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-   -   Area enclosed by a tesseract (4th Dimensional Hypercube) (http://www.chiefdelphi.com/forums/showthread.php?t=88118)

jmanela 27-12-2010 22:29

Area enclosed by a tesseract (4th Dimensional Hypercube)
 
I was reading this book the other day The Time Machine by H.G. Wells, and he described moving through dimensions. I pondered the thought a little bit, and began wondering. If the area of a square is L x W, and the volume of a cube is L x W x H, then what is the area enclosed by a tesseract or a fourth dimensional hypercube. Is it the amount of space it takes up within multiple dimensions? Maybe... What do you think?

DiscoKittyPrime 28-12-2010 01:09

Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
 
Well, if you take the idea from The Time Machine, then you would take time to be the fourth dimension. In which case, the volume occupied by the tesseract would be L x W x H x T with "T" being the amount of time occupied.

BornaE 28-12-2010 01:48

Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
 
W*X*Y*Z

W is the extra dimension in the Minkowski space.

RyanCahoon 28-12-2010 07:24

Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
 
Quote:

Originally Posted by BornaE (Post 988928)
W*X*Y*Z

W is the extra dimension in the Minkowski space.

I would agree this holds for Euclidean 4-space, but is this measure of hyper-volume valid for Minkowski space, since only three of the dimensions are Euclidean?

Quote:

Originally Posted by H. S. M. Coxeter, Regular Polytopes, via Wikipedia
Little, if anything, is gained by representing the fourth Euclidean dimension as time. In fact, this idea, so attractively developed by H. G. Wells in The Time Machine, has led such authors as J. W. Dunne (An Experiment with Time) into a serious misconception of the theory of Relativity. Minkowski's geometry of space-time is not Euclidean, and consequently has no connection with the present investigation.


gblake 28-12-2010 09:21

Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
 
Quote:

Originally Posted by jmanela (Post 988890)
If the area of a square is L x W, and the volume of a cube is L x W x H, then what is the area enclosed by a tesseract or a fourth dimensional hypercube.

It looked like you were going to ask for a formula for a hypercube's 4-dimensional hypervolume (WxXxYxZ), but then you asked about area. Which did you want to learn?

This web site claims to have some answers for you. http://www.physicsinsights.org/hypercubes_1.html

jmanela 28-12-2010 18:26

Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
 
Quote:

Originally Posted by gblake (Post 988949)
It looked like you were going to ask for a formula for a hypercube's 4-dimensional hypervolume (WxXxYxZ), but then you asked about area. Which did you want to learn?

This web site claims to have some answers for you. http://www.physicsinsights.org/hypercubes_1.html

thanks, that site answered my question. I was just confused about what the volume of the tesseract would be.

Karthik 28-12-2010 19:20

Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
 
Quote:

Originally Posted by RyanCahoon (Post 988941)
I would agree this holds for Euclidean 4-space, but is this measure of hyper-volume valid for Minkowski space, since only three of the dimensions are Euclidean?

It's very common for people to confuse the properties a Euclidean Space and a Minkowski Space. Many properties carry over from a Euclidean Space, but some crucial ones do not. When studying Minkowski Space it's critical that the student takes the time to understand the differences and why they exist. The biggest one of course being the differences in the definition of the space's inner product.


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