$$x_1^2 + x_1i(x_2-x_4) + x_2x_4 = \fraccos(\phi),sin(\phi),sin(\phi)\biggr), \phi \in [0,2\pi[$$īut after a long while of calculation it ended up being wrong, because I've got nonsense. Now I'll include the above mentioned calculation: Note 2: As far as I understand, this is a circle that's been rotated 45°'s in the $x_3$, and 45°'s in the $x_4$ direction. When we intersect the sphere with a plane, we apply this definition again, so we must get a circle. Note: The intersection of a 4-dimensional sphere and a plane can only give you a 2-dimensional circle, since by definition a 4D sphere is the collection of points equal distance from the origin. The formula for its surface volume is 2(pi2)(r3). The formula for its interior volume is V (1/2)(pi2)(r4). An interior volume, which is 4 dimensional, and a surface volume which is 3 dimensional. The set of points in (n + 1)-space, (x 1, x 2. A 4 dimensional sphere has two 'volumes'. I'm including the calculation of this intersetion, just to give you a deeper understanding of the problem, however my question only extends to how to obtain the parametric equation of the above circle. a 3-sphere is a 3-dimensional sphere in 4-dimensional Euclidean space. is a plane in 4 dimensions, with the correspondance of To show that this IS in fact a circle, you can solve the following system of equations to obtain their intersection: I need its 4-dimensional parameterization, using a $\phi \in [0,2\pi[$ value. The following rotated circle is given in 4 dimensions: I'm attending a differential geometry course, and I'm stuck at one part of a question that we've been asked.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |