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
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.
Our 3 spacial dimensions are Euclidean (not curved), except where there are strong gravitational fields. No.
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
As far like i know nobody has measured a curved spacetime or curved space predicted by general relativity still 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).
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 dont 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
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.
i know i gotta 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 dont 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.
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.
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)
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 indistinguisable 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.
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.
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.