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Bending of space 
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#1
May409, 06:04 PM

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What exactly is bending of space? Does space bend from position x to position y? If so, what do we call whatever is there in position x (after the bend)? And what was there at position y, before the bend?



#2
May509, 03:43 AM

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Threedimensional space isn't part of a fourdimensional space. It doesn't "bend" in or through a fourth dimension … it only changes its own geometry (curvature). 


#3
May509, 04:11 AM

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#4
May509, 04:42 AM

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Bending of space
There is no 4D space. 


#5
May509, 05:44 AM

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Hi there,
An example might actually enlighten you on this subject. Everyone knows that light (photons) travel in a straight line. Ok, nothing so bad up to now. If this light would pass next to a very massive object (super massive star or black hole), the gravitational pull of this object would also affect the light ray trajectory. Therefore, we call it space bending, just because light follows the curvature of space. Hope this helps a bit. Cheers 


#6
May509, 05:54 AM

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You can also embed a intrinsically curved manifold into a flat higher dimensional manifold to visualize the distorted distances: http://www.physics.ucla.edu/demoweb/...spacetime.html 


#7
May709, 05:28 PM

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#8
May709, 05:46 PM

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It "knows" it's curved/distorted because it uses its own ruler and finds that (for example) the ratio circumference/diameter of a circle is not π. Curvature is an intrinsic property of a space … that means that no outside measurement is required … the space itself "knows" about its own curvature. 


#9
May709, 06:04 PM

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Teeril,
you need some maths to understand intrinsic curvature. You know from that in a 3d Euclidean space the change in distance between two points is given by [tex]ds^2=dx^2+dy^2+dz^2[/tex], coordinates are x,y,z. This is a 'flat' space because the dx^2, dy^2 and dz^2 terms have coefficients that are independent of position. The distances on the surface of a sphere are given by [tex]ds^2=r^2\sin^2(\theta)d\phi^2+r^2d\theta^2[/tex], coordinates phi and theta ( r is constant). This 2D space is curved. As A.T. has said, it's because of distorted distances. 


#10
May709, 06:30 PM

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For the bending of space only, the old analogy of the ball on a rubber sheet works fairly well. In fact, if only consider motion in a plane, the curvature of space around a spherically symmetric, stationary body can be visualized as a Flamm paraboloid embedded in 3 space.
http://en.wikipedia.org/wiki/Schwarz...27s_paraboloid This can be used to compute the spatial contribution to the precession of the perihelion of Mercury. Rindler's book does a good job describing this. The curvature of time contributes the other half to the bending. 


#11
May709, 07:09 PM

P: 216

This is my analogy. Please tell me if it makes any sense.
Let's say, I have a rope of known length l and both the end's coordinates (x1, y1, z1) and (x2, y2, z2). With coordinate (x1, y1, z1) as the center and radius l, I construct a sphere. If the coordinate (x2, y2, z2) is part of the sphere, I can confirm the rope is not curved. If not, the rope is curved. (the (x2, y2, z2) must be always inside the sphere; I cannot think of a possibility (x2, y2, z2) is outside the sphere). What I do not understand, why light has to bend if there's no space? 


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