# Space as “flat” plane

I was watching the documentary Carl Sagan did about gravity (I believe it's quite old though) and wondered about space being "flat" and that mass creates dents in this plane as shown at about 3 minutes in this clip (http://www.youtube.com/watch?v=Y-db4iC0aHw)

Is this simply a metaphor or is it something more than that? Does gravity follow this just instead of being flat the dent is "rotated" in all directions?

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## 3 Answers

The picture by Sagan is somewhat of a simplification of the true geometry - so in a sense that picture is a metaphor. The interpretation given there of Einstein's equations is indeed that of curvature of 4 dimensional Space-Time (not just 3 dimensional space).

However in these higher dimensional geometries there is more than one notion of curvature: Scalar Curvature R; Ricci curvature; Riemann Curvature and Weyl Curvature. The Scalar curvature exists in all dimensions and is simply a function giving the curvature at each point. For an embedded (2-dim) sphere it is R$=2/r^2$ (twice the Gaussian curvature). So for a sphere it is non-zero everywhere on its surface. Higher dimensional spaces use these other Curvature objects (built like generalised matrices and called Tensors) to represent their curvature more precisely than a single number. In higher dimensions the Scalar represents a kind of "average curvature" (at each point).

Corresponding to these different notions of curvature, there are different notions of "flatness" as we will see below.

Now the Einstein equations directly equate the Ricci curvature to something; and in the example shown in the Sagan excerpt it was equated to zero. Ricci$= 0$ is the Einstein Vacuum equation alluded to in other answers, and is appropriate because outside a star there is a vacuum. This has an immediate mathematical consequence that R$=0$ ie the scalar is zero as well! So in this sense the vacuum is flat (called Ricci-flat).

However there is still curvature around in that space, so it is not Minkowski (ie Euclidean) flat. The experimental demonstration of this was the bending of light rays near the Sun (assuming as one does in Einstein's theory that light rays measure the "straight lines"). So where does the curvature come from if it is Scalar and Ricci flat? The answer is that it is not Riemann flat: the tensor Riemann$\neq 0$. However this does not quite explain the origin of curvature here. Expressed very loosely we have the following equation:

Weyl = Riemann - Ricci - R

So the real source of curvature in the Sagan excerpt is the Weyl component of the Riemann tensor: everything else is zero. Now we come to the representation problem that Sagan had: the Weyl tensor is always zero in two and three dimensions. In other words the kind of curvature it represents does not exist in two and three dimensions: only four and above dimensions have this kind of curvature. So it cannot be directly represented on a 2 or 3 dimensional picture.

Instead what Sagan appears to have represented here is the gravitational potential (like in Newton's theory) but expressed as space curvature. It is not completely wrong perhaps, but it is not quite correct and so is just a metaphor.

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Nicely put..+1 – Gordon  Feb 12 '11 at 18:02

Lets look at Einstein's field equations: $G{\mu\nu}=8\pi.T{\mu\nu}$ where $G{\mu\nu}=R{\mu\nu}-1/2g{\mu\nu}R$ The left side is the curvature of space (the metric). The right side is the stress-energy-momentum tensor, which is the totality of what is producing the curvature. $g{\mu\nu}$ is the metric tensor , R, the scalar curvature, and $R{\mu\nu}$ the Ricci curvature tensor, but the terminology doesn't matter here. Simply the equation is saying the the curvature of space (the metric) is produced by what is in the space (pressure, energy etc)
John Wheeler, who always used colorful language said, "Matter tells space how to curve. Space tells matter how to move."

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In the context of general relativity theory the following things happen:

In the absence of matter, spacetime remains flat. The relevant space is a 4 dimensional Minkowski space. It has some similarities with Euclidean flat geometry, which you have learnt in the school. There are also important difference. In an Euclidean space the symmetry group is Euclidean group whereas in Minkowki space it is called a Poincare group. The former space has only space like dimensions, the latter has 3 space and 1 time like dimensions. However the 4 dimensional Minkowski space is also like a table top.

But if there is some mass energy the geometry of the surrounding spacetime no longer remains Minkowskian, It changes to a more general geometry (although locally within an infinitesimal region it remains Minkowskian). The geometry of Minkowski space is analogous/corresponds to geometry of a plane surface and the latter more general geometry is analogous/corresponds to a geometry of a curved surface. In this curved geometry the sum of three angles of a triangle may be more than or less than $\pi$ depending on the curvature. This curvature does not move itself in static case. Only a particle follow the straightest possible path. Since the spacetime is itself curved the particle appears to be moving in a curved path as if by a force.

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You conflate space-time and space in this answer. Some of those "Euclidean" should in fact be "Minkowski". Also, what space you get depends on the observer because different space-like slices of curved space-time can generally differ (contrary to the Minkowski space-time where you'll always get Euclidean space as a slice). – Marek Feb 12 '11 at 7:42

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@Marek: You are right. In fact I made a deliberate attempt to avoid terms like "Minkowski space" and present the affairs in a simple manner assuming (perhaps unjustifiably) the questioner does not already know about special relativity. Secondly I thought putting things in this way may appeal to those members who might be interested in learning relativity but never studied even SR. Maybe, I have made oversimplifications. – user1355 Feb 12 '11 at 8:07
Technically, we don't really know if the massless space-time is flat or curved, because of $\Lambda$ which can be thought to be non zero in even empty space. – Sklivvz♦ Feb 12 '11 at 9:10
@Sklivvz: There is something called Einstein's vacuum equation. Where we simply put $R{\mu\nu} = 0$ i.e. Ricci curvature vanishes. – user1355 Feb 12 '11 at 10:49
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