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#1
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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?
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#2
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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.
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#3
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Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
W*X*Y*Z
W is the extra dimension in the Minkowski space. |
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#4
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Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
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?
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#5
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Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
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This web site claims to have some answers for you. http://www.physicsinsights.org/hypercubes_1.html |
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#6
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Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
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#7
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Re: Area enclosed by a tesseract (4th Dimensional Hypercube)
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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