Visualizing intersecting multidimensional objects.

  • Thread starter Thread starter Keys
  • Start date Start date
  • Tags Tags
    Multidimensional
AI Thread Summary
The discussion explores the intersection of multidimensional planes, specifically how two 2D planes in 4D space that do not share a dimension intersect. It is posited that the intersection would manifest as a point when viewed from either plane, although some suggest there might be no intersection at all. The conversation delves into parametric representations of planes, indicating that in 4D, two planes yield a single point due to having four equations in four unknowns. The parametric form is confirmed to be applicable in higher dimensions, with the dimensionality of the intersection depending on the number of parameters used. Overall, the intersection of non-overlapping multidimensional spaces can be visualized through mathematical representations, affirming the relevance of parametric forms in higher-dimensional geometry.
Keys
Messages
15
Reaction score
0
If we look at 2 intersecting orthogonal planes in 3D, the intersection forms a line if you are "living" on either plane. How would the intersection look if there are 2D planes in 4D where the planes do not share a dimension? For example plane 1 exists on X and Y, and plane 2 exists on Z and T. I'm figuring it must be a point if viewed from either plane. Is this correct? The other answer might be there is no intersection, but I don't think this can be correct.
 
Mathematics news on Phys.org
This belongs in the math forum.

One way to approach it is to consider the representation of the plane in parametric form, such as: X=A+Bu+Cv, where u and v are parameters, while A, B, C are constant vectors and X is a point on the plane.

In 3 space, when you have 2 planes, you will have 3 equations in 4 unknowns (the (u,v) parameters for each plane). The solution is then a line. In 4 space, you have 4 equations in the same 4 unknowns, leading to a single point.
 
Thanks Mathman for your answer. I'm a bit dense, does this parametric form work even higher dimensions? Let's say the last example 2 planes in a 4 space, could be considered 2 different 2 spaces both within a 4 space. By my reasoning a 3 space and 2 space that are both within a 5 space where they don't share any common dimensions, the intersection would appear as a line in both the smaller spaces in question. With two 3 spaces within a 6 space which don't share common dimensions, the intersection in each 3 space would be a plane or a common 2 space (which is not comprised of any of the discrete dimensions which make up the 6 space). Thanks
 
The parametric form works in any number of dimensions. One parameter gives a curve (1 dim.), two parameters a surface (2 dim.), three parameters a solid (3 dim.), etc., where the number of dimensions of the underlying space is the same as the dimension of the vectors.

To get "flat things", the parameters appear as first powers only. Other objects will have higher powers, functions (such as sin, cos), and functions involving products of parameters, etc.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
8
Views
2K
Replies
5
Views
2K
Replies
3
Views
3K
Replies
13
Views
2K
Replies
7
Views
4K
Replies
4
Views
2K
Back
Top