- #1

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1. Is the r in ρ=Rr a unit vector?

2. it shows x^2 + y^2 + z^2 + w^2 = ρ^2 + w^2 = R^2 but isnt ρ=Rr, thus isnt p^2 itself already R^2?

3. Why is w used for the time component? Is this w the angular frequency?

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- Thread starter TimeRip496
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- #1

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1. Is the r in ρ=Rr a unit vector?

2. it shows x^2 + y^2 + z^2 + w^2 = ρ^2 + w^2 = R^2 but isnt ρ=Rr, thus isnt p^2 itself already R^2?

3. Why is w used for the time component? Is this w the angular frequency?

- #2

PAllen

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They seem to be considering a given spatial 3-surface as embedded in a Euclidean 4-space. The w is the extra spatial dimension of this 4-space. Personally, I've never seen it done this way, and I've never bothered reading any texts that introduce differential geometry via embeddings (instead, I prefer coming to grips directly with intrinsic geometry). They soon remove the extra spatial dimension, getting the conventional form of the FLRW metric.

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- #3

ChrisVer

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I don't understand your Q1. [itex]\rho[/itex] is introduced as the "radial" coordinate in the spherical coordinate system [itex](x,y,z) \rightarrow (\rho, \theta, \phi)[/itex]. But be cautious since you have a 4D space. In the analogue of a 2D sphere, [itex]\rho[/itex] would play the role of the radius of the circles that you can draw on the sphere (i.e. [itex]\rho^2_{2Dsphere}= x^2 +y^2[/itex]... and this can vary with [itex]z[/itex]. The circle near the Equator (small z) will have a larger radius than near the pole (large z)... in your example this which is [itex]\rho^2 = x^2 + y^2 +z^2[/itex], varies with [itex]w[/itex]..

Let's suppose that you take a slice of the whole geometric object in consideration with [itex]w=w_0[/itex].

In that case [itex]\rho^2 = R^2 - w_0^2[/itex].

Now you can say that [itex]w_0 = m R[/itex] with m some number. That way : [itex] \rho = R \sqrt{1-m}[/itex]

this [itex]\sqrt{1-m}[/itex] is the [itex]r_0[/itex] which corresponds to the surface of [itex]w=w_0[/itex] you choose.... of course this can vary with [itex]w[/itex] (in some other [itex]w' = w_0'[/itex] you would have a different [itex]m[/itex]).... So in general [itex]\sqrt{1-m}= r[/itex].

and if you want you can also write it as [itex]r=\sqrt{1- (w/R) }[/itex]

The slices again can be "visualized" with slicing a 2D sphere in [itex]z=z_0[/itex] (which would return you back a certain circle with radius [itex]\rho^2 = R^2 - z_0^2[/itex]). In that case [itex]z_0 = R \cos \theta_0[/itex] and [itex] \rho_0 = R \sqrt{1- \cos \theta_0} \Rightarrow \rho = R \sqrt{1-\cos \theta} = Rr[/itex], with [itex]r=\sqrt{1-\cos \theta}[/itex] playing the role of the circle's "radius" a unit-sphere would have..

your rho then would be equal to R in the case that w=0, and that makes sense because in that case you take back the 2D sphere.

No, [itex]\rho^2 = r^2 R^2[/itex]. It's not used as a vector anywhere.thus isnt p^2 itself already R^2?

Let's suppose that you take a slice of the whole geometric object in consideration with [itex]w=w_0[/itex].

In that case [itex]\rho^2 = R^2 - w_0^2[/itex].

Now you can say that [itex]w_0 = m R[/itex] with m some number. That way : [itex] \rho = R \sqrt{1-m}[/itex]

this [itex]\sqrt{1-m}[/itex] is the [itex]r_0[/itex] which corresponds to the surface of [itex]w=w_0[/itex] you choose.... of course this can vary with [itex]w[/itex] (in some other [itex]w' = w_0'[/itex] you would have a different [itex]m[/itex]).... So in general [itex]\sqrt{1-m}= r[/itex].

and if you want you can also write it as [itex]r=\sqrt{1- (w/R) }[/itex]

The slices again can be "visualized" with slicing a 2D sphere in [itex]z=z_0[/itex] (which would return you back a certain circle with radius [itex]\rho^2 = R^2 - z_0^2[/itex]). In that case [itex]z_0 = R \cos \theta_0[/itex] and [itex] \rho_0 = R \sqrt{1- \cos \theta_0} \Rightarrow \rho = R \sqrt{1-\cos \theta} = Rr[/itex], with [itex]r=\sqrt{1-\cos \theta}[/itex] playing the role of the circle's "radius" a unit-sphere would have..

your rho then would be equal to R in the case that w=0, and that makes sense because in that case you take back the 2D sphere.

[itex]w[/itex] is not a time-component. As they mention they use a 4D Euclidean space ... time is not part of an Euclidean space since it comes with an opposite sign in the [itex]ds^2[/itex]. It's in fact the same as [itex]z[/itex] when you would embed a 2D sphere into the 3D Euclidean world; more like a spatial coordinate. In fact it can be connected to a 3rd angle that you need to determine points on your space (as again z is connected to the 2nd angle theta)....3. Why is w used for the time component? Is this w the angular frequency?

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- #4

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Thanks! But there is still one part that I am not too sure. Why is there a need to embed to a higher dimensional space? It is not just this source as I did check out other source and a lot of their derivation involves embedding 3d sphere to a 4d "hyperspace". What does this embed really mean?I don't understand your Q1. [itex]\rho[/itex] is introduced as the "radial" coordinate in the spherical coordinate system [itex](x,y,z) \rightarrow (\rho, \theta, \phi)[/itex]. But be cautious since you have a 4D space. In the analogue of a 2D sphere, [itex]\rho[/itex] would play the role of the radius of the circles that you can draw on the sphere (i.e. [itex]\rho^2_{2Dsphere}= x^2 +y^2[/itex]... and this can vary with [itex]z[/itex]. The circle near the Equator (small z) will have a larger radius than near the pole (large z)... in your example this which is [itex]\rho^2 = x^2 + y^2 +z^2[/itex], varies with [itex]w[/itex]..

No, [itex]\rho^2 = r^2 R^2[/itex]. It's not used as a vector anywhere.

Let's suppose that you take a slice of the whole geometric object in consideration with [itex]w=w_0[/itex].

In that case [itex]\rho^2 = R^2 - w_0^2[/itex].

Now you can say that [itex]w_0 = m R[/itex] with m some number. That way : [itex] \rho = R \sqrt{1-m}[/itex]

this [itex]\sqrt{1-m}[/itex] is the [itex]r_0[/itex] which corresponds to the surface of [itex]w=w_0[/itex] you choose.... of course this can vary with [itex]w[/itex] (in some other [itex]w' = w_0'[/itex] you would have a different [itex]m[/itex]).... So in general [itex]\sqrt{1-m}= r[/itex].

and if you want you can also write it as [itex]r=\sqrt{1- (w/R) }[/itex]

The slices again can be "visualized" with slicing a 2D sphere in [itex]z=z_0[/itex] (which would return you back a certain circle with radius [itex]\rho^2 = R^2 - z_0^2[/itex]). In that case [itex]z_0 = R \cos \theta_0[/itex] and [itex] \rho_0 = R \sqrt{1- \cos \theta_0} \Rightarrow \rho = R \sqrt{1-\cos \theta} = Rr[/itex], with [itex]r=\sqrt{1-\cos \theta}[/itex] playing the role of the circle's "radius" a unit-sphere would have..

your rho then would be equal to R in the case that w=0, and that makes sense because in that case you take back the 2D sphere.

[itex]w[/itex] is not a time-component. As they mention they use a 4D Euclidean space ... time is not part of an Euclidean space since it comes with an opposite sign in the [itex]ds^2[/itex]. It's in fact the same as [itex]z[/itex] when you would embed a 2D sphere into the 3D Euclidean world; more like a spatial coordinate. In fact it can be connected to a 3rd angle that you need to determine points on your space (as again z is connected to the 2nd angle theta)....

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- #6

ChrisVer

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It is one approach... and to me it's intuitive enough [because an analogy to the 2-sphere, which you can visualize, becomes straightforward] ....Why is there a need to embed to a higher dimensional space?

In a similar manner, why would you parametrize a 2-sphere with [itex]x,y,z[/itex] (physical space) and not [itex]\theta,\phi[/itex] (parameter space)? Both works, but at least to me, the 1st was always much easier to visualize. Especially when you add in extra dimensions [as going to a 3D sphere].

Of course you can work [as already mentioned] without embedding, and this becomes more handy for a lot of other applications, but you trade intuition for abstractness and maths.

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I am sorry I dont really understand what you mean by " you first argue that there exists a time slicing such that an observer at rest in the respective reference frame sees a maximally symmetric 3D hypersurface, which he calls "space".". Can you break it down for me so I can understand?

Besides, why can't we explain using embedding? The mathematical approach is through embedding and your above explanation can explain that mathematical approach?

- #8

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What I tried to say in the very brief summary of the argument is just the Cosmological principle, i.e., that there is no preferred location nor a preferred direction in the universe (when you coarse grain over large enough space-time regions, so that all the local specialties of having a solar system, a galaxy, galaxy clusters etc. are averaged over). Then this means in the mathematical language of GR that you can define a family of observers for whom at any time space looks homogeneous and isotropic, and this then leads to the Robertson-Walker line element/pseudo-metric. Each time slice then is a maximally symmetric 3D Riemannian manifold, and these you can find systematically, leading you to the 3D spaces of constant curvature, referring to an open (hyperbolic), flat (also open of course), and hypersphere (closed) universe.

- #9

ChrisVer

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simply a slice means taking a constant value for it... Instead of seeing your whole spacetime as the surface [itex]F(t,x^i)[/itex], you take [itex]t=t_0[/itex] and look at [itex]F(t_0,x^i)=F'(x^i)[/itex], which is what 'he' calls "space". That space has a geometry/is a hypersurface itself that's parametrized by your [itex]F'[/itex] function....I am sorry I dont really understand what you mean by " you first argue that there exists a time slicing such that an observer at rest in the respective reference frame sees a maximally symmetric 3D hypersurface, which he calls "space".". Can you break it down for me so I can understand?

As a rough example, take the ball, each point of which is parametrized [itex]F(r,\theta,\phi)[/itex], the [itex]r[/itex] is the distance from the center and [itex]\theta,\phi[/itex] the angles....

Now if you take a slice [itex]r=r_0[/itex] of the ball, you obtain the 2D sphere (with radius [itex]r_0[/itex]). which is another surface (see next to see that it's parametrized by 2 parameters and not 1, so it's still a surface). Your initial [itex]F(r,\theta,\phi) \rightarrow F(r_0,\theta,\phi) \equiv G(\theta,\phi)[/itex] which also happens to be the parametrization of the 2D sphere as yu can easily imagine.

- #10

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simply a slice means taking a constant value for it... Instead of seeing your whole spacetime as the surface [itex]F(t,x^i)[/itex], you take [itex]t=t_0[/itex] and look at [itex]F(t_0,x^i)=F'(x^i)[/itex], which is what 'he' calls "space". That space has a geometry/is a hypersurface itself that's parametrized by your [itex]F'[/itex] function....

As a rough example, take the ball, each point of which is parametrized [itex]F(r,\theta,\phi)[/itex], the [itex]r[/itex] is the distance from the center and [itex]\theta,\phi[/itex] the angles....

Now if you take a slice [itex]r=r_0[/itex] of the ball, you obtain the 2D sphere (with radius [itex]r_0[/itex]). which is another surface (see next to see that it's parametrized by 2 parameters and not 1, so it's still a surface). Your initial [itex]F(r,\theta,\phi) \rightarrow F(r_0,\theta,\phi) \equiv G(\theta,\phi)[/itex] which also happens to be the parametrization of the 2D sphere as yu can easily imagine.

So the $$pdp+wdw=0$$ is to allow us to find the maximally symmetric 3D hypersurface? And we can have positive, negative and zero curvature cause from $$pdp+wdw=0$$, there are maxima, minima and inflexion points?

- #11

ChrisVer

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That expression is not even needed, so I cannot think of a reason why someone would make use of it... it's clear from the sphere constraint that w=w(p)...

And so it's straghtforward to write dw= (dw/dp) dp = ... giving exactly the same result... You do that because you have the dw^2 in the line element.

How do you see maxima,minia and inflexion points from pdp+wdw =0? Again the whole reason I find the embedding intuitive is because you can view what you are doing with analogies... Is the ρdρ+zdz=0 which holds for the 2D sphere giving you any such information? (At least I can't see it).

http://calclab.math.tamu.edu/~fulling/m467/f09/arcriem.pdf

And so it's straghtforward to write dw= (dw/dp) dp = ... giving exactly the same result... You do that because you have the dw^2 in the line element.

How do you see maxima,minia and inflexion points from pdp+wdw =0? Again the whole reason I find the embedding intuitive is because you can view what you are doing with analogies... Is the ρdρ+zdz=0 which holds for the 2D sphere giving you any such information? (At least I can't see it).

http://calclab.math.tamu.edu/~fulling/m467/f09/arcriem.pdf

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