# Is all curvature the same?

1. Jan 28, 2006

### Mike2

I take it that there are various ways of curving spacetime with a desired value of curvature. Is there an overview or a run-down of the various metrics, their names, where they are applicable, and how they curve spacetime? For example linear acceleration curves space differently from gravity, but an observer cannot tell the difference. Is this true of all the various ways of curving spacetime? Thanks.

2. Jan 28, 2006

### pervect

Staff Emeritus
The single most common metric is probably the Schwarzschild metric. This is the metric outside of any spherically symetric gravitating body. The inside metric of massive bodies is usually not of as much interest as the exterior metric.

The Schwarzshild metric is also the metric of a non-rotating black hole.

The next most common metric is probably the Kerr-Newman metric. This is the metric of a rotating (and possibly charged) black hole.

Both of these metrics cover all of space-time.

We've spent a fair amount of time recently talking about the metric and the coordinate system of an accelerated observer. The best textbook treatment of this case is in "Gravitation", IMO. This is usually called the Rindler metric. (Almost any GR book will cover the other metrics previously mentioned well, but I would still recommend MTW's treatment of the accelerated observer for the sufficiently advanced reader. Unfortunately at least some familiarity with tensor notation is required to fully read this section in MTW, though the notation is explained earlier in the book.).

The accelerated observer has a direction that is apparently "above" and "below" him. He has an apparent plane "below" him where the metric coefficients drop to zero. This plane is essentially an event horizon, much like that of a black hole. (The notion of an event horizon is thus, as can be seen, observer dependent, as a stationary non-accelerating observer does not observe such a horizon). The location of this event horizon also marks the boundary below which the specific coordinate system used for this metric cannot be extended consistently. There is no equivalent problem "above" the observer, the problem occurs "below" him.

The idea that accelerted motion cannot be distinguished from gravity due to an actual mass is known as the "principle of equivalence". There are some subtle points here, for instance tidal gravity is a measurable quantity, one that is zero in an accelerating spaceship and non-zero on a planet.

Probably the most important thing to say is that the metric and its derivatives allows one to calculate whatever quantities are of interest. Quantities of interest include tidal forces (handled by the Riemann tensor, which can be computed from the metric and its first and second derivatives), the apparent direction of gravity for someone with constant coordiantes (these are handled mathematically to "Christoffel symbols, which can be computed from the metric and its first derivative), proper time and time dilation (measured with a local watch as compared to a reference watch) computed from coordinate intervals (handled by the metric itself), etc. etc.

3. Jan 28, 2006

### Mike2

Thank you, pervect, that certainly is a start. Do you know if Wikipedia has an entry listing the metrics and their applications? It would be nice if there were a crip-sheet for easy reference.

My main curiosity is whether any value of scalar curvature for a given metric will yield the same local observations as the scalar curvature for any other metric. And can scalar curvature (no matter what metric produces it) be equated with an acceleration. Thanks.

4. Jan 28, 2006

### pervect

Staff Emeritus
Offhand, I don't know of any collection of metrics. I know there is a classification system of metrics based on their invariants (I recall the name now, Petrov classification) - but I don't use it much personally (this is a reflection on my interests, not on the usefulness of the system!).

As far as the curvature scalar goes, it's not particularly related to acceleration. Christoffel symbols do that job. Christoffel symbols model not only linear acceleration (what one feels as gravity), they model other inertial forces, such as centrifugal / centripetal forces, coriolis forces, etc.

Very very losely, you can think of the Christoffel symbols as being the first derivative of the metric.

Thus "ordinary" gravitational forces are basically the result of $\partial g_{00} / \partial [x,y,x]$, the rate of change of the metric coefficient for time (g_00) with respect to position. The force in the x direction is related to the partial derivative with respect to x, the force in the y with the partial with respect to y, etc.

The actual notion of "curvature", as expressed by the Riemann tensor, is related to the second derivative of the metric, the derivative of the Christoffel symbols. This is related directly to the notion of tidal forces.

Thus you will see that an accelerated observer is not technically in "curved" space-time (the second derivative of the metric is zero, the tidal forces are also zero), though he still feels forces (the first derivative of the metric is non-zero).

When people say that gravity is due to "curved" space-time, they are speaking lose, popular language. Gravity as it is popularly thought of is not due to the curvature tensor, but rather due to the Christoffel symbols.

Last edited: Jan 28, 2006
5. Jan 29, 2006

### pmb_phy

The "gravitational field" is the set of components of the metric tensor (at least as it is described in numerous GR texts and by Einstein). The Christoffel symbols are the components of the field. the metric tensor is equivalent to the gravitational potential while the Christoffel symbols are equivalent to the components of the gravitational force.

Pete

6. Jan 29, 2006

### pervect

Staff Emeritus
That may be a good technical defintion of the gravitational field in some textbooks (and not all textbooks, for MTW's "Gravitation" has a different defintion), but it has absolutely nothing to do with the popular notion of "gravity" as the force which holds a person to the Earth - or the force that a person in an acceleratring spaceship feels.

The metric at the origin of both an accelerating spaceship and a non-accelerating one are exactly the same.

However, the popular notion of "gravity" is not the same in an accelerating and a non-accelerating space-ship - the first spaceship has it, the second doesn't.

Thus the metric cannot describe the popular notion of gravity directly, for the metric is exactly the same for the accelerating and non-accelerating spaceships, while the "felt" gravity is not the same.

(This also implies that no tensor can describe the popular notion of gravity directly, for any tensor that is zero at a point cannot be transformed into one that is non-zero by any change of coordinates).

What field? The gravitational field? Isn't that what I just said?

Which makes the Christoffel symbols (roughly speaking, at least) the first derivative of the metric tensor - gee, isn't that what I just said?

7. Jan 29, 2006

Staff Emeritus
Correct me if I am wrong, but the curvature is described by the curvature tensor (Riemann-Christoffel tensor) which is a mathematical combination of Christoffel symbols and their derivatives. Unlike the C symbols, the R-C tensor is a tensor, and thus generally covariant. So if the curvature vanishes for the unaccelerated frame shouldn't it vanish for the accelerated comoving one, since they are only a diffeomorphism apart?

8. Jan 29, 2006

### pervect

Staff Emeritus
You're quite correct. The Riemann curvature tensor is zero for an accelerating observer. I intended to make this point earlier when I said:

When I say that Christoffel symbols are the "derivative" of the metric, I am speaking very losely, it would be more accurate to say that the Christoffel symbols are formed from the derivatives of the metric.

The Riemann tensor is, as you point out, formed from the Christoffel symbols and their derivatives, which is why I termed it - the Riemann tensor - (again, speaking losely) as "the second derivative of the metric".

Thus the accelerating observer illustrates a situation where we have "gravity" (Christoffel symbols), but no "curvature" in the sense of having a non-zero curvature tensor.

9. Jan 30, 2006

### Mike2

Are you talking about just spatial curvature as opposed to curvature of time in flat space? Is it that a mass produces spatial curvature as well as a curvature in the time coordinate, whereas acceleration is a curvature in the time dimension while keeping the spatial dimensions flat? Thanks.

10. Jan 30, 2006

### pervect

Staff Emeritus
The Riemann curvature tensor measures the curvature of space-time (not just the curvature of space). The Riemann curvature tensor (RCT for the rest of this post) is a coordinate independent object. This basically means that the way in which the RCT varies with changes of coordinate systems is restricted to follow well-known "standardized" laws - so that specifying the components of the RCT in any coordinate system specifies the components in all possible coordinate systems, as the transformation laws are known and standardized.

The Riemann curvature tensor of empty space is zero, regardless of whether one uses the inertial coordinates of an unaccelerated observer, or the accelerated coordinates of an accelerating spaceship (which however must have negligible mass for this thought experiment).

Thus the Rieman curvature tensor is a mathematical entity that models something else than the popular notion of "gravity", in the sense that an observer in an accerating spaceship feels something that acts like gravity, as the RCT is zero, while "gravity" is not zero.

Physically, part of what this mathemtatical entity (the RCT) models is tidal forces (the rate of change of gravity).

The mathematical entity that directly models the notion of gravity "felt" by an accelerating observer is known as the "Christoffel symbols". I've already talked about the relationship between these entities (and the metric) in very general terms - I don't think I can get much more specific without writing down tensor equations.

You might try reading some of Baez's GR tutorial for more detailed info:

http://math.ucr.edu/home/baez/gr/

He also has a good paper on the meaning of Einstein's equations

http://math.ucr.edu/home/baez/einstein/

Last edited: Jan 30, 2006
11. Jan 31, 2006

### Mike2

Let me take one more stab at this before I concede. The GR metric is composed of a spatial part and a time part. And as you say, curvature is related to the second derivative. I assume that means that the metric will be differentiated twice wrt time as part of the equation for curvature. I would naturally think that since acceleration is the second derivative of space wrt time, that this part is not zero. But perhaps I meant the inertial observers consideration of the accelerating observers FOR. The accelerating observer is not moving wrt to his own accerating FOR. Neither is the inertial observer accelerating wrt his own FOR. But only the inertial observer sees that the accelerating observer is accelerating wrt the inertial observers FOR. Someone is accelerating so there is a none zero second derivative of space wrt time somewhere. So doesn't this constitute a none zero curvature? I'm just not sure who is saying what about whom. Thanks.

Last edited: Jan 31, 2006
12. Jan 31, 2006

### pervect

Staff Emeritus
I think part of what's confusing you is that the Riemann curvature tensor is not a single number.

The curvature tensor actually consists of a 4x4x4x4 array which contains 256 numbers (not all of which are unique, however, many of them are constrained by symmetries).

The other part of what's confusing you is that my explanation has been extremely over-simplfiied.

For a slightly less over-simplfied explanation, take a look at Baez's GR course outline:

http://math.ucr.edu/home/baez/gr/outline2.html

Of course, at this point, you may be asking - "what are tangent vectors, tensors, and parallel transport?"

Baez's tutorial does about as good a job of explaining it as any popular description can. Unfortunately, for a really good understanding, you'd need to take a course in differential geometry.

13. Jan 31, 2006

### robphy

I think you mean the Ricci (or Einstein). Riemann is not zero for (say) Schwarzschild, a vacuum solution.

14. Jan 31, 2006

### Mike2

Which one is related to acceleration?

OK, if acceleration does not always imply curvature, then does curvature always imply acceleration? Thanks.

15. Feb 1, 2006

### robphy

[4-]acceleration is associated with the curvature of a worldline ($$a^b=v^c\nabla_c v^b$$)... describing the deviation from being a geodesic.

Riemann curvature $$R^a{}_{bcd}$$, on the other hand, describes the curvature of a manifold-with-connection.. describing the deviation from flatness (path-independence of parallel transport). Ricci curvature $$R_{bd}$$ is a tensor obtained by contracting Riemann. Einstein curvature is a divergence-free combination involving Ricci, its contraction [called the scalar curvature] and the metric: $$R_{bd}-\frac{1}{2}R g_{bd}$$.

Don't confuse worldline-curvature with spacetime-curvature.
One can have an accelerating particle [with nonzero worldline curvature] in Minkowski spacetime [a zero curvature spacetime]. One can also have a freefalling particle [a geodesic with zero worldline curvature] in the exterior Schwarzschild spacetime [a nonflat spacetime with nonzero Riemann curvature but zero Ricci and Einstein tensors].

16. Feb 1, 2006

### hellfire

Yes one can if there are forces different than gravity acting on the particle. I had prepared a similar answer like that but deleted it, because I am not sure if this is what Mike wants to hear. I guess that the question of Mike is rather what happens in situations without other forces than gravity. What is then the relation between acceleration and curvature.

17. Feb 1, 2006

### pervect

Staff Emeritus
What I was trying to say is this:
For the metric represented by the line element

$$ds^2 = (1+gz)^2 dt^2 - dx^2 - dy^2 - dz^2$$

(I've written it out explicitly to avoid confusion)

which is the Rindler metric associated with a uniformly accelerated obsever in a flat Minkowski space-time, i.e. the metric associated with Einstein's elevator experiment

the Riemann tensor is identically zero (all 256 elements of it!).

As Self-Adjoint observed, this can be predicted in advance from the fact that the Rindler metric above is just a different coordiante patch on the Minkowski vacuum (in which the Riemann tensor is obviously identically zero). You can't make a zero tensor non-zero by simply changing the coordinates.

I'll cheerfully agree that the Riemann is non-zero in the Schwarzschld solution (while the Ricci and Einstein tensor are zero), but that isn't what I was trying to say.

The significance of this result is that it illustrates how we can have "gravity" (assuming that we all agree that the passenger in Einstein's elevator experiences "gravity") without having any curvature (a totally zero Riemann).

While I'm on the topic, I'll add something else.

Suppose we want to ask the question "How much gravity does the passenger in the elevator feel, as a function of height (the z-coordinate).

The way in which we proceed is (I believe) the following.

1) First we construct an orthonormal basis of one-forms, to serve as a local "frame" to perform our measurements in. (We measure "gravity" with respect to this local "frame").

In the example above, this ONB of 1-forms would be $d\hat{t}=(1+gz)dt;$ $d\hat{x} = dx$; $d\hat{y}=dy$; $d\hat{z}=dz$

2) We then compute $$\Gamma^{\hat{z}}{}_{\hat{t}\hat{t}}$$ This is our answer. I get -g/(1+gz) for the specific metric above.

This illustrates why I say that gravity is modelled by the Christoffel symbols.

Last edited: Feb 1, 2006