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If the the space-time 'curvature' proposed by Relativity is real, wouldn't the curvature have to be through a higher dimension, since we can't see it? Think of the 2-dimensional grid diagram they use to illustrate gravity sometimes. Now extend that grid into 3 dimensions, creating a lattice. Now place a mass at the center of one of the cubes in that grid. What happens? The cube 'shrinks' a little, dragging the lines to the other cubes (spacetime) with it. That's the spacetime distorion (curvature) caused by gravity. If you think about that models of a tesseract you've seen, one cube inside the other, you see that gravity must be a 4th dimensional vector. It's relatively (!) easy to visualize if you think about it that way.

2007-12-05 18:12:36 · 4 answers · asked by AmigaJoe 3 in Science & Mathematics Physics

The 'Question' is about a 4th physical or spatial dimension, not time. Time can't be treated as a physical vector like the other dimensions.

2007-12-05 19:28:54 · update #1

"You don't need an extra dimension to warp a 3-dimensional spatial lattice. "

-You do if you want to warp space the way gravity does, normal to all other vectors, no?

2007-12-05 21:16:24 · update #2

4 answers

The answer to your question is that curvature is actually an intrinsic property of a shape. Yes, we can measure the curvature of the 2-dim grid by looking at how it sits inside 3-dimensions. But even if you were a 2-dim person living on the grid itself, you'd be able to tell that the grid had curvature. How? By drawing triangles on the grid and measuring the sum of the angles. This fact, that curvature is intrinsic to a surface, is known as Gauss's Theorema Egregium. (Theorema Egregium essentially means 'really amazing theorem' because Gauss found it so surprising.)

And in general, when mathematicians and physicists refer to the curvature of, for example, space-time, we are referring to a property that is intrinsic to space-time itself--in other words, something that can be measured without reference to an ambient space. If this sounds strange, remember that your everyday intuition about the meaning of the word "curvature" is only a very rough approximation of what the concept actually stands for in the math/physics literature. In reality, it's the square of some differential operator associated to space-time. In other words, don't try to conclude too much from these 2-dim grid pictures you see in lay references. You're not getting the full story, unfortunately.

2007-12-05 20:22:09 · answer #1 · answered by robert 3 · 1 0

You don't need an extra dimension to warp a 3-dimensional spatial lattice. You're incorrectly requiring that the basis for your lattice be orthonormal, but it is not necessarily so.

Time is an extra dimension necessary to project the space-time dynamics of objects.

2007-12-06 03:47:47 · answer #2 · answered by Dark Matter Physicist 3 · 0 0

Whats the question?
Time is the fourth dimension, but it is a temporal dimension, as opposed to the other three spatial dimensions.
Gravity distorts time and space, because it changes the geometry of space-time.

2007-12-06 02:52:33 · answer #3 · answered by brownian_dogma 4 · 1 0

Sounds good to me!

2007-12-06 02:20:27 · answer #4 · answered by Anonymous · 0 2

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