# Space curvature

1. May 1, 2013

### Ahmed Samra

When does the arc of the space curvature is large and when is it small?

2. May 1, 2013

### Staff: Mentor

What do you mean with "large/small arc"?
Large masses nearby lead to a large curvature. I don't remember any "arc" used to describe this.

3. May 1, 2013

### Ahmed Samra

I read it in the Internet. Anyway, large masses lead to a large space curvature while small masses lead to a small space curvature right?

4. May 1, 2013

### Staff: Mentor

At the same distance: right.

5. May 1, 2013

### Ahmed Samra

What do you mean by at the same distance?

6. May 1, 2013

### Boy@n

Would you expect it to be in reverse?

I wonder though, is spacetime curving/defining just 3 dimensions?

7. May 1, 2013

### WannabeNewton

I assume you mean space-time curvature, not space curvature. The two are different in general. Anyways, it is not that simple. Remember that in general relativity, the Ricci tensor and Ricci scalar curvature of space-time are related to the energy-momentum tensor so just knowing the mass density is not enough to determine if one mass-energy distribution would generate more curvature than another (if we are looking at a vacuum solution then the curvature is encoded by the Weyl tensor). In the Newtonian limit, the mass density component dominates the other components but this first order approximation does not hold in the general theory.

8. May 1, 2013

### Staff: Mentor

It does matter if you are 1m away from a mass or 1000km.

In addition, see the post above this one.

9. May 1, 2013

### Ahmed Samra

Is there a formula to know how large or small is the space-time curvature? And what is the formula?

10. May 1, 2013

### WannabeNewton

You could, for example, compute the Riemann curvature tensor $R^{a}_{bcd}$ in the coordinate basis using the formulas here:http://en.wikipedia.org/wiki/Riemann_curvature_tensor#Coordinate_expression and then form a curvature scalar such as $R^{abcd}R_{abcd}$. If you do this for the Schwarzschild metric, you will get $R^{abcd}R_{abcd} = \frac{48G^{2}M^{2}}{c^{4}r^{6}}$.

11. May 1, 2013

### Ahmed Samra

Does Jupiter have a larger space-time curvature than earth?

12. May 1, 2013

### Naty1

I don't especially like some of the above answers because mass density is important....If you have a 'ton' of hydrogen gas' versus a 'ton' of lead, you'll get different 'arcs'...that is curvatures,because the gas is spread out, maybe over miles....One of the interesting aspects of 'curvature is that most of celestial [big mass and slow speeds] seems to be curvature of TIME more than that of space.

Spatial curvature is a co-ordinate [observer] dependent property of a given space-time. How fast you move affects what you observe. (SPACE curvature is not coordinate-free; a change of coordinates makes space flat; the only coordinate-free [observer independent] curvature is space–time curvature, which is related to the local mass–energy density or really stress–energy tensor. In other words, spacetime curvature [a gravitational phenomena] is independent of observer motion...how fast you move does NOT affect THAT kind of 'curvature'.

What 'curvature' is....believe it or not.... is NOT generally agreed upon because there are multiple measures of curvature...two are the Ricci and Riemann measures mentioned above and none is agreed upon as 'gravitational curvature'...that is the 'gravitational field' 'g' is not a well defined thing...Curvature is a complex thing mathematically...some like the Christoffel symbol instead as a 'better' measure....

13. May 1, 2013

### WannabeNewton

What notion of "space curvature" are you using that allows you to make such claims? How are you even defining "space curvature"?

14. May 1, 2013

### pervect

Staff Emeritus
Sometimes people are interested in space curvature and not space-time curvature. Some coordinate dependence is implied in the concept of spatial curvature, one needs to define by some means what the surfaces of "constant time' are.

Given such a space-time split:

I'd expect that there should be a similar relation between the angular excess of a polygon drawn in such a space-slice to its area, simliar to the relation on the surface of the earth:

http://en.wikipedia.org/wiki/Angle_excess

angular excess / area = 1/ R^2, where R is some "sectional curvature". A large R implies little curvature. I believe this would be the Gaussian curvature aka sectional curvature.

http://en.wikipedia.org/wiki/Sectional_curvature

In general, this curvature could depend on orientation, I would expect though that if one chose isotropic coordinates, the curvature would be independent of the orientation.

Going to the wiki:

In particular, if ''u'' and ''v'' are ortonormal [[i.e: two orthonormal vectors that define the plane]]

$$K = \langle R(u,v)v,u\rangle$$

Unfortunately, I can't quite figure out the notation they're using for the Riemann to actually do the calculation in isotropic coordinates and confirm that it's independent of the choice of spatial plane.

I think the above might mean $R_{\hat{u}\hat{v}\hat{v}\hat{u}}$ in the notation I'm used to, but I'm not really sure.

15. May 1, 2013

### WannabeNewton

If "space curvature" is being used to refer to the induced curvature on a one-parameter family of space-like hypersurfaces that foliate the spacetime, then the very notion of "space curvature" is utterly meaningless unless such a foliation exists.

IF such a foliation exists, there will be a unit normal field $n^{a}$ to this family and there will be a spatial metric $h_{ab}$ induced on each hypersurface given by $h_{ab} = g_{ab} + n_{a}n_{b}$. Each hypersurface has an extrinsic curvature given by $K_{ab} = h_{a}{}{}^{c}\nabla_{c}n_{b}$. We can relate $K_{ab}$ to the Ricci curvature $R_{ab}$ of space-time using the Gauss-Codacci equations $D_{a}K^{a}{}{}_{b} - D_{b}K^{a}{}{}_{a} = R_{cd}n^{d}h^{c}{}{}_{b}$ where $D_{a}$ is the derivative operator associated with $h_{ab}$ and can be related to $\nabla_{a}$ by $D_{c}K_{ab} = h^{d}{}{}_{c}h^{e}{}{}_{a}h^{f}{}{}_{b}\nabla_{d}K_{ef}$. The Gauss-Codacci equations provide a geometric relation between the extrinsic curvatures of the space-like hypersurfaces to the Ricci curvature of space-time. There is also, of course, the intrinsic Riemann curvature $^{(3)}R^{a}_{bcd}$ of the space-like hypersurfaces given by $h_{ab}$.

For the Robertson-Walker cosmological model, there does exist such a foliation of space-time by a one-parameter family of space-like hypersurfaces with each member of the family having constant sectional curvature $K$ related to its Riemann curvature by $^{(3)}R_{abcd} = Kh_{c[a}h_{b]d}$. This is again a purely geometrical statement. To say such things are "coordinate-dependent" statements is not accurate.

16. May 1, 2013

### pervect

Staff Emeritus
Suppose our space-time is the exterior Schwarzchild space-time using the isotropic chart,

$$g_{ab} = \left(\frac{1-m/2r}{1+m/2r}\right)^2 dt^2 + \left( 1 + \frac{m}{2r} \right)^4 \left(dx^2 + dy^2 + dz^2 \right)$$

We could paramaterize the space part differently of course, but for definiteness I'm going to paramaterize it this way for now.

[add]r is a function of x,y,z, r(x,y,z) = sqrt(x^2+y^2+z^2)

Foilate this space-time by the parameter t, to separate it into spacelike hypersurfaces (a one-parameter foilation).

Such a foilation exists in this case, and I would call that normal vector field $\hat{t}$, I hope the notation isn't confusing.

That would be, in this example
$$h_{ab} = \left( 1 + \frac{m}{2r} \right)^4 \left(dx^2 + dy^2 + dz^2 \right)$$

correct?

is
$$h_{a}{}^{c} = g^{bc} h_{ab}$$
?
or is it
$$h_{a}{}^{c} = h^{bc} h_{ab}$$
??

I think I'm beginning to loose it here, $\nabla_{c}n_{b}$ looks like it should be 4-d, though, and $K_{ab}$ should be 3-d...

Last edited: May 1, 2013
17. May 1, 2013

### WannabeNewton

It isn't confusing at all

Yes indeed that would be the spatial metric on a single hypersurface of $t = const.$.

It is raised by $g^{ab}$ so the first one would the one to use.

I'm not exactly sure what you mean here but in $K_{\mu\nu} = h_{\mu}{}{}^{\alpha}\nabla_{\alpha}n_{\nu}$ if we fix a particular value of $t$ then we would get the extrinsic curvature of a single space-like hypersurface as represented in this coordinate system, but until then it is giving us a way of assigning the extrinsic curvature to any member of the family as $t$ varies. Note however that even though in this particular case we are evaluating all these things in a particular coordinate system, the original relations given were not coordinate dependent.

I think Wald explains these things much better than I could so, since if I recall correctly you have the text, you could take a look at the discussion from page 255 onwards. Cheers Pervect!

EDIT: Also remember to take a look at section E.2 of appendix E where he applies the notion of extrinsic curvature of members of a one-parameter family of space-like hypersurfaces to develop the Hamtilonian formalism. It is insanely awesome if I do say so myself.

Last edited: May 1, 2013
18. May 1, 2013

### pervect

Staff Emeritus
Looking at Wald, it seems that for the example $K_{ab} = 0$. Wald describes $K_{ab}$ as representing the rate of change of the spatial curvature with time - which isn't what I want. I want the curvature itself, not the rate of change with time.

I still think that $R_{\hat{x}\hat{y}\hat{x}\hat{y}}$ should represent the sectional curvature of the x-y plane in my example. But after simplification, I'm getting something like

$R_{\hat{x}\hat{y}\hat{x}\hat{y}} = \frac{ m \left( 3 r z^2 - r^3\right)}{\left( r+m/2 \right) ^6 }$

which is simple, but it's not just a function of r, because of the z dependence.

i.e when z=0, it is $\approx -m / r^3$, but when z >> x,y it is $\approx 2m / r^3$.

19. May 1, 2013

### WannabeNewton

Pervect, $K_{ab}$ measures the rate of change of the spatial metric $h_{ab}$ as one moves along the flow of the unit normal field i.e. $K_{ab} = \frac{1}{2}\mathcal{L}_{n}h_{ab}$ (this result can be proven by evaluating the lie derivative and using the fact that $n^{a}$ is hypersurface orthogonal i.e. $n_{[c}\nabla_{b}n_{a]} = 0$); it is not the rate of change of spatial curvature itself. In this sense, it measures how a space-like hypersurface $\Sigma$, embedded in the space-time $M$, "bends" in $M$. See page 230 of Wald.

20. May 2, 2013

### Naty1

Ahmed....

I'm not at the mathematical level of the several prior posts.....

Here is a quote I kept from a prior discussion in these forums...from a highly regarded textbook [Misner, Thorne, Wheeler]:

wannabe says:

I never thought such a statement was anything but simple....I simply mean, for example, if the ds interval of GR is invariant, how could the space component [by itself] not be?? Or another way I think about it, if there is observer dependent space contraction, how could practically any curvature escape that??

Ahmed: What I think is being asked might be illustrated this way..suppose I see a spherical mass in the rest frame of that mass...then I 'rocket' by at high speed relative to that same sphere...that sphere now appears as an obloid [is that a real word??]....anyway, flattened along my direction of motion, foreshortened a bit....via length contraction.....that's all I meant....

Ahmed:
if you REALLY want to get some detailed insights into curvature, check out this prior discussion....be warned...it IS long, but not heavily mathematical.....

Spacetime Curvature Observer and/or Coordinate Dependent?