The Arc of Space Curvature: Large and Small

[Ahmed Samra]

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

 

[mfb]

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.

 

[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?

 

[mfb]

At the same distance: right.

 

[Ahmed Samra]

What do you mean by at the same distance?

 

[Boy@n]

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?

Would you expect it to be in reverse?

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

 

[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.

 

[mfb]

What do you mean by at the same distance?

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

In addition, see the post above this one.

 

[Ahmed Samra]

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

 

[WannabeNewton]

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

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 $R^{abcd}R_{abcd} = \frac{48G^{2}M^{2}}{c^{4}r^{6}}$.

 

[Ahmed Samra]

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

 

[Naty1]

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

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...

 

[WannabeNewton]

(SPACE curvature is not coordinate-free; a change of coordinates makes space flat

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

 

[pervect]

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.

 

[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.

 

[pervect]

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.

Suppose our space-time is the exterior Schwarzschild 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).

IF such a foliation exists, there will be a unit normal field $n^{a}$ to this family

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

and there will be a spatial metric $h_{ab}$ induced on each hypersurface given by $h_{ab} = g_{ab} + n_{a}n_{b}$.

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

correct?

Each hypersurface has an extrinsic curvature given by $K_{ab} = h_{a}{}{}^{c}\nabla_{c}n_{b}$.

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...

 

[WannabeNewton]

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

It isn't confusing at all

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

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

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

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

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...

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.

 

[pervect]

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 $\approx 2m / r^3$.

 

[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.

 

[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]:

...nowhere has a precise definition of the term “gravitational field” been given --- nor will one be given. Many different mathematical entities are associated with gravitation; the metric, the Riemann curvature tensor, the curvature scalar … Each of these plays an important role in gravitation theory, and none is so much more central than the others that it deserves the name “gravitational field.”

wannabe says:

Quote by Naty1
(SPACE curvature is not coordinate-free; a change of coordinates makes space flat
What notion of "space curvature" are you using that allows you to make such claims? How are you even defining "space curvature"?

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?

https://www.physicsforums.com/showthread.php?t=596224

 

[WannabeNewton]

I never thought such a statement was anything but simple...

The problem is that the notion of "space curvature" is ill-defined as mentioned above. If you have some reference as to some standard definition for "space curvature" for arbitrary space-times then I would much appreciate it but if you mean the curvature of members of a space-like foliation of space-time then that has been discussed above and only applies to space-times where such a foliation exists.

 

[PeterDonis]

if you mean the curvature of members of a space-like foliation of space-time then that has been discussed above and only applies to space-times where such a foliation exists.

It's also worth noting that if more than one such foliation exists, different foliations can have different curvatures for their spacelike hypersurfaces.

*And* it's worth noting that at least one example of "space curvature" that's often discussed, the curvature of surfaces of constant Schwarzschild coordinate time in the exterior region of Schwarzschild spacetime, is not associated with a foliation; surfaces of constant Schwarzschild coordinate time don't foliate Schwarzschild spacetime (because they don't cover the interior region).

 

[WannabeNewton]

Indeed, so as Peter explained and gave examples of, "space curvature" can have so many meanings that it's hard to make general statements by just saying "space curvature".

 

[PeterDonis]

Indeed, so as Peter explained and gave examples of, "space curvature" can have so many meanings that it's hard to make general statements by just saying "space curvature".

Yes, and just to throw yet another distinction into the foodmixer , I believe the quantity $K_{ab}$ that you mentioned, from Wald, is related to the extrinsic curvature--i.e., how a particular spacelike hypersurface curves when it's considered as a submanifold embedded in the 4-D spacetime manifold. (My initial guess is that $K_{ab}$ *is* the extrinsic curvature, but that's just from memory and I haven't checked the relevant passages in Wald.) But there is also intrinsic curvature--basically, what you get if you compute the Riemann curvature tensor of $h_{ab}$ as a 3-metric on the 3-D submanifold defined by a particular spacelike hypersurface.

(To do this, I believe you would have to adopt a coordinate chart which makes the "time" components of $h_{ab}$, considered as a 4-D tensor defined by your equation, $h_{ab} = g_{ab} + n_a n_b$, zero. Then $h_{ab}$ as a 3-metric is just the restriction of $h_{ab}$ as a 4-tensor to the particular spacelike hypersurface, i.e., you just take the nonzero "space-space" components in the "obvious" way to form a 3-tensor. I believe this can be done for any spacetime that can be foliated by spacelike hypersurfaces as you describe, so I don't think there's any loss of generality--i.e., an intrinsic curvature can be defined in this way for any spacelike hypersurface that is part of a foliation.)

 

[WannabeNewton]

Yes Peter every word of what you said is correct and if you do end up looking at the relevant sections of Wald, he will basically describe things in a similar way. Most of it is condensed into the chapters on the initial value formulation and singularities as well as the appendix on the Hamiltonian formulation.

 

[pervect]

Yes, the intrinsic curvature is what I'm after, not the extrinsic curvature.

Rather than talk about $K_{ab}$ let's talk about the topogravitic tensor.

According to Wiki, all we need to define that is a unit timelike vector field.

http://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=544328881

Wald doesn't have much on the Bel decomposition, alas.

The topogravitic tensor should represent the "spatial" parts of the Riemann curvature tensor.

To take a particular example. Let $\hat{t}, \hat{x}, \hat{y}, \hat{z}$ be unit vector fields, and let us use $\hat{t}$ as the required vector field to perform the Bel decomposition.

Then $L[\hat{t}]_{\hat{z}\hat{z}}$ should be the component of the topogravitic tensor that's numerically equal to $R_{\hat{x}\hat{y}\hat{x}\hat{y}}$, as the double dual should be $*R*_{\hat{t}\hat{z}\hat{t}\hat{z}}$, and we contract by $\hat{t}$

I have some more thoughts, but I think this is enough for one post.

 

[WannabeNewton]

Yes, and just to throw yet another distinction into the foodmixer , I believe the quantity $K_{ab}$ that you mentioned, from Wald, is related to the extrinsic curvature--

By the way, just to clarify, I did state that $K_{ab}$ was the extrinsic curvature in post #15 when I first mentioned it.

Also, just to tie into what both you and Pervect said, in the case where we do have a one-parameter family of space-like hypersurfaces $\Sigma_{t}$ foliating the space-time, we can look at the instrinsic curvature of each $\Sigma_{t = \text{const.}}$ by calculating the riemann curvature $^{(3)}R^{a}_{bcd}$ associated with $h_{ab}$. We can calculate $^{(3)}R^{a}_{bcd}$ in the same way as usual i.e. for any one-form $\omega_{a}$ on $\Sigma_{t}$ we have $D_{a}D_{b}\omega_{c} - D_{b}D_{a}\omega_{c} = ^{(3)}R_{abc}{}{}^{d}\omega_{d}$ where $D_{a}\omega_{b} = h^{c}{}{}_{a}h^{d}{}{}_{b}\nabla_{c}\omega_{d}$.

Interestingly enough, one can take that relation and show that $^{(3)}R_{abc}^{d} = h_{a}{}{}^{f}h_{b}{}{}^{g}h_{c}{}{}^{k}h_{j}{}{}^{d}R_{fgk}{}{}^{j} - K_{ac}K_{b}{}{}^{d} + K_{bc}K_{a}{}{}^{d}$ so this gives us an equation relating the extrinsic curvature of each space-like hypersurface to its riemann curvature as well as the riemann curvature of space-time.

Pervect that is a nice decomposition indeed! Do you know of any book(s) that go into more detail on it's geometrical interpretations for the most physically relevant space-times?

 

[pervect]

Pervect that is a nice decomposition indeed! Do you know of any book(s) that go into more detail on it's geometrical interpretations for the most physically relevant space-times?

Alas, I don't. Bill_K might know more, see https://www.physicsforums.com/showthread.php?t=482224

The other person I know who would probably know isn't around anymore (at least not under the name I know him by).

Wiki has a reference to the original paper. Which is in French.

 

[WannabeNewton]

Thank you so much Pervect, I'll take a look at what you linked in detail. I didn't know that this decomposition had any relation to the Bel-Robinson tensor $T_{abcd}$ but according to a paper linked in that thread, it apparently does. Wald does mention the Bel-Robinson tensor in passing as part of an end of chapter problem but doesn't really explain much of its physical significance. He just has you show that it is traceless $T^{a}{}{}_{abcd} = 0$ and that in vacuum it is divergence free $\nabla^{a}T_{abcd} = 0$. I'll dig around with what you linked, thanks again!

 

[pervect]

There's a short bit in MTW - it doesn't give the components names, but it breaks them down in the same way. See exercise 14.14 on pg 360. The 3x3 submatrixes are called E,F, and H

 

[WannabeNewton]

There's a short bit in MTW - it doesn't give the components names, but it breaks them down in the same way. See exercise 14.14 on pg 360. The 3x3 submatrixes are called E,F, and H

Thank you very much for looking that up pervect. I'll take a look at it as soon as I can. Cheers!