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Plotting projections of 4D objects into 3D

Bryan Lettner

Posted 9 years ago

I would like to begin experimenting with parametrically defined 4D objects: { w(r,s,t) , x(r,s,t) , y(r,s,t) , z(r,s,t) } What's the most straightforward way in Mathematica to visualize a projection of 4+ dimensional objects into 3D space? I would love to see a "ParametricPlot4D" or "ParametricPlotnD" function which can do this in a simple way. Anyone else?

POSTED BY: Bryan Lettner

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Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 9 years ago

Hi Bryan, a while ago I was playing around with the same idea. I found the 3D or 2D projection of a 4D lattice quite instructive. Here is my code; it should be self explaining: ClearAll["Global`*"] (rot4D = RotationMatrix[\[Theta], {{1, 0, 0, 0}, {1, 1, 1, 1}}]) // MatrixForm nmax = 1; points4D = Flatten[Table[{ {{x, y, z, t}, {x + 1, y, z, t}}, {{x, y, z, t}, {x, y + 1, z, t}}, {{x, y, z, t}, {x, y, z + 1, t}}, {{x, y, z, t}, {x, y, z, t + 1}}}, {x, -nmax, nmax}, {y, -nmax, nmax}, {z, -nmax, nmax}, {t, -nmax, nmax}], 5]; rotStruct4D[\[Theta]_] = {Red, Line /@ Partition[{#1, #2, #3} & @@@ (rot4D.# & /@ points4D), 2]}; rotStruct4Dx[\[Theta]_] = {Black, Line /@ Partition[{-2.5 nmax, 0, 0} + {0, 1, 1} # & /@ ({#1, #2, #3} & @@@ (rot4D.# & /@ points4D)), 2]}; rotStruct4Dy[\[Theta]_] = {Black, Line /@ Partition[{0, -2.5 nmax, 0} + {1, 0, 1} # & /@ ({#1, #2, #3} & @@@ (rot4D.# & /@ points4D)), 2]}; rotStruct4Dz[\[Theta]_] = {Black, Line /@ Partition[{0, 0, -2.5 nmax} + {1, 1, 0} # & /@ ({#1, #2, #3} & @@@ (rot4D.# & /@ points4D)), 2]}; Manipulate[Graphics3D[{rotStruct4D[\[Theta]], rotStruct4Dx[\[Theta]], rotStruct4Dy[\[Theta]], rotStruct4Dz[\[Theta]]}, ImageSize -> 600], {\[Theta], 0, 2 Pi}] This - of course! - is very simple and rotates the lattice only in a single hyperplane, but one can get a first impression. Regards -- Henrik

Hi Bryan,

a while ago I was playing around with the same idea. I found the 3D or 2D projection of a 4D lattice quite instructive. Here is my code; it should be self explaining:

ClearAll["Global`*"]
(rot4D = RotationMatrix[\[Theta], {{1, 0, 0, 0}, {1, 1, 1, 1}}]) // MatrixForm
nmax = 1;
points4D = Flatten[Table[{
     {{x, y, z, t}, {x + 1, y, z, t}},
     {{x, y, z, t}, {x, y + 1, z, t}},
     {{x, y, z, t}, {x, y, z + 1, t}},
     {{x, y, z, t}, {x, y, z, t + 1}}},
    {x, -nmax, nmax}, {y, -nmax, nmax}, {z, -nmax, nmax}, {t, -nmax, 
     nmax}], 5];
rotStruct4D[\[Theta]_] = {Red, 
   Line /@ Partition[{#1, #2, #3} & @@@ (rot4D.# & /@ points4D), 2]};
rotStruct4Dx[\[Theta]_] = {Black, Line /@        
    Partition[{-2.5 nmax, 0, 0} + {0, 1, 1} # & /@ ({#1, #2, #3} & @@@ (rot4D.# & /@ points4D)), 2]};
rotStruct4Dy[\[Theta]_] = {Black, Line /@        
    Partition[{0, -2.5 nmax, 0} + {1, 0, 1} # & /@ ({#1, #2, #3} & @@@ (rot4D.# & /@ points4D)), 2]};
rotStruct4Dz[\[Theta]_] = {Black, Line /@        
    Partition[{0, 0, -2.5 nmax} + {1, 1, 0} # & /@ ({#1, #2, #3} & @@@ (rot4D.# & /@ points4D)), 2]};
Manipulate[Graphics3D[{rotStruct4D[\[Theta]], rotStruct4Dx[\[Theta]], rotStruct4Dy[\[Theta]], rotStruct4Dz[\[Theta]]}, 
  ImageSize -> 600], {\[Theta], 0, 2 Pi}]

This - of course! - is very simple and rotates the lattice only in a single hyperplane, but one can get a first impression.

Regards -- Henrik

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