# What does curvature of spacetime really mean?

1. Nov 6, 2007

### Yianni

I don't really get GR. Why should curved space and time be a model for gravity? To me, curved space means a observers no longer measure distances as sqrt(x^2+y^2+z^2), but rather, given an x-ordinate, y-ordinate and z-ordinate, the length of the shortest path to that coordinate can be calculated by a different formula. Surely it means that two spatially separated objects which are moving parallel to each other, if they enter a bit of space that is not uniformly curved, will no longer be traveling parallel to each other. In this sense I get why light, for example, is bent by gravity in general relativity. But why do things in a gravitational field actually accelerate in the direction of the most negative gradient (I don't know if this is part of the maths, its just what I've understood from looking at those embedding diagrams or whatever they are called.) Is the previous statement even correct, or does SPACE accelerate into the direction of the most negative gradient, and objects with mass in that space continue to remain at rest in that space, which is moving? And would it be fair to say that mathematically, the curvature of spacetime needn't cause an acceleration of any sort, except that in general relativity, one is always accompanied by the other - you always have an acceleration of space in a gravitational field, and you always have a distortion of spacetime, thus in GR they are one and the same, or have I completely missed the meaning of curved spacetime?

2. Nov 6, 2007

### Chris Hillman

Why Geometrize Gravitation?

You didn't mention your math/physics background or the level of understanding (comprehension of basic principles? mastery of computing with gtr?) which you seek, but I'll guess you are an undergraduate student with at least high school math and some calculus.

In the end, gtr is not terribly difficult or complicated, but many of its principle assumptions and conclusions are sufficiently subtle that it requires substantial effort to learn and appreciate properly the elements of gtr. On the other hand, while there are many beautiful things in sci/math, gtr has been blessed with an unusual number of superb textbooks and even some fine popular books. Among the latter I particularly recommend Wald, Space, Time, and Gravity and Geroch, General Relativity from A to B.

The short answer is that this is a most wondrous and elegant way of incorporating the special feature of "gravitational force" (viz Lorentz force in EM) into the beautiful geometric interpretation of special relativity which was discovered by Minkowski (a leading mathematician who had been one of Einstein's professors in college). See either of the two books for much more about the motivation for this "geometrization" of gravity. A keyword for the "special feature" is "Equivalence principle".

That's the idea. But note we now need to deal with coordinate charts in curved spacetimes, and these can be hard to interpret geometrically, so mathematicians developed various tools to help physicists keep from getting confused by features of the coordinate representation by computing "geometric features" which do not depend upon the coordinates chosen to describe the situation.

BTW, this is not part of gtr but part of the theory of Riemannian or Lorentzian manifolds. Gtr uses Lorentzian manifolds both as the geometrical setting for non-gravitational physics and as its model of all gravitational phenomena.

Right, this goes by "divergence" or "convergence" of initially parallel geodesics. And again, this is not part of gtr but part of the theory of Riemannian or Lorentzian manifolds.

Right, but better say "bending in the large" because in gtr, light rays are in a sense always "straight in the small" (see below). This also explains "gravitational red shift" of signals sent from the surface of a massive isolated object to a distant observer; this is two of the four classical solar system tests of gtr (or if you prefer, of the geometrization of gravitation; other "metric theories of gravitation" predict the same gravitional red shift as gr.

The books I mentioned explain that while Newton's theory of gravitation (in its classical field theory form) does posit a gravitational potential whose gradient gives the direction and magnitude of the acceleration of small objects (independent of their mass, unlike the Lorentz force in EM which depends on the small object's electrical charge), gtr treats gravitation quite differentially, as the curvature of spacetime.

I assume you know that in str, "space" and "time" no longer exist by themselves. Rather, the kinematic (motion) history of all the objects participating in the drama of physics are woven into a geometric picture called "spacetime", in which the "life" of each object is represented by a curve, called its "world line", in a four-dimensional Lorentzian manifold called Minkowski spacetime, which has a distance formula which looks very similar to the Pythagorean theorem but has some critical and easily overlooked differences.

In str, nonaccelerating objects, i.e. those in inertial motion, have world lines which are straight. OTH, if a small object is subjected to acceleration, its world line is bent near the appropriate "events" (points in spacetime), with the amount of bending (path curvature) giving the direction and magnitude of the acceleration. Gtr incorporates this without change, expect that we use the mathematical machinery of the theory of differentiable curves in Lorentzian manifolds.

Then one can say that a free-falling small object near an isolated massive object will have a world line which is a "timelike geodesic" in the spacetime used to model all gravitational phenomena in this scenario. This geodesic will in general be subject to "bending" much like light bending, and it turns out that according to our small object will travel in a quasi-Keplerian orbit, if it has nonzero "orbital angular momentum" wrt the massive object, or else fall straight in. Much as in Newtonian gravitation, but the geometric representation of physics was quite novel in 1915.

No, no, that's nonsense. I don't know why so many PF posters have mentioned the false notion that "space itself" can "accelerate" or "expand" recently.

In any Riemannian or Lorentzian manifold, curvature of the manifold (nonzero Riemann tensor) causes geodesics to converge or diverge and leads to a kind of "bending in the large" phenomena. Also, "bending in the small" means a curve is nongeodesic and then the amount of bending is measured by path curvature. This is quite independent of gtr, but is incorporated into gtr.

In gtr, a gravitational field is more or less identified with the Riemann curvature tensor. This is the mathematical object which in various ways represents, near a given point, how distorted a Lorentzian manifold is at that point from a flat manifold.

3. Nov 6, 2007

### Yianni

Thanks for that, it clarified a lot.

The reason I mention the idea that space itself is moving was because in Brian Greene's book The Fabric of the Cosmos he suggests that if the universe is expanding uniformly on the large scale, then no time dilation should occur between people on distant galaxies assuming their velocities relative to one another are zero other than the velocity caused by the expansion of the universe; i.e. they look at each other and their watches remain in sync. I suppose I mentioned it because I thought the idea behind 'spacetime curvature' was that if everything in a region is accelerating uniformly, then instead of describing those objects as accelerating, you described the space as such. I.e. the space is permeated by theoretical clocks and rulers at every real number along each of the three spatial axis, which are accelerated by gravity, and this is called space. I suppose that idea was just plain wrong though!

I'm only just finishing High School but my maths is pretty decent (I don't mean I actually know that much, I live in Australia and the standard of maths in schools here is quite appalling, but I can pick up on ideas pretty easily). Could you recommend any textbooks for learning both SR and then GTR? Or one for learning SR and then one for at least some sort of light introduction to the maths of GTR? I'm learning some of the ideas in SR from the Feynman lectures, but I think I really need a textbook to get into the nitty-gritty of it. Also, to learn GTR, what maths would be necessary before even opening up the textbook? Obviously some level of calculus and vector maths, but what else? And would the textbook itself explain the "mathematical machinery of the theory of differentiable curves in Lorentzian manifolds," or would this need to be learned before beginning?

Again, thanks for your time... now I better stop wasting my own time and start studying for my exams again :yuck:

4. Nov 7, 2007

### pmb_phy

Your assumptions are correct so far. The formula is related to the metric tensor. The metric you're speaking of the Euclidean metric which is defined as "the interval", which in this case is the spatiual distance, is

$$ds^2 = \sqrt[ (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2]$$

This formula is chosen such that the spatial distance between two points remains unchanged upon a change in coordinate system. In relativity the spacetime interval is

$$ds^2 = \sqrt[ (\Delta ct)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2]$$

That is correct. This phenomena is called geodesic deviation.
The correct term is curved, not bent. And curvature is a sufficient reason for the bending in the spatial path of light. But it is not neccesary. For example, the first derivation of light being deflected in a gravitational field was done in a flat spacetime in an accelerating frame of reference.
Sometimes is just easier to quote a text since the author, in this case, says it quite well. From Introduction to the Theory of Relativity, by Francis W. Sears and Robert W. Brehme. From page 193
Pete

Last edited by a moderator: Nov 9, 2007
5. Nov 7, 2007

### Chris Hillman

You're welcome!

Always good to hear!

Right, well I think of that scenario as a picture, but in words I'd put it like this: the world lines of the dust particles form a congruence (family) of timelike geodesics which fill up the spacetime model (an FRW dust solution of the Einstein field equation). Furthermore, there is a unique family of "spatial hyperslices" for these observers, Riemannian three-manifolds which are everywhere orthogonal to the world lines of the dust. (In general, if the dust were "swirling" it would not be possible to find a family of spatial hyperslices.) For concreteness, here is the "line element" of the FRW dust with E^3 hyperslices:
$$ds^2 = -dT^2 + T^{4/3} \, \left( dx^2 + dy^2 + dz^2 \right), \; 0 < T < \infty, \, -\infty < x, \, y, \, z < \infty$$
Visualizing this requires a four-dimensional figure, but since all spatial directions are equivalent in this model (we say it is "isotropic"), let's just set $z=0$. Then you can visualize the "map" (as in representation of a curved space like the surface of the Earth) --- or, to use the correct term, the "coordinate chart"--- defined by this line element as the upper half space $0 < T < \infty, \, -\infty < x, \, y <\infty$, with the plane $T=0$ representing the Big Bang. Then the world lines of the dust have the simple form $x=x_0, y=y_0$, i.e. "vertical half-lines", and the orthogonal slices have the form $T=T_0$, i.e. "horizontal planes". To study radio signals exchanged by the dust particles, we recall that the world line of a "photon" is a null geodesic, so we can set $ds=0$ in the line element, which defines the "distance" in a small piece of our coordinate chart. Here comes an example of why first year college calculus is fun and useful! From the line element we obtain
$$dT = T^{2/3} \, \sqrt{dx^2 + dy^2} = T^{2/3} \, dr$$
or $dr/dT = T^{-2/3}$, so that the "world sheets" formed by taking all the signals from some event on the world line of some dust particle have the form of a surface somewhat like a paraboloid, whose sides gets steeper as T grows. In flat spacetime the analgous surface would simply be a half-cone with constant slope unity, so the shape of these distorted light cones (as drawn in our coordinate chart) is a consequence of the fact that at the hyperslices for a later time $T_1 > T_0$, a coordinate difference $x_1-x_0$, where $y = y_0, \, T=T_1$, corresponds to a larger distance, $\Delta s = T_1^{2/3} \, \Delta x > T_0^{2/3} \, \Delta x$. This is the Hubble expansion: the dust particles move away from one another, but at a slower and slower rate, as time increases. Note that even though in our chart the world lines appear to be maintaining the same "horizontal distance", appearances are misleading, just as appearances can be misleading in a map of the Earth which typically misrepresents distances.

There is a very nice picture of the "distorted light cones" described above in the excellent popular book The First Three Minutes by Steven Weinberg, BTW. It is quite possible to understand the geometry even before studying differential calculus if one properly interprets such a picture!

The point is that this is coordinate chart is said to be "comoving with the dust particles" in that the coordinate planes $T=T_0$ correspond to the spatial hyperslices orthogonal to the world lines of the dust particles, and also differences $\Delta T$ taken along the world line of a dust particle correspond to intervals of proper time as measured by an ideal clock carried by that dust particle. Indeed, we can imagine that observers riding on a small group of nearby dust particles synchronize their clocks by exchanging light signals, and although this gets a bit tricky, it is clear enough that a successful synchronization would yield values of the coordinate T. (Abstractly, a coordinate is just a monotonic function on some manifold, i.e. which has nonzero gradient. A choice of such functions on some "neighborhood", such that their gradients are nowhere parallel, gives a local coordinate chart defined on that neighborhood.)

If we imagine "time running backwards", then as $T \rightarrow 0$ "from above" we see that the distance between any pair of dust particles (that is, the "horizontal distance" between their world lines) must decrease to zero, and if we compute the density of dust from the Einstein tensor in this solution, we find $\rho = \frac{1}{6 \, \pi \, T^2}$ showing that indeed the density blows up as $T \rightarrow 0$. So we can even say that T is a kind of "universal time" for the dust particles in which "time zero" corresponds to the "Big Bang".

The hyperslices $T=T_0$ in this model have the line element (put dT=0 in the spacetime line element)
$$ds^2 = T_0^{4/3} \, \left( dx^2 + dy^2 + dz^2 \right), \; -\infty < x, \, y, \, z < \infty$$
which is the line element of euclidean three-space, so we can say that the hyperslices in this FRW model are "locally flat", meaning locally isometric to euclidean three-space, $E^3$. In fact, they are globally isometric to euclidean three-space in the model I have described, but with a seemingly simple change (identify the "vertical plane" $x=1$ with $x=0$) we obtain a model with spatial hyperslices which are locally flat but globally isometric to "cylinders". We could also have hyperslices which are locally flat but isometric to "flat tori". If this intrigues you, see Jeffrey Weeks, The Shape of Space.

This cosmological model is highly idealized: the dust particles roughly correspond to galaxies, but of course real galaxies are clumped into "clusters", there are large "voids" free of galaxies, and so on, and the galaxies are not really "pointlike" but have complex and interesting substructures (such as our Solar system). Nonetheless this simple model succesfully reproduces the basic features of the observed Hubble expansion (it does even better at very large distances if we add a small "Lambda terms" to the right hand side of the Einstein field equation to the term representing the dust). It is worth mentioning that it is possible to construct more sophisticated cosmological models which are "perturbations" of our FRW dust model but which attempt to model inhomogeneities in the distribution of galaxies or to otherwise improve on the FRW dust models.

Well, there might be something correct lurking in there, but I hope the above helps in appreciating how specialists in gravitation physics tend to think of this kind of scenario.

Can't be worse than (generic) American schools! (But that's a topic for general discussion). I have the impression you can probably get the right "picture" from the right books.

Linear algebra would be very very useful. Gtr uses the languages of tensors, which are slight generalizations of the "linear operators" of linear algebra. Also, linear operators can be represented by matrices, but this representation depends upon a choice of "basis", which sets up the idea of tensor fields as something you can write down in terms of some choice of coordinates, and also the notion of "orthonormal basis" sets up the idea of "frame fields"

I think the textbook by D'Inverno, Understanding Einstein's Relativity, might be ideal for you; coupled with Feynman's Lectures it just might give you just enough str to get started on gtr. (Maybe not, since I just remembered that Feynman's discussion of Minkowski geometry is a bit murky.) Last I knew, this was available in paper back new for perhaps U.S. $40.00 Alternatively, you could get Edwin F. Taylor and John Archibald Wheeler, Spacetime Phyhsics, Freeman, 1992 (make sure to make a table of the trigonometry appropriate to Minkowski geometry, called "hyperbolic trig" and to compare it with standard high school trig, or "elliptical trig"), and L.P. Hughston and K.P. Tod, An introduction to general relativity, Cambridge University Press, 1990, which cost about U.S.$25.00 new, last I checked. All these books will no doubt be available more cheaply used from InterNet booksellers.

Both the gtr textbooks I mentioned aim to be "self-contained" relative to a standard undergraduate curriculum. I would expect that even a bright high school student with some calculus and some linear algebra will find some topics too challenging at first, but you could still learn a very great deal, I think!

(Feynman's Lectures are a great way to learn physics, BTW! I also really like Blandford and Thorne, Applications of Classical Physics, which is available free at www.pma.caltech.edu/Courses/ph136/yr2004/index.html The authors are leaders in physics; Thorne and Wheeler are also coauthors with Misner of the classic graduate level textbook Gravitation, which you can save for later. And I am very glad to see a high school student who appreciates how much better standard textbooks are for learning topic T than arbitrary websites, which tend to be full of misinformation--- the Cal Tech website I just mentioned being a notable exception!)

Last edited: Nov 7, 2007
6. Nov 7, 2007

7. Nov 7, 2007

### pmb_phy

Even in the presence of a flat spacetime one can change his frame of reference to one in which there is no gravitational field into one for which a gravitational field is present. Things will fall in this field, light will be bent in such a field etc. However there will be no spacetime curvature in such a field. An example of such a field is a uniform gravitational field which is spoken of in Eintein's Equivalence Principle which states

A uniformly accerating frame of reference is identical to (i.e. has the same metric as) a uniform gravitational field.

Pete

8. Nov 7, 2007

### pervect

Staff Emeritus
It's probably worth pointing out (this has been mentioned once or twice in past threads) that space-time curvature has a simple physical interpretation as the presence of tidal forces. The most direct connection is that the tidal forces experienced by an inertial observer are identical to certain components of the Riemann curvature tensor, the abstract mathematical entity which represents curvature in GR (and in Riemannian geometry).

9. Nov 7, 2007

### pmb_phy

Yep. That's what I explained in my new web page - http://www.geocities.com/physics_world/gr/gravity_vs_curvature.htm

I quoted Kip Thorne who said "spacetime curvature and tidal gravity must be precisely the same thing, expressed in different languages."

Pete

10. Nov 8, 2007

### Chris Hillman

Pedantic elaboration of what pervect said

Right, but having said this, to avoid possible misunderstanding, for the benefit of "intermediate students" of gtr, someone should perhaps stress that not all of the components can be identified with the "electric part" and thus with the relativisitic analog of the tidal tensor from Newtonian gravitation. There is also a "magnetic" part, which has no Newtonian analog and which represents (tiny!) spin-spin accelerations on spinning test particles moving in an ambient gravitational field with nonzero "magnetogravitic tensor". Having mentioned "gravity" and "magetism" in the same paragraph, someone should probably say that what I just said refers to a "strong field" (fully nonlinear) machinery analgous to the well-known weak-field machinery known as "gravitomagnetism", or more properly "gravitoelectromagnetism" (GEM); this gets a bit tricky so I won't try to say more. For some old Wikipedia articles see
• tidal tensor in Newtonian gravitation; the gtr analog is the electrogravitic tensor,
• Bel decomposition, which is observer dependent and decomposes the Riemann tensor into three three-dimensional spatial tensors (electrogravitc, magnetogravitic, topogravitic tensors, which have various interesting geometrical and physical interpretations), and which is analogous to the famililar decomposition of the EM field tensor, with respect to some timelike congruence, into two three-dimensional spatial vector fields, the electric and magnetic vector fields,
• Ricci decomposition, which is observer independent and decomposes Riemann tensor into Weyl tensor (completely traceless part) plus a piece built from Ricci tensor (once-detraced part) plus a piece built from the Ricci scalar (scalar part).
I stress that the Bel and Ricci decompositions are purely mathematical and make sense outside the context of physics. However, the physical interpretation of the tensors in question does make use of the EFE. (I cited specific versions of three articles which I read, in fact wrote before I left WP, and thus which I consider to be reasonably reliable (if too sketchy to be truly useful!); more recent versions at any given moment could be much worse or possibly better than the ones I cited! That's just in the nature of the Wikipedia beast; don't assume that WP generally is a reliable source of information; it is not.)

Last edited: Nov 8, 2007
11. Nov 8, 2007

### Chris Hillman

Frame fields, anyone?

Hi, Yianni,

Just noticed something I previously overlooked: with a few changes, this sounds very much like you might be about to independently rediscover the idea of a frame field. That would be cool!

Last edited: Nov 8, 2007
12. Nov 9, 2007

### A.T.

Actually in the simple space time diagram there is a slider "gravity" that creates a uniform gravitational field. Even when you set the initial speed to zero, the object will "accelerate" in space, although it's has a geodesic path in a flat spacetime. The key is that the cone-like spacetime has an inhomogeneous metric (causing gravity), but still can be unrolled in 2D without distortion (so it has no intrinsic curvature)

13. Nov 9, 2007

### Chris Hillman

Some Instructive Exercises on Charts and Frames

"Uniform gravitational field" is a bit tricky in gtr!

Inspired by the EP, many authors take this to mean the Rindler congruence with the following frame field on a Rindler wedge region in Minkowski vacuum:
$$\vec{f}_1 = \frac{x}{\sqrt{x^2-t^2}} \, \partial_t + \frac{t}{\sqrt{x^2-t^2}} \, \partial_x, \; \vec{f}_2 = \frac{t}{\sqrt{x^2-t^2}} \, \partial_t + \frac{x}{\sqrt{x^2-t^2}} \, \partial_x,$$
$$\vec{f}_3 = \partial_y, \vec{f}_4 = \partial_z,$$
$$|t| < x < \infty, \; -\infty < t, \, y, \, z < \infty$$
Here, the first vector field is a timelike unit vector field with path curvature (physically speaking, acceleration vector) $\frac{1}{\sqrt{x^2-t^2}} \vec{f}_2$. Note that this is "uniform" wrt t,y,z but not x.

To understand this better, it is convenient to carry out the coordinate transformation
$$T = \operatorname{arctanh}(t/x), \; X = \sqrt{x^2-t^2}, \; Y = y, \; Z =z$$
which gives the Rindler chart
$$ds^2 = -X^2 \, dT^2 + dX^2 + dY^2 + dZ^2, \; 0 < X < \infty, \; -\infty < T, Y, Z < \infty$$
The Rindler frame field becomes
$$\vec{f}_1 = \frac{1}{X} \; \partial_T, \; \vec{f}_2 = \partial_X, \; \vec{f}_3 = \partial_Y, \; \vec{f}_4 = \partial_Z$$
and the acceleration vector becomes simply $\frac{1}{X} \vec{f}_2$. Same spacetime, same metric tensor, same frame field, same acceleration vector, just written in a new local coordinate chart!

Exercise: work out the geodesic equations, solve for null geodesics, and find the simple geometric characterization of their appearance in the Rindler chart. If you are familiar with the notion of the Fermat metric, compute the Fermat metric for null geodesics in the Rindler chart and recognize them as the geodesics of a famous model of a certain noneuclidean geometry.

Exercise: draw a diagram showing the world lines of two Rindler observers in the Rindler chart, with coordinates $X=X_1, Y=Y_0, Z=Z_0$ and $X=X_2, Y=Y_0, Z=Z_0$ where $X_2 > X_1 > 0$, and draw some null geodesics representing the world lines of radar pips sent out by one and reflected from the other. What is the "radar distance" between these obsvers? Is this notion of "distance in the large" symmetric between the two observers? How does the value of radar distance depend upon $h=X_2-X_1$? How does it compare with "ruler distance"? Can you use the preceding exercise to concoct a third notion of "distance in the large"? Note that this exercise shows that even in flat spacetime, there are multiple operationally significant notions of "distance in the large" and thus of "velocity in the large".

Exercise: The Rindler congruence has acceleration which is constant for each observer in the family, but which varies with X between different observers in the family. Show that the expansion tensor of this congruence vanishes identically, so that the Rindler observers exhibit rigid motion. Can you find a congruence of observers who are also accelerating in the $\partial_x$ direction but which has constant acceleration over the entire family? (This is the Bell congruence.) What is the expansion tensor of this congruence? What can you conclude about a string stretched between two Bell observers with aligned accelerations? Show that both the Rindler and Bell congruences are vorticity free and thus define orthogonal hyperslices. Show these are distinct slicings but that both are locally flat. Can you find a chart which is comoving with the Bell observers in the same way that the Rindler chart is comoving with the Rindler observers? What can you say about the properties of your chart?

Exercise: Read about Weyl's family of all static axisymmetric vacuum solutions of the EFE , e.g. in this review paper. Note that the master functions are static axisymmetic harmonic functions when we write the solution in terms of the Weyl canonical chart
$$ds^2 = -\exp(-2 \, u(z,r)) \, dt^2 + \exp(2\, u(z,r)) \; \left( \exp(- 2\, v(z,r)) \left( dz^2 + dr^2 \right) + r^2 \, d\phi^2 \right),$$
$$u_{zz} + u_{rr} + \frac{u_r}{r} = 0, \; v_z = 2 \, r \, u_z \, u_r, \; v_r = r \left( u_r^2 - u_z^2 \right)$$
Find a transformation from the Rindler chart to the following example of Weyl canonical chart
$$ds^2 = -\left( z + \sqrt{z^2+r^2} \right) \, d\overline{t}^2 + \frac{dz^2+dr^2}{2 \,\sqrt{z^2+r^2}} + \frac{r^2 \, d\phi^2}{z + \sqrt{z^2+r^2}},$$
$$-\infty < \overline{t}, \, z < \infty, \; 0 < r < \infty, \; -\pi < \phi < \pi$$
and express the Rindler frame in this chart. What region of Minkowski spacetime is covered by this chart? What locus in the Rindler chart corresponds to the axis r=0 of this Weyl canonical chart? Can you think of an alternative Weyl canonical chart which also represents a piece of Minkowski spacetime? Can you think of a third alternative?

Exercise: in Newtonian physics, the potential $u(z,r) = a \, z$ is static axisymmetric harmonic and thus defines a Weyl vacuum solution. Find the line element written in the Weyl canonical chart and compute the curvature. Note that this is not locally flat, but is it approximately so? Explain. What do you notice about the behavior of the acceleration vector of static test particles in this chart, as you vary their z coordinate? Does this behavior agree with Newtonian expectation?

Exercise: Recall that in gtr, the "gravitational field" is identified with the Riemann curvature tensor. In a static spacetime, this comes down to saying that the "gravitational field" is identified with the electrogravitic tensor (from the Bel decomposition, evaluated wrt some timelike congruence, of the Riemann tensor), the relativistic analog of the Newtonian tidal tensor. Compute the tidal tensor for a Newtonian uniform gravitational field. Can you find an exact vacuum solution of the EFE such that there exists a family of inertial observers whose electrogravitic tensor has the same geometric properties? What if you add a Lambda term to the RHS of the EFE? Is your solution static? Are the hyperslices orthogonal to your (vorticity-free) timelike geodesic congruence locally flat?

Last edited: Nov 9, 2007
14. Nov 9, 2007

### pmb_phy

I never heard of the term "slider gravity". Can you please define it or quote a reference. Thank you. The rest I agree with.

Best regards