Confused about fourth spatial dimension

  • #1
sandesh trivedi
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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
real radius =r
wrong radius=length of the rope=R
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
:cool:
 

Answers and Replies

  • #2
EnumaElish
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You are assuming your conclusion, is how it seems to me. If Fred is truly a 2-D creature, how does he have any idea of what the radius in 3-D would be? At best he can make a guess at how the curvature is distributed as a random variable, I think.
 
  • #3
selfAdjoint
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EnumaElish said:
You are assuming your conclusion, is how it seems to me. If Fred is truly a 2-D creature, how does he have any idea of what the radius in 3-D would be? At best he can make a guess at how the curvature is distributed as a random variable, I think.

You're wrong Enuma. Two-dimensional Fred can indeed conclude he lives on a sphere (or at least on a manifold that is curved locally) and compute its radius from geometrical measurements like this. This is what is meant by "instrinsic geometry", invented by Gauss for surfaces and extended by Riemann to manifolds of any dimension.
 
  • #4
EnumaElish
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Hmmm. What are our 3 dimensions curved into, then? Is there a measurable, spatial 4th dimension?
 
  • #5
εllipse
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EnumaElish said:
What are our 3 dimensions curved into, then?
Our 3 spatial dimensions are Euclidean (not curved), except where there are strong gravitational fields.

EnumaElish said:
Is there a measurable, spatial 4th dimension?
No.
 
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  • #6
sandesh trivedi
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you guys are not getting it right. what i m trying to say is that fred finds this bcoz of circumference. the circumference he is getting is not satisfying the length of the rope as the radius
 
  • #7
Juan R.
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εllipse said:
Our 3 spatial dimensions are Euclidean (not curved), except where there are strong gravitational fields.

As far like i know nobody has measured a curved spacetime or curved space predicted by general relativity still o:)

In fact, Einstein idea that gravity is curvature is easily shown to be wrong. Take a Schwarzild metric in the limit c--> infinite, curvature is exactly zero, (1 - 2 GM/Rcc) --> 1, but gravity is not because there is a Newtonian potential still.

There exists a well-known principle of epistemology saiyng if A is cause of B then the elimination of A eliminates B.

If there is gravity (B) without curvature (A) then curvature (A) cannot be the cause of gravity (B).
 
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  • #8
pervect
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sandesh trivedi said:
you guys are not getting it right. what i m trying to say is that fred finds this bcoz of circumference. the circumference he is getting is not satisfying the length of the rope as the radius

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
An intrinsic definition of the Gaussian curvature at a point P is the following: imagine an ant which is tied to P with a short thread of length r. She runs around P while the thread is completely stretched and measures the length C(r) of one complete trip around P. If the surface were flat, she would find C(r) = [itex]2\, \pi \, r [/itex] On curved surfaces, the formula for C(r) will be different, and the Gaussian curvature K at the point P can be computed as

[tex]
K = \lim_{r \rarrow 0} (2 \pi r - \mbox{C}(r)) \cdot \frac{3}{\pi r^3}.
[/tex]
 
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  • #9
sandesh trivedi
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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

well i m in high school right now and i don't know much of the mathematics u r talkin about. i was just trying to find the relation between third and fourth dimension.pervect do you have some yahoo messenger where i can ask u a few questions regarding this stuff.but i explained this ant thing a little bit using simple geometry and i m happy that i m gettin original not some bookworm.i know one has to be original especially in physics and mathematics i mean theoretical physics and excuse me if i m gettin very much excited.
now i have this new story about fred of course written by me.
now fred makes some beautiful hemisphere on this spherical surface using some marker. for fred he is making a big circle and for us 3d people it is a hemisphere.he finds the real radius that u already know and finds the surface area.

here
let
real radius=r
length of rope =wrong radius=R
height of third dimension=h=r as it is a hemisphere
please let me know how can i post a diagram along with this
now
R^2=r^2+h^2
R^2=r^2+r^2
R=r*sqrt(2)-------------------------eqn (1)
now when fred tries to know the area he uses his wrong radius R purposely.
for fred area circle=pi*R^2
using eqn(1)
we get area =2*pi*r^2 area calculated by us of hemisphere made by fred.
hence circumference in 2d and 3d is different while area between 2d and 3d remains the same.
next we compare 3d and 4d where i hope we shall get areas to be different and volume to be the same.
what do u think about this pervect and other forum members
 
  • #10
pervect
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What you are doing when you calculate the height is known as finding an "embedding diagram". In this particular case there is a unique embedding diagram that describes the geometry, that turns out to be not the case in general.

Sorry I dont' have any private messenger programs, I just read the forums here when (I'm in the mood.

You might be somewhat interested in

http://www.sff.net/people/Geoffrey.Landis/blackhole_models/paper_blackholes.html

which pushes embedding diagrams via paper models about as far as they can go to explain certain aspects of relativity.

If you really want to understand relativity, though, at some point you'll have to tackle the math.
 
  • #11
sandesh trivedi
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i know i got to learn a lot of maths for special and general relativity.
what i m trying to explain is that draw a right angled triangle
where
R=length of the rope
r=real radius
h=height of third dimension
now
R^2=r^2+h^2
for simplicity i don't consider a small circle drawn by fred on ths sphere but a large circle which from third dimension is a hemisphere.
so the equation becomes simple and we get
R=r*sqrt(2)-------------------------eqn(1)
now according to fred area of the circle made by him=pi*R^2
but according to us its a hemisphere so area=2(pi)r^2
but when we substitute value of R from eqn(1) to pi*R^2
we get
area = pi*(r*sqrt(2))^2
= 2(pi)*r^2
so when we compare 2d and 3d we get circumference to be different and area to be same i.e equal
but when we compare 3d and 4d can we say we will get areas to be different and volume to be same.
help me with this.
 
  • #12
pervect
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Sandesh, basically you've got a good solution, but your approach won't generalize.

Here is an example. Try to imagine an infinite plane surface of constant curvature. By constant curvature, I mean the "Gauss" curvature that was previously mentioned.

A sphere has a constant curvature, but it is not infinite in the sense I mean. Your approach (embedding the plane in a 3-d space) won't work for this problem. You'd need to embed the plane in at least a 4-d space.

If you are still undeterred and start to apply your approach to the embedding of the plane in a 4-d space, you will find that you have an 1 equation with 2 unknowns.
 
  • #13
sandesh trivedi
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i m not gettin it what is wrong with my solution. i know its my incapability to understand that much maths. but can i get the relation between areas volume indifferent dimensions as i m expecting
for example
areas are same but circumference is different between 3d and 4d
and areas are different but volume are same between 4d and 3d
i mean there are somethings which i m not able to explain this computer stuff u also must be experiencing that.do u live in India (bombay)
 
  • #14
fahd
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There is a raging debate going on (and probably will go on forever) about whether the fourth dimension is time, or whether it is a 4th spatial dimension. Your question is not to claim that the fourth dimension is spatial instead of being time. The point is to speculate about what a fourth *spatial* dimension would be like, beyond our three spatial dimensions.

To humans going about their everyday lives, time is fundamentally different than the three spatial dimensions, since only one direction is possible with time. Some have argued that this is only because of our limited perspective. Indeed, time actually behaves like a spatial dimension when you consider it in special relativity. You can get into some quite complex, interesting, and paradoxical discussions about weird things happening with time as a dimension, one of which is what uv mentioned in ur question sanjeev!!
 
  • #15
Juan R.
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fahd said:
There is a raging debate going on (and probably will go on forever) about whether the fourth dimension is time, or whether it is a 4th spatial dimension. Your question is not to claim that the fourth dimension is spatial instead of being time. The point is to speculate about what a fourth *spatial* dimension would be like, beyond our three spatial dimensions.

To humans going about their everyday lives, time is fundamentally different than the three spatial dimensions, since only one direction is possible with time. Some have argued that this is only because of our limited perspective. Indeed, time actually behaves like a spatial dimension when you consider it in special relativity. You can get into some quite complex, interesting, and paradoxical discussions about weird things happening with time as a dimension, one of which is what uv mentioned in ur question sanjeev!!

Hey! Time is not a spatial dimension on SR. Precisely one talk about spacetime not about four-space. Time is different from space as already seen in the metric (-1, 1, 1, 1).

Of course, some authors have misunderstood the concept of time. For example Hawking.

Hawking loves Euclidean and proposes the use of his imaginary time. Then tau = i·t and the metric becomes (1, 1, 1, 1),

ds = SQR[dx + dy + dz + d(tau)]

that is, a pure 4D spatial formulation: time is geometrised. Hawking claims that real time is tau. Perhaps!

But Hawking posterior analysis is incorrect. It is true that tau is indistinguishable from other space-like dimensions in the new metric but if you take Newton equation you may do tau = i·t

dx/dt = -grad(potential)

i dx/d(tau) = -grad(potential)

and again time is different from space, because space dimension is real and time dimension is imaginary. Hawking's geometrisation of time is ineffective. time is differnt from space.
 
  • #16
pervect
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sandesh trivedi said:
i m not gettin it what is wrong with my solution. i know its my incapability to understand that much maths. but can i get the relation between areas volume indifferent dimensions as i m expecting
for example
areas are same but circumference is different between 3d and 4d
and areas are different but volume are same between 4d and 3d
i mean there are somethings which i m not able to explain this computer stuff u also must be experiencing that.do u live in India (bombay)

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 infinitely 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.
 
  • #17
Chronos
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Too many assumptions make for a bad result. It's correct to treat time as a dimension, but not ok to introduce arbitrary factors that restrict the freedom of choice in choosing spatial coordinates. I disagree with Juan, time is clearly a free parameter, but not priveledged. Space without time is meaningless.
 
  • #18
Juan R.
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Chronos said:
Too many assumptions make for a bad result. It's correct to treat time as a dimension, but not ok to introduce arbitrary factors that restrict the freedom of choice in choosing spatial coordinates. I disagree with Juan, time is clearly a free parameter, but not priveledged. Space without time is meaningless.

I think that i emphasized in #15 that time is not another spatial dimension. If was, Hawking did not need introduce his imaginary time concept.

You say it is correct to treat time as a dimension, but you do not specify that is a *dimension* for you. Nor why that hypotesis fails in QM where there is a well defined operator of space but there is not operator of time.

Time is not a dimension like we understand the concept for space. Time is not a "dimension" in quantum mechanics, for example. Time is priveledged because is the "parameter" that defines motion -including motion of space- is restricted by second law, normalizes wave functions, etc.

Precisely, the great problem of Hamiltonian quantum gravity is the problem of absence of TIME (not of space). Time is different.

There is a second law dS/dt > 0

There is not a "spatial" second law dS/dx ?>? 0

Time is also different from a mathematical side. Space can be conveniently represented by real numbers, but time cannot.
 
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  • #19
sandesh trivedi
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aint u guys gettin outta this topic.we r here to talk about spatial dimension not time
 
  • #20
Juan R.
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sandesh trivedi said:
aint u guys gettin outta this topic.we r here to talk about spatial dimension not time

I was replying to fahd and to Cronos. What is your problem?
 
  • #21
pervect
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Here is one of the problems. Time is not the same for all observers. A 4-vector that is purely timelike for one observer (one in which the spatial component is zero) may have both time and space components for another observer (the spatial component which was zero for the first observer will be nonzero for the second). This happens whenever the second observer is moving relative to the first observer.

In Newtonian physics, space and time can be cleanly separated. In relativity, they cannot be so cleanly separated, which is why one talks about the space-time continuum.
 
  • #22
Juan R.
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pervect said:
Here is one of the problems. Time is not the same for all observers. A 4-vector that is purely timelike for one observer (one in which the spatial component is zero) may have both time and space components for another observer (the spatial component which was zero for the first observer will be nonzero for the second). This happens whenever the second observer is moving relative to the first observer.

In Newtonian physics, space and time can be cleanly separated. In relativity, they cannot be so cleanly separated, which is why one talks about the space-time continuum.


Exactly! There exists a spacetime, where time is not a spatial dimension, is just time. This is the reason of the 3+1 decompositions used in both quantum and classical gravity. One of “dimensions” is very different of the others 3.
 
  • #23
hellfire
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I would agree with Juan’s position and insistence. Space and time are both part of spacetime, but in order to have causal and deterministic physics one must choose globally hyperbolic spacetimes which allow for a 3 + 1 decomposition. In such spacetimes a vector pointing in the direction of the time coordinate is always timelike. Thus a “mixing” between time and space is not possible for any kind of local coordinate transformation. Am I wrong?
 
  • #24
DrGreg
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hellfire said:
I would agree with Juan’s position and insistence. Space and time are both part of spacetime, but in order to have causal and deterministic physics one must choose globally hyperbolic spacetimes which allow for a 3 + 1 decomposition. In such spacetimes a vector pointing in the direction of the time coordinate is always timelike. Thus a “mixing” between time and space is not possible for any kind of local coordinate transformation. Am I wrong?
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".
 
  • #25
hellfire
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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".
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)?
 
  • #26
pervect
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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.
 
  • #27
Juan R.
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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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  • #28
sandesh trivedi
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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
x =radius for 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
 
  • #29
Juan R.
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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
real radius =r
wrong radius=length of the rope=R
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
:cool:

Ok, let me talk abut spatial dimensions also :approve:

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)
 
  • #30
pervect
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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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  • #31
sandesh trivedi
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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.
 
  • #32
yukcream
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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 infinitely 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 :confused: ~ 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
 
  • #33
yukcream
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εllipse said:
Our 3 spatial dimensions are Euclidean (not curved), except where there are strong gravitational fields.

No.

the second queston, why no?
 
  • #34
pervect
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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 :confused: ~ 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.
 
  • #35
yukcream
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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?
[tex]K = \lim_{r \rarrow 0} (2 \pi r - \mbox{C}(r)) \cdot \frac{3}{\pi r^3}.[/tex]
 

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