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What would a 3D ball look like (in 3D) if it were partially in 4D space?

  1. Sep 14, 2010 #1
    What would a 3D ball look like if it were partially in 4D space when none of the 4th dimension is visible to the 3D observer?

    The equivalent question in flatland is: What would a 2D filled circle look like in flatland if it was partially in 3D?

    I think that there the answer would be a line, but the direction and length of the line would vary depending upon how the filled circle was tilted in 3D. The endpoints of the line would always be somewhere on the circumference of the circle. A 3D tilt angle could be constructed for any two points on the circle's circumference.

    Dynamically, a filled 2D circle in 3D passing through a different 2D space would appear as a growing then shrinking line. The line would grow to be the diameter of the filled circle and begin and end as a single point. The rate of growth and shrinkage of the line would depend on both the angle between the planes and the relative velocity of the planes. Trying to think about this in terms of volumes makes my head hurt.

    All I got from Google was that I really want to talk about a ball rather than a sphere. (In flatland, for a circle we'd only be talking about the endpoints of the line in 2D rather than the entire line.)
     
  2. jcsd
  3. Sep 14, 2010 #2
    What do you mean by 'partially in 3D' (perhaps I should read flatland)? You can surely embed a 2 dimensional object into a 3 dimensional space, such as a sphere in 3-space.
     
  4. Sep 14, 2010 #3

    Office_Shredder

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    This is basically the same question as take a 2 dimensional slice out of the 3 dimensional ball, and see what you get. Try to think about why that would be true
     
  5. Sep 14, 2010 #4
    Consider a flatland 2D space with a filled circle and its intersection with a different flatland 2D space. Since we live in a 3D volume rather than a 2D plane, we can visualize such an interaction. Statically, the intersection forms a line when viewed from either of the different flatlands. This line cannot be seen in the flatland which contains the circle but can only be seen in the intersecting flatland. Dynamically, the intersection forms a line which begins and ends as a single point and is at its maximum the radius of the filled circle. Obviously it's also possible that the different flatlands might be disjoint, so that there is no intersection, or the different flatlands might be congruent so that the filled circle remains a circle. (The above flatland intersections are stated as fact but are not necessarily correct.)

    Now, instead of that flatland filled circle, similarly consider a ball within a 3D volume and its intersection with a different 3D volume in a 4 dimensional space. The 3D volumes are neither disjoint nor congruent so that the resultant might be more interesting. I expect that the interior of the ball becomes its exterior as the volumes intersect.

    Intuitively, since the intersection of two planes forms a line, it may be that the intersection of two volumes forms a plane so that the ball would only statically form a filled circle, and dynamically a filled circle growing from a point to a maximum filled circle back to a point. The largest filled circle would have the diameter of the ball. If this is the case, I have a follow up question.
     
  6. Sep 15, 2010 #5

    Office_Shredder

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    This is correct. And it extends to other dimensions as well: A popular "visualization" of the four dimensional ball is by imagining it traveling through our 3 dimensional space: It starts off as a point, expands into a ball of full diameter, then shrinks back to a point again
     
  7. Sep 15, 2010 #6
    If you mean projection rather than intersection, wouldn't it be an oval in flatland (and spheroid in 3d)?
     
  8. Sep 19, 2010 #7
    As the intersections described earlier happen in higher dimensionality space, there are several things that can happen: 1) The line remains a straight line, the flat filled circle remains a flat filled circle. 2) The line becomes a curved line, with more curves possible as dimensionality increases. The filled circle becomes the surface of a spheroid, with more bumpiness as dimensionality increases. 3) The intersection of two filled flatland circles becomes multiple straight lines, the intersection of two 3D balls becomes multiple filled circles. 4) The intersection of two filled flatland circles becomes an oval, the intersection of two 3D balls becomes a spheroid. 5) Something else.

    I presume this has already been calculated somewhere. Does anyone know the answer? On the surface of a sphere, two lines can intersect at two points rather than at just one but that doesn't help me to be sure I can visualize which it will be. Options 2 and 3 seem most likely.

    To cesiumfrog, wouldn't a projection result in a distorted circle and a distorted ball? Here I'm thinking that the term projection is used as the resulting shadow from an outside illumination, although it may have a more technical meaning in mathematics. The shadow would be thinner nearer the light source and thicker farther away from it.
     
  9. Oct 7, 2010 #8
    Well, what are the ways that Euclidean n-space and another copy of Euclidean n-space can intersect in Euclidean (n+1)-space?

    Well, it can either be the whole of the n-dimensional space, or an (n-1)-dimensional subspace of it (or no intersection). If we now sat a sphere in one of them, when we've rotated it, we could get the whole sphere back, or we could get a (n-1)- dimensional slice through it's centre, as that's the only subspace still visible from our original "starting" n-dimensional space. Any translation will move this slice around the sphere (or off it), but the two Euclidean spaces will still always intersect completely, not at all, or in an (n-1) dimensional subspace.

    So the sphere will appear either unchanged, as an n-1 dimensional slice of it, or it will disappear completely. Any n-1 dimensional slice of the sphere, as above can be obtained from moving the sphere somewhere.

    That's how I see it anyway! =D
     
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