# Bending of space

1. May 4, 2009

### jobyts

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. May 5, 2009

### tiny-tim

Hi jobyts!

Three-dimensional space isn't part of a four-dimensional space.

It doesn't "bend" in or through a fourth dimension … it only changes its own geometry (curvature).

3. May 5, 2009

### jobyts

If I understand you, in a 3D world, it looks like bending, but what actually happening is a change in geometry in the 4-D space. Please correct if I'm wrong.

4. May 5, 2009

### tiny-tim

No, what is actually happening is a change in geometry in the 3-D space.

There is no 4-D space.

5. May 5, 2009

### fatra2

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. May 5, 2009

### A.T.

"Bending" is the wrong word, as it usually refers to extrinsic curvature. Intrinsic curvature is a distortion of distances: Imagine you fix one end of a string with the length R at a point in 2d-space, and make a full circle with the other end. If you then find that the circumference of that circle is different from 2*PI*R, you conclude that the 2d-space is curved intrinsically. So you don't need extra dimensions to quantify that curvature. You can extend this example to curved 3d-space containing a sphere with the radius R and a surface area different from 4*PI*R^2.

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..._and_general_relativity/curved_spacetime.html

7. May 7, 2009

### Teeril

Think i'm missing something here, would that not still be only possible if the 2d space was "bend" in the 3rd dimension. Otherwise, how can the space "know" it's curved/distorted?

8. May 7, 2009

### tiny-tim

Hi Teeril!

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. May 7, 2009

### Mentz114

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
$$ds^2=dx^2+dy^2+dz^2$$, 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
$$ds^2=r^2\sin^2(\theta)d\phi^2+r^2d\theta^2$$, coordinates phi and theta ( r is constant). This 2D space is curved.

As A.T. has said, it's because of distorted distances.

10. May 7, 2009

### Daverz

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/Schwarzschild_metric#Flamm.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.

Last edited: May 7, 2009
11. May 7, 2009

### jobyts

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 co-ordinates (x1, y1, z1) and (x2, y2, z2). With co-ordinate (x1, y1, z1) as the center and radius l, I construct a sphere. If the co-ordinate (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?