empleat said:
Someone can explain this to me ?
The following visualisation works for me. It may work for you, or not.
Visualise by letting the extra dimension be time, and considering the appearance of the object over a period of time.
An example of doing this that is simpler than a tesseract is a solid hypersphere of radius ##R##. If time is the fourth dimension, the object will appear to we 3D beings as:
- first there is nothing
- suddenly a point appears. Let us label the time at which it appears as time 0
- the point grows as a solid sphere until it reaches a maximum of radius ##R## at time ##R##, then
- it shrinks until it is a single point at time ##2R##, then
- there is nothing from then on
The solid sphere's radius at time ##t## with ##0\le t\le 2R## is ##\sqrt{R^2 - (t-R)^2}##.
Now we'll work up to considering the tesseract.
First, consider how an empty square of side-length L looks to an inhabitant of a 1D world. The world will be a ring, in order to allow the inhabitant to inspect both sides of any solid object and thereby determine its length (since it can only see in 1D, just as we only see in 2D). The 2nd dimension of the square is time for the line-dweller. What they will see over time is:
- first there is nothing
- a line of length L appears for an instant, at time 0
- the line instantly disappears and is replaced by two points where the ends of the line were
- the two points remain in place for time L, then
- a line of length L appears again for an instant, then
- there is nothing from then on
The 'space inside' the square is the space between the two dots, that is there for time L.
Next, consider how a flatlander (2D dweller) sees an empty cube of length L, with time being the 3rd dimension.
- first there is nothing
- a solid square of edge length L appears for an instant, at time 0
- the solid square instantly disappears and is replaced by an empty square in the same place (ie the inside of the square vanishes)
- the empty square remains in place for time L, then
- a solid square of edge length L appears again for an instant, then
- there is nothing from then on
The 'space inside' the cube is the space inside the empty square, that is there for time L.
With that buildup, you can probably guess what a tesseract looks like to us 3D creatures.
- first there is nothing
- a solid cube of edge length L appears for an instant, at time 0
- the solid cube instantly disappears and is replaced by an empty cube in the same place (ie the inside of the cube vanishes)
- the empty cube remains in place for time L, then
- a solid cube of edge length L appears again for an instant, then
- there is nothing from then on
The 'space inside' the tesseract is the space inside the empty cube, that is there for time L.
I hope that helps.
PS:
empleat said:
But i can see space in 2d representation of a cube.
Actually, we
can't see the space inside an empty cube because we only see in 2D. All we can see is its outside. We need to perform some sort of experiment to discover the space, such as insert a probe, cut it open or weigh it. Similarly, if confronted by a tesseract, we'd need to do that sort of thing to distinguish between the solid and the empty cube. We'd need to be very quick with the solid cube, since it's only there for zero seconds. But we could cheat and use a 'thick-walled empty tesseract', for which the walls of the empty-cube slices are k distance units thick, and the beginning and ending appearances of a solid cube last for k time units.
Even a 4D creature could only see in 3D and would need to perform experiments to determine whether the tesseract they saw was empty - like the one we are discussing - or solid.
BTW, for us, a solid tesseract would appear in the same way as an empty one, except that the cube slices would always be solid, rather than empty except at the beginning and end.