Curvation of spacetime

[DParlevliet]

The moon is rotating around the earth because it follows the curvation of spacetime.

So at the moon, how large is the radius of this curvation?

[Strange]

I don't think the curvature can be described by a single (radius) value (because it is the curvature of a four-dimensional surface). It is described by one of the tensors in the Einstein Field Equation (the Ricci curvature tensor, I think).

[DParlevliet]

Then how large is the radius in a 2D plane, approximately?

I suppose it is in all planes about the same.

[studiot]

Excuse me, the radius of curvature exists in a dimension perpendicular to the space in which the curve exists.

 

Since you are talking about 3D space, in which dimension are you suggesting this radius exists?

[Strange]

Since you are talking about 3D space, in which dimension are you suggesting this radius exists?

 

Actually, 4D spacetime. But the curvature is "intrinsic" so it doesn't require a higher dimesnional embedding.

[studiot]

 

Actually, 4D spacetime. But the curvature is "intrinsic" so it doesn't require a higher dimesnional embedding.

 

 

 

Eh?

 

[Strange]

 

 

Eh?

 

 

"Our curved spacetime need not be embedded in some higher-dimensional flat spacetime for us to understand its curvature, or the concept of tangent vector. The mathematics of tensor calculus is designed to let us handle these concepts `intrinsically' -- i.e., working solely within the 4-dimensional spacetime in which we find ourselves. This is one reason tensor calculus is so important in general relativity."

http://math.ucr.edu/home/baez/einstein/node2.html

[studiot]

Yes we can create the more complicated maths on our curved spacetime manifold in a simialr fashion to surveyors creating geodetic maths on the surface of our planet.

 

I note this was abandoned as soon as satellite based GPS became generally available, because employing the higher dimension containg the curvature is easier to calculate in.

 

But that is just calculation, It does not answer the question does the dimension actualy have physical existance?

[DParlevliet]

Going back to the simplified case: I don't need an exact formule, but if our 3D space is curved it must have a radius which can given as a distance. The same way the rotation of the moon has a radius. I am interrested how large that is, approximately.

[Strange]

Going back to the simplified case: I don't need an exact formule, but if our 3D space is curved it must have a radius which can given as a distance.

 

It is not (just) space that is curved, but spacetime. IF it is possible to approximate this with a radius, then you would need four different values.

 

This paper calculates that the difference in the radius of the Earth due to the curvature of space-time is about 4mm

http://www.johnstonsarchive.net/relativity/stcurve.pdf

 

So perhaps you could approximate the radius of curvature as 6x10¹⁰ x radius of the earth or about 100,000,000,000,000 kilometres. I really don't know if that answer makes sense or not.

Edited October 28, 2015 by Strange

[studiot]

 

D Parlevliet Posted Today, 08:46 AM

Going back to the simplified case: I don't need an exact formule,(1) but if our 3D space is curved it must have a radius which can given as a distance. The same way the rotation of the moon has a radius. I am interrested how large that is, approximately.

 

Unfortunately you have not asked a simple question and both you and Strange have mixed up ideas about curvature.

So here is an extract from the linked website, which is about as simple as I can find.

 

I have underlined the relevant parts of the passages in both posts and added numbers to refer to them

 

 

 

Notions of curvature in geometry

Here will be explained the two notions of curvature involved in geometry, and needed in the description of our space-time by General Relativity :

• (2)The extrinsic curvature, for a curve or surface S contained (embedded) in some space E with higher dimension : it is the measure how S differs from being straight in E.
• (3)The intrinsic curvature, measuring how geometric properties of figures inside the given space, differ from those of a "flat geometry" (that contains affine geometry), (4)independently of any other space where it can be embedded.

The extrinsic curvature

When following a curve, its direction (tangent vector) can vary. The angle of deviation of this direction for an arc of a smooth curve S near a given point P, is usually proportional to its length, either exactly (in the case of an arc of circle), or approximately (the approximation getting better as the arc is smaller). The extrinsic curvature of S at P is defined as the ratio of this deviation (proportionality coefficient) :

Limit (for small arcs of S containing P), of (Angle of deviation (in radians) / Arc length).

The curvature of a circle in an Euclidean plane, is the inverse of its radius.

In a sphere, the "straight lines", curves with zero extrinsic curvature, are the "big circles" with the same center as the sphere. On the Earth approximated as a sphere, the equator and meridians are straight, but parallels (circles of latitude) are curved.

We have an intuitive understanding of the curvature of a curve in a plane, by taking the case of a road on a horizontal ground, that a car is following: the curvature of the road near each point is "measured" by the orientation of the wheel that is needed to follow it, and also by the sensation of lateral push when following it at a constant speed.

For a subspace S of E with dimension > 1, the extrinsic curvature of S in E is the measure of how a curve "straight inside S" is curved inside E. It is generally not a single quantity but a multidimensional object, as it gives a direction orthogonal to (the tangent subspace of) S as a (quadratic) function of the chosen direction of curve inside S (precisely, if n = dim S and m = dim E then the extrinsic curvature has n(n+1)(m−n)/2 coordinates). But it may be summed up as a single quantity when it is the same in all directions.

The Gaussian (intrinsic) curvature

The Gaussian curvature is the intrinsic curvature for the particular case of a surface (2-dimensional space), where it is a scalar field.

The Gaussian curvature can be thought of as a "density of angles", as follows.

In any surface (2-dimensional space, only approximately Euclidean in small scales), the sum of angles (in radians) of any triangle, may differ from π. For example, on Earth, a triangle made of a part of the equator and of 2 meridians, has two right angles, and the remaining angle (at the pole) can have any value.

More generally, the parallel transport along any closed curve induces a rotation with angle α, which in the case of a triangle coincides with this difference of angles:

α = (∑ angles) − π.

In the case of a (2-dimensional) sphere with radius r inside a 3-dimensional Euclidean space, this angle α is proportional to the area A of the surrounded region, according to the formula

α = r^-2A

The Gaussian curvature of this sphere is the proportionality coefficient R = r^-2. It equals the square of the extrinsic curvature (r^-1).

 

Spheres are a particular case of curved surface, having a constant Gaussian curvature. In the more general case, the Gaussian curvature of a surface is a (non-constant) field, and the rotation angle α induced by a parallel transport along a closed curve, is the integral of this field over the surrounded surface.

The Riemann curvature

(5)The Riemann curvature is the general case of intrinsic curvature, for spaces with any dimension (higher than 2). It is a field that is not scalar but multidimensional (described by several components when a coordinates system is given). For a 3-dimensional space, the Riemann curvature around each point has dimension 6. For a 4-dimensional space (such as our space-time), it has dimension 20.

 

 

http://settheory.net/curvature

 

Two types of curvature are recognised, extrinsic (2) and intrinsic (3).

 

The statement in your post (1) refers to extrinsic curvature which is always embedded in a higher dimension, hence my query about which direction.

The 'curvature' of spacetime in relativity refers to Gaussian curvature which is intrinsic (3).

 

Intrinsic curvature is additive in regard to the principle curvatures and thus has the property that the curvature achieved by the addition has the same dimension as the principal curvatures, ie the reciprocal of length.

Thus it can be expressed as a radius.

 

Intrinsic edit extrinsic curvature has the property that Strange mentioned in that it is independent of the higher dimension that contains it (4), and can be detected by measurements purely in the dimension of interest (our 4D spactime) and is the type used for relativity without embedding.

This does not mean to say that the higher containing dimension does not exist, just that it is not required for the mathematics.

However intrinsic curvature is a product of principal curvatures and thus its reciprocal does not have the units of length and cannot therefore be expressed by a 'radius'.

 

The curvature that Strange refers to is the extrinsic Riemanian curvature (5), but is this the one you want ?

 

Here is a simpler example of the use of Gaussian curvature.

 

Consider a sting of beads.

Each bead is coloured half red and half blue.

 

Imagine the string is laid out in a circle on a sheet of paper and each bead is oriented so that it appears either all red or all blue from each side of the paper.

A flatland being on that paper could walk round the string and detect the sequence of colours of the beads and could use a primitve rotation function to describe this and could detect torsional rotations about the string axis.

 

This example appears in nature in the orientation of the parts of long chain molecules.

 

It would be really great if ajb were able to offer answers in this post. Differential geometry is his speciality.

Edited October 28, 2015 by studiot

[Strange]

However intrinsic curvature is a product of principal curvatures and thus its reciprocal does not have the units of length and cannot therefore be expressed by a 'radius'.

 

Thanks for that. I was having trouble putting my finger on why the concept of radius was not relevant.

 

 

Intrinsic curvature is additive in regard to the principle curvatures and thus has the property that the curvature achieved by the addition has the same dimension as the principal curvatures, ie the reciprocal of length.

Thus it can be expressed as a radius.

 

I think you mean extrinsic curvature here?

[studiot]

 

strange

I think you mean extrinsic curvature here?

 

 

 

Thanks, good catch +1

[DParlevliet]

So perhaps you could approximate the radius of curvature as 6x10^-10 x radius of the earth or about 100,000,000,000,000 kilometres. I really don't know if that answer makes sense or not.

Perhaps you mean 6x10⁺¹⁰? Then 6x10¹⁰ x 6400 km = 3,8x10¹⁴ km, in stead of 1x10¹⁴ km

[Strange]

Perhaps you mean 6x10⁺¹⁰? Then 6x10¹⁰ x 6400 km = 3,8x10¹⁴ km, in stead of 1x10¹⁴ km

 

You are right (I was originally going to express it as a proportion). However, as noted, it is pretty much meaningless anyway. Except that it gives you an idea that space-time is pretty flat in the vicinity of the Earth.

Edited October 28, 2015 by Strange

[studiot]

Flat means that there is an isomorphism to Euclidian space.

[DParlevliet]

You are right (I was originally going to express it as a proportion). However, as noted, it is pretty much meaningless anyway. Except that it gives you an idea that space-time is pretty flat in the vicinity of the Earth.

Then it is different from what I expected: 9,2 10¹² km (R = c²/g), so I am not sure my calculation is right.

 

But anyway the radius is much larger then the radius of the moon orbit, which graphical presentations of spacetime (like the trampoline-model) suggest (of course you know).

 

So if one limits to our known 3D space, one notice that the moon is moving to the earth, which is not caused by a curvation of the visible 3D space with a radius of the moon orbit, but because the moon is travelling in the fourth dimension at about light speed. Althought the curvation of space-time is very small, the hight speed c results in a movement we can see in our 3D space.

 

I know this is no scientific explanation, but is it good enough to use as explanation for those who know nothing about relativity?

[Strange]

I know this is no scientific explanation, but is it good enough to use as explanation for those who know nothing about relativity?

 

It sounds like incoherent nonsense. For example:

 

"the moon is travelling in the fourth dimension at about light speed"

The "fourth dimension" is time. How can the moon be travelling through time at a speed measured in distance per unit time? It makes no sense.

[DParlevliet]

"the moon is travelling in the fourth dimension at about light speed"

The "fourth dimension" is time. How can the moon be travelling through time at a speed measured in distance per unit time? It makes no sense.

Perhaps. I am trying to figure out here.

It is based on multipling the time axis with (constant) c. Then it has the same dimension of distance as 3D space.

Secondly according Brian Greene in The fabric of the cosmic: "According special relativity the combined speed of movement of every object throught space and its movement throught time is always exact equal the speed of light"

 

So because objects mostly has a low speed, their speed in the time axis is about c.

[Strange]

It is based on multipling the time axis with (constant) c.

 

If you multiply time (seconds) by c (metres/second) you get metres, not speed.

 

 

Secondly according Brian Greene in The fabric of the cosmic: "According special relativity the combined speed of movement of every object throught space and its movement throught time is always exact equal the speed of light"

 

I guess he is alluding to the fact that you can regard the Lorentz transform as a rotation between the time and space axes. To say that this means that everything moves at c seems a little misleading.

[studiot]

I've never read Mr Greene , and from your quotes, I never want to.

 

However I would imagine he is referring to the spacetime interval, ds, which is invariant (the same for all observers) under the Lorenz transformation.

 

It is, however, the summation of the squares that is important.

 

ds² = c²⁽dt)² - (dx)² - (dy)² - (dz)²

Edited October 28, 2015 by studiot

[DParlevliet]

If you multiply time (seconds) by c (metres/second) you get metres, not speed.

Metres means a axis with distantance as dimension, in which in an object can travel at a certain speed.

 

I guess he is alluding to the fact that you can regard the Lorentz transform as a rotation between the time and space axes. To say that this means that everything moves at c seems a little misleading.

Still he mentions that this is "According the special relativity this is a law for all kind of movement" and somewhat further: "The most important fact is that Einstein discovered that these two movements are always complemental"

 

So he states it to be quite fundamental.

[Strange]

Metres means a axis with distantance as dimension, in which in an object can travel at a certain speed.

 

But after (meaninglessly) converting time to a distance you then need another time axis if you want to define speed. So this is all nonsense.

 

Still he mentions that this is "According the special relativity this is a law for all kind of movement" and somewhat further: "The most important fact is that Einstein discovered that these two movements are always complemental"

 

So he states it to be quite fundamental.

 

The Lorentz transform is fundamental. The trouble is that he has come up with what he presumably thinks is a catchy analogy. It is clearly highly misleading and uninformative. (As shown by the fact that you are confused by it.)

 

You would be better off going and looking at the (pretty simple) mathematics of special relativity in order to understand the relationships between time and space.

[DParlevliet]

But after (meaninglessly) converting time to a distance you then need another time axis if you want to define speed. So this is all nonsense.

That is a point.

But also Minkowski diagrams use ct as axis, so it is not that meaningless

I don't mean that ct is a real excisting distance. I don't know. But I also don't know what time (axis) really is.

 

I found here http://physics.stackexchange.com/questions/33840/why-are-objects-at-rest-in-motion-through-spacetime-at-the-speed-of-light a discussion about this. Some agree, some not.

[ajb]

It would be really great if ajb were able to offer answers in this post. Differential geometry is his speciality.

By curvature one is usually talking about the Riemann curvature tensor. Provided this does not vanish then the space-time has 'curvature'. One can look at the components of this tensor in a given coordinate system, but remember that the components do depend on the coordinate system employed (unless the tensor is zero).

 

A little less refined, is the Ricci tensor and less so again is the Ricci scalar.

 

There are other scalar curvatures that are useful; Kretschmann scalar and the Yamabe invariant for example.

 

Anyway, I have not come across anyone speaking about the radius of curvature in the context of general relativity. The closest maybe the sectional curvature.