallel for it to work, so sunlight is a better example than a table lamp. Although I can't imagine any rotation of a simple 3D lattice having a nonrepeating shadow. Perhaps a more complex 3D crystal is necessary?
    - mcswell 4y ago
      Rational/ irrational here depends on the unit of measurement. A full circle (360 degrees) is rational if you measure it in degrees, but irrational if you measure it in radians (it's 2 pi radians).
      - anamexis 4y ago
        By that logic, couldn't any number be deemed rational by just declaring it as a unit?
        function_seven 4y ago
        It just occurred to me that when they’re talking about a “rational angle” they might be talking about a slope that can be specified rationally. If you have a repeating three-dimensional crystal lattice, and a ray of light that is following an irrational slope, then it is guaranteed to intersect one of the cell units or vertices in that lattice. If it did not intersect any nodes, then you would be able to express that slope rationally just by counting the number of vertical nodes over the number of horizontal nodes! I’m assuming an infinite lattice here. For finite ones an irrational slope could still “sneak through”
  - tiler2915072 4y ago
    this might help a little: https://math.stackexchange.com/questions/791848/penrose-tilings-as-a-cross-section-of-a-5-dimensional-regular-tiling https://math.stackexchange.com/questions/791848/penrose-tili... which is just a watered down explanation of the excellent https://tilings.math.uni-bielefeld.de/substitution/fibonacci/ https://tilings.math.uni-bielefeld.de/substitution/fibonacci...
  - rocqua 4y ago
    You need an irrational 'slope'. Like sqrt(2), or pi
cercatrova 4y ago
Look at this image: https://i.pinimg.com/736x/54/1b/1a/541b1afd4a72564f808825b3eecc742e--mughal-architecture-work-lights.jpg https://i.pinimg.com/736x/54/1b/1a/541b1afd4a72564f808825b3e... The window is the lattice, which is regularly ordered. The shadow, however, is distorted, ie each light beam is not the same size as the one next to it.
- abeppu 4y ago
  ... but that window is a 2D lattice, with a 2D shadow. > For every quasicrystal, there's a higher-dimension crystal that it is a shadow of. So what's the 3d crystal whose shadow is the Penrose tiling? The article says it's a "projected slice of a 5D lattice", which I really struggle to visualize. Or perhaps easier, what's the regular 2D pattern of which the Fibonacci sequence is a projection?
  - mcswell 4y ago
    You're struggling to visualize this? "You're just not thinking fourth dimensionally." --Doc Emmet Brown (https://www.youtube.com/watch?v=CUcNM7OsdsY https://www.youtube.com/watch?v=CUcNM7OsdsY)
  - cercatrova 4y ago
    The window is not 2D, it's 3D with depth as well. Now imagine that but with even more detail, like a cube lattice.
  - NamTaf 4y ago
    The system producing the shadow isn't a 2d lattice, because it also involves the sun. Changing the location of the sun relative to the window will change the pattern of the shadow. If you wanted to really convince yourself, add multiple windows such that the shadow is affected by the 'swiss cheese' effect of the holes of all the windows lining up relative to the sun.
  - blix 4y ago
    A projection of the 2D square lattice can produce the binary fibbonacci sequence. This picture I stole from some blog shows this projection: https://grahamshawcross.files.wordpress.com/2012/04/spots.jpg?w=900 https://grahamshawcross.files.wordpress.com/2012/04/spots.jp... -- A 2D square lattice is defined by two perpendicular basis vectors denoting nearest-neighbor distances. Lets put a grid point at the origin at define position as (a,b), where a and b are in units of the basis vectors. That is every integer (a,b) is a grid point that is a hops to the right and b hops up from the origin. Lets also define directions [a,b], where this is the vector from the origin (0,0) to point (a,b). Consider the following operation: we draw an arbitrary line on the lattice then take all the points within some distance of that line and project them onto the line. If you draw the line parallel to either basis vector, i.e. [1,0] or [0,1], you will get a 1D sequence where every point is identical: a 1D grid. This is actually independent of the size of the neighborhood around the line we consider as long as it is large enough to include any points; the projections of more distant points align with closer points due to the symmetry of the lattice. A line at 45 degrees (i.e. [1,1]) produces a similar result. What about some other integer vector? If we draw a line along [5,7], the projected points will no longer be as tidy, but after some distance along the line, we will reach a point equivalent to where we stared: the grid point (5,7). And then (10,14) and (15,21) so on. Every time we hit a new grid point, the pattern of the projection will repeat. The specific pattern between grid points may vary as we change the size of the neighborhood around the line we consider for the projection, but it will always retain the same periodicity. Like the previous cases, as we increase neighborhood size, the projection will stop changing after some critical value as all new points in the neighborhood will line up with previous points. You can see this properties by playing with a piece of graph paper. All rational vectors have parallel integer vectors, and so will have some underlying periodicity. What about an irrational vector, say [e,pi]? After leaving the origin travelling on this path your will ~never~ hit another grid point. Therefore the projections of grid points will be aperiodic. Not only that, as we increase the neighborhood size, the pattern constantly changes: each new point always has a new spot on the projection. However despite being aperiodic, the system is still ordered: you can know where the next point will show up every time. It turns out, if you draw a line along the vector [phi,1], where phi is the golden ratio, and use a neighborhood size of sqrt(2), the projection of the points has another interesting property: there are only two possible distances between projected points on the line, lets call them long (L) and short (S), themselves related by the golden ratio. The pattern of Ls and Ss is itself ordered and aperiodic. Not only that, but a simple substituion relation L->LS and S->L produces a longer sequence that includes the original. This just happens to be the substitution relation for finding subsequent terms in the binary fibbonaci sequence. --- I am sorry for this long-winded explanaton, but I am hoping it helps with visualization without pictures. Perhaps I should write a short blog post with some pictures. The key points are: starting from the origin a rational vector on a grid will always hit another grid point (and then infinite additonal points), which will define periodic relationships will all other grid points to the line. An irrational vector will never hit another grid point, so the relationships are always aperiodic. For a specific choice of projection and vector, you can recover the binary fibbonacci sequence.
vez- 4y ago
This video gives a very concrete example of a non-repeating pattern (Penrose tiling) https://www.youtube.com/watch?v=48sCx-wBs34 https://www.youtube.com/watch?v=48sCx-wBs34
- function_seven 4y ago
  I understand the non-repeating patterns. I just don't see how a regular 3D lattice can produce such a pattern. Unless the light source creating this shadow is a point-source rather than a parallel one? I guess I'm just looking for confirmation on this thought: A parallel light shone through a repeating 3D lattice will always produce a repeating 2D lattice.
  - yazanobeidi 4y ago
    Try projecting it on a surface with curvature. The projected grid spacing will be irregular and follow the curvature.
rocqua 4y ago
My guess is that it has to do with projections at an 'irrational' slope. That would prevent repetition, though I believe it would cause a dense set of points if you project the infinite lattice to a lower dimension.
dataflow 4y ago
I had the same confusion as you, but I'm going to take a guess that it might be analogous to the following: The function n -> sin(n) might be called a "shadow" of t -> sin(t), where k is an integer and t is a real number: namely, it's not periodic, but it's a shadow (projection from reals to integers) of something periodic. Maybe someone can confirm if the analogy is correct here?

mensetmanusman 4y ago

If you angle the 3D lattice at an irrational angle relative to the 2D plane the points will not be perfectly periodic in the plane.

boppo1 4y ago
What is an irrational angle? Is this something I can actually do physically, or is it more of a theoretical math thing? For example, if I'm holding a toy that is a lattice showing the 3d structure of carbon between my dining room table and ceiling lamp, how do I rotate it such that it is irrational relative to my table?
- McBeige 4y ago
  I'm guessing it's irrational as in rational vs irrational numbers. Rational means a fraction of whole numbers, so irrational numbers are those which cannot be represented as such a fraction. A 1/4 turn is rational, a 1/pi turn is irrational. I feel like the light has to be parallel for it to work, so sunlight is a better example than a table lamp. Although I can't imagine any rotation of a simple 3D lattice having a nonrepeating shadow. Perhaps a more complex 3D crystal is necessary?
  - mcswell 4y ago
    Rational/ irrational here depends on the unit of measurement. A full circle (360 degrees) is rational if you measure it in degrees, but irrational if you measure it in radians (it's 2 pi radians).
    - anamexis 4y ago
      By that logic, couldn't any number be deemed rational by just declaring it as a unit?
      - function_seven 4y ago
        It just occurred to me that when they’re talking about a “rational angle” they might be talking about a slope that can be specified rationally. If you have a repeating three-dimensional crystal lattice, and a ray of light that is following an irrational slope, then it is guaranteed to intersect one of the cell units or vertices in that lattice. If it did not intersect any nodes, then you would be able to express that slope rationally just by counting the number of vertical nodes over the number of horizontal nodes! I’m assuming an infinite lattice here. For finite ones an irrational slope could still “sneak through”
- tiler2915072 4y ago
  this might help a little: https://math.stackexchange.com/questions/791848/penrose-tilings-as-a-cross-section-of-a-5-dimensional-regular-tiling https://math.stackexchange.com/questions/791848/penrose-tili... which is just a watered down explanation of the excellent https://tilings.math.uni-bielefeld.de/substitution/fibonacci/ https://tilings.math.uni-bielefeld.de/substitution/fibonacci...
- rocqua 4y ago
  You need an irrational 'slope'. Like sqrt(2), or pi

cercatrova 4y ago

Look at this image: https://i.pinimg.com/736x/54/1b/1a/541b1afd4a72564f808825b3eecc742e--mughal-architecture-work-lights.jpg https://i.pinimg.com/736x/54/1b/1a/541b1afd4a72564f808825b3e... The window is the lattice, which is regularly ordered. The shadow, however, is distorted, ie each light beam is not the same size as the one next to it.

abeppu 4y ago
... but that window is a 2D lattice, with a 2D shadow. > For every quasicrystal, there's a higher-dimension crystal that it is a shadow of. So what's the 3d crystal whose shadow is the Penrose tiling? The article says it's a "projected slice of a 5D lattice", which I really struggle to visualize. Or perhaps easier, what's the regular 2D pattern of which the Fibonacci sequence is a projection?
- mcswell 4y ago
  You're struggling to visualize this? "You're just not thinking fourth dimensionally." --Doc Emmet Brown (https://www.youtube.com/watch?v=CUcNM7OsdsY https://www.youtube.com/watch?v=CUcNM7OsdsY)
- cercatrova 4y ago
  The window is not 2D, it's 3D with depth as well. Now imagine that but with even more detail, like a cube lattice.
- NamTaf 4y ago
  The system producing the shadow isn't a 2d lattice, because it also involves the sun. Changing the location of the sun relative to the window will change the pattern of the shadow. If you wanted to really convince yourself, add multiple windows such that the shadow is affected by the 'swiss cheese' effect of the holes of all the windows lining up relative to the sun.
- blix 4y ago
  A projection of the 2D square lattice can produce the binary fibbonacci sequence. This picture I stole from some blog shows this projection: https://grahamshawcross.files.wordpress.com/2012/04/spots.jpg?w=900 https://grahamshawcross.files.wordpress.com/2012/04/spots.jp... -- A 2D square lattice is defined by two perpendicular basis vectors denoting nearest-neighbor distances. Lets put a grid point at the origin at define position as (a,b), where a and b are in units of the basis vectors. That is every integer (a,b) is a grid point that is a hops to the right and b hops up from the origin. Lets also define directions [a,b], where this is the vector from the origin (0,0) to point (a,b). Consider the following operation: we draw an arbitrary line on the lattice then take all the points within some distance of that line and project them onto the line. If you draw the line parallel to either basis vector, i.e. [1,0] or [0,1], you will get a 1D sequence where every point is identical: a 1D grid. This is actually independent of the size of the neighborhood around the line we consider as long as it is large enough to include any points; the projections of more distant points align with closer points due to the symmetry of the lattice. A line at 45 degrees (i.e. [1,1]) produces a similar result. What about some other integer vector? If we draw a line along [5,7], the projected points will no longer be as tidy, but after some distance along the line, we will reach a point equivalent to where we stared: the grid point (5,7). And then (10,14) and (15,21) so on. Every time we hit a new grid point, the pattern of the projection will repeat. The specific pattern between grid points may vary as we change the size of the neighborhood around the line we consider for the projection, but it will always retain the same periodicity. Like the previous cases, as we increase neighborhood size, the projection will stop changing after some critical value as all new points in the neighborhood will line up with previous points. You can see this properties by playing with a piece of graph paper. All rational vectors have parallel integer vectors, and so will have some underlying periodicity. What about an irrational vector, say [e,pi]? After leaving the origin travelling on this path your will ~never~ hit another grid point. Therefore the projections of grid points will be aperiodic. Not only that, as we increase the neighborhood size, the pattern constantly changes: each new point always has a new spot on the projection. However despite being aperiodic, the system is still ordered: you can know where the next point will show up every time. It turns out, if you draw a line along the vector [phi,1], where phi is the golden ratio, and use a neighborhood size of sqrt(2), the projection of the points has another interesting property: there are only two possible distances between projected points on the line, lets call them long (L) and short (S), themselves related by the golden ratio. The pattern of Ls and Ss is itself ordered and aperiodic. Not only that, but a simple substituion relation L->LS and S->L produces a longer sequence that includes the original. This just happens to be the substitution relation for finding subsequent terms in the binary fibbonaci sequence. --- I am sorry for this long-winded explanaton, but I am hoping it helps with visualization without pictures. Perhaps I should write a short blog post with some pictures. The key points are: starting from the origin a rational vector on a grid will always hit another grid point (and then infinite additonal points), which will define periodic relationships will all other grid points to the line. An irrational vector will never hit another grid point, so the relationships are always aperiodic. For a specific choice of projection and vector, you can recover the binary fibbonacci sequence.

vez- 4y ago

This video gives a very concrete example of a non-repeating pattern (Penrose tiling) https://www.youtube.com/watch?v=48sCx-wBs34 https://www.youtube.com/watch?v=48sCx-wBs34

function_seven 4y ago
I understand the non-repeating patterns. I just don't see how a regular 3D lattice can produce such a pattern. Unless the light source creating this shadow is a point-source rather than a parallel one? I guess I'm just looking for confirmation on this thought: A parallel light shone through a repeating 3D lattice will always produce a repeating 2D lattice.
- yazanobeidi 4y ago
  Try projecting it on a surface with curvature. The projected grid spacing will be irregular and follow the curvature.

rocqua 4y ago

My guess is that it has to do with projections at an 'irrational' slope. That would prevent repetition, though I believe it would cause a dense set of points if you project the infinite lattice to a lower dimension.

dataflow 4y ago

I had the same confusion as you, but I'm going to take a guess that it might be analogous to the following: The function n -> sin(n) might be called a "shadow" of t -> sin(t), where k is an integer and t is a real number: namely, it's not periodic, but it's a shadow (projection from reals to integers) of something periodic. Maybe someone can confirm if the analogy is correct here?