# Confused about fourth spatial dimension

pervect
Staff Emeritus
hellfire said:
Consider a spacetime, with a specific metric, sliced in spatial hypersurfaces. Take an observer located at a point of slide S0. Are you telling me that you can perform a coordinate transformation such that the new time direction is contained in S0 (or has a vanishing normal component to S0)?
No, I don't think that that is possible. What is possible is that the new time direction will include a non-vanishing component of S0 as well as the normal component.

You will wind up with a new coordinate system with a different notion of simultaniety, and different space-like hypersurfaces S'0. In other words, at the same point in the two coordinate systems, the time vector T0 which is the normal vector to S0 will point in a different direction than the time vector T'0 which is normal to S'0.

DrGreg said:
You can't mix them in an arbitrary way, because "spacelike" and "timelike" are different. But you can mix them via a Lorenz transform, which preserves the properties of "spacelike" and "timelike".
That varies is the value of time and space, precisely because ds may be invariant, which implies dt and dx may be not. But in the new system of coordinates after the LT, both space and time continue to be two different concepts. time is 'time' and space is 'space'. The metric is +2 (or -2) again with time diferent of space.

Again in a 3 + 1 descomposition -in both frames conected by the LT- one chooses time like the different parameter of the others three.

If time was equivalent to space one would study, for example, 2+2 decompositions of spacetime, but this is useless.

If the difficult and fundamental problem of time t (there is no equivalent problem with space, which implies that time is not just another spatial dimension) is present in Hamiltonian quantum gravity in one frame, the application of a LT to other frame does not eliminate the problem since again the difficult and fundamental problem of time t' (no of space) is present in the other frame. Of course t =/= t'

As already said if time was a spatial dimension Hawking did not need introduce his imaginary time concept by quantum motives. As already said when he introduces imaginary time, time is "geometrised" in the metric doing it Euclidean, but in the rest of physics -e.g. in the Schrödinger equation- time is transformed into an imaginary quantity whereas space continue being real, doing, again, time looking very diferent of space.

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new story about fred and special guest barney from fourth dimension

well lot of guys like juan,hellfire should better start their own separate thread about fourth dimension time.but guys i m talkin about fourth spatial dimension and so u better pay attention to this. :grumpy:
today we have a special guest barney from fourth dimension.
in my last post i proved that circumferences in for fed and us i.e 2d and 3d is different but the area is the same i.e 2(PI)*r^2
i assume that u have read my other posts on this topic.
Now for us and hyperbeing barney
now same thing
r^2=x^2+H^2--------------------eqn(2)
assuming H=x we get,
r=x*sqrt(2)-----------------------eqn(3)
for us 3d people
area=2(PI)r^2 here put value of r from eqn(3)
and we get
area=4(PI)*x^2
can we say for barney also area is the same and what about volume

sandesh trivedi said:
lets consider a sphere and let there be a two dimensional man and for him he is living on a two dimensional place.we r observing it from third dimension.
now this two dimensional man be fred.
one day fred decides to make a circle using a rope.now start imagining he is on a sphere and he tkaes a rope and he sticks one end of the rope on the surface and the other end on his hands. now he makes a big circle on this sphere and he considers the length of the rope as the radius and tries to find the circumference and he gets some value.remember he is on a sphere so he is getting wrong circumference if he uses length of the rope as the radius but now he some how using measuring tape or formula s=vt fin the real circumference and hence the real radius.since he is in two dimensional place he knows pythagoras theorem and he puts the value of the real radius and wrong radius in the formula
and third dimension = height=h
and R^2=r^2+h^2
voila he finds the height of the third dimension
correct me if i m wrong Ok, let me talk abut spatial dimensions also my friend fred decides to make a circle using a rope. But he does not obtain a wrong result for the radius like you claim because the radius for fred is different of radius for you (if you use like definition of radius that of a flat space which is not obligatory for you).

Still i do not understand like fred obtains the flat-space radius. Since he is in a curved 2D, his s=vt is consistent with radious that he had measured. In fact, he is measuring curved distances.

s(fred) =/= s(you)

v(fred) =/= v(you)

pervect
Staff Emeritus
sandesh trivedi said:
well lot of guys like juan,hellfire should better start their own separate thread about fourth dimension time.but guys i m talkin about fourth spatial dimension and so u better pay attention to this. :grumpy:
You might try the math forums, because your question apparently doesn't have anything to do with relativity at all (being instead about a fourth spatial dimension). Relativity only has three spatial dimensions, and a curved 4-dimensional space-time.

You seem to get grumpy when we keep introducing relativity into your non-relativistic question.

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thanks pervect

pervect said:
You might try the math forums, because your question apparently doesn't have anything to do with relativity at all (being instead about a fourth spatial dimension). Relativity only has three spatial dimensions, and a curved 4-dimensional space-time.

You seem to get grumpy when we keep introducing relativity into your non-relativistic question.
i was also thinking about posting this thread in maths section but i was confused where should i post it.can u please tell me where should i post it.

pervect said:
Let me clarify something. There's nothing technically wrong with the problem solution you posted, it's just that the problem and its solution doesn't have much relevance to General Relativity and how it deals with curvature.

GR deals with curvature not from the viewpoint of someone looking at our universe from outside, but from the viewpoint of someone looking at it from inside. This is known as intrinsic curvature. That's why I stressed that viewpoint, because it is the one that GR uses.

Imagine a 2d surface of constant curvature - you probably imagine the surface of a sphere.

But a sphere requires periodic space and time coordiantes. To use a technical term, it's a compact manifold.

Try to imagine wrapping a piece of paper around a sphere, a piece of paper that's infinite in both directions (a plane). You can't do it in three dimensions.

You can wrap a narrow strip of paper (elastic paper) that's infintely long but finitely wide around the sphere with no problem. But as you try to make the paper wider and wider, eventually you find that the paper has to pass through itself, something that you can't do. (Unless you add extra dimensions).

GR has to deal with non-compact manifolds all the time. You can't simply represent the curved 4-d space of GR as the surface of a 5 dimensional manifold. You need more than 5 dimensions to get the right sort of curvature, just as you need more than 3 dimensions to construct a plane that has a constant curvature everywhere.

This also makes the embedding non-unique. When you embed a n-dimensional maniofld in n+1 dimensions, you can solve the equaitons and find a unique solution, but when you embed a n-dimensional mainfold in n+2 or even higher dimensional space, you find the embedding is not unique.
I have few more question hope you ca answer me!
What is "manifold', what is the difference between manifold and dimensions?
What is non-compact manifolds?
Finally, be honest I am not quite understand the original question ~ once we use a paper to form a sphere, the paper is in 2-D while the shpere is 3-D~ is it means when "curvature" present the case no more within 2-D? If yes, how about the circle in on a plane (2-D) ?

yukyuk

εllipse said:
Our 3 spacial dimensions are Euclidean (not curved), except where there are strong gravitational fields.

No.
the second queston, why no?

pervect
Staff Emeritus
yukcream said:
I have few more question hope you ca answer me!
What is "manifold', what is the difference between manifold and dimensions?
What is non-compact manifolds?
Finally, be honest I am not quite understand the original question ~ once we use a paper to form a sphere, the paper is in 2-D while the shpere is 3-D~ is it means when "curvature" present the case no more within 2-D? If yes, how about the circle in on a plane (2-D) ?

yukyuk
You can think of a two dimensional manifold as composed of pieces of cloth (really finite two dimensional surfaces) that are "glued" or "sewn" together. The "sewing" process has some restrictions, the seams must be smooth.

This generalizes to three or more dimensions - an n dimensional manifold looks like small pieces of R^n "glued" together in a continuous manner.

http://mathworld.wolfram.com/Manifold.html

has a more formal definition of Manifolds.

"compact" is a bit harder to describe informally, but the surface of a sphere is compact (it has a finite area), while the surface of a plane is not compact (it has an infinite area). For a formal definition see

http://mathworld.wolfram.com/CompactManifold.html
http://mathworld.wolfram.com/CompactSpace.html

which includes some examples of 2-d compact manifolds (spheres, torii, klein bottles, etc.)

Any manifold has a dimension, which is the dimensionality of some small piece of the manifold - this is the same no matter which piece of the manifold one considers (this can be proven from the formal defintion).

A circle on a plane would be a 1 dimensional manifold. It would not have a curvature, because you need to have at least a 2-dimensional manifold to define curvature.

pervect said:
We get it. At least the part about the curvature, we get. it appears you may have some question? I don't understand what question you are asking, so far you have simply stated some facts, which are correct.

Fred has found how to measure "Gaussian" curvature, which is an intrinsic sort of curvature first defined/discoverd by (you guessed it) Carl Gauss.

See for instance the Wikipedia article at

http://en.wikipedia.org/wiki/Curvature
Where dose the equation comes from?
$$K = \lim_{r \rarrow 0} (2 \pi r - \mbox{C}(r)) \cdot \frac{3}{\pi r^3}.$$

To Pervect

Thanks so much~ your explanation is so clear~ you help me a lot thxs

yukyuk