Volume of four-dimensional objects.

There are two matters I would like to deal with this morning.

One is “Clifford Pickover`s question” about hyperspace and the other is 1820`s events in Turkey. The first job involves the elucidation of the relationship between dimensionality and volume and the second the impact of events that took place in Europe in the first half of 19th century and their determining effect in the fate of the nascent Greek state.

Since I am undecided, I will channel my thoughts on either of these questions, in confrontation, till either of these questions find a stronger hook to already existing thoughts, mental structures stored in my brain, schema already in place, drawing my thought resources on it and therefore dominate over the other.

I will go through the paragraphs in Clifford Pickover`s book relevant to the question. Let us go in a search of Hinton cubes. I found the Banubula blogspot where a definition of the four-dimensional object, tesseract, is thought provoking:

“A Tesseract is a four-dimensional object which completes the following analogy: Square (2-d) is to Cube (3-d) as Cube (3-d) is to Tesseract (4-d).”

So to take up the challenge, trying to visualize the relationship, the square is one of the 6, obviously flat, sides of the cube, and the cube is one the 6 sides of the tesseract. Each of the sides of the tesseract is a cube analogous to each side of a cube is a square. As the sides of the cube are two-dimensional, the sides of a tesseract are three-dimensional and this analogy can be carried over into the n+1 dimensionality. In the 5th dimension, to a “penteract” with four-dimensional sides and so on.

Deciding about the nature of the sides, it is followed by resolving the space bound by the sides; in cubes, tesseracts and so on. In a cube the six square sides bound space which can be measured in units of volume calculated by multiplying the surface of one side times the height of one of its adjacent sides. E.g. 1cm2 x 1 cm = 1 cm3. In a tesseract the space bound by the six cubes should, according to the rule, be calculated by multiplying its equivalent components, the cubes, in a similar analogy. And these are the volume of one of its sides times the surface of one of its adjacent sides. E.g. 1 cm3 x 1 cm2 = 1 cm5 . What is the significance of this calculation? What is the volume that is measured in cm5 units? Let carry on the calculation further for a “penteract”. This should be 1 cm5 x 1 cm3 = 1 cm8 and further 1 cm8 x 1 cm5 = 1 cm13. Can volume with units other than cm3 exist?

The classic tesseract design accommodate for the sides only but not for the space bound by the sides. The space that is enclosed by the six cube sides. The quantity of volume, quite carefully is defined, as in this website, as “the measure of space taken up by a three-dimensional object”. So the use of the word volume for four-dimensional objects and with cm5 units raises questions.

What is the connection of volume cm5 with volume cm3? Is it possible to convert cm5 to cm3?